General Lattice Theory George Gratzer

G. GRATZER

G E N E R A L LATTICE T H E O R Y

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General Lattice Theory

by

GEORGE GRdTZER

The University of Manitoba

ACADEMIC PRESS NEW YORK SAN FRANCISCO 1978

A Subsidiary of Harcourt Brace Jovanovich, Publishers

Q 1978 Birkhiiuser Verleg, Basel und Stuttgart All Rights Reserved.

No pert of this publication may be reproduoed or transmitted in any form or by any means, electronic or mechanical including photocopy, reoording, or any information storage and

retrieval system, without permission in writing from the publisher.

ACADEMIC PRESS, INC. i i i Fifth Avenue, New York, New York 10003

United Kingdom Edition published by ACADEMIC PRESS, INC. (LONDON) LTD.

24/28 Oval Road, London NW 1 LCCCN 76-22929

Gorge Griitzer, 1978 General Lattice Theory

ISBN: 0-12-2957504 Printed in GDR

To my family, Cathy, Tom, and David, and to the memory of my father, Jdxsef

This Page Intentionally Left Blank

C ONTE NT S

PREFACE AND ACKNOWLEDGEMENTS . . . . . .

INTRODUCTION . . . . . . . . . . . . . . . . .

. . . IX

. . . XI I . FIRST CONCEPTS

1 . Two Dcfinitions of Lattices . . . . . . . . . . . . . . . . . . . 1 2 . How to Describc Lattices . . . . . . . . . . . . . . . . . . . . 9 3 . Some Algebraic Concepts . . . . . . . . . . . . . . . . . . . . 15 4 . Polynomials. Identities. and Inequalities . . . . . . . . . . . . . 26 5 . Free Lattices . . . . . . . . . . . . . . . . . . . . . . . . . 32 6 . Special Elements . . . . . . . . . . . . . . . . . . . . . . . . 47

Further Topics and References . . . . . . . . . . . . . . . . . . 52 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

I1 . DISTRIBUTIVE LATTICES

1 . Characterization Theorems and Representation Theorems 2 . Polynomials and Freeness . . . . . . . . . . . . . . 3 . Congruence Relations . . . . . . . . . . . . . . . . 4 . Boolean Algebras R-generated by Distributive Lattices . 5 . Topological Represent&ion . . . . . . . . . . . . . 6 . Distributive Lattices with Pseudocomplementation . . .

Further Topics and References . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . .

. . . . . . 59

. . . . . . 68

. . . . . . 73

. . . . . . 86

. . . . . . 99

. . . . . . 111

. . . . . . 120

. . . . . . 126

I11 . CONGRUENCES AND IDEALS

1 . Weak Projectivity and Congruences . . . . . . . . . . . . . . . 129 2 . Distributive. St.andard. and Neutral Elements . . . . . . . . . . . 138 3 . Distributive. Standard. and Neutral Ideals . . . . . . . . . . . . 146 4 . Structure Theorems . . . . . . . . . . . . . . . . . . . . . . 151

Further Topics and Refcmnces . . . . . . . . . . . . . . . . . . 158 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . 159

IV . MODULAR AND SEMIMODULAR LATTICES

1 . Modular Lattices . . . . . . . . . . . . . . . . . . . . . . . . 161 2 . Semimodular Lattices . . . . . . . . . . . . . . . . . . . . . 172 3 . Geometric Lattices . . . . . . . . . . . . . . . . . . . . . . . 178

Contents

4 . Partit'ion Lattices . . . . . . . . . . . . . . . . . . . . . . . 192 6 . Complemented Modular Lattices . . . . . . . . . . . . . . . . . 201

Further Topics end R.eferences . . . . . . . . . . . . . . . . . . 218 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . 224

V . EQUATIONAL CLASSES OF LATTICES

1 . Characterizations of Equationel Classes . . . . . . . . . . . . . . 227 2 . The Lattice of Equational Classes of Lattices . . . . . . . . . . . 236 3 . Finding Equational Bases . . . . . . . . . . . . . . . . . . . . 243 4 . The Amalgamation Property . . . . . . . . . . . . . . . . . . 252


VI . FREE PRODUCTS

1 . Free Products of Lattices . . . . . . . . . . . . . . . . . . . . 265 2 . The Structure of Free Lattices . . . . . . . . . . . . . . . . . . 282 3 . Reduced Free Products . . . . . . . . . . . . . . . . . . . . . 288 4 . Hopfien Lattices . . . . . . . . . . . . . . . . . . . . . . . . 298


CONCLUDING REMARKS . . . . . . . . . . . . . . . . . . . . 311

BIBLIOGRAPHY . . . . . . . . . . . . . . . . . . . . . . . . 316

TABLE O F NOTATION . . . . . . . . . . . . . . . . . . . . . 362

INDEX . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365

PREFACE AND ACKNOWLEDGEMENTS

A book that is niore than twelve years in the making has a long history and its final form is shaped by many. It all started in niy formative years as a mathematician, 1955-1961, when I worked with E. T. Schmidt. We often commented upon the need for a two- or three-volume work on lattice theory that would treat the subject in depth. We felt, however, that the time was not ripe for sucha project. Forinstance,no such work would be complete without presenting a t least one example of anondistribu- tive uniquely complemented lattice. We did not know how to doit without reproducing the almost thirty pages of the famous proof of R. P. Dilworth. We also thought that much more had to be learned about free lattices and equational classes of lattices before the project could be attempted.

In 1962, I wrote a proposal for a volume on lattice theory that would attempt to survey the whole field in depth. Apart from doing the research necessary for the proposal, no writing was done on this book. M. H. Stone offered to publish a book on universal algebra in the D. Van Nostrand University Series in Higher Mathematics and I concentrated on that book until the end of 1967.

Maybe because mathematicians in general (or I, in particular) are like hobbits (according to J. R. R. Tolkien [1954]: “Hobbits delighted in such things if they were accurate: they liked to have books filled with things they already knew, set out fair and square with no contradictions.”) or maybe because I felt that the need for an in depth book on lattice theory had not yet been satisfied, I started in 1968 on this book. In the academic year 1968-1969, I gave a course on lattice theory and I wrote a set of lecture notes. The present first two chapters are based on those notes.

This material was augmented by a chapter on pseudocomplemented distributive lattices and published under the title “Lattice Theory: First Concepts and Distri- butive Lattices” in 1971. (This book will be referred to as FC.) The Introduction of this book promised a companion volume on general lattices.

A number of research breakthroughs in the sixties now supplied me with the material (including almost all of Chapters V and VI ) I needed to complete the project. But then it became apparent that a complete revision of my plans was in order.

While back in the late fifties it seemed reasonable to try to give a complete picture of lattice theory, this became patently unfeasible in the seventies. For instance, in 1958 there was one paper on Stone algebras; by 1974 there were more than fifty papers on Stone algebras and related problems. A number of books have appeared dealing with specialized aspects of lattice theory and with various applications.

X Preface ttritl Acknowledgements

Besides, my experience with the writing of Universal Algebra taught me not to stray too far away from my research interests. Thus it was decided that while I try to include all the basic material and research methods, the illustrations will be chosen, as far as possible, from fields in which I have some personal interest.

Another change took place in the publishing field. For the second volume it became desirable to choose a publisher with a greater interest in monographs. The new ar- rangement made it necessary to produce a volume that does not depend on the previous publication. That is why most of the first two chapters of the former book are reproduced here (augmented by a new section, several new exercises, with updated Further Topics and References, and wit,h a new set of Problems), thus giving the reader a self-contained book.

The work on this new book started in 1972 and then continued with an advanced course on lattice theory a t the University of Manitoba in 1973-1974. The lecture notes of this course form the basis of most oE Chapters 111-VI.

I am grateful to my students who took the courses in 1968-1969 or in 1973-1974 and also to my colleagues who attended for their helpful criticisms and for many simplified proofs. A corrected version of the first set of notes was read by R. Balbes, P. Bur- nieister, M. I. Gould, J. H. Hoffman, K. M. Koh, H. Lakser, s. M. Lee, R. Pad- nianabhan, P. Penner, and C. R. Platt. B. Jhnsson, reading the manuscript of F C for the publisher, offered many useful suggestions. The first part was proofread by R. Anto- nius, J. A. Gerhard, K. M. Koh, W. A. Lampe, R. W. Quackenbush, and I. Rival.

Many readers, in particular D. D. Miller, sent me corrections to FC; this made it possible for me to improve some parts of F C that are being reproduced here.

The second set of notes was distributed widely and I am grateful to a11 who offered corrections, in particular, to K. A. Baker, C. C. Chen, M. I. Gould, D. Haley, K. M. Koh, V. B. Lender, G. H. Wenzel, and B. Wolk.

In the proofreading of the present volume I was assisted by M. E. Adams, R. Beazer, K. A. Baker, J. Berman, B. A. Davey, J. A. Gerhard, M. I. Gould, D. Haley, D. Kelly, C. R. Platt, and G. H. Wenzel.

A great deal of organizational work was necessary in the distribution of manuscripts and the collation of corrections; this was faithfully carried out by R. Padmanabhan.

M. E. Adanis undertook the arduous task of getting the manuscript ready for the publisher.

I received help from various individuals in specific areas, including M. Doob (matroids), I. Rival (exercises on combinatorial topics), R. Venkataraman (partially ordered vector spaces), B. Wolk (projective geometry).

Thanks are due to the National Research Council of Canada for sponsoring much of the original research that has gone into this book and to Professor N. S. Mendel- sohn for creating a very good environment for work. Mrs. M. McTavish did an excellent job of typing and retyping the manuscript.

Finally, I would like to thank the members and the many visitors of my seminar who, over a period of eight years, have been lecturing an average of four hours a week, 52 weeks a year, in an attempt to teach me lattice theory. Without their help I could not even have tried.

Despite the improvements so generously offered by so many, I am sure my original work can still be recognized: all the remaining mistakes ere my own.

INTRODUCTION

In the first half of the nineteenth ccntury, George Boole’s attempt to formalize propositional logic led to the concept of Boolean algebras. While investigating the axioniatics of Boolean algebras a t the end of the nineteenth century, Charles S. Peirce and Ernst Schroder found it useful to introduce the lattice concept. Independently, Richard Dedekind’s research on ideals of algebraic numbers led to the same discov- ery. I n fact, Dedekind also introduced modularity, a weakened form of distributivity. Although some of the early results of these mathematicians and of Edward V. Huntington are very elegant and far from trivial, they did not attract the attention of the mathematical community.

It was Garrett Birkhoff’s work in the mid-thirties that started the general development of lattice theory. In a brilliant series of papers he demonstrated the importance of lattice theory and showed that it provides a unifying framework for hitherto unrelated developments in many mathematical disciplines. Birkhoff himself, Valere Glivenko, Karl Menger, John von Neuniann, Oystein Ore, and others had developed enough of this new field for Birkhoff to attempt to “se11~’ it to the general mathematical Community, which he did with astonishing success in the first edition of his Lattice Theory. The further development of the subject matter can best be followed by comparing the first, second, and third editions of his book (G. Birkhoff [1940], [1948], and [1967]).

The goal of the present volume can be stated very simply: to discuss in depth the basics of general lattice theory. In other words, I tried to include what I consider the most important results and research methods of all of lattice theory. To treat the rudimentary results in depth and still keep the size of the volume from getting out of hand, I had to omit a great deal. I excluded many important chapters of lattice theory that have grown into research fields on their own. Ordered algebraic systems and other applications were also excluded. The reader will find appropriate references to these throughout this book. It is hoped that even those whose main interest lies in areas not treated here in detail will find this volume useful by obtaining from this book the background in lattice theory so necessary in allied fields.

In niy view, distributive lattices have played a many faceted role in the development of lattice theory. Historically, lattice theory started with (Boolean) distributive lattices; as a result, the theory of distributive lattices is one of the most extensive and most satisfying chapters of lattice theory. Distributive lattices have provided the motivation for many results in general lattice theory. Many conditions on lattices

XI1 Introduction

and on elements and ideals of lattices are weakened forms of distributivity. Therefore, a thorough knowledge of distributive lattices is indispensable for work in lattice theory. Finally, in many applications the condition of distributivity is imposed on lattices arising in various areas of mathematics, especially algebra.

This viewpoint moved me to break with the traditional approach to lattice theory, which proceeds from partially ordered sets to general lattices, semimodular lattices, modular lattices, and, finally, to distributive lattices. That is why distributive lattices are treated as a first priority in this book. This approach has the added advantage that the reader (or the student in the classroom) reaches interesting and deep results early in the book.

Chapter I gives a concise development of the basic concepts of lattice theory. Diagrams are emphasized because I believe that an important part of learning lattice theory is the acquisition of skill in drawing diagrams. This point of view is stressed throughout the book by about 130 diagrams (heeding Alice’s advice: “and what is the use of it book without pictures or conversations”, L. Carroll [1865]); the reader would be well advised to draw many times more while reading the book. A special feature of this chapter is a detailed development of free lattices generated by a partial lattice over an arbitrary equational class; this is one of the most important research tools of lattice theory.

Chapter I1 develops distributive lattices including representation theorems, congruences, congruence lattices of general lattices, Boolean algebras, and topological representations. The last section is a brief introduction to the theory of distributive lattices with pseudocomplementation. While the theory of distributive lattices is developed in detail, the reader should keep in mind that the purpose of this chapter is, basically, to serve as a model for the rest of lattice theory.

In Chapter I11 we discuss congruences and ideals of general lattices. The various types of ideals discussed all imitate to some extent the behaviour of ideals in distributive lattices.

After giving the basic facts concerning modular and semimodular lattices, Chap- ter IV investigates in detail the connection between lattice theory and geometry. We develop the theory of geometric lattices, in particular direct decompositions and geometric lattices arising out of geometries and graphs. As an important example, we investigate partition lattices. The last section deals with complemented modular lattices and projective geometries.

Chapters V and VI deal with two new areas of investigation. Equational classes of lattices is one of the most promising new fields. I n Chapter V most of the basic facts are presented along with some more specialized methods. Chapter V I grew out of an investigation of free lattices. It intends to prove that almost all the results on free lattices can he obtained within the framework of free products of lattices. I n addition, free products can be used to construct interesting examples of lattices.

The exercises, which number almost 900, form an integral part of the book. The Bibliography contains over 750 entries ; i t is not, however, a comprehensive bibliography of this field. With a few exceptions, it contains only items referred to in the text. The 193 research problems, the “Further Topics and References” a t the end of each chapter, and the Concluding Remarks should be of help to those who are interested in further reading and research in lattice theory.

Introd uc tiou XI11

The abandonnient of the traditional structure of a lattice theory book means that concepts and notations are more evenly introduced throughout. A very detailed Index and the Tshle of Notation should help the reader in finding where a concept or notation is first, introduced.

Finally, the reader will note that t,he symbol is placed at the end of a proof; if a theorem or lemma contains more than one statement, the proof of a part is ended with 0. The abbreviation “iff” stands for “if and only if”. More difficult exercises are marked by *. “Theorem 10” refers to Theorem 10 of the same section, ‘‘Theorem 5.10” refers to Theorem 10 of Section 5 of the same chapter, whereas “Theorem I. 5.10’’ refers to Theorem 10 of Section 5 in Chapter I. Similarly, “Exercise 111.2.6” iiieans Exercise 6 of Section 111.2. References to the Bibliography are given in the form “J. Jakubik [ 1957]”, which refers to a paper (or book) by J. Jakubik published in 1957. Such references as “[1957a]” and “[1957 b]” indicate that the Bibliography contains more than one work by the author published in that year. “R. Wille [a]” refers to a paper by R. Wille that had not appeared in print a t the time the ninnuscript of this book was submitted for publication. I n the list of problems, Problem I. 29 is Probleni 29 of Chapter I ; I. 31 (FC 17) signifies that Problein 31 of Chapter I is a repetition of, or is closely related to, Problem 17 of FC.

\Vinnipeg, Manitoba September 1975

George Griitzer

CHAPTER 1

FIRST CONCEPTS

1. Two Definitions of Lattices

Whereas the arithmetical properties of tho set of reals R can be expressed in terms of additsion and multiplication, the order theoretic, and thus the topological, properties are expressed in terms of the ordering relation <. The basic properties of this relation are as follows.

For all a , b, c E R we have:

(Pl) Reflexivity: a < a. (P2) Antisymmetry : a < b and b<a imply that a = b . (P3) Transitivity : a < b and b l c i m p l y that a l c . (P4) Linearity : a < b or b < a.

There are many examples of binary relations sharing properties (Pl)-(P4) with the ordering relation of reals, and there are even more enjoying (Pl)-(P3). This fact, by itself, would not justify the introduction of a new concept. However, it has been shown that many basic concepts and results about the reds depend only on (Pl)-(P3), and these can be profitably used whenever we have a relation satisfying (Pl)-(P3). Relations satisfying (Pl)-(P3) are called partial ordering relations, and sets equipped with such relations are called partially ordered seta or posets.

To make the definitions formal, let us start with two sets A, B and form the set A x B of all ordered pairs ( a , b) with a E A , b B. If A = B, we write A2 for A x A. Then a binary relation ,g on A can simply be defined as a subset of A? The elements a , b (a , b e A ) are i n relation with respect to e iff (a, b) EQ. For (a, b) Ee, we will also write ueb, or a E b (e). Binary relations will be denoted by small Greek letters or by special symbols.

This formal definition can be compared with the intuitive one: A binary relation ,g on A is a “rule” that decides whether or not aeb for any given a , b E A . Of course, any such rule will determine the set {(a, b) I aeb, a, b € A } , and this set determines e , so we might as well regard e as being identical with this set.

A partially ordered set ( A ; e) consists of a nonvoid set A and a binary relation e on A , such that Q satisfies properties (Pl)-(P3). Note that these can be stated as follows: For all a, b , c E A , ( a , a ) E e ; ( a , b ) , ( b , a ) E e imply that a = b ; ( a , b ) , ( b , c ) E , g imply that ( a , c) Ee.

If e satisfies (Pl)-(P3), e is a partial ordering relation, and it will usually be denoted by 5. Also, a 2 b will mean b a. Sometimes we will say that A (rather than (A ; I))

2 I. First Concepts

is a poset, meaning that the partial ordering is understood. This is an ambiguous although widely accepted practice.

A poset (A; <) that also satisfies (P4) is called a chain (also calledfully ordered set, linearly ordered set, and so on).

If (A; <)isaposet,a,bEA, thenuandbarecomparableif a < b or b s a . Otherwise, a and b are incomparuble, in notation, a (1 b. A chain is, therefore, a poset in which there are no incomparable elements.

Let (A; I) be a poset and let B be a nonvoid subset of A. Then thereisanatural partial order jg on B induced by i : for a, b E B, CI SB b iff a< b; we call (B; sB) (or, simply, (B; I)) a subposet of (A; I).

An unordered poset is one in which a (I b for all a+b. A chain C in a poset P is a nonvoid subset which, as a subposet, is a chain. An

antichain C in a poset P is a nonvoid subset which, as a subposet, is unordered. The length, l (C) , of a finite chain is (C( -1. A poset P is said to be of length n (in

formula, Z(P)=n), where n is a natural number, iff there is a chain in P of length n and all chains in P are of length < n. A poset P is of finite length iff it is of length n for some natural number n.

The width of a poset P is n, where n is a natural number, iff there is an antichain in P of n elements and all antichains in P have < n elements.

Next we define inf and sup in an arbitrary poset P (that is, (P; I)), similarly as it is done for reds.

Let H E P and a E P. Then a is an upper bound of H iff h 5 a for all h E H. An upper bound a of H is the least upper bound of H or supremum of H iff, for any upper bound b of H, we have a< b. We shall write a =sup H or a = VH. (The notations a =l.u.b. H, a = Z H are also common in the literature.) This notation can be justified only if we show the uniqueness of the supremum. Indeed, if a. and al are both suprema of H, then ao<al, since al is an upper bound, and a. is a supremdm. Similarly, al <a,; thus a. = al by (PZ).

Let 0 be the empty set. Let a be the least upper bound of 0, a = sup 0. For every b E P , we have h b for all h E 0 (since there is no such h) and so every b E P is an upper bound of 0. Thus a< b for all b P. We conclude that sup 0 exists iff P has a smallest element. We call sup 0 the zero of P and denote it by 0.

The concepts of lower bound and greatest lower boud or infimum are similarly defined; the latter is denoted by inf H or AH. (The notations g.1.b. H, IIH are also used in the literature.) The uniqueness is proved as in the last but one paragraph. Observe that inf 0 exists iff P has a largest element. We call inf 0 the unit (or identity) and denote it by 1.

The adverb “similarly” in the paragraph introducing infima can be given a very concrete meaning. Let (P; 5) be a poset. The notation a 2 b (meaning b a) can also be regarded as a definition of a binary relation on P. This binary relation 2 satisfies (Pl)-(P3) ; as an example, let us check (P2). If a 2 b and b 2 a, then by the definition of 2 we have b we conclude that a = b. (Pl) and (P3) are equally trivial. Thug (P; 2) is also aposet, called the dualof (P; I). Now, if 0 is a “statement” about posets, and if in 0 we replace all occurrences of < by 2, we get the dual of 0.

a and a < b; using (P2) for

1. Two Definitions of Lattices 3

DUALITY PRINCIPLE. If a statement 0 is true in all psets , then its dual is also true in all posets.

This is true simply because 0 holds for (P; <) iff the dual of 0 holds for (P; >), which is also a poset’.

As an example take for 0 the statement: ,(‘If sup H exists it is unique.” We get as its dual : “If inf H exists it is unique.” The dual of “( P; <) has a zero” is “(P; 2) has a unit”.

It is hard to imagine that anything as trivial as the Duality Principle could yield anything profound, and it does not; but it can save a lot of work.

A poset (L; I) is a lattice if sup {u, b} and inf {a, b} exist for all a, b < L. In other words, lattice theory singles out a special type of poset for detailed in-

vestigation. To make such a definition worthwhile, it must be shown that this class of posets is a very useful class, that there are many such posets in various branches of iliathematics (analysis, topology, logic, algebra, geometry), and that a general study of these posets will lead to a better understanding of the behavior of the examples. This was done in the first edition of G. Birkhoff’s Lattice Theory [1940]. As we go along we shall see many examples, most of them in the exercises. For a general survey of lattices in mathematics, see G. Birkhoff [1967] and H. H. Crapo and 0.-C. Rota [ 19701.

We will take the usefulness of lattice theory for granted (this is a less touchy subject now than it formerly was) and hope that the reader will like it for its intrinsic beauty.

To make the definition of a lattice less arbitrary, we note that an equivalent definition is the following:

H of L. A poset ( L ; () is a lattice iff sup H and inf H exist for any finite nonvoid subset

PROOF. It is enough if we prove that the first definition implies the second. So let ( L ; I) satisfy the first definition and let H E L be nonvoid and finite. If H = { a } , then sup H =inf H = a follows from the reflexivity of I and the definitions of sup and inf. Let H = { a , b, c}. To show that sup H exists, set d =sup {a , a} , e =sup {d , c } ; we claim that e =sup H. E‘irst of all, a < d, b d , and d < e, c < e ; therefore (by transitivity) x s e , for all x c H. Secondly, i f f is an upper bound of H, then a< f , b< f , and thus d < f ; also c f , since e =sup {a, c}. Thus e is the supreinum of H.

f , so that c, d

If H ={a,, . . . , un- , } , n 2 1, then

f ; therefore e

sup {. - * sup {sup {ao, ail , a2} * , afi-i}

is the supremum of H, by an inductive argument whose steps are analogous to those in the preceding paragraph.

By duality (in other words, by applying the Duality Principle), we conclude that inf H exists.

A lattice need not have zero or unit so inf 0 and sup 0 need not exist. This simple proof can be vaned to yield a large number of equally trivial statements

about lattices and partially ordered sets in general. Some of these will be stated as 2 Gritzer

4 I. First Concepts

exercises and used later. TO make the use of the Duality Principle legitimate for lattices, note:

I f ( P ; 5) is a lattice, so is its dual ( P ; 2).

Thus the Duality Principle applies to lattices. We will use the notations

aAb =inf {a, b }

avb =SUP {a, b}

and call A the meet and v the join. In lattices, they are both binary operations, which means that they can be applied to a pair of elements a, b of L to yield again an element of L. Thus A is a map of L2 into L and so is v , a remark that might fail to be very illuminating at this point.

and

The previous proof yields that

(* - - ((aova&m2) * . ) ~ a , - ~ =sup {ao, . . . , a,-l}

and there is a similar formula for inf. Now observe that the right-hand side does not depend on the way the elements a, are listed. Thus A and v are idempotent, coni- mutative, and associative-that is, they satisfy the following :

(Ll) Idempotency : ~ A U =a, ava = a. (L2) Commutativity : aAb = bAa, avb = bva. (L3) Associativity : (ar\b)Ac =ah(bAc),

(avb)vc =av(bvc) .

These properties of the operations are also called the idempotent identities, commutative identities, and associative identities, respectively.

As always in algebra, associativity makes it possible to write

~ Q A U ~ A ’ * ’Aa,-i

without using parentheses (and the same for v ) .

if a< b, then inf {a, b} =a ; that is, ahb =a, and conversely. Thus There is another pair of rules that connect A and v . To derive them, note that

a < b iff ahb=a.

By duality (and by interchanging a and b) we have

a< b iff avb = b.

Applying the “only if” part of the first rule to a and avb, and that of the second rule to aAb and a, we get

(L4) Absorption identities: aA(avb) =a, av(uAb) =a.

Now we are faced with the crucial question: Do we know enough about A and v so that lattices can be characterized purely in ternis of properties of A arid v P

Two coiriiiients are in order. It is obvious that 5 can he characterized by A and v (in fact, by either of them); therefore, obtaining such a characterization is only

1. Two l)efiiiitlions of Latticrs 5

a matter of persistence. More importantly, why should we try to get such a characterization? To rephrase the question, why should we want to characterize (L; <) as ( L ; A , v ) , which is an algebra-that is, a set equipped with operations (in this case, two binary operations)? Note that 5 is a subset of Li, whereas A and v are maps from L2 into L. The answer is simple: We want such a characterization be- because if we can treat lattices as algebras, then all the concepts and methods of universal algebra will become applicable. The usefulness of treating lattices as algebras will soon become clear.

,4n algebra (L; A , v ) is called a lattice iff: L is a nonvoid set, A and v are binary operations on L, both A and v are idempotent, commutative, and associative, and they satisfy the two absorption identities. The following theorem states that a lattice as an algebra and a lattice as a poset are “equivalent” concepts. (The word “equivalent” will not be defined.)

THEOREM 1. ( i ) Let the poset 2 = ( L ; I ) be a lattice. Set aAb =i77f ( a , b}, avb =sup {a, b}.

Then the algebra 2a =(L; A, v ) is a lattice. (ii) Let the algebra 2 = ( L ; A, v ) be a lattice. Set a b i f f ar\b = a . Then 2 p =

( L ; 5) i s a poset, and the poset 2 p is a lattice. (iii) Let the poset 2 =(L; 2) be a laltice. Then (2a)p =2. (iv) Let the algebra 2 =(L; A, V ) be a lattice. Then ( . l P ) a = 2.

REMARK. (i) and (ii) describe how we pass from poset to algebra and back, whereas (iii) and (iv) state that going there and hack takes us back to where we started.

PROOF.

(ii) We set a 5 b to mean aAb =a. Now < is reflexive since A is ideinpotent; (i) This has already been proved. 0

is antisyninietric since a <band b<a niean that aAb =a and bAa = b, which, by the commutativity of A , iniply that a = a ~ b = b ~ a =b; I is transitive, since if a 5 b and b < c, then a =aAb and b = bAc, and so

a =aAb =aA(bAc) = ( A is associative) = ( U A ~ ) A C = a m ,

that is, a c. Thus (L ; <) is a poset. To prove that (L; 5) is a lattice we will verify that m b =inf { a , b} and avb =sup {a, b}. (This is not a definition.) Indeed, aAb 5 a, since

((iAb)Aa =uA(bAa) = a ~ ( u ~ b ) = (uAa)Ab =aAb,

using associativity, commutativity, and idempotency of A ; similarly, aAb< b. Now if c g a and c l b , that is, C A U = C and cAb=c, then cA(aAb)=(cAa)Ab = ~ ~ b = r ; thus a ~ b = i n f {a , a} . Finally, u<avb and b<avb, because

ri = u ~ ( u v b ) and b = b ~ ( a v b ) by the first absorption identity; if a s c and b s c , that is, U = U A C and b=bAc, then avc=(ar\c)vc=c, and bvc=c by

29

6 I. First Concept,s

the second absorption identity. Thus

( U V ~ ) A C = (avb)A(avc) = (avb)A(av(bvc))

= (avb)A((avb)vc) =avb,

that is, a v b s c , completing the proof of avb =sup {a, b}.

to get (iii). (J (iii) It is enough to observe that the partial orderings of 2 and (la)” are identical

(iv) The proof of (iv) is similar to the proof of (iii). The proof of Theorem 1, and even the statement of Theorem 1, are subject to

criticism. To begin with, in the definition of a lattice as an algebra, idempotencp is redundant. The last step of the proof of (ii) can be made neater by first proving that

a = a ~ b iff b =avb.

Theorem 1 should be preceded by a similar theorem for “semilattices.” All these questions will be dealt with in the exercises that follow this section.

Finally, note that for lattices as algebras, the Duality Principle takes on the following very simple form.

Let Q, be a statement about lattices expressed in terms of A and v. The dual of Q, is the statement we get from Q, by interchanging A and v. If Q, is true for all lattices, then the dual of 0 is also true for all lattices.

To prove this we have only to observe that if 2 = ( L ; A , v ) , then the dual of 91’ is ( (L; v , A))”.

Most of the time CD involves < and niaybe 0 and 1 in addition to v and A. When dualizing such a 0 we interchange A and v , replace I by 2, and interchange 0 and 1.

Exercises

1.

2.

3.

4. 5.

6.

7. 8.

Posets

Define x < y to mean x y and x +y. Prove that, in a partially ordered set, x < x for no x, and that x < y, y < z imply that x < z. Let the binary relation < satisfy the conditions of Exercise 1. Define x 5 y to mean z < y or 5 =y. Then show that 5 is a partial ordering. Prove the following extension of antisymmetry: If xo 5 XI I. * *I xn-i IZO, thun z g = x i = * * * =%,-I.

Enumerate all partial orderings on a five-element set. Let 5 be a partial ordering on A and let B be a subset of A. For a, b E B, set a 5 11 b iff a Let A be a set and let P be the set of all partial orderings on A. For e, bEP, set e < a iff aeb implies that mb (a, b € A ) . Prove that (P; I) is a poset. Find an example of a poset in which inf 0 does not exist. Prove that if inf R exists for all nonempty subsets H of a posot P, then sup 0 also exists in P.

b. Prove that < B is a partial ordering on B.

1. Two Definitions of Latticos 7

Lattices as Posets

Prove that the following are examples of posets: (i) Let A = P ( X ) , the set of all subsets of a set X; let Xo 5 Xt mean X o Xi (set

(ii) Let A be the set of all real valued functions defined on X; for f, g € A , set f 5 g

(iii) Let A be the set of all continuous concave real valued functions defined on the

(iv) Let A be the set of all open sets of a topological space; define 2 as in (i). (v) Let A be the set of all human beings; a < b means that a is a descendant of b.

Which of the examples in Exercise 9 are lattices? For those that arc lattices, compute ahb arid avb. Show that every chain is a lattice. Let A be the set of all subgroups (normal subgroups) of a group c f ; for X, Y € A , set X 5 Y to mean X E Y. Prove that (A; 5 ) is alattice; compute X A Y and X v Y . Let (P; 5 ) be a poset in which inf H exists for all H E P. Show that (P; 5 ) is 8

lattice. (Hint: For a, bcP , let H be the set of all upper bounds of {a, b } . Prove that sup {a, b } =inf H . ) Relate this to Exercise 12.

inclusion).

iff f(z) 5 g(z) for all 2 EX.

real interval; define f 2 g as in (ii).

9.

10.

11. 12.

13.

14.

15.

16.

17.

18.

19.

20.

Semilattices as Posets

A poset is a join-semilaftice (dually, meet-semilattice) iff sup {a, b j (dually, inf {a, b } ) exists for any two elrmcnts a and b. Prove that the dual of a join-semilattice is a meet-semilattice, and conversely. Let A be the set of finitely generated subgroups of a group c f , partially ordered under set inclusion (as in Exercise 12). Prove that (A; C) is a join-semilattice, but not necessarily a lattice. Let C be the set of all continuous strictly convex real valued functions defined on the real interval [0, I]. For f, g € C, set f 5 g iff f(z) 5 g(z) for all z [0, 13. Prove that (C; 5) is a meet-semilattice, but not a join-semilattice. Show that the poset (P; 5 ) is a lattice iff it is both a join- and meet-semilattice.

Semilattices as Algebras

Let (A; 0 ) be an algebra with one binary operation 0 . The algebra (A; 0 ) is 8

semilattice iff o is idempotent, commutative, and associative. Let (P; I ) be 8

join-semilattice. Show that the algebra (P; v ) is a semilattice where avb =sup {a, b}. State the analogous result for meet-semilattices. Let the algebra (A ; 0 ) be a semilattice. Define the binary relations 5 and 5 on A a s f o l l o w s : ~ ~ , b i f f a = u o b ; a ( ~ b i f f b = a o b . Provetha t@; i , ) i s a p o s e t , a s t l poset it is a mec.t-semilattice, and ah6 =a o b ; that (A; 5 v) is a poset, as a poset it is a join-semilattice, and avb = a o b ; and that the dual of (A; 5 ,) is (A ; 5 v).

Theorem 1 for Semilattices

Prove the following statements: (i) Let the poset % = ( A ; 2 ) be a join-semilattice. Set avb =sup {a, b } . Then the

(ii) Let the algebra I =(A ; 0) be a semilattice. Set a 5 b iff a o b =b . Then I p = algebra 2P =(A ; v ) is a semilattice.

(A ; 5 ) is a poset, and the poset I p is a join-semilattice.

8 I. First Conccpts

21.

22.

23.

24.

26.

*26.

27.

28.

29.

30.

31.

32. 33.

(iii) Let the posct % =(A; 5 ) be a join-semilattice. Then ( W ) P =%. (iv) Let the algebra % = ( A ; 0) be a semilattice. Then (W)Q =a. Formulate and prove Theorem 1 for meet-semilattices.

Lattices as Algebras

Prove that the absorption identities imply the idernpotency of A and v . (Hint: simplify av(aA(ava)) in two ways to yield a =ava.) Let the algebra (A; A , v ) be a lattice. Define a 2 , b iff ahb =a; a 5 b iff avb =b. Prove that a 5 b iff a 2 ,, b. Prove that the algcbra (A; A , v ) is a lattice iff (A; A) and (A; v ) are semilattices and a = a ~ b is equivalent to b =avb. Verify that if (A; A, v i ) and ( A ; A, v q ) are both lattices, then vl is the same as v,. Are the three identities defining semilattices independent (that is, none follows from the others)? Prove that an algebra (A; A, v ) is a lattice iff i t satisfies

a = ( b ~ a ) v a

and (((aAb)Ac)vd)ve = ( ( ( b A C ) A a ) V e ) V ( ( f V d ) A d ) .

(See J .A. Kalman [1968]. The first definition of lattices by two identities was found by R. Padmanabhan in 1967, Notices Amer. Math. SOC. 14, No. 67T-468, and published in full in R. Padmanabhan [1969]. J. A. Kalman's two identities are slightly improved versions of those of R. Padmanabhan. R. N. McKenzie [1970] pro\-ed the existence of a single identity characteriaing lattices.)

Miscellany

Let e be a binary relation on the set A. The traneitive extension of e is a binary relation 6 defined by the following rule: abb iff there exists a sequence a =xo, x i , . . . , x A = b with xgzi+iA f"' i =0, . . . , n - I . Show that for a reflexive relatione, tho transitive extension e IS apartial ordering relation iff 6 is antisymmetric. Express this condition in terms of e . Let be a reflexive and transitive binary relation (quasi-ordering relation) on the nonvoid set A. Call B A a block iff B * 0, aeb and bea for any a, b C B, and for u C B, b € A , q b and bea imply that b E B. Let P be the set of all blocks and set Bi 5 B2 (Bl, B, EP) iff biebz for some (thus for all) bl E Bl and b2 E B,. Prove that (P; 5 ) is a poset. Let A be a set of sets. Let q b mean that there is a one-to-one map from a into b. What is ( P ; I)? (Notation is that of Exercise 28.) Let the binary operation o on the set A be associative. Give a rigorous proof of the statement that any meaningful bracketing of a,, 0' * * oan-l will yield the same element. Suppose that in a poset bvc, av(bvo), and avb exist. Prove that (uvb)vc exists and that av(bvc) =(avb)vc. Prove that if ahb exists in a poset, so does av(ahb). Let H and K be subsets of a poset. Suppose that sup H, sup K, and sup (H u K) exist. ( H u K is the set union of H and K.) Under these conditions verify that (sup H)v(sup K) exists ltnd equals sup (H u K).

2. How to Describe Latticcbs 9

34. In a poset P define the comparability relation y : For a, b E P , ayb iff a 5 b or b 5 a. Take a sequence at, . . . , ak of elements of P satisfying the following conditions:

(ii) Fornoeleinentsa,bEP,andi,j <k,i+j is a=ai=aj, b = a i + l = a j + i , o r a = a i =

(iii) For no 1 5 i 5 k - 2 is aiyai+2, and neither ak-lya, nor akyaz holds. Prove that k is even. Prove that a binary relation y on a set A is the comparability relation of some poset ( A ; 5 ) iff y satisfies the condition of Exercise 34 (A. GhouilSt-Houri [1962]; see also P. C. Gilmore and A. J. Hoffman [I9641 and M. Aigner [ l969]) .

(i) ai*ai+l ,a iya i+l for i=l , . . . , k - l ; a ~ + a i , a , y a ~ .

=akr b = U i + l = U i .

*35.

2. How to Describe Lattices

To illustrate results and to refute conjectures we shall have to describe a large number of examples of lattices. This can be done by basing the examples on known mathematical structures, as illustrated in the exercises of Section 1. In this section we list a few other methods.

A finite lattice can always be described by a meet table and a join table. For example: let L = { O , a , b, l}.

b O O b b -l-l-l-l- 1 0 a b l

We see that most of the information provided by the tables is redundant. Since both operations are commutative, the tables are symmebric with respect to the diagonal. Furtherinore, $AX =x, xvx =x; thus the diagonals themselves do not provide new information. Therefore, the two tables can be condensed into one.

It should be emphasized again that the part above the diagonal determines the part below the diagonal since either determines the partial ordering. To show that this table defines a lattice, we have only to check the associative identities and the absorption identities.

An alternative way is to describe the partial ordering, that is, all pairs (z, y) with z<y. In the preceding example we get 5 = { ( O , 0), (0, a) , (0, b) , (0 , l), ( a , a), (a, l), ( b , b), ( b , 1 ) s (1, 1)).

10 I. First Concepts

Obviously, all pairs of the form (2, x) can be omitted from the list, since we know that x < x. Also, if x < y and y I z, then x I z. For instance, when we know that 0 < a and a< 1, we do not have to be told that 0 1 1. To make this idea more precise, let us say that in the poset ( P ; <), a covers b or b is covered by a, in notation, a > b or b < a iff u > b and for no x, a > x > b. The covering relation of the preceding example is simply:

= { ( O , 4, (0, a>, (a , I) , (b , 1)).

Does the covering relation determine the partial ordering? The following lemma shows that in the finite case it does.

LEMMA 1. Let ( P ; <) be a finite poset. Then a< b i f f a = b or there exists a finite sequence of elements xo, . . . , x , - ~ such that xg = a , x,,-~ = b, and xi < x i + l , for O<i<n-1 .

PROOF. If there is such a finite sequence, then a =xo<xl< * *<x, , -~ = b, and a trivial induction on n yields a < b. Thus it suffices to prove that if a, < b, then there is such a sequence. Fix a, b C P , a. < b, and take all subsets H of P such that H is a chain in P , a is the smallest element of H, and b is the largest element of H. There are such subsets: {a , b}, for example. Choose such an H with the largest possible number of elements, say, with m elements. (P is finite.) Then H ={xn, . . . , x ~ - ~ } , and we can assume that xo <xl <. - - <x,-~. We claim that in (P; <) we have a =xo <xi < - * * < x,,,-~. Indeed, xi <xi+l by assumption. Thus if xi < xi+l does not hold, then xi < x < xi+l for some x E P, and H u {x} will be a chain of m + 1 element,s between a and b, contrary to the maximality of the number of elements of H.

The diagram of a poset (P; I) represents the elements with small circles o ; the circles representing two elements 5, y are connected by a straight line iff one covers the other; if x covers y, then the circle representing x is higher than the circle representing y. The diagram of the lattice discussed previously is Rhown in Figure 1.

0 Figure 1

Sometimes the “diagram” of an infinite poset is drawn. Such diagrams are always aocompanied by explanations in the text. Note that in a diagram the intersection of two lines does not indicate an element. A diagram is planar if no two lines intersect. A diagram is optimal if t,he number of pairs of intersecting lines is minimal. Figures 1,

2. How to Describe Lattices 11

2. and 3 show planar diagrams; Figure 4 is an optimal but not a planar diagram; Figure 6 is not opt,inial. As a rule, optimal diagrams are the most practical to use.

The methods we will use will be combinations of the previous ones. The lattice of Figure 2 has five elements: 0, a , b, c , i, and b < a , cvb =i, aAC = o . This description

i i

0 0

Figure 2

is complete-that is, all the relations follow from the ones given. 9X3 has five elements: o.a, b , c , i and ar\b=aAc=br\c=o,avb=avc=bvc=i.

In contrast, we can start with some elements (say, a, b, c ) , with some relations (say, b < a), and ask for the “most general” lattice that can be formed, without specify- ing the elements t o be used. (The exact meaning of “most general” will be given in Section 5.) In this case we continue to form meets and joins until we get a lattice. A meet (or join) formed is identified with an element that we already have only if this identification is forced by the lattice axioms or by the given relations. The lattice we get from ( I , b , c and satisfying b <a is shown in Figure 3.

Figure 3

To illustrate these ideas we give a part of the computation that goes into the construction of the most general lattice L generated by a, b, c with b <a. We start

by constructing joins and meets uvc, bvc, (LAC, bAc; note that uvbvc=(nvb)ve = a v c since a v b = a ; similarly, aAb/\c=b/\c. Next we have to show that the seven elements (a, b, c, avc , bvc, UAC, bAc) that we already have are all distinct. Remember that two were equal if such equality would follow from the relation (b <a) and the lattice axioms. Therefore, to show a pair of them distinct, it is enough to find a lattice K with a, b, c E K , a > b, where that pair of elements is distinct. For instance, to show u+avc , take the lattice (0,1,2} with 0 < 1 <2, and b =0, u = I , c = 2 . The next step is to form further joins and meets: bv(uAc), U A ( ~ V C ) . It is easy to see that all the other joins and meets are equal to a given one-for example, bA(aAc)=bAc, U V ( U A C ) = U . Now we claim that the nine elements (a, b, c, avc, bvc, a m , bAc, bv(aAc), ar\(bvc)) form a lattice. We have to prove that by joins and meets we cannot get anything new. The thirty-six joins and thirty-six meets we have to check are all trivial. For instance, av(bv(aAc)) =avbv(wAc) =a, by the absorption identity, since avb = a ; also cA(aA(bvc)) = (cAa)A(bvc) = U A C , since CAU< bvc.

Exercises

1.

2.

3. 4. 5.

G .

7.

8.

9.

10.

11.

12.

Give the join and meet table of the lattice in Exercise 1,9(a) for one-, tn-o-, and three-element sets X. Give the set I for the lattices in Exercise 1. Which is simpler: the meet and join table or 5 ? Describe a practical method of checking associativity in a join (meet) table. Relate Lemma 1 to Exercise 1.27. Let (P; 5 ) be a poset, a, b E P , a < b, and let C denote the set of all chains H in P with smallest element a, largest element b. Let Ho 5 Hi mean Ho c Hi for Ho, H , EC. Show that (C; 5 ) is a poset with zero {a, b}. The poset (Q; 5) is said to satisfy the Amending Chain Condition iff all increasing chains terminate; that is, if xi€&, i =0,1,2,. . . , and xo 5 xi 5 - * * < 2,s ., then for some m we have xm =$,+I =- - *. The element z of Q is mximal if x 5 y (y C Q ) implies that x =y. Show that the Ascending Chain Condition implies the existencr of maximal elements and that, in fact, every element is included in a maximal element (but not conversely). Dualize Exercise 6. (The dual of maximal is minimal and the dual of ascending is descending.) If ( Q ; I) is a lattice and z is a maximal element, then m is the unit. Show that this statement is not, in general, true in a poset. Give examples of posets without maximal elements and of posets with masiinal elements in which not every element is included in a maximal element,. Let (P; I) be a poset with the property that for every a, b E P, a < b, all chains in P with smallest element a and largest element b are finite. Show that the poeet (C; I) formed from (P; 2) in Exercise 5 satisfies the Ascending Chain Condition (see Exercise 6). Extend Lemma 1 to posets satisfying the condition of Exercise 10 (CombineExerciscs 6 and 10). Could the result of Exercise 11 be proved using the reasoning of Lemma 1 ?

2. How to Describe Lat>tices 13

13. 14. 16.

Is the result of Exercise 1 1 the best possible? (No.) Describe a method of finding themeet and join table of a lattice given by a diagram. Are the posets of Figures 4 and 6 lattices?

Figure 4 Figure 5

16. Show that Figures 6 and 7 represent the same lattice.

Figure G

17. Simplify Figure 8.

Figure 7

Figure 8

14

18.

19.

20.

21.

*22.

*23.

24.

25.

I. First Concepts

Simplify Figure 9. What is the number of pairs of intersecting lines in an optimal diagram? (Zero.)

Figure 9

Draw the diagrams of P ( X ) (Exercise 1.9(a)) for 1x1 = 3 and /XI =4. (1x1 is the cardinality of X.) Does P ( X ) have a planar diagram for 1x1 =3? Draw the diagrams of the lattice of binary relations on X (partially ordered by set inclusion) for 1x1 5 4. Describe the most general lattice generated by a, b, c, such that a > b, avc =bvc, aAC =bAC. Describe the most general lattice generated by a, b, c , d , such that a > b > c. (Hint: see Figure VI.1.2.) Show that the most general lattice generated by a, b, c, d , such that a > b end c > d , is infinite. (Hint: see Figure VI.1.6,) Let N be the set of positive integers, L = {(n, i ) I n EN, i =0, 1). Set (n, i ) 5 (m, j ) iff n 5 rn and i 5 j. Show that L is a lattice and draw the “generalized diagram” of L. Draw the diagrams of all lattices with at most six elements.

TO dispel a false impression that may have been created by Sections 1 and 2, namely, that the proof that a poset is a lattice is always trivial, we present Exercises 26-36 showing that the poset T, defined in Exercise 34 is a lattice. This is a result of D. Tamari [1961], first published in H. Friedman and D. Tamari [1967]. The present proof is based on S. Huang and D. Tamari [1972]. See also D. Huguet [1976].

26. Let T, denote the set of all possible binary bracketing8 of x@, * * x,; for instance,

( (zozl)(zzz3)), ( (zo(zlzz))zs), (((zozl)zz)z3)}. Give a formal (inductive) definit’ion of

Replacing consistently all occurrences of (AB) by A ( B ) in a binary bracketing, we get the right bracketing of the expression. For instance, the right brackcting of

(inductive) definition of right bracketing and prove that there is a one-to-one correspondence between binary and right bracketings. Show that in a right bracketing of zozl * * - zn, there is one and only one opening bracket preceding any x;, 1 5 i 5 n.

TO =@O), Tl ={(z@l)}* TZ r{((z@l) x2)r (zOo(zlzZ)))* T 3 ={(zO(zl(zZz3))), (zO((zlz2) z3))9

T,. 27.

((z0(z122))z3) is z0(z1(z2))(23) and Of ((z@1)(zZz3)) is z0(z1)(zZ(z3))* Give a

28.

3. Some Algebraic Cancept,s 15

29.

*30. 31.

32.

33. 34.

*35.

36.

Associate with a right bracketing of x e i * * . x, a bracketing function E: { 1, . . . , n} - { 1, . . . , n } defined as follows: For 1 5 i 5 n there is, by Exercise 28, an opening bracket before xi; let this bracket close following zj; set E(i ) =j. Show that E has the followingproperties: (i) i I E( i ) for 1 5 i 5 n; (ii) i -<j < E(i ) imply that E ( j ) 5 E ( i ) f o r l I i < j < n . Show t,hat (i) and (ii) of Exercise 29 characterize bracketing functions. Let En denote the set of all brackotingfunctionsdefined on (1, . . . , 12). For E, F CE,,, set E 5 F iff E ( i ) 5 P ( i ) for all i, 1 5 i I n . Show that I is a partial ordering, En as a poset is a lattice, and (EhF) ( i ) =inf {E( i ) , P( i ) } . The serniassociative identity applied at the place i is a map ai: T,,-T, defined as follows: I f E = * . . ( A ( B 0 ) ) * * 1, where the first variable in B and C is xi and xj, respectively, then E q = . * * ( ( A B ) C ) * * * ; if E is not of such form, then Eai =E. Let X and 1' denote the bracket,ing functionsassociated with E and Eai, respectively. Show that X ( k ) = Y ( k ) for k +i, i 5 k _< n, and Y ( i ) =j - 1. Conclude that X > Y. Show the converse of Exercise 32. For E, F T , define E < F to mean the existence of a sequence E = X,, Xi,. . . , A-k = F , X i € T n , 0 5i 5 k such that X ; + l can be gotten from X i by some application of t,he semiassociatire law for 0 i < 12. Let' E 5 F mean E = F or E < I". Show t>hat 5 is a partial ordering. Verify that Figure 10 is the diagram of T , and T,. Is t,he diagram of T , optimal? (No.)

Figure 10

Let X,Y € E n and X >- Y. Let E and F be the binary bracketing5 associated with X and Y, respectively. Show that F =Eai for some i . Show that Tn is a lattice for each n 2 0. (In fact, T , z E w )

3. Some Algebraic Concepts

The purpose of any algebraic theory is the investigation of algebras up t o isomorphism. We can introduce two concepts of isomorphism for lattices.

!&), and the map 47 : L,, - L, is an isomorphism iff y is one-to-one and onto, and

The lattices !& = (Lo; <) and gi =(L , ; I) are isomorphic (in symbol, 2"

a b in iff a y < by in &.

The lattices go =(Lo; A, v) and Bi =(L , ; A, v) are isomorphic (in symbol, zo sgl) , and the map y : Lo +L, is an isomorphiam iff pl is one-to-one and onto, and

(aAb)y =aplAbpl, (avb)V =nrpvbpl.

An isomorphism of a lattice with itself is called an automorphism. It is easy to see that the two isomorphism concepts coincide under the equivalence

of Theorem 1.1. However, when we generalize these to homomorphism concepts, we get various new nonequivalent notions. I n order to avoid confusion, each will be given it different name.

From now on we will abandon the precise notation ( L ; A , v) and ( L ; I) for lattices and posets ; we will simply write italic capitals, indicating the underlying sets, unless for some reason we want to be more exact.

Note that the first definition of isomorphism can be applied to any two posets Lo, L1, thus yielding an isomorphism concept for arbitrary posets. Having this concept of isomorphism, we can restate the content of Lemma 2.1: The diagram of a finite poset determines the poset up to isomorphism.

is an n-element chain. Observe that I(&,) = n - 1. If C ={xo, . . . , xap1} is an n-element chain, xo < xI < - then y : i +xi is an isomorphism between En and C . Therefore, the n-element chain is unique up to isomorphism.

LetQ, denote the set (0, . . . , n - l} orderedby 0 < 1 < 2 < - - * < n - 1. Then

< x, - The isomorphism of posets generalizes as follows : The map y : Po --L PI is an isotone map (also called monotone map or order-preserving

niap) of the poset Po into the poset Pi iff a is a semilattice both under A and under v, we get two homomorphism concepts, meet-homomorphism homomorphism) and join-hmamorphism (v-homomorphism). A homomorphism is a map that is both a meet-homomorphism and a join-homomorphism. Thus a homomorphism tp of the lattice Lo into the lattice L, is a map of Lo into L1 satisfying both (aAb)g, =ayr\bg, and (avb)y =uyvbp A homomorphism of a lattice into itself is called an endomorphism. A one-to-one homomorphism will also be called an embedding. (The list of homomorphism concepts will be further extended in Section 6.)

Note that meet-homomorphisms, join-homomorphisms, and (lattice) homomorphisms are all isotone. Let us prove this statement for meet-homomorphisms. Jf y : Lo --c Li, (aAb)g, = uyAbtp for all a , b E Lo, and if 2, y E Lo, x < y in Lo, then x = XAY; thus x y = ( ~ ~ y ) g , = x y ~ y y , and x y s y g , in L,. Note that the converse does not hold, nor is there any connection between meet- and join-homomorphisms.

Figures 1-3 are three niaps of the four-element' lattice L of Figure 2.1 into the three-element chain &. The map of Figure 1 is isotone but is neither a meet- nor a join- honiomorphism. The map of Figure 2 is a join-homomorphism but i R not a meet- homomorphism, thus not a homomorphism. The map of Figure 3 is a homomorphism.

The second basic algebraic concept is that of a subalgebra: A sublattice j ? = ( K ; A, v) of the lattice g = ( L ; A, v) is defined on a nonvoid

subset K of L with the property that a, b c K implies that aAb, a v b < K (A,V taken in g), and the A and the v of 9 are restrictions to K of the A and the v of 2.

3. Scinc: Algebraic Concepts 17

Figuro 1 Figlire 2 Figure 3

To put this in simpler language, we take a subset K of a lattice L such that K is closed under A and v. Under the same A and v, K is a lattice; this is a sublattice of L.

The concept of a lattice as a poset would suggest the following sublattice concept: Take a nonvoid subset Ii of the lattice L ; if the subposet K is a lattice, call K a sublattice of L. This concept is different from the previous one and we will not use it at all; recently some use was made of this in D. Kelly and I. ltival [1975a].

Let A , , L c-4, be sublattices of L. Then n ( A , I 31 € A ) (the set theoretic intersection of A, , Acll) is also closed under A and v ; thus for every H E L, H # 0, there is a sniallest [HI L containing H and closed under A and v. The sublattice [HI is called the aublattice of L generuted by H , and H is a generating set of [HI . If H = { t r , b, c, . . .}, then we write [a , b, c, . . .] for [HI .

The subset K of the lattice L is called convex iff u, b E K , c EL, and a < C< b imply that c E I;. For u, b E L, a < b, the interval [a , b] = {a ] a < a < b} is an important example of a convex sublattice. For a chain C , a, b E C, a< b, we can also define the hulf-open intervals: (u , b] ={x I u <a< b} and [a , b ) = { x I u < x < a } , and the open interval: ( ( I , b ) ={a I u < x < b}. These are also, whenever nonvoid, examples of convex sublattices. A sublattice I of L is an ideal iff i € I and u E L imply that ar\i E I. An ideal I of L is proper iff I +L. A proper ideal I of L is prime iff a, b E L and arrb E I imply that a E I or b E I . In Figure 4, I is an ideal and P is a prime ideal; note that I is not prime.

Figure 4

The concept of a convex sublatttice is a typical example of the interplay between the algebraic and order theoretic concepts. We will now examine these concepts more

Since the intersection of any number of convex sublattices (ideals) is a convex sublattice (ideal) unless void, we can define the convex sublattice generated b y a subset H , and the ideul geiieruted by u subset H of the lattice L, provided that H # 0. The ideal generated by a subset H will be denoted by (HI , and if H ={a}, we write ( ( I ] for ( ( t i } ] ; we shall call (u] a principal ideul.

closely.

L E M M A 1. Let L be a lattice and let H and I be nonvoid subsets of L. (i) I is an ideal i f f a , b E I implies that avb E I , and a E I , x E L, x I a imply that

(ii) I = (HI i f f I is an ideal, H c I , and for all i c I there exists an integer n 2 1 am!

(iii) For U E L , (a]={x I x < a } = { x ~ a 1 X E L } .

XEI.

there exist h,, . . . , hn-l EH such that i<hOv* - ~ v h , - ~ .

PROOF. (i) Let I be an ideal. Then a , b c I implies that a v b E I , since I is a sublattice. I f

x l n € I , then x = x ~ a E I , and the condition of (i) is verified. Conversely, let I satisfy the condition in (i). Let a, b E I . Then avb E I by definition, and, since ar\b<aEI, we also have a A b c I ; thus I is a sublattice. Finally, if x E L and a E I , then aAx<aEI, thus a A x c I , proving that I is an ideal.

(ii) Set I,={i I i<h,v* V ~ L , - ~ for h,, . . . , hnp1 EH and for some integer n 2 I}. Using (i), it is clear that I , is an ideal, and obviously H E I,. Finally, if H J and J is an ideal, then I , c J , and thus I , is the smallest ideal containing H ; that is, I = I,.

0

0 (iii) This proof is obvious directly, or by application of (ii).

the ideal luttice and I,(L) the nugmented ideal lattice of L. Let I ( L ) denote the set of all ideals of L and let I,(L) = I ( L ) u {@}. We call I ( L )

'COROLLARY 2. I ( L ) and I,(L) are posets under set inclusion, and as posets they are 1 attices .

I n fact, I v J = ( I u J ] , if we agree that (a]= 0. From (ii) of Lemma 1 we see that for I , J E I ( L ) , x E I v J iff x < ivj for some i € I , j E J . It does not imtter that we consider only two ideals in this formula. In general?l

VVAI A€A)=(u(IA I A E 4 1 ; that is, any nonvoid subset of I ( L ) has a supremum. Combining this formula with Leniina l(ii), we have:

COROLLARY 3. Let I,, A E A , be ideals and let I = V ( I A I AEA). Then i € I i f f il j,,v. - wjAm-i, for some integer n 2 1 and for aome l o , . . . , A n - I € A , jAiE IJi .

Now observe the formulas:

(a]A(b] = ( a ~ b ] , (a]v(b] = (avb].

Since a+b implies that (a] -+(a], these forniulas yield:

COROUBRY 4. L can be embedded in I ( L ) and also in I,(L), and a +(a] is a n embedding.

The notation (In I REA) stands for a family (of ideals) indexed by A. The operations n, U for families of sets and A, v for families of elements of a poset are defined the same way as they are defined for sets of sets and for subsets of posets, respectively.

3. Some Algebraic Concepts 19

Let us connect homomorphisms and ideals (recall that Qz denotes the two-element chain with elements 0 and 1).

LEMMA 5. (i) I is a proper ideal of L iff there is a join-homomorphism q of L onto B2 such that

(ii) I i s a prime ideal of L iff there is Q homomorphism q of L onto Q2 with I =Oqri. I =Oq-' (the complete inverse image of0, that is, I ={x I xq = O } ) .

PROOF. (i) Let I be a proper ideal and define q~ by x q = 0 if x E I, xq = 1 if x 6 I; obviously,

this q is a join-homomorphism. Conversely, if q is a join-homomorphism of L onto Q2 and I =*-I, then for a, b E I , we have aq =bq = 0 ; thus (avb)g, =ayvbq = OvO=O, that is, avbEI . If a E I , x E L , x I a , then xq<aq=O, that is, q = O ; thus x c I. Finally, q is onto, therefore I +L. @

(ii) If I is prime, take the q constructed in the proof of (i) and note that y can violate the property of being a homomorphism only with a, b ( I . However, since I ie prime, aAb 6 I; consequently (aAb)q = 1 =aqAbq, and so q is a homomorphism. Conversely, let q be a homomorphism of L onto & and let I = O q - i . If a, b( I , thenaq=by=l , thus (aAb)q=q~by=l,andthereforeaAb$I; Iisprime.

By dualizing we get the concepts of dual ideal (also calledfilteri), principal dual ideal, [a ) (principul filter), the dual ideal [ H ) generated by H , proper dual ideal, prime d m 1 ideal (prime filter, or ultrafilter), the lattice B(L) of dual ideals ordered by set inclusion, and Bo(L) =B(L) u {a} ordered by set inclusion. Note that in B(L) (and Bo(L)) the largest element is L ; i f L has 0 and 1, then L = [0) is the largest and (1) = [l) is the smallest element of B(L). Furthermore, for a, b EL we have

[a)A[b) = [avb), and [a)v[b) = [aAb).

LEMMA 6. Let Z be an ideal and let D be a d m 1 ideal. If I n D # 0 , then I n D ia a convex sublattice, and every convex sublattice can be expressed in this form in one and only one way.

PROOF. The first statement is obvious. To prove the second, let C be a convex sublattice and set I = (C], D = [C). Then C E I n D. If t E I n D, then t €1, and thus by (ii) of Lemma 1, t c for some c E C; also, t E D ; therefore, by the dual of (ii) of Lemma 1, t 2 d for some dEC. This implies that t EC since C is convex, and so C = I n D .

Suppose now that C has another representation, C = I , n D,. Since C c Ii, we have (C] c I,. Let a E I, and let c be an arbitrary element of C. Then avo E I , and avc 2 c D,, so avc E Di, thus avc E I , n D, = C. Finally, a< avc E C; therefore, a E (C].

f In the literature, filter means one of the following four concepts: ( i ) dual ideal; (ii) proper dual ideal; (iii) dual ideal of the lattice of all subsets of a set; (iv) proper dual ideal of the lattice of all subsets a set. Further variants allow the empty set as a filter under (i) or (ii). 3 Griltzer

This shows that 11 = (C]. The dual argument shows that D, = [C). Hence the uniqueness of such representations.

An equivalence relation 0 (that is, a reflexive, symmetric, and transitive binary relation) on a lattice L is called a congruence relation of L iff a, = b, (0) and al = bl (0) imply that U , A U ~ = (0) and aOval = b,vbl (0) (Substitution Property). Trivial examples are w , 1, defined by x ~y (0) iff x =y; x - y ( L ) for all x and y. (o is the Greek o and stands for 0; i is the Greek i and stands for identity and 1.) For a € L, we write [a]@ for the congruence class containing a, that is, [a ]@ ={x I x = a (0)).

LEMMA 7. sublattice.

Let 0 be a congruence relation of L. Then for every a E L, [a]@ is a convex

PROOF. Let x, y E [a]@; then x =a (0) and y E a (0). Therefore, ~ A Y = UAU =a (0), and x v y E ava =a (a), proving that [a]@ is a sublattice. If x < t < y, 2, y 6 [a]@, then x ra (0) and y E a (0). Therefore, t = ~ A Y EtAU (O), and t = t v x = ( ~ A u ) v ~ = (tAa)va =a (@), proving that [a]@ is convex.

Sometimes a long computation is required to prove that a given binary relation is a congruence relation. Such computations are often facilitated by the following lemma (G. Griitzer and E. T. Schmidt [1958d] and F. Mseda [1958]):

LEMMA 8. the following three properties are satisfied for x , y , z, t E L :

A reflexive binary relation 0 on a lattice L is a congruence relation iff

(i) x =y (0) iff XAY = x v y (0). (ii) x < y < z , x = y (O), and y ~ z (0) imply that x ~ z (0).

(iii) s l y and x ~ y (0) imply that xAt=yr\t (0) and x v t ~ y v t (0).

PROOF. The “only if” part being trivial, assume now that a reflexive binary relation 0 satisfies conditions (i)-(iii). Let b, c [a, d ] and a =d (0); we claim that b ~c (0). Indeed, a E C E (0) and a I d imply by (iii) that b ~ c = a v ( b ~ c ) = dv(bAc) = d (0). Now bAc I d and (iii) imply that ~ A C = ( ~ A c ) A ( ~ v c ) = d ~ ( b v c ) = bvc (0) ; thus by (i), b =c (0).

To prove that 0 is transitive, let x -y (0) and y = z (0). Then by (i), XAY ~ x v y (a), and by (iii), y v z = ( Y V Z ) V ( ~ A Z ) = ( y v z ) v ( y v x ) = x v y v z (a), and similarly, Z A ~ A X = YAZ (0). Therefore, XAYAZ E Y A Z ~ y v z ~ x v y v z (0) and X A Y A Z 5 y ~ z y v z zvyvz . Thus, applying (ii) twice, we get X A ~ A Z E X V ~ Z (0). Now we apply the statement of the previous paragraph with a. = X A ~ A Z , b =x, c =z, d = x v y v z to conclude that x = z (0).

Let x ~ y (0); we claim that xv t ~ y v l ( 0 ) . Indeed, Z A ~ r x v y (0) by (i); thus by (iii), ( X A Y ) V ~ ~ x v y v t (0). Since xv t , y v t [ ( z ~ y ) v t , x v y v t ] , we conclude, by the first paragraph of the proof, that x v t ~ y v t (0).

To prove the Substitution Property for v , let xo-yyo (0) and z1 =yi (0). Then xovxi ~ r ~ v y ~ ~ y , v y ~ (O), implying that x,vxi =yovyl (0) since 0 is transitive. The Substitution Property for A is similarly proved.

3. Somr Alg.braic Concepts 21

Let C(L) denote the set of all congruence relations on L part,ially ordered by set inclusion (remember t,hat every 0 C C(L) is a subset of L2). As a first application of Lemma 8 we prove

THEOREM 9. be described as follows :

C(L) i s a lattice. For 0, @€C(L) , @ A @ =0 n 0. The join, 0, can

x = y ( O v o ) iff there is a sequence zo = Z A Y , zi, . . . , z,, -~ = x v y of elements of L such that zo< zi < - - * (0) or zi =ziti (0). z,-l and for each i, 0 5 i < n - l , z i =zi

REMARK. C(L) is called the congruence lattice of L. Observe that C(L) is a sublattice of Part(L) (Exercises 45, 46); that is, t,he join and meet of congruence relations as congruence relations and as equivalence relations (partitions) coincide.

PROOF. 0110 =0 n 0 is obvious. To prove the statement for the join, let Y be the binary relation described in Theorem 9. Then 0 Y are obvious. If r is a congruence relation, 0 c_ r, 0 C_ r, and x = y (Y), then for each i, either zi = z , + ~ (@) or Z ~ = Z , + ~ (4)); thus z , ~ z , + ~ (r). By the transitivity of I?, X A Y = X V Y (r); thus x 'TJ (r). Therefore, Y c I?. This shows that if Y is a congruence relation, then Y =ova).

Y is obviously reflexive and satisfies Leniiiia 8(i). If x ( y < z , x =y (Y), and y ~ z (Y), then x = z (Y) is established by putting together the sequences showing x = y (Y) and y =- z (Y) ; this verifies Lemma 8(ii). To show Lemma 8(iii), let x = y (Y), n<y, with x0 , . . . , z , -~ establishing this, and t EL. Then xAt s y n t (Y) and xvt = y v t (Y) can be shown with the sequences z;At, O < i < n , zivt , O < i < n, respectively. Thus the hypotheses of Lemma 8 hold for Y, and we conclude that Y is a congruence relation. e

1Jomomorphisms and congruence relations express two sides of the same phenomenon. To establish this fact we first define quotient lattices (also called factor lrrttices). Let L be a, lattice and let 0 be a congruence relation on L. Let LI0 denote the set of blocks of the partition of L induced by 0, that is,

Y and CD

Set

and

LI0 ={[a]@ I a EL}.

[ a ] @ ~ [ b ] @ = [nAb]@

[ a ] @ v [ b ] O = [ u v ~ ] @ .

This defines A and v on L/O. Indeed, if [a ]@ =[ai]@ and [b]@ = [ a i ] @ , then a = al ( (3) and h = b, (a) ; therefore, nAb = q A b 1 (a), that is, [ a ~ b ] @ = [a1r\bi]@. Thus A and (dually) v are well defined on L/O. The lattice axioms are easily verified. The lattice L/@ is the quotient lattice of L niodulo 0.

LEMMA 10. The m ~ p

is a tmw~oniorpkisna of L onto LIO. To: x + [ x ] O @EL)

3'

22 T. First Conceptjs

REMARK. The lattice K is a homomorphic image of the lattice L iff there is a honiomorphism of L onto K. Lemma 10 states that any quotient lattice is a homomorphic image.

PROOF. The proof is trivial.

THEOREM 11 (THE HOMOMORPHISM THEOREM). Every homomorphic image of a lattice L is isomorphic to a SU&blt? quotient lattice of L. I n fact, if pl: L -L, is a homomorphism of L onto L, and if 0 U the congruence relation of L defined by x = y (0) iff xq = yq, then

an .isomorphism (see Figure 6 ) is given by

L/O s Li ;

y: [z]O -xq, x € L.

Figure 6

PBOOF. It is easy to check that 0 is a congruence relation. To prove that y is an isomorphism we have to check that y (i) is well defined, (ii) is one-to-one, (iii) is onto, and (iv) preserves the operations. (i) Let [z]0 = [y]O. Then x =y (0); thus xq = yq, that is,

([zI@)y, = ([Yl@)Y*

(ii) Let ([x]0)y = ([y]O)y, that is, zpl = y p Then x = y (a), and so [x]O = [y]O. (iii) Let a€&. Since 91 is onto, there is an x E L with zq =a . Thus ([z]O)y =u. (iv) ([z]@~[y]@)y = ([my]@)y = (z~y)q =qAyq = ([z]@)yA([y]@)y. The computation

For a homomorphism q: L +Li (not necessarily onto), the relation 0 described in Theorem 11 is called the congruence kernel of the homomorphism y ; it will be denoted by Ker(q). If L, has a zero, 0, then Oq-1 is an ideal of L, called the ideal kernel of the hommorphim q.

Let L be a lattice and let 0 be a congruence relation of L. If LIO has a zero, [a]@, then [a]@ as a subset of L is an ideal, called the ideal kernel of the congruence relution 0.

I n contrast with group or ring theory all three kernel concepts are useful in lattice theory. Note the obvious connections, for instance, if I is the ideal kernel of q, then it is the ideal kernel of Ker(pl).

The final algebraic concept introduced in this section is that of direct product. Let L, K be lattices and form the set L x K of all ordered pairs (a , b) with a L, b I(.

for v is identical.


Define A and v in L X K “componentwise”:

(a , b,) =(aAal, b A b d J (a, W a 1 , bl) = ( a w , bvb,).

This makes L X K into a lattice, called the direct product of L and K (for an example, see Figure 6).

Figure 6

LEMMA 12. Let L , L,, K , Ki be lattices, L 2 L,, K s K,. Then

L X K L, X K , z K ~ XL,.

REMBRIE. This means that L x K is determined up to isomorphism if we know L and K up to isomorphism, and the direct product is determined up to isomorphism by the factors, the order in which they are given being irrelevant.

PROOF. Let 9: L -L, and y : K + K i be isomorphisms; for a € L and b € K , define (a, b)x = (aq , by). Then x : L x K -. LI x K, is an isomorphism. Of course, Ll x Ki 2 Ki X L, is proved by showing that (a, b) -.(b, a) (a EL,, b E K,) is an isomorphism.

If L , i C I, is a family of lattices, again we first form the Cartesian product of the sets, II(L, I i C I), which is defined as the set of all functions f : I -. lJ (Li I i C I ) suoh that f ( i ) € L , for all iEI . We then define A and v “componentwise”; that is, f A g = h, fvg = k means :

f ( i ) A g ( i ) =h( i ) , f(i)vg(i) = k ( i ) , for all i e I. The resulting lattice is the direct product n(Li 1 i C I ) . If Li = L for all i E I, we get the direct power L’. Letting n denote the set (0, . . . , n - l}, Ln is ( (L X L ) X * * .) X L

n-tlmea (at least up to isomorphism). In particular, if we identify f : 2 + L with ( f (O) , f ( l ) ) , then we get L2 = L X L.

the direct product of the empty family of lattices (respectively, algebras). We shall use the convention that a one-element lattice (respectively, algebra) is

A very important property of direct products is:

THEOREM 13. be a Congruence relation of K . Define the relation 0 x 0 on L X K by

Let L and K be lattices, let 0 be a congruence relation of L, and let 0

( a , b ) = ( c , d ) ( O x < P ) i f f a e c ( 0 ) and b ~ d ( 0 ) .

Then 0 x 0 is a congruence relation on L x K. Conversely, every congruence relation of L X K i s of this form.

PROOF. The first statement is obvious. Now let Y be a congruence relation on L X K . For a, b E L define a = b (0) iff ( a , c) ~ ( b , c) (Y) for some c E K. Let d EK. Joining both sides with (aAb , d) and then meeting with (avb, a), we get (a , d ) E (b, d) (Y); thus (a, c) =(b, c) for some c E K is equivalent to (a , c) =(b, c) for all c E K . Similarly, define for a, b e K, a e b (0) iff (c, a ) ~ ( c , b) (Y) for any/for all cEL. It is easily seen that 0 and 0 are congruences. Let (a, bj ~ ( c , d ) (0 x 0) ; then ( a , x ) = (c , X ) (Y), ( y , b) = ( y , d) (Y), for all x E K and y c L. Joining the two with y = a m and x =bAd, we get (a , b) ~ ( c , d ) (Y). Finally, let (a, b) ~ ( c , a) (Y). Meeting with (CCVC, bAd), we get (a , bAd) ~ ( c , b d ) (Y); therefore, a=c (0). Similarly, b ~d (Q), and so

(a, b) = ( c , d ) (0 X 01, proving that Y = 0 x 0.

In conclusion, we introduce a nonalgebraic concept to balance the false impression that might have been created in this section that all lattice theoretic concepts are algebraic.

A lattice L is called complete if A H and V H exist for any subset H C L. The concept is self-dual, and half of the hypothesis is redundant.

LEMMA 14. 1 attice.

Let P be a poset in which A H exists for ul1 H E P. Then P i s a complete

PROOF. For H E P, let K be the set of all upper bounds of H. By hypothesis, A K exists; set a = A K. If h e H, then k 5 k for all k c K; therefore h a, and a E K. Thus a is the smallest member of K, that is, a = V H .

Lemma, 14 can be applied to Io(L) and C ( L ) . For a further application, let Sub(L) denote the set of all subsets A of L closed under A and v partially ordered under set inclusion. In other words, if A E Sub(L) and A =k 0, then A is a sublsttice of L. Obviously, Sub(L) is closed under arbitrary intersections.

COROLLARY 15. i s a complete lattice. The lattice S u b ( L ) i s a complete lattice.

Io(L) and C(L) are complete lattices. If L has a smallest element, I(L)

Exercises

1 . 2.

3.

4.

5 .

6. 7.

*8. 9 .

10. *11.

12. 13.

14.

16. 16.

17.

18.

19. *20. 21.

22.

23.

24.

Formalize and prove the equivalence of the two isomorphism concepts. Let p: Lo +Ll, y : L1 -Lo be (lattice) homomorphisms. Show that if p(y is tthe identity map on Lo and yp is the identity map on L1, then p is an isomorphism and y =p-1. Furthermore, prove that if p: Lo +Ll, y : Li +Lz are isomorphisms, then SO

are p-1 (the inverse map, p-1: Li +Lo) and py. A one-to-one and onto homomorphism is an isomorphism. Is a one-to-one and onto isotone map an isomorphism ? Find a general construction of meet-homomorphisms and join-homomorphisms that are not bomomorphisms. Find a subset H of a lattice L such tJhat H is not a sublattice of L but H is a lattice under the partial ordering of L restricted to H . Show that a lattice L is a chain iff every nonvoid subset of L is a sublattice. Prove that a sublatt,ice generated by two distinct elements has two or four element,s. Find an infinite lattice generated by three elements. Verify that a nonempty subset I of a lattice L is an ideal iff, for a, b EL, avb € I is equivalent to a, b € I . Prove that if L is finite, t'hen L and I ( L ) are isomorphic. How about Io(L)? Is there an infinite latt,ice L such that L z I ( L ) , but not every ideal is principal? (There is no such lattice; see D. Higgs [1971].) Prove the completeness of Io(L) without using Lemma 14. Prove that C ( L ) is complete by showing the following description of infinite joins: Let H E C ( L ) , @ = V H . Then s=y (a) iff there exists a finite sequence shy= zo, z,, . . . , z , - ~ =zvy, such that zo 5 zi 5 * * * 5 z n - l , and zi =z i+1 (OJ for 0 5 i < n - 1 and for some Oi CH. Let p: L + K be an onto homomorphism, let I be an ideal of L, and let J be an ideal of K . Show that Ip is an ideal of K , and Jp-1 ={a I a EL, ap EJ} is an ideal of L. Is the image Pp of a prime ideal under a homomorphism p prime again? Show that the complete inverse image Pp-1 of a prime ideal P under an onto lattice homomorphism p is prime again. Show that an ideal P is a prime ideal of L iff L - P is a dual ideal-in fact, a prime dual ideal. Let H be a nonvoid subset of the lattice L such that a, bEH implies that avbEH. Then

(HI = { t 1 t < h for some h c H } ; that is,

(HI = u ((hl I EH).

Show that if K is a sublattice of L, then K is isomorphic to a sublattice of I ( L ) . Verify that the converse of Exercise 19 is false even for some finite K . Prove that if !R5 (see Figure 2.2) is isomorphic to a sublattice of I (L) , then ' 2 5 is isomorphic to a sublattice of L. Find a lattice L and a convex sublattice C of L that cannot be represented as [a]@ for any congruence relation 0 of L. State and prove an analogue of Lemma 8 for join-congruence relations, that is, for equivalence relations on a lattice satisfying the Substitution Property for joins. Describe Ov@ for join-congruences.

26

*26.

26. 27. 28.

29. 30.

*31. 32. 33. 34.

36. 36. 37.

38. 39. 40.

41. 42. 43.

44. 46.

46. 47.

48.

49. 60.

I. First Concepts

Find a lattice L such that L nL/O for all 0 *r, and there are infinitely many such 0. Describe the congruence lattice of %s. Describe the congruence lattice of an n-element chain. Describe the congruence lattice of the lattice of Figure 2.3, and list all quotient lattices. Construct 8 lattice that has exactly three congruence relations. Construct infinitely many lattices L such that each lattice is isomorphic to its congruence lattice. Can an L in Exercise 30 be infinite? Generalize Lemma 12 to the direct product of more than two lattices. Show that g5 zL XK implies that L or K has only one element. Show that I(L) is conditionally complete: Every nonvoid set H with an upper bound has a supremum, and dually. I(L) is complete iff L has a zero. Show that the ideal kernel of a homomorphism is an ideal of L. Find an ideal that is the ideal kernel of no homomorphism. Find an ideal that is the kernel of more than one (infinitely many) homomorphisms (congruence relations). Prove that every ideal of a lattice L is prime iff L is a chain. Under what conditions does L x K have a planar diagram? Let L and K be lattices and let q: L +K be one-to-one and onto satisfying {(ahb)q, (avb)q} = { a q ~ b q , aqvbq} for all a, b EL. Let A ESub(L). Show that AqESub(K); that in fact A +Aq is a (lattice) isomorphism between Sub@) and Sub(K). Prove the converse of Exercise 40. Generalize Theorem 13 to finitely many lattices. Show that the first part of Theorem 13 holds for any number of lattices but that the second part does not. Show that the second statement of Theorem 13 fails for (Abelian) groups. For a set X, let Part(X) denote the set of all equivalence relations on X partially ordered under set inclusion. Show that PartlX) is a (complete) lattice. (Equivalence relations and partitions can be identified as explained in Section IV.4 and there we use the notation Part(X) for the set of equivalence relations on X.) Show that C(L) is a sublattice of Part(L). The E”iT8t I8omorphi8na Theorem: Let L be a lattice, let be a congruence relation on L, and let Li be a sublattice of L. If for every a EL there exists exactly one b ELI satisfying a = b (0), then L/O z Li. Let Lo and L, be lattices and let ni: (zo, zi) +xi be the projection map of Lo XLi onto Li, i =0, 1. Prove that if A is a lattice and qi is a homomorphism of A into Li, i =0, 1, then there is a unique homomorphism y : A +Lo x Li satisfying yni =pi, i = o , 1. I n what way does Exercise 48 characterize Lo x Li ? Characterize II(Li I i €1) by projection maps and homomorphisms.

4. Polynomials, Identities, and Inequalities

From variables xo, xi, . . . , x,, . . . , we can form polynomials in the usual manner using A, v, and, of course, parentheses. Examples of polynomials are: xo, x3, zovzo, (z0~z2)v(z3~z0), ( ~ , A ~ , ) V ( ~ ~ A ~ , ) V ( ~ , I A X ~ ) . A formal definition is:

4. Polynomials, Identities, and Iiieqiislities 27

DEFINITION 1. (i) a d (ii): (i) xi E P@), 0 < i < n. (ii) I f p , q E p@), then ( p M , (pvq) E Nn).

The set P(") of n-ary lattice polynomials is the smallest set satisfying

REMARK. We shall omit the outside parentheses and write p lv . * -vpn for (- - - (piv p z ) - - * v p J , and the same for A. Thus we write xovxl for (xovxl) and xovx1vx2 for ( ( zovx~)vxz ) . Note that if n <m then P(*)c P@). It is easy to define formally what it means to substitute in a polynomial p all occurrences of xi by x j ; for instance, subst,itut'ing x4 for xl, p = ( x ~ v x ~ ) A ( z ~ A x ~ ) becomes q = ( x ~ v x ~ ) A ( x ~ A x ~ ) ; observe thatp EP(4)butqQP(4).Ingeneral,forp =p(zo ..., x ~ - ~ ) EP(n), we havep(zio, ..., Pcm) for any m > io, . . . , i,,-l. in terms of such a sequence of symbols we can define a function on any lattice:

By Definition 1, a polynomial is just a sequence of symbols. It is defined because

DEFINITION 2. An n-ary polynomial p defines a function inn variables (a polynomial function, or simply, a polynomial) on a lattice L by the following rules (ao,. . . ,a,,- E L ) : (i) I fp=xi ,p(ao ,..., an-,)=ai,O<i<n.

(11) If p(ao, . . . ,an - 1) =a, q(a0,. . . , an-J = b, uRdpAq= r,pvq = t , then r(ao, . . . , a,n-l) = ..

aAb and t(ao, . . . , a*-l) =avb.

Thus if p = (xo~x , )v (x2vx i ) , then p(a, b, c ) = (anb)v(cvb) =bvc. Definitions 1 and 2 are quite formal but their meaning is very simple.

A polynomial is an n-ary polynomial for some n. We will also use x , y , z, . . . instead of the xi'

Note that if p is a unary (n. = 1 ) lattice polynomial, then p(a ) =a for any a E L. If p is binary, then for all a, b E L, p(a, b) = a ; or for all a, b E L, p ( a , b) = b ; or for all u , b E L , p ( a , b)=ar\b; or for all a, b c L , p ( a , b)=avb.

We shall prove statements on polynomials by induction on the rank. The rank of x i is 1 ; that of pr\q (and of pvq) is the sum of the ranks of p and q.

Now we are in a position to describe [HI , the sublattice generated by H + PI :

LEMMA 3. polynomial p , and for some ho, . . . , ?L,,

a E [ H ] iff a=p(ho, . . . , hn-l) f o r some integer n 2 1 , for some n-ary E H .

PROOF. First we must show that i f a=p(ho, . . . , hn-l), ho, . . . , hn-,EH, then u E [ H ] , which can be easily accomplished by induction on the rank of p. Then we form the set (a I a =p(h,, . . . , hB-l), n 2 1, ho, . . . , hn-l E H } ; observe that it contains H and that it is closed under A and v. Since it is contained in [HI , it has to equal [HI.

COROLLARY 4. l [ H ] l < /HI +No.

PROOF. By Lemma 3, every element of [HI can be associated with a finite sequence of elements of H u {(,),A, v>, and there are no more than ]HI +KO such sequences.

DEFINITION 5. A lattice identity (inequality) i s a n expression oftheform p =q ( p < q ) , where p and q are polynomials. An identity, p = q (inequality, p < q) , holds in the lattice L i f f p(ao, al , . . .) =q(ao, a l , . . .) (p(u,, al, . . .)<q(ao, ai, . . .)) holds for uny a,, al, . . . EL.

An identity p = q is equivalent to the two inequalities p < q and q p, and the inequality p < q is equivalent to the identity p v q = q (and to pAq = p ) . Frequently, the validity of identities is shown by verifying that the two inequalities hold.

One of the most basic properties of polynomials is:

LEMMA 6. p(a,, al , . . .) I p(b,, b i , . . .). Furthermore,

A polynomial (function) p is isotone; that is, if a,< b,, u I b, , . . . , then

XgA' * 'AXn- i<P(Xo? - * - , Xn- i )<Z"V* "VX,-i.

PROOF. We prove the first statement by induction on the rank of p . The first statement is certainly true for p =xi. Suppose that it is true for q a;d r and that p = qAr. Then

P(ao, - . . )A21(4), - . .)=(q(ao, * * .)AT(ao, -))A(q(bof - - - ) A T ( b o , .)) =Ma,, - - -)Aq@o, - * . ) )A(r (ao, * * . ) ~ ~ ( h , ' . I ) = q(aof . . . )A r(ao, . . . ) = p(ao, . . . ) ;

thus p(ao, . . .)<p(b,, . . .). The proof is similar for p =qvr. Since X , A - - AX,-^< xi<x,v- * . V X , - ~ for O ( i < n -1, using the idempotency of A and v , we obtain:

XoA' ' ' A X n - l =p(XoA' ' ' A X n - i , . 9 , XoA' ' ' A X n - l ) < f l ( X o , 21, . . . , X , - i )

- < p(x,v. * 'VX,-1, . . . , XOV' * ' V X , = X O V X I V ' 'VX,- 1 ,

proving the second statement. A simple application is:

LEMMA 7. p = q such that all p i =qi, 0 2 i < n, hold in a lattice L iff p = q holds in L.

Let pi =qi, 0 i < n, be lattice identities. Then there is a aingle identity

PROOF. Let us take two identities, p , = qo and pi =pi. Suppose that all polynomials are n-ary and consider the identity (we use now 2n-ary polynomials formed by substitution, see Remark to Definition 1) :

(N) Po(xo, ' ' ' J xn-l)A$%(xn, * ' ' 9 XZn-l) =qO('o, ' ' Y xn-l)Aql(xn, . . * f x2n-1 ) .

It is obvious that if po =qo and pI =ql hold in L, then (N) holds in L. Now let (N) hold in L and let ao, . . . , a,, -i EL. Substitute xo = ao, . . . , x , - ~ = an-i, x, = - - = xenWl =a,v. * *van- , =a. By Lemma 6 ,

po(a0, - * - , an-J <po(a, * - - , a ) =a =PI@, * * - Y a) ,

4. Polynomials, Ident,ities, and Inc?qualitias 29

whence po(uo, . . . , =po(uo, . . . , U ~ - ~ ) A ~ ~ ( U , . . . , a) , and similarly for q,, q i ; thus (N) yields po(uo, . . . , a n - l ) =qo(ao, . . . , an- l ) . The second identity is derived similarly from (N). The general proof is similar.

The most important (and, in fact, characteristic) properties of identities are given by

LEMMA 8. imuges, direct products, and ideal lattices.

Identities ure preserved under the formation of sublattices, homomorphic

PROOF. Let the polynoniials p and q both be n-ary and let p = q hold in L . If L, is a sublattice of L , then p = q obviously holds in L, . Let v: L - K be an onto homoniorphisni. A simple induction shows that

p(ao, ’ * * J ‘ ,-1)y ‘P(aov9 ’ ’ ’ 9 ‘n-lv)

and proves the similar formula for q. Therefore,

~ ) ( a o ~ , ’ * 9 “n- lp ) =p(‘”, * * * J “n-1)y =q(ao, ’ . ’ J ‘@-1)Y =q(aov> * . 3 a n - l v ) J

and so p = q holds in K . The statement for direct products is also obvious. The last statement is an easy corollary of the following formula. Let p be an n-ary

polynomial and let I,, . . . , I n - l be ideals of L. Then I,, . , . , Ifi-l c I ( L ) ; thus we can substitute the Ij into p : p ( I o , . . . , I n - J is also in I ( L ) , that is, an ideal of L. This ideal can be described by a simple forniula:

p ( I o , . . . , I n - l ) ={z I x<p(io, . . . , i,-,), for some i o c I 0 , . . . in-1

This follows easily from Leninia 6 and the formula in Section 3 describing I v J , by induction on the rank of p .

Now we list a few important inequalities:

LEMMA 9. (i) (XAY)V(NAZ) I XA(YVZ).

(ii) X V ( ~ A Z ) 2 ( X V ~ ) A ( S V Z ) .

(iii) ( z ~ y ) ~ ( ~ A z ) v ( z A ~ ) 2 ( a v y ) ~ (~VZ)A(ZVZ) .

(iv) (XA~J)V(ZAZ) I N A ( ~ V ( Z A Z ) ) .

The following inequalities hold i n any lattice:

REMARK. (i)-(iii) are called distributive inequazities, and (iv) is the nzodular incquulity.

PROOF. We prove (iv) as an example and leave the rest to the reader. Since X A ~ < x, XAZIX, we obtain ( Z A Y ) V ( X A Z ) ~ X ; X A Y < ~ < ~ ( X A Z ) and X A Z < ~ V ( X A Z ) ; therefore, (XAY)V(XAZ) < ~ V ( Z A Z ) . The meet of the two inequalities ( X A ~ ) V ( X A Z ) < X and (SAY)V(SAZ) ( ~ V ( X A Z ) yields (iv).

LEMMA 10. (i) ( z ~ y j v ( X A Z ) = X A (yvz). (ii) (ZV~)A(ZVZ) = x v ( ~ A z ) . (iii) ( Z V ~ ) A Z I xv (YAZ).

Then (i), (ii), a d (iii) are equivalent in any lattice L.

Consider the following two identities and inequality:

REMARK. A lattice satisfying identities (i) or (ii) is called distributive. Note that (i) and (ii) are not equivalent for fixed elements; that is, (i) can hold for three elements a, b, c E L, whereas (ii) does not.

PROOF. Let (i) hold in L and let u, b, c E L ; then, using (i) with x =uvb, y =a, z =c,

(avb)A(avc) = ((avb)Aa)v((avb)Ac)

and using a = (avb)Aa and (i) for x =c, y =a, z = b,

=aV(aAC)V(bAC)=UV(bAC), verifying (ii).

The proof that (ii) implies (i) is the dual of the preceding paragraph. Let (ii) hold in L ; then xv(y~z) = (ZVY)A(ZVZ) 2 ( Z V ~ ) A Z , since xvz Tz, ver-

Let (iii) hold in L. Then, with 2 =a, y =b, z =avc in (iii), ifying (iii).

( W ~ ) A ( U V C ) 5 av(bA(avc)) =av((avc)Ab)

and with z =a, y =c, z = b in (iii)

<av(av(cAb)) = u v ( c A ~ ) .

This, combined with the dual of Lemma 9(i), gives (ii). 0

COROLLARY 11. The dual of a distributive lattice is distributive.

LEMMA 12. The identity

( Z A ~ ) V ( X A Z ) = X A ( ~ V ( X A Z ) )

is equivalent to the condition:

x 2 z implies that ( Z A ~ ) V X =ZA(YVZ).

REMARK. A lattice satisfying either condition is called modular.

PROOF. If X ~ Z , then Z = X A X ; thus the implication follows from the identity. Conversely, if the implication holds, then since x ~ Z A Z , we have ( ~ A ~ ) v ( x A z ) = xA(Yv(ZAz)). 0

4. Polynomials, Tdmtit ies, and Inequalities 31

Exercises

1.

2.

3.

4. 6.

6. 7.

8.

9.

10.

11.

12. 13. 14.

16.

16. 17.

18. 19. 20.

21. 22.

Give a formal definition of substituting all occurrences of xi by xj in a polynomial p . Prove formally the statement of the last sentence of the Remark to Definition 1 and the statement of the first sentence of the proof of Lemma 3. Let EZ = { t h o , . . . , hf i - i } . Prove that aE[H] iff a=p(h,, . . . , hn-i ) for some wary polynomial p. Show that the upper bound in Corollary 4 is best possible if IHI 2 3. Give the best estimates for IHI 5 2. Give a formal proof of Lemma 8. Prove (without reference to Lemma 8) that if L is distributive (modular), then SO

is I ( L ) . Show that the dual of a modular lattice is modular. Prove that L is distributive iff the identity

( Z A ~ ) V ( ~ A Z ) V ( Z A Z ) = ( ~ v ~ ) A ( ~ v z ) A ( z v z )

holds in L. Prove that every distributive lattice is modular, but not conversely. Find the smallest modular but nondistributive lattice. Find an identity p =q charaoferizing distributive lattices such that neither p 5 q nor q 5 p holds in a general lattice. Show that in any lattice

Prove that the following identity holds in a distributive lattice:

V(A(xij I 0 I j < n) I 0 I i < m) =A(V(Zij(i) I O I i < m) I f€P),

where F is the set of all functions from (0, 1, . . . , m - 1) into (0, 1, . . . , n - 1) . Derive (i)-(iii) of Lemma 9 from Exercise 10. Verify that any chain is a distributive lattice. Let L be a lattice with more than one element, let L' = L u {0}, and let 0 < s for all x c L . Show that L' is then a lattice and that an identity p =q holds in L iff it holds in L'. Show that if the identity xo =p(xo, xi, , . .) holds in the two-element lattice, then it holds in every lattice. Prove that P ( X ) is a distributive lattice. Show that Part(X) is distributive iff 1x1 2 2 and modular iff 1x1 5 3. (See Exercise 3.45.) Find an identity that holds in &z but not in '2s. Find an identity that holds in '%V3 bpt not in '2s. Let K and L be modular lattices, let D be a dual ideal of K, and let I be an ideal of L such that there exists an isomorphism 91 of D with 1. Set A = K u (L -1) (disjointunion) anddefinez 5 y,for~,yEA,asfollows:Forx,yEKorx,yEL-1,a: I y retains its meaning; for x EK, y EL -1, let x y mean the existence of& z ED such that x 5 z , zq, 5 y; x 5 y for no 2 EL -1, y EK. Is (A; 5 ) then a modular lattice? (Ph. M. Whitman [1943] and R. P. Dilworth and M. Hall [1944].) I f R and L of Exorcise 20 are distributive, is A distributive? I n Exercise 20, put I = D ={a} and show that any identity holding in K and L holds in A as well.

I. First Concepts 32

23.

*24.

*26.

26.

Examine the statements of Lemma 8 for properties of the form, “If p , =qo, then

Show that ( L ; A , v ) is a lattico satisfying the identity p = q iff it satisfies the identities

and

Pi =q1.”

( W A Z ) V Z =Z

( ( ( Z A p ) A Z ) V U ) V W = (( ( q A 2 ) A Z ) V V ) V ( ( t V U ) A U ) ,

where 2, z, u, o, and w are variables that do not occur in p or q (R. Padmanabhan [1968]). Show that the result of Exercise 24 is best possible: If p=q is an identity *not satisfied by some lattice, then the two identities of Exercise 24 cannot be replacd by one (R. N. McKenzie [1970]). Show that t,he lattice L is modular iff the inequality Z A ( Y V Z ) ~ ~ v ( ( z v ~ ) A z ) holds in L.

5 . Free Lattices

Though it would be quite easy to develop a feeling for the most general lattice (generated by a set of elements and satisfying some relation) of Section 2 by way of some examples, a general definition seems to be hard to accept. So we ask the reader. to withhold judgment on whether Definition 2 expresses his intuitive feelings until the theory is developed and further examples are presented.

The most general lattice will be called free. Since we might be interested, for instance, in the most general distributive lattice generated by a., b, c, satisfying b < ( I ,

it seems desirable to define freeness with respect to a class K of lattices.

DEFINITION 1. Let p , = q j be identities for ic I . The cluss K of afll lattices satisfying all identities pi = q%, i C I , is called an equational class of lattices. A n equational class is trivial iff it contains one-element lattices only.

The class L of all lattices, the class D of all distributive lattices, and the class 1)I of all modular lattices are examples of equational classes of lattices.

Next we have to agree on what kinds of relations to allow in a generating set. Can we prescribe only relations of the form b <a , or do we allow relations of the form unb = c or dve = f ? Lemma 9 can be rephrased to furnish an example in which the four generators a, b , c , d are required to satisfy a v b = a v c = b v c = d . Let us therefore agree that for a, generating set we take a poset P , and for relations we take all a< b that hold in P , all a d = c , where inf {a, b} = c in P , and all a v b = c , where sup {a, b} = c in P. (A more liberal approach will be presented later in this section.)

Now we are ready to formulate our basic concept:

DEPIN~TION 2. Let P be a pmet and let K be an equational class of lattices. A lattice FK(P) is called a free lattice over K generated by P iff the follm‘ng colzditions are satisfied:

(i) pK(P)EK.

5. Free Latt>icrs 33

(ii) €’c FK(P) , and for a , b, c E P , inf {a , b} = c in P i f f a A b = c in F,(P) , and

(iii) [PI = FK(P). (iv) Let L c K and let y : P - L be an isotone map with the properties that i f a , b, c P ,

inf {a, b } = c, in P , then ayAbg? = crp in L , and i f sup {u, b) = c in P , then a y v b y = crp in L . Then there exists a (lattice) homomorphism y : FK( P ) - L extending y (that is , uy = a y for all a c P ) .

sup { n , b } = c in P iff a v b = c in F,(P) .

Let E denote the identity map on P; then the crucial condition (iv) can be expressed by Figure 1. In that and in all such similar diagrams (not in the sense the word is

& P

L Figure 1

used in Section 2), the capital letters represent lattices or poset,s, and the arrows indicate homomorphisms, or maps with certain properties. Such diagrams are usually assumed to be commiitutiere, which, in this particular case, means &y = y , which is (iv). Note that (ii) is also included in the diagram if the arrows are supposed to represent maps as required in (iv). In fact, we could have required in (ii) that the identity niap on P be an embedding of P into F,(P) in the sense of (iv).

If P is an unordered set, [PI = m, we shall write F,(m) for FK( P ) and call it a free lattice on m generators over K. In case K =L, we shall always omit “over L”; thus “free lattice generated by P” shall mean “free lattice over I, generated by P”-in notation, F ( P ) .

It shouldbenotedthat ifb€F,(P), thenby(iii)andbyLemme4.3,b = p ( a oJ. . . ,a , - , ) , where p is a polynon~ial and ao, . . . , P . Thus if the y of (iv) exists (in fact, if q~ is any mapping of P into L and if y : F,(P) - L is any homoniorphisni extending tp)J then we must have

b y =p(a,, . . . , a,-,)w (and since ?,IJ is to be a homomorphismj = P ( ~ o y , - * - > a,b-ly) =P(“oy, * * * 9 %-lqI),

since (iLy =(i i (&y) =a+p

extending y , whence: From this we conclude that there is a t most one homomorphism y : F,(P) --L

COROLLARY 3. The homomorphism y in (iv) is unique.

This corollary is used to prove:


COROLLARY 4. Let both FK(P) and F;",(P) satisfy the conditions of Definition 2. Then there exists an isomorphism x : FK(P) +FB(P), and x can be chosen so that ux = a for all a E P. In other words, free lattices (over K generated by P ) are unique up to isomorphism.

PROOF. Let us use Figure 1 with L=FK(P) and y=e. Then there exist yi: FK(P)+Fg(P) and y2: Ftlf(P)+FK(P) such that &yi=& and E ~ ~ = E . Thus y,y2: FK(P)+FK(P) is the identity map E on P. By the statement preceding Corollary 3, E has a unique extension to a homomorphism BK(P) - FK(P) ; the identity map on FK(P) is one such extension. Therefore, yiy2 is the identity map on FK(P). Similarly, y2y1 is the identity map on FB(P), and so (Exercise 3.2) yi is the required isomorphism.

This settles the uniqueness, but how about existence? Naturally, FK( P ) need not exist. For instance, F,(%,) should be !Jt5, since by (ii) and (iii) of Definition2, FK(P)= = P if P is a lattice; but %&D, so (i) is violated.

THEOREM 6. Let P be a poset and let K be an equatiolaal class of lattices. Then FK(P) exists iff the following condition ie satisfied:

L and for a, b, cE P , inf {a, b} =c (E) There exists a lattice L in K such that P in P iff aAb = c in L, and sup {a , b} = c in P iff avb =c in L.

PROOF. Condition (E) is obviously necessary for the existence of F,(P); indeed, if FK(P) exists, (E) can always be satisfied with L = P,(P) by (i) and (ii) of De- finition 2.

Now assume that (E) is satisfied. In this proof a map q : P + N (NC K) will be called a homommphim iff it satisfies the conditions set forth in Definition 2(iv).

Obviously, in Definition 2(iv) it suffices to consider L with L = [Pq] . Let ( N , y ) denote this situation-that is, NEH, q: P - + N is a homomorphism and N = [ P v ] . Then FK(P) or, more precisely, (F,(P) , E ) has the property that for any (L , q) there exists a y: FK(P) - L with q =ey. To construct F,(P) we have to construct a lattice having this property for all (L , q). How would we construct such a lattice for two?

Let (Ll , ql) and (L2, q2) be given. Form Li X L2 and define a map q: P + Lf X L2 by p y =(pql, pq2) ; set N = [Pq] . The fact that q is a homomorphism is easy to check. A simple example is illustrated in Figures 2-4. Now we define yi : (x , , xz) -xp Ob- viously, for pC P , ppnyi =pvi and yi: N - Li.

P

Figure 2

5 . Free Lat,tices

Figure 3

36

L

Figure 4

If we are given any number of (L, qi), ic I , we can proceed as before and get ( N , q ) ; if one of the (Li,qi) is the (L ,e) given by (E), then (ii) of Definition 2 will also be satisfied. There is only one problem: All the pairs (Li, vi) do not form a set, so their direct product cannot be formed. The (Li,qi) do not form a set because a lattice and all its isomorphic copies do not form a set; therefore, if we can somehow restrict taking too many isomorphic copies, the previous procedure can be followed. Observe that by Corollary 4.4, in every pair ( L , qi) we have

l~il<IPqJil+XO<IPI +X".

Thus, by choosing a large enough set S and taking only those (Li,qi) that satisfy Li G S, we can solve our problem.

Now we are ready to proceed with the formal proof. Choose a set 8 satisfying /PI +Xo=ISI. Let Q be the set of all paiw (M,y) where M c S . Porm A = l l (N 1 ( N , y ) € Q ) , and for each pE P let f p € A be defined by

Finally, set

4 Uriitzer

We claim that if, for all p P, we identify p with fpJ then N satisfies (i)-(iv) of De- finition 2, and thus N = F,(P).

(i) N is constructed from members of K by forming a direct product and by taking a sublattice. By Lemma 4.8, N EH, since K is an equational class.

(ii) Let inf {a , b } = c in P . Then for every ( M , y ) E Q , a y ~ b y = c y , so that f ,((M, y))Afb((HJ y)) =fc((H, v)) ; that is, fQAfb = f c . Since p is identified with f p , we conclude that aAb = c in N .

Conversely, let aAb =c in N , that is, fQA\ fa= fc . Let L be the lattice given by (E) and let E be the identity map on P. We can assume that L = [PI; thus we can form (L, E ) . By Corollary 4.4, ILI IS1 , so there is a one-to-one map a : L +S. Let LI =La and make Li into a lattice by defining

U a A b a = (uAb)a, auvba = (avb)a.

Then L 2 L, and we can form the pair ( L I J q), where a, is the restriction of a to P ( E L). Since Li C_ 8, (LiJ ai) EQ. Now /,.fa = f c yields

fQ((LiJ a d ) A f b ( ( L i J ' 1 ) ) = f O ( ( L i , d) ; that is, aaAba =cu, which in turn gives aAb =c , since a is an isomorphism. By (E), aAb =c in L implies that. inf { a , b} =c. The second part of (ii) follows by duality.

(iii) This part of the proof is obvious by the definition of N. (iv) Take (L , rp); we have to find a homomorphism y: N + L satisfying arp = a y for

a E P . Using ILI < 181, the argument given in (ii) can be repeated to find (LiJ rpi), an isomorphism a : L -+Li such that arpa =arpi for all a E P and L, c S. Therefore, ( Li J Fi) E &* Set

y1 ' f - f ( ( L I J F I ) ) J f E '

Then for a E P, ayi = f Q ( ( L i J TI)) ='ql =aqa.

Thus the homomorphism y =yia-i: N -c L will satisfy the requirement of (iv).

Two consequences of this theorem are very important

COBOLLARY 6. For any nontrivial equational class K and for any cardiml m, a free lattice Over K with m generators, FK(m), exGts.

PROOF. It suffices to find an LEK, X E L such that 1x1 =m, and for x, y e X , x +y, x and y ere incomparable. This is easily done. Since K is nontrivial, there exists an N C K , IN1 > 1 ; thus, G2 is a sublattice of N. By Lemma, 4.8, Qz K ; by Lemma 4.8, (G2)'EK for any set I. Let 111 =m, let L=(Q2)'; for iC1, define fiEL by fi(i) =1, f i ( j ) = O for i + j , and set X ={Ii I i C I } . Obviously, X satisfies the condition.

COROLLARY 7 . exists.

For any poset P , a free lattice (over L) generated by the poset P

PROOF. Take a poset P and define I,(P) to be the set of all subsets I of P satisfying the condition: sup {u, b} c I iff a , b c I . Part,ially ordering Io (P) by set inclusion makes Io(P) a latt,ice. Identifying a with {x I xsa}, we see that Io(P) contains P and satisfies (E) of Theorem 5. The detailed computation is alniost the same as that for Theorem 20, so it will be omitted.

The argument given in the proof of Corollary 3 shows that'whenever L is generated by P , any hoinoinorphism q~ of P has at most one extension to L, and if there is one, it is given by

y: p('0, ' * ' > '%-I) -p(aoTJ ' * ' J ' % - i q ) .

This formula gives a homomorphism iffy is well defined; in other words, iff p(ao, . . . , anmi) = q ( b o , . . . , bm-l ) implies that p(aocp,. . . , c ~ % - ~ c p ) =q(b,q~, . . . , bn$-lcp), for any (1" , . . . ,

This yields a very practical method of finding free lattices and verifying their freeness.

b, ,,..., b , - , c P e n d g , : P - N E K .

THEOREM 8. In the definition of F,(P), (iv) can be replaced by the following condition: If b E F,(P) has two representations, b =p(ao, . . . , u ~ - ~ ) and b=q(bo, . . . , b m - i ) (ao, . . . , aN-i, b,, - - - , b,-iE P), then

P('0, ' ' * J '%-I) =q(bo, ' * Y ' m - 1 )

can be derived from the identities defining K and the relations of P of the form aAb =c and avb = c .

REMARK. cannot use a +b or a +bvc, and so on.

vation will be useful :

Thus, in proving p = q , we can use only the A and v table of P , but we

We illustrate Theorem 8 first by determining F,,(3). The following simple obser-

x v y v z

38 1. First Coiioepts

LEMMA 9. Let x, y, and z be elements o j a lattice L a d let xvy, y v z , zvx be pairwise incomparable. Then {xvy, yvz, z v x } generates a sublattice of L isomorphic to (B2)3 (see Figure 5) .

PROOF. Almost all the meets and joins are obvious; by symmetry, the nonobvious ones are typified by ( (zv~)A(~vz) )v ( (xv~)A(zvx) ) =xvy and ( (XV~)A(YVZ))V(ZVS) = x v y v z . Since y I (ZVY)A(YVZ) and XI (XVY)A(ZVX), we get x v y s ((XVY)A(YVZ))V

((XVY)A(ZVS)), and 2 is trivial. The second equality follows from y< (XVY)A(YVZ).

Note that, for example, (ZVY)A(YVZ) = ( X V ~ ) A ( ~ V Z ) A ( Z V X ) would imply, by joining both sides with zvx, that xvyvz =zvx; thus xvy< zvx, a contradiction. Therefore, all eight elements are distinct.

THEOREM 10. ments (see Figure 6).

A free distributive lattice on three generators, F,(3), hus eighteen ele-

a = (X V y ) A (X V 2) A ( y V 2)

= (X A y ) V (X A 2) V (J‘ A 2)

v 2)

A 2)

X A y A Z

Figure G

5 . F r ~ e Lst,t>ices 39

PROOF. Let x, y , and z be the free generators. The top eight and the bottom eight elements form sublattices by Leinma 9 and its duel; note that (ZAY)V(YAZ)V(ZAS) = ( X V Y ) A ( Y V Z ) A ( Z V Z ) by Exercise 4.7.

According to Theorem 8, we have only to verify that the lattice L of Figure 6 is a distributive lattice, and that if p , q, r are polynomials representing elements of L and pAq = r in L, then pAq = r in every distributive lattice and similarly for v. The first Statement is easily proved by representing L by sets (see Exercise 13). The second statement requires a complete listing of all triples p, q, r with pAq=r. If p , q, r belong to the top eight or bottom eight elements, the statement follows from Lemma 9. The remaining cases are a11 trivial except when p or q is x, y, or z. By symmetry, only p = x, q = y v z , Y = ( X A Y ) V ( Z A Z ) is left to discuss, but then p ~ q = r is the distributive law.

THEOREM 11. (R. Dedekind [1900]). A free modular lattice on three generators, P,(3), has twenty-eight elements (see Figure 7).

Figure 7

PROOF. Let x, y, and z be the free generators. Again modularity is proved by a representation (see Exercise 16). Theorem 10 takes care of most meets and joins not involving xi, yI, zl. Of the rest, only one relation (and the symmetric and the dual cases) is difficult to prove: xi/\?/, = u. This we do now, leaving the rest to the reader. Compute :

xlAyi = ( ( ~ A v ) v u ) A ( ( ~ A w ) v u ) (since u l ( Y A V ) V U )

(substitute u end u ) = ( ( Z A V ) A ( ( Y A V ) V U ) ) V U

= ( ( Z A V ) A ( Y V U ) A V ) V U

= ( Z A ( Y V Z ) A ( Y V ( Z A Z ) ) ) V U

= ( X A Y ) V ( X A Z ) V U = u.

Consider the lattice represented by Figure 8. We would like to gay that it is freely generated by (0, a, b, l} = P, but this is clearly not the case according to Definition 2,

1

0 Figure 8

since sup {a, b} = 1 in P, whereas in the lattice, avb < 1. So to get the most general lattice of Section 2 we have to enlarge the framework of our discussion by introducing partial lattices. Of course, the study of partial lattices is also important for other purposes.

DEKINITION 12. Let L be a lattice, H G L, a d restrict A and v to H a.3 follows: For a, b, c E H , if aAb = G ( d u a l l y , avb = c ) , then we say that in H , aAb (duully, avb) is defined and it e q u a l s c ; i f , for a, b € H , aAb ( d u a l l y , avb)eH, then we say that afrb (dually, avb) is not defined in H . Thus ( H ; A, v) is a set with two binary partial operations. (H; A, v) is called a partial lattice. (H; A , v) is called a relative sublattice of L.

Thus every subset of a lattice determines a partial lattice. The second part of this section is devoted to an internal characterization of partial lattices, based on N. Fu- nayama [ 19631.

We now analyze the way the eight identities that were used to define lattices ((Ll)-(L4) of Section 1) hold in partial lattices:

5. Frec Lat,tices 41

LEMMA 13. (i) UAU exists and UAU =a. (ii) If uAb exists, then bAa exists, and aAb = bAa.

(iii) If aAb, (aAb)Ac, bAc exist, then aA(bAc) eoists, and (aAb)Ac =aA(bAc). If ~ A C ,

(iv) I f a ~ b exists, then av(aAb) exists, and u=av(uAb).

Let ( H ; A, v) be a partial lattice, a , b, c E H .

uA(bAc), aAb exist, then ( U A ~ ) A C exists, and (aAb)Ac =aA(bAc).

PROOF. As an illustration let us prove the first statement of (iii). Let H c L as in Definition 12. The assumption that aAb, (uAb)Ac, bAc exist in H means that in L, a , b, c, aAb, (aAb)Ac, bAc E H . But in L, (aAb)Ac =aA(bAc), and so aA(bAc) EH; that is, aA(bAc) exists in H , and, of course, (uAb)Ac =aA(bAc).

LEMMA 13’. interchanging A and v. Then (i‘)-(iv’) hold an any partial lattice.

Let (i’)-(iv’) denote the statements we get from (i)-(iv) of Lemma 13 by

PROOF. This proof is trivial by duality.

identities for partial lattices. Statements (i)-(iv) and (i’)-(iv’) give the required interpretation of the eight

DEFINITION 14. A weak partial lattice i s a set with two binary partial operations satisfying (i)-(iv) and (it)-(iv’).

COROLLARY 15. Every partial lattice i s a weak partial lattice.

Based on Figure 9, we give the following example of a weak partial lattice that is not a partial lattice.

1

0

Figure 9

Let H = (0, a, b, c, d , e, f , g, h, l}. Consider the lattice @ of Figure 9. Define XAY = z i n H iff x ~ y = z in@. Define x v y =z in H iff either x l y in@ and y = z , or y l x in @ and x = z ; or if {x, y } ={a, c} and z = f ; or { x , y } ={b, d } and z = g ; or {x, y} ={f , g } and z =l . Then (H; A, v) is a weak partial lattice (check the axioms). Now suppose that there exists a lattice L, H E L, such that ( H ; A, v> is a relative sublattice of L. Then 1 = (avc)v(bvd) in L , and thus 1 =sup {a, b, c, d}. Since e >a, b, and h >c, d in L, and 1 re, h, we get 1 =sup {e , h} in L. The fact that e , h, 1 C H implies that evh is defined in H (and equals l), contrary to the definition of (H; A , v) . (Compare this with Lemma 19.)

To avoid such anomalies we shall introduce two further conditions. To prepare for them we prove :

LEMMA 16. Let (H; A, v ) be a weak partial lattice. Then we define a partial ordering relation < on H by “ a l b i f f ahb exists and aAb =a.” If avb exists, then avb =sup {a, b}. I f aAb exists, then aAb =inf {a, b}. Also, a < b i f f avb = b.

PROOF. This proof is the same as the proof of the corresponding parts of Theorem 1.1, except that the arguments are a bit longer.

Note that in a partial lattice sup {a, b} may exist but avb does not. For instance, let L be the lattice of Figure 9, H = { O , a, b, 1). Then sup {a, b} = 1 in H , but a.vb is not defined in H because avbb H.

DEFINITION 17. An ideal of a weak partial lattice H i s a nonvoid subset I of H such that i f a, b C I and avb exists, then a v b E I , and x < a E I implies that x E I . Again we set { x I x< u} = (a]. Dual ideal and [a) are defined dually. I o ( H ) i s the lattice consisting of 0 and all ideals of H (partial ordering i s C), B0(H) i s the lattice consisting of 0 and all dual ideals of H (partial ordering .is s). For K G H , ( K ] is the ideal and [ K ) is the dual ideal generated by K.

COBOLLARY 18. Then I n H i s a n ideal of H provided that I n H =I= 0 .

Let H and L be given as in Definition 12. Let I be a n ideal of L.

LEMMA 19. A n y partial lattice H satisfies the following condition: ( I ) If (a]v(b] = (c] in I,#), then avb exists in H and equals c.

PROOF. Let H and L be given as in Definition 12, let a, b, c E H , and let (a]v (b] = (c] in Io (H) . Set I = (avb],. Then (a]=, (bIa E I n H , thus (cIH = (a],v(b], C (avb],; t’hat is, c< avb. Since a< c and b< c, we conclude that avb =c.

Let (D) denote the condition dual t.0 (I), namely: (D) If [a)v[b) = [c) in Bo(H), then aAb exists in H and equals c.

THEOREM 20. (N. Funayama [1953]). A partial lattice can be characterized as a weak partial lattice satisfying conditions ( I ) and (D).

5. Frro Latticrs 43

PROOF. Corollary 15, Lemma 19, and its dual prove that a partial lattice is a weak partial lattice satisfying (I) and (D). conversely, let ( H ; A , v ) be a weak partial lattice satisfying (I) and (D). Consider the map

Y : - ( ( X I , [ a ) ) 3

sending H into I o ( H ) x50(H), where 3 , ( H ) is the dual of B o ( H ) . This map q is one-to- one. If z v y = x , then ( x ] v ( y ] = (23 in I,@) and [ x ) v [ y ) = [ z ) in B o ( H ) , thus xqvyq = ( x v y ) ~ . Conversely, if xqvyq =xq, then ( x ] v ( y ] = (21 in Io (H) . Therefore, by (I), x v y exists and equals z, so z q v y q = z q implies that x v y =z. A similar argument shows that SAY = z iff X ~ A Y V = zq . Thus we can identify x with xq , getting H C_ L =

I@) x s 0 ( H ) . We have just proved that ( H ; A , v ) is a relative sublattice of L. Let (P; 5) be a poset. We make P into a partial lattice as follows: aAb is defined

iff inf ( ( 1 , b} exists and aAb =inf {a, b}, and similarly for avb.

-

LEMMA 21. ( P ; A , v) i s a pnrtial lattice.

PROOF. Lemma 13’ hold.

It is easy to verify that (I), (D), (i)-(iv) of Lemma 13, and (i’)-(iv’) of

We need some further definitions.

DEFINITION 22. Let ( A ; A , v ) , ( B ; A , v ) be weak partial lattices, q : A --c B. We call q a homomorphism iff whenever m b exists for a, b E A , then aqAbq exists and (aAb)q = ayAbpl,,and the similar condition holds for v. A one-to-one homomorphism q is an embedding provided that ar\b exists iff aqwbq exakts, and the similar condition holds for v . If g~ is onto and q is an embedding, then q is an isomorphism.

Now we are ready again to define the most general lattices of Section 2.

DE~NITION 23. Let % = ( A ; A, v ) be a partial lattice and let K be a n equational class of lattices. The lattice FK(%) (or, simply, PK(A)) is a free lattice over K generated by 9, i f f the following conditions are satisfied:

(i) FK(A) E K. (ii) A c FK(A), and A is a relatirie sublattice of P,(A).

(iii) [ A ] = FK(A). (iv) If L E K and q : A -, L i s a homomorphism, then there exists a homomorphism y :

F,(A) -, L extending q (that is, nq = a y for a A ) .

If P is a poset, then FK(P) is a free lattice over K generated by P, where P is considered a partial lattice as in Lemma 21 ; thus Definition 23 contains Definition 2 as a special case. The general theory sen be developed exactly as i t was in the first part of this section. The final result is:

THEOREM 24. Let % = ( A ; A , v ) be u partial lattice and let K be an equational class. Then FK((lI) exists iff there exists a lattice L in K such that % i s a relative sublattice of L.

As an application we prove the existence of the lattice completely freely generated by a poset. Let P be a poset; we define a partial lattice Pm on P as follows:

XAY = z in P m iff x and y are comparable and z = inf { x , y } ;

z v y = z in Pm iff x and y are comparable and z =sup {x, y}.

DEFINITION 25. F(Pm) ( = F , ( P ) ) is called a lattice completely freely generated by P.

THEOREM 26. For any poset P , a luttice completely freely generated by P exists.

PROOF. A subset A !G P is called hereditary iff x E A and y < x imply that y E A. Let H(P) be the set of all hereditary subsets of P partially ordered under set inclusion. Let P,=H(P) and P2=H(p1) , where is the dual of Pi. Identifying pE P with ( p ] , weget P G Pi ; identifying p E Pi with [ p ) , weget Pi E P,, thus P C P,. Obviously, P2 is a lattice. Let a, by c E P and a v b = c in P,. If a and b are incomparable, then (a ] u (b] (+(c]) is an upper bound for a and b, thus avb < c in P,. A similar argument works for aAb =c. Therefore, P2 1 P satisfies the condition of Theorem 24.

N

CF(P) is the usual notation for P(P) in the literature.

Exercises

1.

2.

3.

4.

Show that an equational class is closed under the formation of sublattices, homomorphic images, and direct products. Show that the class T of all one-element lattices is an equational class. For any equational class K, prove that K 2 T. Let Ki, i ~ 1 , be equational classes. Show that n(Ki I i C 1 ) is again an equational class. Find an equational class A with A z M , D G A Y and A+M, A I D . (Hint: Let A be defined by .

( Z V ~ ) A ( Z V Z ) A ( Z V U ) =ZV ((ZVY)AZAU)V ( (ZVZ)AYAU)V( ( Z V U ) A ~ A Z ) . )

Let K be a nontrivial equational class. Show that K contains arbitrarily large lattices. Let P = { O , a, b}, inf {a, b} =0, and let K be a nontrivial equational class. Show that F K ( P ) s P ( 2 ) . Find an example of an isomorphism I: P K ( P ) - P x ( P ) that is not the identity map on P.

5.

6.

7.

5. Prrc Lattict-s 45

8.

9. 10. 11.

12.

13.

14. 15. 16.

17.

18.

19.

20.

21.

22.

23.

24.

25.

Let P be a poset, let K, N be equational classes, and assume that PK( P) and F N ( P) exist. Prove that if H 3 N, then there exists a homomorphism Q, from F K ( P ) onto FN(P) such that Q, is the identity map on P. Work out a proof of the existence of F ( 3 ) without any reference to Theorem 5. Prove Theorem 8. Formulate and prove the form of Theorem 8 that is used in the proofs of Theorems 10 and 11. Prove that Pm(4) is infinite (G. Birkhoff [1933]). (Hint: Let R be the set of real numbers and let L be the lattice of vector subspaces of R3. Set

a ={(x, 0, x) I x ER}, b ={(O, x , ~ ) I x E R } , c = { ( O , O , Z ) ~ Z E R } , d = { ( X , x , x ) ( z € R } .

Then [ {a , b, c , d} ] is infinite.) Let A and B be disjoint three-element sets. Let L be the set of the following subsets of A u B: all X E A, all AuY, Y c B, all three-element sets 2 with 12 n A1 =2. Prove that (L; c) is a lattice, that A and v are intersection and union, and that thus L is distributive. Figure 6 is the diagram of L (A. D. Campbell [1943]). Represent the lattice of Figure 6 as a sublattice of (6#. Prove that the exponent six is best possible in Exercise 14. Represent the lattice of Figure 7 as a sublattice of L xB3, where L is the lattice of Exercise 13 and a3 is given in Figure 2.2. Show that condition (E) of Theorem 6 is equivalent to the following: For any a, b, c P not satisfying inf {a, b} = c , there exist a lattice L in K and a homomorphism 9: P +L with aghbg +cy, and dually. The statement “ % = ( A ; A, v ) is a partial algebra’’ means that A is a nonvoid set and that A , v are partial binary operations on A. For an n-ary polynomial p , ao, . . . , a,-l EA, interpret p(ao, . . . , anel ) . (When is it defined and what is its value?) I n the weak partial lattice represented by Figure 9, p = X A ( ( X V Y ) V ( Z V Z ) ) is not defined for x =0, y = e , z =h. Verify that in every lattice p =q, where q =x , and q is defined in every partial lattice. An identity p = q hold8 in the partial algebra 1 = ( A ; A , v ) iff the following three conditions are satisfied: (i) If p(ao, . . .), q(ao, . . .) are defined (ao, . . . € A ) , then p(ao, . . .) =q(ao, . . .).

(ii) If p(ao, . . .) is defined, q =qo * 41, where * is A or v , and qo(ao, . . .), qi(ao, . . .) (iii) This condition is the same as (ii) with p and q interchanged. Check that Lemmas 13 and 13’ give this interpretation to the lattice axioms. Let p =( ( ( zv z )v (yvu) )vw)vw and let q =((wvx)v(wvy))v((wvz)v(wvu)). Show that p = q in any lattice. Show that p = q does not hold in the weak partial lattice defined in connection with Figure 9. Let I , and I t be ideals of a weak partial lattice. Set Jo =IouI,, Jn = (z I z I yvz , ~ , Z E J , - ~ } , n = l , 2 , . . . , and J = u ( J i ( i = O , 1, 2 , . . .). Show that J=IovIi . Let the weak partial lattice L violate (I); that is, (a]v(b] = (c], but avb is undefined. Let I , = ( a ] , I , = ( a ] , and cEJ, (see Exercise 22). Generalizing Exercise 21, find an identity p = q that holds in any lattice but not in L. (Exercise 21 is the special case in which n =2.) Prove that a partial algebra % = ( A ; A , v ) is a partial lattice iff every identity p =q holding in any lattice also holds in 1. A honzornorphisrn of a partial algebra 1 =(A; A , v ) into a lattice L is a map q:

are defined, then qo(ao, . . .) L qi(a,, . . .) is defined.

I. First Concepts

26.

27.

28.

29.

30. 31. 32.

33.

34.

35.

36. 37.

38. 39.

40.

41.

A -L such that (ahb)q =aqhbq whenever ahb exists, and the same for v. Prove t,hat there exists a one-to-one homomorphism of OI into some lattice L iff there exists a partial ordering 5 on A satisfying ahb =inf {a, b}, whenever ahb is defined in 91, and avb =sup {a, b}, whenever avb is defined in a. Show that evwy weak partial lattice satisfies the condition of Exercise 26, but not conversely. Let A = {0, a, b, G, l}, 0 5 a, b, c 5 1. For x 2 y define

and define XAY =yAX = X , X V Y =yVX = y ,

aAb =bha=O, avb =bva = I .

Show that (A; A, v) is a partial lattice. Let A = {0, a, b, c, d , 1). 0 4 a, b, c, d 4 1. For x 5 y define

Xhy =yhX = X , XVY =yVX =y, and define

ahb =bha =chd =dhc = 0 ,

avb =bva =cvd =dvc = 1.

Show that (A; A , v) is a partial lattice. Let L and K be lattices and let L n K be a sublattice of L and of K. For x , y E L u I< define X A Y = z iff 2, y , z EL and X A Y = z in L, or x , y, z EK and X A Y = z in K ; define x v y similarly. Is (L u R ; A , v ) a partial lattice? Are the eight axioms of a weak partial lattice independent ? Are the ten axioms of a partial lattice independent ? Define weak partial semilattice and partial semilattice ; prove the analogue of Theorem 20 for partial semilattices. Let A be a weak partial lattice in which ahb exists for all a, bEA. Then A is a partial lattice iff a +(a] is an embedding of A into Io(A) . Let T be the equational class of all one-element lattices. Show that P&4) exists iff IAl=1. Let P=(&)2. Show that for any nontrivial equational class K, P K ( P ) and P K ( P ~ ) exist and that alwaye PK( P) g Pw( P). Determine P ( P ) , P r ( P ) , and PD(P), where P ={a, b, c} and a < 6. Discuss the set of all weak partial lattices on A inducing a given partial ordering on A. Repeat Exercise 37 for partial lattices. Show that a one-to-one-homomorphism of weak partial lattices need not be an embedding. Show that in Theorem 24, the condition “there exists a lattice L in K such that N is a relative sublattice of L’1 can be replaced by the following condition: For all a, b, c E A for which ahb =c does not hold, there exists a lattice L in K, and a homomorphism Q of A into L such that aqhbg, +cq, and the same condition holds for v. Let (A; A, v ) be a’ partial algebra, let K be an equational class of lattices, let L E K , and let M be a relative sublattice of L. Then M is called a maximal homomorphic image of A in K iff there is a homomorphism Q of A onto M such t.hat whenever y is a homomorphism of A into N E K , then there is a homomorphism a: M-N such that Q U = ~ L Prove that a maximal homomorphic image is uniquo up to isomorphism provided it exists.

ti. Bpc.oial Elzments 47

42.

43. 44.

Starting with an arbitrary partial algebra (A; A, v), carry out the construction of Theorem 5 (Theorem 24). Are there posets P o c P such that P(PC) = F ( P m ) t After M. M. Gluhov [1960], a finite partial lattice % is a baeie of a lattice L iff L =B(%), but L=P(a0) for no A o c A . Show that the lattice L =a(%) has more than one basis where 9t is defined as follows: Let (A; 5 ) be the poset given by Figure 10; for x 5 y let Z A ~ =YAX =z; furthermore, the join of any two elements is defined in (0, a, b, c , d, e , f , 1) as supremum, and lvg =gvl =l. (This example. which is due to C. Herrmann, contradicts M. M. Gluhov’s result.)

1

0 Figurc 10

46. Let L and L, be lat,tices, let L =[A], and let p be a map of A into Li. Then p can be extended to a homomorphism y of L into Lf iff p(a,, . . . , a, -1) =q(ao, . . . , a n - l ) implies that p(aop, . . . , an-lp) =q(aop, . . . , an-ly) holds for any a,, . . . , a,-l € A and for any integer n, and pair p, q of n-ary polynomials. Under the hypotheses of Exercise 45 describe y . Characterize K-free lattices using Exercise 45. Under the hypotheses of Exercise 45, show that if A is finite and y is isotone, then q can always be ext,ended to an isotone map of L into Li.

46. 47. 48.

6. Special Elements

A bounded poset is one that has both 0 and 1 . A (0, 1)-homomorphism (of a bounded lattice into another one) is a homomorphism taking zero into zero and unit into unit. A(0, 1)-sublattice of a bounded lattice L is a sublattice containing the 0 and 1 of L. Similarly, we can define {O}-homomorphism, and so on, for lattices and seniilattices.

In a bounded lattice L, a is a complement of b iff aAb = 0 and avb = 1.

LEMMA 1. In a bounded dwtributive lattice, an element can have only one complement.

PROOF. (bor\a)v(b,r\bi) =Ov(b,nbi) = boAb,; similarly, b, =bOAbi, thus b, =b , .

If b, and b, are both complements of a, then bo=bor\l =boA(uvbi)=

Let u, E [b , c ] ; x is a relative complement of a in [b, c] iff aAx = b and avx = c.

48 I. First Concept,s

LEMMA 2. relative complement in any interval containing it.

I n a bounded distributive lattice, i f u has a complement, then it also has (I

PROOF. of a in [b’, c ] , provided that b

Let d be the complement of a. Then x = (dvb)Ac is the relative complement a < c. Indeed,

aAx =aA(dVb)AC = ( (aAd)V(aAb))Ac = (0vb)Ac = b,

and avx =c by duality.

LEMMA 3 (DE MORGAN’S IDENTITIES). In a bounded distributive lattice, i f a, b have complements a’ and b’, respectively, then aAb and avb have complements (aAb)’ and (avb)’, respectively, and

( a h ) ’ = a’vb’,

(avb)’ = a’Ab’.

PROOF. By Lemma 1 it suffices to prove that

(aAb)A(a’vb’) =0, (aAb)v(a’vb’) = 1

to verify the first identity; the second is dual. Compute:

(aAb)A(a’vb’) = (aAbAU’)v(aAbAb’) = o V o =o and

(aAb)v(a’vb’) = (ava’vb’)h(bva’vb’) = 1 ~ 1 = 1.

Acomplemented lattice is a bounded lattice in which every element has a complement. A relatively complemented lattice is a lattice in which every element has a relative complement in any interval containing it. A Boolean lattice is a complemented dist,ributive lattice. Thus, in a Boolean lattice B, every element a has a unique complement, and B is also relatively complemented.

A Boolean algebra is a Boolean lattice in which 0, 1, and ’ (complementation) are also considered to be operations. Thus a Boolean algebra is a system: (B; A, v , ’, 0, l), where A, v are binary operations, ‘ is a unary operation, and 0 , l are nullary operations. (A nullary operation picks out an element of B.) A homomorphism y preserves 0 , l and ’; that is, it is a (0, 1)-homomorphism satisfying (xy)’=z’y. A subalgebra is a (0, 1)-sublattice closed under ’. ’?& will denote the two-element Boolean algebra.

Note that in a bounded distributive lattice L, if b is a complement of a, then b is the largest element x of L with a A x = O . More generally, let L be a lattice with 0; an element a* is a pseudocomplement of a( E L) iff am* = 0, and UAX = 0 implies that x a*. An element can have at most one pseudocomplement. A pseudocomplemented lattice is one in which every element has a pseudocomplement.

The concept of pseudocomplement involves only the meet operation. Thus we can also define pseudocomplemented semilattices.

THEOREM 4. Let L be a pseudocomplernented meet-semilattice, S(L) ={a* I a E L}. Then the partial ordering of L partially orders S(L) and makes S (L) into a Boolean lattice. For n , b ES(L) we haw uAb ES(L), and the join in B(L) is described by

avb = (a*Ab*)*.

REMARK. Even if L is a lattice, the join in L need not be the same as the join in S(L) . This result was proved for coniplete distributive lattices by V. Glivenko [1929] and published in its full generality for the first time by 0. Prink [1962]. Both proofs used special axiomatizations of Boolean algebras to get around the difficulty of proving distributivity. The present proof is direct.

PROOF. We start with the following observations:

a 2 a**. (1) a < b implies that a* 2 b*. (2)

(3) a E S(L) iff a =a** (4) a, b E S ( L ) implies that aAbES(L) . (5 )

For a, b ES(L), sups(L){a, b} = (a*Ab*)*. (6)

a* =a***.

Formulas (1) and (2) follow from the definitions. Formulas (1) and (2) yield a* 2 a***, and by ( 1 ) a* I a***, thus (3). If a E S(L) , then a = b* ; therefore, by (3), a** = b*** = b* =a. Conversely, if a =a**,

then a=b* with b=a*; thus a E S ( L ) , proving (4). If a , b E S(L) , then a =a** and b = b**, and so a 2 (aAb)** and b 2 (aAb)**, thus

aAb 2 (aAb)**; by (l), a ~ b = ( a ~ b ) * * , thus aAbCS(L) . I f xES(L), %_(a, and x l b , then x< aAb; therefore, aAb =infS(L){a, b}, proving (5).

x and b<x (x ES(L)), then a* 2 z * and b* 2 x * by (2); thus by (2) and (4), (a*Ab*)* I z , proving (6).

For a, b ES(L), define avb = (a*Ab*)*. By Formulas (5) and (6), (S(L);A, v) is a bounded lattice. Since for a ES(L), ava* = (a*r\a**)* =0* = 1 and m a * =0, S(L) is a complemented lattice. Now we need only prove that S(L) is distributive. For x, y, z E S(L) , X A Z I ZV(YAZ) and YAZ I SV(YAZ) ; therefore, SAZA(W(YAZ))* = 0 and YAZA(XV(YAZ))* =O. Thus ZA(ZV(YAZ))* 5 x*, y*, and so ZA(ZV(YAZ))* 2 X*AY*. Consequently, Z A (ZV(YAZ))*A (x*~y*)* = 0, which implies that Z A (z*~y*)* 2 (ZV(YAZ))**. Now the left-hand side is z~(zvy) by Formula (6), and the right-hand side is X V ( ~ A Z ) by Formula (4). Thus we get XA(ZVY)<SV(YAZ), which is distributivity by Lemma 4.10.

Other t,ypes of special elements: An element a is an atom if a>O and a dual atom if a < 1 ; it is join-irreducible if a = bvc implies that a = b or a = c; it is meet-irreducible if a = bAc implies that a = b or a =c. An element which is both join- and meet- irreducible is called doubly irreducible. Examples are given in the following exercises.

a* 2 a*Ab*, thus by (2) and (a), a (a*r\b*)* ; similarly, b < (a*Ab*)*. If a

50

1.

I. First Concept,s

Exercises

Find a homomorphism of bounded lattices that is not a (0, 1)-homomorphism and a sublattice that is not a (0, 1)-sublattice. Find a modular lattice in which every element x +0, 1 has exactly m complements. Let L be a distributive lattice, a, b EL. Prove that if a v b and U A ~ have complements, so do a and b. Show that Lemma 2 holds in any modular lattice. In a bounded lattice L, let x be a relative complement of a in [b, 01; let y be a relative complement of c in [x, I]; let z be a relative complement of b in [0, X I ; and let t be a relative complement of z in [z, y]. Verify that t is a complement of a. Let Bo, B, be Boolean algebras and let a, be 8 (0, I}-(lattice) homomorphism of B, into B,. Show that a, is a homomorphism of the Boolean algebras. Let L be a distributive lattice with 0. Show that, I ( L ) is pseudocomp1ement.d. Is the converse of Exercise 7 true? Prove that if in Theorem 4 L is a lattice, then avb in S(L) can be described by

Let L be a pseudocomplemented lattice. Show that a**vb** = ( a d ) * * . Find arbitrarily large pseudocomplemented lattices in which rS(L) = (0, 1). Prove that in a Boolean lattice, x + O is join-irreducible iff x is an atom. Show that in a finite lattice every element is the join of join-irreducible elements. Verify that “finite lattice” in Exercise 13 can be replaced by “lattice satisfying the Descending Chain Condition.” (A lattice L or, in general, a poset, satisfies the Deecending Chain Condition iff xo, xi, x2, . . . EL, xo 2 xi 2 x2 2 - * - implies that x R = x , + l =. * - for some n; see Exercise 2.7). Show that some form of the Axiom of Choice must be used to verify Exercise 14. Prove that if a lattice satisfies the Descending Chain Condition, then every nonzero element contains an atom. The dual of the Descending Chain Condition is the Ascending Chain Condition (see Exercise 2.6). Dualize Exercises 13-16. Show that a lattice satisfies the Ascending Chain Condition and the Descending Chain Condition iff all chains are finite. Find a lattice in which all chains are finite but that contains a chain of n elements for every natural number n. Find a lattice in which there are no join- or meet-irreducible elements. Prove that the Ascending Chain Condition (Descending Chain Condition) holds in a lattice L iff every ideal (dual ideal) of L is principal. Let L be a lattice and let C be 8 chain with ]CI 5 No. Prove that if there is a homomorphism of L onto C, then L contains an isomorphic copy of C. How does this result extend to countable lattices C? Does this result hold for the chain of real numbers? (No.) Let A be a binary operation on L, let * be 8 unary operation on L (that is, for every aEL, a*EL), and let 0 be a nullary operation (that is, OEL). Let us assume that the following hold for all a, b, c EL: ~ h b =bha, (ahb)Ac =aA(bAc), aha =a, O A a =o, ah(ahb)* =ahb*, UAO* =a, (O*)* = O . Show that (L; A) is a meet-semilattice with 0 as zero, and for a11 aEL, a* is the pseudocomplement of a (R. Balbes and A. Horn [1970]). Let L be a pseudocomplemented meet-semilattice and let a, b € L . Verify the for- mdas (aAb)* =(a**Ab)* =(a**Ab**)*.

avb = (avb)**.

2. 3.

4. 5.

6.

7. 8. 9.

10. 11. 12. 13. 14.

15. 16.

17.

18.

19.

20. 21.

22.

23.

24.

25.

26.

27.

*28.

*29.

30. 31.

32.

33.

34.

35.

36.

37.

38.

39.

-

Let L be a meet-semilattice and let a, b EL. The pseudocomplement of a relative to b is an element a * b of L satisfying U A X I b iff x I a * b. Show that a * b is unique if it exists; show that a * a exists iff L has a unit. Let L be a relatively pseudocomplemented meet-semilattice (that is, L is a meet-semilattice and a b exists for all a, b EL). Show that L has 1; for all a, b, c EL: a * ( b * c ) = ( a ~ b ) * c and a * ( b * c ) =(a * b ) * (a * c ) . Furthermore, if L is a lattice, then L is distributive. Let L be a pseudocomplemented distributive lattice. Prove that for each aEL, (a] is a pseudocomplemented distributive lattice; in fact, the pseudocomplement of xE(a] in (a] is X*AU.

Using the notation of Exercise 27, let S(a) denote the elements of the form X*AU,

x I a. Then S ( a ) is a Boolean algebra by Theorem 4. Let V, denote the join in S(a). Show that if x, y ES(a) and x, y ES(b) (a, b EL), then xV, y =xvb y . Let b ES(a). Prove that S ( b ) 5 S(a). (The results of Exercises 28 and29 first appeared in FC.) Show that T,, (see Exercise 2.36) is complemented. Show that in T,, every interval is pseudocomplemented. (Exercises 30 and 31 are due to H. Lakser.) For a finite lattice L, let Irr(L) denote the set of all doubly irreducible elements. Prove that1 ILI 2 2 (l(L)+ 1) - IIrr(L)I. Show that for a finite lattice L,

I(Sub(L)) = IIrr(L)I+l(Sub(L -Irr(L))). A finite lattice L of n elements is diemantlable iff there is achain Li cL, C- - * cL, = L of sublattices satisfying lLil = i . Show that every lattice with at most seven elements in dismantlable. (Hint: use Exercise 32.) Show that for every integer n 2 8 there is a lattice of n elements which is not dismantlable. A finite lattice is Planar iff it has a planar diagram. Show that every finite planar lattice is dismantlable. For every integer n 2 9, construct an 1%-element dismantlable lattice which is not planar.2 If a dismantlable lattice L is not a chain, then it contains two incomparable doubly irreducible elements. Prove that every sublattice and homomorphic image of a dismantlable lattice is dismantlable.3 Is this also true for planar lattices?

*** Let P and& be posets and let a: P -Q and p : Q - P satisfy the following conditions. (G,) For x, y E P, if m I y , then xa I ya. (G,) For x, y EQ, if z 5 y, then xp I yp. (G,) For x E P, map 2 X. (G4) For x EQ, xpu 2 m. Then (a, B ) is a Galois connection between P and &. This concept was introduced in G. Birkhoff [1940]. The results presented in the following exercises are based on G. Hirkhoff [1940] and 0. Ore [1944]. They are

6. Special Elements 51

-

1 Exercises 32-35 are based on I . Rival [1974a]. 2 For Exercises 36-37’, see K. A. Baker, P. C. Fishburn, and F. S. Roberts [1970]. 3 Exercises 38 and 39 are from D. Kelly and I. Rival [1974].

6 Oriitzer

modelled after the correspondence between subgroups of the Galois group and subfields of a separable field extension which is the subject of the classical Fundamental Theorem of Galois Theory. Let L be a pseudocomplemented meet-semilattice. Let P=Q = L and, for %EL, let xu =xp =x*. Show that (u, 8) is a Galois connection. Lot X o and X , be sets and r

40.

41. X o x Xi. For A X,, set

A1 = {zi 1 xi EXl and (zo,xi) Er for some xo € A } ;

for B G X i , set

BO = {xo 1 xo EXo and (xo, xi) Er for some xi E B}.

A -A1 and B -+Bo set up a Galois connection between P(Xo) and P ( X l ) . Let (a, p) be a Galois connection between the posets P and Q . Show that up is a closure map on P and /?u is a closure map on &. (A closure map x on a poset is an isotone map of the poset into itself satisfying x 5 x 2 , xxx=xx. If xx=x, then x is called cloeed.) Let PC and Qc be the set of closed elements in P and Q, respectively (P, &, a, p as in Exercise 42). Prove that if P and Q are complete lattices, then so are PC and Qc, u is an isomorphism of P c with the dual of Qc, and How much of Theorem 4 can be derived from Exercises 40-43 ?

42.

43.

is the inverse of u. 44.

Further Topics and References

Many of the concepts and results discussed in Chapter 1 are special cases of universal algebraic concepts and results. To see this, the reader needs the definition of a universal algebra. An n-ary operation f on a nonvoid set A is a map from An into A ; in other words, if ai, , , , , a, A , then f ( q , . . . , a,) € A . If n = 1, f is called unary; if n = 2 , f is called binary. Since A0 = {a}, a nullary operation ( n = 0 ) is determined by f (0 ) c A, and f is sometimes identified with f ( 0 ) . A universal algebra, or simply algebra, consists of a nonvoid set A end a set F of operations; each f c F is an n-ary operation for some n (depending on f ) . We denote this algebra by % or ( A ; F). Many of the results of Sections 3-5 can be forniulated and proved for arbitrary universal algebras. For inore details, see Chapters 1-4 of the author’s book [1968].

In every poset we can introduce (as suggested by the real line) a ternary relation r called betweenness: r(a, b, c ) iff a < b <c or c a. M. Altwegg [1950] proves that the partial ordering can be defined in terms of betweenness (naturally, up to duality) ; see also J. M. Cibulskis [1969]. In lattices, several betweennem relations are known, see E. Pitcher and M. F. Smiley [1942]. For applications of these relations, see M. F. Smiley and W. R. Transcue [1943] and R. Padmanabhan [1966a].

Lattices and Boolean algebras can be defined by identitics in innumerable ways. Of the eight identities we used to define lattice8 ((Li)-(lA) of Section l), two ((Ll)) can be dropped. Ju. I. Sorkin [1951] reduces six to four, R. Padnianabhan [1969] to two, and finally in R. N. McKenzie [1970] a single identity is found characterizing lattices. J u . I. Sorkin’s identities use only three variables; the others use more. More recently, N. Padnianabhan [1972] has found two identities in three variables char-

b

k'iirt IILT Topics and Iteferences 63

acterizing lattices. It is easy to see that two variables would not suffice. Take the lattice of Figure 5.8 and redefine the join of the two atoms to be 1 ; otherwise keep all the joins and meets. The resulting algebra is not a lattice, but every subalgebra generated by two elements is a lattice. Therefore, lattices cannot be defined by identities in two variables. By nieans of a more complicated construction, A. H. Dia- niond and J. C. C. McKinsey [1947] derive the same conclusion for Boolean algebras. Finite algebras in which every two-generated subalgebra is Boolean have been investigated in R. W. Quackenbush [1974].

A result of A. Tarski [1968] states that, given any integer n, there exists a set of n identities defining lattices such that no identity can be dropped from the set.

By Exercise 4.24, modular and distributive lattices can be defined by two identities. Nicer sets of identities for these cases can be found in M. Kolibiar [1966] ((mv ( b ~ b ) ) ~ b = b and ( ( U A ~ ) A C ) V ( a d ) = ( ( ~ A u ) v ( C A ~ ) ) A U characterize niodular lattices) and M. Sholander [195l] ( U A ( U V ~ ) = a and U A ( ~ V C ) = ( C A U ) V ( ~ A U ) characterize distributive lattices). See also J. RieEan [1958].

One of the most useful axioniatizations of Boolean algebras is that of E. V. Hunting- ton [1904] (observe that a proof of Huntington's result is implicit in the proof of Theorein 6.4) according to which a Boolean algebra is a complemented lattice in which the coniplementation is pseudocomplementation (that is, U A X = 0 implies that x< u'). Contrast this with Corollary VI. 3.8.

One of the briefest axioni systems of Boolean algebras in terms of A and ' is due to L. Byrne [1946]: U A ~ = b ~ u , U A ( ~ A C ) = ( U A ~ ) A C , m b ' =CAC' iff a ~ b =a. Characterization by identities is usually longer. R. N. McKenzie, A. Tarski, and the author observed independently that Boolean algebras can be defined by a single identity (see A. Tarski [1968], and G. Gratzer and R. N. McKenzie [1967]). A thorough survey of the axiom systems of Boolean algebras is given in S. Rudeanu [1963]; see also F. M. Sioson [1964], and S. Rudeanu [1974]. The only known irredundant selfdual axiom system can he found in R. Padnianabhan [1974a].

Boolean algebras originated a s an algebraic formalization of propositional logic. Most introductory logic texts give satisfactory expositions of these ideas; see also P. R. Halnios [1963]. There is a similar relationship between Boolean algebras and the theory of switching circuits; see M . A. Harrison 119651.

P. liibenboini [ 19491 first pointed out that pseudocomplementation can be described hy identities (involving *). A. Monteiro [ 19551 acconiplished the same for relative pseudoconiplenientation; see also H. Balbes and A. Horn [1970a] and R. Balbes and 1%. Dwinger [1974]. This fact is applied in G. Gratzer [1969].

The exaniples of lattice identities we have seen so far seeni to suggest that all such idtwtities are selfdual. The identity ( U A ~ ) V ( ( I A C ) = n , where a = s A ( ( x A ~ ) v ( ~ A z ) v ( z A x ) ) ,

h = ( . v A ~ ) v ( Y A z ) , c = ( Z A Z ) V ( ~ A Z ) , is an example of a nonselfdual identity. This identity holds in a lattice L iff L does not have a siihlnttice isomorphic to the dual of the lattice of Figure V. 2.7; see H. F. J. Lowig [1943].

A diagram of a finite lattice is a graph : however, very little work has been done in lattice theory from a graph theoretic point of view. It is known that a finite distributive lattice is planar iff no element i R covered by more than two elements. Planar niodular lattices are characterized in It. Wille [1974]. It is pointed out in K. A. Baker, 1'. C. Fishburn, and 3'. S. Roberts [1970] that a finite lattice is planar iff i t has order 5'

dimension one or two (a poset (P; I) is of order dimension n iff the relation < is the set intersection of n, but not of fewer, linear orders on P), a result they attribute to J. Zilber; they also point out that the corresponding result for posets does not hold. A graph theoretic characterization of planar lattices is given in C. R. Platt [1976]. A characterization by exclusions of an infinite list of lattices as subposets can be found in D. Kelly and I. Rival [1975a]. (D. Kelly and I. Rival [1974] investigate dismantlable lattices, a generalization of planar lattices.) For a general review of planar lattices see R. W. Quackenbush [1973]. Pairs of lattices whose diagrams are isomorphic 8s graphs are investigated by J. Jakubik [1954], [1954a]. See also J. Ja- kubik and M. Kolibiar [1954].

Automorphism groups of planar graphs are described in L. Babai [1972] and [1973]. L. Babai used these results to characterize automorphism groups of planar lattices (unpublished).

Sublattices suggest a number of problems: For an equational class K of lattices, let fK(k,n) denote the smallest integer such that any lattice in K having at least fK(k , n) elements has at least k sublattices of exactly n elements; write f ( k , n) for fL(k, n) . The function f ( 1 , n) has been investigated: f(1, t ) =t for t < 6 (I. Kaplansky); f ( l , 7 ) =9,f( l , 8) = S , f ( l , 9) 2 17,f(l, n) 2 n +2 for n 2 10 (I. Rival [c]); f(1, n) n4n-3

(R. Wille, unpublished); and f ( 1 , n)< nsn (G. Haves and M. Ward [1969]); see also M. Curzio [1953]. Related questions concerning the spectrum of a finite lattice L (={k I L has a k element sublattice}), the size of maximal sublattices, dismantlabil- ity, etc. have been investigated in a number of papers; see I. Rival [1974e], [c] and D. Kelly and I. Rival [1974] and [1975a].

Infinite lattices have, as a rule, very many large sublattices. T. P. Whaley [1969] proves that if ILI is an infinite regular cardinal, then the lattice L has an infinite chain of sublattices of cardinality ILI.

The intersection @(L) of all maximal proper sublattices of a lattice is called the Frattini sublattice (a misnomer, since @(L) = 0 is possible). If the lattice L satisfies the Ascending or the Descending Chain Condition, then @ ( L ) = 0 iff L is a chain; furthermore, every lattice K can be represented as @(L) for a suitable lattice K; see K. M. Koh [1971]. For planar modular lattices and distributive lattices, the Frettini sublattices are investigated in K. M. Koh [1973a] and C. C. Chen, K. M. Koh, and S. K. Tan [1973], [1975]. It is proved in M. E. Adams [1973] that every distributive lattice is the Frattini sublattice of a suitable distributive lattice.

The lattice of sublattices Sub(L) and the lattice of subsemilattices of a semilattice are investigated in N. D. Filippov [1966] and L. N. gevrin [1964], respectively. They give necessary and sufficient conditions for Sub(L) 2 Sub(Li). Filippov’s results show that if L is distributive (Boolean) and Sub(L) 2 Sub(Li), then L, is distributive (Boolean) : see also I. Rival [1972]. The lattice of subalgebras of a Boolean algebra has been characterized in D. Sachs [1962]; see also G. Griitzer, K. M. Koh, and M. Makkai [1972].

Sub(L) has few of the properties discussed in this book. Sub@) is modular iff L is a chain (K. M. Koh [1973]) and it is semimodular (in the sense of Section IV.2) iff L does not contain a sublattice isomorphic to a2 xa3 (H. Lakser [1973b]).

A new area of study is the investigation of finite sublattices of lattices to prove that many properties can be characterized by the existence of finite sublattices of

Further Topics and References 55

a certain type. Of course, results of this sort are well known for identities; see Theorem 11.1.1 and the discussion of splitting lattices in Section V.2. Recently, similar results were found for properties of many different kinds : three-generated lattices (B. A. Davey and I. Rival [a]), (SD,) and (SD,) of Section VI.1 (B. A. Davey, W. Poguntke, and I. Rival [1975], (W) of Section VI.l in the presence of (SD,) and (SD,) (R. Antonius and I. Rival [1974]). A free lattice also has very many finite sublattices; see H. S. Gaskill [a]. Apparently, an infinite lattice has many more typical finite sublattices than has been suspected before.

By Leninia 1.4.8, identities are preserved when passing from L to I@). In G. Griitzer 119701 the problem is posed: which first-order properties (properties that can be formulated as B first order sentence in the sense of Exercises V.1.23 and V.1.24) are preserved when passing from L to I@), or conversely? The theory of transferable and sharply transferable lattices and semilattices that grew out of this problem will be discussed in the Concluding Remarks. See also K. A. Baker and A. W. Hales [1974] and I. Rival and B. Sands [1975].

A very important property of complete lattices is the Fixed-Point Theorem : Any isotone map f of a complete latt8ice L into itself has a fixed point (that is,

/ ( a ) = u for some a EL) ; in fact, / (a ) = a if a is defined by

-see A. Tarski [1955] end B. Knaster [1928]. A. C. Davis [1955] proved that if this theorem holds in a lattice L, then L is complete. Various generalizations are given in A. Tarski [1955], E. S. Wolk [1957], A. Pelczar [1961] and [1962], S. Abian and A. B. Brown [1961], V. Devid6 [1963], S. R. Kogalovskii [1964], R. Demarr [1964], and B. R. Salinas [1969]. Posets of length one in which every isotone map has a fixed point are characterized in I. Rival [a].

The poset of all partial order relations on a set X has an interesting property recently discovered by B. M. Schein [1972a]: every lattice can be embedded in such a poset as a sublattice. Compare this with Whitman’s result that every lattice can be embedded in a partition lattice (a stronger form of which is Theorem IV.4.4).

It is hard to overemphasize the importance of free lattices; they provide one of the most iniportant research tools of lattice theory. Two typical applications to modular lattices can be found in G. Griitzer and E. T. Schmidt [1961] and G. Grktzer [l966]. We discuss free lattices in great detail in Section VI.2 and we use the method of constructing special lattices FR( P) throughout the book.

Few of the topics in this chapter have been investigated for semilattices. Congruences are one exception (see G.Zacher [1952], D.Papert [1964], R.A.Dean and R.H. Oehmke [1964], R. Permutti [1964], J. Varlet [1965], E. T. Schmidt [1969], R. Freese and J. B. Nation [1973]). Congruences of partial lattices are considered in G. Gratzer and H. Lakser [1968].

Distributivity (see also Section 11.5) and modularity of semilattices is investigated in J. B. Rhodes [1975] and B. M. Schein [1972], see also I. FIeischer 119751. For some other topics on semilattices, see G. Griitzer and H. Lakser [1969b], G. Bruns and H. Lakser [1970], A. Horn and N. Kimura [1971], H. S. Gaskill [1972], K. A. Baker [1973], and K. M. Koh [1971a].

56

1.1

1.2

1.3 1.4. 1.5

1.6.

1.7. 1.8.

1.9.

1.10.

1.11. 1.12. 1.13. 1.14.

I. First Concepts

Problems

(FC 5). Characterize the comparability relations of semilattices and lattices. (See Exercises 1.34 end 1.35. R. N. McKenzie pointed out that neither of these two problems has a first-order solution.) (FC 6). characterize the lattices T, and finite lattices that can be embedded in some T,. (See Exercises 2.26-2.36, 6.30 and 6.31.) (FC 7). Characterize the lattice Sub(L) of all sublattices of a lattice L. Find conditions under which Sub(L) determines L up to isomorphism. (FC 8). For what equational classes K does L C K and Sub(L) s Sub(Li) imply that LI E K? (This is known for K =L, M, D, and T.) For an equational class K, let Sub(K) denote the equational class generated by {Sub(L) I L CK}. Determine all equational classes K, for which Sub(K) =K. Determine all equational classes K, for which Sub(K) =L. Which first-order properties of lattices are preserved under projectivities of lattices, that is, under Sub(L) x Sub(L,)? Which first-order properties are preserved when passing from L to Sub(L), and conversely ? Let Csub(L) be the lattice of all convex sublattices (including 0 ) of the lattice L. Solve 1.3-1.9 with Csub(L) replacing Sub(L). (In 1.8, projectivity, that is, Sub(L) z Suh(Li), has to be replaced by Csub(L) Y Csub(LI). Csub(L) is investigated in K. M. Koh [1972], C. C. Chen and K. M. Koh [1972], and J. W. de Bakker [1967]. See also W. D. Duthie [1942].) Relate the automorphism groups of L, Sub(L), and Csub(L). Characterize planar posets. Determine the order dimensions of planar posets.1 Is there an analogue of the Kelly-Rival theorem for finite lattices of order dimension n (2 n) ?

1.15. Relate the automorphism group of a planar lattice L to the automorphism group of a planar diagram of L (as a graph).

1.16 (FC 10). Define the complexity of a finite lattice as the number of intersections of lines in an optimal diagram. Investigate this concept.

1.17 (FC 11). For an equational class K of lattices, determine the set of all natural numbers n, such that n occurs as the complexity of a finite L E K.

1.18. Determine the Complexity of (&)n and Part(A). 1.19. Compute the complexity of A x B. 1.20. Is there a conneation between complexity and order dimension? 1.21. Investigate first-order properties which are preserved when passing from L to

I(L); and conversely. (See G. Grgtzer [1970] and K. A. Baker and A. W. Hales [ 19741.)

1.22 (FC 14). A (semi) lattice K is celled transferable iff whenever, for a (Remi) lattice L, K has an embedding y in I(L), then K has an embedding y in L. Characterize transferability. (See Concluding Remarks.)

-

1 Solved by D. Kelly, On the dimension of partially ordered sets. Manuscript.

1.23. 1.24.

1.25.

1.26.

1.27.

1.28.

1.29.

1.30

1.31 1.32

1.33.

1.34

1.35 1.36

1.37

1.38 1.39

~

Is a sublattice or the dual of a finite transferable lattice traneferable? If in 1.22 the embedding y can be chosen so as to satisfy the condition: xy € yv iff x y, then K is called sharply transferable. Characterize sharply transferable lattices. (See Concluding Remarks.) I

Find a direct proof that a sublattice and the dual of a finite sharply transferable lattice are sharply transferable. Is a finite transferable lattice sharply transferable? The same question for finite semilat tices. Impose conditions on a class K of partial lattices such that K-transferable and sharply K-transferable become meaningful and include the corresponding concepts for lattices and semilattices. Investigate these new concepts, especially in the light of Problems 1.22-1.26. For an equational class K of lattices, investigate K-transferability and sharp K-transferability. (Results are known for K =L and K = D , see Concluding ltemarks.) Is there an equational class K of lattices, K+L, such that any finite lattice L which is sharply K-trrtnsferable is projective in K ? (FC 16). Describe the equational class of lattices with a single identity in three variables. (FC 17). Find the shortest single identity characterizing lattices. (FC 19). Define and investigate weak partial lattices and partial lattices satisfying an identity. Which lattice identities are equivalent to one of the form p 5 q, where p and q are lattice polynomials and no variable occurs twice in p . (See Lemnia IV.6.11.) (FC20). For a partial lattice A , define a congruence relation asfollows: A binary relation 0 on A is a congruence iff there exists a lattice L and a congruence relation @ on L such that, (i) A is a relative sublattice of L and (ii) @ restricted to A is 0. Give an intrinsic characterization of this concept of congruence relation. (This is known for universal algebras in general; see G. Gratzer [1968], Theoreui 13.3.) (FC 21). Characterize the lattice of all congruence relations of a semilattice. (FC 22). Let L he a meet-semilattice with 0 for which [0, a ] is pseudocomple- niented for each a € L ; let &(a) denote the pseudocomplements in [0, a ] . Char- acterize the family of Boolean algebras {&(a) I a € L}. (See Exercises 6.28 and 6.29.) (FC 23). What is the order of magnitude of f ( k , n)? (Note that f ( k , n) is defined in Further Topics and References. B. Wolk pointed out that the argument of G. Havas and M. Ward [1969] can be modified so as to prove that if L has more than n3” elements, then L has an TL element sublattice in which all but a t most two elements are doubly irreducible. This implies the existence of f ( k , n ) . ) (FC 24). For what equational classes K, do all f H ( k , n) exist? (FC 27). Investigate the lattice of join-endomorphisms of a finite lattice. (See G. Griitzer and E. T. Schmidt [1958].)

Problems 57

-

1 Solved by G. Grat,zer and C. R . Platt, A characterization of sharply transferable lattices. Abstract. Notices Amer. Math. SOC., February 1977.

1.40. Determine those equational classes K of lattices for which every K EK is the Frattini sublattice of a suitable L E K.

1.41. For an equational class K, a partial K-lattice is a relative sublattice of a lattice in K. Examine special cases when characterizations along the lines of Theorem 5.20 can be found.

1.42. Find a class K of lattices such that for all pairs of ordinals tc <p, LE(K) c US(K). The same problem for inverse limits. Can this K (or the K for Ghich I/TK) $: +I/+l(K)) be constructed to satisfy some nontrivial identities? ( F ( K ) (L"(K)) is obtained from K by applying direct limits (inverse limits) a times, see Procblem 26 of G. GrZitzer [1968]. For fixed a and /?, C. R. Platt [1974] constructed a class K of lattices satisfying La(K) +L"+l(Ii) and P ( K ) +LS+l(K).)

-c * c +

CHAPTER 11

DISTRIBUTIVE LATTICES

1. Characterization Theorems and Representation Theorems

The two typical examples of nondistributive lattices are %3 and g5, whose diagrams

Theorem 1 is a striking and useful characterization of distributive lattices; Theorem2 are given in Figure 1.

is a more detailed version of Theorem 1.

i i

0 0

Figure 1

It is clear from Theorem 1 that sublattices isomorphic to g5 or W3 play an important role. We introduce special names and notation for them. A subset A of a lattice L is called a pentagon or diamond iff A is a Rublattice isomorphic to '%~oP m3, respectively. If we say that A ={xo, xl, x2, x3, x4} is a pentagon (respectively, diamond) we also assume that xo +o, xi + a , x2 +b, z3 +c, x4 - i is an isomorphism of A and g5 (respectively, %TI3).

THEOREM 1. A lattice L is dtktributive i f f L does not conhin a pentagon or a diamond.

THEOREM 2. (i) A lattice L is modular iff it does not contain a pentagon.

(ii) A modular lattice L is distributive i f f it does not contain a diamond.

PROOF. (i) If L is modular, then every sublattice of L is also modular; g5 is not modular,

thus it cannot be isomorphic to a sublattice of L. Conversely, let L be nonmodular,

60 11. Distributive Lattices

let a , b, c E L, a 2 b, and let (uAc)vb +aA(cvb). The free lattice generated by c ( , b, and c with a 2 b is shown in Figure 1.2.3. Therefore, the sublattice of L generated by a , b, and c must be a homomorphic image of the lattice of Fig- ure 1.2.3. Observe that if any two of the five elements U A C , (aAc)vb, nr\(bvc), bvc,c are identified under a homomorphism, then so are (aAc)vb and ar\(bvc). Consequently, these five elements are distinct in L, and they form a pentagon. 0

(ii) Let L be modular, but nondistributive, and choose x , y , z c L such that Z A ( Y V Z ) + ( Z A Y ) V ( Z A Z ) . The free modular lattice generated by x, y, z is shown in Figure 1.5.7. By inspecting the diagram we see that u, xi, y l , z,, v form a diamond. Thus in any modular lattice they form a sublattice isomorphic to a quotient lattice of mm3. But m3 has only two quotient lattices: m3 and the one-element lattice. In the former case we have finished the proof. In the latter case, note that if u and v collapse, then so do X A ( Y V Z ) and ( X A Y ) V ( X A Z ) , contrary to our assumption.

Naturally, these results could be proved without any reference to free lattices. A routine proof of (ii) runs as follows: Take x , y , z EL such that X A ( ~ V Z ) * ( X A Y ) V

( X A Z ) and define u, zl, y l , zi, v as the corresponding polynomials of Figure 1.5.7. Then a direct cornputation shows that u, x l , y I , zl, v form a diamond. There is a very natural objection to such a proof, namely: How are the appropriate polynomials found? How is i t possible to guess the result Z And there is onlyone answer: by working out the free lattice.

.

COROLLARY 3. complement in a'ny interval.

A httice L is distributive iff every element has at most one relative

PROOF. The "only if" part was proved in Section 1.6. If L is nondistributive, then by Theorem 2 it contains a pentagon or a diamond, and each has an element with two relative complenients in some interval.

COROLLARY 4. A lattice L is distributive iff for any two ideals I , J of L :

I v J ={ivj I iCI , j C J } .

PROOF. Let L be distributive. By Lemma 1.3.1(ii), if t E I v J , then t l i v j for some i c I and j E J . Therefore, t = (tAi)V(tAi), t ~ i E I and t ~ j E J . Conversely, if L is nondistributive, then L contains elements a, b, c as in Figure 1. Let I = (b] and J = ( c ] ; observe that a E I v J , since a 5 bvc. However, a has no representation as required by Corollary 4, since if a = ivj, i c I , j J , then j < a and j < c. Therefore, j < aAC < b ; thus j E I , and so a = ivj E I , a contradiction.

Another important property of ideals of a distributive lattice is:

LEMMA 5. principal, then so are I and J .

Let I and J be icEeals of a distributive lattice L. If I A J and I v J are

1 . Characterization and Iteprweiit,nt,ion Theorums 61

PROOF. Let I A J =(XI and I v J = ( y ] . Then y = i v j for some i € I and j c J . Set c. =mi and b =xvj ; note that c c I and b c J (since x c I A J = I n J ) . We claim that 1 = ( c ] and J = (b]. Indeed, if, for instance, J +(a] , then there is an a > b, a E J , and {x: a , b, c , y} form a pentagon.

THEOREM 6. Let L be u distributive lattice and let a c L. Then the map

y : x - ( x ~ a , xva ) , x E L ,

is un embedding of L into ( u ] x [ a ) ; i t i s un isomorphism i f a has a complement.

PROOF. cp is one-to-one, since if x y = yy, then x and y are both relative conipleinents of a in the same interval; thus :r = y by Corollary 3. Distributivity also iniplies that cp is R honiomorphisni. ’

If a has a complement, b, and ( 2 1 , w) E (u ] x [ u ) , then for x = (wb)Aw, xv =(u, v); therefore, y is an isoniorphisni.

We start the detailed investigation of the structure of distributive lattices with the finite case.

DEFINITION 7. For a distributive lattice L let J ( L ) denote the set of all nonzero jh-irreducible elements, regarded as u poset under the partial ordering of L. For ( 1 C L set

r ( a ) = { z I x < a , x c J ( L ) } = ( a ] n J ( L ) .

DEFINITION 8. For (I poset P , call A c P hereditary iff x € A and y < x imply that yE.4. Let H ( P ) denote the set of all hereditary subsets partially ordered by set inidusion.

* Note that H ( P ) is a lattice in which meet and join are intersection and union,

respectively, and thus H ( P ) is distributive. The structure of finite distributive lattices is revealed by the following result:

THEOREM 9. Let L be u finite distributive lattice. Then the map

cp: a -r(a)

rk u n isomorphism between L and H ( J ( L ) ) .

PROOF. eleinents ; thus

Since L is finite, every elenlent is the join of nonzero join-irreducible

a= vm, showing that y is one-to-one. Obviously, r (a) n r(b) =r(aAb), and so (uAb)y =agAby. ‘Vhe formula (avb)? = ayvby is equivalent t.0

r(avb) = r ( a ) u r(b).

To verify this formula, note that r(a) u r(b) r(avb) is trivial. Now let x Er(avb). Then x = x ~ ( a v b ) = (xAa)v(xAb) ; therefore, x = xha or x = x ~ b , since z is join-irreducible. Thus x E r(a) or x € r(b), that is, z C r(a) u r(b).

Finally, we have to show that if A < H(J(L)) , then ag, = A for some a E L. Set a=VA. Then r ( a ) a A is obvious. Let x E r ( a ) ; then x = x ~ a = x r \ V A = V ( x ~ y I y E A ) . So x = ZAY for some y C A , implying that x A, since A is hereditary.

COROLLARY 10. The correspondence L -. J(L) makes the class of all finite dislributive lattices with more than one element correspond to the class of all finite posets; isomorphic lattices correspond to isomorphic posets, and vice versa.

PROOF.

Since H(J(L)) is a ring of sets, we obtain:

This is obvious from J ( H ( P ) ) I P and H ( J ( L ) ) 2 L. A subset S of P ( A ) is called a ring of sets if X , Y C S implies that X n Y, X u Y €8.

COROLLARY 11. A finite lattice i s distributive i f f it i s isomorphic to a ring of sets.

If the poset Q is unordered, H(Q)=P(&); if B is Boolean, J ( B ) is the set of all atoms or 0 end therefore J ( B ) is unordered. Thus we get:

COROLLARY 12. all subsets of a finite set.

A finite lattice is Boolean i f f it is isomorphic to the Boolean lattice of

A representation a = xov- * W X , , - ~ is redundant iff a=xov* - - v x ~ - ~ v x , + ~ v * * *VX+,-~,

for some 0 5 i < n; otherwise it is irredundant.

COROLLABY 13. dant representation as a join of join-irreducible elements.

Every element of a finite distributive lattice has a unique irredun-

PROOF. The existence of such a representation is obvious. If

a=xov. ' V X , , - ~

is an irredundant representation, then r(a) = U (r(xi) I O < i < n). Thus x occurs in such a representation iff x is a maximal element of r ( a ) ; hence the uniqueness.

C O R O L L ~ Y 14. length IJ(L)I.

Every maximal chain C of the finite distributive lattice L is of

PROOF. For a < J ( L ) , let m ( a ) be the smallest member of C containing a. Then a+m(a) is a one-to-one map of J ( L ) onto the nonzero elements of C. Indeed, if

1. Characterization and Representation Theorems 63

m ( a ) =m(b) , m(a) >- x , and scC, then z v n = x v b ; therefore, a = (aAz)v(aAb), implying that a x or a 2 b. But a < x implies that m ( a ) < x < m(a), a contradiction. Consequently, a 2 b ; similarly, b 2 a ; thus a = b. Let y c C, y >. z, z c C. Then r ( y ) 2 r ( z ) , and so y = m ( a ) for any a c r ( y ) - r ( z ) .

Let M ( L ) denote the set of nieet-irreducible elements of L excluding 1. Corollary 14 and its dual yield:

IJ(L)I = IM(L)I.

For the modular case see Exercise IV.5.26. The crucial Theorem 9 and its most important consequence, Corollary 11, depend

on the existence of sufficiently many join-irreducible elements. In an infinite distributive lattice there may be no join-irreducible element. Note that in a distributive lattice, a + 0 is join-irreducible iff L - [a) is a prime ideal. In the infinite case the role of join-irreducible elenient,s is taken over by prime ideals. The crucial result is the existence of sufficiently many prime ideals (as illustrated in Figure 2).

Figure 2

THEOREM 15 Let L be a distributive lattice, let I be a n ideal, let D be a dual ideal of L, and let I n D = 0. Then there exists a prime ideal P o f L s u c h t h a t P 1 I a n d P n D = @ .

(M. H. Stone [1936]).

PROOF. most convenient form for this proof is:

Some form of the Axiom of Choice is needed to prove this statement. The

ZORN’S LEMMA. that X has the following property: If C i s a chain in (X; C), then U ( X I X € C) €F. Then X has a maximal member.

Let A be a set and l e tX be a nonvoid subset of P(A). Let us assunie

Let X be the set of all ideals of L that contain I and are disjoint from D. We verify that X satisfies the hypothesis of Zorn’s Lemma. Since I EX, we conclude that F is


nonvoid. Let C be a chain in 2- and let M = U ( X I X E C). If a, b EM, then a E X and bC Y forsomeX, YEC; since Cisacha in , either X C Y or Y E X ; if, say, X’C Y, then a, b E Y, and so avb E Y C_ M, since Y is an ideal. Also, if b I a E M , then a C X ; since X is an ideal, b E X G M. Thus M is an ideal. It is obvious that M r> I and H nD= 0 , verifying that M E F . Therefore, by Zorn’s Lemma, ,T has a niaxitrial element P. We claim that P is a prime ideal. Indeed, if P is not prime, there exist a,, b E L such that a, b( P but a A b E P. The maximality of P yields ( P v ( a ] ) n D + P and ( P v ( b ] ) n D + @ . Thus there are p , q E P such that p v a E D and qvbEU. Then x = ( p v a ) ~ ( q v b ) ED, since D is a dual ideal. Also,

x = (pAp)v(pAb)v(aAq)v(aAb) E P ;

thus P n D =I= 0 , a contradiction.

COROLLARY 16. a6 I . Then there i s a prime ideal P such that P 2 I and a 6 P.

Let L be a distributive lattice, let I be an ideal of L , and let a E L and

PROOF. Apply Theorem 15 to I and D =[a).

COROLLARY 17. prime ideal containing exactly one of a am? b.

Let L be a distributive lattice, a , b e L and a+b. Then there i8 t i

PROOF. Either (a] n [b) = 0 , or [a) n (b] = 0.

COROLLARY 18. ideals containing it.

Every ideal I of a distributive lattice i s the intersection of all prinie

PROOF. Let I, yn (P 1 P 2 I, P is a prime ideal of L). If I+I,, then there is an a E Ii -I, and so by Corollary 16 there is a prime ideal P, with P 2 I and a 6 P. But then a 4 P 2 I, is a contradiction.

THEOREM 19 iff it is i8omorphic to a ring of sets.

(G. Birkhoff [1933] and M. H. Stone [1936]). A lattice i8 distributive

PROOF. For a C L set

Let L be a distributive lattice and let X be the set of all prime ideals of L.

r(a) =(P I a6 P, P E X } .

Then the r(u) form a ring of sets, and a -r(a) is an isomorphism. The details are similar to t,he proof of Theorem 9, except for the first step, which now uses Corol- lary 17.

This result has a very useful corollary.

1. Characterization arid Representation Theorrms 65

COROLLARY 20. Let L be a distributive lattice with more than one element. An identity holds in L i f f it holds in the two-element chain, &.

PROOF. is a sublattice of L, and p =q holds in B2. Conversely, let p =q hold in cS2. Note that g2 = P ( X ) , with 1x1 = 1, and that P(A) is isomorphic to the direct power P(X)IAI. Therefore, p =q holds in any P ( A ) . By Theorem 19, L is a sublattice of some P ( A ) ; thus p =q holds in L.

We can reforniulate Theorem 19 using the concept of a field of sets: A field of sets is a ring of sets closed under set complementation.

Let p =q hold in L. Since {LI > 1,

COROLLARY 21 field of sets.

(M. H. Stone [1936]). A lattice is Boolean i f f it i s isomorphic to a

PROOF. Use the representation of Theorem 19. Obviously r(a’) = X - r(a) , and thus complements are also preserved.

For a distributive lattice L with more than one element, let 9 ( L ) denote the set of all prime ideals of L, regarded as a poset under c. The importance of 9 ( L ) should be clear from the previous results. A topology on 9 ( L ) will be discussed in Section 5.

Some interesting properties of L are reflected in 9 ( L ) . An important result of this type is the following theorem (see also L. Rieger [1949]):

THEOREM 22 L is a Boolean lattice i f f 9 ( L ) is unordered.

(L. Nachbin [1947]). Let L be a distributive lattice with 0 =+ 1. Then

PROOF. Let L be Boolean, P , Q C P(L), and P c Q . Choose a C Q - P. Since a CQ, we have (L’ 4Q, and thus a f t P . Thus u, u’ 6 P , but am’ = 0 P, a contradiction, showing that 9 ( L ) is unordered. This proof in fact verifies that in a Boolean algebra every prime ideal is maximal.

Now let 9 ( L ) be unordered and a EL, and let us assume that a has no complement. Set D ={z I a v x = l}. Then D is a dual ideal. Take D, = D v [ a ) = {z I z 2 dAn for some d C D}. The dual ideal D, does not contain 0, since 0 = dAa, avd = 1 would mean that d is a complement of a.’Thus there exists a prime ideal P disjoint to Di. Note that 1 6 ( a ] v P , otherwise 1 = a v p for some p P , contradicting P n D= 0. Thus some prime ideal Q contains ( a ] v P ; and so P c Q , which is impossible since P ( L ) is unordered.

According to Corollary 18, every ideal is an intersection of prime ideals. When is this representation unique?

THEOREM 33 (6. Hitshinloto [1952]). Let L be a distributive lattice with O + l . Every ideol has a unique representation as a n intersection of prime ideals iff L i s a finite Boolerrn luttice.


PROOF. If L is a finite Boolean latt,ice, then P is a prime ideal iff P = (a], where a is a dual atom; the uniqueness follows from Corollary 13 (or is obvious by direct computation).

Now let every ideal of L have a unique representation as a meet of prime ideals. We claim that I(L) is Boolean. Let I c I(L) ; define

J = n ( P I P E W ) , PPI) .

Then I A J = n ( P I P E P(L)) = (01. If L + I v J , then there is a prime ideal Po 2 I v J , and consequently J has two representations :

nw 1 P P I ) =m n v I p P 1).

Thus L = I v J , and J is a complement of I in I(L) . By Lemma 5, every ideal of L is principal; therefore L s I (L) , and so L is Boolean.

By Exercise 1.6.21, L satisfies the Ascending Chain Condition; thus every element of L other than the unit is contained in a dual atom. Since the complement of a dual atom is an atom, by taking complements we find that every nonzero element of L contains an atom.

If po, pi,. . . ,pn,. . . are distinct atoms in L, then the ascending chain pot POVpi, * * * > PQvPiv' ' 'vz),, - + - does not terminate, and thus L has only finitely many atoms, po, . . . , p , Let a =pov- * . V P , - ~ . If a' + O , then a' has to contain an atom, which isimpossible. Therefore, a'=O, a = l , and L s P(X), with 1x1 =n.

Exercises

1. 2. 3.

4. 6.

6. 7.

8.

9.

10. 11.

12.

Work out a direct proof of Theorem 2(i). Work out a direct proof of Theorem 2(ii). Let K be a five-element distributive lattice. Is there an identity p =p such t'hat p =q holds in a lattice L iff L has no sublattice isomorphic to K ? Does the property stated in Lemma 6 characterize distributive lattices? Let L be a distributive lattice with 0 and 1. Prove that the direct decompositions L =Lo XLl of L are in one-to-one correspondence with the complemented elements of L. Prove that the complemented elements of a distributive lattice form a sublattice. Let L be a distributive lattice with 0 and 1. Let LZXL,XL~ r K o X K i . Show that there is a direct decomposition L %Ao x Al x A 2 x A 3 such that A0 XA, sL",

Extend Theorem 9 to distributive lattices satisfying the Descending Chain Con- dition (see Exercise 1.6.14). Extend Corollary 10 to distributive lattices satisfying the Descending Chain Con- dition. Can Exercises 8 and 9 be further sharpened? Let L be a distributive lattice with 0 and 1. Let Co and Cf be finite chains in L. Show that there exist chains Do 3 Co, Di 2 Ci, such that [Do] = lDi[. Derive from Exercise 11 the result that all maximal chains of a finite distributive lattice have the same length.

A2 x A ~ zLI, A , x A ~ PKO, Ai XA3 z K ~ .

13. 14.

*15.

lG.

17.

18. 19.

20.

21.

22. 23.

24.

26. 26. 27.

28.

29.

30.

3 1 . 32.

33.

34.

35.

36. 31.

*38.

39.

Find examples showing that Exercise 11 is not valid if “finite” is omitted. Prove the theorem “L is modular iff I ( L ) is modular” by showing that “L contains a pentagon iff I (L) contains a pentagon.” Is the second statement of Exercise 14 true for the diamond rather than for the pentagon ? Let L be a distributive lattice, a, b, c EL and a _< b. I s it true that [a, b ] is Boolean iff [ U A C , b ~ c ] and [ U V C , b v c ] are Boolean? For a poset P, let H p ( P ) denote the poset lattices of the form of all subsets of the form (aJu(a,]u. - * u ( u a - i ] , where n is an arbitrary natural number. Show that Theorem 9 holds for H,(P) . Prove that if P is a finite poset, then Hp( P) = H ( P) . Is it true that if Corollary 13 holds for a lattice L with more than one element, then L a H p ( P ) for some poset P? Show that the Ascending Chain Condition implies the Descending Chain Condition for Boolean algebras. Show that Exercise 20 fails to hold for generalized Boolean algebra8 (that is, relatively complemented distributive lattices with 0). Use Exercise 20 to simplify the proof of Theorem 23. Let L be a lattice, let P be a prime ideal of L, and let a, b , cEL. Prove that. if av(bhc) E P , then (avb)A(avc) € P . Using Exercise 23, show that the lattice L is distributive iff, for all x, y EL, x < y, there exists a prime ideal P with 2 E P and ya P. Verify the statement of Exercise 2’4 using Theorem 6. Verify the statement of Exercise 24 using Theorem 1. Let L be a distributive lattice. Then L is relatively complemented iff $‘(L) is unordered. Prove Theorem 16 by well-ordering L, L = {a,, I y < u}, and deciding one by one for each a,, whether a,, E P or a,,@ P . Let L be a distributive lattice with 1. Show that every prime ideal P is contained in a maximal prime ideal Q (that is, P 5 Q, and for any prime ideal X of L, Q X implies that Q = X ) . Let L be a distributive lattice with 0. Verify that every prime ideal P contains a minimal prime ideal Q (that is, P 2 Q, and for any prime ideal X of L, Q 2 X implies that Q = X ) . Find distributive lattice I; with no minimal and no maximal prime ideals. Investigate the connections among the Ascending Chain Condition (and Descending Chain Condition) for a distributive lattice L, for I ( L ) , and for $‘(L). Let L be a distributive lattice with 0 and let IEI(L). Show that the pseudocomplement of I is I* ={a: 1 (X]AI = ( O ] } . Prove tjhat I =I** for any I EI(L) of a distributive lattice with 0 iff L is a generalized Boolean lattice satisfying the Descending Chain Condition. The congruence relations 0 and @ permute iff a = b (0) and b = c (a) imply that u 2 d (a) and d G C (0) for some d. Show t,hat the congruences of a relatively complemented lattice permute. Prove the converse of Exercistx 36 for distributive lattices. Generalize Theorem 23 to distributive lattices without 0 and 1. Let L be a distributive lattice, let utL, let S be a sublattice of L, and let aaS. Show that there exists a prime ideal P and a prime dual ideal Q such that a $ P u Q 3 5‘ provided t)hat n is not the 0 or 1 of L (J. Hashinioto [1962]). Let L be a relatively compkmented distributive lattice. A sublattice K of L is

1. Charact,erization and Representation Theorems 67

6 Griitzer

40.

41.

42.

43. 44.

46.

46. 47.

48.

proper iff K +L. Show that every proper sublattice of L can be extended to a maximal proper sublattice of L (K. Takeuchi [1961]; see also J. Hashimoto [l962] and G. Griitzer and E. T. Schmidt [1958c]). Show that the statement of Exercise 39 is not valid in general if L is not relatively complemented (K. Takeuchi [1961]). (See also M. E. Adams [1973].) Generalize Corollary 13 to infinite distributive lattices, claiming the unique irredundant representation of certain ideals as a meet of prime ideals. If P is a prime ideal of L, then (PI is a principal prime ideal of I (L) . Is the converse true ? Show that Corollary 17 characterizes distributivity. A chain C in a poset P is celled maximal iff, for any chain D in P, C c D implies that C = D. Using Zorn’s Lemma, show that every chain is contained in a maximal chain. Prove that a finite distributive lattice is planar iff no element is covered by three elements. Show that a finite distributive lattice is planar iff it is dismantlable. Let S be a sublattice of the finite lattice L. Then i S can be represented in the form

L - U ([a, b] I a is join-irreducible and b is meet-irreducible). Prove the converse of Exercise 47 for distributive lattices.

2. Polynomials and Freeness

We can introduce an equivalence relation = for lattice polynomials: p F q iff p and q define the same functions in the class D of distributive lattices. More formally, if p and q are lattice polynomials (see Section 1.4.), then p E q iff, for any distributive lattice L and ao, al, . . . L, we have p(ao, . . .) =q(ao, . . .) (see Definitions 1.4.1 and 1.4.2). For an n-ary lattice polynomial p , let [ p ] D denote the set of all n-ary lattice polynomials q satisfying p =q and let P,(n) denote the set of all these equivalence classes, that is, PD(n) = { [ p ] D I p EPcn)}. Observe that, for p , pl, q, qi EP(m), if p =pi and q E qi, then pAq -plAqi and p v q =p lvq i . Thus [p]D.[q]D = [ p ~ q ] D and [p ]Dv[q ]D=[pvq]D define A and v on P,(n). It is easily seen that PD(n) is a distributive lattice and [p ]D 5 [q]D iff the inequality p < Q holds in the class D.

To describe the structure of PD(n), let &(n) denote the dual of the poset of all proper nonvoid subsets of {0,1, , . . , n- l}. Define H(a)=Q1.

THEOREM 1. (i) PD(n) .is isomorphic with H(&(n)).

(ii) PD(n) is a free distributive lattice on n generators. (iii) 2n-2<1P~(n)li ;22~-2. (iv) A finitely generated distributive lattice is finite.

1 Exercises 47 and 48 are from I. Rival [1973].

2. Polyrioiiiinls and 1’ 4 reeness 69

PROOF. (i) A lattice polynomial p is called a meet-polynomial iff it is of the form X , . ~ A * * AX^^.

For J (0, . . . , n - l}, J =+ 0 , set

p J = / l ( x i I i E J ) .

We claim that [pJ ]D < [px ]D ( J , K C_ (0, . . . , n - l}) iff J 2 K. The “if” part is obvious. Now let J 3 K; then there exists an i C K such that ie J. Consider the two-element chain Q2 and substitute xi=O and x j = l for j+i. Obviously, pJ = 1 and p , = O ; thus the inequality p J < p K fails in CS2 and therefore in D.

Every lattice polynomial is equivalent t o one of the form V p J ; because every xi is of this form, the join of two such polynomials is of this form and the same holds for the meet in view of

vp .J iAvPKj Ei V(PJiApKj)

and P J ~ A P K ~ P J ~ -

Next we claim that [p ]D E J(PD(n)) iff it is a [pJ ]D. Since every [ p ] D E PD(n) is a join of sonie [pJ]D, i t suffices to prove that a [pJ ]D is join-irreducible. Let

p J = V ( p J i l i E K j J i G { O j * * . j n - l } ) ;

J E Jifollows from [pJ]D 2 [p,,ID. Now if [pJ]D > [pJJD for all i , then we have J cJ,. Choose j i c J i , j i 6 J for all i E K. I n a2 put xk =o for all k = ji and xk = 1 otherwise. Then pJ = 1, and V ( p J . I iE K ) = O , which is a contradiction. A reference to Theorem 1.9 completes the proof of (i). 0

extended to t h e homomorphism (ii) Let L be a distributive lattice, ao, . . . , an-l E L. Then the map xi-+ai can be

[PID + P ( ~ o , * * J a n - 1 ) ,

proving (ii). a (iii) This proof is obvious from (i). (iv) This proof is obvious from (iii).

0

Figure 1.5.6 is a diagram of PD(3). Boolean polynomials are defined exactly as lattice polynomials except that all five

operations A, v, ’, 0, 1 are used in the fornlation of polynomials. A formal definition is the same a8 Definition 1.4.1 with two clauses added: If I, is a Boolean polynomial, so is p ‘ ; 0 and 1 are Boolean polynomials. An n-ary Boolean polynomial p defines a function in ?i variables on any Boolean algebra B ; p(ao, . . . , an-l) can be defined imitating Definition 1.4.2.

For the Boolean polynomials p and q, set p =q iff, for any Boolean algebra, B and uo, uI, . . . B, we have p(ao, . . .) =q(ao, . . .). Let [ p ] B denote the equivalence class containing p. Observe that p - q is equivalent to the identity p = q holding in the class B of all Boolean algebras.

Let PB(n) denote the set of all [ p ] B , where p is an n-ary Boolean polynomial. It el*

70 11. Distributive Lahtices

is easily seen that [p]B~[plB = [p~q]B, [p]Bv[q]B = [pvq]B, ([p]B)’ = [p’]B, O=[O]B, 1 =[1]B defines the Boolean operations on PB(n), and thus PB(n) is a Boolean algebra.

THEOREM 2. (i) P,(n) is isomorphic to (8.Jzn.

(ii) P,(n) is a free Boo1ea.n algebra on n generators. (iii) IPB(n)l =2Zn. (iv) A finitely generated Boolean algebra is finite.

PROOF. A Boolean polynomial is called atomic iff it is of the forin

X 2 A - * .AX?::,

where ij = O or 1 , xo denotes x and xi denotes 2’ . For every J {0 , . . . , n- 1) there is an atomic polynomial pJ for which ij = 0 iff j E J . The crucial statement is : [ ~ J ~ ] B 5 [pJ1]B iff Jo = J , . Indeed, let Jo + J i . We make the following substitution in Btf2: x i = l if i c J o , xi=O if i $ J o ; thismakespJo=l,pJl=O, contradicting[pJo]B<[pJ,]B.

Let B(n) be the set of all Boolean polynomials that are equivalent to one of the form V ( p J i 1 i E K ) . Then B(n) is closed under v and A , since

V P J ~ A V P I ~ ~ V(PJ~APIJ

and p J . ~ p I =pJ if J i = I,, and PjiA\prk = 0 otherwise. Now we prove by induction on n that xi, xi E B(n) for i < n. &’or n= 1 , x,, x i are

atomic polynomials, so xo, x i € B(1) . By induction, x,= V ( p J i I i E K ) , where the 2)Ji are atomic (n - 1)-ary polynomials; then

a t k

So E X o A ( X f l - i V X i - l ) (XOAX, _ , ) V ( Z o A X i - , )

= V ( p J a A x n - 1 I i E K ) v V ( p J i A x k - , I i E K ) ,

and similarly for x;. Thus xo, x; E B(n) , and, by symmetry, x i , x: E B(n) for all i < 71.

Since (pJ)’ z V(Z: I i e J ) v V ( x i I i $ J ) , we conclude that (pJ)’ E B ( n ) ; therefore B(n) is closed under ’. Thus B(n) is closed under A, v , ’, 0, 1 . Since B(n) includes all x i , i < n, B(n) is the set of all n-ary Boolean polynomials.

Consequently, every [p]B is a join of atomic ones, the [p]B for p atomic polynomials are unordered and 2n in number, implying (i) and (iii). The proof of (ii) is routine (same proof as that of Theorem l(ii)), and (iv) follows trivially from (iii).

We can use Theorems 1 and 2 to characterize free distributive lattices and free Boolean algebras, respectively.

THEOREM 3. Let L be a distributive lattice generated by {ai I i € I } . L is freely generated by the a; iff the validity in L of a relation of the form

A ( % I i E I , ) I Vbi I i E I , )

implies that I , n I, + 0, for I,, I , finite nonvoid subsets of I .

2. Polyiioiiiials and Freeness 71

PROOF. The “only if” part can be easily verified by using substitutions in Qz. For the converse, let F be the distributive lattice freely generated by xi, iE I , and let Q, be the hoiiioniorphism of F into (in fact, onto) L Satisfying xiq = ai for i E I. I t suffices to prove that for the latticc polynomials p , q, 5 qq implies that [p]D -g [q]D. (We think of the elements of F as equivalence classes of polynomials in the xi, i E I.)

Let P E V ( A ( x i I i€Ij) I j € J )

q = A ( V ( x i I i € K t ) I tET) .

v (A (a< I i € ’j) I i E J ) I A (V (aa I i E K , ) I t E TI,

A(U i I i € q < V(0i I iEK,)

and

Then p q 5 qp takes the form

which is equivalent to

for all j E J , t T . By assumption this implies that I j n K,+ 0 for all i E J, t E T ; thus

A ( z i l i E J i ) l V ( x i l i € K , )

for all j € J, t € T ; implying that [p]D I [q]D. 0

THEOREM 4. Let B be a Boolean algebra generattd by {ai j i€ I} . Then B is freely generated by {ai I i E I } iff, whenever I,, Ii, Jo, J , are finite subsets of I with I,, u Ii = J,u Ji and I o n I , = 0, then

A(ui I i€ IdAA(u: . l i € I , ) < A ( a i I i E J o b A ( a i I .L’€Ji)

implies that I. = J o and I , = J,.

PROOF. Again, the “only if” pert is by substitution into B2. On the other hand, clearly B is freely generated by {ai I iEI} iff, for every finite subset K of I, the subalgebra [{ai I i € K } ] is freely generated by {ai I i € K}. By Theorem 2, the latter holds iff [{ai 1 i K } ] has 22IE1 elements, which, in turn, is equivalent to [{a, I i E K } ] having 2IE1 atoms. Using the proof of Theorem 2 and the present hypothesis for I,u I i = K , we can see that the elements of the form A ( a i I i€ I,)AA(a; I i€ Ii) , where I ,u I , = K and I , n I , = 0 , are distinct atoms in [{ai I i E K}], thus completing the proof.

For a simpler variant of Theorem 4 see Exercise 3.43. Now we turn our attention to an important application of polynomials: finding

homom orphisnis.

THEOREM 5. Let the Boolean algebra B be generated by the subalgebra BI and the element a. Let B2 be a Boolean algebra and let y be a homomorphism of 23, into B2. The extensions of q to homomorphisms of B into B2 are in one-to-one correspondence with the elements p of B2 satisfying the following conditiolzs: (i) I f x 6 Bi and x < a , then xy I p.

(ii) If x E B, and x ;1 a, then x y 2 p .

72 11. Dishributive Latticcs

To prepare for the proof of this theorem we verify a simple lemma, in which + denotes the symmetric difference; that is,

x + y = (Z’AY)V(ZAY’).

LEMMA 6. ment a. Then every element x of B can be represented i@ the form

Let the Boolean algebra B be generated by the subalgebra B, and the ele-

x = ( U A X ~ ) V ( U ’ A Z ~ ) , xo, xi E B,.

This representation is not unique. In fact,

(aAxo)v(a’Azi) = (aAyo)V(a’w1), %o,% Yo9 Y l E B,, iff

PROOF. Let Bo denote the set of all elements of B having such a representation. If zE B,, then x=(a~z)v(a’~z); thus B I G B,,. Also a=(aAl)v(a’~O), and so a € B,. Therefore, to show Bo = B, it suffices to verify that Bo is a subalgebra, which is left as an exercise. Now note that for p, q E B, p =q iff pAa = q A a and p ~ a ‘ = q A a ’ ; thus

isequivalentto ( U A Z ~ ) + (UAY,) =O;thatis,a~(x,-,+y~) =O(seeExercise ll), whichisthe same as a < (x,, +yo)’. Similarly, a’ml = a’Ayl iff xi + y1 < a.

(aAZ,)V(U’AZi) = (aAy~)V(a’Ayi) iff UAxo = aAyoanda’Axi = U’Ayi. However, aAz0 = QAyo

PROOF of ”HEOREM 5. Let p be an element as specified and let the map y: B -c B, be defined as follows:

y : ( u A ~ ~ ) v ( u ’ A ~ ~ ) -. @AZ~~~)V(~’AZ~~). By Lemma 6, y is defined on B. It is well defined, because if (UAX~)V(U’AX~) =

( a ~ y ~ ) v ( a ‘ ~ y ~ ) , t h e n x ~ + y l < a l ( ~ 0 +yo)’; thus(q + y ~ ) q I p I (50 +yo)’q,andthere- fore zip, +YiBsP< (.OF +Yoq)’, imPlJ’h3 that (PAxo’fJ)V(P’Wq) = (PAYoq)V(P’AYiq). It is routine to check that ‘y is a homomorphism. Conversely, if y is an extension of Q, to B, then y is uniquely determined by p =up, end p satisfies (i) and (ii).

COROLLARY 7. Let us assume the conditions of Theorem 5. In addition, let B, be complete. Set

50 = V(xq I x E Bi, x I 4 and xi = A (xq I z E Bi, z 2 a).

Then the extensions of Q, to B are in one-to-one correspondence with the elements of the interval [xo, xi]. In particular, there is always at least one such exteasion.

Exercises

1.

2.

Get lower and upper bounds for / P ~ ( n ) l that are sharper than those given by Theorem l(iii). Work out the details of the lest steps in the proof of Theorem 1.

3. Congruence. Relations 73

3.

4.

5.

6.

7.

8.

*9.

10. 11. 12.

13.

14. 15. 16.

17.

18.

Let p~ be an atomic Boolean polynomial. Show that under the substitution = 1 for i EJ and si = 0 for i $ J , we get p~ = 1 and p ~ , = 0 for all Jo =k J. Let f: {0, 1}"-{0, 1). Prove that there is an n-ary Boolean polynomial p that defines the functionf on 'B2 (see Definition 1.4.2). (Let K = { J I J {0, 1, . . . , n - 1) such that f(str, . . . , 2, - 1 ) = 1, wheresi = 1 if i EJ and si = 0 if i $ J } ; set p = v ( p ~ I J € K ) and use Exercise 3.) Let B be a Boolean algebra. Prove that there is a one-to-one correspondence between n-ary Boolean polynomials over B and maps (0, 1}"-{0, 1). In other words, all 0, 1 substitutions take 0 and 1 as values and determine the polynomialp. An algebraic function over a Boolean algebra B is built up inductively from thd variables and the elements of B using A, v, and '. Show that the n-ary algebraic functions are in one-to-one correspondence with maps (0, l}" -, B. Let p be an n-ary algebraic function over the Boolean lattice B and 0 a congruence relation on B. Show that ai=bi(@), i <n, implies that p(ao ,..., a n - i ) Ep(b0 ,...,

Show that the property described in Exercise 7 characterizes algebraic functions (G. Gratzer [l962a]). Use the property described in Exercise 7 to define Boolean algebraic functions over a distributive lattice. Show that for bounded dist.ributive lattices Exercise 6 holds without any change (G. Gratzer [1964]). Show that a free Boolean algebra on countably many generators has no atoms. Show that aA(b+c) =(aha)+ ( U A C ) holds in any Boolean algebra. Let B be the Boolean algebra freely generated by {zi I i €1). Let L be the sublattice generated by {xi I i €1). Prove that L is the free distributive lattice freely generated by {xi I i € I > . Let L and Li be distributive lattices, let L = [ A ] , and let q be a map of A into Li. Show that there is a homomorphism of L into Li extending cp iff, for any pair of finite nonempty subsets A1 and A , of A, A A ~ 5 v A 2 implies that A A ~ P , S V A ~ ~ . (Compare this with Exercise 1.6.46.) State and prove Exercise 13 for Boolean algebras. Interpret Lemma 6 using Exercise 14. Extend the last statement of Corollary 7 to the case in which Bi is generated by B and ao, . . . , a,, -1 E B1, n > 1. Let p and q be lattice polynomials. Since p and q can also be regarded as Boolean polynomials, p =q was defined in two ways in this section: with respect to D and with respect to B. Show that the two definitions are equivalent for lattice polynomials. Define = for lattice polynomials with respect to a class K of lattices. Show that PK(n) EK iff the free lattice over K with n generators exists, in which case PK(n) is a free lattice with n generators.

b , , - l ) ( @ ) .

3. Congruence Relations

Let L be a lattice and let H c L2. We denote by @(If) the smallest congruence relation such that a = b for ell (a , b) H, and call i t the congruence relation generated by H .

LEMMA 1. For any H L2, @ ( H ) exists.


PROOF. Let 0 = A(@ I a = b (0) for all (a,, b) E H ) . Since in the lattic'e C ( L ) the meet is intersection, it is obvious that a =- b (0) for all (a, b) E H ; thus 0 = @ ( H ) .

We will use special notations in two cases: If H={(a , b)}, we write @(a, b) for @ ( H ) . If H =12, where I is an ideal, we write @ [ I ] for @(H). The congruence relation @(a, b) is called principal; its importance is revealed by the following formula.

LEMMA 2. @ ( H ) = V(O(a, b) I (a , b) E H ) .

PROOF. The proof is obvious. Note that @(a, b) is the smallest congruence relation under which u = b, whereas

@[I] is the smallest congruence relation under which I is contained in a single class. In general lattices, not much can be said about 0 ( H ) . In distributive lattices, the

following description of @(a, b) is important (G. Griitzer and E. T. Schmidt [1958d]):

THEOREM 3. Let L be a dietributive lattice, a, b, x, y EL, and let u t b . Then x = y ( @ ( a , b) ) i f f ~ ~ a =YAU a d x v b = yvb .

REMARK. This situation is illustrated in Figure 1.

Figure 1

PROOF. Let 0 denote the binary relation under which x = y (0) iff X A U = Y A U

and xvb = yvb . 0 is obviously an equivalence relation. If x =y (0) and z E L, then

3. Congrueiicc ICdatioiis 75

( S V Z ) A U = ( X A U ) V ( Z A U ) = (yAa)v(zAa) = ( Y V Z ) A U , (xvz)vb =zv(xvb) =zv(yvb) = (yvz)vb; thus xvz = y v z (0). Similarly, X A Z E Y A Z (a), and so 0 is a congruence relation. That a = b (@) is obvious. Finally, let 0 be any congruence relation such that n b (0) and let x = y (0). Therefore, zAa=yAa,xvb=yvb, z v a E x v b (a), and xAb = X A U (0). Then, computing modulo 0, we obtain

and

x = X V ( X A U ) = z v ( y ~ a ) = ( x v ~ ) A ( ~ v u ) = ( X V Y ) A ( X V ~ )

= (zvy)A(yvb) =yv(xAb) =yv(xAa) =yv(y~a) =y ,

that is, x ~y (a), proving that 0s 0.

EXPLANATION. Since a = b implies that (nvp)r\q = (bvp)Aq, we must have x 2 y (@(a, b ) ) if X A ~ = (a,vp)Aq, xvy = (bvp)Aq. It is easy to check that the z, y satisfying the conditions of Theorem 3 are exactly the same as those for which such p , q exist. Thus Theorem 3 can be interpreted as follows: We get all pairs x , y with x = y (@(a, b ) ) and x< y by applying the Substitution Property “twiceJ’. No further application of the Substitution Property or transitivity is needed. An equivalent forin of Theorem 3 is to require zvy I bv ( z A ~ ) and (av (%AY))A (zvy) = ZAY.

Some applications of Theorem 3 follow.

COROLLARY 4. Let I be an ideal of the distributive lattice L. Then x-y ( @ [ I ] ) iff xvy = ( m y ) v i for some i E I . Therefore, I is a congruence class m d u l o @ [ I ] .

REMARK. congruence modulo @ [ I ] .

This situation is illustrated in Figure 2, in which the wavy line indicates

A y ) V i

Figure 2

ROOF. If .rVy = ( z A y ) V i , then x = y (@(xAyAi, i)), X A Y A ~ , i € I J and so x =y(@[I]). Conversely, @ [ I ] = V ( O ( u , w) I u, ~ € 1 ) by Lemma 2. However, 0(u, v)vO(ui, w i ) l O(UAWA~,AW,, ztvwvu,vwi); therefore @ [ I ] = U (0 (u , v ) I u , v E I ) . If z =y(O(u, w)),

u, v E I , US v, then xvv = yvv , and so ( X A Y ) V ( V A ( X V Y ) ) = x v y ; thus the condition of Corollary 4 is satisfied with i = v n ( x v y ) E I . Finally, if a E I and a = b ( @ [ l ] ) , then avb = (aAb)vi, i E I , and so avb c I and b E I, showing that Z is a full congruence class.

COROLLARY 5. a < b I x < y. Then x = y (@(a, a ) ) implies that x = y.

Let L be a distributive lattice, x , y , a, b c L, and let x < y a b or

A very important congruence relation has already been used in the proof of Lem- ma 1.3.5(ii): Given a prime ideal P of the lattice L, we can construct a congruence relation that has exactly two congruence classes, P and L - P. This statement can be generalized as follows: Let A be a set of prime ideals of a lattice L and let us call two elements x and y congruent modulo A iff, for every P c A, either x , y < P or x , y c L - P ; this describes a congruence relation on L. For instance, if A = { P , Q, R}, Qc P, Rc P, we get five congruence classes as shown in Figure 3; the quotient latt.ice is shown in Figure 1.5.8.

This principle will be used often. An interesting application is:

Figure 3

THEOREM 6. Let K be a sublattice of a distributive lattice L and let 0 be a congruence relation of K. Then 0 can be extended to L ; that is , there exists a congruence relation Q, on L such that, for x , y < K, x = y (a) iff z = y (0).

PROOF. Let 9 be the natural homomorphism of K onto KIO, that is, 9: x - [ X I @ ; then, for every prime ideal P of K I 0 , Py-i is a prime ideal of K. Therefore, (Pq1-11 is an ideal of L and [K - P9-i) is a dual ideal of L; thus by Theorem 1.15 we can choose a prime ideal Pi of L such that Pi 2 Py-' and Pi n (K - Pq-1) = 0. For every

3. Congrucncc Ri.lations 77

prime ideal P of K we choose such a prime ideal Pi of L; let A denote the collection of ell such prime ideals. Let 0 be the congruence relation associated with A as previously described. Now for x, y c K, the condition x = y (0) is equivalent to xv = y ~ , and so for every P I C A either 2, y e Pi or x ,y( Pi; thus x = y (a). Conversely, if x = y (a), then for every Pic A either x, y c Pi or x, y @ Pi, and so either xq, y~ c P or XV, yy @ P. Since every pair of distinct elements of KI0 is separated by a prime ideal (Corollary 1.17), we conclude that xq = yy and thus x ~y (0).

It is well known that in rings, ideals are in a one-to-one correspondence with congruence relations. In one class of lattices the situation is exactly the same as in the class of rings.

THEOREM 7. Let L be a Boolean lattice. Then

0 + [0]0

is a one-to-one correspondence between congruence relations and ideals of L.

PROOF. By Corollary 4 the map is onto; therefore, we have only to prove that it is one-to-one, that is, that [0]0 determines 0. This fact, however, is obvious, since u = b (0) iff aAb E a v b (O), which in turn is equivalent to c GO (O), where c is the relative complement of U A ~ in [0, uvb] (see Figure 4). Thus a = b (0) iff c E [0]0.

0 Figure 4

This proof does not make full use of the hypothesis that L is a complemented distributive lattice. In fact, all we need t'o make the proof work is that L has a zero and is relatively coniplemented. Such a distributive lattice is called a generalized Boolean httice.

0 Figure 5

THEOREM 8 (J. Hashimoto [1952]). Let L be a luttice. There is a one-to-one correspondence between ideals and congruence relations of L under which the ideal corresponding to a congruence relation 0 i s a whole congruence class under 0 i f f L is a generalized Boolean algebra.

PROOF The proof of the “if” part is the proof of Theorem 7. We proceed with the “only if” part. The ideal corresponding to co has to be (01, and thus L has a 0. If L contains a diamond (0, a , b, c, i}, then ( a ] cannot be a congruence class, because a = o implies that i = avc = ovc = c, and b = bAi = bAc = 0. But o c (a] , and thus any congruence class containing (a] contains b, and b 4 (a]. Similarly, if L contains a pentegon (0, a , b, c, i} and a congruence class contains ( a ] , then b = 0; thus i = bvc = ovc = c, and so a = aAi = U A C = 0. Therefore, this congruence class has to contain a, and a 4 (a] . Thus, by Theorem 1.1, L is distributive. Let a < b and I = [O]O(a, b). By Corollary 4, @ [ I ] is also a congruence relation of L having I as a whole oongruence class; consequently, @[I] =@(a, b), and so a F b (@[I] ) . Thus, again by Corollary 4, b = a v i for some i c I, and i = 0 (@(a, b)). The latter is equivalent to i vb = Ovb and iAa = OAa. Thus a v i = b and aAi = 0, and so i is the relative complement of a in [0, b].

It is no coincidence. that, in the class of generalized Boolean lattices, congruences and ideals behave as they do in rings. Indeed, generalized Boolean lattices are rings in disguise.

(G. Grlitzer and E. T. Schmidt [1968d]).

THEOREM 9 (M. H. Stone [1936]). (i) Let ‘23 = ( B ; A, v ) be a generalized Boolean lattice. Define the binary operations . and

+ on B by aetting X*y=XAY

and by defining x + y as the relative complement of ZAY in [0, x v y ] (see Figure 5 ) . Then B R = ( B ; +, -) i s a Boolean ring-that is , a n (associative) ring satisfying x2 = x , for all x c B (and, consequently, satisfying x y = y x and x + x =0, for

= ( B ; +, *) be a Boolean ring. Define the binary operatiom A and v on 2, Y € B).

B bY (ii) Let

XAy=X - y

79

and

x v y = x + y +z * y.

Then !BL = ( B ; A , v ) i s u generalized Boolean lattice. (iii) Let B be a generalized Boolean lattice. Then ( @ R ) L = B .

(iv) Let 8 be a Boolean ring. Then ($3L)R =,H.

The proof of this theoreni is purely computational. Some steps will be given in the exercises. The correspondence between Boolean rings and generalized Boolean lattices preserves many algebraic properties.

THEOREM 10. (i) Let I c B,. Then I i s a n ideal of 93, i f f I i s a n ideal of 23:.

(ii) Let 91: B, --c B,. Then 91 i s a (0)-homomorphism of '$3, into B, iff 91 is a hovno-

(iii) %, i s a (0)-sublattice of %I i f f 80" i s a subring of ,H;.

Let B0 and %i be generulized Boolean lattices.

morphism of ,H,K into %p.

The proof is again left to the reader. Congruence relations on an arbitrary lattice have an interesting connection with

distributive lattices :

THEOREM 11 (N. Funayama and T. Nakayama [1942]). Let L be a n arbitrary luttice. Then C ( L ) , the lattice of all congruence relations of L, i s distributive.

PROOF. Let 0, @, y . ' E C(L) . Since @A(@VY) 2 (@A@)v(@AY), it suffices to prove that a = b (@A(@vY)) implies that a -b ( ( @ A @ ) V ( @ A ~ ) ) . So let u -b (@A

( @ v Y ) ) ; that is, a =b (0) and a = b (@vY) . By Theorem 1.3.9, there exists a sequence aAb=z,<- - * ( z , = a v B such that X , = X ~ + ~ (@) or x i = z , + l (Y) for every 0 1 i<n. Since u -b (a), we also have a A b r a v b (O), and so zi=zi+l (0) for every 0 s i<n. Thus for every O l i < n, Z ~ = Z , + ~ (@A@) or zi=zi+i ( ~ A Y ) , implying that ( I -b ( ( @ A @ ) V ( @ A ' r ) ) .

Another property of congruence lattices is given in the following definition.

DEFINITION 12. (i) Let L be a complete lattice and, let a be a n element of L.: Then a i s called compact

(ii) A complete lattice i s ccrlled algebraic i f f every element is the join of compact i f f a 5 V X for some X

elements.

L implies that u < V X , for some finite Xi C_ X .

The name, algebraic lattice, is due to G. Birkhoff [1967], however Birkhoff does not assume completeness. In the literature, algebraic lattices are also called compactly geiierated ltrtlices.

Just as for lattices, a nonvoid subset I of a join-semilattice F is an ideal iff , for a, b E F, we have avb c I exactly if a and b E I. Again, I ( F ) is the poset (not necessarily a lattice) of all ideals of F partially ordered under set inclusion. If F has a zero, then I ( F ) is a lattice.

Using I ( F ) , we give a useful characterization of algebraic lattices:

THEOREM 13. a join-semilattice with 0.

A lattice L i s algebraic i f f it 6 Csomorphic to the lattice of all ideals of

PROOF. Let F be a join-semilattice with 0; we prove that I ( F ) is algebraic. We know that I ( F ) is complete, We claim that for a € F , (a] is a compact element of I ( F ) . Let X c I ( F ) and let

( a ] C V ( I I I E X ) .

V(II I E X ) = { x ~ x I t o v ~ ' ~ v t , L , , t i E l i , l i E X } .

(a1 E V ( I I +Q.

Just as in the proof of Corollary 1.3.2,

Therefore, a<tov- - wtn-i , t iEIi , I i C X . Thus with X i ={Io,.. . ,

Since for any I E I ( F ) we have

'=V((a] I ' E l ) , we see that I ( F ) is algebraic.

Now let L be an algebraic lattice and let F be the set of compact elements of L. Obvioualy,O~F. Let a, bEF and a v b ( V X for some X E L . Then a l a v b l V X , and so a( V X , , for some finite X,, C X. Similarly, b I V X , , for some finite Xi E X. Thus avb< V(Xo u X i ) , and X o u X i is a finite subset of X. So avb E F .

Therefore, (F; v) is a join-semilattice wit,h 0. Consider the map:

p: a - t { x I x E F , x I a } , aEL .

Obviously, p maps L into I ( F ) . B y the definition of an algebraic lattice, a= V a y , and thus y is one-to-one. To prove that p is onto, let I E I ( F ) , a = VI. Then ag, 2 I. Let x cay . Then x< V I , so that by the compactness of x , x( VIl for some finite I, G I. Therefore x E I, proving that ag, c I. Consequently, ag, =I, and so p is onto. Thus p is an isomorphism.

Now we connect the foregoing with congruence lattices.

LEMMA 14. Every principal congruence relation is compact.

PROOF. Let L be a lattice, a, b E L, X c C(L), and

b, vxa

3. Congrueiict: Relations 81

Then a = b (VX), and thus (just as in Theorem 1.3.9) there exists a sequence n =xo, x l , . . . , x n - l =b, in which X ~ - X ~ + ~ ( O ~ ) for some O i E X . Therefore, a ~ b (vX,), where X,, ={Oo, . . . , and so @(a, b ) < (VX,), where X , is a finite subset of x.

THEOREM 15. Let L be a n arbitrary lattice. Then C ( L ) i s a n ulgebruic lattice.

PROOF. For every 0 E C ( L ) ,

0 = V(O(a , b) 1 u E b (0)).

Consequently, this theorem follows from Lemma 14 and Corollary 1.3.15. Combining Theorems 11 and 15 we get:

COROLLARY 16. lattice.

Let L be a n arbitrary lattice. Then C ( L ) is a distributive algebraic

The converse of Corollary 16 is a long-standing conjecture of lattice theory. We shall verify the conjecture in the finite case. This was first established by R. P. Dil- worth. The present proof combines a construction of G. Griitzer and E. T. Schmidt [1962] (Lemma 18) with a result of G. Griitzer and H. Lakser [1968] (Lemma 19). The reader is advised to page over the proof of Theorem 17 at the first reading of this book.

THEOREM 17. such that K is isoniorphic to C(L) .

Let K be u fiwite distributive Zattice. Then there exists a finite lattice L

PROOF. The proof of Theorem 17 is imniediate from Lemmas 18 and 19. We take L = I ( M ) where M is given in Lemma 18. By Lemma 19, C ( L ) z C ( M ) ; by Lemma 18, C ( M ) z K . Since M is finite, so is L.

Let M be a finite poset _such that inf {a, b} exists in M for any a, b M. We define in M: aAb =inf {a, b} and avb =sup {a, b} whenever it exists. This makes 'M into a partial lattice. An equivalence relation 0 on M is a congruence relation iff a,, E b, (0) and al G b1 (0) imply that aOAai = boAbi (0) and that uOvnl = bovbl (0) whenever aOval and b,vbl exist. Then the set C ( M ) of all congruence relations is again a lattice.

LEMMA 18. that inf { u , b} exists for all a, b M and C ( M ) i s isomorphic to K .

Let K be u finite distributive lattice. Then there exists a finite poset M such

PROOF. Take the set M , = J ( K ) u (0) and make it a meet-semilattice by defining inf {a , b} = O i f n+b ( J ( K ) is the set of nonzero join-irreducible elements of K ; see

82

a b

11. Distributive Lattices

0 Figure 6

Section 1), as illustrated in Figure 6. Note that a = b (0) and a +b imply in N o that a rO (0) and b = O (0); therefore, congruence relations of M o are in one-to-one correspondence with subsets of J ( K ) . Thus C(Mo) is a Boolean lattice whose atoms are associated with elements of J ( K ) ; the congruence Qa associated with a E J ( K ) is: z y (Qa) if (2, y} ={a, 0}, and if (x, y}+{a, 0}, then x-y (QJ implies that x=y .

If J ( K ) is unordered in K, then we are ready. However, if, say, a, b J ( K ) , (1 > b in K, then we must have Qa > Qb. The simplest way to make this happen is to use the lattice M ( a , b) of Figure 7. Note that M ( a , b) has three congruence relations, naniely,

0 Figure 7

w , I , and 0, where 0 is the congruence relation with congruence classes (0, b,, b,, b}, (ai, a@)}. Thus @(ai, 0) =i. In other words, ai EO “implies” that bl = O , but b, ~0 does not “imply” that al ~ 0 .

We construct M by “inserting” M ( a , b) in M , whenever a > b in J ( K ) . Figure 8 gives M for the t,hree-element chain.

More precisely, M consists of four kinds of elements: (i) 0; (ii) all maximal join- irreducible elements of K (that is, all a c J ( K ) such that there is no z c J ( K ) with a < x in K); (iii) for any nonmaximal join-irreducible element a of K, three elements: a, ai, a,; (iv) for each pair a, b c J ( K ) with a > 6, a new element, a(&). To simplify the notation, for each maximal join-irreducible element a, we write a =ai =a2. For a, b E J ( K ) with a > b, we set M(a, b) = (0, al, b, b,, b,, a@)}.

Observe that M(a, b) n M(c, d ) = H(a, b) if a =c and b = d ; M ( a , b ) n M ( c , d ) = (0, b, b,, b,} if a+c and b =d; N(a, b) n M(c, d ) = { O , ajl} if n = c and b + d ; M(a, b) n M(c, d ) = (0, 6,) if b = c ; otherwise, M ( a , b) n M ( c , d ) = (0).

3. Congruence Relst8ioris 83

Figure 8

For x, y c M , let us define x y to mean that for some a, b E J ( K ) with a > b, we have x, y c M(a, b ) and x < y in the lattice M(u , b) as illustrated in Figure 7. It is easily seen that x< y does not depend on the choice of a and b, and that < is a partial ordering relation. Since, under this partial ordering, all M(a, b) and M(a, 6) n M(c, d) are lattices and x, y E M , x E M(a, b), and y I x imply that y E M(a, b), we conclude that inf {u, w} exists for all u, w E M .

Now we describe C ( M ) . Let H E H ( J ( K ) ) (notation of Definition 1.8). We define a binary relation 0, on M:

s e y ( @ H ) iff either x, y E U (M(a,, b ) I a, b EH, a > b) u U ((0, ai, a2, a} 1 aEH), or x, yE{ai ,a(b) ,a(c)} , where a > b , a>c , b , c E H , or z=y.

In other words, [ o ] @ H contains all al, a2, a with a C H ; and if a > b, a, b CH, then it also contains u(b). Outside this class the only nontrivial congruences are a(b) =aI ~ a ( c ) , w h e r e a q H , a n d b , c E H , a > b , a > c .

0, is obviously an equivalence relation. The fact that 0, restricted to any M(a, b) is a congruence relation easily implies that 0, is a congruence relation. Given a 0, we get

thus the map H ={a 1 at Eo ( @ H ) } ;

'p: H - O H

is a one-to-one order-preserving map of H ( J ( K ) ) into C ( M ) . To show that 'p is an isomorphism, we have to show that y is onto. So let 0 be a congruence relation of M, and

H ={a 1 ~0 (0)).

Since in M(a, b ) every congruence 0 is determined by the atoms in [0]0, the same holds in M. Therefore, 0 = 0,. Thus H ( J ( K ) ) 2 C ( M ) . By Theorem 1.9, K Y H ( J ( K ) ) , and so K = C ( M ) . 7 Gratzer

84 11. Dist,ributive Lattices

LEnmA 19. Let M be a finite poset with the property thut inf {a , b ) exists for all CI, bEM. Then for every congruence relation 0 there exists exactly one congruence relation 0 of I ( M ) such that for a , b E M , (a] = (b] (0) i f f a = b (0).

PROOF. Since arbitrary nieets exist in M, for every element m c ill, (m] is a (finite) lattice, and so if { x , y } has an upper bound, then x v y exists.

Let 0 be a congruence relation of M . For X C M, set [ X I 0 = U ( [ X I @ I x E X ) ; that is, [ X I 0 = { y I x E y (0) for sonie x c X}. I f I , J c I ( M ) , define I = J (0) iff [ I ] @ = [JIO. Obviously, 6 is an equivalence relation. Let I = J (o), N E I ( M ) , and x E I n N. Then x = y (0) for some y E J , and so x = XAY (0) and SAY E J n N . This shows that [I n N ] 0 c [ J n N]O. Similarly, [ J n N ] 0 c [ I n N ] 0 , so I n N = J n N (0). To show that I v N = J v N (o), recall the description of I v N given in Exercise 1.5.22: Set A , = I u N and, for O<n<w, An={% Ix(tovt, , for some t,,tlcA,-l); then I v N = U(A, I n < w ) . We prove by induction on n that A, c [ J v N ] ~ . For n = O , A , = I u N c [ J ] 0 u N [ J v N ] @ . Suppose that A,-l c [ J v N ] ~ and let x € A , . Then x<tovti for some to, t l€A, - , . Thus t o r u o (0) end tl =uI (0) for some uo, ul E J v N , and so tO=t~Auo (0) and t,=tiAui (0). Observe that tovti is an upper bound for { to~u0 , tiAui} ; consequently, (toAuo)v(tiAui) exists. Therefore, tovt, =

( t f ~ u i ) ) E J v N . Thus x c [ J v N ] ~ . Since I v N = U (A, I n < w ) , we conclude that I v N c [JvNIO.

Siniilarly, J v N [ I v N ] @ , proving that I v N = J v N (0). This completes the verification that

If a = b (0) and x t (a] , then x = X A ~ (0). Thus (a] E [ @ I ] @ . Similarly, (b] E [ ( a ] ] 0 , and so (a] E (b] (0). Conversely, if (a] F (a] (O), then a = bl (0) and al = b (0) for some al I a and bi I b. Forming the join of the two congruences, we get a = b (0). Thus 0 has all the properties required by Lemma 19.

To show the uniqueness, let @ be a congruence relation of I ( M ) satisfying (u] = (b] (0) iff a ~ b ( 0 ) . Let I, J E I ( M ) , I = J (@), and xEI . Then ( x ] n I = ( x ] n J (@), (21 n I = ( x ] , and ( X I " J = ( y ] for some y EJ . Thus ( X I ~ ( y ] (@), and so z ~ y (a), proving that I c [ J ] 0 . Similarly, J E [I]@, and so I = J (G). Conversely, if I = J (O), then take all congruences of the form x = y (O), x E I, y J . By our assumption regarding @, ( X I ~ ( y ] (@), and by our definition of 0, the join of all these congruences yields I = J (0). Thus @ = 0.

More general results along these lines can be found in G. Grlitzer and E. T. Schmidt [1962], G. Grlitzer and H. Lakser [1968], and E. T. Schmidt [1968], [1969], and [ 19741.

(toAUo)V(tiA2(1) (0). Finally, X =XA(toVti) ~XA((toAzlo)V(t11\2,)) (0) and XA((t0AUo)V

is a congruence relation of I ( M ) .

-

Exercises

1. Let L be a distributive lattice and let x, y, U, b EL. Prove that if u 5 11 and I 5 ?/, thvn X A U = Y A ~ and zvb =yvb is equivalent to ( U V p ) A Q =z aid ( b V p ) A Q =!J for some p , g in L.

3. Corigruencc Relations 85

2. 3.

4. 5. 6 .

- 1.

$8. 9.

10.

11.

12. *13.

11.

1.5.

1 ti .

17.

18. 19.

20. 21. 22. 23.

24. 25. 20.

27.

28. i’

Verify Corollary 4 directly. Let K be a sublattice of the distributive lat,t.ice L and let P be a prime ideal of K. Prove that t,here exists a prime ideal Q of I, with Q n K = P. Prove t’hat Corollary 5 characterizes the dist,ribut,ivity of L. Show that Theorem 6 characterizes the distributivity of L. What is the smallest TZ such that Theorem 6 for (KI 5 n characterizes the distributivity of L ? Let L be a sectionally complemented lattice; that is, L has a 0 and all intervals [0, a] are cornplttmc:nt,t:d. Provct that 0 -+ [0]0 is a one-to-one correspondence between congruenFes and certain. ideals of L. Show that the “certain ideals” that appear in Exercise 7 form a sublattice of I ( L ) . Prove that every (principal) ideal of L is of the form [O]@ iff L is distributive. Let L btt a distributive lattice and let I be an ideal of L. Define a binary relation on L: X ’ E ~ ( @ ( I ) ) iff there is no aEL with a (zvy, ~ A Y A U E I , aqI. Prove that @(I) is tht: largest congruence relation of L under which I is a whole congIuence class. Let L br a distributive lat,tice with 0. Provr that therc is a one-to-one correspondence between ideals and congruence relat’ions (in the sense of Theorem 8) iff @[I] =@(I) for all I EI(L). Prove Theorem 8 using Exercises 10 and 11 (U. Ja. Areirkin [1953]). Let L be a lattice and let a he an element of L. Show that every convex sublattice of L containing a is a congruence class under exactly one congruence relation iff L is distributive and all the int,ervals [b, a] (b EL, b 5 a) and [a, c] ( c EL , a 5 c ) are complemented ( G . Gratzer and E. T. Schmidt [1958d]). Derive Theorem 8 (and, also, a variant of Theorem 8 ) by taking a=O (arbitrary a E 15) i n Exercise 13. Lest L be a relatively oomplemented lat,tice, I, J EI(L), and I E J. Prove that, if I is an intersection of prime ideals, t,hen so is J (J. Hclshimoto [1952]). Use Exercises 14 and 16 to get the following theorem: Let L be a relatively complemented lattice. Then L is distributive iff, for some element a of L, (a] is an intersection of prime ideals and [a) is an intersection of prime dual ideals (J. Hashimoto [1952]). Prow that t,he verification of Theorem 9(i) can be rcduced t.0 the Boolean lattice case and that i n this case x+y = ( x A ~ ’ ) v ( z ’ A ~ ) . Let B br a Boolean lattice. Verify that X+ y = (rvy)~(z’vy’). Let, B btt a Boolean lat’tice. Verify that (z+ ~ ) + ~ = ( ~ A ~ ’ A z ’ ) v ( z ’ A ~ A z ’ ) v ( z ’ A ~ ’ A z ) and conclude Lliat + is asBociative. Prove that s(y+ z ) =xy+ zz. Prove Theorem 9(i). Prove Theorem 9(ii). Let \13 be a generalized Boolean lattice. Observt: that for X, y EB, X A ~ is thr! same in \H as in ( t + A ) L (namely, z - y) ; conclude that %3 = (%a)L.

Vcarify Thcorem 9(iv). Verify Tht:orem 10. Show that,, using the concept of a distribut,ivc srmilattice (sce Section 5), Corollary 16 call bo rc:formulated as follows: Let L be an arbitrary lattice. Prove that there exists a distributive join-semilattice F with 0 such that C ( L ) is isomorphic to I ( F ) . Charact,wisv the latt>ice of all idrals of a lattice using the concept of an algebraic lat.ticc-. (Iharactctrizo thu lat,tice of all ideals of a Boolean lattice.


29.

30. 31. 32.

*33.

34. 36.

36.

*37. 38.

39. 40.

41.

42.

43.

Let M be a poset such that ahb exists for all a, b EM. Let 0 and of M . Find a result for M analogous with Lemma 1.3.8. Let M be as in Exercise 29. Generalize Theorem 1.3.9 to M. Let 2M be as in Exercise 29. Is C ( M ) necessarily distributive? Let M be as in Exercise 29. Assume that M satisfies the following conditions: (i) There exists H c M such that for A EH, (h] is a lattice. (ii) For each ideal I that belongs to the sublattice I p ( M ) of I ( M ) generated by the

I with i j 5 hi

Under these conditions, prove Lemma 19 for M, replacing I ( M ) with I 'p(M) (G. Gratser and H. Lakser [1968]).

Exercises 33-36 are from G. Griitzer and E. T. Schmidt [1962]. Let K be a finite distributive lattice and let L be the lattice of Theorem 17 as constructgd in Lemmas 18 and 19. Show that L is sectionally complemented. Let L be given 8s in Exercise 33. Show that the congruences of L permute. Let m be the length of the longest chain in K. Show that the L of Theorem 17 can be constructed so that L is of length 2n - 1 (define a@) only for a > b ) . Let P be a poset. Generalize Theorem 17 for K = H F ( P ) (notation is the same as in Exercise 1.17).

Exercises 37 and 38 are from G. GrZitzer and E. T. Schmidt [1968b]. Show that 8 chain C is the congruence lattice of a lattice iff C is algebraic. Prove that a Boolean algebra B is the congruence lattice of a lattice iff B is algebraic. Let L be a distributive lattice. Show, that a +@[(a]] embeds L into C(L) . Let L be a bounded distributive lattice. Show that for a, b EL, a 5 b, @(a, b) has a complement in C(L) , namely, O(0, a)v@(b, 1). Let L be a bounded distributive lattice. Show that the compact elements of C(L) form a Boolean lattice (J. Hashimoto [1962], G. Grktzer and E. T. Schmidt [ 1958el). Let B be a Boolean algebra generated by X and let L be the sublattice of B generated by X. Show that B is freely generated by X iff L is freely generated by X in D (see Exercise 2.12). Let the Boolean algebra B be generated by X. Show that B is freely generated by X iff AX0 I v X i implies that X,nX,+ 0 for any finite nonvoid X,, &,EX' (compare this with Theorems 2.3 and 2.4).

be congruences

principal ideals, there exist finite {Ai, . . . , h,} and I=(ii] u - * - u (in].

H , {ii, . . . , in}

4. Boolean Algebras R-generated by Distributive Lattices

* The investigations of this section are based on the following result:

THEOREM 1. Every distributive lattice can be embedded in a Boolean lattice.

PROOF. subsets of some set X. Obviously, L can be embedded into P(X).

By Theorem 1.19, every distributive lattice L is isomorphic to a ring of

4. Algebras R-generated by Distributive Lattices 87

DEFINITION 2. Let L be a (0)-sublattice of the generalized Boolean lattice B. Then L i s said to R-generate B iff L generates B as a ring, and if L has a 1 the same element is the 1 of B.

The last condition is equivalent to the following: If V L exists, then VL= V B . Our goal is to show the uniqueness of the generalized Boolean lattice R-generated

by L. The first result is essentially due to H. M. MacNeille [1939]:

LEMMA 3. form

Let B be R-generated by L. Then every a E B can be expressed in the

uo + aI + - * + a, - l , a. I a, I - - I a, - 1, ao, . . . , a, -1 E L.

EXAMPLE. Then L R-generates B.

Let B be the Boolean lattice shown in Figure 1 and let L = (0, a,,, al, a2}.

0

Figure 1

PROOF. Let Bi denote the set of all elements that can be represented in the form u o + - * * + u , & . - ~ , ao, . . . , a,-lEL. Then L E BI , and B, is closed under + and - (since x - y = x + y ) . Furthermore,

and each term a,bi=a,r\bjEL, ao Bi is closed under multiplication. We conclude that B, = B.

Note that L is a sublattice of B ; therefore, for a, b E L, avb in L is the same as uvb in B. Thus avb=a+b+ab, and so

a + b = ab + (avb) = ( a h ) + (avb).

88 11. Distribut)ive Lattices

LEMMA 4. Let L be a distributive lattice with 0. Then there exists u generalized Boolean lattice B freely R-generated by L-that is , a generalized Boolean lattice B witlt the following properties;

(i) B is R-generated by L. (ii) I f BI is K-generated by L, then there i s u ho.mo?norphism y of B onto B ,

that is the identity map on L.

PROOF. (Theorem I.6.24), mutatis mutandis.

The existence of B can be proved by copying the proof of Theorem 1.6.5

LEMMA 5 (J. Hashimoto [1962]). Let B be a generalized Boolean lcittice K- generated by the distributive M i c e L with O. Thea every congruence relation of L lins one and only one extemiofz to B.

PROOF. and 3.10(i), the following statement iinplies the uniqueness of the extension :

The existence of an extension was proved in Theorem 3.6. By Theorems 3.7

If I and J are (ring) ideals of B with I c J, then there are elements a , b EL, u + b , such that a = b (mod J) and a + b (mod I ) .

Indeed, let x ( J -I. By Lemma 3, x can be represented in the form

x=zo+".+x, - , , X O l . . *<xn- l , xu,. . . ,xn-,EL.

If n isodd, then x o = x - x o < x E J , and thus xoEJ; x o + z I + x 2 = z * x 2 E J , therefore x1 + x2 = xo + (xo +xi + x,) E J. Similarly, x3 + x4, xj + xg, . . . E J. Since xo + (xI + x,) + (x3 +x,) + - * * J -I, we conclude that either xo EJ -I, or zzi +x2i+l E J -I for some 2i < n. If n is even, then we obtain zo + xl, x2 +x3, . . . E J (by multiplying z by xi, x3, . . .), and we conclude that for some 2i < n, xqi + x , , + ~ E J - I .

Now if Z ~ ~ + X ~ ~ + ~ E J - I , then xzi=x2i+l(mod J), but x2i+x2i+i (mod I), x2; , xzi+l E L. Finally, if xu E J -I, then xo = O (mod J) and zo + O (mod I).

8

4. Algebras R-generated by Distributive Lattices

THEOREM 6. tive lattice L with 0, then B, and B, are isomorphic.

89

If B, and B, are generalized Boolean lattices K-generated by a distribu-

REMARK. For a distributive latt,ice L with 0 we shall denote by B(L) a generalized Boolean lattice R-generated by L.

PROOF. Let B be a free generalized Boolean lattice R-generated by L (see Lemma 4 ) . Let y be a homomorphism of B onto B, such that y is the identity on L (see Lemma 4(ii)). We want to show that v is an isomorphism. Indeed, if y is not an isomorphism, then the ideal kernel I of y is not 0. Thus by Lemma 5, a = b (mod I ) for some a , b L, a +b. This means that uy = bv, contrary to our assumptions. Similarly, there is an isomorphism y between B and B,. Obviously, y-fy is an isomorphism between B, and B,.

COROLLARY 7. Let Lo and L, be distribiitive lattices with 0 and let y be a (0)-homo- niorpliism of L, onto L,. Then y can be extended to a homomorphisnz of B(L,) onto B(L1).

PROOF. Let 0 be the congruence kernel of y , and let 0 be the extension of 0 to B(L,) (Lemma 5). Then B(L, ) /8 is a generalized Boolean lattice R-generated by L,/0 s L,. Thus B(L,)/@ s B(L,) by Theorem 6, and using this, the proof of Corollary 7 becomes trivial.

COROLLARY 8. Let Lo be a (0)-sublattice of the distributive lattice LI rn-th 0, and assuvne that i f L, has 1 , then Lo is a (0 , 1)-sublattice of Li. Let B denote the subalgebra of B(L,) R-generated by Lo. Then B(L,) 1 B.

PROOF. The proof is trivial. Let Lo and L, be given as in Corollary 8. It is natural to ask the conditions under

which Lo R-generates B(L,). For H C B(L,) , let [HI, denote the generalized Boolean sublattice of B(L,) R-generated by H . We can answer our query by determining

Ll*

LEMMA 9. Let L, and L, be gicen as in Corollary 8. Then Ll n [Lo], i s the smallest sublattice of L, containing Lo that is closed under taking relative complements in L,. Therefore L, 12-generates B(L,) iff the smallest sublattice of Li containing Lo and closed under relatice complementation in L, i s L, itself.

PROOF. It is obvious that Lo G L, il [L,,]R. If a , b, C EL, n [LO],, d EL,, and d is a relative complement of b in [a, c ] , then d = a + b +c EL, n [LO],) since (see Figure 2) d is a relative coniplement of a + b in the interval [0, c]. Thus d E Li n [LO]R. Now

90

C

11. Distributive Lattices

0 Figure 2

suppose that L is a sublattice of Li containing Lo and closed under relative complementation in Ll. If x E Li n [LOIR, then by Lemma 3 we can represent x as

z =a0 +* * * +a,-,, ao, . . . , a,-, E Lo, sol- - *I as-l.

We prove x E L by induction on n. If n = 1, x =ao E Lo E L. If n = 2, then x is the relative complement of a. in [0, a,], 0, ao, ai E Lo, thus x EL. If n = 3, then (see Figure 1) x =ao + ai + a2 is the relative complement of ai in [ao, az] , and so x EL. Now let n > 3 andlet y EL beprovedfor ally=bo+. - +bk-l , bo, . . . , bk-,CLO, b o l e * *< bk- i , and k <n.Note that xE Li andan-, E Loimply that "a,-, =ao +. - +an-, +an+ +an-, =

*+an-, =a,-8 +a,-, +a,-, C Li. By the induction hypothesie, ao+* - - +an-, EL and an-, +a,-, +a,-, C L; therefore, x is the relative complement in Ll of an element (namely, a,-,)of Linaninterval (namely, [ao+* **+an-,, an-, +~, -~+a , - , ] ) in L, and so, by assumption, x EL. Thus L, n [Lola G L.

In Theorem 1 we embedded L into P(X), which is a complete Boolean lattice. The question arises whether we can require this embedding to be complete-that is, to preserve arbitrary meets and joins, if they exist in L.

It is easy to see that not every complete distributive lattice has a complete embedding into a complete Boolean lattice.

a,, +- - * +a,-, E Ll and "van-, = x +an-, +xa,-, =ao +. - - +an-, +an-, +ao+

LEMMA 10 (J. von Neumann [1936/37]). Let B be a complete Boolean lattice. Then B satisfies the Join Infinite Dietributive Identity (JID), l

XAV(Zi1icI)=V(xAXiI i E I ) ,

and its dual, the Meet Infinite Distributive Identity (MID) .

PROOF. X A X ~ ~ X and X A X , . ~ V(xi I i E I ) for any j E I ; therefore, X A V ( X ~ I ; € I ) is an upper bound for { X A X ~ I iC I}. Now let u be any upper bound, that is, X A X ~ < u for all iCI . Then

xi = X ~ A ( X V X ' ) = ( Z ~ A Z ) V ( X ~ A X ' ) I uvx'.

1 Of course, (JID) is not an identity in the sense of Section 1.4 but is only an infinitary analogue of an identity.

4. Algebra8 R-generated by Distributive Lattices 91

Thus XAV(Z~ I ~ € I ) < ~ A ( U W ’ ) = (ZAU)V(XAZ’) = X A U < u ,

showing that Z A V ( Z ~ i E l ) is the least upper bound for { Z A X ~ I iE I } . (MID) follows by duality.

COROLLARY 11. A n y complete distributive lattice that has a complete embedding into a complete Boolean lattice satisfies both ( J I D ) and ( M I D ) .

Easy exaniples show that (JID) and (MID) need not hold in all complete distributive

Our task now is to show the converse of Corollary 11 (N. Funayama [1959]). The lattices.

construction depends on a property of B(L) and on Theorem 1.6.4.

LEMMA 12 is a pseudocomplemented lattice in which

(V. Glivenko [1929]). Let L be a distributive lattice with 0. Then I (L )

I * = { % I x A ~ = O for all i E I } .

Let S ( I ( L ) ) ={I* I I E I (L) } . If L is a Boolean lattice, then S( I (L) ) is a complete Boolean lattice and the map a+(a] embeds L into S ( I ( L ) ) ; this em.bedding preserves all existing meets and joins.

PROOF. The first statement is trivial. Now let L be Boolean. It follows from Theo- rem 1.6.4 that &(I&)) is a Boolean lattice. Furthermore, it is easily seen that for any X c I (L) , the inf and sup of X in S(I (L) ) are A X and ( V X ) * * , respectively, where A and V are the meet and join of X in I ( L ) , respectively. Since A ( ( x ] I x EX) = (inf X I , whenever inf X exists in L, the mapping a +(a] preserves all existing meets in L. Observe that, for x, a E L, xAa‘=O iff z< u , and so (a] = (a’]* ES(I(L)). Now let a = sup X in L and set I = ( X I (= V ( ( x ] I x E X ) ) . To show that x -* (51 is join- preserving, we have to verify that I** = (a], or, equivalently, that I* = (a’]. Indeed, i f 6 E I*, then bAx = 0 for all x E I , and thus x I b’. Therefore, a = sup X < b‘, proving a’ 2 b, that is, b E (a’]. Conversely,let b E (a’]. Then b ‘2 a; therefore, b‘ 2 a =sup X 2 x for all x EX, and so bAx = O for all x E X . This shows that b E I*, proving that I* = (a’]. 0

LEMMA 13. map 6 a complete embedding of L into B(L).

Let L be a complete lattice satisjying (JID) and (MID) . Then the identity -

If n is even let us replace a. by 0 +ao; thus we can assume that n is odd. We cleiiii that for x c L and a c B(L), we have x< a. iff

X A U ~ = xAai,

X A U ~ = x A ~ ~ , . . . , x ~ a , - ~ = x ~ a , - ~ , x<a,-l.

Indeed, let x < u. Then

xaI =xaI(ao + - - - +a,-,) =x(ao +al +al +- * +al) =xuo;

therefore, xAao=~Aal . Thus x(a2+..~+u,-,)=(xa~+xai)+x(a2+~ * - + a , - , ) = xu = x , and so x < a2 + * - +a,-l. Conversely, if xAao = XAal and x < a2 + * * - + u ~ ~ - ~ , then xa =xuo +xal +%(az + a * * +a,-,) =x , proving that x< a. A simple induction completes the proof of the claim.

Let X E L , y=sup X in L , and a E B ( L ) . If x l a for all x C X , then the formulas of the preceding claim hold for all x and a and, by ( J I D ) , for y and a, proving that y I a . ThuA y=sup X in B(L). The dual argument, using (MID), completes the proof.

THEOREM 14 (N. Funayama [1959]). A complete lattice L has a complete embedding into a complete Boolean lattice i f f L satisfies ( M I D ) and (J ID) .

PROOF. Combine Lemmas 10-13. The representation for a c B(L) given in Lemma 3 is not unique in general; the

only exception is when L is a chain. Since this case is of special interest, we shall investigate i t in detail.

Repeating the definition, a Boolean lattice B is R-generated by a chain C with 0 iff B = B(C). This concept is due to A. Mostowski and A. Tarski [1939] and can be extended to distributive lattices as follows :

A distributive lattice L with 0 is R-generated by a chain C ( E L ) with 0 iff C R-generates B(L).

LEMMA 15. Let L be a distributive lattice with 0 and let C be a chain in L, OEC. Then C R-generates L if1 L is the smallest sublattice of itself containing C and closed under the formation of relatilie complements.

PROOF. Apply Lemma 9 to C. An explicit representation of B(C) is given as follows : For a chain C with 0, let B[C] be the set of all subsets of C of the form

(.ol+(a1]+...+(u,-,], O<anla i<-- .<a, - l , an,. . . , a , - , E C ,

where + is the symmetric difference. We consider B[C] as a poset, partially ordered by C. We identify a C with (a] , for a 4 0 , and 0 with 0. Thus C E B[C].

4. Algebras R-gc.nc?rtttvd by Distributive Lattice8 93

LEMMA 16. B [ C ] is the generalized Boolean lattice R-generated by C.

PROOF. The proof is obvious, by construction and by Theorem 6. Note that every nonvoid element a of B [ C ] can be represented in the form

a = (a01 ” (b,, a11 “ * * * u @,-I, a,-i1,

O < a ~ < b i <a1 <*..<bn-i<an-i ,

where the union is disjoint union and the first term (uo] may be missing. An element of the form (x, ylisnothingbut x + y . Thus, ~ ~ = ~ ~ + b ~ + a ~ + . . . + b , - , + a , _ ~ , and so we conclude:

COROLLARY 17. t h e form

I n B(C) every nonzero element a has u unigtie representation in

u =uO + “ I + * * * +~,-l, 0 <ao <* . . < ~ , - 1 , ao, - . , an-1 E C.

The following results show that many distributive lattices can be R-generated by chains.

LEMMA 18. B = [C], for any ?uaxirnul chain C of B.

h’wry finite Boolean lattice B can be R-generated by a chain; in fact,

PROOF. Let B, be the subalgebra of B Ii-generated by C. Using the notation of Corollary 1.14, the lengt,h of C equals I J (B) I ; also, the length of C equals lJ(Bi)\; thus IJ(B)I = IJ(B,)I =n. We conclude that both B and Bi have 2, elements, proving that B=Bi.

COROLLARY 18. i n fact, by any rtruzimul chain of L.

Every finite ditdributire lattice L can be R-generated by a chriin-

PROOF. By Corollary 1.14, C is inaxinial in B. Thus, B = B(C) 2 L.

Let C be a maximal chain in L and let B = B ( L ) . Then I J (L ) [ =IJ(B)I.

THEOREM 20. Every countable distributive lattice L with 0 cun be K-generated by u chain.

PROOF. Let L ={ao, a,, o.,, . . . ,urn, . . .}, a,=O, and let L, be the sublattice of L generated by a(,, . . . , a,. Let C, be a maximal chain of Lo, and, inductively, let C , he a inaximal chain of L, containing C n - l . Set C = U (Ci 1 i <w). Obviously, 0 E C. We claim that C R-generates L. Take a E B(L) ; a = x, + - - * + xm-i, zo, . . . , x,-~ EL. L = U(L, I i < w ) ; thus for some n, xu, . . . , x , -~ E L,, and so a C B(Ln). Since L, is finite, we get B ( L , ) = B(C,); therefore a E B(C,,) B ( C ) , proving that L E B(C).

COROLLARY 21. The correspondence C - B ( C ) maps the class of countable chains with 0 onto tlhe class of countable generalized Boolean lattices. Under this correspondence, {O}-subchains and (0)-homomorphic images correspond to (0)-subalgebras and (0)-homomorphic images.

Note, however, that Co s C, is not implied by B(Co) E B(C,). Much is known about countable chains. Utilizing the previous results, such in-

formation can be used t o prove results on countable generalized Boolean lattices. To help distinguish an important class of chains, we introduce the concept of prime interval. An interval [a, b) is prime iff a< b.

LEMMA 22. Every countable chain C can be embedded in the chain Q of rational numbers. Every countable chain not containing any prime interval is isomorphic to one ofthe intervals (0, i), [0, i), (0, 11, and [0, 11 of Q.

PROOF. Let C={xo, xl , . . , , x,-~, , . .}. We define the map y inductivelyasfollows: Pick an arbitrary ro €0 and set xoq = r,. I f xov, . . . , X , - ~ ~ I have already been defined, we define xnq as follows: Let

Ln = U ((xi91 I xi < xn, i < n), u n = U ( [ x ~ Y ) I xi > xn, i < n)

(&=a or U,=@ is possible). Note that if L,=!=@, then it has a greatest element l,, and if U , =!= 0 , then it has a smallest element u,. If both are nonempty, then 1,< u,. In any case, we can choose an r, EQ with r, 4 L, u U,. We set xnp, = r,. Obviously, q~ is an embedding.

The second part of the proof reduces to the following statement: ' Let C and D be bounded countable chains with no prime intervals. Then C z D. To prove this, let C ={c0, ci, . . .} and D ={do, d,, . . .}. We define two maps:

q ~ : C-Dandy: D-cC,Letusassumethatco=O,c~=1,andthatdo=O,d~=i.For each n < co, we shall define inductively finite chains C,, D, (C, E C, D, C_ D) and an isomorphism vn : C,, -* D, with inverse yn : D, -c C,. Set Co = {co, ci} = (0, i}, Do = {do, d,} = (0, 1) and icpo = i, iyo = i, for i = 0 , l . Given C,, D,, q~,, y,, and n even, let k: be the smallest integer with C,. Define u k = A ( [ c k ) n C,) and lk = V ( ( c k ] n C,). Then I, < ck < uk, and so lkqn < Ukv,. Since D contains no prime intervals, we can choose a d E D satisfying lpp, < d < Uk(P,. Since v, is isotone, d 4 D,. Define Cn+l = C,, u {ck}, Dn+l = D, u {d}, Q),,+~ restricted to C, to be pin, and C ~ Q ) , + ~ =d, Y,+~ restricted to D, to be yn and dy,,, =cE' If n is odd we proceed in a similar way, but we interchange the role of C and D, C, and D,, qn and y,, respectively.

Finally, put q~ = U (q, 1 n < w ) . Clearly, C = U (C, 1 n < co), D = U (D9a I n <o), and is the required isomorphism.

COROLLARY 23. Up to isomorphism there i s exactly one countable Boolean lattice with no atoms and exactly one countable generalized Boolean lattice with no atoms and no unit element, B(Q).

4. Algebras R-generated by Dietributive Lattices 95

PROOF. Take the rational intervals [0, 11 and [0,1). The generalized Boolean lattices in quest.ion are B([O, 11) and B([O, 1)). This follows from the observation that [a, b] is a prime interval in C iff a + b is an atom in B(C). The results follow from Lemmas 16 and 22 and Theorem 20.

THEOREM 24. prime ideals.

Let B be a countable Boolean algebra. Then B has either KO OT 2 ” ~

REMARK. enough, we can give a proof without it.

This is obvious if we assume the Continuum Hypothesis. Interestingly

PROOF. For a Boolean algebra B and an ideal I of B, we shall write B/I for B/O[I] . If J is an ideal of B with J 2 I, then J / I ={ [z ]O[Z] I z EJ} is an ideal of B/I. (This is the usual notation in ring theory.) Let B be a Boolean algebra. We define the ideals I,, by transfinite induction. I , = (01; I, is the ideal generated by the atoms of B ; given I,,, let I be the ideal of B/I , generated by the atoms of BII,, and let y : x +x + I , be the homomorphism of B onto B/I,,; we set I,,,, IF-^. Finally, if y is a limit ordinal, set I,, = U (la I 6 < y) . The rank of B is’ defined to be the smallest ordinal a such that I, = Obviously, the cardinality of a is at most I BI.

CLAIM 1. Let B be countable. If I , + B, then [ P( B)I = 2%

Indeed, if I, +B, then B/I , has no atoms, and thus BII, B(C), where C is the rationalinterval[O, 11. ByLemma5 (see Exercise27), IP(B/I,)l = I9(C)[ = II(C)l =2’0.

CLAIM 2. Let B be countable. ff I , = B, then 1,4a(B)I = No.

Indeed, for y <a, let P,,(B) be the set of prime ideals P of B for which I,,c P , I ,+1 Q P . Since a is finite or countable, it suffices to show that IP,(B)I =KO. If PE,4a,,(B), then, by Corollary 1.16 and Theorem 1.22, we have PvI , ,+ ,=B. It follows that for P,QEP,,(B), P+Q, we have PnIy+l+QnI, ,+ l . Thus P + ( P n [ I r f l l H ) / l y is a one-to-one correspondence of ,4a,,(B) into (in fact, onto) 9([Zy+l ]B/I , , ) ; but, [I ,+l]B/I , , is just the generalized Boolean lattice of all finite subsets of ti, countable set. Therefore, IP,,(B)I =Ko.

In order to avoid giving the impression that most Boolean algebras can be R- generat,ed by chains, we state:

LEMMA 25. Then B is finite.

Let B bc a complete Boolean algebra R-generated by a chain C uritib 0.

96 11. Distributive Letticos

PROOF. Let B = [c], and let the chain C be infinite. We can assume that C contains n subchain {xo, x,, . . . , x,, . . .}, 0 < xo < x1 < * * - < x, < * (or dually, in which case the complements form an increasing sequence). Then define a, = x,) +xi + * * * + x~~ + x ~ ~ + ~ for each n < w . We claim that V(u, I i < w ) does not exist. Indeed, let a be an upper bound for {ai I i<w}. By the remarks immediately following Lemma 16, we can represent each a, by the set (xo, xi] u (xz, x3] u * - - u (z2,,, x ~ , + ~ ] , and we can represent a in the form

a = (bol u @I, bzl " * * - u (bnr-2, bm-11,

where 0 < bo < b, < - * * < b, bi E C for i < m, and the first term, (bo], may be missing. Since a contains each a,, there must exist an n and a j < m such that both (x2,, .c.,~+J and ( x ~ ~ + ~ , x ~ , , + ~ ] are contained in (bjPlJ bi] or in (bo]. Therefore, the interval (x~~+,, x ~ , + ~ ] can be deleted from a, and it will still contain all the a,-that is, ~ + x ~ , + ~ + x ~ , , + ~ is an upper bound for {a, I i<w}, and a + ~ ~ , + ~ + x ~ , + ~ < ~ ~ . We conclude that {ai I i <w} does not have a least upper bound.

Next we consider which chains with 0 can be R-generating chains of a given distributive lattice.

DEFINITION 26. Let L be a distributive lattice with 0 and let C be a chain in L, 0 E C . The chain C is called strongly maximal in L iff, for any homomorphism y of L onto a distributive lattice Ll, the chain Cp? 6s maximal in L,.

LEMMA 27. If L is R-generated by C then C is maximal in L.

Let L be a distributive lattice with 0 and let C be a chain in L, 0 EC.

PROOF. If C is not maximal in L, then we can find a C L, a 6 C, such that C LI { u } is a chain. Write a = u o + a l + . * * + a , - , with O < a o < a l < - - . < a , - , and uiEC for i < n . S i n c e a ( C , n > l . N o w a A a o = a o + a o + ~ ~ ~ + a o , whichisaoif n i s o d d a n d o if n is even. But a. +a and 0 +a, so, since a and a. are comparable, aAaO =a,, and n is odd. Then aAal =ao + ai + * * + a, = ao, contradicting the comparability of (I and u,.

The converse of Lemma 27 is false. The following thcoreni settles the matter.

THEOREM 28. Then C R-generates L iff C is strongly maximal in L.

Let L be a d6tributive lattice with 0 and let C be a chain in L, 0 E C.

PROOF. If C R-generates L, then, for any onto homomorphism y , Cp, R-generates Lq. By Lemma 27, Cp, is maximal in Lq, so C is strongly maximal in L.

Next assume that C is strongly maximal in L but does not R-generate L. Without any loss of generality we can assume that L and C have a greatest element. (Otherwise, add one. Then C u { l} is strongly maximal in L u { l} but does not R-generate L u { i}.) Let B, = B(L) and let Bo = [GI,. By hypothesis, Bo+Bl, so there exists an a E B, - Bo. We claim that there exist prime ideals PI f P, of B1 with Bo n PI = Bo n P,. With

I = ( ( ( I ] n B,,] a i d D = [ u ) , we have I n D = 0, so, by Tlieorcni 1 .Is, there is a prime ideal PI such that I c PI and PI n D = a. Then let I , = (111 and D, = [B,, -PI). Since ( ( I ] n B(, P,, P, n I ) , = o . Then (I E P, - P,. so P , + P,. Because P, n (B,, - P I ) = a, P, n B,, c I’, n B,. Then by Theorem 1.22 (prime ideals of a Boolean lattice are unordered), PI n B, = Y, n B,,, 1)roring our claim.

Xow (again by using Theorem 1.22) we can map B, onto (&)? by a homomorphism y: For n.E PI n P,, ry = ( O , 0); for X E Pi - P 2 , xy = ( O , 1); for xE P2 - PI, xy = ( l , 0); for 26 P, u P,, .ry =(1 , 1). Since C y = ((0, 0), (1, l)} is not maxinial, we conclude that C is not strongly inaxinla1 in L.

PI, it follows that 1, n D , = 0. Let P, be a prime ideal with I ,

COROLLARY 29. L with 0. Then ICI = ID1 and II(C)i = IZ(D)l.

Let C and D be strongly vmxiiii(i1 cliuins of the distributive krttice

PROOF. If L is finite, these conclusions follow from Corollary 1.14. If L is infinite, then [ C ] = [D] = B(L) , and so ICI = ID1 = 1LI. Also, by Lemnia 5 , lS(C)l= 19(B(L))I = ~ S ( D ) ~ , and $’(C) = I ( C ) , S ( D ) = I ( D ) ; hence the second statement.

Corollary 29 is the strongest known extension of Corollary 1.14 to the infinite case. The second part of Corollary 29 is froiii G. Griitzer and E:. T. Schmidt [1957].

Boolean algebras generated by chains were first investigated by A. Mostowski and A. Tarski [ 1!)3Y]. Theorem 20 for Boolean lattices and Theorem 24 were communicated to the author by J. R. Buchi. These results have been known for some time in topology (via the Stone topological representation theorem, see Section 5 ) . Some of the other results appeared first in FC.

Exercises

1. 2.

3. 4.

5 .

6. *7 .

Gi\rt. a drttailrd proof of Lemma 4. Try to describe t,he most general situation to which the idea of t>he proof of Theo- rem 1.5.5 (Theorem 1.6.24) could be applied. Show that Lemma 5 docs not remain valid if the word “generalized” is omitted. Find necessary and sufficient conditions on a distributive lattice L in order that L have a Boolean cxtension B to which every congruence of L has exactly one ex- t,ension. Let L be a distribut.i\re lattice wit,h 0 and define L, to be the latt’ice L if L has a unit elemcmt; let L, be? 1, with a unit added if L does not have a unit element. The Roolerm lattice B[L] R-generated by I, is defined to be B(L,) . Show that if B is any 13oolran lattice, containing L as a sublattice, and B is generated by L under A, v, 1~nd ’, 11~1.11 R is isomorphic to the Boolean lattice R-generated by L. Work o u t Corollarics 7 and 8 for the Boolean lattice R-generated by L. The Complete Inf in i te Distributive Iden.tity is (for I , J 0 ) :

A(V(n; j I j C J ) I i € I ) = V ( A ( a i i , I i€l) I v : I + J ) .

98 11. Distribut,ive Lattices

8.

9.

10.

1 1 .

12.

13.

14.

*15.

16.

17.

18.

19. +20.

21.

22.

23. 24.

26.

Show that this holds in a complete Boolean lattice B iff it is atomic (A. Tarski [1930]). (Hint: apply the ident,ity to A(ava' I a E B ) = 1.) Prove that the Complete Infinite Distributive Identity is selfdual for Boolean lattices. For a, subset A of a lattice L, set

AU = (1: I 1: EL, x is an upper bound of A}

A' = {x I x EL, x is a lower bound of A}.

Prove that this sets up a Galois connection and therefore A, AU =

((Au)r)u, and Az Call an ideal I of a lattice L normal iff I = ( P ) I . Show that every principal ideal is normal. Let IN(L) denote the set of all normal ideals of L. Show that I N ( L ) is a Complete lattice but that it is not necessarily a sublattice of Io(L) . Show that the map: x +(x] is an embedding of L into Ia(L) , preserving all meets and joins that exist in L. (I,(L) is called the MacNeille completion of L ; see H. M. MacNeille [1937].) Let B be a Boolean lattice and let I be an ideal of B. Show that I is normal iff I =I** (for the concepts, see Exercise 10 and Lemma 12). Prove that the Boolean lattice S( I (L ) ) of Lemma 12 is the MacNeille completion of the Boolean lattice L. Show that the MacNeille completion of a distributive lattice need not even be modular. Let L be a distributive algebraic lattice. Show that L satisfies the Join Infinite Distributive Identity. (Thus, for any lattice K, C ( K ) satisfies (JID).) Let L be a distributive lattice, ai, b i c L for i < w and [a,, bo] 2[al, b,] 3- * .. Define

2 A ,

0 =v(O(a,, Ui)V@(bO, bi) I i < w ) .

Show that

@vA(@(ai, b,) I i < W ) +A(@~@(a i , bi ) I i < a).

Use Exercise 17 to show that, for a distributive lattice L, the Meet Infinite Distri- butive Identity holds in C(L) iff every interval in L is finite (G. GrLitxer and E. T. Schmidt [1958b]). For a chain C, introduce and describe B(C) using Corollary 17, Prove the converse of Lemma 18: I f every maximal chain R-generates the Boolean lattice B, then B is finite. Why is it not possible to use transfinite induction to extend Theorem 20 to the uncountable case 7 Let C be a chain with 0 and 1 and let a EC. Define a new order on C: For x, y < a, and for a 5 x, y, let x I y retain its meaning; for x <a 5 y, let y < x; let Ci be the set C with the new order. Then Cl is a chain, and B(C) eB(Cl) , but in general C nCi does not hold. Describe a countable family of painvise nonisomorphic countable Boolean algebras. Give an example of a bounded distributive lattice L with a maximal chain C such that C is not maximal in B(L) (G. W. Day [1970]). Let Lo be the [0, 13 rational interval and let Li be the [0, 11 real interval. Let C = {(x, x) I 0 5 E 5 1, z rational). Then C is a maximal chain in L, xLI. Show that C is not strongly maximal (G. Griitzer and E. T. Schmidt [1957]).

5. Topological Rrprtwntation 99

2G.

27.

Find in Lo x L, of Exercise 25 maximal chains of cardinality No gnd 2% What are the cardinalities of strongly maximal chains? Let L be a distributive lattice with 0 and let B = B(L). Show that P + P n L for P @'(I?) is a one-to-one correspondence between the prime ideals of L and B (use Lemma 5). Let A be an infinite set, B = P ( A ) . Prove that B has maximal chains of cardinality IAl and 2IA1. Construct an example in which the sequence of ideals I, of Theorem 24 does not terminate in finitely many steps. Let C be the [0, 11 interval of the rational numbers. Show that B(C) is F*(Xo).

*28.

29.

30.

5. Topological Representation

The poset P(L) of prime ideals does give a great deal of information about the distributive lattice L, but obviously it does not characterize L. For instance, for a countably infinite Boolean algebra L, P(L) is an unordered set of cardinality Ho or 2N0, whereas there are surely more than two such Boolean algebras up to isomorphism.

Therefore, it, is necessary to endow P(L) with more structure if we want it to characterize L. M. H. Stone [lo371 endowed 9 ( L ) with a topology (see also L. Rieger [1949]). In this section we shall discuss his approach in a slightly more general but, in our opinion, more natural framework.

A join-semilattice L is called dislributiwe iff a 4 bovbi (a, bo, bi EL) implies the existence of ao, a, L, a i l b , i = O , 1, with a =aOvai (see Figure 1). Note that a. and aL need not be unique.

Some elementary properties of a distributive join-semilattice are as follows :

Figure 1

LEMMA 1. (i) If (L; A, v) i s a lattice, then the join-semilattice ( L ; v} i s distributive i f f the

(ii) I f a join-semilattice L is distributive, then for any a, b L there i s a d L with

(iii) A join-semilcrtticc L i s distributive i f f I (L ) , as a lattice, i s distributive.

lattice ( L ; A, V} is dtktribzitiue.

cl< a a d d < b. Comequently, I ( L ) is a lattice.

8 Griitzer

100 T I . Distributive Lattices

PROOF. (i) If (L; A, v)'is distributive, and a< bovbi, then withai =aAbi, i=O, 1, a=aoval.

Conversely, if ( L ; v ) is distributive, and the lattice L contains a diamond or a pentagon (0, a, b, c, i}, then a< bvc, but a cannot be represented as a =aoval with ao( b and ai 2 c, a contradiction.

(ii) a s a v b , thus a =aOvbo, where a,<a, b o g b. Since, in addition, b o < a , bo is a lower bound for a and b. 0

(iii) First we observe that, for I, J c I ( L ) ,

I v J = { i v j I i E I , j c ~ }

follows from the assumption that the join-semilattice L is distributive. Therefore, the distributivity of I (L ) can be easily proved. Conversely, if I (L ) is distributive and a < bovbi, then

(a1 = ( a I ~ ( ( ~ o l v ( b l 1 ) = ~ ~ ~ l ~ ~ ~ o l ~ ~ ~ ~ ~ l ~ ~ ~ i l ~ ~ and so a = aovai, a. E (bo], al < (bi], which is distributivity for L.

A subset D of a join-semilattice L is called a dual ideal iff a c D and x 2 a imply that x E D, and a, b ED implies that there exists a lower bound d of {a, b} such that d < D. An ideal I of L is prime iff I +L and L -I is a dual ideal. Again, let P(L) denote the set of all prime ideals of L.

LEMMA 2. Let I be an ideal and let D be a nonvoid dual ideal of a distributive join-semilattice L. If I n D = 0, then there exists a prime ideal P of L with P 3 I and P n D = 0.

PROOF. In the rest of this section, unless stated otherwise, let L stand for a distributive

join-semilattice with zero. In P(L), sets of the form r (a) = { P I a( P} represent the elements of L. We will

make all these sets open. Let Y'(L) denote the topological space defined on P(L) by postulating that the

sets of the form r(a) be a subbase for the open sets; we shall callJ(L) the Stone space of L. (Exercises 1-22 review all the topological concepts that are used in thissection.)

The proof is a routine modification of the proof of Theorem 1.15.

LEMMA 3. For an ideal I of L, define

T ( I ) =p I PEJ(L) , p P I } .

Then r ( I ) is open in Y'(L). Conversely, every open set U of Y'(L) can be uniquely represented as r ( I ) for 801128 ideal I of L.

PROOF. We simply observe that r ( I ) n r (J ) = r ( I f rJ ) ,

r ( V ( 4 I i E K ) ) = U(rV$ IjEK),

5 . Topologirnl Rrpresentation 101

and r ( ( u ] ) = r (u) . from which it follows that the r ( l ) form the smallest collection of sets closed under finite interaection and arbitrary union containing all the c(a). Observe that a I iff r ( a ) E r ( l ) . Thus r ( I ) = r ( J ) iff a € 1 is equivalent to a EJ, that is, iff I = J. 0

LEMMA 4. open sets.

The subsets of Y'(L) of the form r ( a ) can be characterized as compact

PROOF. Indeed, if a family of open sets { r ( Ik ) I k c K } is a cover for r(a) , that is, r ( n ) c U ( r ( l k ) 1 k E K ) =r(V( I , I k c K ) ) , then u c V ( I , I kEK) . This implies that o c V ( l k I k KO) for some finite KO U ( r ( l k ) 1 k E KO). Thus r ( n ) is compact. Conversely, if l is not principal, then r(1) U(T(U) I a c I ) , but r ( I ) u ( r ( a ) I a E I,) for any finite I, E I.

K , proving that r(a)

0 Fro111 Lemma 4 we inimediately conclude :

THEOREM 5. 27ie Stone spaceb(L) determines L up to isomorphism.

If we want to use Stone spaces to construct new ones in order that new distributive lattices can be constructed from given ones, then we have to know what Stone spaces look like. Stone spaces are characterized in Theorem 8. To prepare for the proof of Theorem 8, we prove Lemma 6.

Let P be a prime ideal of L . Then P is represented as an element of Y'(.L) and also by r(P). The connection between P and r ( P ) is given in Lemma 6 and is illustrated by Figure 2.

Y ( L )

Figure 2

LEMMA 6. logicul clositre of { P } .

For every prime ideal P of L, { P } = b ( L ) -r(P) , where { P } i8 the t q o -

PROOP. By definition of closure,

{P} = {& 1 & r (u ) iniplies that P E r(a)} = {& I & 2 P } =W) -{& I Q P P} =W) -m. 0

a*

COROLLARY 7. If P +&, then {p} +{g} ; in other Words, b ( L ) is a To-space.

PROOF. Combine Lemmas 3 and 6.

of a singleton. In other words: (C)

Lemma 6 also shows that if P is a prime ideal, thenY(L) -r(P) must be the closure

If ' U is an open set with the property that, for the compact open sets Uo and Ui, Uo n Ui U or Ui C_ U , then Y(L) - U = {p> for sonie element P.

U implies that Uo

Now we can state the characterization theorem :

THEOBEM 8. characterized (up to homeomorphism) by the following two properties : (51) 3 0 a T,,-space in which the compact open sets form a base for the open sets. (S2) If F i% a closed set in 3, { U , I kE K } is a dually directed' family of compact

open sets of Y, and UL n F + 0, then n ( U , 1 k E K ) n F =/= 0.

The Xtone space Y' of a distribdive join-semilattice with zero mn be

REMARK. The meaning of condition (51) is clear. Condition (52) is a complicated way of ensuring that (C) holds and that Lemma 2 holds for the join-semilattice of compact open sets of Y'(L).

~ O O F . To show that (Sl) holds, we have to verify that the r(a) , aEL, form a base (not only a subbase) for the open sets ofY(L). In other words, for a, b E L and P Er(a) n r(b), we have to find a, c E L with P Er(c) and r(c) E r(a) n r(b). By assumption, a(Pandb( P ; thus,since Pisprime,thereexistsacEL,c( P,c<a,c<b.Then PEr(c) , r(c) C_ r(a), and r (c) C r(b), as required. To verify (52) for S ( L ) , let F =Y(L) - r ( I ) and uk=r(a,). Thus F={P I P2I ) and U , = ( P I aE( P}. The assumptions on the a, mean that D = {z 1 2 2 ak for some k E K} is a dual ideal ; since u k n F $5 0 , we have r(ab) 0 r ( I ) ; that is, ak( I, showing that D n I = 0. Therefore, by Lemma 2, there exists a prime ideal P with P 3 I and P n D = 0. Then an ( P, and so P E r (aJ for all ii E K. Also P 2 I, thus P ( r ( I ) , and so P E F, proving that P E F n n ( U , I k E K ) , verifying (52).

Conversely, let Y' be a topological space satisfying conditions (131) and (52) of the theorem. Let L be the set of compact open sets of 3'. Obviously, 0 E L and if A , BE L, then A u B c L, and thus L is a join-semilattice with zero. Let

A S BOU B,, with A, Bo, BLEL.

Then A n B, is open, end therefore A n B, = U (Af I j E Ji), i =0,1 , where the '4f are compact open sets. Since A = (A n B,) u ( A n Bi) G U (A: I j E J o u Ji, i =0,1), by the compactness of A we get A E U ( A j I j cJo or jEJI), whereJi is a finite subset of Ji, i=O, 1. Set

A ~ = u ( A ; I jcJ,.), i=o, 1.

1 K =i= 0 and for k, 1 E K , there exists a t CK such that U, E uk n UJ.

5 . Topological Representation 103

.Y

Figure 3

Then A,, A , EL, A = A , u A t , and A , E Bo, A , E B,, showing that L is distributive. It follows from (Sl) that the open sets of 8' are uniquely associated with ideals

of L: For an ideal I of L, let U ( I ) = U (a I a E I ) (keep in mind that an a EL is a subset of 9, as illustrated by Figure 3). Note that for a EL, a E I iff a E U ( I ) .

Now let P be a prime ideal of L , F = 9 - U(P), and let { U , I k C K } be the set of all compact open sets of 9 that have nonvoid intersections with F. Thus, the U, are exactly those elements of L that are not in P. Therefore, by the definition of a prime ideal, given k, 1 E K , there exists t E K with U, c U,, U , G U,, proving that F and {U, I k c K } satisfy the hypothesis of condition (S2). By (S2) we conclude that there exists a p E F n n (U, I k E K ) . If q E F , then for every compact open set U with q E U, we have U n F 4 0 ; thus p E U , proving that {p} = F. Note that 3' is T o ; therefore p is unique.

Conversely, if p E 9 , let I = { Q I a € L , a s Y'-{T}}. Then I is an ideal of L, end 9 -{13) = U ( I ) . We claim that I is prime. Indeed, if U , V < L, U t I , V t I , then U n{p }4=0 , V n { s + 0, and therefore p E U , p € V . Thus, pE U n V and so Un V 9 U ( 1 ) . By (Si) there exists a WE L with W U n 7 and W 9 U ( I ) . Therefore W Q I , and so I is prime.

Summing up, the map 9: P -+p is one-to-one and onto between b(L) and 8'. TO show that gi is a homeomorphism, it suffices to show that U is open i n 9 ( L ) iff Up, is open in$. Since a typical open set i n9 (L) is of the form r ( I ) ( IEI(L)) , and an open set of Y' is of the form U ( I ) , we need only prove that r(I)g, = U ( I ) and U(1)y- i = = r(1)-in other words, that P E r ( l ) iff (Pgi = ) p U ( I ) . Indeed, P Er(I) means that P p I, which is equivalent to U ( P ) U ( I ) ; this, in turn, is the same as

U ( I ) n (3'- U ( P ) ) + 0. Since 9 - U ( P ) =(23} with p = P y , the last condition means that U(Z) n e}+ 0, which holds iff p U ( I ) . (Indeed, if p t U ( I ) , then U ( I ) E U ( P ) , and so U ( I ) n {p} = 0.)

COROLLARY 9 (M. H. Stone [1937]). characterized by (Sl), (S2), and (53)

The Stone space of a distributive lattice is

The intersection of two compact open sets is compact.


COROLLARY 10 (M. H. Stone [1936]). The Stone space Y' of a Boolean lattice (called a Boolean space) can be characterized as a compact Hausdorff space in which the closed open (clopen) sets form a base for open sets. ( I n other words, 9 is totally disconnected.)

PROOF. Let Y'=Y'(B), where B is a Boolean lattice. Then Y ' = r ( f ) , and thus 3' is compact. Let P , Q @ and P+Q and take u E P - Q . Then &Er(u) and PEr(a ' ) ; therefore, every pair of elements of Y' can be separated by clopen sets, verifying that, 9 is Hausdorff. Obviously, 9 is totally disconnected. Conversely, let 9 be compact, Hausdorff, and totally disconnected. Then (Sl) is obvious. (52) follows from the observation that F and the Ui: i C I, are now closed sets having the finite intersection property; therefore, by compactness, they have an element in common. The compact open sets of 9 form a Boolean lattice B, and t h u s 9 is homeomorphic to 9 ( B ) - 0

As an interesting application we prove:

THEOREM 11. Let B be an infinite BooZean lattice. Then 13'(B)I 2 IBl.

PROOF. Let 9 be a totally disconnected compact Hausdorff space. For a , b E Y

U,b n u b , = 0. Now let U be clopen and a E U. Then 3' - U C u (Ub,, I b €3' - U), and so, by the compactness of 3 - U , S - U E U (Ub,, I b E B) for some finite B E 3' - U . Then V , = (u,,b I b E B) is open and a E V , E U. Thus, U E U ( Va I a E U), so by t,he compactnessof U, U~U(V,IaEA)forsomefinite A E U.Therefore, U = u ( V , l a E A ) . Thus every clopen set is a finite union of finite intersections of Ua,b , and 80 there are no more clopen sets than there are finite sequences of elements of 9; this cardinality is 191 provided that lY'l is infinite.

with U*b, fix a pair Of Clopen Sets ua,b and Ub,, such that Ua,b, b E Ub,,, and

@ It might be illuminating to compare this to an algebraic proof; see Exercise 36. Theorem 8 and its corollaries provide topological representations for distributive

join-semilattices, distributive lattices, and Boolean lattices, respectively. It is possible to give a topological representation for homomorphisms. We do it here only for (0, 1)-homomorphisms of bounded distributive lattices.

LEMMA 12. Let Lo and L, be bounded distributive lattices and let q be a (0, 1}- homomorphism of Lo into Li. Then

b(q) : P + Pq- 1

maps Y'(L,) into S(L0); 9 ( q ) is a continuous function with the property that i f U is compact open in Y'(Lo), then U(Y'(q))-' i s compact i n 3 ( L i ) . Conversely, if y:Y'(L,) - 9(L0) has these properties, then y =$'(q) for exactly one q : Lo + Li.

5. Topological Representation 105

PROOF. If U = r (a ) , a E Lo, then

UY'(q)-' = { P I P @(Ll), Pq-' E .(a)}

'{P I PEY'(L,), g p y - 1 ) '{P I PEY'"(L,), a y 4P) = r ( a y ) ,

and so S(v) is continuous, having the desired property. Conversely, if such a y is given and U = r ( a ) , aELo, then Uy-1 is compact open,

and so Uy-1 =r (b ) for a unique b E Li. The map y : a - b is a (0, 1)-homomorphism, andy=Y'(y).

The following interpretation of (52) will be useful. L e t d be a topological space. The Booleanization of Y' is a topological space f B on Y' that has the compact open sets of s" and their complements as a subbase for open sets.

LEMMA 13. a Boolean space.

A compuct topological space Y' satisfies (Sl), (S2), and (S3) i f f zB is

PROOF. Let 9 satisfy (Sl), (S2), and (53). Then Y B is obviously Hausdorff and totally disconnected. To verify the compactness of 2'B, let Z0 be a collection of compact open sets of Y', and let 31 be a collection of complements of compact open sets of 3' such that in 3 = 30 u 3t no finite intersection is void. Because of (S3), we can assume that Z0 is closed under finite intersection. Since members of are closed in 3' and Y' is compact, n (X I X C 3,) = F is nonvoid. Also, for any U E X0 and X G 3 , , U n X is closed inU,andthus U n F = r ) ( U n X ( XcJ1)=/=O. Applying (S2) to F and 3,,, we conclude that n X + 0, which, by Alexander's Theorem (see Exercise 14), proves compactness.

Conversely, if Y'B is Boolean, the compact open sets of Y ' B form a Boolean lattice L. We can easily verify that the compact open sets of Y' form a sublattice Li of L. Thus L, is a distributive lattice, and so, by Corollary 9, Y'=3'(Li) satisfies (Sl) , (s2), and (53).

We close this section with an interesting application. Let Li, i I, be pairwise disjoint distributive lattices. Then Q = U (Li I i E I ) is

a partial lattice. A free lattice generated by Q over the class D of all distributive lattices is called a free distributive product of the Li, i E I. To prove the existence of free distributive products, it suffices by Theorem 1.5.24 to show that there exists a distributive lattice L containing Q as a relative sublattice. This is easily done: Let L be the direct product of the Liu{O), i c I, where 0 is a new zero element of Li. Identify x E Li with f EL defined by f(i) = x, f ( j ) = O for j +i . Then Q becomes a relative sublattice of L.

An equivalent definition is:

DEFINITION 14. Let K be a class of lattices and let Li, i E I , be lattices in K. A lattice L i n K i s calted a free H-product of the Li, i c I , i f f every Li has an embedding E~ into L such that : (i) L is generated by U ( L i ~ i I i c I ) .

Figure 4

(ii) If K is any lattice in K and pli is a homomorphism of Li into K , for i < I , then there exists a homomorphism pl of L into K satisfying vi=eig) for all i E Z (see Figure 4).

For distributive lattices, this is equivalent to the first definition (see Exercise 55). I n most cases we will assunie that each Li is a sublattice of L and that ei is the inclusion map; then (ii) will simply state that the pli have a common extension. Note that in all the cases we shall consider, (i) can be replaced by the requirement that the in (ii) be unique.

If, in Definition 14, K is a class of bounded lattices and all homomorphisnis are assumed to be (0, 1)-homomorphisms, we get the concept of a free K (0, 1)-product. In particular if H =L we get the concept of free (0, 1)-product, see Section VI.3, and if K = D we obtain the concept of free (0, 1)-distributive product.

Our final result is the existence and description of a free (0, 1)-distributive product of a family of bounded distributive lattices.

THEOREM 15 (A. Nerode [1959]). Let L , iE I , be distributive lattices with 0 und 1 . Let 3' = II (a&) I i C I ) (see Exercise 15). Then Y' is a Xtolze space, and thw Y's 9 ( L ) for some distributive lattice L . Any such lattice L is a free (0, 1)-distributive product of the L , i C I .

The proof of Theorem 15 will be preceded by two lemmas. I n these two lemnias a Stone space is a topological space satisfying (Sl), (S2), and (S3).

LEMMA 16. Le tb i , i € I , be compact Stone spaces. Then

PROOF. For U EYi, let E ( U ) ={f I f E IIb0 f ( j ) < U) (see Exercise 15). The compact open sets form a base for open sets therefore, { E ( U ) I U compact open in some Yj) is a subbase for open sets in II(Xi I i C Z ) . Note that all the sets E ( U ) are compact open in IIbi; therefore, V G I-I$$ is compact open iff it is a finite union of finite intersections of some of the E( U). Consequently, declaring the complements of compact open sets to be open (in forming (II3JB) is equivalent to making the complements of the sets E( U ) open. But the complement of E( U ) is E (ai- U ) , and Yi - U is a typical open set of a?. Thus IIYf and (r18'i)B have the same topology.

5 . Topological Representation 107

LEMMA 17. The product of compact Stone spaces is again a compact Stone space.

PROOF. Let Yo i E I, be Stone spaces. T h e n 3 = IIYi is To and compact (Exercises 16 and 21). SinceY: is Boolean (Lemma 13), so is nYf (Exercises 20, 21, and 22). By Lemma 16, YE = nY;, and thus YB is Boolean. Therefore, by Lemma, 13, Y' is a Stone space.

PROOF OF THEOREM 15. Let e i be the i th projection (ei:Y(L) -Y'(Li), f e i = f ( i ) ) . By Lemma 12, there is a unique (0, 1)-homomorphism E ~ : Li - L satisfyingY(e,) = ei. It is easy to visualize ci; think of the elements of Li as compact open sets of Yi; then U E ~ = E( U ) = Ue;'. It is obvious from this that E $ is an embedding.

By applying Y to Figure 4, we obtain Figure 5. Thus the method of defining Y(v) is clear. For z&Y(K) , zY(y) is a member of n 9 ( L i ) , and zJ'(v)(i) =zY(pli) foriE1.

n

Figure 5

To show that this correspondence is indeed of the formY'(q) for some homomorphism 9: L +K, we have to verify that (a) Y(v) is continuous (this statement follows from Exercise 18), and that (b) if V is compact open in 3'(L), then V~'(P))-' is compact open in Y(K). It is enough to verify (b) for V = E ( U ) where U is compact open in some Y(Li ) . Then

Vf(p1)-1 =E(U)Y'(q)-I = Ue; 'Y(v ) - l = U(Y(tp)ei)-l = U Y ( y i ) - l ,

and therefore VY(v)-l is compact open since Y'(qi) satisfies the condition of Lemma 12.

Exercises

The followiiig exercises review the basic concepts and theorems of topology that are utilized in this section.

1. A topological space is a, set A and a collection T of subsets of A, with A E T, closed under finite intersections and arbitrary unions; a member of T is called an open set. Call a set closed iff its complement is open. Characterize the family of closed sets.


2.

3.

4.

5 . 6.

7.

8.

9.

10. 11.

12.

13.

*14.

1 Ti .

16. 17.

18.

19.

A family of nonvoid sets B s T is a base for open sets iff every open set is a union of members of B. Show that for a set A, B s P ( A ) is a base for open sets of some topological space defined on A iff u B =A, and for X , YE B and p EX n Y , there exists a 2 E B with p EZ,Z s X , and 2 Y . A family of nonvoid sets C E P ( A ) is a 8Ubba8e for open sets iff the finite intersections of members of C form a base for open sets. Show that C s P ( A ) is a subbase of some topology defined on A iff U C =A. Let A be a topological space and let X c A. Then there exists a smallest closed set X containing X ; X is the cloeure of X. Show that A, X c Y impliesthat Xc Y, X s X , Xu Y = X u Y, and z=X. Prove that the properties of X given in Exercise 4 characterize it. Show that aEX iff every open set (in a given base) containing a has a nonvoid intersection with X. A space A is a To-8pace iff (8 = {s implies that x = y . Show that A is a T,,-space iff, for every x , y € A , x +y, there exists an open set (in a given base) containing exactly one of x and y. A is a Ti-space iff {x} = { x } for all xEA. A Ti-space is a To-space. Show that A is a Ti-space iff, for x, y € A , x +y, there exists an open set (in a given base) containing x but not y . Let A and B be topological spaces and f : A -B. Then f is called COnth4OU8 iff for every open set U of B, f - i ( U ) is open in A . f is a homeomorphism iff f is one-to-one and onto and if both f and f - 1 are continuous. Show. that continuity can be checked by considering only those f -I( U) where U belongs to a given subbase. Show that f : A -. B is continuous iff f ( X ) cf(x) for all X c A , A subset X of a topological space A is compact iff X c u ( U i I Ui open, i €1) implies that X c u (Ui.1 i €11) for some finite Ii s I. The space A is compact iff X = A is compact. Show that A is compact iff, for every family F of closed sets, if n Fi + 0 for all finite Fi c F, then n F Let A be a compact topological space and let X be a closed set in A. Show that X is compact. Prove that a space A is compact iff, in the lattice of closed sets of A , every maximal dual ideal is principal. Show that a Ti-Rpace A is compact iff it has a subbase C of closed sets (that is, {A - X I XEG} is a subbase for open sets) with the property: If nD = 0 for some D E C, then n Di = 0 for some finite D, c D (J. W. Alexander [1939]). Let Ai, i €1, be topological spaces and set A =II(A, I i €1). For U S Ai, set E( U ) =

{f I f€A, f ( i ) E U } . The product topoEogy on A is the topology determined by taking all the sets E ( U ) as a subbase for open sets, where U ranges over all open sets of Ai for all ic1. Show that the projection map e i : f - f ( i ) is a continuous map. of A onto A,. Show that if the Ai are To-spaces (Ti-spaces), so is A =II(Ai I i €1). A map f : A -. B is open. iff f (U) is open in B for every open U c A. Show that the projection maps (see Exercise 16) are open. Prove that a function f : B-IIAi is continuous iff, for each iE1, eif: B - A i is continuous. A space A is a Hausdorff space ( T2-8pace) iff, for x, y E A with x +y, there exist open sets U, V such that xEU,yEV, Un V = B . Show that: (i) A is Hausdorff iff A = { ( x , x ) I % € A } is closed in A x A .

(ii) A compact subset of a T2-space is closed.

= 0 and that, for X, Y

-

0

0 .

5. Topological Representation 109

20. 21.

22.

23.

24.

25. 26.

27. 28.

29.

30.

31.

32. 33. 34.

36. 36.

37. 38.

39.

40. 41.

42.

43.

Prove that a product of Hausdorff spaces is a Hausdorff space. Show that a product of compact spaces is compact (Tihonov’s Theorem). (Hint: use Exrrcise 14.) A space A is totally disconnected iff, for x, y € A , 2 +: y , there exists a closed open set U with x E U , ?/Ct U . Show that the product of totally disconnected sets is totally disconnected.

Let I arid J be ideals of a join-semilatt,ice. Verify that * * *

IvJ = { t I t _< i v j , i €1, j E J } .

Show that for a join-semilattice L, I ( L ) is a lattice iff any two elements of L have a lower bound. Give a detailed proof of Lemma 2. Prove that every join-semilattice can be embedded in a Boolean lattice (considered a s a join-semilattice). Show that a finite distributive join-semilattice is a distributive lattice. Let L be a join-semilattice and let 0 be a join-congruence, that is, an equivalence relation on L having the Substitution Property for join. Then L/0 is also a join- semilattice. Show that the distributivity of L does not imply the distributivity of LI0. Let F be a free join-semilattice; let Fo be P with a new 0 added. Show t.hat F, is a distribut,ive join-semilattice. t

Let rp bc a join-homomorphism of the join-semilattice Po onto the join-semilattice PI. Show that,, for distributive join-semilattices Po, F l , the proper homomorphism concept is tho one requiring that if P is a prime ideal of Pi, then Prp-1 is a prime ideal of F,. Show that there is no “free distributive join-semilattice” with tho homomorphism concept of Exrrcise 30. Does Theorem 1.22 generalize to distributive join-semilattices? Characterizc the Stone spaces of finite Boolean lattices and of finite chains. Let 3‘0 and 8 1 be disjoint topological spaces; let b =Yo u b i and call U Cz open iff U n b o and U n b l are open. Show t,hat if$,andY‘l are Stone spaces,thenso isb. If, in Exercise 34, .4”i =b(LJ, i =0, 1, t h e n 3 =$(Lo XLJ . To prove Theorem 11, pick an element a ( P , &) E P -& for all P,& Ep(B) with P *&. Show that the a ( P , &) R-generate all of B. In Lemma 12, characterize $(q) for one-to-one and for onto Q.

Determine the connection between the Stone space of a lattice and the Stone space of a sublattice. Call the Stone space of a generalized Boolean lattice a generalized Boolean space; characterize it. (Compactnese of 3 should be replaced by local COmpUCtne88: For every p E b there exists an open set U with p E U and a set V with U I’, such that 17 is compact.) Show that the product of (generalized) Boolean spaces is (generalized) Boolean. Call the join-semilattice L modular iff, for a,b,c EL, a 5 b and b I U V ~ imply the existence of c , EL with cl c and b =avci. Show that a distributive join-semilattice is modular. Show that Lemma 1 remains valid if all occurrences of the word “distributive” are replaced by the word “modular.” Show that the set of all finitely generated normal subgroups of a group (and also the finitely generat,ed ideals of a ring) form a modular join-semilattice.

44. The lattice of congruence relations of a join-semilattice L is distributive iff any pair of elements with a lower bound are comparable (D. Papert [1964], R. A. Dean and R. H. Oehmke [1904]).

* * *

We have seen in this section how TheoremI.6.24 can be used to show the existence of free distributive products. The same method, however, does not apply to distributive lattices with 0 and 1. Nevertheless, the idea of the proofs of Theorems 1.6.6 and 1.6.24 can be used to get a much stronger result on the existence of free products; it is easiest to formulate this result (G. Griitzer [1968], Theorem 29.2) for universal algebras. To do so we have to introduce some concepts.

A type r of algebras is a sequence (no, ni, . . . , n,,, . . .) of non-negative integers, 3' < ~(t), where ~ ( r ) is an ordinal called the order of t . An algebra % of type r is an ordered pair (A; F), where A is a nonvoid set and F is a sequence {fo, . . . , f , , , . . .), y < o(T), wherefy is an n,,-ary operation on A. If ~ ( t ) is finite, ~ ( t ) =n, then we write (A; fo, . . . , f n - l ) for ( A ; P) .

45.

46.

47.

48. 49. 60.

61.

52.

63.

54.

66.

Define the concepts of subalgebra, polynomial, identity, and equat.iona1 class for algebras of a given type t. Show that if K is an equational class, % is an algebra in K, and 8 is a subalgebra of %, then b is in K. Define the concepts of homomorphism, homomorphic image, and direct product for algebras of a given type. Show that an equational class is closed under the formation of homomorphic images and direct products. Let =(A; F ) be an algebra, let ZZr A, and let H * 0 . Show that there exists a smallest subset [EI] of A, [HI 2 H such that ([HI; F) is a subalgebra of $1. (This subalgebra is said to be generated by H.) Show that I[H]] 5 IHI+ IPI+ No. Modify Definition 14 for algebras. Show that the 9 in (ii) is unique. Let 8 and [I: be free K-products of %i, iE1, wit,h the embeddings ei, iE1, and x i ,

i €1, respectively. Show t.hat there exists an isomorphism a : B -C such that q a = x i

and xia-l =ei, for all i €1. Let K bc an cquat.iona1 class of algebras and let %i€K, for iE1. Choose a set S satisfying

IS1 2 ZI4+ PI+ No.

Let Q be t,he set of all pairs (8, ( q ~ I i EI)) such that B c 8, pi is a homomorphism of %*into '$3, and B =[U (A,p, I i €I)]. Form % =lT( 113 ] (8, (qi I i E l ) ) EQ) (direct product), and, for aEAi, define faEA by fa((@, (qi I iE1))) = a y i . Finally, let % be the subalgebra generated by the fa, a € Ai, i €1. Show that % € K, a -fa is a homomorphism ~i of %i into 91, for i €1, and that % is generated by U (Aisi I i €1). Show that ei is one-to-one iff, for i €1, a , b EAi, a +: b, there exists an algebra Q E K and homomorphisms y j : %j -6 such that a y i * byi. Combine the previous exercises to prove the following result. EXISTENCE THEOREM FOR FREE PRODUCTS : Let H be an equational class of ulgebras, 2fiEIi for i €1. A free K-product of ai, i €1, exiets if', for i €1, a, b EAi, a + b , there exist a [I: E K, hornomorphisnzs y j : %j -[I: for j E I such that ayi +: byi. Show that in proving the existence of free distributive products and free (0, 1)- distributive products, w e can always choose Q = CS2, the two-element chain, in applying Exercise 63. Show that the two definitions of free H-product are equivalent for any class li of lattices.

6. Lattices with Pseudocomplementation 111

66.

87.

68.

Show that the free Boolean algebra on m generators is a free (0, 1)-distributive product of m copies of the free Boolean algebra on one generator. Prove that the frer Boolean algebra on m generators can be represented by the clopen subsets of (0, l } m where (0, l} is the two-element discrete topological space. Find a topological representation for the free distributive lattice on m generators (G. Ja. Arehkin [1953a]).

6. Distributive Lattices with Pseudocomplementation

In this section we shall deal exclusively with pseudocomplemented distributive lattices. There are two concepts that we should be able to distinguish: a lattice, (L; A, v), in which every element has a pseudocomplement; and an algebra (L; A, v, *, 0, l),where(L; A, v , 0 , l)isaboundedlatticeandwhere,foreveryaEL, the element a,* is the pseudocomplement of a. We shall call the former apseudocomple- mented lattice and the latter a lattice with pseudocomplementation (as an operation)-the same kind of distinction that we make between Boolean lattices and Boolean algebras. As defined in the exercises of Section 5 , a pseudocomplemented lattice is an algebra of type (2,2), whereas a lattice with pseudocomplementation is an algebra of type (2,2, 1 ,0 ,0) . To see the difference in viewpoint, consider the lattice of Figure 1. As a distributive lattice it has twenty-five sublattices and eight congruences; as a lattice with pseudocomplementation it has three subalgebras and five congruences.

Figure 1

Thus, for a lattice with pseudocomplenlentation L, a subalgebra Ll is a (0, 1}- sublattice of L closed under * (that is, u E Ll implies that a* E Li). A homomorphism

is a (0, 1)-homomorphism that also satisfies (mp)* =x*Q). Similarly, a congruence relalim 0 shall have the Substitution Property also for * : a = b (0) implies that u* = b* (0).

A wide class of exaiiiples is provided by

THEOREM 1. Any complete lattice thut sutisfies the Join Infinite Distributiue Identity (JID) i s (L pseudocomplemented distributive lattice.

PROOF. Let L be such a lattice. For a EL, set

a* = v(Z I XcL, a A X =o). Then, by (JID),

a m * = a ~ V ( z 1 UAX =0) = V(anz I UAX =0) = VO =O.

Furthermore, if U A X =0, then z<a* by the definition of a*; thus a* is indeed the pseudocomplement of a .

COROLLARY 2. Every distributive ulgebraic lattice is peudocomplemented.

PROOF. Let L be a distributive algebraic lattice. By Theorem 3.13 and Lemma 5.1 we can assume that L =I(&) , where S is a distributive join-semilattice with. 0. Let I, I j c I ( S ) , for j c J . Then I A I ~ G I A V ( I ~ 1 jEJ) for any k E J , and thus

V ( I A I ~ I i € J ) G I A V ( I ~ I j E J ) .

To prove the reverse inclusion, let a E I A V ( I ~ 1 j EJ), that is, a E I and a E V(Ii I j E J ) . The latter implies that a I t ,v* * Ovt,,, where ti E Ijl, . . . , t , E 15n, jl, . . . , j,, E J. Thus acI j l v - * ~ V I ~ ~ and so, using the distributivity of I (L) , we obtain

completing the proof of the (JID). The statement now follows from Theorem 1. Thus, the lattice of all congruence relations of an arbitrary lattice and the lattice

of all ideals of a distributive (semi-)lattice with zero are examples of pseudocomplemented distributive lattices. Note that for I E I ( K ) ,

I*={z I X E K , z A i = O for all i E I } .

Also, any finite distributive lattice is pseudocomplemented. Therefore, our investigations include all finite distributive lattices.

The first class of distributive lattices with pseudocomplementation, other than the class of Boolean algebras, to be examined in detail was the class of Stone algebras. A distributive lattice with pseudocomplementation L is called a Stone algebra iff it satisfies the Stone identity :

a*va** = 1.

The corresponding pseudocomplemented lattice is called a Stone lattice. To understand the meaning of this identity, define the skeleton of L:

S(L) ={a* I a EL}.

The elements of S(L) are called skeletal. By Theorem 1.6.4, (S(L); A, v, *, 0, 1) is a Boolean algebra. For a Stone algebra L, S(L) is a subalgebra of L:

6. Lattices with PseudoconiplemPiltatiori 113

LEMMA 3. conditions are equivalent :

(i) L i s u Stone algebru. (ii) ( m b ) * =a*vb* for a , b E L.

(iii) a, b ES(L) implies that avb c S ( L ) . (iv) S(L) i s a subalgebru of L.

For u distributive lattice with pseudocowiplententutiort L , the following

PROOF. The proofs that (ii) implies (iii), that (iii) implies (iv), and that (iv) implies (i) are trivial. Now let L be a Stone algebra; we show that a*vb* is the pseudocomplement of aAb. Indeed, (aAb)A(a*vb*) = (aAbAu*)v(aAbAb*) =OvO =O. If (aAb)Ax =0, then (bAx)Aa=O, and so b/\x(a*. Meeting both sides by a** yields bAxAa**< a*Aa** = 0 ; that is, xAa**Ab = O : implying that xr\a** b *. By the Stone identity, a*va**=l, and thus x = X A l =xA(a*va**) = (XAU*)V(XAU,**)<U*~~* .

This is already enough to yield the structure theorem for finite Stone algebras (G. Gratzer and E. T. Schmidt [1957a]):

COROLLARY 4. A finite distributive lattice i s a Stone lattice i f f i t i s the direct product of finite distributive denEe lattices, that is , finite distributive lattices with only one atom.

PROOF. By Leninia 3, a Stone lattice L has a complemented element ag{O, l} iff S(L)+{O, l } ; thus the decomposition of Theorem 1.6 can be repeated until each factor Li satisfies S(Li) =(O, l}. I n a direct product, * is formed componentwise; therefore, all the Li are Stone lattices. For a finite lattice K with S(K)={O, l}, the condition that K has exactly one atom is equivalent to K being a Stone lattice.

Another significant subset of a Stone algebra is the dense set:

D(L) ={a I a* = O } .

The elements of D(L) are called dense. We can easily check that D(L) is a dual ideal of L and that 1 E D(L) : thus D(L)

is a distributive lattice with 1. Since ava* I$ D(L) for every a E L, we can interpret the identity

a = a**A(ava*)

to mean that every a E L can be represented in the form a = bAc, where b ES(L), c E D(L). Such an interpretation correctly suggests that if we know S(L) and D(L) and the relationships between elements of S(L) and D(L), then we can describe L. The relationship is expressed by the homomorphism y ( L ) : S(L) -B(D(L)) defined

q ( L ) : a +{x I x E D(L), x 2 a*>. by

THEOREM 5 Let L be a Stone algebra. Then S(L) i s a Boo1ea.n a.lgebra, D(L) .is a distributive lattice with 1, and q(L) i s a (0, 1)- homomorphisnz. of S ( L ) into B ( D ( L ) ) . The triple (S(L) , D(L), q (L) ) characterizes L u p to isoniorphism.

(C. C. Chen and G. Grltzer [1969a]).

W Figure 2

PROOF. The first statement is easily verified. For a E S(L) , set Fa=(xI x**=a}.

The sets {F, laES(L)} form a partition of L; for a simple example, see Figure 2. Obviously, Fo={O} and Fi =D(L). The map z+xva* sends Fa into F, = D ( L ) ; in fact, the map is an isomorphism between Fa and aq(L) c D(L). Thus x E Fa is coinpletely determined by u and xva*Eaq(L)-that is, by a pair (a, z), where u E S ( L ) , z E aq(L)-and every such pair determines one and only one element of L. To complete our proof we have to show how the partial ordering on L can be determined by such pairs.

Let X E F , and yEF,. Then x<y implies that x**Iy** , that is, ash. Since x l y iff

and since the first of these two conditions is trivial, we obtain:

uvx 5 av y and xva* < yva*,

x S y iff a l b and xva*<yva*.

Identifying x with (xva*, a) and y with (yvb*, b), we see that the preceding conditione are stated in terms of the components of the ordered pairs, except that yva* will have to be expressed by the triple.

Because q(L) is a {O,l}-homomorphism and a is the complement of a*, we conclude that a y ( L ) and u*q(L) are complementary dual ideals of D(L). Thus, every z can be written in a unique fashion in the form z =ze,Azi, where zeaEaq(L) and ziEa*q(L). Observe that e, is expressed in terms of the triple. Finally, yva*= yvb*va*=(yvb*)e,. Thus for u € a q ( L ) and vEbq(L), we have (u ,u )<(v , b)iff a<b anduswe,.

6. Lattices with Pseuclocomplemeiitatioii 115

This theorem shows that for Stone algebras, the behavior of the skeleton and the dense set is decisive. This conclusion leads us to formulate the goal of research for Stone algebras and for all distributive lattices with pseudocomplementation as follows:

A problem for distributive lattices with pseudocomplementation is considered solved if it can be reduced to two problems: one for Boolean algebras and one for distributive lattices with 1.

By applying Zorn’s Lemma to prime dual ideals of a lattice with 0, we obtain that every prime dual ideal is contained in a maximal prime dual ideal, or, equivalently, we get that every prime ideal contains a minimal prime ideal P, that is, 8

prime ideal P such that Qc P for no prime ideal Q. Minimal prime ideals play an important role in the theory of distributive lattices with pseudocomplementation a8

illustrated by the following result :

THEOREM 6 (G. Griitzer and E. T. Schmidt [1957a]). Let L be a distributive lattice with pseudoco?npleinenta,tion. Then L is a Stone algebra i f f PvQ = L for any two distinct rninirrjal prime ideals P and Q.

PROOF. Let L be a Stone algebra and let P and Q be distinct minimal prime ideals. Choose ( I E P-Q (note that P C Q , since Q is minimal, and Q+P, hence P - Q + @ ) ; since ~ ~ ~ * = O c Q , a ( Q , a n d Q i s p r i m e , weobtain thata*cQ.

L - P is a maximal dual prinie ideal, hence by the dual of Corollary 1.18, it is a maximal dual ideal of L. Thus ( L - P)v[a) = L and so 0 = a A x for some xE L - P. Therefore, u* 2 xE L - P and so a*4 P.

Hence (I* Q - P. Similarly, a** E P -Q, and so 1 = a*va** E PvQ, yielding PvQ = L.

To prove the converse (for this proof see J. Varlet [1966]), let us assume that L is not a Stone algebra and let aEL such that a*va**+l. Let R be a prime ideal (see Corollary 1.17) such that a*va**E R.

R e claim that ( L - R)v[a*) += L. Indeed, if ( L - B)v[a*) = L, then there exists an x E L - R such that X A U * = ~ . Then a** > x c L - R, hence a**EL- R, aoontradiction.

Let F be a iliaxima1 dual prinie ideal containing ( L - R)v[a*) and, similarly, let G be a maximal dual prime ideal containing ( L - R)v[a**). We set P = L --P and

Then P and Q are minimal prime ideals. Moreover, P i & , because a*EF =L- P Q=L-G.

and hence u* 6 P ; thus a**€ P, while a**aQ. Finally, P , Q c R, hence PvQ+L.

We conclude this section by’ proving two representation theorems for Stone algebras which correspond to the two representation theorems for distributive lattices given in Section 1 . The proofs we present use the Subdirect Product Representation Theorem of G. Birkhoff [1944]. Direct proofs are possible but we shall present a proof that can be generalized to other equational classes of distributive lattices with pseudocomplementation.

I n the remainder of t,his section “algebra” means universal algebra. For the purposes of t,his book, the reader can substitute “lattice” or “lattice with pseudocomple- 9 Grrtzer

mentation” for “algebra”. Just as for posets and lattices, we write A for the algebra % = ( A ; F) if there is no danger of confusion.

DEEINITION 7. u, v E A such that u +v and u = v (0) for a12 congruences 0 > w.

An algebra A i s called subdirectly irreducible iff there exist elements

0

In other words, A has at least two elements and C(A) ={o> u [@(u, v)) . An equivalent form is:

COROUARY 8. f o r i c I ) implies that Oi=o for some i€ I .

The algebra A is subdirectly irreducible i f f w = A(@$ I i E I ) (Oi E C ( A )

EXAMPLE 9. A distributive lattice L is subdirectly irreducible iff ILI =2.

PROOF. If IL1=1, then L is not subdirectly irreducible by definition. If IL( =2, then obviously L is subdirectly irreducible. Let ILI > 2. Then there exist a, b, c EL, a < b <c. We claim that @(a, b)A@(b, c ) =o, which by Corollary 2 shows that L is not subdirectly irreducible. Let x y (@(a, b)A@(b, c ) ) . By Theorem 3.3 this implies that x v b = yvb and xAb = y ~ b ; thus x = y by Corollary 1.3.

EXAMPLE 10. B2 is the only subdirectly irreducible Boolean algebra.

PROOF. 2. If I BI > 2, then B has a direct product representation, B = Bi x B,, I B1l, I B,I 2 2 (use Exercise 1.5) ; thus B cannot be subdirectly irreducible.

Let B be Boolean. The statement is obvious for I BI

We shall need a simple universal algebraic lemma:

LEMMA 11 Let L be a n algebra and let 0 be a congruence relation of L. For any congruence Q, of L such that Q, 2 0, define the relation Q,/@ on L/O by

[ X I @ = [y]O (0) iff 2 = y (@).

Then cP/0 i s a congruence of LIB. Conversely, every congruence Y! of ,510 can be (uniquely) represented in the form Y =@I0 for some Congruence Q, 2 0. I n particular, the congruence lattice of L/O is isomorphic with the dual ideal [O) of the congruence lattice of L.

(The Second Isomorphism Theorem).

PROOF. and it has the Substitution Property. To represent Y, define Q, by

We have to prove that @/0 is well-defined, it is an equivalence relation,

x E y (@) iff [XI@ = [y]0 (Y).

6. Lattices with Pscudocomplemttritation 117

Again, we have to verify that Q, is a congruence. @/0 =Y follows froin the definition of 0. The details are trivial and left to the reader.

Equational classes of universal algebras can be introduced by defining polynomials and identities, just as in the case of lattices. However, in the next theorem the reader can avoid the use of this terminology by substituting for “equational class” the phrase “class closed under the formation of subalgebras, homomorphic images, and direct products”. (This does not make the result more general, see Theorem V.1.3.)

THEOREM 12 (The Subdirect Product Representation Theorem. G. Birkhoff [1944]). Let K be an equational class of algebras. Every algebra A in K can be embedded in a direct product of subdirectly irreducible algebras in K.

PROOF. For u , b c A , a +b, let 3E denote the set of all congruences 0 of A satisfying u + b (0). Since w E 2, % is not empty. Let C be a chain in X. Then 0 = U (0 I Q, C C) is a congruence, a + b (O), and thus every chain in % has an upper bound. By Zorn’s Lemma, Z has a maximal element Y ( a , b) . We claim that A / Y ( a , b) is subdirectly irreducible; in fact, u = [u ]Y(a, b ) and w = [ b ] Y ( a , b) satisfy the condition of Defi-. nition 7. Indeed, if 0 is any congruence of A / Y ( a , b) , @+w, then by Lemma 11, @ = Q , / Y ( u , b ) . Since@+w, weobtainQ,>Y(a, b),andsoa=b(Q,) .Thusu=v(@),as claimed.

Let B = fl ( A / Y ( u , b) I a , b E A , a +b) ; then B is a direct product of subdirectly irreducible algebras. We embed A into B by y : x +is, where f , takes on the value [ x ]Y(u , b) in the algebra A / Y ( a , b ) . Clearly, y is a homomorphism. To show that y is one-to-one, assume that fx=iv. Then x =y (“(a, b ) ) for all a , b e A , a+b . Therefore, x = y ( A ( Y ( a , b ) I u , b E A , a + b ) , a n d s o x = y .

We got a little bit more than claimed. If we pick x E A / Y ( a , b) , then x = [ y ] Y ( a , b) for some y E A. Thus there is an element in the representation of A whose component in A / Y (a, b) is x; such a representation is called subdirect.

COROLLARY 13. subdirect product of subdirectly irreducible ulgebras in K.

In un equationul class H, every ulgebra can be represented a8 a

Observe how strong Theorem 12 is. If combined with Example 9 it yields Theo-

The reader should note that subdirect representations of an algebra A are in one-to- rem 1.19; when combined with Example 10 we obtain Corollary 1.21.

one correspondence with families (Oi I i e l ) of congruence relations of A satisfying

A ( O i l i E I ) = w .

A subdirect representation by subdirectly irreducible algebras corresponds to such families (Oi I i E I ) which satisfy in addition the property that all Oi are completely meet-irreducible (an element 0 is completely meet-irreducible iff 0 = I j EJ) implies t,hat 0 = Qj for some j c J ) . Thus Lemma 11 and Theorem 12 combine to yield 0.

COROLLARY 14. meet-irreducible elements.

Every congruence relation of an algebra is a meet of completely

Let Gi denote the three-element chain (0, e, l} (0 < e < 1) as a distributive lattice with pseudocomplementation.

THEOREM 15. Stone algebras.

U p .to isomorpliism, %z and Gi are the only subdirectly irreducible

PROOF. '& and Gi are obviously clubdirectly irreducible (the congruence latt'ice of GI is a three-element chain).

Now let L be a subdirectly irreducible Stone algebra. By Lemnia 3, S(L) is a subalgebra of L. By definition, ILI > 1. If IX(L)I > 2, then S(L) is directly decomposable and therefore so is L. Thus IS(L)I =2, that is, S(L) = { O , l}. If ID(L)1 > 2, then there exist congruences 0 and CD on D(L) such that @A@==O on D ( L ) (by Example 9). Extend 0 and @ to L by defining (0) as the only additional congruence class. We conclude that L is subdirectly reducible.

Thus S(L) = { O , l} and so L = D ( L ) u (0) and ID(L)I < 2, yielding that L z B 2 or L E G i .

COROLLARY 16 (G.Gr&tzer [1969]). Every Stone algebra can be embedded in N

direct product of two- and three-element chains.

PROOF. Combine Corollary 13 and Theorem 15.

whether every Stone algebra can be embedded in some I ( P ( X ) ) ? Every distributive lattice can be embedded in some P(X). 0. Frink [1962] asked

THEOREM 17 (G. Griitzer [ 19631). A distributive lattice with pseudocomplemeii- tation L is a. Stone algebra i f f it can be embedded into ~ o m e I ( P ( X ) ) .

PROOF. I ( P ( X ) ) , and therefore any of its subalgebras, is a Stone algebra by Co- rollary 2.

It is obvious that the class of Stone algebras that can be embedded into some I ( P ( X ) ) is closed under the formation of direct products and subalgebras. Hence by Corollary 16 it is sufficient to prove that and GI can be so embedded. For %2

this is obvious; To embed GI, take an infinite set X and embed GI into I ( P ( X ) ) as follows :

0 -4 01,

1 -P(X).

e + { A I A c X and A is finite},

It is obvious that this is an embedding.

6. Lattices with Pseudocomplementation 119

Exercises

1. 2.

3.

4.

5 .

6.

7 . 8. 9.

10.

11.

12.

13.

"14.

16. 16.

l i .

Show that every bounded chain is a pseudocomplemented distributive lattice. Let L be a lattice with 1. Adjoin a new zero 0 to L: Li = L u {0} , 0 < z for all 1: EL. Show that L, is a pseudocomplemented lattice. Call a lattice with 0 deme iff 0 is meet-irreducible. Show that every bounded dense lattice K is pseudocomplumentod and that every such lattict: can be constructed by the method of Exercise 2 with L = D ( K ) . Find an examplu of a complete distributive lattice L that is not pseudocomplemented. Prove that if L is a complete Stone lattice, then so is f ( L ) . (Hint: I* =(a], where

Show that, a dist)ributivc pswdocomplemented lattice is a Stone lattice iff (avb)** =

a**vb** for a, b EL. Find a small set of identities characterizing Stone algebras. Let L be a Stone algebra. Show that S(L) is a retract of L. Let L be a St,one algebra, a, b € S ( L ) , and a s b . Prove that x - + ( z v a * ) ~ b embeds P , into Fb. Let B be a Boolean algebra. Define BLsl B2 by (a, b) E BIZ] iff a 5 b. Verify that B"] is a sublattice of B* but not a subalgebra of BZ. Show that BIZ] is a Stone lattice. Let, L be a. pseudocomplemunted distributive lattice. Show that, for a, b EL,

=A(%* 1 x E f ) . )

and (avb)* =a*Ab*

(a A b ) ** = a* * A b**.

Prove that a prime ideal P of a Stone algebra L is minimal iff P as an ideal of L i8 generated by P n S ( L ) . Show that a distributive lattice with pseudocomplementation is a Stone algebra iff every prime ideal contains exactly one minimal prime ideal {G. Griitzer and E. T. Schmidt, [1957a]). Prove that a poset & is isomorphic to the poset of all prime ideals of a Stone algebra, iff (i) every element of Q contains exactly one minimal element, (ii) for every minimal element 111 of &, the poset {1: 1 1: > m, x EQ) is isomorphic to the poset of all prime ideals of some distributive lattice with 1 (C. C. Chen and G. Griitzer [ 1969bl). Give a detailed proof of the Second Isomorphism Theorem. Prove Corollary 1 G directly.

Exercises 17-29 are from C. C. Chen and G. Gratzer [1969a]. Exercises 30 and 31 are from C. C. Chen and G. Gratxer [1969b]. Let B be a Boolean algebra, let D be a distributive lattice with 1, and let q be a (0, 1)-homomorphism of B into a ( D ) . Set L = {(z, a) I a EB, x Eaq} and define (2, a) 2 (y, b) iff a 5 b and 1: 5 ye,, where [yea) = U ~ A [ Y ) .

Verify the following formulas : (i) If aEB and d c D , then de,=d iff dEup.

(ii) dpa 2 d for a E B, d ED. (iii) de,Ade,, =d for aEB, d ED (where a' is the complement of a in B). ( i V ) @ , e b = e a h b for a, bEC.

120

18.

19. 20.

*21. 22.

23. 24. 26. 26.

27. 28. 29.

30. *31. *32.

33.

11. Distribubive Lattices

Prove that: (i) d@,Ad@b =deavb for a, b E B, d ED.

(ii) de, hb =de,vdeb for a, b E B, d ED. Show that L is a poset under the given partial ordering. For (x, a), ( y , b ) EL, verify that (x, a)A(y, b ) = (XQbhyQ,, ahb). Show that (5, a;)v(y, b) = ( ( X e v h y ) V ( X h Y @ , t ) , avb). 1

For (2, a), ( y , b), ( z , @EL, let

and

Compute A ; show that B =(d, (avo)A(bvc)), where d =dovdlvdZvds, and do =

X @ b , w A Y e a A c r A z , d i = x @ b h c ' A Y e a h c A Z ~ d 2 = X e b ~ c A Y @ a h c ' A Z , and d3=x@bv c A Y @ b v eAZ@a'vb'*

Show that do 2 d, and do 2 d , ; therefore, d =d,vde. Show that L is distributive. Show that L is a Stone lattice. Identify bEB with (1, b) and d ED with (d , 1). Verify that S ( L ) = B, D(L) = D , and q(L) =q. In other words, we have proved the Construction Theorem of C. C. Chen and G. Grateer [1969a]: Given a Boolean algebra B , a distributive lattice D with 1, and a (0, 1)-homomorphism p: B +.0(D), there e d a t a a Stone algebra L whose triple i a

Describe isomorphisms and homomorphisms of Stone algebras in tcrms of triples. Describe subalgebres of Stone algebras in terms of triples. For a given Boolean algebra B with more than one element and distributive lattice D with 1, construct a Stone algebra with S(L) z B and D(L) z D . (S (L ) and D ( L ) are independent.) Show that a Stone slgebra L is complete if S(L) and D(L) are complete. Characterize the completeness of Stone algebras in terms of triples. Show that a distributive lattice with pseudocomplementation L has the Congruence Extension Property: for a subalgebra K of L and for a congruence relation 0 of K, there exists a congruence relation CP of L extending 0, that is, for a, b EK, a = b (0) iff a = b (a) (G. Grlitzer and H. Lakser [1971]). Let L be a lattice or a lattice with additional operations. I f a, b EL and [b, a] is simple, then the "(a, b) (defined in the proof of Theorem 12) is unique.

(B , D, q>.


A number of papers are devoted to the subject of guaranteeing the distributivity of a lattice by imposing equations on a fixed generating set. For instance, if L is generated by a, b, and c, then L is distributive iff XA(YVZ) = ( ~ A Y ) V ( Z A Z ) and its dual holds for any permutation x, y, x of a , b, c and (aAb)V(bAC)V(CAU) = ( a v b ) ~ (~VC)A(CVU), by a result of 0. Ore [1940]. These seven conditions (there are only seven by commutativity) are independent but if L is assumed to be modular, all seven are equivalent, as it is evident from Figure 1.6.7. Generalizations can be found in M. Kolibiar [1972] and S. Tamura [1971], [1971 a], and for the modular case in B. Jbns- son [1955], R. Musti and E. Buttafuoco [1956], and R. Balbes [1969].

Purt#her Topics and References 121

For some special classes of lattices, Theorem 1.1 and 1.2 have various stronger forms that claim the existence of very large or very small pentagons and diamonds. For instance, a bounded relatively complemented nonmodular lattice always contains a pentagon as a (0, 1)-sublattice. The same is true of the diamond in certain complemented modular lattices; such results are implicit in J. von Neumann [1936/37]. If the lattice is finite modular nondistributive, then it contains a diamond in which a, b, c cover o and i covers a, b, c. If L is finite and nonmodular, then the pentagon it contains can be required to satisfy a t b.

It seems hard to generalize the uniqueness of an irredundant join-representation of an element of a finite distributive lattice. The best generalization is that of K. P. Dilworth and P. Crawley [1960] to distributive algebraic lattices in which every proper interval has an atom. See the survey article by R. P. Dilworth [1961] and S. Kinugawa and J. Hashimoto [1966]. Some results on and references to the modular and semimodular cases can be found in Chapter IV.

By Theorem 1.19, every distributive lattice is a subdirect product of copies of CC2. Distributive lattices that arc subdirect products of copies of Q, are charrtcterized in F. W. Anderson and It. L. Blair [1961].

For a finite distributive lattice L, what is the smallest k such that L is a subdirect product of k chains? For a L, let n, be the number of elements of L covering a. Then k = max (n, I a L} . This is an easy application of a well-known combinatorial result of R. P. Dilworth [1950] (first discovered by T. Gallai in 1936) according to which a poset P is a union of at most k chains iff P is of width I; k. See also H. Tverberg [ 19671.

The size of maximal sublattices of a finite distributive lattice is investigated in I. Rival [1973] and [1974]; maximal (0, 1)-sublattices of finite Boolean lattices are considered in H. Sharp [1968] and D. Steven [1968]. (0, 1)-sublattices of finite Boolean lattices and finite topologies are closely related. See also R. P. Stanley [1973].

Theorems 1.9 and 1.19 can be generalized by using ideals other than prime ideals In this connection see Theorem IV.5.17 and G. Birkhoff and 0. Frink [1948].

The problem of determining IP,,(n)l goes back to R. Dedekind [1900]. For some older and some recent contributions (including the tabulation of IFD(%)] for n 1 7 ) see R. Church [1940], E. N. Gilbert [1954], G. Hansel [1967], D. Kleitman [1969], R. Church [1966] (this gives the correct value for lFD(Y)l), and F. Lunnon [1971].

Infinitary Boolean polynomials are considered in H. Gaifman [1964] and A. w. Hales [1964]; they prove that free complete Boolean algebras do not exist.

Free distributive lattices have many interesting properties. All chains ere finite or countable (the proof of this is similar to that of Theorem VI. 2.10). If a and H are such that XAY =a for all z, y E H , x+y, call H a-disjoint. I n a free distributive lattice ell a-disjoint sets are finite, see R. Balbes [1967].

All chains of a free Boolean algebra are also finite or countable, see I. Reznikoff [ 19631 and A. Horn [ 19681.

A more general form of Theorem 2.5 can be found in R. Sikorski [1964]; for the universal algebraic background see Theorem 12.2 in G. Griitzer [1968].

Some properties of @[I] can be generalized to certain ideals of a general lattice, see Chapter 111.

Let L be a lattice and let f be an n-ary function on L, that is, f : Ln +L. We say that

f has the Congruence Substitution Property iff, for every congruence relation 0 of L and a,, biC L, 1 5 is n,ai = b,(O), 1 5 is n, imply that /(ail . . . , a,) =f (b , , . . . , b,) (0). On a Boolean algebra B, a function has the Congruence Substitution Property iff it is a Boolean algebraic function, that is, a Boolean polynomial in which elements of B are substituted for some variables. Functions satisfying the Congruence Substitution Property on a bounded distributive lattice were described in G. Griitzer [1964].

The method given in Theorem 3.9 is not the only one used to introduce ring operations in a generalized Boolean lattice. G. Grlitzer and E. T. Schmidt [1958d] prove that ring operations + , - can be introduced on a distributive lattice L such that + and * satisfy the Congruence Substitution Property iff L is relatively complemented. Furthermore, + and are uniquely determined by the zero of the ring, which can be an arbitrary element of L.

Whether every distributive algebraic lattice is isomorphic to the congruence lattice of some lattice is one of the longest-standing problems of lattice theory. The method used in Section 3 for the finite case can be easily extended to infinite algebraic lattices in which every element is a finite join of join-irreducible elements. A further extension of this result is in E. T. Schmidt [1962] and [1968]. (In reading the two papers, the reader should disregard the Theorem and Lemmas 9 and 10 of the first paper.) See also E. T. Schmidt [1969], [1974], [1975] and J. Berman [1972].

Congruence lattices of distributive lattices are not considered in this text became their characterization problem is trivial by Lemma 4.5: A lattice is isomorphic to the congruence lattice of a distributive lattice iff it is isomorphic to I ( B ) , where B is a generalized Boolean lattice; I ( B ) , by Theorem 3.13, is characterized as a distributive algebraic lattice in which the compact elements form a relatively complemented sublattice.

Algebraic lattices originated in A. Koniatu [1943], G. Birkhoff and 0. Frink [1948], L. Nachbin [1949], and J. R. Buchi [1952]. The original definition in G. Birk- hoff and 0. Frink [1948] is as follows: (i) L is complete; (ii) in L every element is the join of join-inaccessible elements; (iii) L is join-continuous. An element a of L is join-accessible iff there is a nonvoid subset H of L that is directed (for x, y C H there exists an upper bound z E H), VH =a, and a 6 H; otherwise, a is join-inaccessible. L is join-continuous iff for any a C L and directed H G L, we have UAVH = V(ar\h I h E H). The significance of algebraic lattices for universal algebras is discussed in the Concluding Remarks.

Interestingly, it is sufficient to formulate conditions (i) and (iii) of the previous paragraph for chains only. In other words, a lattice L is complete iff /\C and V C exist for any chain C of L ; and a (complete) lattice L is join-continuous iff UA'\/C = ~ ( ~ A C I C E C ) for any chain C of L. These statements are immediate consequences of the following result of T. Iwamura [1944]. Let H be an infinite directed set. Then H has a decomposition H = U (H, I y <a), where each H, is directed; for y < 6 <: a we have H, E H,, and for y < a we have IH,, I < H. Extensions of lattices preserving (iii) are considered in D. A. Kappos and F. Papangelou [1966].

The results of Section 4 were presented in FC without the assumption that L has a 0. The necessity of this assumption is pointed out in R. D. Byrd and R. A. Mena [1976] and R. D. Byrd, R. A. Mena, and L. A. Troy [1976]. These papers give a detailed analysis of what happens if this assumption is dropped.

W. Hanf [1976] proves that there is no algorithmic way to find a generating chain in all countable Boolean algebras. There are, however, always generating chains of a rather special order type; see R. S. Pierce [1973].

The problem of completions of a lattice has been extensively studied. The standard method is the MacNeille completion (see Exercises 4.9-4.14, which are from H. M. MacNeille [ 19371). This method is described in detail in G. Bruns [ 19621 ; see also Y. Sanipei [1953]. One of the important shorteoinings of this method is that it does not preserve identities, not even distributivity; see M. Cotlar [1944] and N. Funayama [1944]. See also P. Crawley [1062], R. P. Dilworth and J. E. McLaughlin [1952], and W. H. Cornish [1974]. What is preserved for complemented modular lattices is examined in J. E. McLaughlin [1961].

If we define a completion e of a lattice L as any complete lattice e containing L as a sublattice such that all infinite meets and joins that exist in L are preserved in z, we can ask whether there is aky distributive completion of a distributive lattice. The answer, in general, is in the negative; see P. Crawley [1962].

For Boolean algebras, set representations preserving all joins and meets, or joins and meets of certain types, play an important role, especially in relationship with infinite distributive identities. The larger part of the book by I%. Sikorski [1964] is devoted to this problem. For distributive lattices these questions become even inore coniplicated-see, for instance, G. Bruns [1959], [1961], and [1962a]; C. C. Chang and A. Horn [1962]; R. P. Dilworth and J. E. McLaughlin [1952]; A. Horn [1962]; J. Jakubik [1972]; and G. N. Raney [1952], [1953]; see also R. Balbes and Ph. Dwinger [1974].

Cardinalities of maximal chains in Boolean algebras are considered by J. Jakubik [1957], [1958] and G. W. Day [1970]. See also Exercise 4.28.

R. S. Pierce [1972] gave a deep analysis of the structure of countable Boolean algebras. A generalization of Boolean algebras generated by chains can be found in R. W. Quackenbush and H. C. Reichel [1975].

The method used in establishing Lemma 4.22 was employed by B. Jbnsson [1956] to construct “universal” lattices, that is, lattices containing a large class of lattices as sublattices (see also B. Jbnsson [1960] and M. D. Morley and R. L. Vaught [1962]).

For a distributive lattice L, S ( L ) is a poset and, by Section 5, a topological space. These two approaches are married in H. A. Priestley [1972] where S ( L ) is considered, and characterized, as an ordered topological space. This approach proved very convenient in applications; see H. A. Priestley [1972], [1974], and [1975]; see also M. E. Adanis [1974], [1975], [1976], B. A. Davey [1973], [1974], T. P. Speed [1972a]. T. P. Speed [1972] characterizes S ( L ) for bounded L by the property tha t ‘ 9 ( L ) u (0, 1) is a profinite poset. See also M. E. Adams [1975], R. Balbes [1971], B. A. Davey [1973], T. P. Speed [1972a] and [1974].

A solution to the word problem of free (0, 1)-distributive products is given in G . Griitzer and H. Lakser [1969]. In the same paper it is proved that if m is regular cardinal and L,, i E I, are bounded distributive lattices satisfying IC1 <m for any chain C in any L,, then the same holds in the free (0, 1)-distributive product. I n particular, if all Li satisfy the Countable Chain Condition, so does the free (0, 1}- distributive product; see also Sectioii 12 of FC. M. E. Adams and D. Kelly [a] has just extended this result to free products of lattices. See also B. Jbnsson [1971].

The analogous problem for a-disjoint sets is considered in H. Lakser [1973a] and M. E. Adams and D. Kelly [b]. An interesting description of free (0, 1)-distributive products can be found in R. W. Quackenbush [1972a].

A weaker form of Corollary 6.16 can be found in T. P. Speed [1969]. Subdirect products of lattices have an interesting property due to G. M. Bergman

(see K. A. Baker and A. F. Pixley [a]): a subdirect product L of L,, . . . , L, is determined up to isomorphism by its projections onto Li x L , 1 I i < j h ; n. Subdirect products of ideal lattices are investigated in R. Freese [1975].

The interdependence of the various “finiteness conditions” for distributive lattices is investigated in S. P. Avann [1964] and S. Rudeanu [1964].

Boolean algebras with only the trivial automorphism are constructed in M. Kritetov [l951], B. J6nsson [1951], and L. Rieger [1951]; Katetov’s example has 2x0 elements, the others are very large.

Some categorical concepts have been studed for distributive lattices. For a class K of algebras, PEH is called projectiwe in K iff, for any A, BCK, homomorphisms y : A - B and p : P -c B, if y is onto,then there is a homomorphism a : P -A satisfying ay =p. Free algebras are projective.

R. Balbes [1967] proves that a finite distributive lattice L is projective in D iff the join of any two meet-irreducible elements is meet-irreducible. See also R. Balbes and A. Horn [1970b], G. Griitzer and B. Wolk [1970]. R. Engelking [1965] has shown that a subalgebra of a free Boolean algebra need not be projective; but every countable Boolean algebra is projective, see P. R. Halmos [1961].

I is injective in K iff, for any A, B K , embedding y : B -t A and homomorphisin p : B--I, there is a homomorphism a: A - I satisfying ya =p. Injective Boolean algebras are the complete ones, see P. R. Halmos [1961]; and injective distributive lattices are the complete Boolean lattices. For more details on categorical concepts for distributive lattices, see Section 13 of FC.

R. Dedekind found the distributive identity by investigating ideals of number fields. Rings with a distributive lattice of ideals have been investigated by E. Noether [1927], L. Fuchs [1949] (who names such rings arithmetical rings), I. S. Cohen [1950], and Chr. U. Jensen [1963]. Equational classes of rings with distributive ideal lattices were considered in G. Michler and R. Wille [1970] and in H. Werner and R. Wille [1970]. E. A. Behrens [1960] and [1961] considered rings in which one-sided ideals form a distributive lattice. Rings with a distributive lattice of subrings were classified in P. A. Freidman [1967]. In this context G. M. Bergman’s work on the distributive- divisor-lattice of free algebras should be mentioned, see Chapter 4 of P. M. Cohn [1971].

For groups the corresponding problem is much simpler: The subgroup lattice of a group G is distributive iff G is locally cyclic (see 0. Ore [1937] and [1938]). H. L. Silcock [1977] has proved that every finite distributive lattice is isomorphic to the lattice of normal subgroups of a suitable group.

The distributivity of congruence lattices of lattices has a number of important consequences. B. J6nsson [1967] discovered that many of these results hold for arbitrary universal algebras with distributive congruence lattices. His results have found applications that go far beyond lattice theory-they have already been applied to 1 attice-ordered algebras, closure algebras, nonassociative lattices, cylindric algebras,


monadic algebras, lattices with pseudoconiplementation, primal algebras, and multi- valued logics.

The foregoing examples show the central role played by distributive lattices in applications of the lattice concept.

Except for V. Glivenko's early work [1929], the study of pseudocomplemented distributive lattices started only in 1956 with a solution of Problem 70 of G. Birkhoff [1948] giving a characterization of Stone lattices by minimal prime ideals (a. Gratzer and E. T. Schmidt [1957a]; for a simplified proof see J. Varlet [1966]), see Theo- rem 6.6. The idea of a triple was conceived by the author as a tool to prove Frink's conjecture (see 0. Prink [1962] and Theorem 637) . This attempt failed and as a result triples were not utilized until C. C. Chen and G. Griitzer [1969a] and [1969b]. Frink's conjecture was solved using the Compactness Theorem in G. Gratzer [1963] (see Theorem 6.17) and the proof was simplified in G. Bruns [1965] and G. Gratzer [1969]. An interesting generalization can be found in H. Lakser [1971]. Projective and injective Stone algebras are investigated in R. Balbes and G. Griitzer [1971]; see also R. Beazer [1975]. See also Section8 16 and 17 of PC.

Let B, denote the equational class of distributive lattices with pseudocomplementation satisfying the identity

(L,) ( z ~ A - - *AZ,)*V(X:A. - -AX,)*V* * * V ( Z ~ A - * . A Z ~ ) * = 1

for n >_ 1. Then B, is the class of Stone algebras. K. B. Lee [1970] has proved that B,, - 1 5 n w , is a complete list of equational classes of distributive lattices with pseudocomplementation, where B- i is the trivial class, Bo is the class of Boolean algebras, and B, is the class of all distributive lattices with pseudocomplementation. Moreover,

B-, c B , c B l C' . . cB, C * * 'CB,.

In H. Lakser [1971] and G. Gratzer and H. Lakser [1971] and [1972] most of the structure theorems known for Stone algebras have been generalized to the classes B,. I n these papers the anialgamation class of B, (in the sense of Section V.4) is also considered. See also Chapter 3 of FC where many of these results are presented in detail.

I n more than 30 papers T. Katriiitik investigates various generalizations and specializations of distributive lattices with pseudocomplementation and establishes triple constructions for inany classes. These classes include relative Stone algebras, generalized Stone algebras, pseudocomplemented posets, generalized Stone join- semilattices, distributiie pseudocomplemented meet-semilattices, modular pseudocomplemented lattices, modular pseudocomplemented lattices satisfying Z*VX** = 1, double Stone algebras, and so on. Free algebras, equational classes, injective hulls and many other questions are considered. The reader is referred to the papers of T. Katriiltik listed in the Bibliography and to R. Balbes and A. Horn [1970], R. Balbes and Ph. Dwinger [1974], G. Bruns and G. Kalmbach [1971] and [1972], A. Day [1970a], H. Lakser [1970] and [1973], J. Varlet [1963], [1966], and [1968], P. V. It. Murty and V. V. R. Rao [1974], U. M. Swamy and V. V. R. Rao [1975].

Relatively pseudoconiplemented distributive lattices (often called Heyting algebras) arise from nonclassical logic and were first investigated by T. Skolem about 1920.


For a detailed development, see H. B. Curry [1963], which contains all the important rules of computation with u * b. Injective8 and projective8 were investigated in the work of R. Balbes and A. Horn [1970a] and A. Day [1970a]; see also A. Horn [1969] and [1969a], W. C. Nemitz [1965].

In connection with nonclassical logic, many algebras emerged that are distributive lattices endowed with some additional structure. The best known of these, the Post8 algebras, happen to be Stone lattices. In fact, a Post algebra is a free (0, 1)-distributive product of a finite chain and a Boolean algebra, in which the elements of the chain are regarded as nullary operations; see E. L. Post [1921], P, C. Rosenblooni [1942], G. Epstein [1960], T. Traczyk [1963] and [1967], C. C. Chang and A. Horn [1961], G. Rousseau [1970], It. Balbes and Ph. Dwinger [1971] and [1971 a], and M. Mandelker [1970].

For some other types of distributive lattices with additional operations, see N. D. Belnap and J. H. Spencer [1966], J. M. Dunn and N. D. Belnap [196S], H. Rasiowa and R. Sikorski [1963], and J. Varlet [1968].

'Two books have appeared recently dealing with distributive lattices with additional operations. R. Balbes and Ph. Dwinger [1974] deals with them from a lattice theoretical viewpoint while H. Itasiowa [ 19741 investigates their usefulness to logic.

Some of the results for pseudoconiplemented distributive lattices referred to above have been extended to ideals of general distributive lattices; see R. Beazer [1975a], W. H. Cornish 119721, [1974a], B. A. Davey [1974a], M. Mandelker [1970].

Some results do not require the distributivity of the whole lattice; for instance, to prove that I ( L ) is pseudocomplemented it is sufficient that L be 0-distributive, that is, a r \ b = a ~ c = O implies that aA(bvc)=O, for u, b, GEL. This &nd some related concepts have been in the literature for 20-30 years. Here are some recent references: W. H. Cornish [1974b], M. F. Janowitz [1975], J. Varlet [1968a].

Problems

11.1 (FC 28). 11.2 (FC 30).

11.3 (FC 31).

Find short one-identity axioms characterizing Boolean algebras. Characterize the autoniorphisni groups of Boolean algebras. (B. Jtjnsson has proved (unpublished result) that the finite automorphisni groups of Boolean algebras are exactly the finite symmetric groups. See also R. N. McKenzie and J. D. Monk [1973].) Let B and C be infinite Boolean algebras and let m be a cardinal number satisfying m 2 I BI and ICI. Does there exkt a Boolean adgebra D containing B as a subalgebra and such that the autonio1:phisiii groups of C and D are isomorphic? (This is a variant of F C 31. Call a Boolean algebra rigid iff it has a one-element automorphism group. F. W. Lozicr [1969] constructs a rigid Boolean algebra of cardinality 2M for every infinite cardinal m. V. Kannan and M. Rajagopalan [1971] announce that there exist rigid Boolean algebras of any infinite cardinality. In September, 1975, J. D. Monk communicated to me the following partial solution using some unpublished results of R. N.

I’rohlcms 127

11.4 (FC 33).

11.5 (FC 34).

11.6 (FC 35).

11.7 (FC 36).

11.8 (FC 37).

11.9 (FC 38). 11.10 (FC 40).

11.11 (FC 41).

11.12 (FC 42).

11.13 (FC 43).

11.14 (FC 44). 11.15 (FC 45).

11.16 (FC 46). 11-17.

11.18.‘

II.1Y.t 11.20.

~~ _ _

McKenzie and S. Shelah, based on R. N. McKenzie and J. D. Monk [1973]: (GCH) thcre is a cardinal number p =p(B , C ) , such that there is a Boolean algebra D as required for all m 24.) Investigate further the poset P (L) of prime ideals of a distributive lattice. Characterize P(L) under the additional a&sumption that L has a unit and/or a zero. (If L has a unit, then every chain has a supremuin. T. Tan pointed out that the converse does not hold& For which classes K of finite lattices can we claim that if L EK and L is nonmodular, then L contains a “minimal” pentagon? Characterize the congruence lattices of lattices. (Conjecture : distributive algebraic lattices). Characterize t hc congruence lattices of sectionally complemented lattices. (Conjecture : distributive algebraic lattices). Determine f,(k, 72).

Investigate the cardinalities of inaximal chains in complete Boolean algebras. What can be proved about the cardinalities of maximal chains in a complete and at oniic Boolean algebra without the Generalized Continuum Hypothesis? For instance, do they forin a convex set? In a free (0, 1)-distributive product there are two normal forms for an element : normal V-representation and normal A-representation. (See G. Gratzer and H. Lakser [1969] and also Section 12 of FC.) Compare the number and form of the terms in a normal V-representation and normal /\-representation of an element in a free (0, 1}- distributive product. For a bounded distributive lattice L, let F(L) and F,(L) denote the lattice of all functions and all n-ary functions on L, respectively, with the Congruence Substitution Property. To what extent do F(L) and F,(L) determine the structure of L? Characterize F(L) and F,(L). Describe the functions with the Congruence Substitution Property on unbounded distributive lattices. Study problems 11.13 and 11.14 in the unbounded case. For a natural nuinher n, let g ( n ) denote the smallest integer such that for every distributive lattice D of length n there is 6 lattice L of length a t most g(n) satisfying C ( L ) 2% - 1 by Exercise 3.35. S e e also J. Berman [1972] and E. T. Schmidt [1975].) Let K be a lattice and let G be a group. Does there exist a lattice L satisfying C(L) C ( K ) and Aut(L) rG? (Aut(L) is the automor- phisiii group of L.) In 11.18, if K is finite, can L be chosen to be finite? Is there a fixed type of algebras such that every algebraic lattice is

D . Determine g(n). (g(n)

1 These two problems havr been solved by A. Urquhart, A topological rapresciittltioil t1woi-y for lattices. Algebra Universalis 8 (1978).


11.2 1.

11.22.1

11.23.

11.24.

11.25 (FC 53). 11.26 (FC 58). 11.27 (FC 59). 11.28 (FC 60). 11.29 (FC 61).

11.30 (FC 62).

11.31 (FC 64).

11.32 (FC 66).

11.33 (FC 67).

11.34.

isomorphic to the congruence lattice of an algebra of that type? Can this type be (2)? (See Concluding Remarks.) Is the statenient “There exist two bounded distributive lattices in which every a-disjoint set is countable but there exists an uncountable n-disjoint set in their free (0, 1)-distributive product” equivalent to the Souslin Hypothesis? (See M. E. Adams and D. Kelly [b].) Prove that for every infinite distributive lattice A there exist distributive lattices B and C such that the free distributive products of A, B and A, C are elementarily equivalent but B and C are not elementarily equivalent. (Two lattices are elementarily equivalent iff a first-order sentence holding for one, also holds for the other. For A finite such B and C do not exist, see B. Jbnsson and Ph. Olin [a].) Prove that a distributive lattice L is equationally compact iff L is complete and satisfies (JID) and (MID). Let K be an equational class of lattices such that the word problem of FK(KO) can be solved. Give a solution to the word problem of (bounded) free K-products. Describe the projective Stone algebras. bharacterize free B,-products (n > 1). Describe Amal(B,) for n > 2. Determine the projective5 in B,, n > 1. Let n>O. Determine all injectake structures in the sense of J.-M. Marande [1964] in the category B,. For every identity I for distributive lattices with pseudocomplementation, there exists a first-order sentence @ ( I ) such that I holds for L iff @(I) holds for $’(L). (In K. B. Lee [1970], the sentence for (L,) is given as follows: “Every element contains at most n minimal elements.” For n = O use Theorem 11.1.22.) Is there a natural class of first-order sentences properly containing all identities for which the same statement can be made? Investigate the lattice of implicational classes of distributive lattices with pseudocomplementation. For a distributive lattice L, define the number n(L) as follows: Let n(L) be the smallest integer n such that I (L) EB, if L has a zero and Io(L) EB, if L does not have a zero. Classify and investigate distributive lattices according to the value of n(L). (The case n = 1 was considered in T. Katrid&k [1966] and [1968].) Let K be an equational class of algebras satisfying HS(K,) =ISH(Ki) for all KI K. Does K satisfy the Congruence Extension Property? Does the ausumption that the congruence lattices of algebras in K are distributive make any difference? Find maximal classes K c D such that for Li, L,EK, 9 ( L , ) = 9 ( L 2 ) implies that L, s L,. (See M. E. Adams [1975].)

,

1 This was solved by Ph. Olin, Elementary properties of free products. Abstract. Notices Amer. Math. SOC. 23 (1976), 735-El.

CHAPTER 111

CONGRUENCES AND IDEALS

1. Weak Projectivity and Congruences

Let a , b, c, and d be elenients of a lattice L ; if, for any congruence relation 0 of L, a = b (0) implies that c =d (O), then we can say that “a = b forces c = d”. It is necessary to understand “forcing” in order to study the structure of congruence relations of lattices; this will be accomplished in the present section.

In Sect,ion 1.3 we proved that a = b (0) iff a n b E a v b (0) and so it is enough to deal with pairs of comparable elements. To simFlify our notation, let a/b denote an ordered pair of elements a , b of a lattice L satisfying b i a ; alb is called a quotient of L. (This notation obviously imitates quotient groups: GIH.) cld is a subquotient of a/b iff b d < c < a ; we call a /b a proper quotient iff b < a. If b < a then a/b is called a prime quotient. We write w/b = c/d iff a = c and b = d.

Now take a look at Figures 1 and 2 permitting the degenerate cases: a = c and b = d. In both cases (t = b (0) iff c =d (0) because XAY = X (0) iff y = m y (0). In

Figure 1 Figure 2

either case we shall write alb-cld, and say that alb is perspective to cld. I f we want to show whether the perspectivity is “up” or “down”, we shall write a l b p c l d in the first case and alb \ cld in the second case.

If for some natural number n there exist alb = eo/fo, e l / f i , . . . , e,/f,, =c/d such that e,l f ,-ei+l/fz+,, i = O , . . . , n-1, then we shall say that alb is projective to cld, and write a l b =c /d . (Note that a/b z a l b with n = 1 and a/b-c/d implies that alb xc ld . ) Yrojectivity is the trnnsitive extension of perspectivity. Observe that alb Y c l d and a = b imply that c =d.

130 111. Congruelices end Idcrtls

Figure 3 Figure 4

The concept of projectivity is sufficient for the study of “forcing” in many large classes of lattices (for instance, in the class of modular lattices, see Section IV.1). In general, however, we have to introduce somewhat more cumbersome notions : weak perspectivity and weak projectivity.

Consider Figures 3 and 4 permitting the degenerate cases ai = c and b, = c l . If a = b (O), then al = b, (0) by Lemma 1.3.7; since aJbi -c/o?, a = b (0) iniplies that c ~ d (0). Let us say that cld is weakly projective into alb iff we can get from c]d into a/b in finitely many steps as described in Figures 3 and 4. Figures 3 and 4 then should describe the concept cld weakly perspective into alb. It turns out however that a special case of Figures 3 and 4 suffices to describe the same.

d and c = avd ; similarly, eld p w a / b iff c 2 a and d = b Ac

(see Figures 5 and 6). If c/d p alb or cld \, aJb, then c/d is weakly perspective into alb, in symbols, cld-, alb. If for some natural number n and cld =eo/fo, e i / f l . . . . , en/fn=a/b we have eilflhW e i + , l f i + l , i = O , . . . , n - 1 , then c/d is weakly projective into alb, in notation, c l d z , alb. (Note that cld is weakly projective into cld with n = 1.) Weak projectivity is the transitive extension of weak perspectivity. Observe that neither weak perspectivity nor weak projectivity is a symmetric relation. The concept of weak perspectivity is due to R. P. Dilworth [1950 a].

The notational system we use is geometrically motivated. Many papers (starting with G . Gratzer and E. T. Schmidt [1958d]) use the notation [b , a] - [d , c ] or alb -cld

We write cld 4, alb iff b

cld -w alb c/d \ alb

Figure 5

1. Weak Projectivity end Congruences 131

for c/dZw a/b; t,hese notations appear to be more natural from a universal algebraic point of view and have the advantage of emphasizing the nonsymmetric nature of weak projectivity.

The following elementary observation shows the equivalence of the various forms of the definition of weak projectivity.

LEMMA 1. ditions are equivalent:

(ii) There i s nn integer m and there are elements eo, . . , , em-l EL such that

Let L be a lattice, a , b, c, d E L, b $ a, and d I c. Then the following con-

(i) cld i s weakly projective into alb.

pm(a, eo, . * * 9 em-1) = C a d ~ m ( b , eo, . . > em-1) =d, where the polynomial pm i s defined by

Pm(X, YO, * * * 9 ~ m - 1 ) * . ( ( ( x A Y o ) v Y I ) ’ Y ~ ) ~ * * * *

(iii) There i s nn integer n and there are quotients cld = e& eJ f& eilfl, e;/f; . . . , e Jf, =aJb such that e:/fl i s n subquotient of ei+llfi+i and ei/fi-ei/fi, for i = O , 1,. , . , n - 1 .

REMARK. Condition (iii) is the definition of weak projectivity using Figures 3 and 4.

PROOF. Figures 5 and 6 are special cases of Figures 3 and 4 respectively; hence (i) implies (iii). Now let (iii) hold. Then for each i =0, 1, . . . , n - 1, either ei/fi 7 e;/fi or e i / f i \ e:/f:. In both cases,

ei=p4(ei+l, e:, f:, ei, f i ) ,

fi=P4(fi+1, e:, f : , ei, f i ) .

Repeating these steps n times we get (ii) with m < 4n. Thus (iii) implies (ii). Finally, let (ii) hold. Then

aAeo/bAeo 7 , alb,

(aAeo)vel/(bAeo)Vel L, aAeolbAeo,

and so on; in m steps this yields that c/d=, alb.

We shall write cld =, a/b iff Lemma 1 (ii) holds with rn = k. This is slightly artifioial. It corresponds to requiring that the series of k weak perspectivities end with 7 , and that 7, and \, alternate throughout.

Intuitively, “a = b forces c = d” iff cld is put together from pieces each weakly projective into alb. To state this more precisely, we describe @(a, b) , the smallest congruence relation under which a = b (see Section 11.3).

k

THEOREM 2 (R. P. Dilworth [ 1950 a]). Then c ~d (@(a , b) ) i f f for some sequence c = eo 2 el 2 * * 2 em = d we have

Let L be a lattice, a,, b, c, d EL, b 5 a, d < c.

ejlei+i=w alb, for j = O , . . . ,m-1 . 10 Grltzer

132 111. Congruences and Ideals

PROOF. Let @ denote the following relation on L: x = y (0) iff x v y =c and XAY =d satisfy the condition of the theorem. We first prove that @ is a congruence relation by verifying the conditions of Lemma 1.3.8. 0 is reflexive since for any c L w e get

c/c p w avclbvc I,,, alb.

It is also obvious that if al 2 a2 2 a3 and al = a2 (a), a2 = a3 (a), then al = u:) (a). Indeed, take the sequences establishing uI =a, (@) and a2 = a3 (a) ; putting the two sequences together we get a sequence establishing ai = a 3 (@).

Nowlet CEO!(@), c > d , and /EL. Let c = e o 2 e l > - - . 2 e m = d bethesequence establishing c ~ d (@), that is, e i / e i+ lxw alb for i = O , , . . , m - - l . ThencAf=e,,Af> el/ \ / 2 - - 2 e,Af =dAf and eiAflei+lAf fwei le i+lxw alb, hence eiAf/ei+lAfx,,, alb for i = 0, . . . , n - 1 ; this proves that cAf =dAf (@). Similarly, cvf E d v f (@).

is a congruence relation. a EE b (.@) and so 0 is a congruence relation under which a = b. Now let 0 be any congruence relation satisfying a = b (0). It is easy to see that for x > y and u 2 w, x = y (0) and ulv wW xly imply that u = v (0). By a trivial induction, x = y (0) and u/vxw x / y imply that u = w (0). So finally, let c = d (a), established by cvd = e0 2 el 2 * * - 2 em =cAd. Since ei/ei+lxw a/b we conclude that ei=ei+l (@), for i = O , . . , , m -1. Therefore, by the transitivity of 0, we obtain that c = d (0). This proves that (D is the smallest congruence relation under which u = b, and so @ =@(a, b).

L? To compute @ ( H ) , the smallest congruence relation 0 under which a =- b (0) for all ( a , b) E H , we use the formula (Lemma 11.3.2)

Thus by LemmaI.3.8,

Let L be a lattice and H

= V ( @ ( a , b) I (a, b> EH),

and we need a formula for joins:

LEMMA 3. a = b ( V ( O i I i E I ) ) i f f there i s a sequence zo = aAb I zl 5 - - - each j with 0 < j < n there i s a n ii E I satisfying zi = zi+l (04).

Let L be a lattice and let Oi, i C I , be congruence relations of L. Then z, = avb such that for

The proof of Lemma 3 is the same as that of Theorem 1.3.9, namely, a direct

By combining' Theorem 2 and Lemma 3 we get : application of Lemma 1.3.8.

COROUARY 4. Let L be a lattice, let H L2, and let a , b L with b< a. Then a ZE b ( 0 ( H ) ) i f f for s m e integer n there exists a sequence a = co 2 ci 2 - - - 2 c, = b such that for each i, with 0 5 i <n, there exiats a (d, e) EN mtisfying

ci/c,i d ve/dAe.

COROLLARY 5. Let L be a lattice, let I be a n idea,l of L, and let a , b E L, b < u. Then a = b ( @ [ I ] ) i f f for some integer n there exists a sequence a = cg 2 cI 2 - - . 2 c, = h such thut for euch i, with O( i <n, there exist d, e E I with e l d satisfying ~ ~ / c ~ + ~ = ~ d / e .

1. Wmk Projwtivit,y and Congruences 133

0

Figure 7 Figure 8

Recall that, an ideal I is called the (ideal) kernel of a congruence relation 0 iff I is a congruence class modulo 0 (Section 1.3).

COROLLARY 6. u congruence relation iff

Let L be a lattice and let I be u n idenl of L. Then I i s a kernel of

u/b=, c / d , u L, and b, c, d I imply that a E I .

PROOF. Combine Corollary 5 with the observation that I is a congruence kernel of sotiie congruence relation iff I is the kernel of @[I] .

In distributive lattices every ideal is the kernel of some congruence relation, in fact this property characterizes distributivity. In general lattices we shall introduce various classes of ideals that are congruence kernels for which @ [ I ] can be nicely described. This will be done and applied in Sections 2-4.

In a sense, weak projectivity describes the structure of congruence relations of a lattice. It is not surprising, therefore, that many important classes of lattices can be described by weak projectivities. We give two examples.

To introduce the first class, the class of weakly modular lattices, we need a lemma, for niotivation.

LEMMA 7. Let L be a lattice, ( I , b, c , d E L, b <a , d <c, and let a l b x , c/d. If L i s modulur or i f L i s relatively complemented, then there exists a proper subquotient c'ld' of cld such that c ' / d ' x w alb.

PROOF. Let alb = eolfo -," ell / , -w * * - - en/fn =c /d ; we prove the statement by induction on n. By duality, we can assume that en-l / fa- l p w c / d and by theinduction hypothesis there exist eL-i >fk- i , such that e ~ - i l f ~ - i = w alb and f ,- tSfi-i < e:&-, i ert-1-

10.

Let L be modular. As in Figure 7, define d'=dvf;-, and c' =dveL-,. Then d'veL-, = (dvfk-,)vei-, = d v ( f ~ - , v e ~ - , ) =rive;-, =c'. By modularity, d'Aek-., = (dvfk-,)Ae;-, = (since &-,< ek-,) = ( d A e ~ - , ) v f ~ - , = f n - i v f ~ - l =I;-,. Hence c'/d'- e ~ - , / f ~ - , . Thus d' <c' and c'ld' -ek-,If~-,=, alb; that is, cfld'=:w alb.

Next, let L be relatively complemented. Let x be the relative complement of I;-., in [ f , e i - ,] and define c' = d v x , d' = d (see Figure 8) . Then e(, -,If: - , \ xlf, - , f c'ld' and so c'ld' ~ e ~ - , I f k - . , . We conclude, as before, that d' <c' and thatc'/d'=w alb,

DEFINITION 8 (G. Griitzer end E. T. Schmidt [1958d]). A lattice L i s called weakly modular i f f a, b, c, d L, b <a, d < c, alb=:, cld imply the existence of a proper subquotient c'ld' of cld satisfying c'ld' =, aJb.

COROLLBY 9. modular.

Every modular and every relutively complemented lattice i s weakly

The import'ance of the class of weakly modular lattices will be illustrated in

To introduce the second class of lattices we again need a lemma. Sections 2 and 4.

LEMMA 10. Let L be a lattice and let ai, bi E L, bi < ai, for i = 1, . . , , n. Then

A(@(a. bi) I i= 1 , 2 , . . . , n) + w

i f f there exists a proper quotient alb of L such that

alb=:, ailbi, for i = 1, 2, . . . , n.

PROOF. I f such a, b exist, then a=b ( @ ( a , b J ) , for i= 1,2 , . . . , n. Hence u = b (A(@(ai, b,J I i = 1,2, . . . , n)) and so A(@(ai, bi) I i = 1, 2, . . . , n) + w . We prove the converse by induction on n in a somewhat stronger form: if v < 2~ and u = v (A(O(ai, bi) I i= l , 2, . . . , n)), then there exists e proper subquotient alb of u/v such that a l b z , a J b , for i = 1,2, . . . , n. For n = 1, apply Theorem 2. Assuming the statement proved for n - 1 we get a'/b' a proper subquotient of ulv satisfying a'lb'z, ai/bi, for i = 1, . . . , n - 1. Since u = w (@(a,, 6,)) we get a' = b' (@(a,, bJ) . Hence by Theorem 2 we obtain a proper subquotient alb of a'lb' with a l b x , anlb,. Since a l b x , a'lb' we also have a/b=sw ai/bi, for i = 1 ,2 , . . . , n - 1.

Since in a subdirectly irreducible lattice w is meet-irreducible we conclude :

COROLLARY 11. Let L be a subdirectly irreducible lattice and let be proper quotients of L, for i = 1, . . . , n. Then there exists a proper quotient alb of L satisfying albw, ai/bi, i = 1, . . . , n.

In fact, Corollary 11 holds for any lattice in which w is meet-irreducible.

1. Weak Projrct,ivity and Congruelices 136

DEFINITION 12 (R. Wille [1972]). subset of L. Then P is called primitive i f f

Let L be a lattice and let P be a finite nonvdd

A ( @ ( z , x v y ) I G Y E P , x * . V Y ) * o .

Combining Lemma 10 and Definition 12 we obtain

COROLLARY 13. thut a.plbp=:tu cvdlc for all c , d C P , c =i= cvd.

P is priniitice iff there exists a proper quotient ap/bp of L, such

Observe also that by Corollary 11 every finite nonvoid subset of a subdirectly irreducible lattice is primitive.

It will be pointed out in Chapter V that important classes of lattices can be defined by the property of not containing a primitive subset (poset) isomorphic to a member of a given collection of posets.

We study weak projectivities to enable us to obtain descriptions of quotient lattices such as L/O(a , b). Sonietimes however L/0 can be identified very simply within L : if there is a sublattice Li of L having one and only one element in every congruence class, then LIO s L i . Such congruence relations are called representable. It happens much more often that we can get a meet-subsemilattice or a join-subsemilattice L, of L having one and only one element in every congruence class; in such cases we call 0 meet-representable or join-representable, respectively.

LEMMA 14. congruence class of @ has a mininial element, then 0 is join-representable.

Let L be a lattice and let 0 be a congruence relation o/ L. If every

PROOF. Let Li be the set of minimal elements of congruence classes of 0; then Ll contains exactly one element of each congruence, class. Now let a, b c Ll. We show that avb is the snisllest element of [avb]@=[a]@v[b]O. Indeed, if c < a v b and c Eavb (O), then aAC = a (0) and bAc = b (0). Since c <avb we get that UAC <a or bAc < b, say QAC < a . Then a ~ c < a and aAc c [a]@ contradict that a € L,.

Thus if L satisfies the Descending Chain Condition, then every congruence relation is join-representable, and dually.

Exercises

1. For all pairs of quotients xly, u1.u of

2.

investigate when “xzy forces u ~ v ” . Repeat the investigation for XU3. Show that if L is sectionally complemented, that is, L has a zero and for every aEL [0, a] is complemented, then in order to learn the congruence structure of L it is sufficient t o consider ‘‘a E b forces c Ed’’ in the special case b =d =O. Prove that, if a lattice is modular or relatively complemented, then alb xw c/d iff alb xc’ld’ for some subquotient c’ld’ of cld.

3.


4. 5.

6.

7.

8.

9. i 0.

11.

12.

13.

14. 16.

16.

17.

18. 19.

20.

21.

22.

23.

Show, by an example, that the conclusion of Exercise 3 is false for lattices in gencAral. Let L be a distributive lattice and let a/b kW c /d . Prove that there exists a quotient elf and a subquotient c'ld' of c/d such that

a / b p elf \c'/d'.

Let p =p(z,, . . . , xn-l) be a lattice polynomial, and let L be a lattice and a l , . . . , an-lEL. Consider the algebraic function q(z) =p(z , a l , . . . , an-l) . Under what conditions on p is it true that for all quotients a/b of L,

q(a)/n(b) =w a/b.

(This is true for p =. * - ( ( ~ , A Z ~ ) V Z ~ ) A *

Referring to the proof of Lemma 1, axiomatize those properties ofp, which make it possible to define pm from pz to obtain generalizations of Lemma 1 to some equational classes of algebras. Show that pm(x, yo, . . . , yi, yi, yif2,. . . , ym-l) does not depend on z, yo, . . . , yi-1 (that is, the value of pm(a, e,, . . . , ei, e i , e i + 2 , . . . , em-l) is independent of a, eo, . . . , e i -1) .

Prove that p,(z, c, d , c , d , . . .) =z for any d Rephrase and prove Lemma 1 using qm =. - ((z v Y,)A yl) v- * *. Show that -",,

m defined in terms of qrn, may differ from -w defined by pm. Show that in any sequence of weak perspectivities, any number of steps f w and h, can be put in. Observe that in any nonredundant sequence of weak perspectivities, f w and h uI have to alternate. Find examples to show that m cannot be bounded in Theorem 2. (Make all your examples planar modular lattices.) Find examples to show that in Theorem 2, aw cannot be replaced by E w for any natural number 7 ~ . (Make all your examples planar modular lattices.) Prove that an ideal I is a kernel iff I is the kernel of @[I]. Let L be a lattice, and let I and J be ideals of L with J E I. Assume that J is a kernel in the lattice 1 and that I is a kernel in the lattice L. Is J a kernel in L ? Show that the ideal kernels of the lattice L form a sublattice Ic(L) of I (L) , the lattice of all ideals of L. In fact, Ic(L) is closed under arbitrary joins and under all meets existing in I ( L ) . Let L be a distributive lattice, let I be an ideal of L, and let a/b be a quotient of L. Then a = b (@[I]) iff a/b -a'/b' for some quotient a'lb' of I. Show that Exercise 6 characterizes distributivity. Show that Exercise 17 characterizes distributivity. In Exercises 20-23 (which are due to R. Wille [1972]) let L and L' be lattices and let cp be a homomorphism of L onto L'. Let a'lb' be a quotient of L' and let c/d be a quotient of L with a'/b' zW ccp/dcp. Prove that there is a quotient a/b of L with a/b =w c / d , acp =a', and b p = b'. Let ci/cli, . . . , cn/dn be quotients of L such that cia, = C ~ Q =- * * =cnp and dlpl =dzcp = * * = d , p Show that there is a quotient a/b of L satisfying a/b =w cJdi and aglbcp =

czp/dicp, for i = 1, 2, . . , , n. Let P' be a primitive subset of L', and let P be a subset of L such that cp restricted to P is a (poset) isomorphism of P with P'. Show that P is primitive in L. Let P be a primitive subset of L, let ap, b p be given as in Corollary 13, and let app =+ bpp . Show that Pcp is primitive in L' and that cp restricted to P is a poset isomorphism of P with Pq.

* by Lemma 1.)

tn z 5 c.

1. Weak Projectivity and Congruences 137

24.

25.

26. 27.

28.

29. 30.

31.

32. 33.

34.

36. 36.

Let L be a lattice and let Y be a congruence relation of L. Verify the formula :

@([a]Y, [ b ] Y ) =(@(a, b )vY) /Y .

Let ai, bicL, for i = 1, . . . ,PO. Show that

Use Exercises 24 and 25 to verify Exercises 22 and 23. Show that for ideals I and J of a lattice L

@[I] v@[J] =@[IvJ] .

Let L be a lattice and let J i , i€I, be ideals of L. Prove that

V(@[Ji] I i €1) =@[v(Ji I i € I ) ] .

Show that @[I]A@[J] = @ [ I h J ] does not hold in general. Verify

@[I]A@[J] = @ [ I d ]

for ideals I and J of a distributive lattice. Show that the formula of Exercise 30 holds also under the condition that every ideal is the kernel of at most one congruence relation. Show that Corollary 11 does not hold for infinitely many ai/bi. Show that in a finite lattice (or in a lattice in which all chains are finite) every congruence relation is meet- and join-representable, Find a congruence relation 0 of the lattice of Figure 9 which is not representable. (Observe that 0 is both meet- and join-representable.) Let L be a finite lattice and let L / O ~'9.2~. Show that 0 is representable. Give a formal proof of the statement that if L, represents the congruence relat,ion 0 of L, then L/0 s Li.

Figure 9


37. A congruence relation 0 of a lattice L is order-repreeentable iff there exists a subset H L mtisfying [HI@ = L and

a 5 b in L iff [a105 [b]O in L/0, for a, b CH.

Show that H as a subposet of L is a lattice and H EL/@. Prove that every meet-representable congruence relation is order-representable. Let L be a lattice and let 0 be a congruence relation of L. Verify that if L / 0 is finite or countable, then 0 is order-representable. Find a chain C, a lattice L, and a congruence relation 0 of L, such that L/O SC but 0 is not order-representable.

38. 39.

40.

2. Distributive, Standard, and Neutral Elements

The three types of elements of a lattice mentioned in the title of this section were discovered by 0. Ore [1935], G. Griitzer [1969], and G. Birkhoff [1940a], respectively. It turned out later that all three can be defined using distributive equations only. This may be a coincidence, but it could be considered as a confirmation of the principle that distributive lattice theory provides the foundation of general lattice theory.

DEFINITION 1. Let L be a lattice and let a be an element of L. (i) The element a is called distributive iff

a v ( x A y ) = ( a v z ) A ( a v y ) ,

for all x, y 6 L. (ii) The element a is called standard iff

x A ( a v y ) = ( X A U ) V ( X A Y ) ,

for all x , y C L . (iii) The element a is called neutral iff

( a A X ) V ( X A y ) V ( y A a ) = ( a V X ) A ( z V y ) A ( y V a ) ,

lor all x , y EL.

For instance, in 0, i , a, and c are distributive; 0, i, and a are standard but c is not standard; only o and i are neutral. In %R3 only o and i are distributive; they are also standard and neutral.

We can also dualize these definitions and define dually distributive elements and dually standard elements. Observe that the concept of neutrality is selfdual.

Of course, every element of a distributive lattice is distributive, standard, and neutral.

Various useful equivalent forms of these definitions are given in the three theorems that follow.

2. Distributive, Standard, and Neutral Elements 139

THEOREM 2 (0. Ore [1936]). following conditions on a are equivalent:

Let L be a lattice and let a be a n element of L. The

(i) a i s distributive. (ii) The map

y :x - .avx (xEL)

is a homomorphism of L onto [a) . (iii) The binary relation 0, on L defined by

x ~ y (0,) i f f a v x = a v y

i.s (L congruence relation.

REMARK. (i) and (ii) are equivalent since y is a homomorphism iff it preserves meets, while (ii) and (iii) are equivalent because 0,=Ker(y). A more formal proof follows.

PROOF. (i) implies (ii). y maps L into [a) for any element a of L ; in fact y is always onto since by = b for all b 2 a. The map y is always a join-honiomorphism since

q v y y = ( a v x ) v ( a v y ) = a v ( x v y ) = ( x v y ) y .

In view of Definition l(i), if a is distributive, we also have

x y ~ y y = (avx)A(avy) = a v ( x ~ y ) = ( x A Y ) ~ ,

and so y is a homomorphism.

gruence relation. @

Therefore,

(ii) implies (iii). 0, is the kernel of the homomorphism 9 and therefore 0, isacon-

(iii) implies (i). Since x v a = (avx )va , therefore x E a v x (0,). Similarly, y r a v y (0,).

XAY E (avx)A(avy) (0,).

By the definition of 0, we get

aV(xAy) =aV((aVx)A(avy)) = (aVX)A(aVy),

proving (i).

THEOREM 3 (G. Griitzer and E. T. Schmidt [1961]). crn element of L. The following conditions are equivalent:

Let L be a lattice and let a be

(i) a i s standard. (ii) The binary relation 0, on L defined by

x EE y (0,) i f f (XAy)Vai = x v y for some ai a

is u congruence relation. (iii) a i s distributive and, for x, y EL,

aAX = a ~ y and a v x = a v y imply that x = y .

PROOF. (i) implies (ii). Let a be standard and let 0, be definedas in (ii). WeuseLemma 1.3.8 to verify that 0, is a congruence relation. By definition, 0, is reflexive and x ~ y (0,) iff X A Y ~ X V Y (0,). If x l y l z x = y (@,), and y z z (0J, then xval = y and yvaz = z for some al, az<a. Hence xv(alva2) = z , showing that x s z (0,) since a1vaz < a. Finally, let x< y and x = y (@,), that is, xva l = y with u1 < a. For any t E L, (xvt)vai = yv t , hence x v t = y v t (a,). To show the Substitu- tion Property for meet, observe that

y At y = xva, 5 x v a ,

and so, using the fact that a is standard we get

YAt = (YAt)A(XVU) = ((yAt)AX)V((yAt)Aa) = (XAt)VU*,

where u2 = YAtAa < a. Thus the conditions of Lemma 1.3.8 are verified and 0, is a congruence relation. @

(ii) implies (iii). Let us assunie that the 0, defined in (ii) is a congruence relation. We can show that a is distributive just as in Theorem 2. Now let U A X = m y and avx=avy . Since y ~ a v y (@,), meeting both sides with x and using a v y = n v x we obtain

X A Y = X A ( U V Y ) =XA(aVX) = x (0,).

Thus x = ( x A ~ ) v ~ for some al<a. Also, a l < x , hence U ~ < U A X = U A ~ , and so ai I XAY. We conclude that x = XAY. Similarly, y = X A Y , and so x = y. @

(iii) implies (i). Let us assume (iii), let 2, y € L , and define b = m ( a v y ) and c = (xAa)v(xAy). In order to show b=c, by (iii), it will be sufficient to prove that uAb = ~ A C and avb =avc. To prove the first,

U A X I aAc < (using c I b) < aAb = aAXA(UVy) = U A X ,

hence iiAc=uAb. To prove the second, we compute using the fact that a is distributive:

avb =aV((xA(avy))) = (UVX)A(aVy) =aV(XAy) =aV(xAa)V(xAy) =avc.

THEOREM 4. Let L be a lattice and let a be a n element of L. The following conditions are equivalent:

(i) a i s neutral. (ii) a i s distributive, a is dually distributive, and aAx=aAy and a v x = a v y imply

(iii) There i s a n embedding y of L into a direct p r d u c t A X B where A has a 1 und B

(iv) For any x , y L, the sublattice generated by a, x , and y is distributive.

x = y for any x, y C L .

has a 0 and a y = ( l , 0).

REMARK. The equivalence of (ii)-(iv) is due to G. Birkhoff [1940a]; in fact, G. Birk- hoff used (iv) as a definition of neutrality. It was conjectured in G. Griitzer and E. T. Schmidt [1961] that (i) is equivalent to (ii)-(iv). This was proved in G. Grlitzer [1962], J. Hashimoto and S. Kinugawa [1963], and Iqbalunissa [1964].

2. Distribiit ive, Standard, and Neutral Elements 141

PROOF. (i) implies (ii). Let a be neutral. Then

(*) U V ( Z A ~ ) =XA(UVY) for x 2 u.

Indeed,

civ(x~y) = ( u A ~ ) v ( x A ~ ) v ( ~ A u ) = (by Definition l(iii))

= ( U V X ) A ( X V ~ ) A ( Y V ~ ) =xr\(avy).

To show that (L is distributive, compute:

UV(ZAY) = w v ( a ~ x ) v ( x ~ y l ) v ( y ~ a ) =(by Definition l(iii))

= UV((UVZ)A(W~)A(~VU))

(apply (*) to u, uvx, and ( x v ~ ) A ( ~ v ~ ) )

= (a v 4 A(av((xvY 1 A (?Pa) ) 1 (apply (*) to a, yva, and xvy) = (am) A (y va) A (UVZVY ) = (avx) A (a vy ),

as claimed. By duality we get that a is dually distributive. Finally, let U A X =UAY and uvx =avy. Then

2 = X A (a V X ) A ( U vy )~ (xvy ) = X A ( ( u A ~ ) v ( ~ A ~ ) v ( u A ~ ) )

= ZA ( (a AZ)V(X/\Y)) = (~AZ)V(XAY) = ( ~ A ~ ) v ( u A ~ ) v ( ~ A ~ ) .

Since the right-hand side is symmetric in x and y we conclude that x =y. 0 (ii) implies (iii). Let (ii) hold for a and define A = (a] and B = [a). Let

g, : x +(.xAu, ma).

Since ( I is distributive and dually distributive, 9 is a homomorphism of L into A x B (by Theorem l(ii) and its dual). The map 91 is one-to-one, since if xg, =yrp for x, y E L, then

( x A U , xVa) =(yAU, yVa),

that is, XAU =YAU and xva =yvu; thus x =y by (ii). So g, is an embedding, ng, =(a, u ) , and a is the unit element of A and the zero of B.

(iii) implies (iv). The following three statements are obvious: (iv) holds for the zero and the unit in any lattice; (iv) holds for (ao, at) in A o X A i iff it holds for ui in A , for i = 0, 1 ; if a is an element of Lo, Lo is a sublattice of L,, and (iv) holds for u in L , , then (iv) holds for a in Lo. (iii) along with these three statements proves (iv). Q

(iv) implies (i). Obvious by Exercise 1.4.7.

properties of distributive, standard, and neutral elements. The results stated in Theorems 2-4 make it possible to verify the most important

THEOREM 5. (i) Every neutral element i s standard. (ii) Every standard element i s distributive.

(iii) Every standard bnd duully standard (dually distributive) element i s neutral. (iv) If a i s distributive or standard, then the 0, in Theorem 2(iii) and Theorem 3(ii),

respectively, agrees with @[(a]] .

PROOF. (i) and (iii). By Theorem 4(ii) and Theorem 3(iii). @

(ii) By Theorem 3(iii). (iv) If u = a (0) for any u 5 a and a congruence relation 0, then a v x = a v y implies

that x = x v ( a n x ) = xva = p a = y ~ ( a ~ y ) = y (O) , hence for the relation 0, given by Theorem 2(iii) we have a,( 0. Similarly, if (zr\y)vni = x v y for some ai a, then XAY = ( X A Y ) V ( U A X A Y ) = (xr\y)va, = x v y (O) , and so x = y (0); hence for the relation 0, of Theorem 3(ii) we obtain 0,s 0.

For a principal ideal (a] =I let @ [ I ] be denoted by 0,. Then by Theorem 5(iv), the 0, of Theorem 2(iii) and the 0, of Theorem 3(ii) are indeed the 0, just defined in the case a. is distributive or standard. Hence 2(iii) and 3(ii) are definitions of distributive and standard elements, respectively, in terms of the properties of 0,.

As we have already seen, the converse of Theorem 5(i), as well as that of Theorem 5(ii), is false. Two wide classes of lattices in which the converse holds are the class of modular lattices and the class of relatively complemented lattices.

THEOREM 6 (G. Griitzer and E. T. Schmidt [1961]). lattice. Then an element of L i s distributive i f f i t is neutral.

Let L be a weakly modular

Before proving the theorem we verify a lemma connecting the distributivity of an element with weak projectivity.

LEMMA^. Let L be a lattice and let a be a n element of L. The elementaisdistributive iff u<z<a<y<x and xlyz:, zlu imply x = y .

PROOF. Let a be a distributive element, u< z< a < y s x , and x/y=., zlu. Since u 2 z 5 a, we get u = z (0,), hence x E y (en). By Theorem 2(iii), x = x v a = y v a = y. Let us now assume that a is not distributive. Then there exist b, c E L such that av(bAc) $5 (avb)~(avc ) . Since a = aAbAc (0,) we obtain

(UVb)A(aVC) = av((avb)A(avc)) = U V ( ( ( U A ~ A C ) V ~ ) A ( ( ~ AbAc)vc))

=UV(bAC) (0,)

so by Corollary 1.5 there exist x, y , u EL satisfying, with z = a ,

u < z ( a ( (avb)A(avc)<y < z _ ( a v ( b ~ c ) , x / y z w z / u .

2. Distributive, Standard, atid Neutral Elements 143

PROOF OF THEOREM 6. Let L be a weakly modular lattice, and let a be a distributive element. If a is not dually distributive, then by Lemma 7 there exist u < z < a < y < x sat,isfying z l u ~ . , xly. By weak modularity, there exists a proper subquotient xl /y l of x / y satisfying x I / y i ~ w zIu, contradicting by Lemma 7 that a is distributive.

Now let U A X =UATJ, a.vx = a v y , and x+y for some x , y € L. Proceeding just as in the step “(ii) implies (iii)” in the proof of Theorem 3, we conclude that

and dually x = y (@(a, aAx))

x = y (@(a, a v z ) ) .

Applying Theorem 1.2 to x = y (@(a, U A X ) ) we get a proper subquotient xi/yl of x v y / x ~ y satisfying xi/yiXw n / U A X . Since x = y (@(a, a v x ) ) we also have xi = yi (@(a, a v x ) ) , and again by Theorem 1.2 weget aproper subquotient x2/y2 of xJyi such that x2Jy2=:w. avxJa. Using the weak modularity of L, we obtain a proper subquotient x3/y3 of avxla satisfying x3/y3xW x2Jy2. Hence

xs/y3zw X2/y2~,u X ~ / Y I X , ~ fl/aAX,

contradicting Lemmrt 7. By Theorem 4(ii), a is neutral.

COROLLARY 8. In a weakly modular lattice every standard element i s neutral.

THEOREM 9. Let L be a lattice. (i) Let D denote the set of all distributive elements of L. Then a and b c D imply

(ii) Let S denote the set of all standard elements of L. Then a and b ES imply that

(iii) Let N denote the set of all neutral elements of L. Then a and b E N imply that

that avb E D .

aAb and avb €8.

aAb and a,vb E N .

PROOF. (i) Let a , b E D and conipute:

(av b) v(xAy) = LC v b v (a ~y ) = rt v ( (b vx) A (b v y ) ) = (a Vbvz) A (a Vb Vy ) , s o a v b c D . 0

(ii) h t a, b €8. First we do the join: x / \ ( ~ V b V y ) = (xAa)V (xr\(bvy)) = (X /\a) V (x/\b) V (XAy) = (Z A (a v b ) ) V ( x Ay), proving that nvb E 8. Now we verify the formula

@,A@, = @, ,,,,, where 8,, a,, and GIOhb are the relations described by Theorem 3(ii). Since @,A@* 2 a,,,, is trivial, let x G y (@,A@,). Then x G y (a,), and so (xAy)val = x v y for some a l g a . We also have X G ~ (0,) and therefore a l = a l ~ ( x v y ) = a , ~ x ~ y (ab). Thus for some bi<b we have al=(aiAxAy)vbi. Now (xAy)Vbl= ( x A ~ ) v ( o , A x A ~ ) v ~ , = (xAy)Vai = x v y ; since b, < b and bI < al a , we obtain bI u A b ; these verify that x = y (0, ,,,).


Figure 1

This formula shows that if a, b E S , then the relation OahB of Theorem 3(ii) is a congruence relation, hence aAb ES by Theorem 3. 0

(iii) Let a , b EN. By Theorem 5, a, b E S . Hence by (ii), aAb E S. By Theorems 3 and 4, to show that aAb E N we have to prove only that aAb is dually distributive. Since a and b are dually distributive, they are distributive e1enient.s of the dual of L, hence by (i), avb in the dual of L is dist.ributive and so aAb in L is dually distributive.

Pigure.4 shows that a, b E D does not imply that arAb € D.

Exercises

1. 2.

3.

4.

Let L be 5 bounded lattice. Show that 0 and 1 are distributive, standard, and neutral. Let L be the lattice of Figure 2. Then (a] is the kernel of some congruence relation but a is not distributive. If a is a distributive element of L, then LIO, s [a). To what extent does L/O, z [a) characterize the distributivity of a? Find distributive elements which are not dually distributive.

Figure 2

2. Distributive, Standard, and N~iit>ral Eltmmts 145

5.

6.

7. 8. 9.

10.

11.

12.

13. 14.

15.

16.

17. 18.

19. 20.

21.

22.

Investigate the relation 0, of Theorem 2(iii) as a congruence relation of the join- semilattice. Let L be a lattice, let p be a homomorphism of L onto L’, and let L, be a sublattice of L’. Let a be an element of L satisfying apEL,. Show that if a is distributive in L, then ap is distributive in L,. Prove the analogue of Exercise G for standard and neutral elements. Let a be a distribut’ive element of L. Show that (a] is a distributive element ofl(L). Prove the analogue of Exercise 8 for standard and neutral elements. Show bhat in Theorem 3(iii) (and in Theorem 4(ii)) the condit,ion can be weakened by assuming that IC 2 y. Verify that tho element a of a lattice L is standard iff x 2 avy implies that x = ( z A ~ ) v ( x A ~ ) (G. Griitzer and E. T. Schmidt [196l]). Prove that t,he join of two distributive elements is again distributive by verifying the formula @av@b=@avb.

Find classes of lattices in which the meet of two distributive elements is distributive. Show that the map a - 4 3 , for standard elements is an embedding of the sublattice of standard elements into the congruence lattice. Let L be a lattice and let a, b, c L. Show that a, b, and c generate a distributive siiblattice iff for any permutation IC, y, z of a, b, c we have

Z A ( ~ V Z ) = ( Z A ~ ) V ( Z A Z ) .

SV(yA.2) =(ICVy)A(ZVZ),

( Z A ~ ) V ( ~ A Z ) V ( Z A Z ) = ( z v y ) ~ ( y v z ) ~ (ZVZ)

(0. Ore [ 19401). Let L be a lattice and let a, b be standard elements of L. Show that foranycEL, a, b, and c generate a distributive sublattice. Do three distributive elements generate a distributive sublattice ? Verify directly that in a modular lattice distributive element =standard element = neut,ral element.

Verify the conclusion of Exercise 18 in a relatively complemented lattice. An element of a modular lattice is neutral iff it has a t most one relative complement i n any interval containing it (G. Griitzer and E. T. Schmidt 119611). Let L be a modular lattice and let a E L. Then a is neutral iff a is neutral in every interval of the form [UAIC, av z ] , for zEL. Show that the conclusion of Exercise 21 fails in weakly modular lattices. (See Figuru 3.)

Figure 3

146

h

111. Congruences atid Ideals

Figure 4

23. Show that Exercise 21 also fails for weakly modular lattices and standard elements. (See Figure 4.)

24. Let L be a bounded relatively complemented lattice. Then an element a of' L is neutral iff it has a unique complement (J. Hashimoto and S. Kinugawa [l963]).

26. Show that the N of Theorem 9 is the intersection of the maximal distributive sublattices of L (G. Birkhoff [1940a]).

26. Prove Theorem 9(iii) using Theorem 4(iii).

3. Distributive, Standard, and Neutral Ideals

The t,hree types of ideals mentioned in the title of this section derive naturally from the concepts introduced in Section 2.

DEFINITION 1. An ideal I of a lattice L is called distributive, standard, or neutral, respectively, iff I is distributive, standard, or neutral, respectively, as an element of I ( L ) , the lattice of all ideals of L.

TO establish a connection between the type of elements and the type of ideals they generate we need a lemma:

LEMMA 2. mials. Let us assume that for all a E I there i s a b E I such that a 5 b and

Let L be a lattice, let I be an ideal of L , and let p and q be wary polyno-

PROOF. In Section 1.4 (in the proof of Lemma 1.4.8) we proved the formula:

p(Io , . . . , In-l) ={z I 2 < p(io, . . . , in -1) for some io E 10, . . . , in -i E In-i.c. ?

3. Distributive, Standard, and Neutral Ideals 147

COROLLARY 3. Let L be a lattice and let a E L . Then a is distributive, standard, or neutral, respectively, i f f (a] as a n ideal is distributive, standurd, or neutral, respectively.

The main characterization theorems can be proved similarly to the proofs of Theorems 2.2-2.4.

THEOREM 4. on I are equivalent: (i) I is distributive. (ii) @ [ I ] can be described as follows:

Let L be a lattice and let I be an ideal of L. The following conditions

s=y ( @ [ I ] ) i f f z v i = y v i for some iCI.

PROOF. If in (ii) we put ( x ] v I = ( y l v l in place of zvi =pi, then the equivalence can be proved as in Section 2 since x =y (@[I]) in L iff ( X I = ( y ] (0,) in I(L) . Now, if z v i = y v i for some i E I , then ( s ] v I = ( y ] v I is obvious. Conversely, if ( x ] v l = ( y l v l , then x( yvi0 and y< x v i , for some io, il E I. Therefore, xvi= y v i with iovil=iEI.

THEOREM 5.1 Let L be a luttice and let I be a n ideal of L. The followingconditions on I are equivalent:

(i) I i s standard. (ii) The equality

(a]A( Iv (b]) = ( ( a ] ~ l ) V ( a A b ]

holds for all a , b EL. (iii) For any ideal J of L,

I v J = { i v j I i c I a n d j E J } .

(iv) @ [ I ] can be described by

z = y ( @ [ I ] ) i f f ( x n y ) v i =xvy for some i c I .

1 This result and all tho other unreferenced results in this section are based on G. Gratzer [1959] and G. Grtitzer and E. T. Schmidt [l961]. 11 Gratzer

(v) I is u distributive ideal and, for all J, K I(L),

IAJ = I A K and I v J = IvK imply that J = K .

PROOF. The equivalence of the five conditions can be verified as in Section 2. Only (iii) is new. But (iii) follows from (ii) : if I satisfies (ii) and a E I v J , then a i v j for some iE I and j € J ; hence (a] = (a]~(Iv( j ] ) = (by (ii)) = ((a]AI)V(aAj]. Therefore, a< ilvjl, where ii E ( a ] d and jl I at$ Consequently, a = ilvji. Finally, observe that when proving the analogue of “(i) implies (iv)” it is sufficient to use (iii).

TIIEOREM 6. Let L be a lattice and let I be an ideal of L. The following conditions on I are equiualent :

(i) I is neutral. (ii) For all i, k E L,

( I A (ill v ( ( i l ~ (kl 1 v ((4 1\11 = ( Iv (il ) I\ ( (il v ( kl ) A ((kIv4 - (iii) For all J, K E I(L) , I , J , and K generate a distributive sublattice of I(L). (iv) I is distributive and dually distributive, and, for all J , K E I (L) ,

IAJ = I A K and I v J = IvK imply that J= K .

PROOF. We can verify that (i) is equivalent to (ii) by u&ng the argument of Lemma 2. The rest of the proof is the same as in Section 2.

Observe that every distributive ideal I of a lattice L is the kernel of @ [ I ] . Indeed, if iEI ,aEL, and i = a ( @ [ I ] ) , then i v j = a v j for some $ € I , thus a S i v j E I , and so a E I. This explains why L/@[I ] can always be described. This description isespecially simple if I is principal, I = (a]. Then @[(a]] = 0, is a representable congruence relation and [a) represents a,, hence

L/O, s [a) .

.

If I is not principal, then we describe I (L/O[I] ) as follows:

THEOREM 7. Let I be a distributive ideal of a Za,ttice L . Then I (L /@[I ] ) is isomorphic with the lattice of all ideals of L containing I , that is, with [ I , L ] i n I (L) .

PROOF. Let Q] be the homomorphism x + [ x ] 0 of L onto LIO. Then the map y: K -+Kq-l maps I (L /@) into [ I , L ] . To show that this map is onto it is sufficient to seethat [ J ]O=J forall J 2 I . Indeed,if j E J , a € L a n d j = a ( @ [ I ] ) , thenjvi=avi for some i E I and so a < i v j E J and a E J, as claimed. Since y is obviously isotone and one-to-one, we conclude that it is an isomorphism.

CO~OLIARY 8. LIO[I] is determined by the interval [I, L] of I ( L ) , provided that I is a distributive idenl of L.

3. Diatrihutirc., St,andard, and Neutral Ideals 140

PROOF. We know (Section 11. 3) that I (L/O[I]) determines L/O[I] up to isomorphism.

Call a congruence relation 0 distributive, standard, or neutral, respectively, iff 0 = @ [ I ] where I is an ideal which is distributive, standard, or neutral, respectively. The following result was first established by J. Hashimoto [1952] for neutral congruences and by G. Gratzer and E. T. Schmidt [1961] in its present form.

The congruences @ and Y of a lattice L are said to permute (or are permutable) iff @vY =@ * Y where @ * Y is the binary relation defined by

a 3 b (0 * Y) iff there exists a c E L with a = c (0) and c = b (Y). An equivalent definition is that 0 andY permute iff @ - Y =Y - 0, which is, in turn, equivalent to @ - 'V being a congruence relation.

THEOREM 9. A n y two standard congruences of a lattice permute.

PROOF. Let @ and Y be standard congruences of the lattice L, that is, @ = @ [ I ] and Y = 0[J], where I and J are standard ideals of L. Let a G b (@) and b E C (Y), a, b, c EL. We want to show that there exists a d EL satisfying a = d (Y) and d = c (0).

If x < y ( z , x = y (@), and y z (Y), then y = x v i for some i c I and z = y v j for some j c J. With u = x v j we have z = uvi, .hence x E u (Y) and u z z (0).

Applying this observation to a = avb (@), avb ~ a v b v c (Y), and to c iz bvc (Y), bvc = avbvc (a), we obtain elements e , f L satisfying a sze (Y), e = avbvc (a), (i<e<avbvc, and c=f (@), f E a v b v c (Y), c l f l a v b v c . Set d =er\f. Then

a = e = er\(avbvc) = eAf =d (y),

c ~ f = f ~ ( a v b v c ) r f r \ e = d (@).

Theorem 9 is significant because in a wide class of lattices all congruences are standard. Recall, that a lattice L is sectionally complemented iff L has a zero, 0, end all intervals [0, a] of L are complemented.

THEOREM 10. standard.

In a eectionally complemented lattice L all congruence relations are

Yaoo~. Let 0 be a congruence relation of L and let I = [ 0 ] 0 be the ideal kernel of 0. Let a, b c L and a = b (0). Let c be the complement of ar\b in [0, avb]. Then

c=cr\(avb)~cr\(ar\b) = O (@),

hence c €1. Thus a = b (0) iff (ar\b)vi = avb for some i c I . By Theorem S(iv), this shows t,hat I is standard and 0 = @ [ I ] .

COROLLARY 11. Let L be a weakly modular sectionally complemented lattice. If L satisfies the Ascending Chain Condition, then all congruences of L are neutral, in fact, of the form 0, where a i s a neutral element. 11.

PROOF. Let 0 be a congruence relation of L. By Theorem 10, 0 =@[I] where I is a standard ideal of L. By the Ascending Chain Condition, I = (a]. By Corollary 3, a is standard. Since L is weakly modular, by Theorem 2.6, a is neutral. Hence 0 = 0, with a neutral.

Exercises

1. 2. 3. 4.

6. 6.

7.

8.

9. 10.

*ll .

12. 13.

14. 16.

16.

Prove that an ideal generated by a set of distributive elements is distributive. Show that an ideal generated by a set of standard elements is standard. Do the analogues of Exercises 1 and 2 hold for neutral ideals ? Verify that the converse of Exercise 2 does not hold. (Hint : Consider the lattice of Figure 1.)

0 0

0

Figure 1

Prove Corollary 3 directly. Consider the lattice L as a sublattice of I ( L ) under the natural embedding r -+(r]. Show that every congruence relation of L can be extended to I ( L ) . Characterize those congruence relations of I ( L ) which are extensions of congruences of L. For any ideal I of a lattice L relate the congruence relation @ [ I ] of L with the congruence relation 01 of I (L) . Characterize a standard ideal I of L in terms of the congruence relation 01 on I ( L ) . Show that in Theorem 6 it is sufficient to assume that condition (iii) holds for principal ideals. Construct a lattice L and an ideal I of L such that Theorem S(v) holds for all principal ideals J and K yet I is not standard (Iqbalunissa [1966]). Show that in Theorem 6(v) end Theorem 6(iv) we can assume that J 3 K. Show that the congruence relation 0 and Q, of the lattice L permute iff, for all a,b,cELwitha<b <c,a=b(@),and b=c(@), thereexistsadELsatisfyinga<d I c , a E d (@), and d G C (0). Use Exercise 13 to reprove Theorem 9. Show that I -.@[I] is an isomorphism between the lattice of standard ideals and the lattice of standard congruence relations. Construct a lattice L and standard ideals Ij, j E J , of L such that 1 = A ( I ~ I j EJ) 4 0 and I is not a standard ideal. (Hint: Consider the lattice of Figure 2.)

4. Structure Theorems 161

\ -. \

Figure 2

17.

18.

Let L be a lattice and let I and J be ideals of L. If I is standard and I A J , I v J are principal, then J is principal. For a lattice L and standard ideal I of L, let L/I denote the quotient lattice L/O[I ] . Verify the First Isomorphism Theorem for Standard Ideals: For any ideal J of L, IAJ is a standard ideal of I and

IvJlI eJ /IAJ.

19. Prove the Second Isomorphism Theorem for Standard Ideals: Let L be a lattice, let I and J be ideals of L, J is standard iff IIJ is standard in LIJ, and in this case

I, and let J be standard. Then I

L/I EZ ( L / J ) / ( I / J ) .

20. State and verify the Second Isomorphism Theorem for Neutral Ideals (J. Hashi- mot0 [1952]).

4. Structure Theorems

Let A and B be bounded lattices and L = A x B. Then a =(l, 0) and b = ( O , 1) are neutral elements, and they are complementary. Conversely, if L is a bounded lattice, a , b L, a and b are complementary elements, and a is neutral, then, by Theo- rem 2.4, the map

y: 5 -+ (XAa, z V U )

embeds L into (a] x [a). Observe that y is also onto: if (a, w) (a] x [a), then zy = ( u , w) for x = u v ( w ~ b ) . Indeed, the first component of zy is (uv(wAb))Aa= ( u A n ) V ( v A a A b ) = 21, since u a , aAb =0 , and a is dually distributive. Similarly, xva=uv(wr\b)va=uv((wva)~(avb))=w, sinceaisdistributive, u l a < w , andavb=l .

THEOREM 1. The direct decompositions of a bounded lattice L into tuo factors are (up to i8omorphism) in one-to-one correspondence with the complemented neutral elements of L.

The proof given above is slightly artificial. Its essence will come out better if we

Let L be a lattice with 0, and let L = A x B. Then both A and B have 0 so we can consider lattices with 0.

set I ={(a, 0) I a €4, J = {( 0, b) I b E B}.

It is easily seen that I and J are ideals of L, a +(a, 0) is an isomorphism of A and I, b +(O, b) is an isomorphism between B and J ,

I A J = { O } , I v J = L.

The last equality holds because (a, b) =(a, O)v(O, b). For the same reason, every ideal K of L has a unique representation of the form

K =I ivJ i , Ii E I , J i E J , I,, J i € I ( & )

(with I, = I n K and Ji = J n K ) and every such pair (I,, J,) occurs (indeed for I< =I,vJ1). Therefore, the map

K - ( K n I , K n J )

is a direct product representation of I(L) and so I and J are neutral ideals by Theorem 2.4.

Conversely, let I and J be complementary neutral ideals of L. Then, for any x EL,

x E (x] = (x]A(IvJ) = ( ( X I n I )v ( (x] n J ) ,

and so x< iv j , i € (x] n I , j E (XI n J . But i< x, j < x, hence x = iv j . So every elernent x of L has a representation of the form x=iv j , i E I , j c J . This representation is unique since if x = ivj, i E I , j E J, then

( X I n I = (ivj] n I = ((i]v(j]) n I = ( i ] v ( ( f n I ) = (i],

and similarly (51 n J = ( j ] .

Since every pair (i, j ) occurs in some representation we obtain that

x 4, j>

is an isomorphism between L and I x J . Generalizing slightly we get the following:

THEOREM 2. correspondence between direct decomposition

Let L be a lattice with 0. There is (up to isomorphism) a one-to-one

L = A, X * * * X Aa-i

and n-tuples of neutral ideals (Io, . . . , In-l) which satisfy

I , A I ~ = O for i + j and Iov- * *vIf i - i =L.

COROLLARY 3. Let L be a lattice with 0. Let L have two direct decompositions:

L=Ao X . * * XAn-l,

L=BoX- - * X Bm-l.

Then L has a direct decomposition

L =Co,o X Co,1 x * x C 0 , n - l x CI.0 x * * * X C1,n-i X * . X.Cm-l,o X * * * Xem-1,n-l

such that

A , z C ~ , , ~ X . .-xC,-,,,, for O l i < n ,

Bi Y Ci,, X * * * X Ci,n-I, for 0<i<m.

u nd

PROOF. with the two decompositions as in Theorem 2. Then

Let ( I o , . . . , I n - 1 ) and (Jo, . . . , J,-J be the neutral ideals associated

( I o ~ J o , I i ~ J O , - - * I ~ - ~ A J o , * * * , IoAJm-1, * * , In-lAJm-1)

is again a sequence of neutral ideals satisfying the conditions of Theorem 2. (We only need Theorem 2.9 to help verify this.) The direct decomposition associated with this sequence will yield the decomposition required.

A lattice L is called directly indecomposable iff L has no representation in the forin L = A x B where both A and B have more than one element.

COROLLARY 4. Let L be a lattice with 0. If L has a representation

L = A, x * * X An -1,

rohere each Ai, 0 < i < n, i s directly indecomposable, then for any other decomposition

L = BOX*. * X Bm-i

of L into directly indecomposable factors we have that n = m and that there e&ts a permutation a of (0, . . . , n - l} such that Ai B,, for all 0 2 i < n.

Results analogous to Corollaries 3 and 4 also hold for general lattices, see the exercises.

In trying to sharpen Corollary 4 to a structure theorem there are two difficulties we have to overcome. A lattice need not have a decomposition into directly indecomposable factors. Directly indecomposable lattices are hard.to accept as " building blocks "; one would rather have simple lattices, that is, lattices with only the two trivial congruences,w and L.


The first difficulty is easy to overcome by chain conditions. Observe that if A and B are lattices with 0, C = { O , ai, a2, . . . , an} is a chain in A, and D = { O , b,, . . . , b,} is a chain in B, then

= { ( O J O), (ai, O), * * * > (an, O ) , ‘ I ) , ‘ ’ ’ > ( a f i > b,)}

is a chain in A x B of length n +m. This easily implies that l ( A x B) = l ( A ) + l ( B ) for lattices A and B of finite length.

LEMMA 5. Let L be a bounded lattice. If L is of finite length, then L is isomorphic to a direct product of directly &decomposable lattices.

The passage from ‘I directly indecomposable ” to ‘ I simple ” requires much stronger hypotheses.

THEOREM 6. Let L be a sectionally complemented weakly modular lattice. If L is of finite length, then L can be represented as a direct product of simple lattices.

PROOF. First observe that if L z A x B and L satisfies the hypotheses of this theorem then so do A and B ; only the statement about weak modularity has to be verified and it follows from the observation that

(q, a2)/( b ~ , b2) xW (ci J c2)/(4, d2) implies that qlb l xw Gildl.

By Lemma 5, L 2 Lt X * X L, where all Li are directly indecomposable. By the observation above, all Li srttisfy the hypotheses of this theorem. So it is sufficient to prove that if L satisfies the hypotheses of the theorem and L is directly inde- composrtble, then L is simple. Indeed, if 0 is a congruence relation of L, then 0 = @[Z] where I is a standard ideal by Theorem 3.10. By the chain condition, I.= (a], and a is standard by Corollary 3.3. By Theorem 2.6 a is neutral. Since L is complemented, by Theorem 1 (a] is a direct factor of L. But L is directly indecomposable, hence a = O or a = 1, yielding 0 =w or 0 = I , verifying that L is simple.

Since modular lattices and relatively complemented lattices are special cases of weakly modular lattices we obtain two famous structure theorems at3 special cases:

COROLLARY 7 (The Birkhoff-Menger Theorem, G. Birkhoff [1935b] and K. Menger [1936]). Let L be a complemented modular lattice. If L is of finite length, then L is isomorphic to a direct product of simple lattices,

We shall see in Chapter IV that these simple lattices are exactly g2 and the nondegenerate projective geometries of finite dimension.

COROLLARY 8 If L i s of finite length, then L is isomorphic to a direct product of simple lattices.

(R. P. Dilworth [1950]). Let L be a relatively complemented lattice.

If L is a direct product of finitely many simple lattices Li, . . . , L,, then by Theo- rem 1.3.13

C(L) s C(LJ x * * * x C(L,) z ( Q k ,

that is, C(L) is a Boolean lattice. This led G. Birkhoff to suggest (in G. Birkhoff [1948]) the study of lattices L for which C(L) is Boolean. We are going to present a solution to this problem.

Let us call a congruence relation 0 of a lattice L separable iff, for all a, b EL, a < b, there exists a sequence a =xo <x, < * - - <x, = b such that for each 0 1 i < n , x.=x. I - , (0) or u = w ( 0 ) for no proper subquotient u/v of xi+JxP

THEOREM 9 (G. GriCtzer and E. T. Schmidt [1958d]). For a lattice L, C(L) i s Booleait iff L is weakly niodular and all congruences of L are separable.

PROOF. Let C ( L ) be Boolean. Let

a/b=., cld, a , b, c, d E L, a + b ;

we wish to find a proper subquotient c’ld’ of c/o? satisfying c’/d’xW a/b. Let 0 =@(a, b) and let 0’ be the complement of 0. Then 0v0’ = L and so c G d (Ova’). By Theorem 1.3.9, there is a sequence d =xo <xi < * - - < x, = c such that, for each 0 < i < n ,

x.=x. $ - or X~=X~+~(@’).

If, for all O < i < n , X~=X~+~(@’), then c ~ d (0’); since wlb=.,c/d, a ~ b (0’). But 8 =@(a , a) and so n b (0). We conclude that a = b (@A@’), that is, a = b (w), contradicting u =k b. Therefore, xi = xi+l (0) for some 0 I i < n . Applying Theorem 1.2 to xi=xifI (@(a, b ) ) , we obtain a proper subquotient c’/d‘ of satisfying c ‘ / d ’ = , u/b, proving that L is weakly modular.

Now let 0 be a congruence relation of L and let a , b E L, a < b. Then a G b ( O V ~ ’ ) , where 0’ is the complement of 0, and so by Theorem 1.3.9 there exists a sequence u =xo <.r, <- - - <x, = b such that, for each O I i < n ,

x.=x. $ - 2 + 1 (0) or xi=xi+i (0’).

We clsini that the saine sequence establishes the separability of 0. Indeed, if for some i , ~ ~ + x & + ~ (a), and u,wE[x~,x~+~],u=w (a), then xi=xi+l (0’) and so u ~ v ( ( O ’ ) , implying that u e w (@A@’), that is u = w.

To prove the converse, let us start out with a weakly modular lattice L and a congruence relation 0 of L. We claim:

The binary relation Q, defined on L by

u = b (0) iff ZL = w (0) for no proper subquotient u/w of avb/aAb

is a congruence relation; in fact, CD is the pseudocoinplement of 0. We prove the first part of this claim by verifying that 0 satisfies conditions (i)-(iii)

of Lemma 1.3.8. Condition (i) is clear by the definition of 0. To verify (ii), let a , b,

c E L, a < b 2 c, u E b (a), and b = c (0) ; we wish to show that a = c (0). Assume to the contrary that a + c ( @ ) ; then U E V (0) for some proper subquotient 21/21 of clu. Since U E V (@(a , c ) ) and @(a, c) =@(a, b ) v 0 ( b , c ) , by Theorem 1.2 and Lemnia 1.3 there exists a proper subquotient x/y of u/v satisfying x/yxw c/b or x/yzw bla, say, xlyw, c/b. Therefore, b < c . By weak modularity, there exists a proper subquotient cl/bi of c/b satisfying c l /b i zw x/y. But x, y E [v, u] and u G v (a), so x = y (0). Hence ci = b l ( 0 ) , contradicting ci, bi E [b, c ] and b = c (0). This shows that (ii) holds.

To verify (iii), assume to the contrary that a, b, c E L, a< b, a = b (a), but U A C + ~ A C (a); then there exists a proper subquotient u/v of bAc/aAc satisfying v = u (0). Then u/v=:, bAc/aAcz,,, bla, hence u / v x w b /a ; so a < b and by weak modularity bl/al=zw U J V , where bJai is a proper subquotient of bla. Since U E V (0) we obtain b, E ui (a), contradicting a = b (0). By duality, uvc = bvc (a).

This shows that 0 is a congruence relation. 0110 =o is trivial. Now if @AY = m , a = b (Y), u, v E [aAb, avb ] , and u E v (a), then also u = v (Y), hence u =v, showing that Y c 0. This completes the proof of the claim.

Now to complete the proof of the theorem, let L be weakly modular and let) the congruences of L be separable. Let 0 be a congruence of L. Let Q, be the congruence constructedin the claim. For a, b E L, a < b, let a =xo < xi < - - < x, = b be the chain establishing the separability of 0. Then for each O < i < n , either x i ~ x i + l (O), or, by the definitions of separability and of 0, s ~ E x ~ + ~ (@). Therefore, c r ~ b (Ova), and so 0 v 0 = I . This shows that 0 is a complement of 0. Since C(L) is distributive (Theorem 11.3.11) this proves that C(L) is Boolean.

We get a large number of corollaries from Theorem 9 yielding that C(L) is Boolean for special classes of lattices. Some of these will be discussed in the exercises.

The centre of a bounded lattice L is the sublattice Cen(L) of compleniented neutral elements. It is obvious that 0, 1 ECen(L) and Cen(L) is a Boolean lattice. In some chapters of lattice theory it is important to know when Cen(L) is a complete lattice. (If this is the case L can often be " coordinatized " over Cen(L).) This was established for continuous geometries by J. von Neumann [1936/37].

Let L be a complete lattice and K E L. Then K is a complete sublattice of L iff AX and VX CK for all X E K (AX and VX are formed in L).

THEOREM 10. Let L be a complete, sectionally complemented, dually sectionully complemented, weakly modular lattice. Then the centre of L is a complete sublattice of L.

PROOF. relation

Let XECen(L). Set u=r \X . Then (a] is the kernel of the congruence

A (0, I .2: E X).

Hence, by Theorem 3.10, (a] is standard; by Corollary 3.3, a is standard. Now Co- rollary 2.8 tells us that a is neutral. Since a is complemented, a E Cen ( L ) .

By duality, we obtain VX ECen(L).

COROLLARY 11 mented lattice i s n complete sublattice.

(M. F. Janowitz [1967]). The centre of a complete relatively comple-

This result has soiiie interesting implications concerning direct decompositions of complete relatively conipleiriented lattices; see S. Maeda [1966].

Exercises

1.

2.

3. 4.

6.

6.

7. 8. 9.

10.

11.

12.

13.

14.

15.

16.

17.

Show that the representations of a lattice L with 0 as a direct product of two lattices are (up to isomorphism) in one-to-one correspondence with pairs of ideals (I, J) satisfying I n J = (0) and every element a of L has exactly one representation of the form a = i v j , i €1, j E J. Let L = A , x A 2 . Define the binary rulation Oi on L by (al , a,)=@*, b,) (0i) iff ai=bi (i =1, 2 ) . Show that 0 1 and 02 are congruence relations of L, 0 1 ~ 0 2 = w , and Olv02 =L. Show that 0, and O2 of Exercise 2 are permutable. Prove that the representat,ions of a lattice L as a direct product of two lattices are (up to isomorphism) in a one-t,o-one correspondence with pairs of congruence relations (01, 0,) which are complementary and permutable. Show that if in Exercise 4 we pass from two to n factors, then we get an n-tuple of congruences (a,, . . . ,on) such that @'A* * - ~0~ = w , (@,A. - A @ ; - I ) V @ ~ = ~ ,

and @,A. * . A @ ~ - I and 0i permute, for i =2, . . . , n. Can we replace in Exercise 5 the condition " @,A* m ~ 0 i - i and 0i permute " by Use Exercise 5 to verify Corollary 3 for arbitrary lattices. Verify Corollary 4 for arbitrary lat,tices. Relate Exercise 5 to Theorem 2. Let B be the Boolean lattice R-generated by the rational interval [0, 11. Show that, for any natural number n, B has a representation as a direct product of n lattices L,, . . . , L,, ILil > 1 for i = 1, . , , , n, but B has no representation as a direct product of directly indecomposable lattices. Construct a lattice which is not of finite lengt'h but every chain in the lattice is finite. Statements 5-8 of this section deal with lattices of finite length. Which of these statements remain valid for lattices in which every chain is finite? Prove that every congruence relation of a finite lattice or of a lattice of finite length is separable.' Verify that if in a lattice L, for every a, b EL, a < b, there is a finite maximal chain in [a, b], then all congruence relations of L are separable. This holds, in particular, if the lattice is locally finite, t>hat is, if all intervals are finite. Let L be a distributive lattice, a, b, al , a,, . . . EL, a =al < a, < a3 < * - - < b. Then 0 =V(O(a2i- i , ati) I i = 1, 2, . . .) is not a separable congruence relation. For a distributive lattice L, C ( L ) is Boolean iff L is locally finite. (Use Exercises 14 and 16. J. Hashimoto [1952].) The separable congruences of a lattice L form a sublattice of C(L) .

any two distinct Oi and Oj permute " 1

1 Exercises 13-19 are based on G. Griitzer and E. T. Schmidt [1958d].

18.

19.

20.

Let L be a lattice with 1. A neutral congruence reletion @ [ I ] is separable iff I is principal. Let L be a complemented modular lattice. Then C(L) is Boolean iff all neutral ideals of L are principal (Shih-chiang Wang [1963]). Find a complete lattice L and a sublattice K of L such that K is a complete lattice but not a complete sublattice of L.


The properties of weak projectivity and of the congruences @[I] are discussed in greater detail in G. Gratzer and E. T. Schmidt [1958d]. For universal algebras A. I. Mal'cev introduced similar concepts; the difference is that while for lattices it is sufficient to consider " unary algebraic functions " of a special form (namely, . . . (( (2AaO)Vai)Aa2) - * .), for universal algebras in general we have to consider arbitrary unary algebraic functions (see, for instance G. Griitzer [1968]). The polynomial p 2 is also of special interest ; identities of pz that hold for lattices imply that the congruence lattices of lattices are distributive. A general condition for the distributivity of congruence lattices of algebras in a given equational class of algebras can be found in B. Jbnsson [1967].

Weak modularity is a rather complicated condition. It can be somewhat simplified for finite lattices: a finite lattice is weakly modular iff a l b x , cJd and a > b 2 c > d imply the existence of aproper subquotient c f / d f of cfd satisfying cp /d 'Xw alb (G. Gratzer [1963a]).

In general, if in a lattice L any interval contains a finite maximal chain, then it is sufficient to consider weak projectivities of prime quotients, see for instance N. Fu- nayama [1942], J. Jakubik [1956a] and D. T. Finkbeiner [1960]. Under this condition, L is simple iff any pair of prime quotients are projective. In general, no bound can be put on the number of perspectivities. However, if L is a relatively complemen-

ted lattice of length n, a, b >. 0 and a10 x b/O, then a/O x b/O, where 12 < 2 - (n + 1) , see J. E. McLaughlin [1951] and [1953].

The investigation of the number of projectivities becomes important in the study of equational classes of lattices, see qection V.3 and C. Herrmann 119731.

Local conditions implying the projectivity of any pair of prime quotients can be found in F. A. Smith [1974].

Modular lattices and relatively complemented lattices share a property stronger than weak modularity. In P. Crawley and It. P. Dilworth [1973] a lattice 'is said to have the Projectivity Property iff whenever a /b is weakly projective into cld, then alb is projective to some subquotient c'ld' of cld.

Weak modularity seems to be a very natural concept. It appears quite surprisingly in some results. The following result of Iqbalunissa [1966] is a good illustration of this point: if in the lattice L every congruence relation is neutral, then L is weakly niodular.

A lot more information on standard elements and ideals can be found in G. Griitzer [1959] and G. Griitzer and E. T. Schmidt [1961]. Among the topics discussed in

1: 1 k

Problems 169

these papers are the Isomorphism Theorems, the Zassenhaus Lemma, the Jordan- Holder Theorem for standard ideals, and the Schreier extension problem. It is also shown that in a finite modular lattice exactly the neutral ideals satisfy the First Isomorphism Theorem. Some of these results were inspired by J. Hashimoto [1952].

In the proof of Theorem 4.9 we describe the pseudocomplement 0 of a congruence relation 0 of a weakly modular lattice. Iqbalunissa [1966] observes that the converse also holds: if in a lattice L the relation 0 given by any congruence relation 0 (as described in the proof of Theorem 4.9) is always a congruence relation, then L is weakly modular.

For some additional results on standard ideals see M. F. Janowitz [1964], [1964a], and [1965], in which some types of ideals, more general than standard ideals, are discussed. Congruences of relatively complemented lattices, in particular a description of @(a, b), is given in M. F. Janowitz [1968]. Some counterexamples are given in Iqbalunissa [1965] and [1965a].

Lattices whose congruences form a Boolean lattice are discussed also in T. Tanaka [1952], J. Hashinioto [1957], P. Crawley [1960]. See also D. T. Finkbeiner [1960] and J. Jakubik [1955]. Lattices whose congruences form a Stone lattice are studied in M. F. Janowitz [1968] and [1968a], and Iqbalunissa, [1971].

Standard elements in the lattice of all subsemigroups of a group are studied in S. G. Ivanov [1966].

It is pointed out in S. Maeda [1974], that if L is the dual of the lattice of ell Ti- topologies on an infinite set, then L has infinitely many standard elements, but only the elements 0 and 1 are neutral.

The concept of a standard ideal has recently been extended to convex sublattices in E. Fried and E. T. Schmidt [1976].

Distributivity of a pair of lattice elements is investigated in P. G. KontoroviE, S. G . Ivanov, and G. P. Kondrajov [1965]; see also F. and S. Maeda [1970]. We shall consider modular pairs of elements in Section IV.2.

A great deal of useful information on congruence relations of lattices, in particular about regularity and permutability, can be found in J. Hashimoto [1963].

There is a connection between projectivity and representability. If L/@ is projective (in a class containing L), then @ is representable. A. Day [1973] considers a weaker concept implying the representability of @ for finite L.

Problems

111. 1. Generalize the concepts of distributive and neutral ideals to convex sublattices. (For standard convex sublattices, see E. Fried and E. T. Schmidt [1975].)

111.2. Let L =Io 2 I , 2 * * * 3 I,, =I be a descending sequence of ideals. If Ii+i is a standard ideal of I j for j = 0, . . . , n, then I iEc a standard ideal of order n of L. Investigate standard ideals of order 2 (order n). Under what conditions do they forni a sublattice?

160 111. Coiigruences and Ideals

111. 3. An ideal I of a lattice L is said to satisfy the First Isomorphism Theorem (G. Gratzer and E. T. Schmidt [19&3d]) iff (IvJ)/@[I] s J / @ [ I n J ] for any ideal J of L (under the natural isomorphism). Investigate this concept and relate it to standard ideals.

111.4. Under what conditions do the ideals of a weakly modular lattice form a weakly modular lattice?

111.6. Investigate lattices whose congruences form a Stone lattice. 111.6. For n > 1, investigate lattices whose congruence lattices, as distributive 1a.ttices

111.7. Develop structure theorems for lattices all of whose congruences are stand- with pseudocomplementation, belong to B,.

ard (distributive, neutral).

CHAPTER Iv

MODULAR A N D SEMIMODULAR LATTICES

1. Modular Lattices

The modular identity is unquestionably the most important identity apart from the distributive ident,ity. In this section we examine the most important consequences, of modularity.

THEOREM 1. For a lattice L, the following conditions are equivalent: (i) L i s modular, that is,

x 2 z implies that Z A ( ~ V Z ) = (XAY)VZ.

(ii) L satisfies the shearing identity 1:

ZA(YVZ) =ZA((YA(ZVZ))VZ).

(iii) L does not contain a pentagon.

REMARK. In Section 11.1 we have already proved the equivalence of (i) and (iii). The importance, or convenience, of the shearing identity is that it can be applied to any expressions of the form S A ( ~ V Z ) without any assumptions. Observe also the dual of the shearing identity:

xV(y AZ) = Z V ((yV(ZA2)) AZ).

PFCOOF. (i) implies (ii). Since xvz > z , by modultwity, ( ~ A ( X V Z ) ) V Z = ( y v z ) ~ ( z V z ) , and so

SA( (~A(ZVZ))VZ) = ZA( (~VZ)A(ZVZ) ) = ~ ~ ( y v z ) .

(ii) implies (iii). In 'ill5 (ii) fails; indeed,

a ~ ( c v b ) =aAi=a and ~ A ( ( c / \ ( ~ v b ) ) V b ) =aA((cAa.)Vb) =aAb = b .

Thus (ii) implies (iii). 0 (iii) iniplies (i) as in Section 11.1.

The most iinportant form of modularity is the following:

1 This identity was named by I. Halperin.

162

THEOREM 2 a, b c L . Then

IV. Modular and Semimodular Lattices

Let L be a modular lattice and let (The Isomorphism Theorem).

yb: z +xAb

is an isomorphism of [a, avb] and [a/\b, b]. The inverse isomorphism is

ya: x +xva. (See Figure 1 .)

Figure 1

PROOF. It is sufficient to show that, for x E [a, avb], x(pby)a =x. Indeed, if this is true, then by duality, x y # b = x for all x E [aAb, b]. The isotone maps yb and ya thus compose into the identity maps, hence they are isomorphisms, as claimed.

So let x E [a, avb]. Thus xybya = (xr\b)va. Since x E [a, avb], we have a I: x and so modularity applies :

x%ya = (xAb)va =xA(bva) =x ,

because x < avb.

COROLLARY 3. Let L be a modular lattice and let a, b € L. Then for x , y € [ u ~ b , b] we have

a v ( x ~ y ) = (aVx)A(aVy).

Most applications of the Isomorphism Theorem are like Corollary 3: the special form of the isomorphism is used. There are two important applications where the form of the isomorphism plays no role.

Let us write a . a) for a - - b implies that avc -< - bvc, for a, b, c E L. The Lower Covering Condition is the dual.

THEOREM 4. Condition.

A modular lattice satisfies both the Upper und the Lower Covering

1. Modular Lattices 163

x, A-%

Figure 2

PROOF. Let L be a modular lattice. Let a, b, c EL and a< b. If avc = bvc we have nothing to prove. If avc+bvc, then a $ bvc, and so aA(bvc) = b. Applying the Isomorphism Theorem to a and bvc we obtain

[b , a] 2 [bvc, avc].

Since [b, a] is a prime interval, so is [bvc, avc], that is, bvc < avc. By duality, we get the Lower Covering Condition.

The second application deals with representations of elements. I n Corollary 11.1.13 we proved that in a finite distributive lattice the irredundant representation of en element as a join of join-irreducible elements is unique. This is obviously false in m3. But we have the following result :

THEOREM 5 (The Kurok- Ore Theorem, A. G. Kuroi3 [1935], 0. Ore [1936]). Let L be u modular lattice and let a E L. If a =xov* * -vxnn-l and a = yov- * *vy,-l are irredundant representations of a as joins of join-irreducible elements, then for every xi there i s a yi such that

a =xov* * - V X ~ - ~ V ~ ~ V X ~ + ~ V - - * V X , - ~

u.nd n = m.

PROOF. Figure 2.) Since yov* *vy,-i =a we obtain that

Let us prove the first statement say for xo. Let xg =xIv* - -vxn:n-i. (See

(.gvy,,)v(x$JYl)v. - .V(XEVYm-i) =a,

where x6vy0. . . . , E [xg, a]. By the Isomorphism Theorem,

[xg, 01 [XOAL& % I , 12 Griltzer

164 IV. Modular and Semimodular Lattices

and the image of a under any such isomorphism is xo. But xo is join-irreducible in L, and, therefore, in [ X ~ A X ; , xo]; thus a is join-irreducible in [x;, a] . Hence xivy; =a for some j , proving the first statement.

Now let a=zov* * *vzk-l be an irredundant representation of a as join of join- irreducibles with k minimal. Applying the statement just proved to a = z0v- * - V Z ~ - ~ ,

zo, and a =xov- - -vx,-l, we obtain that a=xi0vziv* * V Z ~ - ~ . This representation is irredundant, otherwise the minimality of ii would be contradicted. Repeating this we eventually obtain a=x . v* a However, a=xov- * * V X , - ~ is an irredundant representation and so {jo, . . . , j k - l } = { O , . . . , n- l}. This shows that n < k . Thus k = n (=m).

Weak projectivities in modular lattices can be described rather precisely. Let us call a sequence of perspectivities:

90

~ 0 1 ~ 0 N Xi/yi N. * * ‘ ~ x n l y n

alternating iff, for each i with 0 <i <n, either xi-llyi-l p xilyi L xi+llyi+l or x$-l/yi-l 4 x i /y ip xi+i/yi+i. If, in addition, xi =xi-ivxi+l in the first case and yi= yi-lAyi+i in the second, we call the sequence normul. (See Figure 3.)

Figure 3

THEOREM 6 (G. Griitzer [1966]). Let L be a modular lattice, and let alb and c/d be quotients of L. Then a/b xw cld iff there eqists a normal sequence of perspectivities n/b = xo/yo - xi/yi - * a * - x,/y, where x,/y, is a subquotient of c/d.

~ O O F . It follows from the Isomorphism Theorem that if x / y p u / v and xi/yi is a subquotient of xly, then xilyi 7 ui/vl where u, = ylvv and oi = xlvv. This, tho dual statement, and a simple induction show that if a/b xw c /d , then there is an alternating sequence of perspectivities

aJb=xo/yo-. * .-x,/y,

where xn/yn is a subquotient of c/d. Starting with this sequence we show by induction on n that there is a normal sequence of perspectivities

( I Jb = uo/Vo - U I I V ~ - * * ‘ - tI,/V, = x,Jy,.

This statement is obvious for n I 1. By duality, we canassume that un-l/vn-i \ u,Iv,. By the induction hypothesis, there is a normal sequence

ulb = uol~o N * * Nun - llvn -1 =xn - l lyn -1.

We define a new sequence as follows:

alb = U O I ~ O ~ ~ ~ I ~ ~ - * * L un-21vn-2 7 u n - 2 ~ x n l ~ n - 1 ~ ( ~ , - 2 v ~ n ) L XnIy,.

We claim that this is a normal sequence of perspectivities. The following formulas have to be checked; (i) and (ii) are the perspectivities, and (iii) the normality:

(i) un -21vn -2 7 U, -zvxJvn - 1~ (un - 2 ~ 3 , ) ; (ii) u , - ~ v x ~ / v ~ - ~ A ( u ~ - ~ v x ~ , ) L XnlYn;

Since Vn - 2 V, - 1 A(Un -2VXn) 5 V n - 1,

... (111) vn -2 =vn - ~ A ( v , - 1 ~ ( u n - 2 ~ 2 % ) ) *

Un -2A(Vn - IA(U,-~VX,)) = Vn-2

is obvious. Now compute :

u,-~v(v, - l ~ ( ~ n - 2vzn)) (by modularity and u, - 2 I u,n -2vxn)

= (en -2vvn- 1)A(Un - 2 ~ 5 , )

- un -2VXn 9 -

verifying (i). The proof of (ii) is similar. For (iii), observe that v,,-~< u ~ - ~ v x , , end V , - ~ A V , - ~ = v , - ~ , hence

V, - 3 A (Vn - i A ( U n -2VXn) ) = ( V n - 1)A(U, -2VXn) = vn -2'

This completes the induction step.

three consecutive quotients. We can do even more than this; we can determine the sublattices generated by

THEOREM 7 (0. Gratzer [1966]). Let L be a modular lattice and let xo/yo f xl/yi \ x2/y2 be u normu1 sequence of quotients. Let A be the sublattice of L generated by xo, x i , x2, yo, y,, and ys. Then either A i s distributive, in which case A is the lattice of Figure 4 or some quotient tliereof and we also have

X d Y O L ( X o A X 2 ) 1 ( Y o ~ ! h ) f X21Y2,

or A is. not distributive and A i s the lattice of Figure 5 or some quotient not collapsin the diurtiond in A .

PROOF. A is generated by the elenientsxo, y l , and x2, hence A is isomorphic to a quotient lattice of P,(3), see Figure 1.5.7. The generators satisfy the relation xovyl = y,vx2 =xOvx2 (the last one conies in because of normality). The corresponding quotient lattice is given in Figure 5. A quotient of this is distributive iff the diamond is collapsed, yielding Figure 4. 12'

4

Yo Yz

Figure 4 Figure 5

COROLLABY 8. Let L be a modular lattice and let alb 2 c/d in L. There is a shortest normal sequence of quotients alb =xolyo -xJy, - * * -x,/y, =cld. Then either n g 2 , or for each i, 0 2, then the sublattice generated by xi-l, xi, xi+l, yi-l, yi, yi+l cannot be distributive, otherwise we could turn the down arrow up and the up arrow down and get a sequence of length n - 1. Thus these elements generate the lattice of Figure 5, or its dual, or a quotient lattice of Figure 6, or its dual. The only nontrivial quotient lattice of the lattice of Figure 6 is given by Figure 6.

As a typical example of the advantages of using the shearing identity we consider independence in the sense of J. von Neumann [1936/37]. This plays a very important role in the applications of lattice theory to direct decompositions of groups and rings and also in continuous geometries.

Figure 6

DEFINITION 9. Let L be a lattice with 0. A subset I of L - {O} is called independent i f f for any two finite subsets X , Y of I we have

V X A V Y = V ( X n Y) .

COROLLARY 10. A subset I of a lattice L i s independent i f f

q:x-vx is a n isomorphisnt between [I] and the generalized Boolean lattice of all finite subset8 of I .

PROOF. V X v V Y = V ( X u Y ) and so if I is independent then q is a homomorphism. If q is not one-to-one, then V X = V Y for some X + Y , X , Y E I , and X , Y finite. Let, say, X

a = a A v Y = V { a } A v Y = v ({a} n Y ) = V 0 = 0,

Y and a E X - Y . Then a< V Y , a ( Y . Therefore,

a contradiction. Thus q is an isomorphism. The converse is obvious. A singleton {a} ( a EL -{O}) is always independent. {a, b} is independent iff aAb =O

(a, bcL-{O} and a+b). For {a , b, c} to be independent (a, b, cEL-{O} and a, b, c all distinct) we have to require that

m ( b v c ) =0, bA(avc) =0, cA(avb) = O

(avb)A(avc) = a , (bvc)A(bva) = b, (cva)A(cvb) =c.

For modular lattices, fewer relations will do:

THEOREM 11. L - { O } is independent i f f

Let L be a niodular lattice with 0. Then an n element set {ai, . . . , a,} E

(aiv.**Vai)Aai+l=O, for i = 1 , 2 , . . . ,n-1.

PROOF. We obtain the necessity of the condition by setting X ={ai,.. . , ad}, Y ={ai+,}. Now assume that {ai, . . . , a,,} satisfies the condition. Let X, Y c { a I , . . . , a,fi}, X n Y = 0. We claim that

V X A V Y=o.

Indeed, let a k c X u Y with k maximal. Let, say, akc Y. Apply the shearing identity to V X A V Y with s = V X , y = a k , a n d z = V ( Y - { a k } ) :

vxAv y = v X A ( a k v V ( -{ak)))

= V X A ( ( a k A ( V X V V (y-(uk})))vv ( y - { a k } ) )

= v X A v ( y - { a k h

since a,r\(V X v v ( Y -{ak})) (alv- - 'Vak-I)Aak = O .

Proceeding thus we can eliminate all the ui belonging to X u Y , getting V X A V Y = V 0 = O . Now in the general case,

V X A V Y = V X A ( V ( X n Y ) v V ( Y - X ) )

= V ( X n Y ) v ( V X A V ( Y - X ) ) (by X n ( Y - X ) = 0) = V ( X n Y ) .

(by modularity)

We conclude this section with three important I ‘ sublattice theorems ”.

THEOREM 12 (J. von Neumann [1936/37]). Let L be a modular lattice and let a, b, cE L. The sublattice of L generated by a, b, and c is distributive iff a ~ ( b v c ) = (aAb)v(uAc).

PROOF. By inspection of the diagram of P,(3) (see Figure 1.6.7). If a, b, and c are the generators and aA(bvc) = (aAb)v(aAc) (0) where 0 is a congruence relation, then 0 collapses the only diamond and so P,(3) /0 is distributive.

One can view the definition of modularity as requiring that any sublattice generated by three elements, two of which ere comparable, has to be distributive. This is true in general for any two chains:

THEOREM 13 (G. Birkhoff [1940]). Let L be a m d u l a r lattice and let Co and Ci be chains i n L. The sublattice of L generated by Co u C, is distributive.

PROOF. Since a lattice is distributive iff every finitely generated sublattice is distributive, it is sufficient to verify this result for finite Co and Ci. Let Co ={ao, . . . , aln- i } andCi={bo,. . . , b n - ~ } , a o < . * * 4 2 ) > - - - > a(r ) 2 1 and =(/?(1), . . . , p( r ) ) , 1 </?(I) <p(2) < - * * </?(r) n. Let A denote the set of ell elements of L of the form x(r, ct, B). Then Co, Ci G A, since ai=aiAbn and bi =u,Abi. It is also easy to see that A is closed under join: if we form x(r , a, /?)vx(s, a’,/?’), then if we have both a,Abj and ajAbk, we can eliminate one by absorption; if ai < ai, b, < b, end we have aiAbn and apb , , the former is eliminated by absorption; what is left can be written in the form z(r’, ct”, /?”).

Now we prove by induction the formula

~ ( ‘ 9 Q, B) =au(i)A(bg(i)vau(~))A’ * .A(bp(r-i)vaa(r))AbS(,,

and its dual. The case r = 1 is obvious. Let us assume that this formula and its dual have been verified for r - 1. Now compute:

w, a, /?) = (aa ( i )~bp( i ) ) v* * *v(~a( l )Ab&¶( , ) ) =

(observe a,(,) 2 ( u , ( ~ , A ~ ~ ( ~ , ) v - * * v ( u , ( ~ ) A ~ ~ ~ ~ ) ) and apply modularity)

= a=(1)~('~(1)~(aQ(2)~bs(e))v. . *V(a=(r)Abb(r)))

(observe bS(,.) 2 bscl,v. * and apply modularity)

=~=(~)A('S(I)V(~Q(~)A'B(~))~. * .V(a=(,-l)A'8(,-1))va=(r))A'g(r)

(apply to the expression in parentheses the dual of the formula for r - 1 )

= ~ Q ( ~ ) A ( ' s ( ~ ) v ~ Q ( z ) ) A ( ~ S O ~ ~ , ( ~ ) ) A . * .A(bscr- 1)Va=cr))A'p(r),

completing the proof. Now we easily see that A is a sublattice. Indeed, x ( r , a , / ? ) ~ x ( s , a ' , p ' ) can be

rewritken by the above formula as a meet of b,va j ; but by the dual of the above formula the result can again be put in the form x(r , a, p ) .

SO we conclude that A is the sublattice generated by Co u Ci and therefore ]A1 <the number of expressions of the form x ( r , a, /?).

Now consider the set X = [ O , n + l ] X [ O , m + l ] , let F be the sublattice of P(X) generated by

a i = { ( x , y) I y < i + l } , for i = O , I , . . . , m - I ,

bj={(z,y) I x < i + l } , for i = O , I , . . . , n - 1 ,

and let Co = {ao, . , . , a,,,-.l} and C, = {bo, . . . , bn-i} . I n this lattice

z ( r , a , / ? ) = U ( { ( x , ~ ) l ~ ~ a ( i ) + l , ~ < ~ ( i ) + l ) } l i = l , 2 , ~ . . J ~ )

and so all these represent different elements. Consequently, F is the free modular lattice generated by Co u C,. Since F is distributive, any modular lattice generated by C, u C, is distributive.

Finally, we generalize Theorem 11.1.6 to modular lattices:

THEOREM 14. generated by [ u A ~ , a] u [ a ~ b , b] i s isomorphic to [ u A ~ , a ] X [ a h , b] .

Let L be a modular lattice and let a , bcL . Then the sublattice of L

REMARK. In the distributive case [ [ a ~ b , a ] u [ a ~ b , b ] ] = [ a ~ b , avb]; this does not hold for modular lattices as exemplified by B3.

PROOF. The isomorphism we set up is

Q, : (x, y) -+xvy, for x E [ a h , a] and y C [ a h , b] .

Using the formula for the dual of x(r , a, p ) in the proof of Theorem 13 we obtain:

XVy = (aVy)A(bVX).

Hence U A ( X V ~ ) = U A ( U V ~ ) A ( ~ V X ) = u A ( ~ v ~ ) = ( a ~ b ) ~ = x

and bA(xvy) =y, proving that cp is one-to-one. rp is obviously onto and preserves join. Let 2, q E [a& a] and y, yi E [ a d , b]. Then

(~VY)N~,VYl) = (avY)A(bvr)A(avY,)A(bvXl)

= (av (Y AYi )) A (bv (XAX, )) = (%A%, 1 v (Y AYl) , = (UVY)A(UVY, )A(~VZ)A(~V~~) (use Corollary 3)

proving that cp is an isomorphism.

Exercises

1.

2.

3.

4. 6.

6.

7. 8. 9.

10.

11.

12.

13. 14. 16.

16. 17. 18.

19.

Show that a lattice L is modular iff it satisfies the identity

( z V ( Y ~ z ) ) ~ ( Y V z ) = ( z h ( Y V z ) ) V ( Y h z ) .

Let p = (ZVY)A(YVZ)A(ZVZ) , do =avbvc, and define, for n 2 0, d,+l =p(avd,, bhd,,, C A ~ , , ) . Show that the identity di =d2 is equivalent to modularity but d2 =d, is not (B. Wolk). A finite lattice L is modular iff i t does not contain a pentagon (0, a, b, c, i } satisfying a t b. Can the numbers of covering pairs in Exercise 3 be increased? Let L be a lattice and a, b EL. Then pb: z +xhb maps [a, avb] into [ahb, b], and

Using the notation of Exercise 6, prove that {z I x E[ahb, b], zvapb =z} is isomorphic t o {z ] zE[a, avb], XTbva =z}. (Exercises 6 and 6 are due t o W. Schwan [1948], [1948a], [1949], and [1949al.) Does Corollary 3 characterize the modularity of a lattice? Find a nonmodular lattice satisfying the Upper and Lower Covering Conditions. Show that the Kuroi-Ore Theorem fails in the first lattice of Figure 2.1. Prove that in the Kuroi-Ore Theorem the representation a =zov- - .vz i - ivyp Z ~ + I V * * wzn-l is always irredundant. Using the notation of the Kuroi-Ore Theorem, show that there is a permutation n of (0, . . . , rn - I} such that a =zov . *vzi-lvyi,pzi+lv- * ~ z , - ~ holds for all i, 0 _< i < ra (R. P. Dilworth [1946]). Can the Kurog-Ore Theorem be sharpened so as to require that all zi can be simul- taneously replaced by yr? (No, inspect Figure 3.4b.) Does Theorem 0 characterize modularity? Prove Theorem 7 directly, that is, without reference to p ~ ( 3 ) . Show that for finite modular lattices i t is sufficient to consider projectivities of prime intervals. Simplify Theorems 0 and 7 for prime intervals. Extend Exercise 16 to locally finite lattices. Show that Exercise 16 does not generalize to nonmodular lattices. Show that to define independence in modular lattices i t is sufficient to assume that VXAVY = 0 for X n Y = 0 , X , Y c I , X and Y finite. Let L be a complete lattice and let I be a set of compact elements of L. Then I i s independent iff X + v X is an isomorphism between the Boolean lattice of all subsets of I and the complete sublattice generated by I.

va: X-XVa maps [ahb, b] into [a, avb]. Show that pbvapb=pb and vapbva=va.

Figure 7

20.

21. *22.

23.

24.

25. 26.

27.

28.

29. 30.

31. 32.

33.

34.

Let L be a complemented modular lattice and let a, = O < a , < - - - < a, = 1 be elements of L. Let bi =al , let bz be a relative complement of a, in [0, a2], . . . , let b, be a relative complement of a,-l in [0, a,]. Then B = { b i , . . . , b,} is an independent set. Prove the converse of Exercise 20. Let L be a modular lattice and let C,, Ci, and Cz be chains in L. The sublattice generated by Co u Ci u Cz is distributive iff ah(bvc) = (ahb)v(ahc) for all a ECO, bECi, and cECz (B. J6nsson [1955]). Let K be an equational class of lattices and let P be a poset such that P w ( P ) exists in the sense of Definition 1.6.2. Let A be the class of allfinite lattices L such that L is generated by some homomorphic image of P (in the sense defined in the proof of Theorem 1.5.6). Let us assume that there is an L EA such that IL( 2 IL'I for any L'CA. Is it true that P x ( P ) exists and is finite, in fact, is it true we can take

Prove that in a modular lattice L, the sublattice generated by [ahb, a] u [ahb, b] equals [ahb, avb] for any a, b EL iff L is distributive. Find equations to replace " ah(bvc) =(ahb)v(ahc) " in Theorem 12. Find a direct proof of the statement that in a modular lattice every distpibutive element is neutral. Show that in a distributive lattice a/b=c/d and cld a subquotient of alb imply that alb =c/d. Let C be a bounded chain. Define a subset M of C3 by (x, y, z ) EM iff X A Y =YAZ = ZAZ. Then M as a subposet of C3 is a lattice. For a, PEM, a@ in M is the same as ah@ in C3. Compute avg. Show that the lattice M of Exercise 28 is a modular lattice. Show that in M of Exercise 28, (1, 0, O)/ (O , 0, 0) =(x, 0, O ) / ( O , 0, 0) for any x > 0 provided that C is infinite. (Exercises 28-30 are due to E. T. Schmidt [1962].) Show that the identity d2=d3 of Exercise 2 fails in the lattice of Figure 7 (B. Wolk). Show that a finite modular lattice is dismantlable iff it has breadth two or less (D. Kelly and I. Rival [1974]). Verify that if L is a finite modular lattice satisfying

P K ( P ) =L?

then L has an m element sublattice for all m _< ILJ. (I. Rival [b]. Hint: Use Exercise 1.6.32.) Does the fact that va of Theorem 2 is one-to-one for all a, bCL characterize the modularity of L ?

172 IV. Modular a,nd Semimodular Latttices

2. Semimodular Lattices

A lattice is called semimodular iff it satisfies the Upper Covering Condition, tha,t is,

a -.: b implies that avc < bvc or avc = bvc.

Examples of semimodular lattices include finite modular lattices and some important lattices from geometry (see Section 3). Two further examples are given in Figure 1.

Figure 1

Let L be a lattice with 0. We define the height function: for a EL, let h(a) denote the length of a longest maximal chain in [0, a] if there is a finite longest maximal chain; otherwise put h(a,) = OD. If L is of finite length, 0 1 h(a) < o, for all a E L . We first show that for semimodular lattices of finite length, h(a) is the length of any maximal chain in [0, a].

THEOREM 1 (The Jordan-Holder Chain Condition. 0. Ore [1943]). Let L be a lattice of finite length. If L is eemimodular, then any two maximal chains of L are of the same length.

PROOF. Let C = {ao, . . . , un}, 0 =ao < * - <a, = 1, be a maximal chain of length n. We prove that any other maximal chain is of length n by induction on n. If n< 1, the statement is trivial. Let us assume the statement to hold for length < n . Let C’ ={bo, bl, . . . , bm}, 0 = b, < b, < - * - < b, = 1 be another maximal chain in L. If al = b,, then in the semimodular lattice [ai) the maximal chain C -{ao} is of length n - 1, so the maximal chain C’ -{bo} has to be of length n - 1, therefore n = m . If q * b , (see Figure 2) , then let C” be a maximal chain in [alvbl) and let k be the length of C”. Because of semimodularity, aivbi >al and aivbi> bi. Hence C” u {al} is a maximal chain of length k + 1 and C -{ao} is a maximal chain of length n - 1 in [a,); thus k + 1 = n - 1. Similarly, k + 1 = m - 1, hence n =m.

Theorem 1 suggests that semimodularity can be characterized in terms of the height function (the equivalence of (i)-(iii), and that they imply (iv) can be found in G. Birk- hoff [1933], see also G. Birkhoff [1967]):

2. Semimodular Lattices 173

C'

O=a,,=b,

Figure 2

THEOREM 2. equivalent for all u, b, c E L :

Let L be u lattice of finite length. The following conditions on L ure

(i) L i s semimodular. (ii) If a +b, u und b cover aAb, then avb covers a and b.

(iii) I f a

(iv) h(a) +h(b) >k(ar\b) +h(avb).

b and C i s n muximd ch in in [a, b] , then { X V C I x c C} i s a maximal chain in [uvc, bvc].

PROOF. (ii) implies (i). Let a < b. If c s n or avc 2 b, then bvc >avc. - If c % a and a v c z b, then bl\(avc)=a. Let a = a o < u l < a 9 - <a,=avc be a maximal chain in [ a , nvc]. Since b and al cover a and b+ai, bvai covers ai. An easy induction shows that bva, covers ui for i= 1,2, . . . , n. Thus bvu, covers an, that is, bvc covers U V C . @

(i) iniplies (iii). Obvious, since if x < y in C, then cvx < cvy. (iji) implies (iv). (iii) obviously implies semimodularityand so, by Theorem 1, we can

assunie that the Jordan-Holder Chain Condition holds. Let C be a maximal chain in [aAb, b]. By the Jordan-Holder Chain Condition, the length of C is h(b) -h(aAb). By (iii), D = { U V X I x E C} is a iiiaxinial chain in [a, avb]. The length of D is a t most the length of C, that is, h(b) -h(ar\b); on the other hand, the length of D is h(avb) -h (a ) by the Jordan-Holder Chain Condition, and so

h ( b ) - h(aAb) 2 h(avb) - h(a) ,

which was to be proved. (iv) implies (ii). Certainly, (ii) holds in (01. By induction on h(z) , let xcL and let

(ii) hold in (y] for all y < x. Let a , b E (XI, a , b >. uAb, and a +b. Then a < x , so (ii)

Figure 3

holds in (a] and (ii) implies (i), hence (a] is semimodular. Thus h(a) -h(a/\b) = 1. Hence by (iv).

h(avb) < /&(a) + h(b) - h(a/\b) = h(b) + 1,

Using Theorem 2 it is easy to check that some constructions yield semimodular lattices. We give one example. Take a semimodular lattice L of length n and pick a k < n. Let L, be the set of all x L with h(x) k along with 1. It is easy to check that Lk is a semimodular lattice of length k + 1. The result of this construction with L = ( Q 4 and k =2 is shown in Figure 3.

Another application of Theorem 2 is

and so avb t b. Similarly, avb >- a.

COROLLARY 3. equivalent :

Let L be a lattice of finite length. The following conditions OR L are

(i) L is modular. (ii) L satie-fiees the Upper and the Lower Covering Conditions.

(iii) h(a) + h(b) = h(ahb) + h(avb).

PROOF. We know from Section 1, that (i) implies (ii). Theorem 2 and its dual (to be more precise, the dual of 2(iii)) yield (iii). Now assume (iii). If L is not modular, then L contains a pentagon (0 , a, b, c, i}. Thus

and h(i) = h(b) + h(c) - h(o)

h(i) =h(a) +h(c) -h(o) ,

implying h(a) = h(b), a contradiction. Independence in semimodular lattices can be easily handled only for atoms.

THEOREM 4. the following conditions on I w e equivalent:

Let I ={al, . . . , u,} be a set of n atonis o f a sen~inLodular lattice. Then

(i) I i s independent.

2. Semimodular Latt,icns 175

(ii) ( a I v - - * v u , ) A c ~ ~ + ~ =0, for i = 1,2 , . . . , n - 1. (iii) Ii(a,v. Sva,) =n.

PROOF. (i) implies (ii) is obvious. @ (ii) implies (iii). We prove by induction on i that h(alv. * *va i ) =i. This is true for

i= 1. If h(a,v. * *vai) =i, then by (ii) and semimodularity,

alv* * -Vai<alv. * -VaiVai+l

and so h(alv. - .vaivai+l) = h(a,v* * .vai) + 1 =i+ 1. 0 (iii) implies (i). We show that the condition of Corollary 1.10 holds for I. It follows

from (iii), that h ( V X ) = 1x1 for any X G I so the map v: X -. V X is one-to-one. Obviously, is join preserving. Now take X , Y E I , and let a = V X A V Y and b = V ( X n Y ) . Then u 2 b and by Theorem 2(iv)

h ( V X ) + h ( V Y ) > h ( a ) + h ( V ( X u Y ) ) ,

I X I + I Y I 2 h ( a ) + I X u YI.

h ( u ) < 1x1 + I YI - IX u YI = IX n Yl .

h(a) 2 h(b) = IX n YI,

so by (iii),

We conclude that

On the other hand,

hence h(a) = h(b) and a = b. Let A be a set of atoms. Then G ( G A ) spans A iff for every a € A there is a finite

G', c C such that a< VG,. The following generalizes a result familiar from any first course on vector spaces.

THEOREM 5. Let A be rc set of atoms of a semimodular lattice, let I be a n independent subset of A , trnd let G 3 I span A . Then there i s a.n independent subset J of A such that G 2 J 2 I nnd J spans A .

PROOF. G. If C G X and C is a chain, then UC is again independent, since independence is tested with the finite subsets of U C and every finite subset of U C is also a finite subset of some X C. Hence by Zorn's Lemma (Section 11.1) there is a maximal independent subset J of A with I E J G. We wish to show that J spans A . It is sufficient to show that J spans G. Indeed, let g CG. If g C J there is nothing to prove. I f g EQ - J , then J u { g } is not independent, and so there is a finite J l E J , such that J l u {g) is not independent, that is, by Theorem 4(iii) and by semirnodularity,

Let 8 be the set of all independent subsets X of A with I E X

h ( V ( J I U { 8 } ) ) < IJil + I *

Since h ( V J , ) = IJII, we obtain that V ( J I u (9 ) ) = V J i , that is, g ..: V J i , proving that {g} is spanned by J .

=(XU a) A b

Figure 4

The definition of semimodularity was given for arbitrary lattices but it is obvious that it is not very useful for lattices without many prime intervals : lattices without prime intervals are always semimodular. Various attempts have been made to rectify this situation, that is, to come up with a definition agreeing with semimodularity for lattices of finite length, that also selects an interesting class of lattices which are not of finite length.

DEFINITION 6 (L. R. Wilcox [1939] and S. Maeda [1965]). Let L be a lattice. A pair of elements (a , b) of L i s called modular (see Figure 4), in notation, aMb, i f f

x < b implies that X V @ A ~ ) . = (xva)Ab.

The lattice L i s called M-symmetric i f f aMb implies that bMa for any (c, b C L.

Now it is obvious that L is modular iff aMb for all a, b E M. In the proof of [aAb, b] s [a, avb] in the Isomorphism Theorem we only used that aMb in L and bMa in the dual of L, hence

COROLLARY 7. Let L be a lattice and a, b C L. If aMb in L and bMa in the dual of L, then [aAb, b] ES [a, avb]. This isomorphism i s given by ya whose inverse i s qb.

The notation yb and ya are from Section 1 (see Figure 1.1). Half of this conclusion still holds if only aMb is assumed.

LEMMA 8. (i) aMb.

(ii) i s onto. (iii) ya i s one-to-one.

Let L be a lattice and let a, b < L. The following conditions are equivalent:

PROOF. (i) implies (ii). If xE[aAb, b] , then x l b, and so

(avx)yb= (aVz)Ab =xv(aAb) =x. 0 (ii) implies (iii). Let x , y E [aAb, b], qua =yya, and x+y. Then ( z v y ) y a =xya = yya

and x or y < x v y , say x < xvy . There exists a z E [a, avb] such that zq)b = x. We 1

2. Brminiodular Lattices 177

(iii)

have z >x and z 2 a, hence z 2 x v a =xya = (xvY)~, . Thus z > x v y and so zyb 2 x v y > x , a contradiction. @ irnp1ie.s (i). Let x< b. Then

(Xv(al\b))y, = xv(ur\b)va = X ~ U ,

((xva)Ab)ya = ( (xva)nb)va =xva ,

hence, using that ya is one-to-one, we obtain aMb. To verify that ( (xva)Ab)vu = x v a , observe that xva 2 a and x v u 2 (xva)r\b, hence xva 2 ((xva)Ab)va. More- over, if t > (xva)r\b and t > a, then t > (xva)r\b > x (since b >x) , and so t 2 a v x .

The following result is implicit in L. R. Wilcox [1939].

THEOREM 9. symmetric.

Let L be u lattice of finite length. Then L is semimodular i f f L i s M -

PROOF. Let L be a semimodular lattice of finite length. We shall prove that aMb iff h ( a ) +h(b) =h(aAb) +IL(avb),

from which M-syininetry trivially follows. Using the notation of “(iii) implies (iv)” in the proof of Theorein 2, the length of

C is h(b) -h(uAb) and the length of D is h(avb) -h(a). So if aMb, then ya is one-to- one, (CI = IDI, and h(b) -h(aAb) =h(avb) -h(a). Conversely, if a M b fails, then ya is not one-to-one, and C can be chosen so a5 to include x , y E [aAb, b] , xya = yya. Then ID[ < ICI and we obtain h(b) -h(aAb) > h(avb) -h(a).

To prove the converse we do not have to assume that L is of finite length. So let L be an M-symmetric lattice, let a , b, c L, and let b >a. If bvc =avc we have nothing to prove. If bvc > avc, then p u t d = avc and we have bAd =a, bvd = bvc. We have to prove that bvd >d. Indeed, let bvd > x 2 d. Then x 2 b and so bAx = a and bvx = bvd. Since b>bAx, yb as a map from [ x , xvb] into [xAb, b] is an onto map and so, by Lemma 8, we obtain xMb. By M-symmetry, bMx, which means by definition that for any y 2 x ,

yv(bAx) = (yvb)/\x. Let y = d , then we obtain

d = d v ( b ~ x ) = (dvb)Ax = x , that is, bvcFavc.

Exaniples of M-symmetric lattices not of finite length inciude the lattice of all closed subspaces of a Banach space and the projection lattice of a von Neuiiiann algebra.

Exercises

1. 2.

Show that a lattice L is semimodular iff z > X A ~ implies that mvy > y. Modify the proof of t,he Jordan-Holder Theorem. Assume only that C‘ is a chain and rk < m, and derive a contradiction. What conclusion can be drawn from this proof?

178

3.

4. 5 .

6. 7.

8.

9. 10. 11.

12.

13. 14.

15.

16.

17.

18.

19.

20.

21.

22. 23.

IV. Modular and Semimodular Latt,ices

Let A, B, and C be sets of atoms of a semimodular lattice. Show that if A spans B and B spans C, then A spans C. Show that (i)-(iii) of Lemma 8 are equivalent to (iv) y)&b is the identity map. Let L be a semimodular lattice. Prove that if p and q are atoms of L, a E L, and a < avq 5 avp, then avp =avq (Steinitz-MacLane Exchange Axiom). Let L be a lattice. Show that all sublattices of L are semimodular iff L is modular. Show that direct products and convex sublattices of semimodular lattices are again semimodular. Prove that a homomorphic image of a semimodular lattice of finite length is again semimodular. Investigate the statements of Exercises 6-8 for M-symmetric lattices. Show that Part(A), the lattice of all partitions on a set A, is a semimodular lattice. Show that the lattice of all congruence relations of a semilattice is a semimodular lattice (R. Freese and J. B. Nation [1973]). Let L be a modular lattice and let I be an ideal of L. Then L' =(L -I) u (0) is a lattice. Under what conditions is L' semimodular 9 Is the lattice L' of Exercise 12 always M-symmetric? Let L be a lattice. Prove that Sub(L) is semimodular iff L is a chain (K. M. Koh [1973]). A poset P is called graded iff an integer-valued function h can be defined on P satisfying, for x, y c P:

E < y and h(z)+l =h(y) iff x<y.

Show that P is graded iff every interval of P is of finite lcngt(h and satisfies the Jordan-Holder Chain Condition. Let L be a lattice. Show that Sub(L) is graded iff the dual of Sub(L) is semimodular, which in turn is equivalent to the condition that L has no sublattice isomorphic to &,x&, (H. Lakser [1973b]). Prove that a lattice of finite length is semimodular iff it is graded and satisfies h(z)+h(y) 2 h(xhy)+ h(xvy) (G. Birkhoff [1933]). The Infinite Jordan-Holder Chain Condition holds for a lattice L iff for any two maximal chains C1 and C, of L we have lCil = ICzl. Show that this fails for the lattice L =A XB, where A is the [0, 11 real interval and B is the [0, 13 rational interval. Show that the Infinite Jordan-Holder Chain Condition holds for the Boolean lattice of all subsets of a set A iff A is finite. A lattice L is said to satisfy the MacLane Condition iff for a, b, c EL satisfying a 1) b and a h b < c I a there exists a d E L satisfying ~ ~ ( b v d ) =c. Show that the MacLane Condition implies the Upper Covering Condition. Prove that a lattice of finite length is semimodular iff it satisfies the MacLane Condition. (Exercises 20 and 21 are from S. MacLane [1938].) Find M-symmetric lattices that fail to satisfy the MacLane Condition. Find lattices satisfying the MacLane Condition that are not M-symmetric.

3. Geometric Lattices

Just as most lattices arising out of algebraic examples are algebraic (Section 11.3, see also Concluding Remarks), most lattices arising out of geometry are geometric in the following sense :

3. Geometric Lattices 179

DEFINITION 1. A lattice L is called geometric i f f L i s semimodular, L i s algebraic, und the compact elements of L apre exactly the finite joins of atoms of L.

Equivalently, L is complete, L is atomistic (that is, every element of L is a join of atoms), all atoms are compact, and L is semimodular.

This concept was introduced by G. Birkhoff [1935a], influenced by H. Whitney [1935], for lattices of finite length. Geometric lattices with no restriction on length were introduced and investigated by S. MacLane [1938] under the name exchange lattices. The name, geometric lattice, is due to M. L. Dubreil-Jacotin, L. Lesieur, and R. Croisot [1953]. G. Birkhoff [1967] and some others retain in their definition the requirement that the lattice be of finite length. Many authors call these lattices matroid lattices. (For combinatorial definitions of matroids, see the Further Topics and References for this chapter.)

Later in this section we shall investigate the connection between geometric lattices and geometries. We start out, however, by investigating the lattice theoretic consequences of this definition.

COROLLARY 2. (i) Let L be a geometric lattice. Then the set F of elements of finite height i s a n

deal of L. The lattice F is semimodular and every element of F i s a finite join of atoms. L is isomorphic to I ( F ) , the lattice of all ideals of F.

(ii) An interval of a geometric lattice i s again a geometric lattice.

PROOF.

(i) Let a, b c F, GEL, and c<a. Then h(c)< h(a) , hence cE F. By Theorem 2.2, h(avb) < h(a) +h(b) -h(aAb) and so avb E F. Thus F is an ideal. The second statement of (i) is obvious. The third statement follows from the proof of Theorem 11.3.13. 0

(ii) An interval [a, b] of an algebraic lattice L is again algebraic. Moreover, {avp I p an atom of L, p $ a and p < b} is by semimodularity the set of all atoms of [a, b]. The rest is easy.

Now we utilize the theory of independence.

LEMMA 3. Let L be a geometric lattice. Every element a of L i s a join of a n independent set of atoms, in fact, of any maximal independent set of atoms of L contained in (u,].

PROOF. Let A be a maximal independent set of atoms of L contained in (a] (A exists by Theorem 2.5 with I = 0 and G the set of all atoms in (a]). By Theorem 2.5 and the compactness of atoms, A spans all the atoms in (a]. Since L is atomistic a = V A .

THEOREM 4 is corrtplemcnted. In fact, it i s relatively complemented. 13 Gritzer

(G. Birkhoff [1935a] and S. MacLane [1938]). A geometric luttice L

180 IV. Modular and Semimodular Lritticee

PROOF. Let a c L, let A be a maximal independent set of atoms in (a], and let K be the set of all atoms not contained in (a]. Then by Theorem 2.5 there is a ma,ximal independent set I of atoms of L satisfying A c I c A u K. Set b= v(I -A). Since VI=1 by Theorem 2.5, we obtain a v b = i . Let c = a ~ b . If c+O, t,hen there is an atom p < c . Since p < VI and p 1 V(I -A), by the compactness of p there exist finite I , c I and I2 c I -A such that p < VIi and p VI,. By the definition of independence,

p I VI,AVI, = V(I i n I?) = V 0 = 0,

a contradiction. Hence aAb =0, proving that L is complemented. The second statement follows from the first and from Corollary 2.

Our next task is to prove the structure theorem of geometric lattices due to F. Maeda [1951] and in its sharper form due to U. Sasaki and S. Fujiwara [1952].

THEOREM 5. indecomposable geometric lattices.

Every geometric lattice i s isomorphic to a direct product of directly

This theorem is augmented by a characterization theorem of direct indecomposability. Let us call the elements a and b perspective, in symbol a-b, iff they have a common complement, that is, a v x = bvx = 1 and aAx = bAx = 0 for some element x. The reader should not confuse the perspectivity of elements with perspectivity of quotients. (See Exercise 13.)

THEOREM 6. perspective.

A geometric lattice i s directly indecomposable i f f uny two atoms are

For the proof of Theorem 5 we need a lemma.

. LEMMA 7. Let L be a geometric lattice. If a , b, x c L, aAx = bAx =0, and uvz = bvx, then a-b. If a, b e L, a-b , and h(a) , h(b) <-, then there i s an x E L satisfying aAx = bAx =0, a v x = bvx, and h(x ) <-.

PROOF. Let aAx=br\x=O and avx=bvx. Let y be the relative complement (which exists by Theorem 4) of a v x = b v x in [x , 13. Then aAy=aA(avx)Ay= aAx = O and a v y = a v x v y = 1 ; similarly, bAy = O and bvy = 1, that is, a- b. To prove the second statement, let UAy = bAy = 0, a v y = bvy = 1, and observe that a 1 bvy and h(a), h(b) <-, hence, by the compactness of a there is an xl< y , h(xi) <- satisfying a < bvxl. Similarly choose x2 I y, h(x2) < 03 satisfying b 22 am,. Then x =xivx2 will satisfy the lemma.

PROOF OF THEOREM 5. iff a=xo-x l - - - * N X , =b for some xi, . . . , x , - ~ EL. For an atom p of L we define

R ( p ) = {q I q is an atom of L and p x q}

For a, b EL, let US write u zz b, and call a and b projective,

3. Ckoinrtric Lattices 181

and 2 = { R ( p ) I p is an atoni of L}.

For ~ € 2 , x = R(p) , we set c(x) = c ( p ) = Vli’(p). We are going to show that L is isomorphic to II([O, c(x)] I ~ € 2 ) . Every element a of L is a join of atoms, hence if we set a R ( , ) = V ( ( a ] ~ R ( p ) ) , then

To establish that a = V(a,(,, I W E Z ) , where a,(,)G(P).

a -(a,,,, I N P ) €2)

is the required isomorphism it is sufficient by Exercise 21 t o show that every a € L has a unique representation in the form

(6 = V(aRcp) I R(p)EZ) with ~ , , p ) ~ c ( p ) .

Indeed, let u have two such distinct representations :

a = V(X,(,) I R 4 E ~ ) = V(Yncp, I R(P) E n X n ( p ) , Y,(,,lC(P)

and for a t least one atom, say q, ~ , ~ ( ~ ) + y ~ ( ~ ) . By taking the join of the two representations we can assume that x , (~ )< y,,,) and SO x ~ ( ~ ) < yR(@).

We now need the following simple statement: If A u {p} is a set of atoms of L and p < V A , then p - r for some r E A . Indeed, by the compactness of p, p < V A , for some finite A , E A . Choose A,

minimal. Then p - r for any rEA,, because if x = V (A, - { T } ) then r v x = V A , ; p 5 x (since A , is minimal), hence h ( p v x ) =h(x) + 1 = h(VA, ) and SO ~ V X = V A , . By minimality, A, is independent, hence T A X = 0 and p x , hence PAX =O. Thus p - r by Lemma 7.

It follows from this statement that there is an atom t E R(q) such that t < Y ~ ( ~ ) and 1 $ x,(~). It also follows that if we set

a, = X N q ) and a2 = V(X,(,, I P * q),

then u=u1va2 and alAa2=yn(q)Aa2=0. So we obtain elements t , al, and a2 satisfying

t m , =0, uiAa2=0, t j a l v a 2

(see Figure 1). Let xi be the relative complement of tva1 in [ai, aiva2] and xz the relative complement of a2Ax1 in [0, a2]. We claim that

AX^ = X ~ A X , = O and t v x , =x2vx1

and therefore t -x2 by Lemma 7. Indeed,

182 IV. Modular and Semimodular LRttices

0 Figure I

Now observe that x1 < tvxi =x1vx2. Therefore, for any atom u I x2, we have xIv?i = tvxi and, of course, x i ~ u = O . Hence t - u by Lemma 7. Now recall that tE R(q) , so t - u implies that u E R(q). But u I a2 = V ( X ~ ( ~ ) I p +q), hence u E R ( p ) for soiiie p+q, a contradiction.

Finally, we have to show that [O,c(p)] is directly indecomposable. If it is not, then by Theorem 111.4.1, it has a pair a, b of complemented neutral elements, a , b ( { O , c(p)}. So we can choose atoms u, v of [0, c(p)] such that u< a and w < b. Since u FZ v in L, u x v in [0, c(p)] (mnce [0, c(p)] is a direct factor). Taking the homomorphism q: x ~ X A U of [0, c(p)] onto [0, a] , we obtain that uq x v q . But u = z q , ty =0, and u x0 imply that u = 0, a contradiction.

It is clear from the proof of Theorem 5, that indecomposability is equivalent with the projectivity of atoms. Therefore to prove Theorem 6 it suffices to prove the following statement :

THEOREM 6'. In a geometric lattice, perspectivity of atoms is transitive.

PROOF. and y such that

Let p , q, and r be atoms of L, p-q, q-r ; by Lemma 7, we can choose 2

PAX = qAx = 0, p v x = qvx, h(x) = m,

qAy=rAy=O, qvy=rvy, h(y) =n.

Choose z so as to minimize h(z). Let {sI, . . . ,a,} be an independent set of atoms in. ( X I , and { t i , . . . , t,} in ( y ] , and let xi =slv- * . V S ~ - ~ V S ~ + ~ V - * 'vs,, for 1 ism. We claim that p x,vq. Otherwise, p 5 xivq and so xivp I xivq. However, h(xi) =

m - 1 and therefore h(x,vp) = h(x,vq) =m, implying that x ivp =xivq and then xi could replace x , contradicting the minimality of h(x). So h(pvqvxJ = 2 +m - 1 = m + 1, and

pvqvx i = p v x =qvx , for 1 i s m .

Now by Theorem 2.4 {si, . . . , sm, p } is an independent set of atoms so, by Theorem 2.5, we can choose a subset { t i , . . . , t i } of { t i , . . . , t,} such that

{P , 51, * - * , s,, t ; , * * * 9 t i }

is independent. Set y‘ =t;v. * .vt;. Then { p , sl, . . . , s,, y‘} and also {q , sl, . . . , a,, y‘} are independent. Therefore, using the definition of independence, we obtain

y’ = (XVy’)A(qVXiVy’)A. * *A(qVX,Vy’).

Since rAy=O, we have r s y ; hence r s y ‘ and, therefore, either r s x v y ’ or r s qvxivy’, for some 1 < i s m . Set a =xvy ’ if r 5 xvy’ or a =qvxivy’ if r 5 qVxivy’. Then

rAa=O and r s p v a

(the latter because r_( q v y < p v x v y = p v x v y ’ =pvqvxivy‘ ) . Finally, since r 5 a, we obtain p %a, that is,

pAa = 0,

thus a s a v r s a v p and a < a v p , and 00

avr = u v p ,

proving that p - r . If the geometric lattice L is of finite length, then L is said to be finite dimensional.

In this case, Theorem 5 follows from Lemma 111.4.6. With some effort, the general case could be reduced to the finite dimensional case. However, without the present proof of Theorem 5 it would be more difficult to prove Theorem 6.

Theorem 111.4.6 could be used to prove the result that every finite dimensional geoiiietric lattice is isomorphic to a direct product of simple geometric lattices. The factors we obtain from Theorems 5 and 6 are such that any two atoms are perspective. Now if L is a geometric lattice of finite length in which any two atoms are perspective, then L is simple. Indeed, if 0 is a congruence relation of L and 0 =+w, then there exist ( I , b EL, a < b such that a = b (a). Choose an atom p such that p < b and p 5 a. Then p ~0 (a), since p = p ~ b r p A a = O (0). Now let q be any atom of L. Since p - q , p and q have a common complement x . Then

Pi0 f 11% L qIQ,

and so q = O (0). Since k ( 1 ) <-, 1 = q i v - * ‘ vq , for some finite set of atoms { q i , . . . , q,} and qi = 0 (0), i = 1, . . . , n, hence 1 = 0 (0). Thus 0 = I and L is simple.

Exactly the same proof yields that if L is a geometric lattice in which any two atoms are perspective, then L is subdirectly irreducible; the only atom @ of c(L) contained in every O+w is the congruence @=0(13, 0), where p is any atom. If a EL and A(u) <-, then a ~0 (@). In general, however, L is not simple.

B y a geometry we mean a set A certain subsets of which are called subspaces (or flats) such that certain basic properties hold; we shall list these properties in the next definition.

DEFINITION 8. itself eatiafying the following conditions :

(i) X Ex; if X E Y, then X E P; (ii) O= 0 , and {5)={x) for ~ E A .

A geometry ( A , -) is a set A and a function X +x of P ( A ) into

=x. . . . , . - -

(iii) I f x E X u { y } , but x ( X , then y < X u { z } . (iv) I f x C x, then z Exi for some finite Xi C X .

(i) means that (A, -) is a closure space. A subset X of a closure space is called closed iff x=X. It follows simply form (i), that x is the smallest closed set containing X.

LEMMA 9. A} is u complete lattice with respect to set incEusion, and A Y = n Y for Y E L(A, -). Conversely, if L c P ( A ) is closed under arbitrary intersection, then

Let (A, -) be a closure space. Then, L(A, -) ={x 1 X

X = n ( Y j ~ z x and YEL}

defines a closure space ( A , -) such that a set is closed i f f i t belongs to L.

PROOF. Let Y E L ( A , - ) a n d Z = n Y. Then Z C X forall X E Y , s o Z c X = X , and therefore z c n Y =Z. Since 1; C z always holds, Z =z. The remaining steps in the proof of Lemma 9 are very easy and will be left to the reader.

, Thus a closure space is completely determined by the lattice L of closed sets. In case of a geometry we have @} ={x} meaning that the elements of A can be identified as atoms of L, hence L determines the geometry even if L is known only up to isomorphism.

For geometries, a closed set is called a subspace and 1 is called the subspace spanned by X. Thus (iv) means that if x belongs to the subspace spanned by X, then x belongs to the subspace spanned by some finite Xi EX. A closure space satisfying (iv) is usually called algebraic.

LEMMA 10. A lattice L is algebraic iff L ia isomorphic to the lattice of closed sets of an, algebraic closure space.

PROOF. Let L be an algebraic lattice. By Theorem 11.3.13, L s I (B) where F is a join-semilattice with 0. Take A = F and for X C_ A define x = (XI, the ideal generated by X. The verification of (i) and (iv) for (A, -) is contained in the proof of Theorem 11.3.13. Then I (P) is the set of closed sets and so L 2 I ( P ) by assumption.

Conversely, if ( A , -) is an algebraic closure space and L is the lattice of closed sets,

then it is easily seen that X C L is compact iff X =Xi for some finite Xi E X, from which it follows immediately that L is algebraic.

Condition (iii) makes it possible to define the rank or dimension of a subspace. Let r(a)=O andr({a})=l. If Xisasubspace,r(X)=n, a n d y t X , then r(Xu{y})= )L + 1. (iii) implies that this defines r uniquely.

Finally, (ii) states that 0 is a subspace and so is every singleton {x}. If we drop this last condition we get what is known as a pregemetry from which a geometry can be obtained by identifying any two elements, x, y of A , for which {y} = &}.

THEOREM 11. Let (A , -) be a geometry. Then L = L(A, -) is a geometric lattice. Conversely, if L is a geometric lattice, A is the set of atoms of L, and for X E A , X i.9 the set of atoms spanned by X (in the sen8e defined in Section 2), then ( A , -) i8 a geoiiretry and L Y L(A, -).

PROOF. Let (A, -) be a geometry. Then by Lemma 9 L = L(A, -) is algebraic. Since X is compact iff X =xi for some finite Xi X, we see that X is compact iff X is a finite join of atoms (for x E A , {x} is an atom by 8(ii)). It remains to show that L is semimodular. Let X, Y € L, Y = X u {x}, and x(X. We claim that X < Y. Indeed, if Z € L and X c Z c Y, then there isa z €2 -X andz E Z c Y = X u {x}, and so by 8(iii), x C X u { z } c 2. Thus, Y = X u {x} 2; we conclude that Y =Z, proving X < Y. Now let UEL. Then YvU=YuU=XuUu{x} and XvU=XuU, hence either x E X u U and so XV U = Y v U or x 4 X u U and in this case Xv U < Y v U.

Conversely, let L be a geometric lattice, let A be the set of atoms of L, and let x be the set of atoms spanned by X for X c A. It is immediate that (A, -) is a closure space. 8(ii) and 8(iv) are also clear by definition. To verify 8(iii), let x E X u {y} and x g x . Since {y} is an atom, by semimodularity, Xu{y}=~v{y}>X; thus X C X u { x } ~ X u { y } implies that Xu{x}=Xu{y}; and so y€Xu{x}.

Now let rp be the map X -. VX for X c A, X E L(A, -). Then rp inaps L(A, -) into L. Since every element of L is a join of atoms, X c Y iff V X < V Y. Therefore,

is onto, one-to-one and both rp and 9-1 are isotone. Hence 9, is an isomorphism. The lattice of subspaces thus completely determines the geometry and vice versa.

The diagrams of geometric lattices are as a rule too complicated to be of any use. Many times, however, we can draw a picture of the geometry associated with the lattice.

Let us call an element of height 2 a line. A line is the join of any two distinct points in the line. Some geometries (for instance, projective geometries, see Section 5 ) are conipletely determined by their lines, some are not. When drawing the picture of a geometry we try to represent lines by straight lines or by curves. Lines having exactly two points are not drawn as a rule. Figures 2a, 3a, and 4a are pictures of geometries and Figures 2b, 3b, and 4 b are the corresponding lattice diagrams. Observe the simplicity of geometries compared to the intricate diagrams of the lattices.

Closure spaces satisfying the additional requirements 8(ii)-8(iv) can be found in

_ _ _ ~ _ _ _ ~


Figure 2 s Figure 2b

Figure 3a Figure 3b

Figure 4a Figure 4b

3. Geoinet,ric Lattices 187

Figure 5

large numbers, especially in algebra, geometry, and combinatorics. Here is en example from combinatorics.

A gruph is a set G with a fixed set E (the edges) of two- element subsets of G (un- oriented graph without loops). Figure 5 shows a diagram of a graph. Two points a and b (elements of G) are connected with a straight line segment iff {a, b} E.

Let a , b CG and A c E. We shall call a and b A-connected iff there is a sequence of edges {xo, xl}, {xI, z2}, . . . , {x,.-~ x,} C A such that a =xo and x,, = b.

Now we define a geometry, called the edge geometry, on E : for {a, b} C E and A E E, let {n, b} E 2 iff a and b are A-connected.

THEOREM 12. ( E , -) is a geometry.

PROOF. All the properties of a geometry obviously hold except perhaps 8(iii). Let u and b be A-connected, then there is a connecting sequence {x,,, xl}, . . . , {xnVl, 2,)

for which n is minimal. We claim that no edge can occur twice in this sequence. Indeed,let O < i < j < n - l and{xi,xi+l}={xi,xi+i}. Now if ~ ~ = x ~ a n d x ~ + ~ = x ~ + ~ , then by dropping {xi+l, xi+2}, . . . , {xi, xi+1} from the sequence we get a shorter connecting sequence. If xi and xi+l =xi, then we can drop all of {xi, xifl}, .. . ,

Now we prove 8(iii). Let z X u { y } but x $1, where x = { a , b} and y ={c, d } . Let eo, . . . , eae1 c X u {y} be a shortest sequence connecting a and b. Since x $1, one of the eo, . . . , must be y. By the observation above, exactly one of eo, . . . , en- , , say e j , equals y. But then ei+l, . . . , e n - l l x, e,, , . . , e i - l will connect c and d , hence

For instance, if we take the graph of Figure 6, then Figure 2 a shows the edge geo-

{xj, "L;. + 11.

yEXu(x). 0

metry and Figure 2 b the lattice, called the edge lattice, associated with it.

Figure 6

188 IV. Modular and Semimodular Latt,ices

To conclude this section we discuss two results of a coinbinatorial nature. Let L be a finite geometric lat,tice. Let E(i ) be the set of elements of L of height i

and Wi = IE(i)l.

Observe that each E(i) is an antichain of L. ]El for any antichain

X of L. The following two theorems will establish the size of a maximum sized antichain for some classes of geometric lattices. The first is credited to R. P. Dilworth and L. H. Harper in K. Baker [1969].

An antichain E of L is a maximum sized antichain iff 1x1

THEOREM 13.

maximum sized antichain of L has max ( W i I 0 1 i< Z(L)) elements.

Let the finite gemnetric lattice L have the property that every element of height i is covered by ai and covers bi elements (ai and bi depend only on i). Then a

PROOF. i < n) = W,; then E(E) is an antichain with W , elements. We have to prove that if S is an antichain, then IS1 5 Wk.

For xEL, let s(x) denote the number of maximal chains going through x. It is obvious, that if h(z) =i, then s(z) =b , - - bi - ai - - une1, hence s(2) depends only on i. Since every maximal chain has exactly one element of order i, we obtain that s(2) W , = s where 8 is the number of maximal chains in L. Using Wi( W, we obtain

Let n = 1(L) and let max (Wi I 0

Finally, a maximal chain goes through at most one element of the antichain S. Therefore,

that is, IS1 I W,.

iier [1928]) stating the conclusions of Theorem 13 for P ( X ) .

that

Theorem 13 as well as the next result was motivated by Sperner's Lemma (E. Sper-

A finite geometric lattice L of length n is unirnodal iff for some integer L we have

wi < ' ' * Wk-l< WE 2 WE+, 2 * ' ' 2 Wn-1-

If X and Y are subsets of L we say that there is a matching between X and Y iff there is a one-to-one map y : X + Y (or q : Y -X if 1x1 2 I YI) such that x is coni- parable with xy for all xEX. The following is G.-C. Rota's approach to the problem of finding maximum sized antichains:

THEOREM 14. - i < n, then a maximum sized antichain of L has W , elements.

Let L be a finite geometric lattice of length n. If L is unimodal ( Wi 2 .< W , 2 * * * 2 Wn-l) and there is a matching between E(i) and E(i + 1 ) for all

PROOF. For a subset S of L let

t (S)=max ( h ( x ) -h(y) I x, YES)

be the thickness of S. Let S be an antichain of L. We shall prove IS1 < W , by induction on t(S).

If t(S) = 0, then S C E( i ) for some i, hence IS1 I W i < W E . Now let t (S) > 0 and assume the inequality for antichains with a smaller thickness. Since t(8) > 0, there is an x €23 with h(x) +k, say h(x) < k. Let i=niin (h(x) I x ES), and set S =So u S , where S, = S n E( i ) and So =S -Si. By assumption, i < k , and so IE(i)l< IE(i + 1)1, hence there is a matching qj: E ( i ) - E ( i + 1). Define

S‘=S,uS,q.

S,, and S,qj are disjoint’, because if x E 8, and x E Sp, that is, 2 = yqj with y E S,, then x, YES, x+y, and x and y are comparable, contradicting that S is an antichain. Thus IS‘[ = ISI. Moreover, S’ is an antichain; indeed, if x, yes ‘ , x+y, x and y are comparable, then x € S o and y ES,qj, that is, y =zv for some z ESi. Since h(y) = i + 1, we have h(y) h(x), therefore x < y is impossible. Thus y <x, contradicting z < y andzljx.

Thus S’ is an antichain, t(S’) = t (S ) - 1, and so

IS1 = IS” I w,.

1.

2. 3.

4.

6.

6.

7.

8.

Exercises

Let F be a semimodular lattice with 0 in which every element is a finite join of atoms. Show that I ( F ) is a geometric lattice. Is the converse of Exercise 1 true? Call a lattice L atomic iff L has a 0 and for every a EL, a +0, there is an atom p 5 a. A lattice L is (upper) continuous iff

aAVD=V(aAx I X E D )

for any (upper) directed subset D of L ( D is directed iff for all 2, y ED there is a z ED satisfying 2 5 z and y 52). Prove that a lattice L is geometric iff L is atomic, relatively complemented, cont,inuous, and semimodular (G. Birkhoff [ 19481 and F. Maeda [1951]). Show that in Exercise 3 “ semimodular ” can be replaced by the condition that if a and b coVer ahb, then a v b covers a and b. Prove that i n Exercise 3, “ scmimodularity ” can be replaced by “M-symmetry ” (F. and S. Maeda [1970]). Let, B’ be a semilattice and let C ( F ) be the lattice of all congruencp relations of F. Prove that C ( F ) is geometric iff C ( F ) is atomistic. Let I( be a subfield of the field K’. Let L be the lattice of all relatively algebraically closed subfields nf 2 K. (M is relatively algebraically closed iff a polynomial with coefficient,s i n Xi which factors in K’ also factors in M.) Prove that L is a geometric lattice. Prove that the lattice L of Exercise 7 is modular iff the transcendence degree of K’ over K is at most two (S. MacLane [1938]).


9.

10.

11.

12.

13. 14. 15.

16.

17.

18.

19. 20.

21.

22.

23.

24. 25. 26. 27. 28. 29. 30.

Let G be an abelian group. Let L denote the lattice of all subgroups H with the property that GIH (the quotient group) has no element of finite order excepting zero. Prove that L is a geometric lattice. Consider the concepts of p-independence and p-basis in fields of characteristic p (see, for instance, B. L. van der Waerden [1931]). Can these be viewed as independence in a suitable geometric lattice? Let L be a geometric lattice. Show that a - b in L iff aAx =bhx and avx-Ibvz for some x EL Let a and b be atoms in the geometric lattice L. Show that a-b iff there exists a finite independent set of atoms I such that b E l and for I, I, a I VI , iff I , = I . Show that if a-b , then a102 l/x\ b/O for some x . Prove that in a geometric lattice a - b iff a/O-x/y-b/O for some quotient x /y . Relate a % b and a/O = b/O in a general lattice, in a geometric lattice, and in a modular geometric lattice. Let L be a modular geometric lattice. Show that L is simple iff L is a directly indecomposable lattice of finite length. Let I and J be maximal independent sets of atoms of a geometric lattice. Prove

Show that in a geometric lattice L, for every element a there is a smallest element c(a) of the centre of L satisfying a 5 c(a). Prove that in a geometric lattice a-b iff c(a) =c(b) (with c(z) as in Exercise 18). A lattice L is called left-complemented (L. R. Wilcox [1942]) iff L is bounded and for a, b EL there exists bi 5 b such that

that III = IJI.

avbi =avb, aAbi =0 , and aMb,.

Is every geometric lattice left-complemented ? Let L be a complete lattice and let X a EL has one and only one expression of the form

L. Then L %II((z] I x EX) iff every element

a = v ( a , I X E X ) ,

where a; 5 m for all z EX. Define the isomorphism of two geometries. Prove that two geometries are isomorphic iff the associated geometric lattices are isomorphic. Define a circuit of a graph as a set of edges ( 2 0 , zi}, . . . , {xn -1, zn}, {xn, zo}. Define the subspaces of the geometry on the edges of the graph in terms of the circuits. What role is played by the minimal circuits? Draw the lattice associated with the graph of Figure 5. Does the lattice associated with a graph determine the graph (up to isomorphism) ? Generalize Theorem 13 to posets. Prove Sperner’s Lemma using Theorem 13. Prove Sperner’s Lemma using Theorem 14. Does Theorem 13 or Theorem 14 apply to finite projective geometries? Consider the graph G, of Figure 7. Let L,, be the edge lattice of Gw If A EL, and {a, b} € A , then A is determined by {i I {a, i} and { b , i} € A } . If {a, b } $ A , then A is determined by {i [ {a , i } or {b , i} € A } and a choice function selecting one of {a, i ) and {b, i}. Conclude from this, that

Wk =Z k (;)+ (k: l).


Figure 7

31. Show that in L,, there is an antichain contained in E(0) u E(7) that has more elements than anyE(k). (Exercises 30 and 31 are due to R. P. Dilworth and C. Greene [1971].)

Let P be a finite poset. The Mobiuafunction p is an integer valued function defined on P x P by the formulas :

p(z,z) = 1 for zEP;

(See A. F. Mobius [1832] and L. Weisner [1935].) Establish the Mobius inversion formula: i f f and g are real valued functions on a finite partially ordered set P and g(x) =Z(f(y) I y 5 z) for every x € P , then

32.

f(z) = W Y ) P ( Y Y 5) I 2/ 5 4. 33. Let P and Q be finite posets with Mobius functions pp and pg, respectively. Then

the Mobius function p p X ~ of P XQ satisfies

P P X Q ( ( z 9 y>, ('% .>I =!LP(Z, u) pQ(Y, .)

for x, uEP and y, v € Q . Show tjhat the Mobius function p of the Boolean lattice B is given by 34.

p ( x , y) =( - 1 ) h ( r ) - h ( ~ )

for z 5 y, where h(z) is the height of z in B. Let ,u be the Mobius function of a finite lattice L and let 2, y, z EL. 35.

( i ) If z 5 y a n d y is not the join of elements covering z then p(z , y) = O (Ph. Hall [ 1 93 41).

(ii) If x 5 z I y thrn

(L. Weisntjr [1935]). 36. Show that t,he Mobius function p of a finite distributive lattice L is given by:

p(z , y) =O, if y is not the join of element,s covering 2; p ( z , y) = ( - l)", if y is the join of n, distinct, elements covering x. (Hint: Use Exercises 34 and 35(i).) Let p be the Mobius function of a finitc! geometric lattice L and x, y E L, z I y. Thcn p(c , y) + O ; p(x , :y) is positive if h ( y ) -h(x) is even; p(x, y) is negative if h ( y ) - h ( x ) is odd ((!.-C. 1iot.a L19641). (Hint: Apply Exercise 35(ii).)

37.

4. Partition Lattices

A partition of a set A is a set n of nonempty pairwise disjoint subsets of A whose union is A . The members of z are called the blocks of z. A singleton as a block is called trivial, If a and b (a , b c A ) belong to the same block we write a = b (n) or anb.

Given a partition n we can define an equivalence relation E by (x, y ) E E iff x = y (n). Conversely, if E is an equivalence relation, then n ={[a]& I a A} is a partition of A. Thus we set up a one-to-one correspondence between partitions and equivalence relations. Partitions are easier to visualize so we shall use them; if the reader prefers equivalence relations, he can use them throughout.

Part(A) will denote the set of all partitions of A partially ordered by

n o ( q iff x = y (no) implies that x = y (q).

Part(A) with this partial ordering (which corresponds to set inclusion for the corresponding equivalence relations) forms a complete lattice, called the partition lattice (or equivalence lattice) of A.

We draw a picture of a partition by drawing the boundary lines of the blocks. Then n o l n l iff the boundary lines of ndi are also boundary lines of no (but no may have some more boundary lines). Equivalently, the blocks of n1 are unions of blocks of no. (See Figure 1.)

LEMMA 1. Part(A) is a complete lattice and x = y (A(ni I i E I ) ) i f f x = y (nJ, for all i € I ; x = y (V(ni I i c I ) ) i f f there is a natural number n, io, . . . , in-1 E I , and xo3 . . . , X ~ + ~ E A such that x=xo, Y = X , + ~ , and X ~ = X , . + ~ (n+, for O I j K n . The zero of Part(A) has only trivial blocks and the unit has only one block, namely A. The atoms of Part(A) are the partitions with exactly one nontrivial block, and this block has exactly two elements. The dual atoms of Part(A) are the partitions with exactly two blocks. no<q in Part(A) i f f x i t3 the result of replacing two blocks of no by their union. In Part(A), [n) i s isomorphic with the partition lattice of the set n. In Part(A), (n] t3 isomorphic with the direct product of all Part(X) where X ranges over the nontrivial blocks of n,

x,: - Jto 5 n1 no:- and---

Figure 1

4. Partition Latt,ices 193

PROOF. (i) This is implicit in the proof of Theorem 1.3.9.

(ii) and (iii) are trivial. (iv) If .z 2 4 then each block of E is a union of blocks of z. Thus 6 defines a partition

of the blocks of z, that is, of n. This sets up the required isomorphism. (v) If E<n, then for each nontrivial block X of n, 4 defines a partition Ex of X.

Since a. block of 5 which intersects X is completely within X, the map 5 -(Ex [ X is ;t nontrivial block of n) sets up the required isomorphism.

(J @

Froni Lemma 1 we easily conclude:

THEOREM 2 (0. Ore [1942]). Part(A) i s CL simple geometric lattice.

PROOF. By Lemma 1 (i), Part@) is complete. It follows from l(ii) that every element is a join of atoms and, by the formula for join, the atoms are compact. If ?to and nl satisfy the condition of l(iii), then so do nov6 and nlv5 unless novE =nlvE. There- fore, Part(A) is semimodular. This shows that Part(A) is geometric.

be atoms. By l(ii), R and 5 have only one nontrivial block each, say {a, b } and {c , d} , respectively. Let t be a partition with two blocks as follows: if { u , b } n { c , d } = @ , the blocksoft are A - { u , c } and{a,c};if { a , b } n { c , d } = { e } , then the blocks of t are A - { e } and {e}. In either case, t is a complement of ~t and 5 and so z-5. By the discussion in Section 3, Part(A) is simple if A is finite. Let A be iniinite and let 0 be a congruence relation of Part(A), 0+:w. Then all atonis are congruent modulo 0 to the zero of Part(A). Since A is infinite we can split A up into two disjoint sets A, and A, of the same cardinality. Let y be a one-to-one map of A, onto A,. Now we define some partitions of A. Let n have the two blocks, A, and A,. Let ti have exactly one nontrivial block, Ai (i=O, 1). The blocks of t are all the sets of the form {x, q~} where xEA,. Then these partitions form a sublattice of Part(A); a diagram of this sublattice is given in Figure 2, where no and nl denote the zero and unit of Part(A), respectively.

Let a be any atom st. Then a=no (a), hence 7cGItva (0). Since n<nl, we obtain that nva =7ci and so 7cl =~d (0). Taking the meets of both sides with t and joining with ti we get Ei=nl (a), i=O, 1, hence no=Eo/\EI =ni (0). Thus 0 = L and Part(A) is simple.

call a partition IG finite iff all blocks of n are finite and only finitely many blocks

Let z and

Figiirc 2

are nontrivial. Then it is clear that n is finite iff it is a finite join of atoms (recall l(ii)). Since Part(A) is geometric, this is furthermore equivalent to n being compact.

Let Partfi,(A) be the set of all finite partitions. This is obviously an ideal of Part(A). By Corollary 3.2 we obtain:

COROLLARY 3. Part,,(A) is an ideal of Part(A) and the following isomorphism holds :

I(Part,,(A)) Y Part(A).

The importance of partition lattices lies in the fact that they are as universal for lattices as permutation groups are for groups. Ph. M. Whitman [1946] proved that every lattice L can be embedded in some partition lattice. One drawback of this result isthgt the joins of the partitions representing elements of L have to be computed by l(i) which can be rather complicated. This was corrected by B. Jbnsson [1953]. To state Whitman's theorem as improved by Jbnsson, we need some concepts.

A representation of a lattice L is an embedding a of L into some Part(A); a is called of type 3 iff for all a, b E L and x, y C A

x E y ( (avb)a) iff there exist zI, z2, z3 E A such that- x =zi (aa) , zl = z 2 (ba), z2 ~ 2 3 (aa) , and 23 =y (ba).

In other words, type 3 means that for the partitions aa, a L, the join formula in 1 (i) always holds with n 5 3. The number 3 appears arbitrary but we shall see in Theorem 5 that 3 cannot be reduced to 2 unless L is modular.

THEOREM 4. Every lattice has a type 3 representation.

PROOF. A meet-representation a of a lattice L is an embedding of L as a meet-Remi- lattice into some Part(A) ; in other words, a is one-to-one and preserves meets but not necessarily the joins. We start out by describing a simple way of constructing meet-representations of a lattice.

Let A be a set and let L be a lattice with 0. A distance function 6 maps A% onto L and it satisfies the following rules:

(i) 6 is symmetric, that is, 6(x, y) =S(y, z), for all x, y € A ; (ii) 6(x , x ) = 0, for all x A ;

(iii) 6 satisfies the triangle inequality, that is, d(x, y)vd(y, z ) >6(x , z), for all x, y, z E A. For instance, if L is a lattice with 0 and A = L, then putting 6 ( x , y) =xvy for

x +y and 6 ( x , x) = 0 yields a distance function. With a distance function 8 we can associate a meet-representation a=a(6) as

followa: for a EL and x, y E A, let z -y (aa) iff S(x, y) I a. Indeed, aa is a partition. z = y (aaAba) iff x = y (aa) and x = y (ba), which is equivalent to d(z, y) < a and 6(z, y ) I b ; we obtained S(z, y) I a A b , which is the same as x =y ( ( m b ) a ) . Thus aaAba = (aAb)a and a is a meet-homomorphism. To show that a is one-to-one, take a, b c L and a+b; we can assume that a s b . Since 6 is onto, there are x, y € A such that d(z, y) =a. Then x E y (aa). Since a b, z = y (ba) fails to hold, and so aa +ba.

4. Partition Lattices 195

Figure 3

With a distance function 6: A2 -c L and a quadruple (2, y , a, b) (2, y E A and a, b EL) satisfying 6(z, y ) < avb we associate a new set

A* = A U { Z ~ , Z ~ , 23)

with three distinct elements zi, z2, z3 not in A and a new distance function a*: (A*)2 -L defined as follows (see Figure 3) : 6* is symmetric, satisfies 6*(x, x) =0, for all x E A*, and for zi, v A , 6* ( t i , v) =S(zi, w ) ;

6*(z,, 23) =a, 6*(zl, z2) =b, and 6*(z1, 23) =avb;

for u € A ,

d*(u, zl) =S(u, z ) v a , 6*(u, z2) =B(u, x )vuvb , and 6*(u, z3) =B(u, y )vb .

It is easy to verify that 6* is a distance function by virtue of 6(x, y ) 5 avb.

ing zy, y,, € A , ay, by € L, and S(x,, y,) < a,vb,,. For all y < x we define: Now let {(x,, y,,, a,,, b,) [ y < x } be a well-ordered sequence of all quadruples satisfy-

A, = A*, do = 6* ; this construction being performed with the quadruple (x,, yo, ao, b,) ; A, = (U (A , I E <?))*, 6, = (U (6, I 5 <?))*, this construction being performed with

Summarizing, we obtain : For every set A and distance function 6 there exist a set A+ 2 A and a distance

avb

the quadruple (s,,, y,,, a,, by). Finally, A+ = U (A,, I y < x ) and 8+ = U(6,,1 y < x ) .

function 6+ on A+ extending 6 such that, for all x , y E A and a, b E A , if S(x, y) then there exist zl, z2, z3 A+ such that

6+(x, zl) =6+(z2 , z3) =a and d+(z1, z2) =8+(z3, y ) =b.

Now to prove Theorem 4 embed the lattice L into a lattice K having a zero. Let A , = K and define do as above: $(x, y ) = x v y if x+y and d0(x, x ) =O. Again we define inductively sets and distance functions:

Obviously, 6 : A2 - K is a distance function. Let a =a(d) be the meet-represent,ation associated with 6. If x = y ((avb)a), that is, if 6(z, y) 5 avb, then x, y E A, for some n <a, hence there exists q, z2, 23 E A (q, z2, 23 E A,,,) such that 6(z, zl) =6(zz, 2,) = a and &zl, z2) =B(z3, y) = b, and so

x=zl (aa) , z1=z2 (ba), 2 2 2 2 3 (aa) , and Z ~ = Y (ba).

Thus x = y (aavba), proving that (avb)a =aavba. Therefore, a is a representation, in fact, a type 3 representation of K and hence also of L.

We shall give three important applications of this result.

COROLLARY 5 (Ph. M. Whitman [1946]). all subgroups of some group.

Every lattice can be embedded in the lattice of

PROOF. Let a : L -Part(A) be a representation of L. Let G be the group of all permutations of A. To an element a E L, we assign the subgroup H a generated by all transpositions (xy) such that x = y (aa). (The transposition (xy) is the permutation interchanging x and y and leaving all other elements of A fixed.) It is easy to see that (xy) €Ha iff x = y (aa), hence Ha determines aa. Thus a - H a is one-to-one. HaAHb= H a A b is immediate. A transposition (xy) belongs to HaVHb iff (xy) = (xlyi) (xzyz) * * (x ,~,) , where xi = yi (aa) or xi = yi (ba) , which is equivalent to x = y (aavba). Thus HaVHb=HaVb.

A well-known problem asks for the characterization of finite lattices that can be embedded in finite partition lattices. The next result shows that identities cannot be utilized in the solution of this problem.

COBOLLARY 6 (D. Sachs [1961]). then it holds for all lattices.

If an identity holds for ull finite partition lattices,

PBOOF. If an identity holds for all finite partition lattices, then it also holds for all Partfh(A) (because every principal ideal of Part,,(A) is isomorphic to a principal ideal of a finite partition lattice), and so by Lemma 1.4.8 and by Corollary 3 it also holds for I(Partfi,(A)) s Part(A). Since every lattice can be embedded into some Part(A), this identity has to hold for all lattices.

All finite distributive lattices can easily be embedded in finite partition lattices. There is only one result in the literature providing such embeddings for a more comprehensive class of lattices.

b

COROLLARY 7 (A. W. Hales [1970]). embedded in a finite partition lattice.

Any finite sublattice of a free lattice can be

PROOF. Let n be a positive integer and let L be a finite sublattice of P(ii) , the frce lattice with free generators x i , . . . , x,. Since a finite sublattice of a free lattice is isomorphic to a sublattice of some F(n), there is no restriction of generality. Let

4. Ptwtit ion Lattices 197

u, bcP(n) , a+b. Let a =&I,. . . , xn) and b =q(xI,. . . , xn) for some n-ary polynomials p and q ; we shall consider the identity E : p = q. Since a +b, the identity E fails in P(n). By Corollary 6, it fails in some finite Part@), that is, there exist al, . . . , a, E Part@) such that p(ai, . . . , un)+q(al, . . . , an). Extending the map xi-ai (i= 1 , . . . , n ) to a homomorphism we obtain the following statement:

For every a, b € F ( n ) , a+b, there exist a finite set A(a, 6) and a homomorphism y (n , b ) of F ( n ) into Part(A(a, b) ) such that

a y @ , b) 4 b y @ , b ) .

Now let A be a disjoint union of the sets A(a, b ) , a , b E L and for x E L define the partition xv on A by

ti = w (xrp) iff 21, u E A ( a , b) . for some a, b E L and u = v (xy(a, b) ) .

It is easily seen that y is an isomorphism of L into Part(A). The proof concludes by the observation that A is finite since L is finite and all A(a, b ) are finite.

Let us call a representation a : L -Part(A) of type 2 iff for all a, b c L and x, y € A

z =g ( (avb )a ) iff there exist z,, z , c A such that x =zi (aa) , z1 = z 2 (ba), and z 2 - y (aa) .

THEOREM 8 (B. J6nsson [1963]). ,modular.

A lattice L has a, representation of type 2 iff L i8

PROOF. Let L have a representation a : L-Part(A) of type 2 . Let a, b, GEL and I ( 2 r . Since ( a / \ b ) v c < a ~ ( b v c ) in any lattice, we wish to show that aA(bvc)< ( a ~ b ) v c . s o let x, y € A and let x -y ( (aA(bvc) )a) , that is,

x = y (aa) and x =y ( (bvc)a) .

Since Q is a type 2 representation, there exist elements zI and z2 such that

z =z1 (ca), z1 = z 2 (ba), and z2 -y (ca).

Using c < a , we obtain that z i z x (aa), x-y (ua), and y=z2 (aa); thus zl=zz (ua). Also, z1 E z2 (ba), hence

z1 = z 2 ((aAb)a).

Thus z = y (((aAb)vc) a) , implying that (nA(bVC))a< ((UAb)VC)U. Therefore, U A ( ~ V C ) ( U A ~ ) V C , as claimed.

Now to prove the converse, let L be a modular lattice and let K be L with a zero adjoined. Then K is again modular.

For every distance function 6 : A2 -. K and (5, y, a, b ) satisfying 6(x , y) < avb we construct a set A* = A u {zl, z2} having two more elements than A and a distance function d* defined as follows (see Figure 4): 14'

Figure 4

8*issymmetric,satisfiesd(x, z) =OforallxEA*,andd*(x, y) =d(x, y)forx,yEA;

d*(z,, 22) = (d@, y ) v a W ;

for u € A , P ( u , zi) =d(u, z ) v a and d*(u, z2) =d(u, y )vu .

The triangle inequality is again trivial excepting the “triangle” (21, zI, z.,}. We have to verify t,he inequality

6*@, Zl)Vd*(Zl, 22) 2 8 * ( u , 22).

Indeed, d*(u, zi)v6*(zl, z2) =B(u, x ) v a v ( ( d ( x , y)vu)Ab)

(by modularity) =d(u, z)v(((d(z, y )va )A(avb) )

(by triangle inequality) (by absorption, using 6(x , y )<avb) =d(u, z)vd(z, y ) v a

Now we proceed the same way as the proof of Theorem 4. Under what conditions can we improve the “2” of “type 2” to 11 Call a representa-

28(u, y ) v a =B*(u, z2).

tion a: L +Part(A) of type 1 iff, for x, Y E A and a , b E L,

z = y ((avb)a) iff there exists a z E A such that z ~z (aa) and z =y (ba).

This is equivalent to the statement that aa and ba are permutable. Lattices having a type 1 representation satisfy a special identity which is not a consequence of the modular identity.

DEFINITION 9 (B. J6nsson [1953]) 1. Let zo, xi, x2, yo, yl, y2 be variables. We define some polynomials:

zij = ( x i v x j ) ~ ( ~ i v ~ . j ) , 0 < i < i < 3, Z =ZmA(Z02VZ12).

i See also M. Schutzenberger [1945].

4. Partition Lattices 199

The arguesian identity i s

(%VY/O)A ( 5 1 V Y l )A(X,VY2) I ((ZVXi) ~ ~ O ) V ( ( ~ V Y I ) * Y O ) ~

A lattice satisfying this identity i s called arguesian.

REMARK. This identity is a lattice theoretic form of Desargues’ Theorem. To understand the meaning of the identity, the reader should read the discussion of Demrgues’ Theorem and its connection with the arguesian identity in Section 5.

THEOREM 10 (B. Jbnsson [1953]). Any lattice Iiuuing a type 1 representation i s a rguesia n.

PROOF. Let u: L +Part(A) be a, type 1 representation of t,he lattice L. We shall show that Lu is arguesian (this is sufficient since L 1 La). Let a,,, ai, a,, b,,, bi, bl E La; thus a,,, . . . , b, are partitions of the set A. If x, y c A satisfy

5 = Y ((aovb,,)~(aivbi)~(azva,)),

then (see Figure 5 ) there exist uo, q, u z c A such that

x G ui (ai) and ui -y (bi), i = 1,2,3.

Let cii = (a,vai)r\(b,vbj) for 0 5 i < j < 3. Then for all such i and j , ui E u, (c& set c =colA(c,,2Vc,2); then uo = u1 ( c ) . So x uo ((cVai)Aao) and uo -y ((cvbi)&), therefore

5 = Y (( (cvai 1 Aao) v ( (CVbi) &)L proving that La is arguesian.

!)I5 is not arguesian. If we take then z = a , however, the identity demands a

arguesian.

Every arguesian lattice is modular. Indeed, to see this it is sufficient to show that and substitute xo = yi = c, xi =x2 =yo = b, yl =a,

It follows from the results of Section 5 that there are modular lattices that are not b, which is false.


Exercises

1. 2. 3. 4.

Compute IPart(A)I for IAl 5 6. Show that Part(A) is modular iff IAl 2 3. Let A c B . Find embeddings of Part(A) into Part(B). Let A{, i €1, be pairwise disjoint sets and let ai be a partition of Ai, i €1. Define a partition u on A =u (Ai 1 i €1) by

u=w(a) iff u, wcAi for some ic1 and u=w (ai).

Show that this defines an embedding of II(Part(Ai) I i €1) into Part(A). With Ai, i € I , and A as in Exercise 4, is there an (0, 1}-embedding of II(Part(Ai) J ic1) into Part(A)? Is there an embedding of II(Part(AJ I i €1) into Part(rI(Ai I .i € I ) ) ? Let A be a set and let B =A u { z } ( zgA) . For X

6.

6. 7. A define a partition a ( X ) on B by

x =y ( a ( X ) ) iff 2, y EX u {z}.

Show that X + a ( X ) is an embedding of P ( A ) into Part(B). Using Exercise 7, show that every distributive lattice has an embedding into a partition lattice. Show that every finite distfibutive lattice can be embedded in a finite part,ition lattice. Let % be a finite (universal) algebra. Define the concept of a congruence relation of an algebra. The congruence relations of 9.l form a lattice, C(%). Verify that the congruence lattice of a finite algebra can be embedded in a finite partition lattice.

11. Following 0. Ore [1942], for a, BEPart(A) we define a graph G(a, B ) on the set

a u j3; an edge {X, Y} of U(a, B) is a block X of a and a block Y of B such that X n Y =k 0. Show that the blocks of ad@ are maximal connccted subgraphs of

Let u, pEPart(A). Prove that the modular equation

8.

9.

10.

8) . 12.

aA(BVY) =(aAB)vy

holds for all y 2 u ( y EPart(A)) iff G(a, B ) is a tree (that is, a graph without cycles) (0. Ore [1942]). Let B, yEPart(A). Show that 13.

a N 8 v r ) = (aAB)v(aAy)

holds for all aEPart(A) iff B and y are permutable (0. Ore [1942]). Let a, yEPart(A). Prove that 14.

aV(8Ay) = ( a v B b ( W 4

holds for all @EPart(A) iff a is the zero of Part,(A), or a 2 y, or y has only one nontrivial block and y 2 a (0. Ore [1942]). Show that the zero and the unit are the only distributive elements of Part(A). Let D be a distributive ideal of Part(A). Show that every element of D is contained in another element of D with only one nontrivial block. Also show that v D is a distributive element of Part(A). Show that Part(A) has no distributive ideal.

16. 16.

17.

5 . Complemented Modular Lattices 201

18.

19.

20. 21.

*22.

Usc Theorem 111.3.10 and Exercise 17 to show that Part(A) is simple (0. Ore [ 19421). Define type 72 rcprcsrntations. Find a rcprcsentation which is not of type 7) for any 72 < 0. Find a representation which is not associated with a distance function. Disprove the converse of Corollary 7. Disprove the convcbrsc of Theorem 10 (R. J6nsson).

5. Complemented Modular Lattices

The key to the investigation of complemented niodular lattices is provided by modular geometric lattices and projective spaces. We begin with some properties of modular geometric lattices.

LEMMA 1. geometry.

Let L be u w d u l a r yeoinetric lattice und let ( A , -) be the associated

(i) For p , q E A and p +q, p -q i f f there exists a n r E A such that r +a, q and r 2 pvq.

(ii) Let X Thuv the elements { O , p , q, r , pvq} form a diumond.

A . Then X is a aubspuce i f f p , q E X implies that r E X for ull r S p v q .

PROOF. (i) Let z be a complenient of 11 and q and define r = (pVq)Ax. Then

= (pVX)A(pVq) = 1 A(pVq) =pVq. p v r =pV((pVq)AZ) (by niodularity and p l p v q )

Similarly, qvr =pvq. Obviously, pAr =qAr = O . Hence

h(r) =h(pvq) + h ( O ) - h ( p ) = 1.

Thus r A . Since r +O and r A p = rAq = 0 , we have r +p, q.

prove by induction on n that

p iv - - - ~ p ? ~ ,

@ (ii) Let X ( c A ) satisfy the condition that r < p q and p , q EX imply that r EX. We

r E A , r and pl, . . . , p , E X imply that r E X.

This isobvious for n = 1 and by assumption for n =2. Let us assume this statement for n - 1 and let r , pi, . . . , pn be given satisfying r<piv. * *vp,. We can assume that {pi, . . . , pn} is independent, since otherwise r l p i , v . - -vpik for some {ii, . . . , i,}c{ 1, . . . , n} and so r E X by induction. With a =p2v* -vp,, we have then rva<plvu and h(rva) =h(p , va )=n , hence rva=piva. If ria, then r EX by the induction hypothesis, so we can assume that r a. Thus rAa = p l ~ a = O . By the proof of (i), t=(ptVr)Aa is an atom and, by the induction hypothesis, t X . Since t , pl E X and r < tvpi we conclude that r E X . This completes the proof of the “if” part of (ii); the “only if” part is trivial.

Figure 1

LEMMA 2. Under the conditions of Lemma 1 , the geometry (A , -) satisfies the following property (see Figure 1 ) :

The Pasch Axiom. If p , q, r , x , y are points, x l p v q , y < qvr, and n+y, then there is a point z such that z< (pvr)A(xvy) .

REMARK. Apart from degenerate cases, the Pasch Axiom requires that if a line x v y intersects two aides, pvq and qvr, of the triangle { p , q, r}, then i t interaects the third side, pvr. If x +p, q and y +q, r , then we shall also have z +p, r .

PROOF. If 1{p, q, T, x, y}I < 5, then we can always choose a z E { p , q, r , x , y } satisfying ZI (pvr)r\(xvy). If r l p v q , again one can choose z = x . So let 1{p, q, r , x , y].l=5, and let r S p v q . Then h(pvr) =h(xvy) =2 and h(pvqvr) =3.

Thus h((pvr)A(xvy)) = h(pvr) + h(xvy) - h(pvrvzvy) .

Now observe that pvqvxvy =pvqvr, hence

h((pvr)r\(xvy)) =2 +2 -3 =1,

and SO z = (pvr)A(xvy) is the required atom. We can see from Lemmas 1 and 2 that the important properties of a geometry

associated with a modular geometric lattice are formulated only with pointg and lines. This pattern is followed in the next definition:

DEFINITION 3. Let A be a set and let L be a collection of subsets of A. ( A , L) a's called a projective space i f f the following properties hold:

(i) Every 1 EL has at least two elements. (ii) For any two distinct p , q E A there i s exactly one 1 C L satisfying p , q C 1. (iii) For p , q, r , x, y E A and l , , 1, E L satisfying p , q, x Eli and q, r , y el , , there exist

z E A and I, , l4 EL sa,tisfying p , r , z E 1, and x , y , z E 14.

Let us call the members of A points, and those of L, lines. For p , q c A , p+q, let p + q denote the (unique) line containing p and q ; if p = q set p + q = {p}.

5. Complemented Modular Lattices 203

A set X A is called a linear subspace iff p and q E X imply that p + q c X . If X and Y are linear subspaces define

X+ Y = U(z+y I x E X and y e Y ) .

LEMMA 4. If X and Y are linear subspaces of u projective space, then so is X + Y .

PROOF. Take the points p , q, r such that r Ep +q and p , q E X + Y . We wish to show that r € X + Y . Now if p , q E X or p , q E Y , then r E X u Y X + Y , since X and Y are linear subspaces. If p E X and q Y (or the other way around), then r c X + Y by the definition of X + Y . Therefore we can assume that p B. X u Y and so there exist p x E X and p v E Y such that p ( p x + p y . Now we distinguish two cases. Firstly, if q E X u Y , say q X , then consider q +p,, q + p , p +px , and r +p, (see Figure 2 ) . We can assume that q , p , p , , pu, and r are all distinct, hence r + p , has a point in coninion with q + p and p +p, , hence by 3(iii) there is a t EX such that t c q + p x and t Er +p,. If t =p,, then p E X , a contradiction. Hence t +pV and PO r c t +pg E X + Y . Secondly, let q B X u Y . Again we apply 3(iii), this time to p + p x , p + q , p x + q , and p V + r (see Figure 3). Hence there is a point t E p x + q such that r E t + p , . By the first. case, t E X + Y . Again by the first case, r € X + Y .

Now we come to the crucial step:

THEOREM 5 . lattice.

The linear subspaces of a projective space form a modular geometric

PROOF. Since the intersection of any number of linear subspaces is a linear subspace again, we have a closure space ( A , -). For X A , the closure x can be described as follows: set X o = X , X i = X + X , ... , X n = X n - l + X n - l , . . .; then x = u ( X i [ i = 0, 1 , 2 , . , .). It follows immediately, that ( A , -) is an algebraic closure space, and so the linear subspaces form an algebraic lattice and for the linear subspaces X and Y , X v Y = X u Y .

X Y

Figure 2

r Figure 3

304 IV. Modular and Scamimodular Lat,tices

If X , Y , and 2 are linear subspaces and X 2 2, then XA( Y v Z ) 2 ( X A I’)vZ obviously holds.

Now let ~ E X A ( Y V Z ) , that is, p c X and p E Yv2. Since p E Y v Z = Y +Z, there exist pl/ Y and pc €2 such that p E p , +pz. From X 3 2, it follows that p and p , I$ X . If p=p,, then pEZ, and so pC((Xr\Y)vZ. If p+p , , then p , E p + p , E X . Thus p , E X A Y and p , € 2 ; therefore, p E ( X A Y ) v Z . This proves that the latt,ice is modular.

3.8(i), (ii), (iv) have been verified or are obvious and (iii) follows from modularity. @

We have proved somewhat more than stated. We also obtained that the linear subspaces define a, geometry whose subspaces are exactly the linear subspaces.

Call a geometry projective if the associated geometric lattice is modular.

THEOREM 6. There i s a one-to-one correspondence between projective spaces (defined by points and lines) and projective geometries (defined as geometries with rtiorlulur subspace lattices). Under this correspondence, linear subspaces of projective spaces correspond to subspaces of projective geometries. -

Applying the results of Section 3 we obtain that every modular geometric lattice is a direct .product of indecomposable modular geometric lattices, which by Lemma l( i ) are characterized by the property that in the associated projective space evemy line hus at least three points; such projective spaces (and projective geometries) are called nondeyenerate.

COROLLARY 7. Every modular geometric lattice can be represented us a direct product of modular geometric lattices which are associated with , nondeyeii erate projective geometries.

The most important example of a modular geometric lattice associated with a, nondegenerate projective geometry is the following.

Let D be a division ring (that is, an associative ring with unit in which u-1 exists for any a E D , a + O ) and let m > 0 be a cardinal number (finite or infinite). We take a set 1 with 111 =m and we construct V = V(D, m), an m-dimensional vector space over D , as the set of all functions f : I - D such that f(i) = O for all but a finite number of i C I : we define h = f + g and k=af ( a E D ) by h ( i ) = f ( i ) + g ( i ) and k ( i ) = u f ( i ) , respectively. U is a submodule of V iff U is nonempty and f , gE U imply that f +ag E U , for all a E D. The submodules of V form a lattice denoted by L ( D , tn). It is easily seen that L ( D , m) is a modular geometric lattice. The atoms of L ( D , m) are exactly the submodules {a! I aED} where f is any element of V which is not the identically zero function fo.

The submodules {uf I a E D } (f +fo) form the points of the associated projective geometry. If {af I aCD} and {ag I a E D } are distinct points, then the points of the line through these two points are

{ a ( 4 + Y 9 ) I aED} , x, Y ED, w/o . f ( +f0) will be called a representative of the point {a/ I a ED}.

5. Complvmentad Modular Lattices 305

This geometry is nondegenerate: i f f and g are representatives of two distinct points, then f - g represents a third point of the line.

This projective geonietry is typical of projective geometries in which Desargues’ Theorem holds.

Let us call a set of points X collineur iff X c 1 for some line 1. A triplet (a,, a,, a,) of noncollinear points is a triangle. Two triangles (a,, al, a,) and (b,, bl, b2) are perspective with respect to the point p iff a,+b,,a,+ai+bl+bj for O ( i , j<3, and thc points p , a, , b, are collinear for i = O , 1,2 . They are perspective with respect to a line 1 iff co,, c,,, c,, I, where cl j is the intersection of a, + ai and b, + bj. The classical Desargues’ Theorem states that if two triangles are perspective with respect t o a point, then they are perspective with respect to a line.

Now we shall prove that the arguesian identity of the last section is a lattice theoretic formulation of Desargues’ Theorem. See B. J6nsson [1953], [1954], [1959 b] and M. Schutzenberger [1945].

THEOREM 8. identity i f f Desuigues’ Theorem holds in the associuted projective geometry.

Let L be a modular geometric lattice. Then L satisfies the arguesian

The proof of Theorem 8 will follow immediately froni Lemmas 9-11.

LEMMA 9. the utotns of L i f f Desurgues,’ Theorem holds in the associated projective geometry.

Let L be a modular geometric lattice. Then the urguesiun identity holds for

PROOF. Let us substitute the atoms a,, a l , a,, b,, b,, b, in the arguesian identity. Let us form the c , ~ , 0 1 i < j l 2 , c = c ~ ~ A ( c ~ , v c ~ , ) , and p = (a,vbo)A(aivbl)A(u2vb2), see Figure 4. We have to verify that

p I ( ( c vu?) Aao)v( (cvb,)Ab,).

Let us assume first that u,, a,, a2 and b,, b,, b, are not collinear, al+bl, i = O , 1, 2, and that the triangles (a,, a l , a,) and (bo, bi, b,) are perspective with respect to the point p . Then the cli are all distinct atoms and they are collinear iff c = C ~ , A ( C ~ , V C ~ ~ )

is an atoni, in fact, c =coi, otherwise c =O. If c =cot, then ((cvb,)~ao)v((cvbl)r\ai) is the line uUvui, and so p( ((cvbo)r\ao)v((cvbl)r\a,), otherwise c = O and (cvb,)Aa, = (cvb,)r\al = 0 and p $0. Thus, under the conditions stated, the identity holds iff the geometry satisfies Desargues’ Theorem.

Now observe that if the above conditions are not satisfied, then p < ((CVU,)AU,)V

((cvb2)Ab0) always holds. There are several cases to consider ( p = O , 1) is a line, u,, a t , u, are collinear, b,,, b,, b, are collinear, a , = bi for some i, and so on) but all are trivial.

LEMMA 10. which each variable x, (0

Let L be u niodulur geoitietric lattice. Let p be a n n-ary polynomial in i < n ) occurs at most once. Then for a n atom a and elements

b”,. - . ) bn-i EL, a l p ( b 0 , - - * 9 ba-1)

206 IV. Modular and Semimoduler Lrbtticr.s

Figure 4

holds i f f there exist ao, . . . , an- L such that

a < p(ao, . . . , an-l) , ni< bi for all 0 < i < n, ulzd a.i is a n atom if bi +O.

PROOF. We use induction on the rank of p . Let p =.xi. Then u< bi and so we can take ai=a. Now let us assume the statement to hold for p , =po(xo, . . . , x n - * ) and pi =pl(yo, . . . , Y ~ - ~ ) , where {xo,. . . , x ~ - ~ } n {yo, . . . , Y ~ - ~ } = 0. First, let p = p o ~ p i . Now if a l p @ , , . . . , bn-l , b ; , . . . , b k d 1 ) =po(bo,. . . , bn-l)Api(b&. . . , then (I < po(bo, . . . , bnv1) and a < po,(bG, . . . , b&-l). So by the induction hypothesis, we can choose a i l bi and a; bi such that ai is an atom if bi+O and uj’ is an atom if bj’ +O, and a l p o ( a o , . .. ,a%.-JApi(a&. . . ,ah- l ) =p(ao, . . . (Observe that if xi = yi, then we would not know how to choose ai =.I.) Second, let p =povpl. Then a<po(b,,, . . . , bn-l)vpi(b& . . . , bk-l) . If po(bo,. . . , bn- l ) =0, then a l p l ( b h , . . . , bb-,), hence we can choose the a: by the induction hypothesis. We choose ai as an arbitrary atom - < bi if bi+O and let ai = O if bi =O. If pl(b;, . . . , b;-J = O we proceed similarly. Now if po(b,, . . . , bn-l) +O and pl(b;, . . . , 4 0 , then Lemma 4 and Theorem 6 imply that there exist atoms a’<po(bo,. . . , bn-l) and a”<pi(b&. . . , bh- l ) such that a l a ’ v a ” . Applying the induction hypothesis twice we obtain uo, . . . , an- l , a”, . . . , u ; - ~ suchthat ui(bi(O<i<n),al<bE (O<i<m), a i i s ana tomi f bi+O, a: is an atom if bi=/=O, and a’<po(ao,. . . , unPl), a ” l p l ( u & . . . , am-l). Thus

,

completing the induction.

5 . Cnmplcmc~nt~cd Modiilur Lattices 207

LEMMA 11 (G. Gratzer and H. Lakser [1973a]). Let us assume that we cun write the lattice identity E in the form p q, where p and q are lattice polynomials and each variable occurs in p at most once. If E holds for the atoms and the zero of a modular geometric lattice L , then E holds for L.

PROOF. Let b,, . . . , b p P l L ; we want to verify that

l j ( b o , * 2 bn-1)<q(bo7 * * * J bn-l)*

It is sufficient to show that a < q(b,, . . . , b, -1) f o r any atom a < p(bo, , . . , bfl So let n be an atom and a<p(bo, . . . , bnPl) . By Lemma 10 we can choose u ; I bi such that the aL are atoms or zero, ui( bi for O < i < n , and

a l p ( a 0 , . . . ,a,-l).

P("0, * * - , an-J i !I(",,, * * * , an-,).

By assumption, E holds for atoins and zero, hence

Using these two inequalities and the fact that q is isotone, we obtain

r r iP(" , ,* . . . . C i , l - l ) < q ( " , . * * , un- l )<q(bo, * - * , bn-l) . 0

PROOF OF THEOREM 8. If Desargues' Theoreni fails in the projective geometry, then by Lemma 9 the arguesian identity E fails in L. Conversely, if Desargues' Theorem holds in the projective geonietry, then by Leniiria 9 E holds for the atoms of L. A trivial step extends this to the stateiiient that E holds for the atoms and the zero of L. Observe that E is of the foriii required in Lemma 11, hence E holds in L.

('OROLLARY 12. luttice L(D, m).

Desurgues' Theoreiu holds in the projective space associated with the

O ROOF. With a subiiiodule U we associate an equivalence relation a ( U ) on V = V(U,nt) defined by fa (U)g iff f - g c U . Then obviously U - a ( U ) is a type 1 representation of L(D, m), hence by Theorem 4.10 L ( D J m ) is arguesian. Thus by Theorem 8, Desargnes' Theorenr holds in the projective space associated with L(D,n t ) . 0

The next result is a well-known theorem of geometry; it is the classical proof of Uesargues' Theorem in a three-dimensional space.

THEOREM 13. Let L be u modular geometric lattice. Let us assume that the projective geometry associcited with L i s nondegenerate. If the length of L i s at least 4, then L is arguesian.

PROOF. Let the triangles (ao, ai, a2) and (bo, b,, b2) be perspective wit,h respect to the point p . A plunc~ u is an element of height 3. u = a o ~ u I ~ a 2 and @ = bovb1vb2 are planes.

Let us assume that a+@. Since a v b = p v u =pv@, h ( a v p ) =4, and SO h(aq‘?) = h ( a ) + h ( @ ) -h(av@) = 3 + 3 -4 = 2 ; thus UA@ is a line. The lines a,vui and b,vbi are in the plane pvupv(ii, hence h(cZj) = h(aiv/i7) + h(b,vbj) - h(pva,vaj) = 1, and so cd is a point. Since a,vaj is in the plane u and b,vbi is in @, chi is in a@. Therefore, c,),, co2, and cI2 are collinear.

Now let us assume that u =@. Since the length of L is a t least 4, a is not the unit element. Take a 7~ < L, n >u. Let 1 be a relative complement of 0: in [ p , n]. Then It ( I ) = h ( n ) + h ( p ) - h(a ) = 4 + 1 - 3 = 2, so 1 is a line. The projective geometry associated with L is nondegenerate, hence we can choose two distinct points p’ and 1)’’ on

Now define d,= (p’va2)A(p’’vb2), i = O , 1,2. It is easily seen that (do, d l , d,) is a triangle and the plane 6 =dO~divd2+a, @. Furthermore, (a,, ai, a2) and (do, d,, d2) are perspective with respect to p‘, and (b,, bl, b2) .and (do, dl, d2) are perspective with respect to 11’’. Thus we can apply the first case to conclude that (a,vai)A(d,,vdl) and also (bovbl)A(d,vdl) are in 1’ =aAd =/?AS; but this implies thah (a,vai)A(b,vb,) E l ’ , hence col 1’. Similarly, co2 and ci2 1’. This shows that the two triangles are perspective with respect to the line 1’.

Before we come to the classical Coordinatization Theorem we prove one more lemma.

1, P’ S’P, P’’=kP.

LEMMA 14. A projective space satisiying Desargues’ Theorem also satisfies the converse.(or-“dual”) statement :

If two triangles are perspective with respect to a line, then they are perspective with respect to a point.

PROOF. Let us use the notation of Figure 4. We now assume that col, cO2, and c12 are on a line 1 and we do not have the point p . Define p as the intersection of the lines ai + b, and a? + b,. To show the two triangles perspective with respect to p we have to verify that a,, b,, and p are collinear.

Consider the triangles (a2, b2, cO2) and (u,, b,, coJ; they are perspective with respect to the point c12. By Desargues’ Theorem, the intersections of the corresponding sides, a,, b,, and p , are collinear.

Now we are ready to state the classical Coordinatization Theorem of Projective Geometry.

THEOREM 15. Let L be a directly irreducible arguesian geometric lattice of length at least three. Then there exist a division ring D, unique u p to isomorphism, and a unique cardinal number m such that L = L(D, m).

REMARK. For lattices of finite length this was already included in 0. Veblen and W. H. Young [1910] (of course, in a form appropriate for projective spaces). Some authors claim that this proof is not complete and that no complete proof appeared until J. von Neumann [ 1936/37]. The condition of finite dimensionality was eliminated in 0. Frink [1946].

5. Complemented Modular Latt,icc.s 209

a b a + b

Figure 6

The method described below is “ von Staudt’s algebra of throws ”. A more modern (maybe less intuitive) proof can be found in R,. Baer [1952]; see also E. Artin [1940].

SKETCH OF PROOF. A complete proof of this result is too long and of not enough interest for a lattice theory book. However, the mere statement that there is such an isomorphism 9 : L L(D, m) is insufficient for workers in lattice theory. So we choose a middle course: we are going to construct D and 9 but will not provide detailed proofs.

Let I be an arbitrary line in L ; we fix three distinct points of 1 and call them 0, 1 , and m. We set D = I -{-} and we define addition and multiplication on D.

Let a, b D. Fix two distinct points p and q not in L such that 0, p , and q are collinear (see Figure 5) . Then form the points r = (aVp)A(qv-) and s= (pv-)A(bVq). Then set

~1 + b = ( rVs )A l .

It is easy to see that a + b does not depend on the choice of p and q. Indeed, let p’ and q’ be another pair of distinct points not in 1, such that 0, p’ , and q’ are collinear and let us form r’ and s‘ as before. Then ( p , q, r ) and (p’ , q’, r’) and also ( p , q, s) and (p’ , qf, s’) are perspective with respect to the line I ; 80 by Lemma 14, they are perspective with respect to a point, in fact, the same point u, that is

u,p,p’; a , q , q ‘ ; u , r , r ’ ; u,s,s‘

are all collinear, respectively. Hence ( r , s, q) and (r’, d, q’) are perspective with respect to u, therefore by Desargues’ Theorem, they are perspective with respect to a line Z’. But (svq)r \ (s ’vq ’ ) = b Eland ( r vq )A( r ’ vq ’ ) =cQ €1, hence.! =Z‘. Therefore, ( r v s ) A ( r ‘ v d ) € I , that is,

( r V 8 ) d = ( r ’ V d ) A Z ,

showing that ( I + b is independent of the choice of p and 4. To define trb, choose two distinct points p and q not in 1 such that p , q, and 00 are

210 IV. Modular arid Serniniodular Lattices

Figure G

collinear (see Figure 6). Then we form the points r = (iVp)A(qVb) and s = (Ovr)A(pV(/) and we define

cxb = (qVS)Al.

Then D is a division ring with 0 as a zero and 1 as the unit element. Now we choose a maximal independent set I of atoms of L. Let 111 =tn and we

consider the lattice L(D, m) where each atom is represented by a function f : I -r D which is 0 a t all but a finite number of elements; we wish to associate with each r of L such a function d,: I -D.

Fix p , t c I , p+t and set 1 =pvt. We choose a point t , on 1, t,+p and t,+t. We define a division ring D on 1 with p as a zero, t, as the unit, and t as infinity.

Now for any q I , q +p and q +t, fix a point q, of the line pvq, q1 +p and q, +q. The map

pg: x ~ ( x v ( ( t v q ) A ( t , v q l ) ) ) A z

is one-to-one and onto between p v q and E and it fixes p and takes q1 into pI. We define d,: I - D for any atom r of L. First, let rsV(I-{p}). Let r < V I l .

where I , is a minimal (finite) subset of I satisfying this. By assumption, p E I,. We define d,(p) = 1 (the unit of D), dr(s) = 0 (the zero of D) for s 4 I,, and for s E I , - {p} ,

dr(8) = ( ( r v V ( 1 1 - { ~ } ) ) A ( P v ~ ) ) Y ~ E D*

Second, if r l V ( I - {p}) , then r l V I i , where I, is a minimal (finite) subset of I - { p } with this property. By assumption, ~41,. Let s be any point on pvr , s+p, s + r . Then s % V ( I - {p} ) , hence d, has already been defined. Now we define

d,(x) =Cl,(x) if xfp and d,(p) = O .

The claim is then that r -4, maps the atoms of L onto the set of functionsrepresent- ing the atoms of L(D,m) in such a manner that collinearity is preserved. Hence this can be extended to an isomorphism cp of L and L(D, m).

Combining Theorems 13 and 15 we obtain

3 . Complrnit.iit,ed Modular Lattices 21 1

COROLLARY 18. Let L be u directly indecomposable modulur geometric lattice of length ut least 4. Then L 2 L(D, m) with a suitable division ri%g D and cardinal mtt rber ni.

The following embedding theorem is the most important embedding result for complemented modular lattices :

THEOREM 17 (0. Frink [1946]). Every complemented naodular lattice L can be ewbedded in u modular geometric lattice K . This embedding can be chosen to be a (0, 1)-embedding and K cult be chosen to satisfy all the identities of L.

REMARK. The last. observation was made in B. J6nsson [1954].

PROOF. By Zorn’s Lemma, every proper dual ideal of L can be extended to a nmximal dual ideal.( Let M denote the set of maximal dual ideals of L. For x c L, we set

R ( z ) = { D l z E D and D E M } .

Take the dual of a ( L ) and let A be the sublattice generated by M. Then K = I ( A ) is a modular geometric lattice and if we identify DEM with the principal ideal it generates, M is the set of atoms of K . Thus there is a projective space ( M , -) associated with K . In fact, we can define (M, -) directly as a projective space:

D E D, +D, iff D 2 D i n D,.

We claim that x - + K ( x ) is a lattice embedding of L into the lattice of subspaces of ( M , -), that is, into K .

This claim proves the theorem, since this is a (0, 1)-embedding; furthermore, by 1,einnia 1.4.8 and its dual, any identity that holds in L holds in the dual of B(L), and therefore in A and K = I ( A ) .

R(z) is a subspace: If Do, D I E R ( x ) and D Z D O n D,, then xED,nDi, hence z E D, and so D C R(x).

K(xr\y) = R(z) n R(y) : D E R(x) n R ( y ) iff x, y ED, which is equivalent to X A Y E D, that is, DC R(xAY).

If x +y, then R( r ) +.ll(y): Let y ;e x, and let xi be a relative complement of x ~ y in [o, XI. Let D be a maximal dual ideal containing xI. ( D exists by Zorn’s Lemma since L has a 0.) Then D C R(x) but D $ R(y), otherwise y, XI ED which contradicts

H(xvy) = H(x)vR(y): We already know that R is isotone so it is sufficient to prove that if D E R(zvy), then D E R(x)vR(y). This is trivial if x or y is 0. So let x, y+O. By Lemma l(ii) we have to prove the following:

If x, y L , x, y + O , and D is a maximal dual ideal satisfying xvy ED, then there exist maximal dual ideals E and F satisfying

XAYl = 0.

xCE, yCF, E E F F D .

i A rna.vimnl tho1 ideal of L is proper (+L) by definition. 15 Gratzer

312 IV. Modular and Scinimodular Lattices

This is obvious if x or y E D ; if x D , choose E = D and let F be any maximal dual ideal F containing y . So let x, y 4 D.

First, let us assume that XAY = 0. Let D, = D n [y) . Since y Q Dl we have [y) 3 D,. Coniputing in B ( L ) , we obtain

[ W 4 Y ) = [X)ADI( = [ X V Y ) )

since inB(L) the A is set intersection; and so by modularity

[ X M Y ) X[X)VDI.

Therefore, the dual ideal generated by x and Di is proper. Thus there exists a maximal dual ideal E 2 [x) vD, ; in particular, x c E. We repeat this trick. Let Ei = E n D. Then D I E , (since E and D are both maximal) and

[ ~ ) A E , = [y) n E n D (use E 2 [y) n D = Dl) = [ y ) n D.

Hence by modularity,

[Y 1 " D 3 [Y )VEl*

Therefore, there exists a maximal dual ideal F 2 [y)vEl ; in particular, y c F. Now since F 2 El = D n E , we conclude that F is on the line spanned by D and E

and since D, E , F are all distinct (E =+F F since x E , y c F , and SAY = O ) , D is on the line spanned by E and F , that is, D 2 E n F , which was to be proved.

Second, if x ~ y + O , then let y1 be a relative complement of X A Y in [0, y]. Then X A Y , = O and xvy1 =xvy , hence there exist inaxiinal dual ideals E and F satisfying x E E , y I E F , and D 2 E n F. Obviously, y c F so E and F will do for x and y .

We can combine the previous results to obtain the following embedding theorems :

COROLLARY 18. Every complemented modular lattice can be embedded in a direct product of lattices L(D,,m,) and subspnce lnttices of projective planes that do not satisfy Desargues' Theorem.

PROOF. This is immediate from Theorem 17, Corollary7, and Theorems 13 and 1.5.

COROLLARY 19. A complemented modular lattice L can be embedded in a direct product of lattices L ( D , mi) i f f L is arguesian.

PROOF. If L has such an embedding it is arguesian by Corollary 12. If L is arguesian, then by Theorem 17 L can be embedded in an arguesian geometric lattice. The direct factorization of Corollary 7 gives us arguesian directly indecomposable geometric lattices, which, by Theorem 15, are of the form L(Di, mi).

Suniniarizing these we obtain a result of B. J6nsson [1960a]:

THEOREM 20. L ure equivulent :

L e t I; be ( I complenienfed ?nodular laltice. The following conditions on

(i) L can be embedded in a direct product of L(D, ,m,) . (ii) L r a 9 ~ be embedded in a n arguesian geometric luttice.

(iii) L can be embedded in the lattice of all subgroups of an ubeliun group. (iv) L has a representation of type 1. (v) L i s urguesian.

PROOF. (i) is equivalent to (v) by Corollary 19. The equivalence of (i) and (ii) was proved in the proof of Corollary 19. (i) implies (iii) since elements of L are represented by subgroups of the additive group of IID,. (iii) implies (iv) was observed in the proof of Corollary 12. (iv) iniplies (v) by Theorem 4.10.

Theorem 20 explains the significance of the arguesian identity for complemented iiiodular lattices. It is interesting that a weaker version of the arguesian identity holds for all conipleniented modular lattices.

Let & : p ( X [ > * * * , x,)<q("1, * - * ) X G )

be the arguesian identity. Choose three new variables 5, y, z and define u = ( X V Y ) A ( Y V Z ) A ( Z V X ) ,

w = ( X A Y ) V ( Y A Z ) V ( Z A X ) ,

Z ~ = X ~ A ( ( U A X ) V V ) , i = l , 2 , . . . , 6 .

THEOREM 21 (G. Gratzer and H. Lakser [1973a]). The identity

E : p(Z1, . . . , 2g)vw

Itolds in tiny complemented modular lattice.

q(2,, . . . , ijj)vw

PROOF.

We claiiii that if a lattice has property (P), then it satisfies the identity E .

cases to consider :

Consider the following lattice property: (P) If (0, ( I , b, c , i} ifi a diamond, then (a] is an arguesian lattice.

Let t , r , s L and let u s consider the substitution x = t , y = r , z =s. There are two

Case 1. (tvr)r\(rvs)A(svt) = (i!Ar)v(rAs)v(sAt) =a. Then Zi =x,Aa and so 6 becomes

p(X1Aa, . . . , z6Aa)Va<q(xlAcc, . . . , xGAa)Va.

But p ( r I A ? , . . . , XGAU) <p(a , . . . , a) = a and siiililarly for q, so 6 becomes ass, a triviality.

Case 2. i = (Ivr)A(rvs)A(svt) +(tAr)v(rAs)v(sAt) =o . Then it follows from the diagram of FM(3)) that o and i along with a = (iAt)VO, b = (ifvr)vo, c = (iA8)VO form a diamond. Thus property (P) applies and (a] is arguesian. Since XlAa,. . . , XgAa E (a] we obtain

J j ( X i A U , . . . , %,/\a) Q ( X l A U , . . . , ZGAU) ;

joining both sides by o we get that t? holds in L. 1B*

214 IV. Modular and Somimodular La.t ticrs

Next we claim that property (P) holds for the two types of lattices listed in Co- rollary 18. Naturally, (P) holds for an L(D, rn) because L(D, m) is arguesian. Let L be the subspace lattice of a projective plane. If {o, a, b, c, i} is a diamond, then a is either a point or a line. In either case, (a] is arguesian.

By Corollary 18, every complemented modular lattice K can be embedded in a direct product of lattices having property (P) and therefore satisfying 1. Thus K also satisfies E .

It is eaay to construct a niodular lattice in which .5 fails. Let L be the subspace lattice of a projective plane in which Desargues) Theorem fails. By our discussion in Lemma 9, there are seven atoms of L, a,, ai , a2, b,, b,, b,, and d , such that

p(a0, a1, a2, bo, biY b,) = d , da,, "i, a?, b", b,, b2) =o. Let e be a complement of d ; e is a dual atom of L. Let K = L u {b, c, i} such that with o=e and a = 1 (the unit of L), (0, a, b, c, i} is a diamond. It is immediate that K is a modular lattice. Substitute xi = a,, x, = al, x2 = a2, xg = b,, x4 = b,, 2; = b,, x = a ( = l ) , y = b , and z=c. Then u=i , v=o, (uAx)Vo=a(=i) , hence x i = ? ; , for i = 1, 2 , . . . , 6. Thus i yields dvo<Ovo, that is, 1 < 0, which is not true. We conclude :

COROLLARY 22 (R. P. Dilworth and M. Hall [1944]). that cannot be embedded in a complemented modular lattice.

There i s a modular ltrtfice

The modular lattice, K we constructed above is of length 4. The next result shows that this is best possible.

THEOREM 23. Every modular lattice B of length at most three can be embedded in a complemented modular lattice C. The embedding can always be constructed to be 11

{0, 1)-embedding and C can be chosen to have the length of B.

We shall prove Theorem 23 using the concept of a projective plane.

DEFINITION 24. of A satisfying:

(i) Every 1 E L has at least two elemente. (ii) For any two points p , , p l E A , p,+pl, there is exactly one ICL satisfying p,,

(iii) For any two distinct lines lo and l , , there is exactly one p E A satisfying p €2, and

A projective plane ( A , L) is a set A a d a collection L of subsets

Pi € 1 -

P E 1,.

A partial plane satisfies (i)-(iii) with " a t most one " replacing '' exactly one " in (ii) and (iii). Observe that the first two of the three propsrties already define a partial plane.

It is easily seen that a projective plane is a projective space.

3 . Coniplt~mrntocl Modular Lattices 215

T>EMMA 25. every partiul plane ( A , L) there is a projective plane (A’ , L‘) such that A every 1 E L there is un 1‘ E L’ satisfying 1

Every pa.rtial plane can be embedded ise a projective plane, that is, for A‘ and for

1’.

PROOF. Let ( A , L ) be a partial plane. Define

A+=Au{p(Z ,m) I1,rnCL and l n m = @ } ,

L+ = (1 u {2J(l, m) I m EL, 1 n m= 0} I EL} u {{I), q} I p , P E A , P*P> I), q‘E 1

for no 1 E L} ,

where p(1, m) = p ( m , I). The new points defined are distinct from each other and from the points in A .

Now define A o = A , L o = L , A , + l = ( A , ) + , L , + l = ( L , ) + , f o r n < w , and

A’= U ( A , I n <o).

Lines are defined as follows: let 1, be a line in ( A f i , L,); then we obtain thelines&+, for k 2 T I :

lk + 1 = lk u {2 ) (1k ) m) I m E Lk, 1, n m = 0)-

L’ is the collection of all sets of the form u(Zk I k 2 n ) . It is obvious that (A’ , L’) satisfies the requirements.

PROOF OF THEOREM 23. If the length of B is less than 3, the statement is obvious. So let B be of length 3. We define a partial plane ( A , L) as follows: the points of (-4, L) are of two kinds, namely, the atoms of B and for every a E B, h(a) =2, which contains exactly one atom, a new point p(a). For every a E B, h(a) =2, containing at least two atoms we define a line { p I p is an atom of B, P S I } ; if aE B, h(a) = 2 , and (I contains exactly one atom p , then we define the line { p , p(a)}. It is clear that A , L) is a partial plane. Embed ( A , L) into a projective plane (A’ , L’) by Lemma 25

and let C be the subspace lattice of (A’ , L’).

Exercises

1.

2.

3.

4. 6 .

l h d a gcwnetry ( A , -) in 11 hicli X is a subspace iff r 5 povp,vp, and p,, p i , p , EX imply that r E S , but ( A , -) is not a projective geometry. Why do we need x *y in the Pasch Axiom? Phrase the Pasch Axiom so that this assumption can be dropped. For points p and q of a projrctivr gcometry, define p’q iff there is a third point r 5 pvq. Show that this relation is an equivalence relation. Derive Corollary 7 from Exercise 3. Prove by dircct computation that the projective geometry associated with L(D, ttt) satisfies Drsargues’ Theorem.

216 IV. Modular and Semimodular Lstticcs

6.

7.

*8. 9.

*lo.

11. 12.

13.

Show that the Pasch Axiom is equivalent to the following implication holding for all points: '

I f a. 5 alvaz and a i ~ ( a o v a z ) 5 a3va4, then a0 I (agA(ar,V (a,A(a,vaz))) )v((aoVadA (((a1Vao)A~z))V ((a3v (a1 A (a"V%)))Aa4)).

Prove that the implication of Exercise 0 holds for all subspaces of a projeot,ive space. Prove that the D constructed in the proof of Theorem 15 is a division ring. Let K and L be modular geometric lattices and let p be a one-to-one map from the atoms of K onto the atoms of L. Prove that p can be extended to an isomorphism iff

a I bvc iff ag, 5 bpvcp

holds for any atoms a, b, c of K . I n the proof of Theorem 15, verify that for atoms r, 8, f of L, r 5 svt iff ad,+ ad,+ cd, = 0 holds for some a, 6, c 6 D. Combine Exercises 9 and 10 to prove L z L ( D , m) in Theorem 15. Show that the identity E is not self-dual and that the dual of E could also be used in Theorem 21. The projective plane constructed in Lemma 26 is the free projective plane grnerated by the partial plane. What does " free " mean for projective planes?

Figure 7

5 . Coniplerntdnttd Modular Lat,tirc.s 217

\ U

I /

14.

1 A.

16.

17.

18.

19. 20.

Figure 8

Investigate the lattice theoretic properties of the embedding of the subspace latticc of ( A , 1,) int,o t,hc subspace lattice of <A+, L + ) in the proof of Lemma 25. Why could Lvrnma 25 not be ext,ended to provide embeddings of modular lattices of length 4Y Let L be a modular lattict: of length 3 or 4. Let a be the join of all atoms of L and b thc meet of all dual atoms of L. Consider the situations shown in Figure 7. Show that t,hey (along wit8h their duals) suggest a complete classification of all modular lattices of Imgt'h 3 or 4 with the exception of distributive lattices and of complemented modular lattices (B. J6nsson [ 19591). Develop a dualit,y theory for project,ive planes. The dual of ( A , L ) is (L, A dual), where, for aEA, a dual={Z I aEZ}. Show that for a finite projcctivc plane ( A , L),

IAl= JLI.

What is thtb " dual " of Desargues' Theorem ? On the Euclidean plane fix a line 1 (see Figure 8) . Let u be a line and let a be the angle determined by 7c and 1. A refracted l ine is defined &s follows: if 0 5 ci n/2, then form the line u, with angle a/2; below 1 the refracted line is u, above 1 it is a,. Otherwise, u is the refracted line. Define a projective space as the Euclidean plane with a line a t infinity; the lines are the refracted lines with a point at infinity and the line a t infinity. Prove that this defines a projective plane in which Desargues' Theorem fails.

Figure 9

218 IV. Modular and Semimodular Lat’tices

*21. Pappus’ Theorem is said t,o hold in a projective geometry iff whenever the points a,,, a,, a2 and b,, b,, b2 are collinear, and all six are in the same plane, and ci j= = ( ~ , v b j ) ~ ( a i v b ~ ) (0 2 i <j < 3), then col, co2, and ci2 are collinear. (See Figure 9.) Prove that if in a projective space Pappus’ Theorem holds, then the division ring constructed in Theorem 16 is commutative. Prove the converse of Exercise 21. Use the Coordinatization Theorem to show that, LI finite projective geometry of length four or more and its dual are isomorphic. Prove that in a finite complemented modular lattice the number of k-element subsets of atoms whose join is the unit is equal to t,he number of k-element subsets of dual atoms whose meet is the zero. In a finite modular lattice let L, (U,) denote the set of all (a+ 1)-element subsets { p , qI,q2,. . . , q,} such that everyqicoversp ( p covers every qi) . Show that lLnl == I U,l for every positive integer n. (Hint: use Exercise 24.) Spply Exercise 25 t o establish Dilworth’s covering theorem for modular lattices (R. P. Dilworth [1964]): in a finite modular lattice the number of elements covered by precisely k elements is equal to the number of elements covering precisely k elements. In particular, in a finite modular lattice the number of join-irreducibles is equal to the number of meet-irreducibles. (This proof is due to B. Ganter and I. Rival [1973].)

*22. *23.

24.

25.

26.


For finite niodular lattices L, the posets J ( L ) and M ( L ) do not determine L aa they do for finite distributive lattices; but they have many interesting properties. The most celebrated one is a result of R. P. Dilworth [1954]: IJ(L)I = IM(L)I. (See Exer- cises 5.23-5.26 and, for a generalization, S. P. Avann [1958a].) By R. Wille [1976], J(L) and M ( L ) have the same width and, by B. Ganter and I. Rival [a], the same length. In this connection one should mention a result of R. Wille [1974a]: a finitely generated modular lattice of finite length is finite.

The Isomorphism Theorem (Theorem 1.2) is obviously equivalent to modularity. The statement : [a, avb] s [ a ~ b , b], for all a, b EL, characterizes the modularity of L only in special cases: for finite lattices (M. Ward [1939]) and, more generally, for algebraic lattices (P. Crawley [1959]).

The niost important result on join-representations of an element of a finite niodular lattice is the Kuroh-Ore Theorem (Theorem 1.5). It holds for a, finite semimodular lattice L iff L is locally modular, that is, for every a L, [va, a] is modular, where v,, = A(s 1 a>x); R. P. Dilworth [1941 a]. For generalizations to infinite lattices see the survey article R. P. Dilworth [1961] and Chapters 5 and 6 of P. Crawley and R. P. Dilworth [1973].

I n a modular lattice L, a =xov* 0 -VX,-~ is a direct join representation iff {so, . . . , is independent. Then Theorem 1.5 holds for direct join representations in

modular lattices of finite length; this is a famous theorem of 0. Ore [1936], much deeper than Theorem 1.5. See also H. Have1 [1955] and P. Crawley [1962a]. Such join representations are connected with direct factorizations of groups, rings, loops,

Flirt hvr Topics and References 219

and so on. A4 number of papers are devoted to this topic started in 0. Ore [1937], [1938] and A. G. Kurog [1943], [1946]; see M. Grayev [1947], A. H. LivGc [1962], E. N. MoCulskii [1961], [1962], [1968].

An important new line of research in modular lattices was started in I. M. Gelfand and V. A. Ponoinarev [1970]. In this paper all 4-generated (generated by a 4-elenlent set) subdirectly irreducible modular lattices of subspaces of a vector space are determined. This inspired R. Wille, his coworkers, and others to start a systematic search for all 4-generated subdirectly irreducible modular lattices. Partial results of this search can be found in A. Day, C. Herrmann, and R. Wille [1972], R. Freese [1973], C. Herrmann [1973], C. Herrmann, C. M. Ringel, and R. Wille [a], C. Herrmann, 11. Kindermann, and R. Wille [a], A. Mitchke and R. Wille [1973], G. Sauer, W. Sei- bert, and It. Wille [1973], R. Wille [1973], [1973a].

Many of the harder computations in these papers are based on the D,-lemniaand Wj-leiiima of R. Wille [1973]. Two illustrations of these results are given below.

,4 modular lattice conipletely freely generated by a finite poset P (that is, FM(Pm)) is finite iff P does not contain a four-element antichain nor does it contain the poset

[1973a]). This trivially yields an earlier result of R. Wille [1973] determining when a free modular lattice generated by a set of chains is finite.

Let P={o, u, b, c, d, i}, o < u , b , c , d < i , a r \ b = a A c = u A d = b A c = b A d = c A d = o , trvb =(IVC =uvd = bvd =cvd = i; let '$ denote the corresponding partial lattice. Then there are exactly two modular lattices (up to isomorphism) generated by P and containing as a relative sublattice: (m, (which is the same as (m3 but with four atoms) and an infinite (subdirectly irreducible) lattice (A. Day, C. Herrmann, and R. Wille [1972]).

Several attempts have been niade to learn something about the structure of free modular lattices (see, for instance, K. Takeuchi [1951 a] and [1959]). An interesting result on sublattices of PM(4) can he found in R. Freese [a].

A proper subclass of the class of finite sublattices of free modular lattices is the class of finite lattices projective in M. The distributive members of this class are described in A. Mitchke and K. Wille [1976].

Although it is still not known whether there is an algorithm deciding when two polynomials agree in a free modular lattice, the word problem for modular lattices has been shown to be recursively unsolvable in G. Hutchinson [1973a] and [a]. His most recent result is that there exists a five-generated finite partial lattice Q defined by a single equation such that there is no algorithm to decide when two polynomials agree in FM(Q). This may be best possible: if the list of known 4-generated subdirectly irreducible modular lattices is complete, then " five " cannot be reduced to " four ') in Hutchinson's results.

G. &I. Bergman [l969] and A. P. Huhri [1972], [1972a] defined an important class of modular lattices. A modular lattice L is called n-distributive iff the i'dentity

{ X O , YO, YI, Z O , 21)) where YO <yt, 20 < Z I and z0llyi, zollzj, yilIzi, for i, j = O , 1 (R. Wille

xvA (y, I O<i<n)=A(:cvA(yi 1 0 1 j i l t , j+i) I O I i l n )

holds in L. An intwesting result of A. P. Huhn states that a modular lattice L fails to he ,!-distributive iff L contains a sublattice B s (&)"+I with bounds o and i and an element (( which is a relative coinplenient of all dual atonis of B in [0, i]. This

220 I\‘. Modular and Scmimodular Lattices

result has many applications, see, for instance, A. P. Huhn [1975], B. Freese [a], and Section 2 of C. Herrmann [1973].

Theorem 11.3.17 is also true for modular lattices, that is, every finite distributive lattice is isomorphic to the congruence lattice of a suitable modular lattice, by E. T. Schmidt [1974]. The basic construction used to establish this result is further applied in E. T. Schmidt [a] in which a finitely generated simple modular lattice without a prime interval is exhibited.

An equational class K of algebras is called congruence modular iff, for all %EK, the congruence lattice of (21 is modular. Conditions for this can be found in A. Day [1969]. R. Freese [1975a] exhibits an identity E , a weaker form of the arguesian identity, which holds in every congruence lattice of a congruence modular equational class, but E is not implied by the modular identity.

There are t w o related conjectures. 1 (It. N. McKenzie). If a nontrivial lattice identity E holds in all congruence lattices of algebras in an equational class K , then K is congruence modular. 2 (J. B. Nation). If a nontrivial lattice identity E can lie expressed by a Mal’cew type condition (in the sense of G. Griitzer [1970a]), then E implies the modular identity. Some results concerning the first conjecture can he found in J. B. Nation [1974] and A. Day [a]. For some further results and an excellent review of this field see B. J6nsson [1974a].

Necessary and sufficient conditions for the validity of the Jordan-Holder Chain Condition in a finite poset P is given in 0. Ore [1943]; see also S. MacLane [ 19431. A cell C in P is C ={o, ai, . . . , a,, bl , . . . , bm, i}, where (0, al, . . . , a,, i} and (0, h i , . . . , b,,, i} are maximal chains in the interval [o, i] of P and inf -(ak, bi} = 0, sup {ak, b j } = i for k = 1, . . ., n, j = 1, . . . , m. Ore’s condition is: for every cell C , n = m holds (n and m depend on C ) . This is obviously satisfied in semimodular lattices, where, for every cell, n = m = 1. This result has been rediscovered (and published!) half a dozen times since 1943.

The Jordan-Holder Chain Condition for modular elements is discussed in T. Fuji- wars and K. Murata [1953].

Some 23 definitions of semimodularity are discussed in R. Croisot [1951] and [ 1952/53], especially their interdependence with and without chain conditions.

For further results on M-symmetric lattices, the reader should consult F. andS. Mae- da [1970]; see also L. R. Wilcox [1939], [1942], [1944]; S. Maeda [1965], [1969]; and It. Padmanabhan [1974].

For a generalization of semimodularity, see 0. Tamaschke [ 19601. For more general discussions of independence in semimodular lattices, the reader

should consult R. Y. Dilworth [1944] and D. T. B’inkbeiner [1951]. A sublattice of a semimodular lattice is not necessarily seminiodular. In fact,

1%. P. Dilworth proved that every finite lattice can be embedded in a finite geoinetric lattice; this result was published in P. Crawley and R. P. Dilworth [1973]. For earlier related results, see It. P. Dilworth [1941 a] and D. T. Finkbeiner [l960a].

The treatment of geometry from a lattice theoretic viewpoint started in the twenties by K. Menger; his work was continued by J. von Neumann and G. Birkhoff, end later by 0. Prink, L. R. Wilcox, F. Maeda, and others. The reader should consult B. J6nsson [1959b] for a general review of this field; see also the book G. Zappa [1952]. Apart from projective geometries, there are many kinds of geometries that

can be classified from a lattice theoretic viewpoint : these include classical concepts such as uffine geometriw and planur geometries (Xis a subspace iff ($1, q, r } c X for all 11, q, r E X ) as well as some new ones, such as the “n-stufige Geonietrie ” of It. Wille [1967], which has a particularly beautiful lattice theoretic characterization. For other lattice theoretic results of a geonietric nature, see the books M. L. Dubreil- Jacotin, I,. Lesieur, and It. Croisot [1953], F. Maeda [1958], F. and S. Maeda [1970]; the papers of F. Maeda listed in the Bibliography; W. Prenowitz [1948] and U. Sasaki [1952] and [1953].

Geometric lattices, originally called mutroids, developed from a branch of cornbi- natorics.

The study of matroids, in the combinatorial sense, originates with the famous Seven Bridges of Konigsherg probleni solved over two hundred years ago by L. Euler. Euler demonstrated that a graph is a disjoint union of circuits iff each vertex has an even number of edges incident to it. Thus those subgraphs of a graph which are disjoint unions of circuits are closed under syiuinetric difference and given two circuits with nonvoid intersection, the symmetric difference will always contain a circuit. This motivates the circuit definition of a matroid :

Let E be a finite set and C , the circuits, a collection of nonvoid subsets of E. Then ( E , C ) is a mutroid iff ( C l ) No circuit properly contains another. (C2) If C , and C , are distinct circuits and e EC, n C,, then there is a circuit C , such

There exist many niatroids that do not correspond to the circuit structure of it

graph. As an example, consider a chain grozip, that is, a set of niaps from a finite set E to an integral domain 12 which is closed under pointwise addition and scalar multiplication. For any one of these niaps, define the support of the map as the set of those elements of E that do not get mapped onto zero. If the mininial supports (with respect to inclusion) are defined to be the circuits on E , then a matroid results.

An equivalent approach to niatroids is by the rank definition: given a finite set E and an integer valued function r on P ( E ) , ( E , r ) is a matroid iff, for A E and e, el, e2 E E , ( I t l ) r ( 0) = O . (R2) r ( A ) I r ( A u {e}) < r ( A ) + 1. (R3) r ( A ) = r ( A u{e,}) = r ( A ~ { e , } ) implies that r ( A ~ { e , , e 2 } ) =?-(A).

Let ( E , r ) be a matroid and E‘={eEE I r ( { e } ) = O } . Then for any A E E , r ( A ) = r(A -E’) and hence ( E - E’, r ) is a matroid with essentially the sanie structure as ( E , r). For z, y c E , let z = y iff ~ ( { z , y}) = 1. Then is an equivalence relation and if [el], . . . , [e,] are equivalence classes, then r([el] u * - * u [ e , ] ) =r({el, . . . , e,}) irrespective of the choice of representatives. Thus we may identify related elements with no real loss of generality. Having done this, for X

that C, c (C, u C,) -{e}.

E define

x = { e E I r ( X u {e}) = r ( X ) } .

A geometry has been formed and the rank is completely determined by its definition on the closed sets. Hence the structure of the matroid can be determined from the geometric lattice of subspaces of the geometry.

Matroids have conibinatorial application in areas other than the study of the circuit structure of graphs. One such application is in transversal theory. Given sets A,, . . . ,A , ,

222 IV. Modular and Somimodular Lattices

a set X c E = A , u . . - u A,,, is called a partial trunswerscxl iff there is a one-to-one imp : X +{ 1 , . . . , m } satisfying x AJv, for x c X . The partial transversals then satisfy the independence axioms for the set E . It immediately follows, for example, that all maximal partial transversals have the same cardinality. Further results concerning the theory of transversals can be found in the definitive book of L. Mirsky [1971].

Another application of niatroid theory is to the labeling of edges of graphs. Any set of labelings that are closed under addition and scalar niultiplication form a chain group and hence a niatroid. If the edges are labeled with elements from the integers iiiodulo two, the set of labelings such that the sum of the labels on edges incident to any given vertex is zero forms a chain group which yields the ordinary circuits of the graph as its matroid. More generally, a chain group can be formed by using labels from an arbitrary integral domain and including in the chain group those lahelings such that for every circuit C in the graph the sum, S(C) , of the labels on the edges of C is even, i.e., there is an element x in the integral domain such that 22 = S ( C ) . An element in this chain group may be formed by assigning a value g(w) to each vertex and then defining f ( e ) =g(w,) +g(wn) where w, and w2 are the two vertices incident to e . In topological terms this is the usual coboundnry operotor. The set of all labelings that can be formed in this manner forms a chain group, and, as a matroid i t has the same rank as the original chain group. Hence every chain group element arises in this manner. If the integral domain has characteristic two, and a labeling in this chain group exists whose support is all of E , then in the associated vertex assignment no two joined vertices have the same value, If we call these assignments colours, then we are looking a t the classical graph theoretic colouring problem. Thus these chain groups can be used to evaluate the chromatic numbers of graphs. Further considera- tions of this type are found in the book of H. H. Crapo and G.-C. Rota [1971].

The interested reader should consult the two books referred to above and W. T. Tutte [1959], [1965], [1971] and R. J. Wilson [1973].

Many papers have been published on maxiniuni sized antichains in posets, mostly in geometric lattices. For a finite poset P and maxiinuin sized antichains A and B, set A 5 B iff for every u E A there is a b c B such that a 5 b in P. Then the n~aximum sized antichains form a distributive lattice by R. P. Dilworth [1960]; for a simple proof and an application, see R. Freese [1974].

Many generalizations of Sperner’s Lemma are known; see D. A. Drake [1971], 31. J. Klass [1974], D. Kleitnian, M. Edelberg and D. Lubell [1971], D. Lubell [1966]. It is still not known whether Sperner’s Lemnia holds for finite partition lattices. Nevertheless, many nice results have been proved for finite geometric lattices about the numbers W,. Let L be a finite geometric lattice of length n. Then Wl < wk for k = 2 , 3 , . . . , n - 1 (J. G. Basterfield and L. M. Kelly [1968]) and W1 = MTn-i iff L is modular (C?. Greene [1970]); furthermore, for 1 < k < n - 2 ,

wi + - - * + W , I W n - k + * - - + Wn-i, and equality holds iff L is modular (T. A. Dowling and 1%. hl. Wilson [1975]). For finite modular geometric lattices wk = W n - k for all k (It. I?. Dilworth [1954]). For further results see T. A. Dowling and R. M. Wilson [1974].

The proof of these results utilizes the Mobius function, see Exercises 3.32--37 and H. H. Crapo [1966], [1968/69].

Maximum sized antichains in (&)k are determined in I. Rosenberg [1967]. The connections of geometric lattices and coinhinatorics are further explored in

H. H. Crapo and (2.4. liota [l970] and [1971]. Partition lattices were first studied in detail in 0. Ore [1942]. Characterizations of

partition lattices can be found in It. P. Dilworth [1944], U. Sasaki and S. Fujiwara [1952a], and D. Sachs [1961a].

“Which finite lattices can be embedded in a finite partition lattice?” is one of the oldest open problems of lattice theory. Corollary4.6 tells us that we shallnot obtainan answer by looking at identities. K. A. Baker observed that the same holds for all universal sentences: if a universal sentence 0 holds for all finite lattices, then 0 holds for all lattices. (The proof is easy; it uses that a lattice L can be embedded in an ultra product of its finite partial sublattices.) A. Ehrenfeucht, V. Faber, S. Fajtlowicz, and J. Mycielski [1973] consider the problem of finding, for a finite lattice L, the smallest integer p ( L ) such that L can be embedded in Part(X) with 1x1 =p(L). They obtain results for some special classes of lattices.

Lattices having type 1 representations are characterized in B. J6nsson [1959a] based on some ideas of l i . C. Lyndon [I9501 and [1956].

An equivalent form of the arguesian identity that formally resembles Desargues’ Theorem can be found in G. Gratzer, B. Jbnsson, and H. Lakser [1973]. Applying this characterization it is proved that if { o , ( I , b , c, a’} is a diamond in a modular lattice, then the arguesian identity holds in [o , u ] ; this is also used in G. Griitzer and J. Sichler [1975]. By B. Jbnsson [1959], every arguesian lattice of length 1 4 has a type 1 representation.

If a lattice L can he embedded in the lattice of all normal subgroups of a group, then L has a type 1 representation. The converse is false by B. Jbnsson [1954]; the c~ounterexaniple can be chosen to have length 5.

Representations of type 2 and 3 and congruence lattices of universal algebras are considered together in G. Griitzer and E. T. Schmidt [1962] and G. Griitzer and W. A. Lampe [a]; more about this in the Concluding Remarks.

A very simple proof of Whitman’s result, in the stronger form due to B. Jbnsson (Theorem 4.4), is given in S. K. Thomason [1970] for finite lattices; in fact, the representation found is recursive, This is applied in S. K. Thomason [1970a] to prove that every finite lattice is isomorphic to a sublattice of the lattice of degrees of unsolvability. The reader should not find it difficult to generalize the reasoning of 8. K. Thomason [1970] to arbitrary lattices.

An n-partition (n 2 1) of a set A satisfying [A 1 2 n is a collection of a t least n element subsets of A such that every n-element subset of A is contained in exactly one member of the collection, Let Part,(A) denote the set of all n-partitions of A ; note that Partl@) = Part(A). For P , Q E Part,(A), let P < Q iff every member of P is contained in a member of &. This makes Part,,(A) a lattice, see J. Hartmanis [1956] and [1961]. Hartmanis’ main result is that every finite lattice L can be embedded in some finite Part2(A). For other related results, see J. Hartmanis 119591 and B. Frontera-MarquBs [ 19641.

The major step in the generalization of the Coordinatization Theorem was made in J. von Neunrann [ 1936/37] replacing the division ring by a regular ring, L(D, m) by the lattice of principal right ideals of a regular ring, and directly irreducible arguesiarl

32.1 IV. Modular and Semimodular Lattices

geonietric lattices by conipleinented modular lattices with a hornogeneous basis with 4 or inore elements (a homogeneous basis of n elementsis (Il, . . . , u R such that ( a l , . . . ,a,,) is independent, aiv. - .va, = 1, and ai-aj for i + j). Von Neuniann’s Coordinatization Theorem is considered by many the deepest lattice theoretic result.

Von Neumann’s conditions have been relaxed in many ways. The reader should consult I. Amemiya [1957], I. Aniemiya and I. Halperin [1959] and [1959a], K. D. Fryer and I. Halperin [1956] and [1958], B. Jbnsson [1960a], B. Jbnsson and G. S. Monk [1969]; the books F. Maeda [1958], L. A. Skornjakov [196l], the survey article I. Halperin [1961] and the review by I. Halperin of B. Jbnsson [1960a] in the Mathe- matical Reviews.

M. Hall [1954] contains some useful information on projective planes; see also B. Artniann [1969] and R. Wille [b]. For a rather unexpected connection between projective planes and weakly associative lattices, see E. Fried and V. T. S6s /1975].

Complemented modular lattices play an important role as building stones of all modular lattices of finite length; see C. Herrmann [1973c].

Problems

IV. 1.

IV .. .2.

IV. 3.

IV. 4.

IV. 5. IV. 6.

IV. 7.

IV. 8.

IV. 9. IV.10. IV.11.

IV.12. IV.13.

Investigate those finite modular lattices L, for which there is 8 inatching between J ( L ) u (0) &nd M(L) u (1). (I. Rival.) Investigate those finite modular.latticea L, for which there is an isomorphisni y of the posets J ( L ) and M ( L ) that is also a matching. (I. Rival.) Does every finite (countable) modular lattice L have an embedding in a 4-generated modular lattice K ? If L is finite, can K be chosen finite? Is M generated by FM(4) (in the sense of Section V.1)? In other words, if an identity E fails to hold in a modular lattice, is there a consequence E’ of E such that E’ has only four variables and E’ also fails in some modular lattice? Is M generated by its finite members (members of finite length)? Which are the finite (simple) sublattices of FM(4) and FM(X,)? (For P&) see Concluding Remarks.) Is FM(Xo) isomorphic to a sublattice of FM(4) or of F,(n) for some n < w ? Are there n, rn <w, n <m, such that FM(m) is isomorphic to a sublattice

Is the word problem for F,(4) decidable? Is the word problem for FM(Xo) decidable ? I s the word problem in four variables decidable for modular lattices? Describe all 4-generated subdirectly irreducible modular lattices. A modular partial lattice is a partial lattice obtained froin a modular lattice as in Definition 1.5.12. Show that there is no finite set of identities characterizing the modularity of a partial lattice. (See Exercise 1.5.20 for the concept of validity of an identity in a partial lattice.) I s the modularity of a partial latticc undecidable? Investigate modular lattices L with the following property: if alb is a subquotient of c/d and alb =:Id, then alb =cld. Is it possible to define in,such L a “distance function” with values in a distributive lattice?

of FM(n) ?

Problems 225

lY.14.

IV . 1 5.

IV.16. IV.17.

IV.18.

IV.19.

IV.20.

IV.21.

IV.22.

IV.23.

IV.24.

IV.25.

IV.26.

IV.27.

IV.28.

IV.29.

IV.30.

IV.31.

IV.32.

Characterize the congruence lattices of inodular lattices. (Distributive algebraic lattices.) Find a coniiiion generalization of the Kuro8-Ore Theorem (Theorein 1.5) and the Ore Theorem (on direct representations). (For modular lattices of finite length this is done in E. T. Schmidt 119701.) For algebraic lattices, is M-syniinetry equivalent to semimodularity ? Investigate the. construction of ExLrcise 2.12 from thg point of view of M - sy niiric t ry . For infinite sets A and B, are Part(A) and Part(B) elementarily equivalent? If IAl 2 IBI, is there an elementary embedding of Prtrt(A) into Part (B)? Characterize the congruence lattices of semimodular lattices. (Distributive algebraic lattices.) Characterize the congruence lattices of weakly modular lattices. (Distributive algebraic lattices.) Let the lattice L have a type 1 representation. Does L have a type 1 representation associated with a distance function ? Let L be a lattice having a type 1 representation. Does every hoiiioinorphic image of L have a type 1 representation? Show that there is no finite set of identities defining the equational class generated by lattices having type 1 representation. Let L be an algebraic lattice having a type 1 representation. Does L have a type 1 complete representation? Let L be an algebraic lattice having a type 1 representation. Is there a (finit- ary) algebra i)l with permutable congruences such that L is isomorphic to the congruence lattice of a? Let K be an equational class of algebras. If there is a nontrivial lattice identity holding in all congruence lattices of algebras in K, is then K congruence modular (R. N. McKenzie)? If the validity of a nontrivial lattice identity, E , for congruence lattices of algebras in an equational class is a Mal'cev type condition, does th tn E

imply modularity (J. B. Nation)? Find a Mal'cev type condition equivalent to the validity of the arguesian identity for the congruence lattices of the members of an equational class of algebras. To what extent is an equational class H of lattices determined by {C(L) I L E K} ? Check K = D, N5, MJ, M3,3, etc. Let the lattice L have an embedding into a complemented modular lattice. Does every homomorphic image of L have such an embedding? Find an effective algorithm that turns configurational conditions for geometries into equivalent identities for the subspace lattices. Can every finite lattice be embedded in a finite partition lattice? I n particular, does the dual of a finite partition lattice have such an embedding?'

1 A 1cttc.r received by the author in January 1977 from P. Pudlak and J. Tdma contains a remarkable proof that every finite lattice can be embedded in a finite partition latt,icr.

226 IV. Moclular and Srinimoclular Lat ticcs

IV.33. For a finite lattice L, let p ( L ) denote the smallest integer such that L can be embedded in Part(A) with ] A [ c( i , 2) > > c(i, k J , i = 1 , 2 , . . . , n, such that if i =k j , then c( i , p ) is a complement of c ( j , a). Compute p(L(k, , . . . , k J ) . (This is done for n = 2 in A. Ehrenfeucht, V. Faber, S. Fajtlowicz, and J. Mycielski [1973].) Is it true in Part(A), A finite, that any maximum sized antichain is of the form E ( i ) ? Is every finite lattice isomorphic to the congruence lattice of a finite algebra? Is every finite type 1 lattice isomorphic to the congruence lattice of a finite algebra with permutable congruences? Investigate Problems IV.26-IV.28 for infinitary algebras.

IV.34.

IV.35.

IV.36. IV.37.

IV.38. IV.39. IsthereanequationalclassKof infinitaryalgebrassuchthatM ={C(%) 1 % K}?

CHAPTER v EQUATIONAL CLASSES O F LATTICES

1. Characterizations of Equational Classes

In this section we shall discuss the basic properties of equational classes of lattices. Of the four characterizations and descriptions given, three apply to arbitrary equational classes of universal algebras ; the fourth is valid only for those equational classes of universal algebras that are congruence distributive (that is, the congruence lattice of any algebra in the class is distributive). For the sake of simplicity, all these results are stated and proved only for lattices.

For a class K of lattices, let Id(K) denote the set of all identities holding in all lattices of K ; if K ={L} we write Id(L) for Id(K). (The same convention will be used for all “operators” as well; if X is an operator, that is, X is any one of El, S, P, Equ, P,, P,, Si, I, then we write X(L) for X(K) if K ={L} . ) For a class C of identities, let Mod( Z) denote the class of all “lattice models of C”-lattices in which all the identities of C hold. By definition, K is an equational class of lattices iff K=Mod(X) for some set C of identities, or equivalently, iff

K =Mod(Id(K)).

Let E : p = q be an identity where p and q are n-ary lattice polynomials. Let K be an equational class and let FK(n) be the free lattice over K with n free generators, uo, . . . , un-l; F,(n) exists by Corollary 1.5.6. Let FK(Ko) denote the free lattice over K with K O free generators, uo, u l , . . . , 11,) . . . .

Tlet us suppose that E holds for the free generators of F,(n), that is,

p(u,, * * - , l l n - , ) =PI(% - - 9 U,-1).

Let L he any lattice in K and let (to, . . . , a,& -1 EL. Then the map ui +ai, i = O , 1, . . . , u - 1 , can he extended to a hoiiioniorphisni v of F,(n) into L, and so

P((QJ, . . . , <+I) =p(u,v, - * * , un-1v) =p(uo, * * 9 Un-i )Y

=q(uo,. . . , ~ C , - ~ ) ~ ) = Q ( U O V , . . . , univ) =duo, - . . , ~ ~ - 1 ) )

that is, E holds in L. Thus E C Id(K) iff E E Id(F,(n)).

If we make no restrictions on the arity of p and q we obtain:

E E Id(K) iff E E I d ( F K ( & ) ) .

This nieans that K is completely deiermined by FK(Ko). 10 Griitzer

228 V. Equational Classes of Lattices

THEOREM 1. (i) There is a one-to-one correspondence between equational classes of lattices and

(up to isomorphism) free lattices with KO generators, wherein H corresponds to

(ii) A lattice L is a free lattice with KO generators over some equational class H of lattices iff L has a countable generating set U such that every map U -I, can be extended to an endomorphism of L.

FK(KO)’

REMARP. (i) sets up the correspondence between equational classes and countably generated free lattices while (ii) describes which lattices occur in this correspondence. It follows from the results of Chapter V I that the U in (ii) is uniquely determined as the set of doubly irreducible elements of L.

PROOF. (i) has already been proved. The “only if” part of (ii) is trivial. To prove the “if” part, let L and U be given as in (ii). We wish to construct an equational class K such that L is free over K. Let K be the class of all lattices A such that any map U -,A can be extended to a homomorphism of L into A. Let C be the set of all identities E : p = q such that p(uo, ui, . . .) =q(uo, ui, . . .) holds in L for all uo, q,. . . E U .

We claim that K = Mod(C). Indeed, arguing as at the beginning of this section, we obtain H C Mod(X). Conversely, let A E Mod(C) and let a : U + A be a map. If E : p = q is an identity and p(uo, ui, . . .) =q(uo, ui, . . .) for uo, ui, . . . E U (with all ui distinct), then E E C. Since A E Mod(C) we obtain that E holds in A and therefore p(uoa, . . .) = q(uoa,. . .). Thus (see Exercise 1.5.45) a can be extended to a homomorphisni and so LEH.

This proves that K = Mod(C), so H is an equational class. By the definition of H, L is free over H.

Let uo, ui, . . . be the free generators of F(Ko) and let vo, vl, . . . be the free generators of FR(KO). Then ui -,vi, i =0,1 , . . . extends to a homomorphism Q, =Q,,; let 0 = 0, be the kernel of ‘p. Since, by Theorem 1, the equational class K is determined by FK(KO) and FK(K0) is determined by 0, we conclude that H is determined by 0. If we can ascertain which congruences arise this way, we shall have another description of equational classes.

Call a congruence relation 0 of a lattice L fully invariunt iff a = b (0) implies that aa = ba (a), for all a, b E L and for all endoniorphisms a of L.

THEOREM 2 (B. H. Neumann [1962]). equational classes of lattices and fully invariant congruence relations of F(K,).

There is a one-to-one correspondence between

PROOF. Let K be an equational class and let Q,, 0 (=OK), ui, v i he as desmibed above. We show that 0 is fully invariant. Let a he an endomorphism of P(K,,), let a, b P(K,,), and let u = b (0).

Let p be the endomorphism of FK(KO) extending v i - , u t a ~ . Since u E F(X,), a =

1. Characterizations of Equational Classes 229

p ( ~ " , . . . , un-l) for some integer n and for some n-ary polynoniial p. We compute:

~ v P = P ( u o , * 3 u n - i ) ~ P = ~ ( u ~ @ , - * - 9 Un-1@) = P ( V O ~ , - * - 3 v N - I B )

=p(uOaq, . . . , un-1aq4 =p(uO, . . . , =say,

and similarly for b. Thus,

and therefore aa = bu (0). Now let 0 be a fully invariant congruence relation of F(Xo). We shall show that

F(XO)/O satisfies the condition of Theorem l(ii). If u i = u j (0) for some i + j , then it is easily shown that 0 = I and so F(&)/@ is the one-element lattice. So let us assume that ui+ui (a), for all i + j . We take U=([u ,]@ 1 i = O , 1,. . .} and a : U -F(Ho)/O. Let [ui]Oa=[ai]O, i = O , 1 , . . . Then ui+ai, i = O , 1,. . . , can be extended to an endomorphism fl of F&). Since 0 = Ker(cp) is fully invariant, we have Ker(v) Ker(fl9) and so xq~ -x& extends a to an endoniorphisni of F(K,) /@.

We have aheady proved in Section 1.4 that an equational class is closed under the formation of homomorphic images, sublattices, and direct products. The converse, which is due to G. Birkhoff [1935], is the third description of equational classes.

aup, = aq$ = bq$ = bay,

THEOREM 3. formution of homomorphic irnuges, sublattices, and direct products.

A cluss K of lattices is u n equational class iff K i s closed under the

REMARK. The direct product of an empty family of lattices is the one-element lattice and therefore if K is closed under the formation of direct products, then K is not. the empty class. Observe also, that if K is closed under the formation of homomorphic images, then K is closed under the formation of isomorphic copies.

PROOF. Let K be closed under the formation of homoniorphic images, sublattices, and direct products. If K consists of one-element lattices only, then K can be defined by the identity: x,, =xi and so K is equational. Now we can assume that H contains a lattice of more than one element. Therefore, we conclude just as in Section 1.5 that F,(m) exists for any cardinal m.

Let L E Mod(Id(K)) and take an P,(m) E K with ILI + K O =m. We denote by U the set of free generators of FK(m). Since ILI + KO = I UI, there is a map a of U onto L. Let p ( ~ ~ , ul, . . . , u n - l ) =q(uo, ui, . . . , u ~ - ~ ) hold in FK(m) with uo, . . . , u , -~ E U. Without loss of generality we can assume that the ui are all distinct. Therefore, p =q E Id(F,(m)) (as argued at the beginning of this section) and so, by the freeness of F,(m), 21 =qc ld(K). Because L c Mod(Id(K)), p = q E Id(L) ; in particular, p(u,a,. . . , un-la) =p(uOa, . . . , i(n-la). This shows that a satisfies the hypothesis of Exer- cise 1.5.45 and can therefore be extended to a homomorphism ,!? of F,(m) onto L. Thus L is a homomorphic image of a meniber of K, and so L E K.

The converse was proved in Lemma 1.4.8. To obtain a slightly different version of this result, we introduce sonie notation.

For a c l a s ~ K of lattices, let H(K), S(K), and P(K) denote the class of all homomorphic images, sublattices, and direct products of members of K, respectively. 10.

230

COROLLARY 4 (A. Tarski 119461). smallest equational class containing K.

V. Equational Classes of La1 t ices

Let K be a class of lattices. Then HSP(K) i s the

PROOF. We start out by observing three formulas for any class K of lattices: (i) SH(K) E HS(K).

(ii) PH(K) E HP(K). (iii) PS(K) c SP(K).

To prove (i), let L c SH(K). Then there is an A c K and a homomorphism u of A onto a latt,ice B containing L as a sublattice. Set C = (x I x c A and xu c L}. Then C c S(K) and L c H(C). Hence L E HS(K), proving (i).

To prove (ii), let L c PH(K). Then there exist Ai c K and a homomorphism ui of Ai onto B , for i c I , such that L= n(Bi I i c l ) . It is clear that L is a hoinoniorphic image of II(A, I i c I ) c P ( K ) , proving (ii).

The proof of (iii) follows that of (ii) with Ai a sublattice of Bi. To show that HSP(K) = H I is an equational class using Theorem 3, we have to

KI, and P(Kl) E K,. Indeed (using (i)-(iii) and that verify that H(K,) c K,, S(K,) XX(K2) =X(K2) for Xc{H, S, P} and any class K2) :

H(K,) =HBSP(K) = HSP(K) =Hi ,

S(K1) = SBSP(K) E HSSP(K) = HSP(K) = K,,

P(K1) =PHSP(K) G HPSP(K) E HSPP(K) = HSP(K) = Kl.

If K2 is any equational class containing K, then K, =HSP(H2) 2 HSP(K) =Hi .

Let Equ(K) denote the sniallest equational class containing K. Then Corollary 4 can be suininarized in the formula Equ =HSP.

For a class K of lattices, let Ps(K) denote the class of all lattices that are isomorphic to a subdirect product of members of K. The following variant of Corollary 4 is not especially useful, but the construction used in the proof has found some applications.

.

COROLLARY 5 (S. R. Kogalovskii [1965]). A class K of lattices is an equational cltrss iff K is closed under the formation of homomorphic images and subdirect products. Moreover, for any class K of lattices,

Equ(K) =HPs(K).

REMARK. In other words, Equ = HP,,

PROOF. the following inequality : (iv) S(K) E HP,(K).

To verify (iv), let L c S(K), that is, let L be a sublattice of some A H. Now take A' with 111 = K O and form a sublattice B c A' defined as follows: for f c A', let f E B

We leave it to the reader to verify that both statements follow readily from

1. Charactrrizations of Equational Classcs 231

iff f( i) = u, for soiiie a E L and for all but finitely many i E I ; a map q~ : B -c L is defined by iiiapping f to this a . It is trivial that BEPs(A) and L c H ( B ) , concluding that

We have obtained two formulas for Equ, namely, Equ = E S P = HP,. Their usefulness is, however, somewhat limited when applied to describing members of equational classes of lattices. A great improvement of these formulas is possible for lattices and we shall proceed to develop this.

Let L,, i E I , be lattices, let I=?= 0 , and let I, be a dual ideal of the lattice P ( I ) of all subsets of I . For f , g E H(L, I i E I ) , we set

L E H P s ( K ) . 0

I(/, g ) ={i I i c I and f(i) = g ( i ) } .

( I in I ( / , y) stands for “identical”.) We introduce a congruence relation 0 = @(a) on L = II (L , j i E I ) . For f , g E L , let

f = g (0) iff I ( / , g) €3.

(View nieiiibers of a) as “sets of measure one”. Then f ~g (0) iff f and g are identical on a set of iiieasure one.) 0 is obviously reflexive and symmetric. I f f , g , h E L, f - g (a), and g =lz (M), then I ( / , g) and I (g , 11) so that I ( / , h ) 2 I(/, g ) n I(g, h ) €1,. Thus I ( / , h ) €a) and f E /L (0). If f , g, h EL and f - 9 (@), then I ( f ~ h , g A h ) 2 I ( / , g ) €3 so that I ( f A h , g A h ) € a ) and f A h - g A h (0). Similarly, f v h z g v h (0). Thus 0 is a congruence relation and we can form the lattice L/O, denoted by n (Li I i E I ) , a, called a reduced product of ( L , I ic I ) . When 1, is a prime dual ideal (also called, ultra filter)-briefly, 1, is prime over I-, n (L, I ic I ) is called an ultra product (also called, prime product); see J. E6s [1955]. For a class K of lattices, P,(H) will denote the class of all lattices that are isomorphic to an ultra product of members of K.

a

We start out by proving some elementary properties of these constructions :

LEMMA 6. Let I be (c nonvoid set and let 3 be a dual ideal of P ( I ) . Let J €a, J * 0, and define &‘ = { X n J I X €a}. Then i$ is u dual ideal of P ( J ) and if 1, i s prime so is 8. k’urtliermore, for any family (Li I i c I ) of lattices

IIa(LiI i E I ) r I I (L , I i E J ) . i5

]’ROOF. The statenients concerning i$ are trivial. Let n denote the homomorphism /+IJ7 the restriction of f to J , and let 9: n ( L i I i E I ) - H a ( L i I i c I ) and y : n ( L i I I.€ J ) - IIg(Li I i c J ) be the natural homomorphisnis. 0 =@(a) is the kernel of q~. Let Q, he the kernel of ny. Then f = g (0) iff I ( f J , gJ) E 27; or equivalently, J n I(/, g) €3. In view of J €a, this is equivalent to I ( / , g) €1,. Thus 0 =@ and the isomorphism follows.

COROLLARY 7 . principul, 1, = [ J ) , then for any fa,rtiily (Li 1 i c I ) of lattices,

Let I be a nonvoid set and let B be a dual ideal of P ( I ) . If a i.9

IIa(Li I i E I ) = II(Li ] i c J ) ;

232 V. Equational Classes of Latatices

in particular, if B is principal and prime, 3 = [ { j } ) , then

I13(Li I i E I ) 2 Li.

LEMMA 8. be finite lattices. Let (Li I i E I ) , I =k 0 , be a family of lattices each Li being one of L o , . . . , L,,-i, Let 3 be prime over I . Then there is a j , 0 < j < n, such that

IIB(Li 1 i E I ) Y L j .

Let Lo, . . . , L,

PROOF. We can assume that the lattices L o , . . . , L,,-l are pairwise nonisomorphic. Thus if we define Ii = { i I i E I and Li = Li} for 0 j < n, then I,, . . . , I, - are pairwise disjoint and I , u - - * u In-1 = I. Since B is prime, there is exactly one j with 0 < j < n such that Ii EB. Applying Corollary 7 to I j , we obtain

where 8 = {X n Ii I X €a} is prime over I j . For a E L,, let f , E II (Li I i E I i ) be defined by f ,( i) =a , for all i E I. Then for a, b E Li, a =i= b, we have I ( j a , f b ) = 0 4 8, and therefore fa+fb(@(8)) . Thus a: a + [ f a ] @ ( 8 ) embeds Lj in II (Li I i E I j ) . I n fact, a is an isomorphism. To see that, let f E II (Li I i E I j ) . Then for each a E L we define I j , , = { i 12’ E I j and f(i) =u}. Since the Ij,a are pairwise disjoint and u ( I j , , 1 a E Li) = I j , we conclude that there exists exactly one a E Lj such that I j sa E 8. Now I ( / , f a ) = Ij ,a € 8 , thus f = 1, ( O ( 8 ) ) and a is onto. Therefore, II (Li I i c I i ) s LP

8

8 Now we are ready to state the formula of B. Jbnsson [1967] for Equ:

THEOREM 9. For u class K of lattices,

Equ(K) =PsHSPu(K).

PROOF. By the Subdirect Product Representation Theorem it is sufficient to prove that if A is a subdirectly irreducible lattice and A EEqu(K), then A E HSP,(K). By Corollary 4, AEHLIP(K) and therefore there exist AiEK, i C I , a sublattice B of II(Ai I iE I ) , and a congruence relation @ on B such that B/@ 1 A.

We claim that there is a 3 prime over I such that @(a) restricted to B is contained in 0. Indeed, if we have such a 3, then by the Second Isomorphism Theorem, B/0 is a homomorphic image of B/@(B),, which is, in turn, a sublattice of II(Ai I iEZ) /@(B)=IIB(Ai I i E I ) . Thus, A ~ B / @ ~ E S P , ( K ) , as required.

For J C I, set 0, = @ ( [ J ) ) , that is, for f , g E n(Ai I i € I )

f -9 (0,) iff I ( / , 9 ) 2 J .

Observe that @(a) = u (@,I J €3) ; consequently, to find B we have to look for 3 in

8 ={J I J E I and (Oj)B<@}.

1. Characttrrizatioris of Equational C~RSSCR 233

8 has the following three properties: (i) I C 8 a n d 0$i%‘;

(ii) J o c 8 and Jo E J , G Z imply that J, E 8 ; (iii) M , N c Z a n d M u N E 8 i n i p l y t h a t M c 8 o r N E 8 .

prove (iii), let M and N 8, =o and 0, = L , so (i) is obvious. If J o C J , , then OJ0 2 OJ,, proving (ii). To

I. It is trivial that OMUN and

( @ M U N I B = ( @ M ) B A ( @ & ) B .

Now let M u N E i%‘. Then

( @ M ” N ) B < @,

( @ M ) B A ( @ i V ) B S @.

that is,

Since B/@ is subdirectly irreducible, @ is meet-irreducible in C(B) . So using that C ( R ) is distributive we conclude that that is, M or N C 8 , proving (iii).

Now let B be a dual ideal of P(Z) maximal with respect to thep rope r tyBc i5 . We show that B is prime. By (i), 0 $B so B is proper. If B is proper but not prime, then there exists a J C I such that J$B and I - J @ . If J n J ’ E 8 for every J’ EB, then by (ii) B and J would generate a dual ideal contained in 8, contradicting J @ and the maximality of 3. Thus there exists a J o c B such that J n Jo$i%‘. Similarly, there exists a Ji €3 such that (Z - J ) n J i $8. Then

@ or

J,, n J , = ( J fl ( J o n J1)) u ((1 - J ) n ( J o n J1 )I, contradicting (iii). ThusB is prime.

of Theorem 9 is the following: Let Si(K) be the class of subdirectly irreducible lattices in H. An equivalent form

COROLLARY 10. For a class K of lattices,

SiEqu(K) c HSP,(K).

We shall illustrate the power of Theorem 9 with two simple applications taken

If H is a finite set of finite lattices, then by Lemma 8 P,(K) is, up to isomorphic from B. J6nsson [1967].

copies, the same as H, so we conclude: b

COROLLARY 11. Let K be a finite set of finite lattices. Then

SiEqu(K) G HS(K).

Observe that HS(K) is, up to isomorphic copies, a finite set of finite lattices.

234 V. Equational Classcs of Lattices

COROLLARY 12. lAj

Let A and B be finite nonisomorphic subdirectly irreducible lattices. If lBI, then there exists an identity E holding i n A but not holding i n B.

PROOF. 11, B 4 Equ(A) and so some identity holding in A must fail in B.

B $ HS(A) since I BI 2 ] A I and B is not isoniorphic to A . Hence by Corollary

Exercises

1. 2.

3.

4.

5,

6. 7.

8.

9.

10.

11 .

12.

13.

14. 16.

16.

17. 18. 19.

20.

Show that K -Id(K) and C +Mod@) set up a Galois connection. Prove that K is an equational class iff K =Mod(Td(K)). For a set C of identities, S =Id(K) for some class of lattices iff C =Id(Mod(C)). Characterize sets of identities I: that are of the form Id(K) for some class K of lattices. Find properties of equational classes K of lattices satisfying Id(K) =Id(FK(?t)) for some integer %.

Let L be a lattico and let U be a generating set) of L. Show that if L is free ovt’r K with U a13 a free generating sot, then the same holds over Equ(K). Reprove Theorem l(ii) using Exercise 6 and Theorem 3. Prove that the elements of a free generating set are doubly irreducible (B. J6nsson [1971].) Let R be an equational class of lattices. Let p be the natural homomorphism of F(Ko) onto E I K ( K ~ ) , and let 0 be the kernel of ~ 1 . We assign to 0 an equational class Ki over which P(KO)/O is free. Prove that K =Ki . Let 0 be a fully invariant congruence relation of F(No). We assign to 0 an equational class K, and to K a fully invariant congruence relation Oi of F(Ko) as in the proof of Theorem 2. Prove that 0 =Of. Let L be a lattice. Show that the fully invariant congruence rclations of L form a complete sublattice of C ( L ) . Prove that the lattice of fully invariant congruence relations of a lattice L is a distributive algebraic lattice. Characterize the compact elements. A congruence relation 0 of a, lattice L is celled invariant iff a-b (0) implies that aa=ba (0) for any automorphism a of L. Do the invariant congruences form a lattice, if so, is this lattice distributive or algebraic? Let ( A ; 3’) and ( A ; B) be algebras and P c B. Show that the congruence latticc of ( A ; a) is a complete sublattice of the congruence lattice of ( A ; F). Derive Exercises 10-12 from Exercise 13. An equational class K of lattices is generated by a lattice A iff K =Equ(A). Show that every equational class of lattices is generated by a lattice. An equational class K of lattices is Zocallyfinite iff every finitely generated member of P is finite. Prove that an equational class generated by a finite lattice is locally finite. (Do not use Theorem 9 or any of its consequences. This result is true even in classes of algebras for which Theorem 9 fails.) Is the converse of the statement of Exercise 16 true? Prove that P$I(K) For a class K of lattices, let PR(K) denote the class of reduced products (up to isomorphism) of members of K. Prove that PR(H)cPsPu(K) (G. Gratzer and H. Lakser [1973]). An implication is a sentence (allvariablesareuniversallyquantified) :p&o,. . . ,z, -1) =

HPs(K) for any class K of lattices.

1. Charactrrizatio:is of Equational Classes 235

21.

* 2 2 .

*23.

24.

25 .

*26.

2 7 . 28.

'9.

30.

31.

32.

33.

qO(.ql, . . . , x , ~ -1) and * * * and p, ,b- l (q l , . . . , x n - l ) =qm-I(xo,. . . , x ~ - ~ ) imply that pttL(xO, . . . , q - 1 ) =q,(z,,, . . . , z r l - l ) . (For instance, the semidistributive law of Chaptrr V I is an implication: Z A ~ = Z A Z implies that ZAY = z A ( ~ v z ) . Any ideiit,ity is an implicat,ion.) An impZicutio~~~2 C ~ U S S is the class K of all lattices satisfying a set of i~nplications. For a class K1, the class of all isomorphic copies of members of K, is clenoted I(Kl) . Prove that ISPE(K) is the smallest implicational class containing R (A. I. Mal'cev [1966] and G. Griitzer and H. Lakser [1973]).

Let a, be prime over I . An ultra,power L' of a lattice L is an ultra product na,(L.i I i C I )

siich that Li = I, for all i €1. Prove that L can be embedded into L Provc? that, rvi'ry latt,iccx L can be embedded into an ultra product of all finitrly generstcd sublatticrs of L. (Hint: let I be the set of all nonempty finite subsets of 1,. For J C I , lot L, be t,he sublattice gc.iierated by J . Choose a 2, prime over I such that for each J €1, { K I KEI and K 3 J } €3.) Lt.t 11s dcafine (first>-order) formulas: (i) for wary polynomials p and q, p =q is a for- nul la; ( i i ) if @ is a formula, so is T @ (read: not @); (iii) if @" and @, are formulas, t,licsIi so is @,v@, (read: @" or a,); (iv) if @ is a formula, so is ( 3 s k ) @. Then @"A@, (ao and is intjroduced as ~ ( ( T @ " ) v ( - T @ ~ ) ) , (zk) @ is introduced as ~((Ixk)-i@), and so on. Define inductively what it means for a formula @ to be satisfied by ccrtain elements of a lattice L. 11t:finc. precisely a free variable x which is not "bound" by a quantifier 32. A sentence is a formula without free variables. Find sent,enccs expressing that the latt,ice L has at most or exactly n elemrnt,s. For a finik lattice K construct a sentence @ such that @ holds for a lattice L iff L has a sublattice isomorphic to K. Yro\-e t,hat a sentence @ holds for an ultra product na,(Li I i € I ) iff { i I @ holcls in

Prove Lemma 8 using Exercise 26. A model of a set C of sentences is a lat,tice L in which all sentences @ EC are satisfied. Prove the Compucttzess Theorem: Let C br a set of sentences. If every finite subset of C has a modt.1, then C has aniodel. (Hint : for every nonemptyfinite R c C choose a model I&. Let I be tho set of all finit.c noncmpty subsets of C and choose a a, prime ovt'r I containing all sets of the form {X I X C l and X z J } , nhrre J C I . Then L =n Let C and C, be sets of sentences. C implies C, iff every model of C is also a model of C,. X is equiaulent to C, iff C implies C1 and Ci implies C. Prove that if C is equivalent, to {a}, thvn there is a finite C, E C that is equivalent to {@}. (Use the Com- pactness Theorem.) Call a lattice L finitely subdirect2y irreducible iff w is meet-irreducible in C ( L ) . Prove that Corollary 10 holds also for the finitely subdirectly irreducible members of

Let YX4 bc the subspace latt,ice of a projective line with four points. Show that >I4 =Equ(9Jt,,) 1 M 3 =Equ(qR,) and for every cquat,ional class K, M, 3K2M3 implies that K = 81:). Find an eriuational class N of lattices such that NxN, =Equ(\)15), N D i)I, and NxKzN, implies that I< -N, for any equational class K. Show that Equ(K) =SHPS(K) if K is one of the classes of lattices listed below:

( i i ) K = (9RX }, where 9Rxo is the subspace lattice of a projective line with countablp

a, I a -

Li} €3 (J. LO$ [1955]).

(Ln ~ R €1) is a modrl of C.) 3

Equ( 10.

( i ) K = {LJ, wherc L is a finite lattice;

0 many points;

236 V. Equational Classes of Latticrs

(iii) for each prime number p we choose a field F , of charactcristicp and H = {L13 I p is a prime number}, where L, is the lattice of subspaces of the projective plane coordinatized by F,.

34. Let K be an equational class of lattices. Prove that

for any cardinal m. Prove the Compactness Theorem for any type of universal algebras. Find applications of this result to lattices that go beyond the Compactness Theorem of Exer- cise 28. (Hint: use a type with A and v and infinitely many constants.)

FK(m) CIsP(FK(KO)),

35.

2. The Lattice of Equational Classes of Lattices

Let KO and Hi be equational classes of lattices. Then KO n HI is again an equational class and

KO n K, = Mod(Id(H,) u Id(&)).

There is also a sniallest equational class K,vK, containing both KO and Hi, namely,

HovKl =Mod(Id(H,) n Id(&)).

All equational classes of lattices form a lattice with the lattice operations H o ~ K , = K,nK, and K0vHI. (Axiomatic set theory does not permit the formation of a set whose elements are classes. Thus, formally, one cannot form the lattice of equational chases. It is easy to get around this difficulty. For instance, replace an equational class H by Id(K), which is a subset of the countable set of all lattice identities. Then we form the lattice of all subsets of the form Id(H) of the set of all lattice identities.)

The zero of this lattice is T (the trivial equational class) consisting of all one- element lattices. Let H be an equational class of lattices different from T. Then there is a lattice L in K wit<h ILI > 1. Therefore L has g2 as a sublattice and. so g2 c K. By Corollary 11.1.20, Equ(@,) = D the class of all distributive lattices. Thus K 2 D. We have just verified that D is the only atom of the lattice of equational classes of lattices and every nonzero member contains D. Now let K be an equat,ional class of lattices properly containing D. Then there is a

nondistributive lattice L in H. By Theorem 11.1.1, 'iR5 or 9J13 is a sublattice of L, hence 'iR5 or 9J13 c K. Set

N 5 = E qu ('iR5) , M, =Equ(9Jl3).

Then H z N 5 or K ZM,. Thus D is covered by exactly two equational classes, N5 and M,, and every equational class properly containing D contains N, or M3, see Figure 1.

The correspondence set up in Theorem 1.2 between equational classes of lattices and fully invariant congruences of F(Ko) is an isomorphism between the dual of the lattice of equational classes of lattices and the lattice of fully invariant congruences of F(X0).

2. The Lattice of Equational Classes of Latt,ices 237

L

6 T

Figure 1

THEOREM 1. The lattice of equational classes of lattices is distributive and dually algebraic. The dually compact elements are exuctly those equational classes that can be defined by finitely wiciny identities.

PROOF. Observe that an identity p = q corresponds to forming the sniallest fully invariant congruence under which p(zq,, u,, . . .) =q(uo, u l , . . .). Thus the compact fully invariant congruence relations correspond to finite sets of identities.

The distrihutivity of the lattice also follows from the following formula (B. J6nsson [ 19671) :

THEOREM 2. Let KO ctnd K, be equational classes of lattices. Then,

Si(K,vHi) = Si(K,) u Si(KI).

PROOF. Apply Corollary 1.10 to K = K , u K l : if LEK,vKl and L is subdirectly irreducible, then L HSP, (KO u K,)? that is, L c HS(Ll), where Ll is an ultra product of lattices from KO u K,. Let L, = nB(Li I i E I ) , where Li E K,,'u K,. Set Ii = { i I i € I and Li c K j } , for j = O and 1. Then I, u I , =I EB, hence I, or I, EB. If I j €3, then by Lenima 1.6, Ll EP,(K,) =Hi. Thus we conclude that Ll €KO or L,EK,.

Figure 2

Thus K -. Si(K) is analogous to the set representation of distributive lattices. (It is easy to turn this into a set representation, see Exercise 2. ) In particular, we should note that if Si(K,) c Si(Ki) and up to isomorphisni Si(K,) has only onc more niember, then K,-tK,. To illustrate this, consider the latticesgx,, and 9x3,3 of Figure 2 , arid the equational classes M, and M3,3 they generate, respectively. By Corol- lary 1.11, Si(M,) C HS(rrJz,) and Si(&J C HS(rrJzm,,,). So up to isomorphic copies, Si(M,) ={(I2, ma, ';n,} and Si(M3,3) ={(I2, 9X3, (Jn3,s}. Thus, M3-tM, and M3<J13,3 in the lattice of equational classes of lattices (see Figure 1) . Also, M3 = M 4 ~ M 3 , 3 . ' All the equational classes considered above are of finite height.

LEMMA 3. T h e collection of equational classes of lnttices that can be pnerccted by ( I single finite lattice i s a n ideal of the lattice of equutimml classes of lattices. T h i s ideal contains only elements of finite height.

PROOF. If the equational class K is generated by a finite lattice L, then by Corol- lary 1.11, up to isomorphism Si(K) is a finite set of firiitelattices. Thusif KO c K, thenup to isoniorphisni Si(K,) must be a subset of this finite set, hence there are only finitely many such KO. All the statements of this corollary now follow immediately.

From the observations niade above it should be clear that the join-irreducible elements of the lattice of equational classes of lattices are connected with equational classes generated by a single subdirectly irreducible lattice. The following result states a number of connections of this sort (It. N. McKenzie [1972]):

THEOREM 4. Let K be a n element of the lattice of equcitional classes of lattices. (i) If [K) i s u conipletely prime dual ideal (that i s , V ( K i I i E I ) E[K) impZies thrrt

K i E [K), for some i I ) , then K can be generrrted by (L f ini te subdirectly irreducible lut tiee .

(ii) If K can be generated by u finite subdirectly irreducible lattice, then K is completely join-irreducible.

(iii) If K i s completely join-irreducible, then K can be generuted b y a subdirectly irreducible lattice.

(iv) If K can be generated by a subdirectly irreducible lattice, t h e i ~ K i s join-irre- dzieible.

2. The Lattice of Equational Classes of Latticcxs 239

PROOF. (i) Let K,, denote the equational class generated by the partition lattice on an

n-element set. By Corollary IV.4.6, V(K, 1 n = 1,2,3, . . .) = L 3 K ; since [K) is conipletely prime, K, 1 K for some integer n. Thus, by Lemma 3, K can be generated by finitely many finite subdirectly irreducible lattices. Since K is join-irreducible, K can be generated by a single finite subdirectly irreducible lattice. @

(ii) If K is generated by the finite subdirectly irreducible lattice L and K = KovKl, then by Theorem 2, L E Si(K,) or L E Si(K,) implying K =KO or K =K, . Thus K is join-irreducible. By Lemma 3, K is of finite height, hence K is Completely join-irreducible. @

(iii) Any equational class K is of the form K = V(Ko I KO c K and KO is generated by a subdirectly irreducible lattice). Thus if K is completely join-irreducible, K can be generated by a subdirectly irreducible lattice. @

(iv) We proceed as in (ii) by reference to Theorem 2. The converse statements of (i)-(iv) all fail, see I<. N. McKenzie [1972] and Exercises. Figure 1 can also be used to illustrate the very important concept of splitting due

to IL. N. McKenzie [1972]. A pair of equational classes (K,,, K,) is said to be splitting iff for every equational class K,, either K, E K,, or K, E K,. For instance, (M, Nj) is a splitting pair. ((L, T) is a trivial splitting pair.) Equivalently, (KO] and [K,) are prime, KO p HI, and (K,,] u [HI) = (L]. Obviously, KO determines K,, and conversely. Since [K,) is a completely prime dual ideal, K, can be generated by a finite subdirectly irreducible lattice. Finite subdirectly irreducible lattices that arise this way are called splitting lattices and they are characterized in R. N. McKenzie [1972].

How big is the lattice of equational classes of lattices? Since there are only & identities, there are a t most 2'0 equational clauses. Now we shall show by construction that there are exactly 2'uequational classes of lattices.

Let II be the set of prime numbers and, for p c n, let Lp be the subspace lattice of the projective plane coordinatized by the p-element field (that is, the Galois field, G F ( p ) ) . For a subset S c n set

K(S) =Equ({L, I iES}). We claim that AS can he recovered froni K(S) , in fact,

L,EK(S) iff p E S .

Obviously, if p €23, then Lp c K(S). Now let Lp c K(S) and p GS. Since Lp is subdirectly irreducible we can apply Corollary 1.10: L,cHSP,({L, I iES} ) , that is, L,EHS(L), where L = II (L, I i E I ) , a is prime over I , and each L,, i c I , is an Lj with jES. Each L, is a complemented modular lattice of length three, in which any two atoms are perspective. Since these properties can be expressed by (first-order) sentences, by Exercise 1.26, L has the same property and so, by the results of Section IV.5, I, is the subspace lattice of a nondegenerate projective plane. Each L, satisfies Desargues' Theorem and, by Theorem IV.5.8, this property can be expressed by a sentence. Thus L satisfies Desargues' Theorem and, by the Coordinatization Theorenr, I, can he coordinatized hy a division ring D . We assumed that p a s so each L,, i E I ,

a

240

Figure 3 Figure 4

b. Equational Classes of LBttices

Figure 6 Figure 7

Figure 9 Figure 10

Figure 6

Figure 8

Figure 11

2. The Lattice of Equational C1assc.s of Lattices 241

i8 coordinatized by a division ring not of characteristic 1). This again can be expressed by a forniula, hence D is not of characteristic 11.

L p c H S ( L ) , that is, L has a sublattice L' such that Lp is a homoinorphic image of L'. Since both L and Lp are of length three and a proper homomorphic image of a modular lattice of length three is of length less than three, we conclude that we can assunie that Lp = L', that is, Lp is a sublattice of L. This is a clear contradiction: we can introduce x + y for points x and y in Lp using only elements of Lp; thus for points x and y of Lp, x + y in Lp is the same as x + y in L. But in Lp we have p x = O while in L, p * x + 0 if x +0, which is impossible.

Thus there are at least as many equational classes of lattices as there are subsets of II, which number 2'0.

THEOREM 5. There are 2'0 equational classes of (modular) lattices.

This result was proved by K. A. Baker [1969a] (on whose example the above discussion was based), R. N. McKenzie [1970] (without modularity), and R. Wille [1972].

The following method of constructing equational classes of lattices is due to It. Wille [1972].

Let 9 be a set of finite posets. We denote by Equ(9) the class of all lattices that do not contain an isomorphic copy of a member of 9 as a primitive subset.

THEOREM 6 (R. Wille [1972]). tional class of lattices.

For any set of finite posets 9, Equ(9) i s an equa-

PROOF. This is clear from Exercises 111.1.22 and 111.1.23.

be verified using Theorem 6. The reader should verify that the construction in the proof of Theorem 5 could

Exercises *

1.

2.

3.

4.

5 .

Consider the lattices of Figures 3-11 and their duals (fifteen in all). Show that each one generates an equational class covering N5 (R. N. McKenzie [1972]). Let A be a fixed countable set. For an equational class IS, let SiA(K) denote the set of all subdiiwctly irreducible lattices L in K satisfying L A. Prove that K -, SiA(K) is a slit representation of the lattice of all equational classes of lattices. (Hint: F K ( X o ) can be recovered from SiA(K).) D0t.s tho representation of Exercise 2 preserve infinite meets and joins of equational claosies? Provv t,liat an oquational class K of lattices can be generated. by finitely many finite 1atticc.s iff K caii be generated by a single finite lattice, which, in turn, is equivalent t,o t.111. statemcwt, t,hat, all subdirect.ly irreducible lattices in K are finite and there arc' only finitely many nonisornorphic subdirectly irreducible lattices. Show that, I, can bc genrratcd by a snhdirectly irreducible lattice and L is completely join -rt:diieible.

. . .

Figure 12

Prove that the equational class generated by the lattice of Figure 12 is join-irreducible but it cannot be generated by a single subdirectly irreducible lat’tice (R. K. McKenzie [1972]). Let KO and Ki be the equational classes of lattices generated by all projective planrs satisfying Desargues’ Theorem and Pappus’ Theorem, respectively. Prove that KO and K, have the same finite members (K. A. Baker [1969a]). Find an equational class of lattices that is not generated by its finite members (K. A. Baker [1969a] and R . Wille [1972]). Find a sublattice of the lattice of equational classes of lattices that is isomorphic to the lattice of all subsets of a countable set (K. A. Baker [1969a]). Let K be an equational class of lattices. If K +L, then K < K O for some equcttiod class KO (B. J6nsson [1967]). (Hint: Consider KvEqu(L), where L is a finite lat,ticr. not in K.) Prove that L is join-irreducible (B. J6nsson [1967]). (Hint: If K c L , then the partition lattice on an infinite set does not belong to K.) There is no equational class of lattices covered by L (B. J6psson [1967]). Show that every proper interval of the lattice of equational c1asst.s of lat,ticee contains a prime interval. Let KO and K, be cquational ciasses of lattices. If K O c K , and KO can be defined by finitely many identities, then there exists an equational class K satisfying KO -<K K1. (Hint: use Theorems 1 and 11.3.13.) Provc that in Exercise 14 we cannot require that KO Prove that SiEqu(9) consists of those subdirectly irreducible lattices in which no P €9 can be embedded. Let 9 = (65). Describe Equ(9) . A lattice L is said to be of breadth n iff n is the smallest integer with the property that for every finite X c L there exists a Y S X such that IYI I n and V X =v Y . Choose a finite set of finite poscts 9 such that E q u ( 9 ) is generated by all subdirectly irreducible lattices of breadth at most n. Let 9 consist of the n-element unordered poset. Describn E q u ( 9 ) using the concept of width. Prove that N, contains all lat,tices of width two (0. T. Nelson, Jr. [1968]).

K < HI.

6 .

7.

8.

9.

10.

11 .

12. 13.

14.

16. 1F.

17 . 18.

19.

*20.

3. l’inding Equational Bases 243

3. Finding Equational Bases

An equational basis of a class K is a set of identities C. such that Equ(K) = Mod(X). This C is of special interest if it is irredundant (that is, Equ(K) fMod(XI) for any Xi c C) or finite. There is not much one can say about the problem of finding equational bases in general. However, if K has some special properties, then there is hope for some meaningful results.

A universal disjunction of equations, or briefly UDE, is a sentence of the form (recall, v stands for ‘‘ or ”)

(20) * * * (xt1-1) (ft)(xo, * * * 7 xn-1) =go(xoo, - - * 9 xn-i)V* * *

Vfm-l(zg, * * * 9 x n - 1 ) =gm-i(xo, * * - ) Zn-i ) ) ,

where fo , , . .., f m - l and go, . . . , gmP1 are polynomials. I n the sequel, we shall omit the quantifiers and we shall assume that f l < g i holds in any lattice. Examples abound: every identity is a UDE. The following lemma yields examples of a different kind. (In the formula V stands for the disjunction of the terms.)

LEMMA 1. Let P = { ( i o , . . . , a n - 1 > be a poset. Set g,= V ( x i I ai<ai) and

@(‘) : v(gi =gi”gk I ‘ k 5 ‘ i ) . . Then @ ( P ) i s n UDE which holds for a lattice L iff L has no subset isomorphic (as a poset) with P.

PROOF. If Q = {b,,, . . . , bn- l } c L is isomorphic with P and b, --at, 0 < i < n, is an isomorphism, then setting x, = b,, 0 < i < n , we find that g,(b,, . . . , b n - l ) = b,. Thus if u p p a , then g E $ g , ; and so g,*g,vg,. Since all the terms of @ ( P ) fail, @ ( P ) itself fails in L. Conversely, if @ ( P ) fails in L, then there are elements bo, . . . , b, - ,CL such that gE(b,), . . . , bn . - j ) $ gt(bo, . . . , bne1) whenever aE g a,. Since elk< a, obviously implies that g,(b,, . . . ) bn- , )<g , (bo , . . . , bn- l ) by the definition of g,, we conclude that ( I ~ -g,(b,, . . . , b n - l ) is an isomorphism of P with

{go(bg, * - - 9 bpi-,), * * * , gn-l(b,) * - * 9 bn-1)}* 0 Another iniportant UDE is

@ ( n ) : V ( x , = x i 1 O < i < j < n )

which holds for L iff 1LI < n. (To follow the inequality convention introduced above we should replace “xZ = xi” by ‘ ‘ X ~ A ~ =x1vx, ”.)

Now let 11s take an arbitrary UUE:

@: V(f,(zo, * * . , zn-l ) =g1(x0, . * * , x,&-1) I O l i < m )

and consicier the following statement : 17 Grutrer

In rE the variables are x, x,, . . . , xnP1 and yf, 0 1 i < m, 0 < j < k. We claim that S(0 ) holds in a subdirectly irreducible lattice L iff @ does. Let @ hold in L. Substituting xi = ai for 0 2 i < n, we obtain the elements fl(u,, . . . ,

for some i . Then two successive elements substituted into pzm agree and so, by a trivial property of p,, (see Exercise III.1.8), rk does not depend on x , verifyingS(@).

Now let Q, fail in L. Then there exist a,, . . . , EL such that fl(ao, . . . , a n - l ) < gz(ao, . . . , a n - l ) for O<i<m. Since L is subdirectly irreducible there are, by Corollary 111.1.1 1, a, b E L satisfying b < a and a/b zW gl(ao, . . . , a,- l)/ft(uo, . . . , a, - l). Thus for a suitable integer k and elements cg, . . . ,

(tn-& and gl(ao, . . . , an-') for 0 1 i <m. By 0, fi(ao, . . . , a,-J =gz(ao, . . . , %-A

c L (0 < i < m ) i

i

pk(g,(ao, * * * 3 an- l ) , c:, . - * 9 c k - 1 ) = a )

p,E(f,(ao, * * * 9 a n - l ) , Cf, . - * ) c k - l ) =b.

Hence rk with x i = ai (0 < i < n) and yf = cf (0 < i < m and 0 5 j < k) takes the form rk =$)2k(x, a, b, . . . , a, b).

But it is evident that r 4 = x on the interval [b, a] (see Exercise 111.1.9) and so rk is dependent on x. Thus S(@) fails in L.

The stateiiient that Tk does not depend on x is equivalent to the identity: E ~ : rk(x, . . . ) = r k ( z , . . .)

where z is a variable distinct from x , xL , and yf. Set X ( 0 ) = { E ~ j k = 1 ,2 , . . .}. Now we are ready to state a solution to our problem :

THEOREM 1 (K. A. Baker [1971]). Let Q, be u UDE. Then C ( 0 ) i s un equational basis for Mod(@). For a set 52 of UDE-s, U(C(Q,) I Q,cR) is an equational basis for Mod (a).

PROOF. Let KO be the equational class defined by u (C(Q,) I Q, € 5 2 ) and let Hl be the equational class generated by Mod(52). If L c Si(Ko), then by the discussion above L E Mod(52) c HI. Hence KO c HI. Conversely, if L c Si(Ki), then by Corollary 1.10, L cHSP,(Mod (a)). But 52 is preserved under ultra products (Exercise 1.26) and, obviously, by the formation of sublattices and honiomorphic iniages. Hence L E Mod@) and so, by the discussion above, L c Mod( U (C(Q,) 1 Q, c R)) = K,. Thus Hi K,,, proving that KO =Kl .

We can recast this proof using the following concept: a set or sequence of quotients of a lattice L is (&)bounded iff there is a proper quotient weakly projective (in k steps) into each quotient. Coro11ary111.1.11 and Exercise 111.1.21 (dightly sharpened) then read:

3. Finding Equational Bases 246

LEMMA 2. (i) I n a subdirectly irreducible lattice any finite set of proper quotients i s bounded. (ii) Let q~ be a ho?noniorptiism of the lattice L onto L’, let a,/bo, . . . , alL-l/b,-l be

quotients of L, and let uov/bov, . . . , a,-lv/bn-lv be k-bounded. Then a,/bi, . . . , (r,Jbt, i s ( k + 2)-bounded.

Xow for a UDE Q,: Vf i = g i , let, ak(Q,) =ak denote the sentence stating that for no substitution is golf,,, . . . , gm-l / f , , l - l k-bounded. We can write this down simply by

requiring that if u/v xw g i l l i for all i, then u = w. Then observe: k

LEMMA 3. ( i ) @ implies ak und t lk implies ut for any t

(ii) If cil l ap, k = 1,2, . . . , hold in a subdirectly irreducible lattice, then so does Q,. (iii) xk i s preservcd under the forniation of direct products and sublwttices. (iv) For m y sct R of UDE-s,

Equ(Mod(R))=Mod({a,(Q,) I ( D E L I and k = 1 , 2 , . . .}).

k.

1’ROOP.

(i) and (iii) are trivial. @ (ii) restates Lemma 2(i). @

(iv) follows from Corollary 1.10 and (iii). The reader should have no difficulty relating the sentence uk to the identity ek. We can use these ideas to give a simple proof of

THEOREM ‘4 (E. N . McKenzie [1970]). A n y finite lattice has a finite equational busis.

REMARK. Baker [1974].

The present proof is due to C. Herrmann [1973] which is based on K. A.

PROOF. is a finite equational basis for Mod(Q,(n)) (the class of at niost 11 element lattices), then we can easily find a finite equational basis for L : let A be a finite set of (up to isomorphism) all finite lattices N satisfying N ( Equ(L) and IN1 < n ; for each N E A choose an identity holding in L but not in X ; then C u {eg I N A } is a finite equational basis for L.

Any lattice in the class K = Mod(Q,(n)) has two properties: (i) i t is defined by an at most n2-ternied UDE:; (ii) every bounded set of at most n2 quotients is n2-bounded (hecause there are a t iiiost n2 quotients in the lattice). Thus the following lemma coinplrtes the proof of Theorem 4:

Let L have n eleiilents. If

LEMMA 5. propwtic’s ;

Let K 60 ( I cluss of lattices and let m be an integer with the f o l l o w i n g two

(i) K =Mod(X) ioherc 2: i s a finite set of ut most rn-termed UDE-s. l i ’

246 V. Equational Classes of Lathicm

(ii) There is an integer r such that in every subdirectly irreducible luttice LEK every

Then K has’a finite equational basis. bounded set of m quotients is r-bounded.

PROOF. Let em,r denote the sentence expressing that any set of m quotients that is (r + 1)-bounded is r-bounded. A trivial induction shows that e,,, implies that any bounded set of m quotients is r-bounded.

Let L E Equ(K). Then L is a subdirect product of subdirectly irreducible lattices Li, i c I , where each Li c K by Corollary 1 .lo, since UDE-s are preserved under the formation of ultra products, sublattices, and homomorphic images. Let c + d and cld =zm ai/bi, for 0 j < m, in Li for some i E I , where yi is the projection of L onto L,. By the second hypothesis, holds in Li and so, by Lemma 2(ii), {ai/bi 1 0 1 j <m} is (r +a)-bounded. Thus holds in any L EEqu(K).

By Lemma 3(iv), Equ(K) is defined by {ak(@) I @ c s2 and 12 = 1,2, . . .}. Hence this set of sentences implies em,r+2 and so by the Compactness Theorem (Exercise 1.29) finitely many ak(@) imply em,r+2. Let A ={ak(@) I @ ER and k = 1,2, . . . , t } imply and let us further assume that t 2 r +2. We claim that A defines Equ(K). Indeed, A holds in Equ(K) by Lemma 3, hence Mod(A) 2Equ(K) . Conversely, A includes ar+2(@) for any @ E R, and A implies ena,r +2. But C C , + ~ ( @ ) and em,r+2 imply ai(@) for any i ; thus A implies a;(@) for any i and any @ E R , proving that Mod(A) E

We have proved that A is equivalent to Id(K) and A is finite, hence, by the Com- pactness Theorem, A is equivalent to some finite C Id(K). C is a finite equational basis for K.

The proof of Theorem 4 as exhibited above does not give a finite equational basis; it only proves that there is one. I n R. N. McKenzie [1970] and in K. A. Baker [1974] more complicated arguments are presented which actually construct a finite equational basis.

In some cases a finite equational basis for a finite lattice can be found using the following method. Let L be a finite lattice. Let us assume that we have constructed the lattices Lo, L,, . . . such that (i) Equ(L) is covered by each Equ(L,) and (ii) if K is an equational class of lattices and Equ(L) c K, then L, E K for some i . Find lattice identities q,, . . . , which hold in L but for which E,. fails in Li. Then {E,,, . . . , E,,

is a finite equational basis for L. We shall illustrate this method with an example.

j < m in L. Then cy, * d y i and cyildyi =zw aj/bi, for 0

Equ(K).

THEOREM 6. and M3,3. If K is any equational class of modular lattices and M , c K, then M4 or M,,,EK. ,

In the luttice of all equational classes of lattices, M, is covered by M, K

REMARK. This result was proved in G. Griitzer [1966] under the additional hypothesis that K is generated by 8 finite lattice. This hypothesis wag removed in B. J6nsson [1968] where t,he following corollary was also stated.

3. Eliiiding Equational Bases

i

247

0

Figure 1

COROLLARY 7. identity

An equational basis for M, i s given by the modular identity and the

x A ( y V Z ) A ( z V W ) A (w v y ) 1 ( x ~y ) V ( X A Z ) v ( X AW).

By the discussion above, to prove this corollary it is sufficient to see that this

The proof of Theorem 6 is based on the following two lemmas. identity holds in m3 but fails in m4 and n3,3; this is left t o the reader.

LEMMA 8. a < xa < i and set (see Figure 1)

Let L be a modular lattice and let {o, a , b, c, i } be a diamond in L. Let

xb = bv(xaAc), xc =cv(xaAb), nnd 0, = ( f lAb)v( f lAc) .

Then {ol, xu, xb, xc, i} is a diamond.

PROOF. xbvxC = i obviously hold. Also,

Since a 4 xa < i, b < xb < i, and c < xC < i, t'he relations xavxb = xavxC =

xaAxb =xar\(bv(xa,r\c)) = (xaAb)V(xaAc) =ol,

and similarly, X ~ A X ~ = ol ; finally,

xbAxC = (bv(xu/\c))/\(cv(x'rb))

= ((bV(x"/\c))Ac)V(xa~b) = (b/\c)v(xur\c)v(zuAb) = ovol = oi.

LEMMA 9. Let L be a subdirectly irreducible inodular lattice which has no sub-

lcrttice of which 9.J13,8 i s a homomorphic image. Let a lbgc ld in L such that alb has no proper subquotient projective to a subquotient of cld in less than n steps. Then 1 1 1 3 .

4 PROOF. Let xolyo zx x41y4 such that no proper subquotient of xolyo is projective to a proper subquotient of x,ly, in less than four steps. By duality we can assume that xolyo f xilyi L x2/y2 f x31y3 L x4/y4. By (the proof of) Theorem IV.1.6, we can assume

that this is a normal sequence. Let pi/qi be a proper subquotient of xi/yi, for 0 < i l 4 ,

qi+l} cannot generate a distributive sublattice for i= 1, 2, or 3. Indeed, if one does, as for instance {po, qo, pi, ql ) p2, qa} generates a distributive sublattice, then po/qo L poAp2lq0Aq2 f p 3 / q 3 \ p4/q4 by Theorem IV.1.7, contrary to the hypot.hesis. Thus the sublattice generated by {xi-l) yi-l, xi, yi, xi+l, yi+l} is isomorphic to a homomorphic image of Figure IV.1.6 not collapsing m3 or to the dual of such a lattice (see Figure 2 ) .

Now we claim that oIvil=il, using the notation of Figure 2. Let oIvi2 +i1. Since il > o1 and ii (=x,) > i2, we must have i, > olvi2 =xC. Thus by Lemma 8, there are elements xb, xa such that A, = { x ~ A x ~ , xa, 9, xc, i,} is a diamond and bl I xb << i,, al < xa < i,. Since il/bi i.Ja2, there is an element ya satisfying il/xb L i2/ya. Again by Lemma 8, we find the elements yb and yc satisfying b, yc < i,, and A , = {ya/\yb, ya, yb, ye, i2} is a diamond. Thus il/xb \ i21ya and since i, < xc < i,, we also have X ~ I X ~ A X ~ L i2/ya. This last' relation means that A , u A, sm3,3 or if xc =k i, (and x b ~ x C + ya), then ( A , u A2)/@(xc, i2) z .~ ,~ , a contradiction.

such that pulqo 7 pilq1 L p2Iq2 7 333143 L p4lq4. we claim that {Pi - 1 ) qi - 1) ~ i > q i ) 1'i t 1,

yb < i,, c2

Therefore, olvi2 = i,. By duality, 0,Ai2 = 02, that is

Figure 2

3. Finding Equat,ional Basra 249

and similarly, i2/03 i3/03.

By normality o2 = 0 1 ~ 0 3 , and by definition i2 = i l ~ i 3 . Thus {ol, 02, 03, i,, i2, i3} is contained in the sublattice generated by {q, i,, 03, i3} which is distributive by Theorem IV.1.13. We conclude, by Theorem IV.1.7, that

il/ol ilvi3/01v03 L i 3 / 0 3 ,

XOlYO f XIVO3IYIV?l L XdY,,

which trivially implies that

contrary to the hypothesis.

PROOF OF THEOREM 6. Let K be an equational class of modular lattices such that K 1 M:, and '?Ill4, 4 K. In order to show that K = M,, it is sufficient to verify that if L E Si(K), then L is a sublattice of '532,. Assume to the contrary that L C Si(K) but L is not a sublattice of 'Jn3. Since '532,, 'Jn3,34K, we must have 'B4, (m3,3(HS(L).

Obviously, I LI > 2. If L is of length 2, then L must have B4 as a sublattice, a contradiction. Thus L has a chain c, < c l < c 2 < c3 of length three. Since L is subdirectly irreducible, @(co, c , ) A @ ( c , , c2) +t.to and so, by applying Theorem 111.1.2 twice, we obtain a proper quotient x/y weakly projective into c2/cl and into cl/co. By Theo- rem IV.1.6 and by the symmetry of projectivity, there is a proper subquotient a / b of cr/cl and a subquotient c/d of cl/co such that alb =.Id. Choose these a, b, c , d so that n

is mininial in a / b c c / d . Then a, b, c , d and L satisfy the conditions of Lemma 9 and therefore n < 3. We claim that n = 3. Indeed, in any lattice L we cannot have a > b 2

c > d and a / b k c / d with n 5 2 . (Proof. If a / b 7 p/p L c /d , then d =CAq 2 cAb = c. If u/b \ p / q 7 c/o!, then a = b v p 5 bvc = b. The case n = 1 is trivial.)

xl/yi, then xI > yl 2 c > d

and xI/yi x cld which is impossible as noted in the previous paragraph. Thus xolyo L

Applying the same arguments to @(c3 , C ~ ) A @ ( U , b ) , we obtain zo/uo L zl/ul 7 z2/u2 L z,/u,, where zo/uo and z3/u3 are proper subquotients of c3/c2 and alb, respectively. By making the trivial replacenients x,/yo by z3/u3, xl/yl by XIAZ~IXIAU~, x2/ya by y2v (xl Az3)/y2v (xl Al l3 ) , and x3/y3 by x3A(y2V(xI A Z ~ ) ) / X ~ A ( y2v(x1 A U ~ ) ) , we can assume that xo/yo equals z3/u3. Thus by Theorem IV.1.7, there are diamonds A j = { o j , ai, hi, ci, $1, j =0 , 1, such that a / b \ io/ao and n/b 7 a,/ol. We conclude that al/ol \ i,,/ao and so A , u A , is a sublattice of which W3,3 is a homomorphic image, a contradiction. @

Now let alb = x,/y, N xl/yl - x2/y2 -x3/y3 = c / d . If alb 2

"IIYi f X2IYa L X,lY,.

Exercises

1. A universa l sentence is a sentence Y of the form (q,) * * (xn-l) CD, where there is in CD no quantifier and no variable other than xo, . . . , ~ ~ - 1 . Show that every universal serit)ence in which no negation or implication occurs is equivalent to a finite set of UDE-a.

2.

3. 4. 8. 6.

7.

8.

9.

*lo.

*11.

12.

13.

14.

16.

16, 17.

18.

*19.

20.

Show that a UDE is preserved under the formation of sublettices and homomorphic images. Are all UDE-s preserved under direct products? Is every UDE equivalent to some O ( P ) ? Let P be the n-element unordered set. Compare O ( P ) with @(n). Let P be 8 finite p08et. Show that Equ(Mod(O(P)) =Equ({P}) as introduced in Section 2. In the proof of Theorem 4 the statement is used that in an n-element lattice every bounded set of proper quotients is nz-bounded. Can nz be improved in this statement? Write out the formulas Irk and ern,r to prove formally that these are indeed (first- order) formulas. An algebra (A; A , v ) is called a weakly aeeociatiue lattice (WA lattice) iff the following are setisfied : the two binary operations setisfy the idempotent, commutativo. arid absorption identities, and

x S z and y s z imply that z v y s z

and its dual hold, where LC 5 b meana that a = a ~ b or equivalently, avb = b (E. Fried [1Q70] and H. L. Skala [1971]). Show that all the results of this section hold for WA lattices (K. A. Baker [lD74]). Takethe three-element WA-lattice T = { O , 1,2} defined by 0 5 1 5 2 5 0. Findefinite equational besis for T (E. Fried and G. Griitzer [lQ73]). If L is e modular lattice of length n, then any bounded set of proper quotients is

k-bounded, where k 5 -- + 2 (C. Herrmann [1Q73]). ([x] stands for the largest

Let Mn denote the C ~ W S of modular lattices of length a t most n. Prove that I n has a finite equational basis. (For n = 2 this is due to B. J6neson [1968]; for n = 3 this is due to D. X. Hong [lQ72]. For general n this is due to K. A. Baker; a reference to this fact and 8 proof of this result b a e d on Exercise 11 is due to C. Herrmann [1973].) Show that a finite equational basis for Ma, for notation Bee Exercise 12, is given by the modular identity and

integer 5s.) KI

Z A ( Y V ( Z A U ) ) V ( Z A U ) 5 ~ V ( Z A Z ) V ( Z A U )

(B. Jbnsson [1Q68]). (Hint: Use Lemma Q and the reasoning in the proof of Theorem 6.) Let L be a subdirectly irreducible lattice of length at most 3. Show that any bounded set of proper quotients is in fact &bounded (C. Herrmann [1973]). Let L3 denote the clees of ell lattices of length a t most 3. Prove that L3 hes a finite equational basis (C. Herrmann [1D73]). Show that Lemma 8 can be derived from (and is implicit in) Theorem IV.1.7. Using the notations of Lemma 8, project x'hto [o, a], [o, a], [o, c] obtaining xar z b , zc, respectively. Show that {o, x,,, xb, xc, oi} is a diamond. The results of this section depend heavily on Theorem 1.9. Howevnr, Theorems 4 and 6 use only Corollary 1.11. Show that the latter can be proved with no reference to ultra products, using only the primitive sets of Section 111.1 (R. Wille [ 19721). Let L be a subdirectly irreducible lattice of width not greater than four. Then either the lattice of Figure IV.3.4b or one of the eight lattices of Figures 3-10 is a homomorphic image of B sublattice of L (R. Freese [1972]). Let M(4) denote the class of all modular lattices of width four. Prove that M(4) has a finite equational basis (R. Freese [1972]). (Hint: use Exercise 19.)

3. Finding Equational Bases 251

Figure 3 Figure 4

Figure 6 Figure 7 Figure 5

Figure 8 Figure 9 Figure 10

252 V. Equational Classes of Lat,tices

21. Show that to prove Theorem 6 for an equational class generated by a finite lattice it is sufficient to analyse projectivities of prime quotients. To what extent would this simplify the proof? Let L be a modular lattice and let n be a positive integer. Call L .n-distributive iff the following identity holds:

22.

Z h v ( y i I 0 5 i 5 R) =v(2hv(2/i 10 5 i 5 It, i *j) I 0 <j I n ) .

(The following form of the identity is easier to visualize:

X A V Y ~ = V @ A v yi).) i j i+ i

Prove that L is n-dist,ributive iff it satisfies the following identity in the variables 201 z1, * * * , 2, + 1 :

A ( V z i ) = V ( A ( V x i ) ) i i k i f k i+j,k

(G. M. Bergman [1969] and A. P. Huhn [1972]). For a positive integer n. we define a partial lattice P, as follows: Pn = B IJ {w}, where B is a 2n+i-element Boolean lattice with bounds 0 and 1 ; for 2, y E B, ZAY

and xvy are defined as in B; W h l = 1 h W = W and W A X = X A W = O for xCB-{l}; wvO=Ovw=w, wvd = d v w = l if d is a dual atom of B or if d = l ; wvx and xvzu are not defined if x E B and 0 < h ( z ) < n. Prove that a modular lattice L is n-distributive iff L does not contain P , as a relative sublattice (A. P. Huhn [1972a]). For a positive integer k and a division ring D, form the lattice L = L ( D , k). How is k determined by the smallest integer n such that L is n-distributive? An equational basis for N, is provided by

23.

24.

*25.

X A ( Y V U ) A ( Y V W ) I ( Z A ( Y V ( U A V ) ) ) V ( X A U ) V ( Z A W )

ZA(Y V ( U A (zv'u))) = (%A ( Y V ( U A Z ) ) ) V ( S A ( (ZAY)V(UAW)) )

(R. N. McKenzie [1972]). Consider the three-element algebra M = ( { O , 1 , 2 } ; -) with one binary operation such that 0 is a zero (0 . 2 =x * 0 =0) and 1 * 2 = 1 , 2 1 = 2 2 =2, 1 * 1 =O. Prove that 42 has no finite equational basis (V. L. Murskii [1965]; see also R. C. Lyndon [1954]).

*26.

4. The Amalgamation Property

For a class K of lattices (or of algebras, in general) it is very important to know how members of the class can be glued together to obtain a larger member of the class. Such properties are known as amalgamation properties. We shall mention three of them.

A V-formation in K is a pair of lattices Bo and Bl in K with a lattice A E K which is a sublattice of both Bo and B,. More precisely, a V-formation is a quintuplet (A, Bo, B1,vo,q1) such that A, B,, BIEK and vi is an embedding of A into Bi, for i =0, 1. This V-formation is amalgamated by (yo, yi, C) iff C E K, yi is an embedding of Bi into C, for i=O, 1, and ~ o y o = y l p l (see Figure 1). The V-formation is strongly amalgamatedby (yo, yI, C) iff, in addition, Boy0 n Biyi= Aqoyo(= Avlyi).

4. The Andgarnation Property 253

\

B.

Figure 1

A’class K is said to have the (Strong) Amalgamation Property iff every V-formation can he (strongly) amalgamated.

K is said to have the Weak Amalgamation Property (also called the Embedding Property) iff for any Bo, B1 EK there exists a CEK into which both B, and B, can be embedded (this is the “special case” A = 0 of the Amalgamation Property). This is of little interest for lattices: any equational class of lattices has the Weak Amalgania- tion Property. Indeed, C =BOX Bl will do with the embeddings yo: x +(x, bi), y l : .2: -+(AO, x), where bi is a fixed element of Bi, for i=O, 1.

L has the Strong Amalgamation Property. To see this take a V-formation (A, B,, B,, ylr, TI); we can assume that B,n B1 = A and A is a sublattice of B, and B,. On the set P = B,, u BI we define a partial ordering as follows:

For a, b € Bi, a< b in Piff a< b in Bi (i=O, 1 ) ; for a € Bi and b e Bj, i+= j , a< b in P i f f a g c i n B i a n d c g b i n BiforsomecEA(i,jE{O,l}).

It is easy to check that P is a poset. Then

avb =sup {a, b} aAb =inf {a, b},

turns P in to a partial lattice (see Lemma 1.5.21) of which A, B,, and B, are sublattices. Thus by Theorem 1.5.20, P can be embedded into a lattice, proving the Strong Amalgamation Property for L.

D does not have the Strong Amalgamation Property. Indeed, let B, = B, = (&#, A =cS3 (A = { O , a, l}), andletq+,=yl =p, be given byOp,=(O, 0), ap,=(l, 0), lp, =( l , l ) . Let (w,,, yI, C ) stronglyamalgamate(A,Bo, B,,Q;,,~I~) with CED. ThenBoyon Biy1 = A and so (0, l)yO+(O, l)yl. Both of these elements are relative complements of (l,O)y, (= (1 , 0 ) ~ ~ ) in the interval [(0, O)yo, (1, l ) ~ , ] of C, contradicting C ED and Corollary 11.1.3.

However, D has the Amalgamation Property. This we shall prove shortly after soiiie general remarks.

Let K be an equational class of lattices (or of algebras, in general) and let & = (A, B,, B,, p,,, 9,) be a V-forniation in K. We assume that Bon B, = 0. We associate with & a congruence relation 0 = @(&) as follows:

Let F = FK(Q) be the free lattice over K freely generated by the set BO u -& where B L = { x I xE Bi}, for i=O, 1, and the sets B,, BI, B,, Bl are pairwise disjoint. We set

_ -

-~

0, = v (@(ap,,, a%) I a E A )

2 54 V. Equational Classes of Lattices

and

Oi = V (O(ahb, E d ) I a, b E BJv V ( @ ( a x , iiv6) I a, b E B J ,

for i = 0: 1. Note that E;n6 and iiv6 are to be performed in F. Now we define

0 = oAvOovol.

Obviously, 0 is the smallest congruence relation of F such that

ai: x - [ X I @ , i = O , 1,

is a homomorphism of Bi into F/0 (because O i l 0) and yoao =ylai (because 0,s 0). Thus (ao, al, F/O) amalgamates Q iff a. and aI are one-to-one. Therefore, if a. a d ai are one-to-one, then Q can be amalgamated. Conversely, if (yo, yl, C) (C E K) aina 3 ga- mates Q, then inap Bou Bl into C by

- -

Z+xyi for x E B , i = O , 1.

This map extends to a homomorphism B of F into C and obviously Ker(b) 2 0, and $ =xyi, for x E Bi. Thus ai followed by the natural homomorphism of F / 0 into C equals yi which is one-to-one by assumption. So ai is one-to-one, for i = 0, 1. We have proved (Theorem 1, Corollaries 2-5, and the application to the proof of Lemma 12 are from G. Grlitzer [1976;]) :

THEOREM 1. Let K be a n equational class, let Q =(A, B,, Bl, yo, y , ) be a V-formation in K, and let F and 0 be constructed as above. Then Q can be amalgamated in K i f f , for i = O or 1 and x, yE B ,

k = 5 (0) implies that x = y.

COROLLARY 2. An equational class K has the Amalgamation Property i f f for any V-formation (A , Bo, Bi,y0,y1) in K and x, yC Bo, x + y , there exist a CEK and homomorphisms y,,: Bi -. C such that yoyo = ylyl and xyo + yyo.

PROOF. In order to prove the ai (introduced in the proof of Theorem 1) one-to-one it is sufficient to have honioniorphisms yi separating a fixed pair of distinct elements. The necessity is obvious.

COROLLARY 3. Let K be an equational class and let Q =(A, B,, B,, yo, y l ) be a V - formation in K . If Q cannot be amalgamated in K , then there are finitely generated subalgebras A’ of A and Bl of B,, i = O , 1, such that &‘=(A’, Bh, B;, yh, 9;) cannot be amalgamated in K , where y i is the restriction of yi to A’, i = O , 1.

PROOF. Let F and 0 be given as in Theorem 1. If Q cannot be amalgamated in K, then there are iE(0 , l} and x , y E Bi such that x =I= y and 5 = (0). By Theorem 111.1.2 and Lemma 111.1.3, we can select finite subsets A* of A and Bf of B , i =0, 1, such

4. The Amalgamation Propert,y 255

that in computing x = (0) we use only elements of A* u B$ u BT. Thus we can set A’ =[A*], BI = [Bf], i = 0 , 1. A trivial application of Theorem 1 shows that Q’ cannot be anialganiated in K.

Call a V-forniation ( A , B,, B,, vo, vl) finitely generated (finite) iff A, B,, and B, are finitely generated (finite).

COROLLARY 4. An equational class K has the Amalgamation Property i f f every finitely generated V-formation in K can be amalgamated in K .

An equational class K is called locully finite iff every finitely generated member of Ii is finite. It is easily seen that K is locally finite iff all F K ( n ) are finite (n < w ) . This always holds if K is generated by a single finite lattice L. Let K , , denote the class of finite nienibers of K.

COROLLARY 5. Let K be a locully finite equational class. Then K has the Atnalgamu- tion Property i f f all finite V-formations in K cun be amalgamuted in K , or equivalently, i f f Kfi,, has the Anialganiution Property.

The last equivalence follows froni the observation that if Q =(A, B,, B,, v,, 9,) is finite and can be anialganiated in K, then some quotient of FK(Q) will amalgamate Q ; but FK(Q) = FK( & u sl) is finite since K is locally finite.

Now we return t,o D. By Theoreni 11.2.1, IFD(n)l < 22n and so D is locally finite. 9

(‘OROLLARY 6. D has the A~tuilgutnation Property.

PROOF. Let Q = ( A , Bo, B,, q,, ql) be a finite V-formation in D and let x, y E B,, x + y . Set C =E2 and let yo be a hoinoinorphisni of B, into E2 such that xy, +yyo (see Corollary 11.1.1 1) . Let P be the ideal kernel of yoyo. If P = A, defineyl: B, -a2 by zyL = O for all xE B, . If P = 0 , defineyl: B, +E2 byxyl = 1 for all xC Bi. If P =i= A and Ps; 0 , then P = ( u ] where a =t= 1 is a nieet-irreducible element in A. Thus there is a unique b E A such that b >a. By Corollary 11.1.13, there is a meet-irreducible element 1) in B, such that a 2 p and b $ p . We then define y , : BI -Q2 by xyl = 0 iff x p and xy, = 1 iff x s p . It is obvious that (yo, yl, E2) satisfies the conditions of Corollary 2, thus Q can be amalganiated in D.

To further illustrate the usefulness of Corollary 2, for a class K of lattices (or algebras, in general) define A C K to be injective in K iff for any B, C K, B a sublattice of C , any hotiiomorphisrn of B into A can be extended to a honiornorphism of C into A.

COROLLARY 7 (R. S. Pierce [1968]). Let K be a n equational class. If any member of K can be embedded in a n injective member of K , then K has the Amalgamution Property.


PROOF. We apply Corollary 2. Let C be an injective member of K into which B, can be embedded; let yo be this embedding. Then xyo=kyyo. Set A ' = A y l . Then yi-%,uo is a homomorphism (in fact, an embedding) of A' into C, thus this homomorphism can be extended to a homomorphism y1 of B, into C. Obviously, yoyo= 9IYI. 0

A further application of Corollary 3 will be given a t the end of this section. A typical application of the Amalgamation Property is the sublattice theorem of

The next two theorems are negative. They show that certain equational classes of free products of lattices discussed in Section VI.l.

lattices do not have the Amalgamation Property.

THEOREM 8.1 then K does not have the Amalgamatim Property.

Let K be an equational class generated by a finite lattice. If K I D .

PROOF. If K is generated by a finite lattice L, then by Corollary 1.11, Si(K) c HS(L). Thus no subdirectly irreducible lattice in K has more than n = ILI elements.

Since KID, there is a nondistributive lattice in K and thus by Theorem 11.1.1. Y15 or 'JJ3 E K.

If Y15EK, Y15={o, a, b, C , i } as in Figure 11.1.1, then consider the V-formation (a2, !R5, as, yo, cpl), where Q2 ={O, l} and w0 =o, lvo =i , Oyl = b, lyi = a . Let (yo, y1, A ) amalgamate this V-formation. Then A will have the lattice A , of Figure 2

Figure 2

as a sublattice. I n fact, A , is the union of the two images of !J15 in A . Observe that A , is again subdirectly irreducible and ]All =8. Then we take the V-formation {a2, 'ill5, 81, yo, yl) with Oyo = 0, 1yo = i , Oyi = b, 1ql =a . The union of the images of '32; and A , will form a sublattice A , which is again subdirectly irreducible and I A,I = 11. Proceeding by induction we obtain the subdirectly irreducible lattice A , K with I A,] = 5 + 3k. If k is large enough so that 5 + 3k > n, this is a contradiction.

If Fm3EK, Fm3 ={o, u, b, c , i } , then we take (a2, FJJE3, '13t3, rpo, yl) where Oyo = u , ly,] =i , Oyl = o , lyl =c. Then the union of the images is again a sublattice of the amalgam, thus obtaining 'JJ3,, EK. Proceeding the same way we obtain a sequence of simple modular lattices of increasing size, leading to a contradiction as above.

i A. Day, S. D. Comer, and S. Fajtlowicz. See G. Griitzer, B. J6nsson, and H. Lakser [ 19731.

4 . Thc A4~nalgamatJioii Property

THEOREM 9. 11 does not have the Aniulyanrntion Property.

257

REMARK. In fact, no equational class K of modular lattices has the Anialganiation Property unless K =T or D, as proved in G. Gratzer, B. Jbnsson, and H. Lakser [1973].

PROOF. Let us asmine that DZ has the Anialganiation Property; under this assumption we prove a few embedding theorems.

1. Any modular lattice A can be embedded in a bounded modular lattice B which has a five-element chain.

This is trivial. If A has no five-element chain we add new zeros knd ones until it has a five-element chain. If A is not bounded, add bounds. (J

2. Any bounded modular lattice A can be embedded in a modular lattice B having the 6ame bounds and having the property that for every a c A, a +0,1, B has a diamond (0, a, b, c, l}.

To prove this, let A - (0, l} = (u,, I y < u} and we define an increasing sequence of modular lattices A,, y < u with the same bounds0 and 1. Set A, =A. If A,, y <6 (S <a), have already been defined, set A; = U (A,, I y <6).

Let E3 = (0, 1,2} and consider the V-formation Q, =(&, m3, &, v,, v1) defined by Opl0=o, lyo=a ,2yo=i ,0vi=0 , ly l=a , ,2v l=1 . Let (y ,,yi, A) amalgamate Q,. By forming the interval [Ovoyo, 2voy,,] we obtain A,. It is obvious that with a =as, A , has the property required of B. We define B = U (A , 1 y <a), which obviously has the required properties. 0

3. Any bounded modular lattice A can be embedded in a modular lattice B having the sanie bounds as A and having the following properties: (i) For every a c B, u + 0, 1, there is a diamond (0, a , b, c, l} in B.

(ii) B is a simple complemented modular lattice. Indeed, set A , = A , and if A , is defined, then define A I L + I as the lattice constructed

from A, in 2. Then B = U (A, I n < w ) obviously satisfies (i). Let 0 be a congruence relation of B, o =!= 0. Since B is a compleniented modular

lattice, B is relatively complemented; hence (see Theorem 111.3.10) there is an a c B, such that a + O and n ~0 (0). If a + 1, then by (i) there is a diamond (0, a, b, C, l}. Since %13 is simple, we obtain 0 = 1 (a), that is, 0 = L . If u = 1, again 0 = L. Thus B is simple. 0

Now we are ready to prove the theorem. Let A c M. By the embeddings 1-3 we can embed A into a simple complemented modular lattice B having a five-element chain. By Theorem IV.5.17, B can be (0, 1)-embedded into a modular geometric lattice C. C is directly indecomposable because it has a simple (0, 1)-sublattice (naniely, B). C has a five-element chain because B has one. Thus Corollary IV.5.16 applies and C z L(D, m). Thus by Lemma IV.5.9 and Corollary IV.5.12, the arguesian identity holds in C. Since A is a sublattice of C, the arguesian identity holds in A .

Since A was an arbitrary member of M and the arguesian identity does not hold in M, this is fi contradiction.

Since the Anialgamation Property faih for so many equational classes of lattices

it seems reasonable to ask to what extent does it hold in general. The answer is rather surprising.

Following G. Griitzer and H. Lakser [1971], for a class K of lattices (or algebras, in general), let Amal(K) be the class of all those A E K for which all V-formations in K of the form (A,. , .) can be amalgamated in K. Obviously, K has the Amalgamation Property iff Amal(K) = K . For any equational class K of lattices Amal(K) + 0. Indeed, QIEAmal(K). (Argue similarly to the proof of the Weak Amalgamation Property for K.)

Call a subclass K, of K cofinal in K iff for all A c K there is an extension B E K1.

THEOREM 10 (M. Yasuhara [1974]). in K.

Fm a n equational class K, Amal(K) i s cofinal

The proof of this result will be given in the next two lemmas. Some of the ideas of Lemma 11 originate in A. Robinson [1971], [1971a]. The present form of Lemma 11 is a slight variant of a lemma of M. Yasuhara and it is due to M. Makkai.

For a class K letEc(K) stand for the class of all A c K having the following property: For any extension B E K of A and for any finite X A and Y c B, there is an

em bedding p: [XU Y]-A

fixing X (that is, xp =x for all x c X).

LEMMA 11. For an equational class K, Ec(K) i s cofina.2 in K.

PROOF. To facilitate the proof we introduce some concepts. For A E K, (p, X, C) is a triple over A in K iff X is e finite subset of A, C is a finitely generated member of K, and p: X - C is a homomorphism of the relative sublattice X of A into C. (This last condition means that if x ~ y = z or xvy = z for x, y, z E X , then ~ p ~ y q =zq or xpvyq =zp in C , respectively.) The triple (p, X, C) is realized over A by B iff B E K, B is an extension of A, and there is an embedding y of C into B such that xpy=x for a11 xEX.

Now let (p,,, X,,, C,,), y <a, list all the triples over A. We define two increasing sequences of lattices as follows: & = A ; if A,, has been definedfor y <6(<a), then set x, = U (A,, [ y < 6 ) . Regard (p,,, X,, C,) as a triple over A,. If it is realized in H define A, as any member of K realizing it. If it cannot be realized in K, define A, = 2,. Then set A(') = U (A,, I y <a). We define

and

Since K is an equational class, A* E K. We claim that A* ( Ec(K). Indeed, let B E K be an extension of A*, end choose finite subsets X c A* and Y E B. Define C = [X u Y ]

4. The Amalgamation Property 259

and let q: X -A be the identity map. Then ( y , X, C) is a triple over A* which can be realized by B. Since X c A* and X is finite, X A(n) for some n <o. Hence (pl, X, C) can be regarded as 8 triple over A@). Thus (9, X, C) occurs as some (998, X,,C,) in the list of all triples over A(fi). I n thed-th step of the constructionof A(n+l) = (A(%))(l), we view (pl, X, C) as a triple over (A@)),; we observe that ( y , X, C) can be realized by B, and so (A@)) , realizes (q, X, C), that is, there is an embedding y : C - (A@)) , such that xyw = x for x EX. However, xy =x for all x EX and therefore, x q y =xy = z for all 5 E X . Thus y is an embedding of [X u Y ] into A* keeping X fixed, proving A*EEc(K).

-

LEMMA 12. For an equational class K, Ec(K) G Amal(K).

PROOF. Let A CEc(K) and consider a V-formation Q =(A, B,, BI, qo, yi) in K. We can assume that A C Bo, A G B,, and yo =yi is the identity map on A. W e wish to show that Q can be amalganiated in K. Assume to the contrary, that Q cannot be amalgamated in K. Then, by Corollary 3, there exist finite subsets X A, Yo 5 B, and Yi c B I , such that &’ = ( [ X I , [Xu Yo] , [Xu Y,], y; , &) cannot be amalgamated, where 91; is the restriction of yi to [Xu Y,], for i = O , 1. Since AEEe(K) there exist enibeddings yi: [X u Yi] + A keeping X fixed (i=O, 1). Thus for x E X , xy iy i=z , for i=O, 1, and so ag iy i=a for all a E [ X ] . This shows that (yo, y l , A) amalgamates Q‘, which is a contradiction.

Exercises

1.

2. 3.

4. 6.

6. 7. 8.

9. 10.

11. 12.

13.

Show that the class of all Boolean algebras does not have the Weak Amalgamation Property. Show that the class of all groups has the Strong Amalgamcttion Property. Investigate which equational classes of pseudocomplemented distributive lattices have which Amalgamation Property. Show that Lfi, has the Strong Amalgamation Property. Define the St-timel(K) as the analogue of Amsl(K) for the Strong Amalgamation Property. Determine St-smel(D). Show that “C EK” can be changed to “C E Si(K)” in Corollary 2. Did we use the Axiom of Choice in verifying Corollary 69 Let L be injective in K. If A is an extension of L in I(, then L is ti retract of A, that is, there is a homomorphism pl: A -L such that xpl -x for all z EL. Show that any retract, of a complete Boolean lattice is a complete Boolean lattice. A distributive lattice L is injective in D iff L is a complete Boolean lattice (P. R. Helmos [1963]). (Hint: use Exercises 8, 9 and Corollary 11.2.7.) Prove Corollary 6 using Corollary 7. Find further examples of the phenomenon observed twice in the proof of Theorem 8, namely that for some special V-formation, if (yo, yi, C) amalgamates (A, B,, Bi, qo,qi) then Boyo u B,y , is a uniquely determined sublattice of C. Show that thwe are 2xo equational classes of modular lattices not having tho Amalgemotion Property.

18 Griitzer


14.

16.

Let K be an equational class. Then for every A EK we construct an extension BEAmsl(H). Give an upper bound for IB!. Let K be an equational class having the Amalgamation Property. Let A, ECK, B an extension of A, and let a be an automorphism of A. Show that i there exist an extension C K of B and an automorphism /3 of C extending a. Let K be an equational class of lattices. Then &2cAmal(K) iff there is a D CAmal(K) with ID1 > 1. Let K be an equational class of lattices. Prove that if I ( L ) EAmal(K), then L E Amsl(K). Let K =Equ(L), where L is a finite subdirectly irreducible lattice. Then L ~Arnal (K) .

18.

17.

18.


H, S, P, P,, Pa, and so on, are examples of operators: an operator X assigns to a class K of lattices (algebras) another class X(K) of lattices (algebras). If X and Y are operators, so is their product: XY defined by XY(K) = X(Y (K)). Thus we can easily define the semigroup generated by a set of operators. The standard semigroup of a class K of algebras is the one generated by H, I, S, and P, equality defined with respect to K. For references see S. D. Comer and J. Johnson [1972] which also lists all the known results for lattices, distributive lattices, Boolean algebras, and so on ; one problem they leave unresolved is our Problem V.1.

Theorem 2.1 is very trivial but it has a number of useful consequences: the lattice of equational classes of lattices satisfies the Meet Infinite Distributive Identity and every proper interval contains a prime interval ; properties shared by all dually algebraic, distributive lattices. An additional property is that no element covers more than countahly many elements, since the lattice then would have more than 2% elements.

For a more detailed discussion of reduced products and ultra products, see T. Frayne, A. C. Morel, and D. S. Scott [1962/63] and G. Griitzer [1968].

Splitting can be defined relative to an equational class. Apart from the general concept, only splittings in M and E have been studied to some extent, where E is the equational class generated by the subspace lattices of all projective planes. For instance, '%4 does not split in M but the subspace lattice of any projective plane splits in E. For a discussion of these and related results see R. Wille [b]; see also A. Day [1973] and D. X. Hong [1972].

Wille's method of constructing equational classes (Theorem 2.6) is based on his two earlier papers, R. Wille [1969] and [1969a]. For a refinement of Wille's method, see K. A. Baker [1974a].

By a result of R. N. McKenzie [1970], lattices can be defined hy a single ident'ity and obviously so can T; no other equational class of lattices can be defined by a single identity. By R. Padmanabhan [1969], any equational class of lattices defined by a finite set of identities can be defined by two identities.

i Exercises 16-18 arc from G. Gratzer, B. Jbnsson, and H. Lalcser [1973]. Exerciw 15 is implicit in B. H. Neumann and H. Neumann [1962].

Furthcr Topics mid References 261

Combining these results with a result of A. Tarski [1968] we obtain: L and T have an irredundant equational basis of n elements for any 1 n <w. If

the equational class K of lattices has a finite equational basis and K +L, T, then K has an irredundant equational basis of n elements iff 2 5 n < w.

Lattices are quite exceptional in that Theorem 3.4 holds. A finite algebra with no finite equational basis was found by R. C. Lyndon [1954]; this was improved in V. L. Murskii [1965] in which a three-element example is found (this is optimal, see E. L. Post [1921]) and P. Perkins [1969] in which the example is a six-element semigroup. For additional references and for the main result in the field (namely, that Theorem 3.4 holds in any congruence distributive equational class), see K. A. Baker [ 19741.

Let KO and K, be equational classes; the identities of K,vKi are those that are shared by KO and Hi. Thus KO and Kt may have finite equational bases, while K,vK, may not have a finite equational basis. For lattices thisnegative result has been verified by K. Baker (mpublished) Find in B. Jbnsson [1974]. Some positive results are also given in B. J6nsson [1974]; for instance, MvN, has finite equational bases.

Equational bases are known for many equational classes of lattices (and other reiated algebras); see especially, K. A. Baker [1969a], [1971], [1974], [1974a], and [a], S. D. Comer and D. X. Hong [1972], A. Day [1973], R. Freese [1973] and [b], E. Fried and G. Grlitzer [1973], C. Herrmann [1973] and [1973a], C. Herrmann and W. Poguntke [1974] and [a],U. X. Hong[l972], G. Hutchinson [1973 b], B. J6nsson [1968], M. Makkai [1973], M. Makkai and G. McNulty [a], R. N. McKenzie [1970], [1972], and [1973], R. Padmanabhan [1968], [1969], [1972], and [ 19744, M. Schutzenberger [1945], and It. Wille [b]; see also R. Padmanabhan [1966] and D. H. Potts [1965].

K. A. Baker’s method of finding equational bases discussed in Section 3 is exploited in great detail in K. A. Baker [1974]; for further results and applications, see K. A. Baker [1974a] and [a]. C. Herrmann [1973] and M. Makkei [1973] are further variants on the same theme; see also G. Griitzer and H. Lakser [1971a].

Theorem 3.6 has been extended in B. J6nsson [1968] and D. X. Hong [1972]. Related questions are considered in C. Herrmann [1973] and in the papers of R. Freese.

The development of the Amalgamation Property is described in B. J6nsson [1965]; probably, it was B. Jbnsson’s work more than any other influence that convinced the algebraists of the importance of this property. The Amalgamation Property for lattices and posets is noted in B. Jbnsson [1956]; it is observed for Boolean algebras in A. Daigneault [ 19591.

The results of Section 4 on the Amalgamation Property are mostly negative. A further negative result of G. Griitzer, B. Jbnsson, and H. Lakser [1973] states that if K is an equational class of lattices, K M, and H is not arguesian, then g2 $ Amal(K). This was strengthened in G. Gratzer and H. Lakser [1973a].

The existentially complete algebras of M. Yasuhara [1974] are different from members of Ec(H). Thus Lemma 4.11 is slightly stronger and Lemma 4.12 is slightly weaker than the corresponding results in M. Yasuhara [1974].

The connections of the amalgamation properties and the interpolation theorems of logic are examined in P. D. Bacsich [1975]; see also P. C. Eklof [1974].

Finite menibers of Amal(M,) are described in E. Fried, G. Griitzer, and H. Lakser [1971]. 18’

Problems

v. 1. v. 2.

v. 3. v. 4.

V. 6.

V. 6.

v. 7.

v. 8. v . 9. v.10.

v.11.

v.12. v.13. V.14.

Is Equ(K) = SHPS(K) for any class K of lattices? Describe those sets of operators introduced in this chapter thak generate finite semigroups for lattices. Characterize splitting lattices in M. The method of R. N. McKende [1972] to give an equational basis for g5 has been modified in S. D. Comer and D. X. Hong [1972] to obtain an equational basis for 'B3 Explain the similarities and determine whether the method has other applications (for instance, to yield equational bases for covers of N5). For an equational class K of lattices, define i(K) (i,(K)) as the smallest integer n such that FK(n) is infinite (and gene-

rates K) ; and if there is no such integer, then i(K) = - (i,(K) = =) ; e(K) (ef(K)) as the smallest integer n such that every finite member of I(

can be embedded into an n-generated (finite) member of K ; and if there is no such integer, then e(K) =- (ef(K) =-); as the smallest integer n such that every countable member of K can be embedded in an n-generated member of K ; and if there is no such integer, then c(K) = -; (f,(K)) as the smallest integer n such that F,(n) contains FK(n+l ) (FK(KO)) as a sublattice; and f(K) = - (f,(K) =a) if there is no such integer.

Are there any relationships amongst these apart from the obvious ones (e(K)<ef(K), c(K)<fi(K), and SO on)? Compute the functions introduced in Problem V.5 for particular equational classes of lattices: N5, M,, M,, M, and so on. (m, is the subspace lattice of a projective line with n points and M, =Equ('B,).) For an equational clam K of lattices and positive integer n, define H,,= EqU(FK(n)). Characterize I ={n I Kfl=kK,+l}. (Conjecture: I ={1,2,3,. . . , k } or I = {1,2, . . . , k, . . .}. For universal algebras in general, see B. Jbnsson, G. McNulty, and R. W. Quackenbush [1976].) Find an LEAmal(M), ILJ > 1. Does there exist an equational class K satisfying D c K c M and E2 E Amal(K) ? Does there exist an equational class K c L having the Strong Amalgamation Property? How many such equational classes exist? (Conjecture: finite or countably infinite.) Show that there are only oountably many equational classes of lattices having the Amalgamation Property. Describe Amal(M,). Investigate Amal(Equ(comp1emented modular lattices)). Is it true that in the lattice of equational classes of lattices every proper interval contains an atom? 1

c(K)

f(K)

1 Yes, see A. Day [b].

Problems 263

V.15.

V.16. V.17. V.18. V.19.

V.20. V.21.

In the lattice of equational classes of lattices, are the elements of finite height exactly the equational classes of the form Equ(L), where L is a finite lattice? Find all covers of Equ(L) where L is finite planar. Find all the covers of N5.1 Is there an equational class of lattices with uncountably many covers? 2

Is there a nontrivial lattice identity holding in the lattice of equational classes of commutative semigroups? Find a finit,e equational basis for IMvN,.~ Let KO, Kt be equational classes of modular lattices. Is it true that if KO and K, have finite equational bases, then so does KovKi?

1 The covers of N, are those listed in Exercise 2.1. This was proved for equational

2 Yes, there are such equational classes. An example was given by R. Wille. 3 Solved in B. J6nsson, The variety covering the variety of modular lattices. Manuscript.

classes generated by a finite lattice by I. Rival and in the general case by B. J6nsson.

CHAPTER VI

FREE PRODUCTS

1. Free Products of Lattices

The formation of a free product of a family of lattices is one of the most fundamental constructions of lattice theory. This specializes to the construction of free lattices, which form a class of lattices that is probably the closest rival of the class of distributive lattices in the richness of its structure. Also, free products provide a very useful tool for the construction of pathological lattices.

This is made possible, in part, by the Structure Theorem of Free Products (Theorem 10 below). Since the theorem is based on a large number of long inductive definitions, we shall present first a short intuitive description of its contents. We ask the reader to follow it up with a careful reading of the theorem bearing in mind that the final result is very simple and most efficient in use.

Figure 1

Let A and B be lattices and let L be a free product of A and B, in symbols, L = A * B; for the present discussion this should mean that A and B are sublattices of L, A u B generates L, and L is the “most general” lattice with these properties, as in Section 1.5. Figures 1 and 2 provide two simple examples: Figure 1 is the diagram of 6 2 * 61 and Figure 2 that of G3 * (Xi.

266 VI. Free Products

\ c

4

c(

Figure 2

Since L is generated by A u B, every element of L can be written in the form

p(%, * * . 9 ' f i - 1 , bo, * ' * J 'm-l), ' 0 9 ' ' ' 9 u,-i '09 ' * ' > 'm-1 E B,

where p is en (n+m)-ary lattice polynomial. For instance, if A and B are as given in Figure 3, then

bz9 a 3 v b i , ( ~ 3 & ) v(uOA'O), (('5AbO) V U 3 ) A (('ZAb0) v'zvb3)

are examples of elements of L. Observe, however, that expressions of this sort that appear to be very different may, in fact, denote the same element of L. For instance,

a i V ( ( U 3 V b 3 ) A ( U 4 V b 3 ) ) = U5Vb3.

Theorem 10 will help the reader to construct much more complicated examples.

1 . Free Products of Lattices

A

Figure 3

267

The problem then is how do we decide when two expressions represent the same element of L. The key to this problem is the observation that if

p =p(ao, . . . , an bo, . . . , b,,,-i) I a

for some a E A, then there is a smallest element of A with this property. In fact, this smallest element (called the upper cover of p in A) is easy to compute knowing only p , A , and B.

So now our plan is the following: define formally the free product, expressions of the form p(ao, . . .), upper and (dually) lower covers, present the algorithm deciding p(ao, . . .) 5 q(a,, . . .), and finally prove that this algorithm works.

For this whole section, let L,, i c I , I * 0 , be a fixed family of lattices; we assume that Li and Lj are disjoint for i, j E I , i =k j . We set Q = U (Li I i E I ) and we consider Q a poset under the following partial ordering:

For a, b e & , let a g b i f f a, bEL, for some iEI and a<b in Li. A free product L of the L , i C I , is a free lattice F(Q) (=EIL(Q)) generated by Q, in

the sense of Definition 1.5.2. Or, equivalently,

DEFINITION 1. Let K be a n equational class of lattices and let L, L,EK for i E I . The lattice L i s a free K-product of the lattices L , i E I , i f f the following conditions are satisfied:

(i) Each Li i s a sublattice of L and for i, j E I , i =+ j , Li and Li are disjoint. (ii) L i s generated by U(Li 1 i E I ) .

(iii) For any lattice A E K and for any family of homomorphisms yi: Li -A, i E I , there exists a homomorphism y : L - A such that ip on Li agrees with yi, for all i c I .

For K =L, L is called a free product. The next definition is a slight adaptation of Definition 1.4.1.

DEFINITION 2. smallest set satisfying (i) and (ii) : (i) X c P ( X ) .

Let X be an arbitrary set. TlLe set P ( X ) of polynomials in X i s the

. (ii) I f P, q E P ( X ) , then (PA!?), ( P V ) E P ( X ) .

268 VI. Free Produots

The reader should keep in mind that a polynomial is a sequence of symbols and equality means formal equality. As before, parentheses will be dropped whenever there is no danger of confusion.

In what follows, we shall deal with polynomials in Q = U (Li I i E I ) . Let a, b, c E L , avb =c. Observe that as polynomials in Q, avb (which stands for (avb)) and c are distinct.

For a lattice A , we define Ab=Au{Ob, 1 9 , where Ob, 1"A; we order Ab by the rules :

O b < x < l b , forall xEA.

x < y in Ab iff x < y in A , for x , y E A .

Thus Ab is a bounded lattice. Note, however, that Ab =i= A even if A is itself bounded. It is important to observe that 06 is meet-irreducible and l b is join-irreducible. Thus if aAb =Ob then either a or b is Ob, and dually. This will be quite important in sub- sequent computations.

DEFINITION 3. Let p P(Q) and i € I . The upper i-cover of p, in notation, ~ ( ~ 1 , is an element of (L,)b defined a8 follows: (i) For a € & , we have aCLj for exactly one j ; if j = i , then a'O=a; if j + i , then

(ii) (pAq)(*? =p(i)Aq(i) and (pvq)({) =p(i)vq(i), where A and v on the right side of these a(9 = 16.

equations are to be taken in (Li)b.

The definition of the lower i-cover of p , in notation,

If there is some danger of confusion, we shall write pLi and p

An upper cover or a lower cover is proper if it is not Ob or lb. Observe that, however,

is analogous, with Ob

for p(i) and replacing l b in (i).

respectively.

no upper cover is Ob and no lower cover is lb.

=i

COROLLARY 4 . Por any p e p ( & ) and i C I we have that

p(i) IP(",

and if p ( i ) and p(j) are proper, then i = j.

PROOF. If p EQ, then p =p(i) =p(i) for p EL, and p(%) = o b < 1, for 2,e hi, SO the first statement is trie. If the first statement holds for p and q, then

(PAq)(t) =P(i)Aq(i) I p("Aq(i) = (PAq)'",

and so the first statement holds for p A q and similarly for pvq. To prove the second statement it is sufficient to verify that if p(i) is proper, then p(i) is not proper for any j + i . This is obvious for p CQ by 3(i). If p =qAr and p( i ) is proper, then both q(i) and r(i) are proper, hence q(j) = lb, and so p(j) = lb. Finally, if p =qVr

1. Free Products of Lattices 269

and p(i j is proper, then q(() or r(i) is proper, hence q(i) = lh or r( i) = ib, ensuring p(l) = ( 1 ) (i) = 1b. (I vr

Finally, we introduce a quasi-ordering of P(Q).

REMARKS. In (C), C stands for Cover; W stands for Ph. M. Whitman. Each Whitman condition assumes a A or v on the left or right of c ; at most two of these conditions may be applicable to a particular p c q. Note also, that if p and q EQ, then only (C) can apply.

Definition 5 gives essentially the algorithm we have been looking for. For p , q < P(&), it will be shown that p and q represent the same element of the free product iff p c q and q G p . We shall show this by actually exhibiting the free product as the set of equivalence classes of P(Q) under this relation. To be able to do this we have to establish a number of properties of the relation c. All the proofs are by induction and will use the rank r (p) of a polynomial p EP(Q): for p € & , ~ ( p ) = 1 ; r(pAq) = r(pvq) = T(P) + dq) .

LEMMA 6. Let p , q, r E P(Q) and i E I .

(ii) p E q implies that pCi) < q(i) and p ( i )< qCi). (iii) p E q and q c r imply that p c r.

(i) P c P.

PROOF.

(i) Proof by induction on r (p) . If p EQ, then p E Li for some i E I. Hence p = P ( ~ ) by 3(i), and so p E p by (C). Let p = q A r . Then q C q and r c_ r by induction hypothesis. By (,,W), qAr C q and q A r c r , hence by (W ,,), qAr E qAr, that is, p c p . We proceed similarly if p = qvr.

(ii) If p E q by (C), then p(j)< q(j) for some j E I. We conclude on the one hand, that p(i,<q(i, and p(j)<q(i) by Corollary 4, and on the other that p(j) and q(j ) are proper; thus again by Corollary 4, p(i) and q(0 are not proper for i + j , hence p( i ) = Ob, q(i) = 16, and so p( ( )< q(o and p ( * ) I &i).

Now we induct on r ( p ) +r(q) . If p , q EQ, then only (C) applies to p E q ; in this case and also in the induction step if (C) is applied, we obtain the result by the last paragraph. Now if p = p o ~ p i and (,,W) applies, then p,, c q or pl c q, say po G q. Then ( ~ ~ ) ( ~ ) < q ( ~ ) and (po)(i)<q(i), hence

Q

~ ( i ) = ( P o ) ( i ) A ( P i ) ( i ) c PO)(^) Q ( i )

and similarly for upper covers. Q

(iii) If p c q by ( C ) , then p( i )<q(o for some i E I. By (ii), q(i,< r(i), hence p(i)< r ( i ) and so by (C), p c r. This takes care of the base of the induction, p , q, r EQ since then only (C) applies. We induct on the sum of the ranks of p , q, and r. If p E q by ( C ) , p c r has already been proved. If p c q follows from (,,W), then p = p o ~ p f and p0 C q or pi E q. Thus, by the induction hypotheses, po E r or pi C r , and so by ( W), p0r\pl = p E r . If p E q follows from ("W), then p =povpi, po E q, and pi c q, and so again po G r and pi E r, implying povpl = p C r by ( W). If q G r follows from (C), (W ,), or (W"), we can proceed dually (that is, by interchanging A and v) . Only two cases remain; since the second is the dual of the first, we shall state only one: q=qoAql, (w,) applies to p G q , and (,,W) is applied to q E r (observe that ("W) is not applicable). Then p C qo and I , c ql , and qo r or qt E r. Hence p c qi c r for i = O or 1, hence by the induction hypotheses, p E r.

By Lemma 6, the relation c is a quasi-ordering and so (Exercise 1.1.28) we can define :

p = q iff p c q and q E p (p ,qEP(Q)) .

R ( p ) = { q I BEP(Q) andp=q} (PEP(Q))*

R(Q) = {W) I P E P(Q)).

R ( p ) l R ( q ) iff p c q .

In other words, we split P(&) into blocks under the equivalence relation p ~ q ; R(Q) is the set of blocks which we partially order by 1.

LEMMA 7. R(Q) is a lattice, in fact,

R ( p b R ( q ) = R ( P 4 ) and R ( P W = R(p)vR(q) .

Furthermore, if a, b, c, d E L , i E I , and aAb = c , avb =d i n Li, then

R(a)AR(b) = R(c) and R(a)vR(b) = R(d) ;

and if z, y EQ, z =!= y, then R ( z ) f R(y).

PROOF. pAq E p and p A q C q by p C p , q E q, and ( A W). If r G p and r C q, then r C pAq by (W ,,) ; this argument and its dual give the first statement.

c c a and c E b is obvious by (C), hence R(c)< R ( a ) and R(c)( R(b) . Now let K ( p ) I R(a) and R(p) 5 R(b) for some p EP(Q). Then p G a and p E b, and so, by Lemma 6, p(i)< a($) = a and p ( i ) g b(i) =b. Therefore, p(")I c = c t i ) and thus p C_ c by (C). The second part follows by duality. Finally, if R ( z ) = R ( y ) (2, y EQ), then .2: E y. Since only (C) applies, ~ ( ~ ) < y ( ~ ) for some i E I. Thus d i ) and y(i) are proper and so x = di), y = y(o. We conclude that z I y ; similarly, y I x. Thus x = y.

By Lemma 7 a -c W), a E L ,

1. Free Products of Lattic(?s 271

is an embedding of Li into R(Q), for i c I. Therefore, by identifying a < L i with R(a) we get each Li as a sublattice of R(Q) and hence Q R(Q). It is also obvious that the partial ordering induced by R(Q) on Q agrees with the original partial ordering.

THEOREM 8. R(Q) is a free product of the Li, i E I.

PROOF. i E I. We define induct,ively a map

as follows: for pEQ there is exactly one i E I with p EL,; set py =pyi ; if p =pOApi or p =povpi, and if poy and piy have already been defined, then set py = p o y ~ p I y or py =poyvply, respectively. Now we prove:

l(i) and l(ii) have already been observed. Let y i : L , + A be given for all

y : P(Q) + A

LEMMA 9. Let p CP(Q) and i€ I. (i) If p ( , ) is proper, then p(gy I py. (ii) If p(i) is proper, then py < ~ ( ~ ) y .

(iii) p z q iwiplies that p y < qy for p, q E P(Q).

PROOF. (i) If p EQ and p ( i ) is proper, then p E Li. Hence p =p(,) and SO p(,)y s p y is obvious.

If p =qAr, then p ( i ) =q(i)Ar(i) so q(i) and r ( i ) are proper. Thus q(i)y<qy and r(i)y < ry by induction ; hence p(i)y = q(i,yAr(i)y I qy Ary = (qAr)y =py. If p =qvr, then qCi, or r(i) is proper. If both q(<, and r(i) are proper, we proceed as in the previous case. If, say, q(i) is proper and r,i) =Ob, then p(i)y = (q(i,vr(i))y =

q(i)YIqYYIY. 0 (ii) This follows by duality from (i).

(iii) If p c q follows from (C), then, for some i < I, p(i)<q(i,. Thus p(i) and q(i) are proper. Therefore, py < p("y by (ii), p("y < q(i)y because p(i) and q(;) EQ, and

This takes care of p , q E Q and the first case in the induction step. If p~qfol lowsfroln (,,W), thenp = p O ~ p l , wherepoGqorpi Gq. HencepoyYqy or piy<qy, therefore p y =poyApty<qy. If p c q follows from ("W), (W,,), or (W"), the proof is analogous to the proof in the last case.

0

q ( i p I qy by (i), inlPlYing PY I w.

Now take a p CP(Q) and define WP)V = PY -

9 is well-defined since if R ( p ) = R(q) (p, q EP(Q)), then p E q and q E p - Hence, by I,emnia 9, py

(B(l))AR(q))T =R(pAq)v = (PAP)Y ' I nyAW =R(p)yAR(q)Q! and siinilarly for v, we conclude that y is a homomorphism. Finally, for 1) E Li , i E I,

qy and qy I py, and so py =qy. Since

R(P)g, =PY ' P 9 i by the definition of y , hence y restricted to Li agrees with yi.

Lemma 6(ii) implies that if p = q (p, qEP(&)), then, for all i E I , ~ ( ~ ) = q ( ~ ) and p ( i ) =q(’). Hence we can define

=qn and = P ( ~ ) .

All our results will now be summarized (G. Griitzer, H. Lakser, end C. R. Platt [ 19701) :

THEOREM 10 (The Structure Theorem of Free Products). Let L , iE I , be lattices and let L be a free product of the Li, i E I . Then for every a E L and i E I , i f some element of Li i s contained in a, then there i s a largest one with this property, a(i). If a = p(ao, . . . , a%-l), where p i s an n-ary polynomial and a,, . . . , an-l E U(Li I i€ I ) , then a(i) can be computed by the algorithm given in Definition 3. Dually, a(i) can be computed. b =q(bo, . . . , bm-J, a,, . . . , a,,-l, bo, . . . , bmWl E U (Li I a E I ) , we can decide whether a< b using the algorith,m of Definition 5.

For a, b E L, a =p(a,, . . . , an-l) ,

The idea o f the proof of Theorem 10 goes back to Ph. M. Whitman [194l] and R. P. Dilworth [1945].

We should comment on the use of the term “algorithm” in Theorem 10 and elsewhere in this section. (C) of Definition 5 brings in covers and so the procedure described in Definition 3. This procedure is an algorithm insofar as the structure of L, is described in an effective way. Thus if we consider the free product of finitely many finite lattices then we really deal with an algorithm. In the general case, algorithm should be interpreted intuitively and not be given the precise meaning assigned to it in mathe- niatical logic.

The existence of covers is not special to L, as it was observed in B. Jbnsson [1971]. To prove their existence let L be a free K-product of Li, i E I . Fix an i E I and for j c I defineyj:Lj+(Li)* asfollows:yiistheidentitymapon Li; for j+i ,ayj=Obfor all a E Lj. B y Exercise 1.4.14, ( L J a E K and so by the definition of free K-product, there is a homomorphism y: L-(L$ extending all yi, jC I. For a E L , a(i)=ay. Now if b EL, b < a, then by < a y , that is, b a, then a($) is the largest such element. Otherwise a y = 0,; indeed, if a y + o b , then b = a y E L , a contradiction in view of b=ay<a. (We uae the fact that a y ( a for all aEL since this holds for all a € U(Li I i E I ) . )

Free products (in fact, free K-products) of two lattices satisfy a special condition which is due to G. Gratzer, H. Lakser, and C. R. Platt [1970]:

a(<). Thus if there is b E Li, b

THEOREM 11 (The Splitting Theorem). Let the lattice L be a free product of the lattices Lo and L,. For every a c L , if a o ) i s not proper then u(1) i s proper a i d conversely.

Thus L = (Lo1 ” W1)l

where the union i s a disjoint union. I n other words, for every element a of L either a is conta.ined in some element of Lo or it contains a n element of Li.


PROOF. Hence (Lo] u [L,) = L. Now Theorem 10 yields the statement.

(Lo] u [L,) is a sublattice of L containing a generating set, namely, Lo u L,.

COROLLARY 12. Lo * LI can be represented as a disjoint union of four convex sublattices: the convex sublattices generated by Lo and L1, respectively, (Lo] n (Ll] , and [Lo) n [L,).

The next result is a trivial application of Theorem 10.

COROLLARY 13. Let K i be a sublattice of Li, for i c I , and let L be a free product of the Li, i c I . Let K be the sublattice of L generated by U ( K c I i € I ) . Then K i s a free product of the Ki, i€ I .

PROOF. This is obvious since for p , q EP(U (Ki [ i E I ) ) we have p C_ q iff it follows from Definition 5 and in applying Definition 5 we use only elements of the K,, i E I . Thus, by the proof of Theorem 8, K is a free product of the K, i E I .

This result is a special case of a result of B. J6nsson [1961] and [1965]:

THEOREM 14. Let K be a n equational class satisfying the Amalgamation Property. Let A , B, CEK and let C bt a free K-product of A and B. Let A, be n sublattice of A , let B, be a sublattice of B, and let C' be the sublattice of C generated by A , u B I . Then C' is a free H-product of A , and B,. (See Figure 4.)

Figure 4

PROOF. Let C" be a free K-product of A, and B,. Thus there exists a homomorphism is the identity on Ai and Bi. A, is a sublatticeof A and

of C"; thiis by the Anialgamation Property there is a lattice D in K containing A and C" as sublttttices, A , c A n C". Similarly, Bl is a sublattice of D and B, and thus there exists a lattice E in K containing B and D as sublattices, BIG B n D

of C" into C' such that

(see Figure 5). Since C is a free product of A and B, there exists a homomorphisin q1 of C into E such that q+ is the identity on A and B. Let p, be the restriction of q1 to C'. Then p, maps C' into C", qq is the identity on Ai and B1, and p , ~ is thus the identity on C'. Similarly, ~ p , is the identity on C", so p, is an isomorphism between C'and C".

Figure 5

There are three properties that play an important role for free lattices. They are the following:

(W) (SD,) (SD,)

m y ( uvv

u =XAY =ZAZ implies that u =XA(YVZ) .

u =xvy = X V Z implies that u =zv(~ 'Az ) .

implies that [my, uvv] n (x, y , u, v} =+ 0.

W again stands for Ph. M. Whitman. (W) can be rephrased: ZAY< uvv implies that x< uvv or y < uvv or x ~ y < u or x ~ y < 01. Observe, that a free product need not sat,isfy (W); indeed, X A Y < uvv may hold in one of the factors or on account of (C).

Let us say that a subset A of a lattice L satisfies (W) iff (W) holds in L for x, y, z, u E A. The next result (G. Gratzer and H. Lakser [1974]) shows that many subsets of a free product have (W).

THEOREM 15. be a subset of (Li)b satisfying (W). Let A be a subset of L satisfying

Let the lattice L be a free product of the lattices Li, i c I . Let A , iE I ,

and

for all ic I . Then A satisfies (W) in L .

PROOF. Let 2, y, u, v c A , x = R(p) , y = R((I), u = R(r), v = R(s), where p , (I, r , s E P(&), Let XA~J<UVU. Then pAqErvs. If (C) applies, then, for some iEl , (pAq)"<

275 1. Free Products of Latkices

(rvs)(i), hence p(i)Aq(i)S r(i)vs(i) in A+ Therefore, pci)< r(i)vs(i), or q(i)< r(i)vs(i), or p(i),-, q (0 <r(i), or p ( i ) ~ q ! i ) < s ( ~ ) . Again by (C) we conclude that P E T V S , or q !Z rvs , or pAq C r, or P A P E 9. (,W) and (W " ) give exactly the same conclusions. We conclude therefore that (W) holds in A.

Observe how strong (W) really is. For instance, it follows immediately, that there is no element in A which is doubly reducible.

Naturally, every a e L has infinitely many representations of the form p = Pbo, * * > U , - ~ ) < P ( Q ) (ao, . . . , a n d l EQ). If r(p) is minimal we call p a minimal representation of a and we call p a minimal polynomial. Using the notation of Theorem 10, the next result (H. Lakser [1970a]) tells us how to recognize a minimal representation :

THEOREM 16. Let p EP(Q). Then p is a minimal representation iff p EQ, or i f p = pave - - V P , - ~ , n > 1, where no pi i s a join of more than one polynomial and conditions (i)-(v) below hold, or the dual of the preceding case holds.

(i) Each pi is mininial, 0 I; j < n. (ii) For each O<j' < n , p j Q p o v - * *vpi-lvpj+lV* * *vp,-,.

(iii) If 0 2 j < n, r(pi) > 1, i E I , then (pi)(%) F p(i) in (LJb. (iv) If pi =pi"\pi' (0 < j < n and p i , pi' E P(Q)) , then p,! Q p and pi' g p. (v) I f P i , p k E L i ( O ~ j l k < 1 2 n n d i E I ) , then j = k .

PROOF. Consider the polynomials qi =pov- . ~ p ~ - ~ v p , ! v p ~ + ~ v . * ~ v p , - ~ , where pi = p j and r(p;) <r(pi), qii =pov* - 'VPj-lVPj+lV' - -vpn-l , qiii=pov. - 'vpj-~v(Pj)(i 'vpi+1v' .VPn_ l ,

q . 22) =pov. * .vpj-1vp;vpj+,v- * *vpn-1, qv =POv' ' 'vpj-lvpj+lv' 'vpk-lvpk+lv' * 'vpn-lvq, '

where q E L, and q ' p jvpk in Li. It is obvious, that if condition (a) fails ( i<z< w ) , then p =qz and r(qz) <r (p ) ,

contradicting the minimality of p. To prove the converse we show that if p, p e p ( & ) , p z q , p satisfies (i)-(v), and q is

minimal, then r (p) = r(q). This gives us an algorithm to reduce any polynomial to a minimal one and at the same time verifies the converse. Indeed, if p was not minimal, there would be a q wit3h p E q , r ( p ) > r (q) , and q minimal, contradicting the above statement,.

So let p and q be given as specified, p =pave * * ~ p ~ - ~ ) n > 1. Firstly, we claim that r(q) > 1. Indeed, if r(q) =1, then qE L , for some i € I . Thus,

by Lemma S(ii), ~ ( ~ ) = p ( i ) = q . At most one component of p is in &. Indeed, if pi, pk E&, j' $: k , pi E Lij, pk E Lik, then by (v), ij =+ ik, and so one of them ia not i, say i =k iP But then p i i j p , hence P,~,, is proper, which contradicts by Corollary 4 the fact that p(i) is proper. Thus there is a j with 0 < j' < n and r(pJ > 1. Then ( ~ ~ ) ( ~ ) < p ( ~ ) =q and so (pi)(i) is proper, and (pj)(i)<q = P ( ~ ) , contradicting the fact that p satisfies (iii).

Secondly, we claim that p is not of the form qoo"ql. Indeed, let q =qoA\Qi. Consider q ~ p . If (C) applies, then ~ ( % ) < p , ~ , = q ( i ) , hence q c . 1 ) =q(i)=q, contradicting r(q) 1 1 19 Griitzer

and the minimality of q. If ( AW) applies, then qo C p or q I E p . Obviously, p E qi, thus if, say, qo E p , then p Eqo, contradicting the minimality of q. Finally, if (W,) applies, then q c p o v - *vpn-2 or q C p n - l . The first possibility yields pn- l CPOV. - * \ / P , , - ~ while the second gives pn-l = p (and therefore po E p l v . - - ~ p ? , - ~ ) , both contradicting the fact that p satisfies (ii).

Thus q is of the form q=qov* * *vqm-1, m> 1,

where no q is the join of two polynomials. Next we show that there are functions

f : {o , l , ..., n - ~ } + { O , l , ..., m - I } , g:{O,l, ..., m-1}-{0,1, ..., n - I }

satisfying the following conditions: (a) g ( f ( j ) ) = j for 0 1 j < and f ( g ( j ) ) = j for 0 1 j <m. (b) If O < j <n and r(pi) > 1, then qftn -p i and r(qfcn) =r (p i ) , and similarly for any

(c) If p j < L, (0 j < n and i E I), then qfCj, EL, and similarly for qi < Li. Let 0 1 j < n and r (p j ) > 1. Then pi G q. If (C) is applicable, then ( ~ ~ ) ( ~ ) < q ( ~ ) , for

some ic I; since q(i) =p!$,, we obtain ( p # i ) ( ~ , ~ ) , contradicting condition (iii) for p . (,,W) is not applicable either because it would contradict that p satisfies (iv). Hence only (W,) is applicable. Thus p i i q o v . * -vqm-z or p j c q m - l . Continuing this argument we conclude that pj C qf(i) for some 0 < f ( j ) < m. If qf(i) E Li, for some i < I, then pi E qrcn implies that (p,.)(Q< qfiCi, in L, thus (pj)(i)< q f ( j ) I q(+ = P ( ~ ) , contradicting the fact that condition (iii) holds for p . Therefore, r(qrci,) > 1.

0 j < m. with r(qj) > 1.

Since q is minimal it satisfies (i)-(v), hence we can define the function g. Thus

pi E %(rCi) C pu(f(i)),

and so by (ii), g ( f ( j ) ) = j and pi = qf,i,. Similarly, f ( g ( j ) ) = j and q,. ~ p , ( ~ , . Now let 0 < j < n and r(pj) = 1. Then p,. E Li for some i C I. Since p ,.E q we get that

q(i) is proper. Since q(i, = (qo)(i,v. - . ~ ( q , - , ) ( ~ ) , some (qi)(o must be proper. By renumbering qo, . . . , qmdl we get that (q,)(i) is proper iff 0 I k < t , where t 5 m. Thus

If r(qs) > 1 for all 0 I < t , then r(pg(#)) > 1 and (P,,(,))(~) = (qs)(i). Therefore, g ( s ) * j for all O<s < t , and so

Q(i) = (!IO)(i,V. * *v(qt-l)(i) .

P j I (!IO)(~V. . v ( q t - l ) ( i ) S (POV' . 'vPj-lVPj+lV' . .VPn-i)(iy Thus p j ~ p o v * * *Vpj_1Vpj+1V. . 'vpn-l, contradicting the fact that (ii) holds for p .

Consequently, we can choose O < f ( j ) < n such that r(qfli,) = 1 and is proper, that is, such that qf(i, E Li. Similarly, we define g( j ) for 0 < j < m and r(qi) = 1. It is obvious that (a), (b), and (c) are satisfied.

Now (a) implies that f arid g are one-to-one, hence n =m. Since r ( p j ) =r(aici ,) , we conclude that r ( p ) =r(q).

A subset A of a lattice L is said to 8uti8fy (SD v ) iff (SD v ) holds for all x, y, z E A. The following result (G. Griitzer and H. Lakser [1974]) establishes for (SD,) what was done for (W) in Theorem 15.

THEOREM 17. Let the lattice L be a free product of the lattices Li, i C I . For each ic I, let A i be a subset of (Li)b satisfying (SD,). Let A be a subset of L such that A(i) E A , for all i c I . Then A satisfies (SD,).

PROOF. Let z = R ( p ) , y = R(q), z = R(s) E A, x v y = x v z , and let u =uov* * -vu,-l, n 2 1, be a representation of p v q ~ p v s satisfying (i)-(v) of Theorem 16. We show that for each j with 0 2 j < n and r(ui) > 1 we have uj p or uj C q. Consider u j c pvq. If (C) applies, then ( u ~ ) ( ~ ) < (pvq)( i ) = U ( ~ ) , contradicting 16(iii). If (,W) applies, that is, if uj=u;Au;' and ui E p v q , we get a contradiction with 16(iv). Thus only (W,) can apply, yielding ui C_ p or ui C_ q. Similarly, u j G p or ui C s, hence ui E pV(qAs). Now take a j with 0 < j < n and r(ui) = 1. Then ui = yi) for some i € I. Therefore,

(q ( i )m( i J by ( S D , ) . Thus, U ~ C ~ V ( Q A S ) by (C). Since u i C p v ( q ~ s ) is proved for nil j with 0 < j < n, we conclude that uOv* * -vu,-l C pV(qAS), and SO u =SV(YAZ) ,

as clainied.

~ ( i ) =~o(1)v~(i) and ?((,) = P ( ~ ) v s ( ~ ) in (Lila and P(i), q(i), s(i)EAi, and 80 uj<P(i)v

By dualizing Theorem 17 we can obtain the analogous result for (SD,). We conclude this section by proving that the Common Refinement Property holds

for free products (G. Griitzer and J. Sichler [1975]).

THEOREM 18. free product of (A,n B j I i, j = O , 1, A i n Bi+ 0).

Let L be a free product of A, and Al and also of Bo and B,. Then L is a

PROOF. Let a C A,. We claim that

E (A, n Bo) " { O b h

Indeed, since L is generated by B, u B,,

a =p(bO,O, b O , l , * . ' 7 b l ,O , ' 1 . 1 7 . ' *), where bo,,, bo,l, . . . C Bo and blFo, bl,l, . . . < B,. Computing the lower A,-covers and observing that ado =a, we obtain

a =P((bO,O)Ao' (bO,l)Ao, - * * 3 (b1,O)A"' (bl,l)Ao' - - Forniing lower B,-covers in the original expression for a we get

"no =p(b,J,07 B O , l , * * 9 Ob, Ob> * . .), and from the previous forniula:

flJlo = (aA0)Bo = P(((bO,O)AJBo~ * - * 7 ((bl,O)Ao)Bo, * * *)*

But bl,D,a 2 (b,,,,Jd0 and 80 Ob = ( ~ I , ~ ) B ~ 2 ( ( ~ I , ~ ) A ~ ) B ~ , ( ( b i , a ) ~ O ) ~ o =Ob- Thus

10'

and so

aBo=p((bO,O)do> (bO,l)Ao, * * * J Ob, O*J ; * .) E A , " {@}a

By definition, aE0 E Bo u {Ob}, hence aBo E (A, n Bo) u {Ob} . qimilarly, for i, j E (0, l}, aEi E (Ai n Bi) u {Ob} . It follows immediately, that

a =p((b0,0)&,9 (bO, l )& ' ' * J (bi,0)du9 (bl,l)diJ ' *) B i ) " {Ob)*

Now a simple induction on the rank of a polynomial proves that for aEAo and the polynomial p of smallest rank representing a in the form

a=p(ao,.. . ,a,&-l),aoJ ~ ~ ~ ~ a ~ - ~ ~ ( A o n B ~ ) u ( A ~ n B ~ ) u { o b } ~

no ai is Ob. We conclude that

A0 E [ (A , n Bo) u (A , n BdI- Thus

L = [ A , uAl] =[U (A,n Bi ] i, j = O , l)].

Applying Corollary 13 twice we get

A , = (A, n B,) * (A, n B,) and A, = (Al n Bo) * (A, n B l ) ,

Bo = (Ao n Bo) * (A, n B,) and B1= (A, n 231) * (A, n Bl),

hence ~ = ( A , n B o ) ~ ( A , n B , ) * ( A , n B o ) * (Al nB,)

(to be more precise, drop all Ain Bj= a), and this is the common refinement of Ao*Ai=Bo* Bi.

1. 2. 3.

Prove that E2 * El is indeed the lattice of Figure 1. Show that E3 * El is the lattice of Figure 2. Find an infinite descending chain in g6 the two chains. Define

G1. (Hint: Let uo < a1 <a2 <a3 and b bo

C l = ((%A (alvb))v(a3Ab))A ( b 3 A (aovb)) V a l )

0, =( (~ , - i~u2)v (a ;3~b) )~ ( (o ,_ lh (a ,vb) )~ut ) .

end, for n > 1,

Then cl >c, > c 3 > . - . .)

1 . Free Products of Lattices 279

Figure 6

VI. Free Products 280

4.

6.

6.

7. 8.

9.

10.

11.

12. 13.

14.

16.

Construct an infinite descending chain in E2 * E2. (Hint: Let a. <a, and b, < h i be the two chains. Define

ci = ((aiA(aovb,))vbo)A((bl A(aivb0))vad and, for la > 1,

Cn = ( (cn - ihai ) V b o ) h ( (cn - lAbi)vao).)

A * B is finite iff A or B is the one-element lattice and the other is a chain of not more than three elements (Ju. I. Sorkin [1352]). Let al < a2 and bi < b2 be two chains. Introduce the following n o t d o n (we Figure 6) :

Ai=az , B,=bz, A;=al , B;=bi

and, for n > 1,

A, =a,h(alvBn-l), Bn =bzA(biVAn-I)

0, =alvB,

Dn =bivAn

P , =AnvBn

Qn =CnhDn

Mi =aivbi

M , =(a2Abz)vaivbi

v, = b2h ( (a,hb2)va,vbi)

vz =(a,hb2)v(bzA(aivbi))

V3 =b2A(alvbi)

w, =a2A((a2Ab2)va,vbl)

WZ = (a2Abdv(a2A(aivbil)

W , =a2h(aivbl). \

Prove that these polynomials and their duals (denoted by ’) represent distinct elements and all the elements of 6 2 * E2 as shown on Figure 6 (H. L. Rolf [1958]). Show that Figure 7 gives a description of E4 * El (H. L. Rolf [1968]). To what extent can Theorem 10 be simplified for free products of chains. (Hint: replace (C) in Definition 6 by “for some i €1, p, q E Li and p Define the concept of a free (0, 1)-product of bounded lattices. Develop the theory of free (0, 1)-product. Let L be the free (0, 1)-product of the lattices (L$, i €1. Show that L - (0 , 1) is a sublattice of L and it is a free product of Li, i €I. Let L be a free product of the Li, i€I. Show that aELi is join-irreducible in L iff a is join-irreducible in Lp (Hint: use only covers in the proof.) Show that Exercise 11 holds for free R-products in general (B. J6nsson [1971]). Let L be a free product of the Li, iEI, and let ‘pi: Li + A be an isotone map of L , into a lattice A, for $ € I . Show that there is a n isotone map ‘p: L -+A extending the ‘pi (Ju. I. Sorkin [1962]). In a free product of chains, the minimal representation of an element is uniquely determined up to commutativity and associativity. Define the canonica2 repreeentation p of an element of a free product in the following way: p EQ; or p is as in Theorem 16, all pi are canonical and if pi E Lj, then pi =p(j,;

q in L:’.)


Figure 7

VI. Free Products 282

16.

17.

18.

a.nd dually. Show that the canonical representation is uniquely determined up to commutativity and associativity (H. Lakser [ 1970al). Find a pair of lattices A , B, at least one of which is not a free product of chains and such that every minimal representation is canonical in A * B (H. Lakser [1970e]). Let L be a free product of Li,iEI. Under what conditions is L-Li 8 sublattice of L? Let I be the disjoint union of Ij, j € J . Let Li, iEI, be lattices and let Aj be a free product of Li, i E I p Then A is a free product of Li, i € I , iff A is a free product of Aj, j € J .

2. The Structure of Free Lattices

By comparing the definition of a free lattice F(m) (see Section 1.6) with the definition of a free product (see Section 1.6 and Definition 1.1) we observe:

A lattice is free iff it is a free product of a family of one-element lattices. Thus the results of the previous section can be specialized to describe the structure

of free lattices.

DEFINITION 1. For a set X and p , q € P ( X ) , set p q iff it follows from x C_ x, for x E X , and from (,,W), (vW), (WJ, and (WV).

Again we set p = q iff p C_ q and q ~ p , R(p) = { q I p = q and q EP(X)}, R(X) =

Now we can state the celebrated result of Ph. M. Whitman [1941]: { W P ) I p € W ) } , and R(p) < R(q) iff p 5 q.

TZIEOREM 2. R(X) is a free lattice on 1x1 generators. I n other words, Definition 1 provides an algorithm to decide whether a , for i € I. It is sufficient to show that if p C q by (C), then p E q by Definition 1. We prove this by induction. This is obvious if p , q E X . Now let r@) +r(q) > 2, q(i) for some i ( I, that is, p( i ) =q(ci, = i . If p =pOApi, then (p0~pi)(') = (p0)(Q~(pi)(i), hence (po)(i) = i or (pi)(i) = i . Thus (po)(i) < q(i) or (pi)(i)< q(i), hence by the induction hypothesis, po C q or pi 5 q, and so by (,,W), p E q. We proceed similarly in the other three cases.

From Theorem 1.16 we learned that minimal polynomials differ only in some Li component. If all lLil = 1, they have to be identical.

TIIEOREM 3. A minimal representation of an element is unique up to commutativity and associativity. Let p E P(X). Then p is minimal iff p EX, or if p =pov- - *vP,-~, n > 1, where no p j ia a join of more than one polynomial and conditions (i)-(iii) below hold, OT dually.

(Ph. M. Whitman [1941]).

2. The Structure of Free Lattices 283

Figure 1

(i) Each pi is minimal, 0 < j < n. (ii) For each j with 0 1 j < n , pi$pov. - ~ v ~ ~ - ~ v ~ ~ + ~ v ~ * *~p?,-,.

(iii) If pi = p i A p i ’ ( O < j < n and pi , pj’ E P(X)), then p i Q p and pi’ Q p .

PROOF. In the special case considered, l.l6(v) is made superfluous by 1.16(ii). Also, l.l6(iii) does not apply since (P~)(~), p(;)EL,, hence = P ( ~ ) . The remaining conditions of Theorem 1.16 are identical with (i)-(iii) above.

A minimal representation of an element in a free lattice is called a cunonical form of the element in the literature.

Now we shall study in some detail the structure of F(3) . Let xo, x1 and x2 be the free generators, X ={xo, xl, x2}. The zero and unit of F(3) are xOAz1Ax2 and z0vxIvx2, respectively. By Lemma 1.5.9, xovzl, xovx2, and xlvx2 generate a sublattice isomorphic to (Q2)3 . Since F(3) ={xo} * { x ~ A x ~ , xi, x2, x1vx2} (see Exercise 1.18), the Splitting Theorem (Theorem 1.11) gives that for any z c F ( 3 ) either ~ 2 x 0 or $1 x1vx2. Similarly, x 2 xi or X < x0vx2; and x 2 x2 or x ~ s 0 v x 1 . From this we can infer that rOvxL < xovzivx2, ( x o v ~ 1 ) ~ ( x o v ~ 2 ) -< xovxl ; these, and the symmetric cases give the nine coverings at the top of Figure 1. Let us prove the second covering: if (ZoVZl)A(ZoVX2) < t < xOvzl, then t 5 xOvz2, hence t 2 x1 ; similarly, t 2 xo, hence t 2 xovzl, proving that t =xovxl.

If t > xo, then t $ xo, and so t 2 x l ~ x 2 ; hence xo -< x o v ( z l ~ ~ 2 ) =xova; by symmetry and duality this accounts for six coverings in the middle of the diagram. The other six are typified by xova >- (xova)Ab, which is again a t,rivial consequence of the Splitting Theorem.

Again, easy applications of the Splitting Theorem yield that all elements of rank 4 or more lie in one of the intervals: [xova, (XoVZl)A(ZoVZ2)], its two symmetric intervals, the three dual intervals, and [a, b]. Thus to complete the proof of Figure 1 we have to verify that

[xova, ( ~ o V ~ 1 ) ~ ( ~ 0 ~ ~ 2 ) 1 = [%ova, xovbl u {(~ovzl)~(qlv”2)}. The trick (due to P. Gumm) is to verify first, by induction on r ( x ) , that if x E [zova, (xoVxl)A(xoVx2)], then x is comparable with x0vb. Condition (W), and therefore Theorem 5 is needed in this step. Then zovb < (xovxl)~(xovx2) follows easily. The details are left to the reader.

A word of warning about Figure 1 : it gives a ‘‘ generalized diagram ” of F(3) but it does not contain all the partial ordering or all the meets and joins. For instance, ( z ~ A ~ ) v ( z ~ A ~ ) (the uo defined below), an element of [a, b], is not shown. Nevertheless, Figure 1 can be very useful in visualizing F(3) .

We obtain mme additional coverings in F(3) from a result of R. A. Dean [1961a]:

THEOREM 4. For any c E [xov(q~x2) , (zovzi)A(xovx2)] we haee that

CA(XiVX2) < (CA(XlVZ2))VXo.

PROOF. These two elements are obviously distinct since they belong to disjoint intervals. If C A ( X ~ V X ~ ) < t l (cA(xlvx2))vxo and t 2 x o , then obviously t = (CA

2. The Structure of Fret? Lat,tices 285

(xiv~.,))vxO. Otherwise, f I xIvz2, hence t I ( ( C A ( Z ~ V X ~ ) ) ~ X ~ ) A ( X ~ V ~ ~ ) I cA(zIVx2), and so t = C A ( X ~ V X . , ) .

Apart froiii the Splitting Theorem the most important properties of a free lattice are given in the next result, which is due to Ph. M. Whitman [1941] and B. J6nsson [1961]:

THEOREM 5. Conditions (W), (SD,), and (SD,) hold i n any free lattice.

PROOF. Ai=(L i )b and the result follows from Theorems 1.15 and 1.17.

We illustrate the power of these conditions by two results.

If lAil =1, then (Ai)*=&, hence it satisfies all three conditions. So choose

THEOREM 6 (B. J6nsson [1961]). A sublattice A of finite length of u free lattice (of (my lattice satisfying (SD,)) i s finite.

PROOF. the statement is trivial. Now let A be of length n and let a be an atom of A. Then

Let A be of length n. We prove the statement by induction on n. If n = 1 ,

A = [ a ) u {X I U A X =O}.

By (SD,), {z I U A X = 0 } is a sublattice. Hence A is a union of two sublattices of length less than n , and so A is a union of two finite sets. Thus A is finite.

A sublattice of a free lattice need not be free. The next result shows how free sublattices can be found.

THEOREM 7. Let the lattice L be generated by X . If X is irredundant, that is,

X< V Y implies X E Y for any x ~ X and finite Y E X ,

and dually, then L i8 a free luttice iff L satisfies (W).

PROOF. The necessity of these conditions follows from the Splitting Theorem and from Theorem 5. Now let these conditions hold. With every a c L associate the set S ( p ) of all p P(X) which represent a. If a =S(p) , b =S(q) , then a( b iff p c q by Defi- nition 1. This is easily seen except if p = p , , ~ p ~ and q =qovql in which case it follows from (W). Thus L is a free lattice.

Now consider in F(3) the elements

2 b = (%A (Xi VX?) ) v (XI A (ZoVX2)) , 211 = (XoA ( z i V X 2 ) ) V(X2A(XoV$i) 1, '+ = (XoV (XI As*)) A (21 V (XoAX2))

213 = (XOV ($1 AX.,)) A (X2V @oA%)).

286 V.I. Free Products

It is an easy computation to show that U ={uo, ul, u2, ug} is irredundant. Hence P(3) x F ( 4 ) as a sublattice. Now define yo =uo, a. =ui, bo =UZ, co =US. If Y,, a,, b,, and c, are defined, construct Y,+~, a,+l, b,+l, and c , + ~ from a,, b,, and c,, just a8

the ui were constructed from xo, x i , and x2. We claim that Y ={yo , yl, y2 , . . .} is an irredundant set. To see this we prove by

induction that yo, . . . , y,, a,, b,, c, generate a free lattice on n +4 generators. This is true for n =O. Assume it for n. Then

[YO, * * 9 Yn, Yn+l, an+,, bn+i, cn+11

is the sublattice generated by [ y o , . . . , y,] and [yn+,, a,,,, b,+l, c , + ~ ] in [ y o , . . . , y,, a,, b,, c,] = [yo, . . . , y,] * [a,, b,, c,]. Thus by Theorem 1.14,

[Yo, * . * , ~ n , ~ n + l , a , + l , b n + ~ , ~ , + i l ~ ? ~ ~ , . . - , ~ n l * I ~ n + i , a , + i , b , + i , c , + i l Y F ( n + 1) * P(4) 2 F(n + 5).

This proves the statement. Thus Y is irredundant and so, by Theorem 7 , [ Y ] rF(No). So we have proved :

THEOREM 8 (Ph. M. Whitman [1941]). P(3) contains P(No) as a sublattice.

A free generator of a free lattice is doubly irreducible, and of course no other element is such. Hence we obtain:

THEOREM 9. lattice and permutations of its free generating set.

There is a one-to-one correspondence between automorphisms of a free

This is used in proving the following result of F. Galvin and B. J6nsson [1961].

THEOREM 10. Every chain of a free lattice i s countuble.

PROOF. Let X be the free generating set of L. We can assume that X is not count able. Let X o be a countable subset of L, let Lo = [X , ] , end let C be a chain in L.

For a , b C L let a = b mean that a y = b.for some automorphism 'p of L. Obviously = is an equivalence relation. Now if a E L, then a f [ Y] for some finite Y s X. Take a permutation p of X with Yp c X,, and let q be the automorphism of L extendingp. Then a q C Lo. Hence there are no more equivalence classes than JLol = No. Therefore the proof will be complete if we show that for any a , b C C, a + b, we have a + b. Indeed, let a < b and aq = b for some automorphism y . For some finite Y E X, a f [ Y ] . There is a permutation p of X such that xp =xy for x E Y and xp = x for x EX - ( Y u Yy). Let p be the automorphism of L extending p . Then ay = b and y is of some finite order n (as a group element). Thus,

a < ay < a@ < - * - < ayfl =a , e contradiction.

is the following result of R. P. Dilworth [1945] and R. A. Dean [1956]: A somewhat unexpected dividend of the study of the structure of free lattices

2. The Structure of Free Lattices 287

THEOREM 11. maximal elements.

There exists a three-generated partial lattice with infinitely many

PROOF. Let xo, xl, x2 be the free generators of F(3) and define yo =xo,

and a n =xi v ( Y ~ ~ A ( Y -,v22 ) 1 *

It is easy to check that [a,, uq, . . .] Y P(N,). For a polynoinial p, define the component subset of p , Komp(p), as follows:

Komp(xi) = { x i } ; K ~ ~ i i p ( p ~ ~ p , ) =Koinp(po) u Komp(pi) u (po~pl} and similarly for PoVPi-

If a, c P(3) , let a = R(p) where p is the canonical polynomial representing a. Define Komp(u) ={ R(p) I p E Komp(p)}. Then Komp(a) E F ( 3 ) . Regard Komp(a) as a relative sublattice of F(3) . Then Komp(a) becomes a partial lattice. Komp(a) is generated by {xo, q, G,} n Konip(a).

Now define A = U (Komp(n,) I n = 1,2, . . .). All the elements of A are easily enumerated and we observe that ai, u2, . . . are maxinial elements of A.

COROLLARY 12 (Ju. I. Sorkin [19M] and R. A. Dean [1956]). Every countable luttice can be embedded in u three-generated luttice.

PROOF. Let L ={bo, b,, b?, b,, . . .} and, with the partial lattice A of Theorem 11, form the set B = L u A (we assume that L and A are disjoint). We define a partial ordering 4 :

for u,v c L or u,v A , u s v iff u<v in L or A, respectively; a2n+l<b,~anda2,+,<bn;for u c A a n d v c L , u<viff there &revo,. ..,v,-,EL such that (i) V ~ A - - . A W , - ~ < V in L; (ii) for each i, O<i<n, there is a b, such ,

that b,<vi in L and ~ < a ~ ~ + ~ , u I a , j + 2 in A ; for u E A and v EL, v u never holds.

Now it is easy to check that B is a partial lattice, b , = a z n + l ~ a z n + 2 , for n =0, 1,. . . , and so B is three-generated. Every partial lattice B can be embedded in a lattice C and obviously, [B] in C is three-generated and contains L as a sublattice.

Exercises

1. Let L be a lattice completely freely generated by tho poset P. Show that an algorithm is provided for deciding whether a 5 b in L by the rules: if a, b E P, a b iff a < b in P ; and by the rules (AW), (vW), (WA), and (Wv) (R. P. Dilworth[1945]). Let L be a free product of Li, i C 1 , and let Q =u(Li I icl) (disjoint union). Show that L 1s completely freely generated by Q iff all Li are chains.

2.

288

3. 4.

6.

6.

7 . 8.

9.

10.

11.

12.

13. 14. 15.

16.

17. 18. 19.

VI. Free Products

Prove that ( B p is a sublattice of a free lattice iff n 5 3 . Let L be a modular lattice. Verify t,hat if L is a sublat,tice of a free lattice, theu L is distribut,ive. Let the lattice L be of length n. Show that if L is a sublattice of a free lattice, then

Let the lattice L be generated by a finite set X and let US assume that z $V(X - {x}) for some x EX. Then for any a 2 z we have that

ILJ 5 2%.

W V ( X - {z}) < (ahV(X - {z}))vz.

Let m 2 it,,. Show that there is no covering in P(m). Let nt 2 No and let a, b € P ( m ) , a < b. Show that [a, b] has a sublattice isomorphic

Let f (n ) =max {ILI 1 L is a sublattice of a free lattice and L is of length n}. Then (vF)n s f (n ) (B. J6nsson and J. E. Kiefer [1962]). A lineay decomposition Ai, i CI , of a lattice A consists of a. chain I, sublattices Ai of A, for i €1, such that if i, j €1, i < j, then a < b in A for all a EAi and b € Bi atid A =U(Ai I i c1) . Show that if all A$ are sublattices of a free lattice and I is countable, then A is also a sublattice of a free lattice. A distributive lattice D is a sublattice of a free lattice iff D has a linear decomposition A;, i €1, such that 111 5 it,, and each A < is either &, or (&#, or of the form Lr, X C where G is a countable chain (I?. Calvin and €3. J6nsson [1961]). Let A be an algebra with the following properties: a partial ordering 5 is definrd on A ; A has a generating set X such that every pelmutation on X can be extended to-an'isotone automorphism of A.'Pl'ove that' every chain in A is countable. Enumerate in canonical form all the elements of A in the proof of Theorem 1 1. Prove that B in the proof of Corollary 12 is a partial lattice. Is t,here a general lernnia about " gluing together " two partial lattices which cont,ains the construction of Corollary 12? Use component subsets to prove that if a, b €$'(it,,), a +b, then there is a honiomorphism g, of P(Xo) onto a finite lattice such that ag, + b p (R. A. Dean [1956]). Derive Exercise 16 from Corollary IV.4.6. Prove that every finit.e lattice can be embedded in a finite three-generated lattice. Using the notation of Figure 1, let y € P ( 3 ) , ei 5 y i < di, i =0, 1, 2. Prove that [yo, yi, y2] is a proper sublattice of P(3) isomorphic to F ( 3 ) .

to P(m).

3. Reduced Free Products

Let L , i < I , be bounded lattices and let L be a free (0, 1)-product of the Li, i c l (see Definition 11.5.14 and the discussion following it). As we shall see, a pair of elements x, y is cmplementa.ry in L (that is, XAY = 0 and xvy = 1) iff either they belong to some Li and they are complementary in Li or if there exist elements xo, xi, yo, yl in some Li such that zo( x 5 yo, zl y < yi, and {zo, xi}, {yo, yI} are complementary in Li. We shall describe a, construction in which there are many more complements than in the free (0, 1)-product, but in which we can still keep track of the complements. We call this construction the reduced free product. Several applications will be given in this section and the next.

3. Rcduced Free Products 289

In the discussion below let L , i E I, I$- 0 , be pairwise disjoint bounded lattices and & = U ( L i l i c I ) .

DEFINITION 1. A C-relaticn b on Li, i c I , is ( I set of two element subsets of Q such that i f {a, b} c b, then there exist i, j E I , i =k j , satisfyi,ng a E Li -{Oi, 1 i } and b E Lj - {Oj, lj}.

DEFINITION 2. product of the Li, i E I , i f f the following conditions hold:

(i) Each Li, i E I , is a (0, 1)-sublattice of L and L = [ U (Li I i € I ) ] . (ii) If {a, b} c b, then a , b is a complementary pair in L.

(iii) I f , for i E I , yi is a (0, 1)-homomorphism of Li into the bounded lattice A , and {a, b} c b (aELi , b E L j ) implies that uvi, byi are complementary in A , then there is u homomorphism v of L into A extending all the vi, i E I .

Let b be a C-relation on L , i c I . A lattice L is a b-reduced free

Using the technique of Section 1.5, the existence of a b-reduced free product is proved if we find a lattice satisfying (i) and (ii). Such a lattice is

{0,1} u u ( L i - { o ; , l i } I i€O with the obvious partial ordering. The uniqueness of b-reduced free products can be established as in Section 1.5. The next result shall give a description of a 8-reduced free product based on the Structure Theorein of Free Products. This description is then used to describe the complementary pairs in a b-reduced free product.

Theorems 4 and 5 appeared in their present forin in G. Griitzer [1973], but earlier versions can be found in G. Griitzer [1971 a] and C. C. Chen and G. Gratzer [1969]. The gerni of the idea can be traced back to R. I?. Dilworth [1945]. .

DEFINITION 3. on the rank of p : (i) For p EQ, p ES i f f pE Li-{Oi, li} for some i E Z .

(ii) For p = q A r , p E S i f f q,r E S and the following two conditions hold:

We define a subset S of P(Q); for p E P(Q), p €8 i s defined by induction

(i i ,)pEOifornoiEI; (ii2) 21 G ZAY for no {x , y } E i?.

(iiii) l i s p f o r n o i c I . (iii2) x v y E p for no {x, y } E 8.

(iii) For p = qvr, p E S i f f q,r S and the following two conditions hold:

Now we set

and partially order L by L = { O , 11 u {Wp) I p ES},

O < W P ) < 1 for p € S ,

R ( p ) I H(q) iff p ~ q .

If we identify a E Li with R ( a ) , thc n we g t t the setup we need:

THEOREM 4. L is a i3‘-reduced free product of the Li, i E I .

PROOF. L is obviously a poset. To show that L is a lattice we have to find the meet and the join of R ( p ) and R(q) in L (p, qES) . We claim that R(p)AR(q) = R ( p ~ q ) if pAqES and otherwise R(p)nR(q) = O .

Oi for some i E I or u E SAY for some {x , y} E i3‘, then u ( S . We prove this by induction on the rank of u. If u E Q and u E Oi, then u =OitS. If u EQ and u c XAY for some {x, y } E .!?, then ucLi , xELk, yEL,, k+n, u<x and u g y ; these imply that i=lc and i = n , a contradiction. If u = u ~ A u ~ , and uo or ul ( S , then u ( S by 3(ii); if uo, ul ES, then u 4s by 3(iii) or 3(ii2). Finally, if u = uovul, then uo E Oi or ui s Oi in the first case, and uo c XAY or ui c XAY in the second case, and so uo or ui ( S implying u 4 S by 3 (iii).

To verify this claim it is sufficient to prove that for any u E P(&), if u

Dually, R ( p ) v R ( p ) = R ( p v q ) iff pvq E X ; R ( p ) v R ( q ) = 1 otherwise. Now it is obvious that a-+R(a) is a (0, 1)-embedding of Li into L. So after the

identification 2(i) becomes obvious. 2(ii) is clear in view of 3(iil), 3(ii2), and our description of meet and join in L.

Let K be the free product of the Li, i E I , as constructed in Section 1. Then L - (0, 1} K. We define a congruence 0 on K :

0 = V ( @ ( z , Oi ) I iC I , 2 < O i ) V V ( 0 ( X , li) I i E I, x 2 li)

V v ( @ ( X , UAW) I 2 1 UAW, { U , W) E g ) V V ( @ ( X , UVW) [ X 2 U V W , {U, W } c 6 ) .

In other words, 0 is the smallest congruence relation under which all Oi and , u A v (u, v E tY) are in the congruence class which is the zero of K / 0 and dually. We claim that

K/O s L.

To see this, it is sufficient to prove that every congruence class modulo O except t,he two extremal ones contains one and only one element of S.

Let ci be the identity map as a map of Li into L. Then there is a map pl extending all E ~ , i c I, into a homomorphism of K into L. Observe, that R(p)pl = R(p) for all p CS. Indeed, p =p(ao, . . . , where ao, . . . , an-l CS n Q; hence

R(P)F=R(PY)=R(P(~o, * * * 2 un- l )q )

=R(P(~oT, * * * 9 an-ly)) = R ( P ( ~ o , * * * 9 an-1)) =R(P), since aoy = ao, . . . , a, . - l ~ = an-l.

Let 0 be the congruence kernel of pl. Since L satisfies 2(i) end 2(ii), 0 < 0. Now if p , q ES, and R(p)pl = R(q)q, then R ( p ) = R(q). In other words, R ( p ) 3 R(q) (0) implies that R(p) =R(q) . Therefore, the same holds for 0. This proves that there is at most one R ( p ) in the nonextremal congruence classes of 0. To show “at least one” take a p EP(&) such that R ( p ) +Oi (0) and R(p) + li (0) (for any/all i E I) ; we prove that there exists a q €19 such that R ( p ) = R(q) (0).

Let PEL,, for some icI. Then, by assumption, p*Oi and li; hence we can take

take q = q o ~ q l . Otherwise, by 3(ii), qOghql =Oi(0), hence p = O i (0), contrary to our assumption. The dual argument completes the proof.

q =p. Let p =powl, R(po) = Wqo) (a), Wpi) = WqJ (a), where qo, qt €8. If q O w l €8,

3. Reduced Free Products 291

Thus we have proved that K/O s L. Now we are ready to verify 2(iii). For each i E I , let tpi be a (0, 1)-homomorphism

of Li into the bounded lattice A . Since K is the free product of the Li, ic I , there is a homomorphism y of K into A extending all the yi, iCI . Let Y be the congruence kernel of y. It obviously follows from the definition of O that O I Y . Therefore, by the Second Isomorphism Theorem,

[ X I @ ' x y

is a homomorphism of K/O into A . Combining this with the isomorphism L r K/O as described above, we get a (0, 1)-homomorphism tp of L into A extending all the % i E 1 . 0

THEOREM 5. the Li, i E I . Then there exist ao, bo and at , b, such that

Let a, b be a complementary pair in a Y:-reduced free product L of

ao<a<al and bo<b(bi

such that either {a,, bo}, {a l , b,} E Y: or, for some i E I , a,, bo and al, bl arecomplementary p i r s in L,, and conversely.

PROOF. and ai , b, are complementary in L, hence

aAb ( a, r\bl = 0,

The converse is, of course, obvious. I n either cam, by Definition 2, a,, b,

avb 2 aovb, = 1,

and so u and b are complementary in L. Now to prove the main part of the theorem, take p , q €8 such that a = R ( p ) and

b = R(q) are complementary in L. Then pAq violates 3(iil) or 3(ii2) and p v q violates 3(iiil) or 3(iii2). The four cases will be handled separately.

CASE 1. pAq violates 3(&) and p v q violates 3(iiil). Hence, for some i, j € I , p A q G Oi a,nd l i c pvq. Thus in the free product K of the Li, ic I , ( p ~ q ) ' ~ ) = Oi and (pvq)(+ = li. Note that q(0 is proper, because otherwise P ( ~ ) = O ~ , that is, p c O i , contradicting p E S. Similary, qu, is proper. This is a contradiction unless i = j , in which case we can put a, = P , ~ ) , bo =q(i,, al =p(O, b, =q(n and these obviously satisfy the requirements of the theorem.

CASE 2. pAq violates 3(ii,) and p v q violates 3(iii2). Hence there exist i € I and {x, y} € Y: such that

pAqCOi and x v y C p v q .

Let x E Lj and y E Lk, j , k E I , j + k. Then i 4: j or i + k ; let us assume that i + j . Since (pr\q)(i)=O, and p(i)nq(l)=O,, we conclude, just as in Case 1, that p(i) and qCi) are proper. From i + j we conclude that pCi, and q(i) are not proper, that is, p,i, = qci, = O b . Thus (pvq),j , =p,nvq(n =Ob, contradicting that p v q a x e Li. Case 2 cannot occw. 20 GrPtzer

CASE 3. as Case 2 does.

pAq violates 3(iiz) and p v q violates 3(iiil). This leads to a contradiction just

CASE 4. { u , , b,} E g such that a, E Li, b, E L , a1 E L k , bl EL,, i, j , k , n E I , i =t= j , k =t= n,

pAq violates 3(ii2) and p v q violates 3(iii2). Then thtre exist {u,, b,} E t: and

a,vb, C p v q and pAq E a,Ab,.

We conclude, as above, that'

a0 I P(i)VQ(i), bo s P(j)Vq(j)) p(k)Aq(k) < ai , ~ ( ~ ) r \ q ' ~ ) I bi .

Therefore, p( i ) or y(i) is proper, p ( j ) or is proper, p(k) cr q(k) is proper, end p(n) or p(n) is proper.

We cannot have both p(i) and q(i) proper, because then neither p ( k ) nor q(k) could be proper unless i = k ; and similarly, i = n, contradicting k * n.

So we can assume that p(i) is proper and qCs = O b , and therefore p ( i ) = (pvq)( i ) > uo. We cannot have pu, proper because then qO).=Ob and pU, 2 b,; thus p 1 aovbo. Now k * i and k * j yields a contradiction (neither p ( k ) nor qck) could be proper), hence k = i or k = j . Let k = i (the case k = j is similar). Since i * j , q(i) is not proper, hence p(a) is proper and p") = (pAq)")< a,. But i + n, hence p(,) is not proper and so q(n) must be proper. Since q(i) is proper we conclude that n = j and q( j ) = (pAq)(j) < b,. To sun1 up, we obtained

a, I P ( ~ ) , P ( ~ ) I a1 , b,) I qCs, q") I bl , hence

00 E E QI, b,) E q C b,,

as required. Let us say that a bounded lattice A has no compuruble complements iff A contains

no pentagon (0, a, b, c, l}. Let Comp(A) stand for the set of complementary pairs in A.

A C-relation Y: is said to have no compuruble complements iff {a,, b,}, {ai, b,} E g, uo< Q, and b,< b, imply that a,, =a, and bo = b,. The following result is immediate from Theorem 5.

COROLLARY 6. Let Li, i C I , be a family of lattices with no comparable complements and let Y: be a C-relation on the Li, i I , with no comparable complements. Let L be u .!"-reduced free product of the L , i E I . Then

Comp(L) = Y: u U (Conip(Li) I i E I ) ,

and L has no compara,ble complements.

Now we are ready for our first application.

3. Reclucrd Frat: Products 293

THEOREM 7 ( C . C. Chen and G . Griitzer [1969]). Let L be u bounded lattice i n which every element hus ut niost one complement. Then L has u (0, 1)-embedding into u uniquely complemented lattice K (that i s , into a lattice K i n which every elevient has exactly one complement).

PROOF. If L is complemented, then set K = L. Otherwise let L =Lo. We define by induction the lattice L,. If Lnv1 is defined let I n - i be the set of noncomplemented elements of Ln.-l. For i c l n - l let Li=(ui}b. Define the C-relation bnP1 on the fainily {Lnp1} u (Li j i c 1 9 t - l ) by the rule

{u , b}E bn-l iff {u , b} ={i, u i } for some i C l n - l .

Let L, be the b,,-,-reduced free product. Since

L=L,GL1 E L,G. *

and all these containnients are (0, 1)-embeddings, we can form K = U (Li I i E I ) . Now observe that LO is a lattice with no comparable complements. By induction,

if this is known of L,- , , then it is true for L, since bn-i has no comparable coniple- nients and so Corollary 6 applies. Therefore, again by Corollary 6,

Conip(L,) = Conip(L, -1) u

Thus L, is a t most uniquely conipleniented and every element of Ln-l has a complement in L,. It is now obvious that K is uniquely complemented.

Every lattice L can be embedded in a lattice in which every element has a t most one coniplement, namely into L*. Thus as a special case of Theorem 7 we get the celebrated result of R. P. Dilworth [1945]:

COROLLARY 8. Every luttice can be embedded into u uniquely complemented lattice.

For a graph G (or, more precisely, (G; E ) ) let (L, I aEG) be a family of lattices, where L, = { O , , ( I , l,,} is a three-element lattice. Define the C-relation b on (La I a E G )

hy (z, y } E b iff {x, y } E E .

Let L(G) denote the b-reduced free product of (L , I uEG). Some examples are given in Figures 1 - 3.

For a bounded lattice L, let Endo,l(L) denote the monoid (that is, seniigroup with identity) of all (0, 1)-endomorphisms of L. For a graph G, End(G) denotes the monoid of endoniorphistns of G ; a map p: 0 -G is an endomorphism iff (a , b) CE implies that (upl, bp) EE. Observe that G c L(G) and, in fact, G generates L(G) as a (0, 1)-suhlattice (that is, G u (0, l } generates L(G)). Therefore, every pl E End(G) has a t niost one extension p to a (0, 1)-endoniorphism of L(G).

COROLLARY 9. to u { O , l}-endo.Nzorphism of L(G). If pl is onto so i s p.

Every endoniorpliism pl of u graph (;r ILUS exactly one extension p

'LO'

294 VI. Free Product,s

Figure 1

0 0

G

Figure 2

A G

Figure 3

PROOF. This is clear from Definition 2(iii) with A =L(G). For an integer n 2 2, an n-cycle of a graph U is an n-tuple of elements (ao,. . . ,a,, -J

with{ao, at}, . . . , {a,-2, {a,-l, a,}EE.

THEORBM 10. contained in some cycle of odd length. Then End(U) is isomorphic with Endo,l(L(G)).

Let U be a graph satisfying the property that every element of G is

PROOF. the hypotheses of Corollary 6. Therefore,

It is obvious that the lattices and the C-relation used in forming L(G) satisfy

Comp(L(G)) = E u (0, l}.

By our assumption, every element of G is the endpoint of an edge, and G is recognized in L(a) as the set of complemented elements, other than 0 and 1. Since a (0,1}- endomorphism y takes a complementary pair into a complementary pair, we conclude that @ c U u (0, l}. Let gy E (0, l} for g EU, for instance gy = 0. By assumption, there is a cycle of odd length (go,. . . , gzn) with g=go. This means that {go, g,}, . . . , {g2n-1, gPn}, and {gZn, go} are complementary pairs. Thus

!7v =goy -0, - g l y = i , gsp=O, g3y=1, * * * , g 2 n - I ~ = 1 , gZnl/l=O, and gOV=',

a contradiction, This shows that every (0, 1)-endomorphism w is the unique extension of a map q~ of a into itself; this q~ is obviously a, graph endomorphism. Thus the map q~ +gj is the required isomorphism of End((?) with Endsl(L(G)).

As it is shown in the Exercises, every monoid is the endomorphism semigroup of a graph in which every element lies on e cycle of odd length. Thus we conclude e result of G. Griltzer and J. Sichler [1970]:

THEOREM 11. group of a bounded lattice.

Etiery monoid can be represented as the (0, 1)-endomorphism semi-

Exercises

The following exercises should provide the reader with the necessary background in category theory (theory of concrete categories, graph theory) riecessary for Theorem 11 a d for the results of the next section. References will be given in the Further Topics elid References. The present sequence of exercises is based on unpublished lecture notes of J. Sichler.

In what follows, we shall deal with systems (A; R), (A; Ro, . . . , R,-I) , (A; R,, . , , , R,, . . .), where A is a nonempty set and R and the Ri are binary relations. Given two systems of the same type, say (A; R,, . . . , R,, . . .) and (B; So,. . . , S,, . . .) a map p,: A -. B is called a hornomrphiern iff

(a, b) E Ri implies that (ap,, by) ESi, for all i =O, 1, . . . An endornorphiern of (A; R,, . . .) is a homomorphism of (A; Ro, . . .) into itself. The endomorphism semigroup End((A; Ro, . . .)) is defined as before. ( A ; Ro, . . .) is rigdd iff the only endomorphisni is the identity map.

Let (A; 5 ) be a well-ordered set with unit. Let A, consist of all a € A that ere cofinal with w, that is, for which there is a sequence a1 < * * * <a,, < * * such that a =V(ai I i = 1, 2, . . .). For each a € A , fix such a sequence (ai, ay, . . .). Let Ro be the relation <, and let (x, y) E Ri, for e' = 1, 2, . . . , iff y €A: and x =yi (that is, x is the i-th member of the sequence associated with y). Let p, be an endomorphism of (A; R,, R,, . . .). Prove that x (xp, for all x ~ A . Let a < ap, for some a € A . Set a, =ap,, a, =alp,, . . . , and b =V(a, I R =1,2, . . .). Prove that b q =b. Prove that b,p, =b,, for all n = 1, 2, . . . Conclude that (A; R,, Ri, . . .) is rigid. Given t i system (A; Ro, R,, . . .) we construct a new one (B; So, 8,) as follow6: B = A X N ( N = (0, 1, 2, . . .}),

1.

2.

3. 4.

( ( x , ~ ) , (y,m))cSo iff x=y and m=n+l, or

x = y and n=O,m=2;

(+A R), (y, m))ESj iff A =m and (2, y)ERn.

Prove that End((A; RU, R,, . . .)) sEnd((B; So, Sl)) and if A is infinite, then JAI =PI.

M. Free Products 296

6.

6. 7. 8.

9.

10. 11.

12.

13.

Given a system (A; R,, R l ) we construct a new one (B; S) as follows: B =A ~ ( 0 , 1 . 2, 3, 41,

((a,i) ,(b,j))ES iff a = b and j=i+l, or

a = b and i = O , j=4, or

i = O , j = 2 , and (a ,b)ER, , or

i = 2 , j = 4 , and (a,b)ERi.

Prove that End((A; R,, R,)) rEnd((B; 8)) and if A is infinite, then Prove that for each infinite cardinal m there exists a rigid (A; R) with IAl =m. Prove the statement of Exercise 6 for finite cardinals. Let (A; (Ri I i E 1 ) ) be a system without any restriction on IZ(. Let ( I ; R ) be a connected rigid graph. We define a new system (B; R,, R,) as follows: B =A XI,

=IBl.

.

( (a , i) , (b, j)) E R, iff a = b and ( i , j) E R,

((a, i), (b , j))E R , iff i =j and (a, a)€ Ri.

Prove that End((A; (Ri I i E I ) ) ) sEnd((B; R,, Ri)) . Let M be a monoid. For every aEM define R,={ ( z , ax) I z E M } . Show t,hrtt, End((M; (R, I a E M ) ) ) r M . For any monoid M , find a system (A; R ) such that End(@; R ) ) 11cl. For a system (A; R) and B E A we construct B new system (A; R, S.B) as follows: (2, y) € 8 ~ iff x +y and x, y E B or 2, ye B. Show that if (A; R ) is rigid, then the new systems are mutually rigid, that is, if 9 is a one-to-one homomorphism of one system to another, then the two systems are the same and Prove that there are 2m mutually rigid systems of cardinality m, where m is any infinite cardinal. Consider the graph (a; E) of Figure 4. Let (A ; R ) be a system satisfying (a, a) E R for no aEA. We construct a new graph (H; F) by “replacing each (a, b ) c R by a copy of (a; E) with the pair (4, 8) signifying (a, b)”. Formally,

is the identity map.

H = ( R X { l , 2,3, 6 . 6 , 7 , 9, 10})~A,

Figure 4

14.

15.

16.

17.

18.

19.

20.

21.

22.

23. 24.

26.

26.

27.

28.

and for each e =(a, b ) € R , ( e , I), (e , 2), ( e , 3), a, (e , €9, (e, S ) , (e , 7), b, (e, 9), (e, 10) should form a subgraph isomorphic with (G; E). Discuss the connections between End(@; R)) and End(@; F)). A triangle {a, b, c } in a graph ( G ; E) is a so t of thrce elements of G such that {a, b}, {b, c ) , and { c , a) EE. A graph (G; E) is triangle connected iff for any a, b EG, a = b or there is a sequence To, . . . , T,-1 of triangles such that aET,, b € T , - l , and Tin Ti+i + 0 , for i =0, . . . , n -2 . Prove that for each infinite cardinal m there are 2m mutually rigid triangle connected graphs. Prove that for each infinite cardinal m, there are 2m mutually rigid connected graphs in which every element lies on a cycle of length 7. Let M be a monoid and let m be an infinite cardinal satisfying [MI 5 m. Prove that there are 2m pairwise nonisomorphic graphs (G; E) of cardinality m satisfying End((G; E)) aM. Prove that the L of Theorem 4 is a g-reduced free product of the L;, i €1, by verifying directly 2(iii). (Hint: pattern the proof after the proof of Theorem 1.8.) A bi-uniquely complemented lattice L is a bounded lattice in which every element z+O, 1 has exactly two complements. Define the concept of a free bi-uniquely complemented lattice and prove the existence and uniqueness (up to isomorphism) of a free bi-uniquely complemented lattice on m generators.1 Let F , and F , be free bi-uniquely complemented lattices and let Xo and X i be the free generating sets of Po and F , , respectively. Let u be a one-to-one map of X , onto X i . Let IX,I = (Xi[ = N,,. Show that there are 2’0 isomorphisms of Fo and Fi extending u. I n a bounded lattice L the complementation i s transitive iff whenever x, y and y, z are complementary pairs, then either z = z or 2, z is also a complementary pair. Prove that every lattice can be embedded into a bi-uniquely complemented lattice with transitive complementation. Let L be a bounded lattice. Under what condition is there a (0, 1)-embedding of L into a bi-uniquely complemented lattice with transitive complementation? Show that a relatively complemented uniquely complemented lattice is Boolean. (Hint: use Exercist? 1.6.5.) Prove that, a modular uniquely complemented lattice is Boolean. Show that an atomic uniquely complemented lattice is Boolean (T. Ogasawara and U. Sasaki [1949]). (Hint: For an element a, let R(u) denote the set of atoms - < a ; then a -R(a) is a set representation.) Let L be a uniquely complemented lattice and let a>- b in L. Prove that ahb’ is an atom, where b’ is the complement of b (R. Padmanabhan). (Hint: if aAb’ =0, then b‘ has two complements: if 0 < 1: < aAb‘, then bva’ has two complements.) A lattice L is called relatively dtomic iff, for all a, b EL, if a < b, then there exist x and y satisfying a 2 z< y 5 b. Show that a relatively atomic uniquely complemented lattice is Boolean. (This includes t.he result of V. N. SaliI [1972] that every uniquely complemented algebraic lattice is Boolean.) A lattice with complementation ( L ; A , v, ‘, 0, 1) is a bounded lattice in which for each a EL, a’ is a complement of L. An endomorphism Q is a lattice endomorphism satisfying (ap)’=a’p for all aEL. Prove that for any monoid M there is lattice with complementation whose endomorphisms semigroup is isomorphic to M (G. Griitzer and J. Sichler [1970]). In Theorem 11 and in Exercise 27, how many pairwise nonisomorphio lattices of a given cardinality can be constructed satisfying the requirements?

1 Exercises 18-21 are based on C. C. Chen and G. Griitzer [1969].

VI. Free Products

4. Hopfian Lattices

A lattice L is hopfian iff every onto endomorphism is an automorphism. Equi- valently, a lattice L is hopfian iff L s L/O implies that 0 =IN, that is, L is not isomorphic to a proper quotient of itself. The hopfian property is similarly defined for groups, rings, and algebras, in general. The first definition of the hopfian property also applies for graphs; one should note however, that for a graph a one-to-one and onto endomorphism need not be an automorphism. For graphs, an automorphism has to be defined as an invertible endomorphism.

Obviously, every finite lattice (algebra, graph) is hopfian. I n fact, the hopfian property is a generalization of one of the most important characteristics of finiteness.

Closest to finite lattices are finitely presented lattices, that is, lattices of the form F(Q) (see Section 1.5) where Q is a finite partial lattice. (It is easily seen that these are the lattices that can be described by finitely many generators and finitely many relations, so this concept agrees with “finitely presented” as it is used for groups and semigroups. )

THEOREM 1 (T. Evans [1969]). Every finitely presented lattice is hopfian.

The proof of Theorem 1 is contained in the following three lemmas.

LEMMA 2. finite ldtices.

A finitely presented lattice can be represented as LC subdirect product of

PROOF. It is sufficient to prove that if Q is a finite partial lattice, a, b C a(&), and a i b, then there exists a finite lattice K and a homomorphism q : F(Q) + K such that aq =i= by. Just as in Section 2, we can define the component subsets of a and b and set

Q’ =Q u Komp(a) u Komp(b).

Regarding &’ as a partial lattiae, obviously P(&) =P(&’). By (the proof of) Theo- rem 1.5.20, Q’ can be embedded into some finite lattice K. This embedding extends to a homomorphism q of F(Q’) = P(Q) into K. Finally, a q +bq~, since q is one-to-one onQ’.

LEMMA 3. Let L be a finitely generated lattice and let 0 be a congruence relation of L. If LIO is finite, then there ex&% a fully invariant congruence relation @ of L such that Q, 0 and L/@ .iS finite.

4. Hopfiaii Lattices 299

It is easily seen that @ is fully invariant. Since the identities used to construct @ all hold in LJ0; we conclude that CP < 0. Finally, L/@ is finitely generated and LICP E K which is a locally finite class, hence L/@ is finite.

A subdirect representation of a lattice L is associated with a family (Oi I i E I ) of congruence relations of L such that A (Oi I i c I) =o. If all Oi, i c I, are fully invariant, we call this subdirect representation fully invariant.

LEMMA 4. product of hopfian lattices. Then L is hopfian.

Let the lattice L have a representation as a fully invariant subdirect

PROOF. Let (Oi I i E I ) be a family of fully invariant congruence relations of L satisfying A (ai I i E I ) =w. Let y be an onto endomorphism of L. Since Oi is fully invariant for i E I, we can define

yi: [a]Oi-.[ay]Oi,

which is an endomorphism of LIO,. Obviously, all yi are onto, hence they all are automorphisms. Now if a, b E L and aq, =by , then, for all i C I, ([a]Oi)rpi = ([b]Oi)yi , hence [ a l e i = [b]O,. Since A (Oi 1 i E I ) =w, we conclude that y is one-to-one.

R. Wille [1975] exhibits a finitely generated lattice that is not hopfian. An earlier example of a finitely generated lattice that is not finitely presented (or presentable, to be niore precise) is FM(4) (see T. Evans and D. X. Hong [1972]); of course, FM(4) is hopfian.

The free product of two finitely presented lattices is again finitely presented. What about hopfian lattices?

THEOREM 5 (G. Griit,zer and J. Sichler [1974]). whose free product is not hopjian.

There exist hopjian lattiees Lo and Li

The proof of Theorem 5 is based on the construction of the lattice L(G) from the graph G introduced in Section 3 and a lattice theoretic result (Theorem 6) which is due to J. Sichler [1972] and H. Lakser [1972]. I n fact, the result we shall prove will be stronger than Theorem 5 and it will be based on J. Sichler [1975].

For graphs Gi( =(a,; Ei)), i € I , we form the lattices L(G,) (with bounds Oi, l i ) and the free product L of the L(Gi), i c I.

A triangle of Gi is a three-element set {a , b, c } such that {a , b}, {a , c } , {b, c} c Ei. If { [ I , b, c } is a triangle of Gi, then {O i , a, b , c , li} is a diamond. We call this sublattice the diamond associated with a triangle.

THEOREM 6. Any diamond in L i s associated with a triangle of some G , i c I .

PROOF. for some j c I .

Let M = {o, a, b, c, i } be a diamond in L. Let us first assume that M E L(Gi),

300 VI. Free Product,s

C~ASII 1 . o =Oi and i = li. Then {a, b}, { a , c}, {b, c} E Comp(L(Ci)), hence by Corol- lary 3.6, {u, h} , { a , c } , {b , c } EE, , and indeetl, M is associated with the triangle ( ( I , b, c} of Gi.

CASE 2 . o =Oiand i < li. Recall that L(Gi) -{Oil li} E F(Gi) (the free latticegenerated by the set Gi), hence a, b, c , iEF(Gi) , and there is a congruence relation @ on %(Gi) surh that F(Gi)/@ Y L(Gi) and [ X I @ +x under this isomorphism for all x EL(G,) - (Oj, lj}. Since i = a v b = u v c in F(Ci), by (SD,), i = a v ( b m ) . But br\c=Oj, that is, [bAC]@ is the zero of S(G,)/@. Thus

[i]@ = [a]@v[b/\c]@ =[a]@,

and we conclude a = i, a contradiction. Hence this case is not, possible.

CASE 3. o > Oi and i = li. This is impossible just as Case 2.

CASE 4. diamond in a free lettice.

each j c J,

o>Oi and i < lj. Then M F(Gi) which is impossible since there is no

L and consider, for Thus Theorem 6 is proved for sublattices of L(Ci). Now let M

M(j) =@(j) I x E MI.

Since x --c xu, is a homomorphisni, M,i, M or M , j , is a singleton. If, for all j E J. M(i ) is a singleton, then Theorem 1.17 yields that (SD,) holds in M , a contradiction. Hence there exists a j E Z such that M ( j ) is a diamond and it is therefore associated with a triangle of Gj, in particular,

o . =Oj and i . = li. (7) ( 3 )

By duality, there exists a Is E Z such that M,,) is a diamond; in particular,

o(,) = 0, and = 1,.

This yields immediately that j = k and o =Oi , i = li. Since a,i,vb(i, = li, a(j)Ab(i) = O . qj,< dj), and h(i)< b(i), we conclude that qi, and di) are complements of bti, in L(Gi). But L(Gi) is a lattice with no comparable complements, hence aci, ,a(i)=a cL(G,). Similarly, b, c E L(Gi). Hence M E L(Ci) and M is associated with a triangle of Gi.

Let Z be a set of a t least two elements. For each i c Z , a lattice Li will be constructed such that the free product of all Li, i E Z , is not hopfian but the free product of L,, i E Z’, is hopfian for all 0 + Z ’ c Z . The special case IZI = 2 is Theorem 5 .

Let N ={l, 2, 3, . . .} and we consider maps pl: N -Z . The map pl is eventually constunt iff there exists an integer n such that np, = (n + 1)pl = a - - . Let M ( Z ) denote the set of all eventually constant maps of N into Z . By &!,,(I) we denote the set of all maps pl: ( 1 , . . . , n} -I and M,(Z) = U (M,(Z) I n = 1,2, . . .). For pl E &!(I), n E N , and for q c Mm(Z), 1 2 1 1 7 2 , let pl, be the restriction of pl to (1 , . . . , n}. Finally, for pl E M , ( I ) we write n =U(pl) and FV(pl) =npl ( D for domain and FV for final value).

1’

For a graph G (=(C; E ) ) , u , b c G , we say that a and b are triungle connected iff ( I = b or there is a sequence T,, . . . , T,% of triangles of G such that u c To, b E T,, and T , n T , =t= 0, for i = 0 , . . . , n - 1. Any graph can he decomposed into a disjoint union of triangle connected components. G is triungle connected iff it has a single coinponent .

Now choose a cardinal m > \M(Z) l , a triangle connected graph Go, and for each a E M ( Z ) choose a triangle connected graph G, such that IG,I = IG,I =m and the graphs Go and G,, a E AZ(Z), are pairwise disjoint and mutually rigid (Exercise 3.14). We fix the elements u,,EGo and tr,EG, for all a € M ( Z ) .

For every n c N and y E M,(Z) we define a graph Gv :

GlY = U (G, I a E M ( I ) and u, = y ) ,

EY = U ( E , I a E M ( I ) and a, = y ) u ( ( ~ 0 , a,} I a E M(Z)} .

I n words, Gq is a disjoint union of all G, such that a, = y and we add the edges connecting the distinguished element of GI, with the distinguished element of the G,. Observe that Gv is connected and that GI, and the C, are the triangle connected components of G9.

Let x be a homoniorphisni of Gq into G l W (yCM,(Z) and y E M , ( Z ) ) . Then the triangle connected components of G9 have to be mapped into the triangle connected coniponents of Gv'. Thus if x c M ( Z ) with a, = y or a = O then G J Z G,, where p E M ( Z ) and l , = y or p = O . In view of the mutual rigidity of these graphs we must have a = p and x is the identity on G,. So we conclude, that if u E M ( I ) and a, = y , then a,,, = y ; this is possible iff n a g ? , and am=/?. Thus there is a homomorphism of G* into Gv' iff D ( v ) 2 U ( y ) and yucW, = y . Moreover, in this case there is exactly one honioiiiorphisni which is the identity map on G,, and on all G, (a, =pl).

Now we are ready to define L , , for i E Z : let L, be a free product of all L(GV) satisfying F V ( y ) = i .

Let L be a free product of all L, , i c Z ; in other words, L is a free product of all L(Gq), q~ c M w ( I ) . We verify that L is not hopfian. We define a map p :

/!I restricted to G/ is the homomorphism of G9 into G V n - l , where n = D ( y ) > 1 ; p on Gv is the identity if D(p) = 1 . Thus p extends to a homomorphism y of L(GT) into L(GVn-') if n = D ( y ) > 1 and to the identity niap y on L(GV) if D ( y ) = 1 . Since L is a free product, y extends to an endoinorphisni 6 of L.

Observe that the iniage of /!I covers all W. Indeed, if uEG', then uEG, for soiiie a E M ( Z ) with a,, =cp, where 11 = D(~,I) . Then p maps u EG, G Gun+' onto CI CG, G Gv.

Thus 6 is an onto endoniorphisni of L. j3 is not one-to-one since every elenlent of any Gq' with D ( y ) = 1 is the image of two elements: of itself and of a suitable GW with U ( y ) = 2. Thus 6 is not one-to-one. We have proved that L is not hopfian.

Now let 0 + I ' c Z and let L' be the free product of the L,, i c Z ' . We have to show that L' is hopfian.

Ohserve that I;' is the free product of all L ( W ) satisfying F V ( y ) E 1'. T,et 6 he an onto endoniorphisni of L'. We shall verify that 6 is the identity niap,

i iiiplying that L' is hopfian. Assume to the contrary that 6 is not the identity map. Then there is an uCG9

( F V ( y ) EI' , D ( y ) = n ) such that (I +ad. Since N is an elenient of a triangle, i t is an

302 M. Free Products

atom of the associated diamond M . If 1 Mp,l = 1, then the bounds of L(GV) are collapsed by 6, hence all of L(GV) is mapped by d onto a single element. Otherwise, Mp, r M , hence, by Theorem 6 , ad E GI for some y M,,,(Z). Since GV is connected, all of Cp is mapped by 6 into GV, Hence D(q) 2 D(y) and y = r i ~ ~ where m = D ( y ) < D(p,), since D ( y ) = D ( y ) implies that p, = y , contradicting the rigidity of GV.

In either case we see that GP n GP6 has a t most one element. Let i c I -I‘ and consider the map 8: N -I defined by b, =p, and k/ l= i for k > n. Then E M ( I ) and G, c GV. Let GF =G, - (G. n GV6). Since 6 is ont,o, every a EGF must be in the image of some Gv; but i t cannot come from the homomorphism of GV into GP since it would iinply D ( y ) = m > n = D ( y ) and p, =ty, contradicting the definition of L’. (GV is not in L’ since F V ( y ) = igZ’ . ) Thus a c G j must come froin a GV collapsed by 6 onto a. However, 1G;l =m and IM(Z)l <m, a contradiction. Thus 6 must be the identity map.

Exercises

1.

2.

3. 4. 6.

6.

I. 8.

9.

10.

11. 12. 13.

14.

15.

Prove that for any nontrivial equational class K and infinite cardinal m, P K ( m ) is not hopfian. Prove that for any equational class K of lattices and natural number n, P K ( ~ ) is hopfian. Find a one-to-one and onto endomorphism of a graph that is not an automorphism. Can the graph in Exercise 3 be chosen to be finite? Define a finitely presented lattice as the “most free” lattice generated by zo, . . . , zn-l satisfying the “relations” . . . , x,,-i) =qi(z0, . . . , ~ ~ - 1 ) . i = 1, . . . , m, where pi and p i are n-ery polynomials. Show that this is equivalent to forming an P(&) with a suitable partial lattice &. Let P(&) 2 &’a& where & and &’ are partial lattices. Show that F(Q) PP(&’) provided that &’ is generated by &. Formulate and prove the converse of Exercise 6. For a lattice L and equational class K of lattices, we constructed in the proof of Lemma 3 a congruence relation CD =CD(K). Prove that @(K) is fully invariant. Let @ be a fully invariant congruence relation of a lattice L. Prove that CD =CD(K) for some equational class K. Let L be a fully invariant subdirect product of the Li, i €1. For an onto endomorphism p, write down a formula for Kerp, from which Lemma 4 can be derived. What is the analogue of Lemma 4 for graphs? Show that the free product of two finitely presented lattices is finitely presented. Prove that L(GouBi) is a free (0, 1)-product of L(Qo) and L(Ui) , where Uouff, is the disjoint union of the graphs Go and Q,. Find two bounded hopfian lattices whose free {0, 1)-product is not hopfian (G. Griit- zer and J. Sichler [1974]). Let Go and Bi be hopfian graphs. Prove that a free product of L(G0) and L(Gi) is hopfian.

Further Topics and Iieferences 303


The idea of the covers of elements in a free product goes back to R. P. Dilworth [1945]. It becomes more explicit in C. C. Chen and G. Gratzer [1969] in which L*F(H,) is considered. The four W conditions, are, of course, from Ph. M. Whitman [1941]; this should be apparent from Theorem 2.2.

The Structure Theorem has its limitations. Despite many attempts, it could not be used to verify the Common Refinement Property for free products. .

The Splitting Theorem is first formulated for F ( n ) in Ph. M. Whitman [1941]: for a E F ( n ) , if (L 2 x1 does not hold, then a xqv* - 'vx,; proof by induction on r(a) using Theorem 2.2. Theorem 1.11 is much inore general; still, i t is a very special case of a complete triviality :

The splitting Theorem. Let L be a lattice and L =[XI. Then the prime ideals of L are in one-to-one correspondence with subsets Y of X satisfying Y * 0 , X - Y + 0 , and

VY12 AX,, for any finite nonvoid Yi c Y and Xl X - Y. If X is finite, this condition reduces

A very special case of the Coninion Refinement Property for free products was proved in A. Kostinsky [1971]. The Coninion Refinement Property holds for free K-products for any equational class K of lattices by G. Gratzer and J. Sichler [1975]. B. Jbnsson and E. Nelson [1974] verify the same result for regular equational classes of algebras, that is, equational classes defined by identities in which the same variables occur on both sides. In G. Gratzer and J. Sichler [1975] examples are exhibited of lattices that cannot be represented as free products of freely indecomposable lattices ; the same result holds also relative to any equational class K of lattices K + T.

t o V Y Z A ( X - Y ) .

Let g(L) = min { 1x1 I X generates L} ,

by G. Gratzer and J. Sichler [1974a], the formula

9(A * B ) = d A ) + g ( W

holds; the same formula holds in any equational class K * T of lattices. Many of the observations made about F ( 3 ) hold for all F,(3), where K +T is any

equational class of lattices. An interesting example is xOva > X, >sod. This even holds for K =D, where (I = b . Many intervals of P ( 3 ) contain isomorphic copies of F ( 3 ) . The same problem for YK(n) is considered in J . Bernlan [1972a]; this probleni is closely connected with the Aiiialgamation Property.

Computing F,(n) iiiay be very difficult even when i t is finite. For instance, FN5(3) has 99 elements and a rather complicated structure, see A. G. Waternian [1967]. A rather effective niethod is worked out in R. Wille [a] which lends itself well to computer prograiii tiling.

A curious consequence of Theorem 2.7 is that if K is an equational class of lattices, K +T, and FK(Ko) satisfies (W), then K =L.

Corollary 2.12 suggests that there are very many three-generated lattices. Indeed, i n 2'. Crawley and K. A. Dean [1959], itl is proved that there are 2'0 three-generated

30.1 VI. Free Products

lattices. Call a lattice L m-iiniwersal iff ILI =m and every lattice of cardinality a t lllost m is isomorphic to a sublattice of L. Thus there are no H,-universal lattices. By B. J6nsson [1956], X,-universal lattices exist for all CI > 0.

Corollary 2.12 has been generalized to WA lattices in E. Fried and G. Gratzer [a]; in this case “two-generated” suffices.

For an empty C-relation a 8-reduced free product becomes a free (0, 1}- product; the structure theorem for this special case is quite easy. This suggests an alternative development for this chapter. Start with establishing the structure theorem for free (0, 1)-products. Derive from this the development of Section 1 by observing that the free product of the L,, i c I , is the sublattice [U(Lj I i c I ) ] of the free (0, 1)-product of (L$, i c I. Still one more approach to Section 1 is the one in B. J6nsson [1971]: Once we know the Structure Theorem, it is possible to give a direct proof without the use of polynomials, equivalence classes, and so on. This approach is the most econoniical but not necessarily the most illuminating.

The basic idea of Section 3 is in Definition 3. The credit for this should go to It. P. Dilworth [1945] which contains a significant special case of this construction. This was generalized to L * F(X,) in C. C. Chen and G. Gratzer [1969] under the constraints of Corollary 3.6.

An alternative proof of Corollary 3.8 is given in P. Crawley and I<. P. Dilworth [ 19731.

The background of Corollary 3.8 is quite interesting. Results, such as the one in E. V. Huntington [1904], suggested that Boolean lattices can be characterized as uniquely complemented lattices. There was a widely accepted conjecture in the thirties supported by a growing body of results of the type, that if a lattice L is uniquely complemented and has property X, then L is Boolean. For X =modular, and X = relatively complemented, this appears to have been known already in the thirties to G. Birkhoff and J. von Neumann; for X =finite dimensional, see R. P. Dilworth [1940a]; for X =complete, atomic, and dually atomic, see G. Birkhoff and M. Ward [ 19391. Then came R. P. Dilworth’s result showing that without some additional property the conjecture is false. Thus the search for weaker and weaker properties X continued. For X =atomic, see T. Ogasawara and U. Sasaki [1949], see also ,J. E. McLaughlin [1956]; for X =the niap u +a’ is order-inverting, see G. Birkhoff [1948]; for X =algebraic, see V. N. Salii [1972]; see also Exercises 3.25 and 3.26. It. Pad- nianabhan has some related unpublished results: for X we can take L € MvN,; for X we can also take L (Equ(K), where K is a finite lattice satisfying (SD,) or an iniplira- tion which is due to E. Fried and G. Gratzer:

X A Y = X A Z implies that XV(YAZ) = ( X V ~ ) A ( Z V Z ) .

Many of the results of this chapter can be extended to m-complete lattices (lattices in which A X and V X exist for all nonvoid subsets of X with 1x1 <m). Some of these results can be found in P. Crawley and It. A. Dean [ 19591 and B. J6nsson [ 19621. The Structure Theorem of Free Products and its applications have not been published for the in-coniplete case.[

1 Now, however, see the abstracts by G. Griiteer and D. Kelly, Prvc rn-products of lattices. Notices Amcr. Soc. January, February, April 1977.

Some covering relations in F ( 3 ) are shown in Figure 1 ; sonie more can be found in Ph. M. Whitnian [1941] and [1942] and in It. A. Dean [1961a]. In H. N. McKenzie [1972] an algorithm was found deciding whether p > q holds in F ( 3 ) . Sonie of McKenzie’s results combined with a new theorem of A. Day [b] yield the astonishing fact that every proper interval of F ( 3 ) contains a prime interval!

The solution to the word probleni for C F ( P ) , the conipletely free lattice generated by a poset P , is very siniilar to Theorem 2.2: except, of course, that the relations we have to start with are ( I I b where ( I , b E P and (I 5 b in P. M. E. Adanis and D. Kelly [a] prove that the free product of L,, i E Z , can be embedded into CF(U (Li I i E I ) ) . This is the crucial step in proving that free products preserve chain conditions.

For various generalizations of the free product construction see K. Babes and A. Horn [1967], H. A. Dean [1964], %. TAadzianska [1974], and H. Lakser [1968].

Condition (W) is implicit in Ph. M. M’hitnian [1941]; it was first explicitly stated in B. J6nsson [1961]. A remarkable result of It. Freese [1975] states that a lattice L with no infinite chains satisfies (W) iff L is a retract of some B ( l ( F ( n ) ) ) .

Condition (W) plays a role in the characterization of finite (R. N. McKenzie [1972]) and finitely generated (A. Kostinsky [1972]) projective lattices.’ In B. A. Davey and B. Sands [a] it is proved that for a lattice L with no infinite chains (W) is equivalent to the projectivity of L in the class of all lattices with no infinite chains.

B. Sands has examples showing that without any chain condition this result does not hold.

T. Kucera and B. Sands have recently considered finite lattices L such that for all finite lattices K , the set of all honioniorphisnis of L into K forrn a lattice under pointwise ordering. Again (W) plays a role in this investigation.

On the connections of (W) and transferability, see the Concluding Remarks. An easy, but illuminating property of (W) is pointed out in K. A. Baker and A. W. Hales [1974]: (W) holds for a lattice L iff it holds for I ( L ) .

The characterization problem of finite sublattices of a free lattice goes back, a t least, t o Ph. M. Whitnian (19411. In B. J6nsson [196l] and B. J6nsson and J . E. Kiefer [1962] it became appartnt that anything that can be proved for finite sublattices of a free lattice follows already from (W), (SD,), and (SD,). The obvious converse, conjectured by B. Jbnsson, is still open. For an early review article on this field, see R. A. Dean [1961]. The distributive case is settled in F. Galvin and B. J6nsson [1961]. For some additional coniriients on recent results see the Concluding Remarks and H. S. Gaskill [a].

F ( 3 ) has many rionfinite sublattices. It was observed in P. Crawley and R. A. Dean [1959] that for every countable poset P , C F ( P ) is a sublattice of F ( 3 ) . It. N. McKenzie [1973] discnsses some open problems related to free lattices.

Elementary equivalence and free products are considered in B. J6nsson and l’h. Olin [a]. They prove that if K +T is any equational class of lattices, then free K-products do not preserve elementary equivalence. This contrasts with Boolean algebras, s w I’h. Olin [a].

Autoniorphisni groups of lattices are, up to isoniorphisni, arbitrary groups. This

1 Pro,jt.ctivib l t i t , t i m s X art’ characterized in It. E’rrese and J. B. Nation, Projective lattices. Abstrwt. Noticvs Amw. Math Soc. 43 (1976): A-478.


special case of Theorem 3.11 goes back to G. Birkhoff [1946]; in fact, the same result is proved there for distributive lattices. Moreover, for finite groups the lattice constructed is finite. Small lattices with given automorphism group are considered in R. Frucht [1948] and [1950]. R. N. McKenzie and J. Sichler have some related results for lattices of finite length. Two sample results: every group is the autonior- phism group of a lattice of finite length; for every lattice L, there exists a lattice K , such that End(L) rEnd,,,(K) and if L is finite or finite length, then so is K , where End(L) is the endomorphism semigroup of L. See also J. Sichler [1972].

Theorem 3.11 is one of a large body of results representing monoids as endomorphism monoids of various types of algebras. All these results are based on P. Vopgnka, A. Pultr, and Z. Hedrlfn [1965] proving the existence of rigid relations and on Z. Hedrlfn and A. Pultr [1964] proving the representation for graphs. See also Z. Hedrlin and A. Pultr [1966] for the case of algebras with two unary operations; Z. Hedrlin and J. Lambek [1969] for the case of semigroups and for an alternate proof of the existence of rigid graphs. The result on triangle connected graphs is a special case of a result of P. Hell [1972].

Theorem 4.6 combines a result of J. Sichler [1972] with a very special case of a result of H. Lakser [1972]. The result of J. Sichler states that the diamonds in an L((3) are all associated with the triangles of G. H. Lakser investigates the simple sublattices of a free product L of lattices Li, i c I ; in particular, he proves that if S is a simple lattice, and 8 cannot be embedded in any L , i c I , then S cannot be embedded in L. In addition, he obtains that L contains a diamond iff some Li contains a diamond and L contains a diamond which is not in one of the Li, ic I , iff some Li contains YJ13 x& as a sublattice.

Problems

VI. 1. VI. 2.

VI. 3.

VI. 4.

VI. 5.

VI. 6.

Let X be a finite smallest generating set of A * B. Is it true that X G A u B? 1

Does some forin of the Common Refinement Property hold for free (0, 1}- products? Investigate this problem also in an arbitrary equational class of lattices. For which equational classes of lattices does Sorkin’s Theorem (Exercise 1.13) hold? Find a condition implying the Common Refinement Property for free K-products for an equational class K of algebras including the case of equational classes of lattices and of regular equational classes. Which properties of homomorphisms Bi +B, are preserved by the induced homomorphism A * Bi + A * B2? (See B. J6nsson and Ph. Olin [a].) For lattices A and B, does A * A Y B * B imply that A Y B? More generally does A, * * * * An% Bl * - * * * B,, A, .%A,, and Bi r- * *Y B, imply that A i z B B , ?

1 No, it is not true. See M. E. Adams, Generators of free products of lat,tices. Algebra Universalis 7 (1977) : 409-410.

Problems 307

VI. 7.

VI. 8.

VI. 9. VI.10.

VI.11.

VI.12.

VI.13.

VI.14.

VI.15.

VI.16.

VI.17. VI.18.

VI.19.

VI.20. VI.21.

VI.22.

Does A, * - - - * A , * C 1 B1 * * * * B, * C , A , 2- - - s A,, and Bi 2. - Y B, imply that A , * C Y B, * C ? Investigate Problems VI.6 and VI.7 for free (0, 1)-products and free K- products. Give a structure theorem of free M3-products. Same problem for N5. Let K be an equational class of lattices and let us assume that there exists an algorithm deciding identities in K. Is there an algorithm (relative to A and B ) deciding the structure of the free H-product of A and B? Let K be an equational class of lattices for which the word problem for PK(HO) is solved. When can we conclude that Theorem 1.14 holds for K? Is there any equational class K 2D of modular lattices for which the conclusion of Theorem 1.14 holds? For which equational classes K of lattices is it true that free K-products preserve the m-chain condition ? (m is an infinite uncountable regular cardinal; the ni-chaiiL condition means that all chains are of cardinality <m.) Investigate equational classes K of lattices in which the following holds: let Li c K, i c I , be a disjoint family of lattices and let L be a free K-product of L, , i c Z ; then L has an order embedding (that is, one-to-one and isotone map) into FK( u (Li I i c Z)"), where u (Li I i c I ) is regarded as a poset. For which equational classes K is S(PK(Ho)) closed under the formation of free K-pi oducts ? Give necessary and sufficient conditions for an interval [ p , q ] of F(3) to contain a copy (infinitely many copies) of P(3) . Does P ( 3 ) have intervals of length n, for n =3,4 , . . .? Is a finite lattice isomorphic to a sublattice of F ( 3 ) iff it satisfies (W), (SD,), and (SO,)? For p , q F(&), is it decidable whether there is an endomorphism g, of F(Ho) satisfying pg, =q? For which equational class K of lattices does FK(Ho) satisfy (SD,)? Does P,(4) have a proper sublattice isomorphic to F,(4)? I n general, if K is an equational class of lattices, n < Xo, and PK(n) is infinite, does FK(n) then have a proper sublattice isomorphic with FK(n)? Find a common generalization of the McKenzie-Koatinsky Theorem and the Davey-Sands Theorem.

VI.23 (FC 1). Characterize the endomorphism semigroups of lattices. VI.24.

VI.25.

VI.26.

Describe the endomorphism semigroups and (0, 1)-endomorphism semigroups of complemented lattices. Characterize the endomorphism semigroups and the (0, 1)-endomorphism semigroups of complete lattices. What about the semigroups of complete endomorphisms ? Characterize the {0,1}- endomorphism semigroups of finite lattices, of lattices of finite length, and of lattices without infinite chains. (Consider also the categorical versions of problems VI.23-VI.26.) 1

1 Every finite monoid is the (0, 1)-endomorphism semigroup of a finite lattice. For this result and others concerning the second and third part of this problem, see M. E. Adams end J. Sichler, Bounded endomorphisms of lattices of finite height. Manuscript. 21 OrStzer

VI.27. Let ill be a nionoid, let m be an infinite cardinal, and let [MI <m<2IMI. How many pairwise nonisomorphic bounded lattices exist whose (0, 1}- endomorphism semigroup is isomorphic to M ? (If m < [MI , the answer is 2m; see G. Gratzer and J. Sichler [1970].)

Give a concrete example of a nondistributive uniquely complemented lattice.

Does there exist a complete nondistributive uniquely complemented lattice ? Does every uniquely complemented nondistributive lattice contain F(3) as a sublattice? (An affirmative answer would also settle the next problem.) Does there exist a uniquely complemented nondistributive lattice satisfying a nontrivial lattice identity? For any infinite cardinal m, prove the existence of m-complete uniquely complemented nondistributive 1attices.i Can they be made join- or meet- continuous, or both?

VI.33. Investigate finitely presented lattices. (See G. Griitzer and H. Lnkser [1975].)

VI.34. Is it true that the cardinality of a basis of a finitely presented lattice is uniquely determined?

VI.35 (FC 12). VI.36. Let L be an infinite lattice. Show that with finitely many exceptions if L

is finitely presented, then it contains F(3) as 8 sublattice. VI.37. Describe sublattices of finitely presented lattices. VI.38. Describe all finite partial lattices '$I for which F(%) is finite. VI.39. For an equational class K of lattices, define finitely K-presented. Investigate

Problems VI.33-VI.38 for finitely K-presented lattices. VI.40. Let K be an equational class of lattices and let n <o. Show that if K + L

and PK(n) is infinite, then FK(n) is not finitely presented. (For K = M and n =4 this has already been verified; see T. Evans and D. X. Hong [1972].)

VI.41. Under what conditions does Theorem 4.1 hold for finitely K-presented lattices?

VI.42. Describe hopfian distributive lattices. VI.43. Improve Theorem 4.6 by requiring that one or both of Lo and L, be finite,

finitely presented, finitely generated, or countable. (It follows from Theo- rem 4.1 that both of LO and Li cannot be finite or finitely presented.) Prove that the free product of two bounded hopfian lattices is hopfian again. (If this is true, then in Problem VI.43, both Lo and L, cannot be finitely generated.) In which equational classes H does Theorem 4.5 hold (Lo, Ll EH and we form the free K-product).z

VI.28 (FC 26).

VI.29 (FC 26).

VI.30.

VI.31.

VI.32.

I s the automorphism group of a finitely presented lattice finite?

VI.44.

VI.45.

1 This much has been proved by G. Gratzer and D. Kelly, Free m-products of lattices. Abstract. Notices Amer. Math. SOC. January, February, April 1977.

2 Theorem 4.5 has been proven to hold in all nontrivial equational classes, see M. E. Atlams and J. Sichler, V-free produots of hopfian lattices. Math. Z. 151 (1976): 259-262.

V1.46.

VI.47.

Under what operators is the class of hopfian lattices closed? For instance, is the direct product of hopfian lattices hopfian? For what equational classes K , do there exist two cohopfian lattices in K whose direct product is not cohopfian? (A lattice L is cohopfian iff L is not isoiiiorphic to any of its proper sublattices.) Investigate first-order properties @ which are preserved when passing from a lattice to its ideal lattice, or when passing from the ideal lattice to the lattire, or bot,h ways. (See Concluding Remarks.) Let L be a lattice satisfying (SD,), (SD,), and (W). Prove that if ILI > 1, then L contains a prime ideal. Find an equational class of lattices K =kL such that a finite lattice is projective in K iff it is embeddable in FK(X,,).

VI.48.

V1.49.

VI.50.

[Sote added in January 1978. During the last ten iiionths solutions were found for a niiiiibcr of problenis proposed in this book.

Recer,t results of C. R. Ylstt, €3. Sands, arxl the author show that, for finite lattices, transierahle = sharply transferable, solving the first half of 1.26. This also yields nn aifirniative solution to 1.23. A recent paper of A. Day contains the proof deiiinnded by 1.25. All equational classes have the property described in 1.40, as proved by M. Adairis and J. Sichler. W. A. Lanipe solved 11.20 in the negative. A negative solution to IV.26 was given by V. Polin. R. Freeae and B. J6nsson proved that the arguesian identity is equivalent to niodularity for congruence lattices of an equational class; IV.28 follows from this. IV.35 was disproved by E. Rodney Can- iield.]

21.

CONCLUDING REMARKS

What is lattice theory? Where to draw the line between lattice theory proper and its allied fields? Does a theorem characterizing a lattice associated with some object outside of lattice theory belong to lattice theory?

These questions are rather academic but they require some response and, in fact. decisions in a book of this kind.

The first question is really answered by this book just like the question, “What iR an animal?” is best answered by a tour of a zoo.

The second question is more difficult. For instance, the theory of Boolean algebras develops by an interaction of lattice theory, topology, axiomatic set theory, and logic. A complete bibliography of lattice theory would be rather difficult to compile on account of jurisdictional disputes.

The third question, however, is a simple one for me: a theorem characterizing the subgroup lattices of abelian groups belongs to the theory of abelian groups and not to lattice theory.

A contentious case in point is the representation theory of algebraic lattices (complete lattices) as subalgebra lattices and congruence lattices of universal algebras (infinitary universal algebras). I n my opinion, these results belong to univereal algebra and not to lattice theory. However, in view of the close affinity between the workers in lattice theory and in universal algebra, I digress to outline this field.

join-seniilattice with 0. For a c C , introduce on C the binary operation Let L be an algebraic lattice. Then by Theorem 11.3.13, L I ( C ) , where C is

f&, Y) =aA(xvY)-

Then the subalgebras of ( C ; {fa I a c C } ) are exactly the ideals of C, hence the subalgebra lattice is isomorphic to L. Thus we obtain that every algebraic lattice is isomorphic to the subalgebra lattice of some algebra. The converse is even easier. Thus we have obtained a result of G . Birkhoff and 0. Frink [1948]. In the same paper they conjecture that, up to isomorphism, congruence lattices of algebras can also be characterized as algebraic lattices. This is proved in G . Gratzer and E. T. Schmidt [1962]. For the briefest known proof, see P. Pudlhk [b]. A rather detailed history of this field, including the case of infinitary algebras and the connections between congruence lattices of algebras and type 2 and type 3 representations of lattices, can be found in the paper G . Gratzer and W. A. Lampe [a]. A detailed discussion of this material with proofs would require a monograph of about 400 pages.

312 Concluding Remarks

The congruence lattice of the algebra ( A ; F ) is a sublattice of Part(A); if a lattice has the property that whenever it is represented as a (0, 1)-sublattice of sonie Par.t(A), then that sublattice is the congruence lattice of some algebra ( A ; F) , then the lattice is called strongly representable. Such lattices are described in R. W. Quackertbush and B. Wolk [1971] and P. Pudlbk [a].

Another case in point is the investigation of the lattice X ( X ) of all topological spaces defined on a set, where for the topologies S , and S , on X we set Si<S2 iff every S, open set is also S, open. Saniple results: X ( X ) is a conipleniented lattice: see A. K. Steiner [1966] and [1966a].

This point of view was introduced by G. Birkhoff [1936] and it pernieates the topology book of R. Vaidyanathaswaniy [ 19471 (reprinted in 1960).

Along with Z ( X ) , about a dozen lattices are being considered in topology. For an excellent survey of this field (with 124 references) the reader should consult R . E. Larson acd S. J. Andima [l975]; see also J. Rosickjr [1975].

The closed (and also the open) subsets of a topological space X form a lattice L ( X ) , and this lattice determines the topological space (provided it is TI). These lattices were characterized by H. Wallnian [1938]. Thus certain parts of topology can be studied within lattice theory-the theory of conipactifications beicg one exainple (H. Wallman [1938], 0. Frink [1964]). The set of all continuous real function? 011

a topological space X also forms a lattice C ( X ) under poir.twiue ordering. The lattice C ( X ) determires X if X is compact Hausdorff (I. Kaplnnsky [1947]; see also 1,. Gill- inan and M. Jerison [1960]).

A topologicul lattice L is a lattice equipped with a topology such that A acd v are continuous functions. The theory of topological lattices, the study of which waq started by A. D. Wallace and his students, is quite extensive. The methods used are mostly topological rather than lattice theoretical in nature. The reader should cotisult L. W. Anderson [1961] for an early review of this field.

Incontrast with topological lattices, intrinsictopologiesareconsideredon arbitrarylat- ticee. Intrinsic topologies are topologies defined on a lattice in ternis of the partial ordering. For instance, the interval topology takes the intervals as a subbase for closed sets. For niore details, see 0. Frink [1942], G. Birkhoff [1962], and B. C. Rennie [1951].

The study of partially ordered topological vector spaces was pioneered by the works of F. Riesz, H. Freudenthal, S. Kakutani, L. V. KantoroviE, M. G. Krein, H. Na- kano, and G. Birkhoff. The motivation came froni a variety of sources-the theory of Herinitian operators in Hilbert spaces, C* algebra theory, choquet boundary theory, and the inescapable fact that partially ordered topological vector spaces abound in analysis. The goal is to study how the presence of a partisl order ielation compatible with the algebraic structure of a vector space over the field of real numbers can substantially influence the topology. The earliest expositions of the subject are L. V. KantoroviE, B. Z. Vulih, and A. G. Pinkser [1950], B. Z. Vulih [ 19671, and H. Nakano [1950]. The modern theory of partially oIdered topological vector spaces is characterized by the major role played by the concept of duality (just as in the tnodern theory of topological vector spaces) and there are a number of recent books stressing this aspect of the subject: A. L. Peressini [1967], G. Janieson [1970], Yau-chuen Wong and Kung-fu Ng [1973] and M. Dohoux [1974]. These books have extensive bibliographies for the interested reader.

Concluding Remarks 313

The important work C. Caratheodory [ 19561 (English translation : C. Carathkodory [1963]) initiated a study of finitely additive functions and measures, and the integrals they generate on rings of elements of Boolean algebras. The concept of integral was made completely algebraic. Thus he created what may, in analogy to J. von New mann’s study of Continuous Geometries, be called “pointless” integration. Rediscov- ering Caratheodory’s work and using the potent tools of topological Riesz spaces (partially ordered linear topological spaces which are also lattices with respect to the partial order), D. H. Fremlin [1974] gives a thoroughly modern account of this field.

There are many publications in quantum mechanics that border on lattice theory. A typical question is whether the lattice of propositions of quantum mechanics is modular, as postulated by G. Birkhoff and J. von Neumann [1936]. For some of the recent problems discussed in the literature, see the books: J. M. Jauch [1968] and G. W. Mackey [1963] and the papers: P. D. Finch [1969], J. M. Jauch and C. Piron [1963] and [1969], P. Mittelstatdt 119721, and C. Piron [1964]. It may be of some interest to note that of the two volume work V. S. Varadarajan [1968] and [1970] on quantum mechanics, a substantial part of the first volume deals with lattices.

Lattices and posets appear in other areas alongside with an algebraic structure and some or all the operations relate in some way to the partial ordering (for instance, they are isotone). For a survey of this field, see L. Fuchs [1963].

Many generalizations of the lattice concept appear in the literature. However, only one has been developed to some depth: weakly associative lattices, introduced independently by E. Fried [1970] and H. L. Skala [1971]. Here is a partial bibliography of this new field: P. Erdos, E. Fried, A. Hajnal, and E. C. Milner [1972]; E. Fried [1973], [1974], [a], [b], and [c]; E. Fried and G. Griitzer [1973], [1973a], and [a]; E. Fried, G. Grhtzer, and R. W. Quackenbush [a] and [b]; E. Fried and V. T. S6s [1975]; K. Gladstien [1973]; G. Griitzer and H. Lakser [1971a]; H. L. Skala [1972]: R. W. Quackenbush [1972].

There is one topic that I would have liked to see included in this book, unfortu- nately, its late development made it inlponsible to discuss it in detail. This is the theory of transferability.

Let 11s call a lattice K transferuble iff, for any lattice L, if K has an embedding v into I ( L ) , then K has an embedding y into L. Tf, in addition, this y can always he chosen to satisfy

uyEap but a y e b v for any b < a

then K is called shurply transferable.

in 1966). There are only a few results on transferability: These coccepts originate in G. Griitzer [l970] (which is the text of a lecture given

A transferable lattice has co doubly reducible eleiiicct (G. Griitzer [1970] and [ 19741). Every finite projective lattice is transferable (K. A.Baker and A. W. Hales [1974]). A finite’ transferable lattice satisfies (SD,) and (SDv) (H. S. Gaskill, G. Griitzer, and C. R. P h t t 119753).

I This result is true even when “finite” is clropped, see G. Gratzer and C. H. Platt, Two ernbedcling theorems for lattices. Abstract. Notices Anier. Math. Sor. April 1977. Seth also f o o t w t c to l’robleiii 1.24.

314 Concluding Remarks

I t is obvious from Corollary IV.4.3, that if K is transferable and a € K , then (a] is finite.

T. Tan [1975] found the first example of a transferable lattice that is not sharply transferable: &z~&,, where 6, is the chain of natural numbers. No such finite lattice is known.

It is primarily due to H. S. Gaskill’s work that sharp transferability has been lnuch further developed. To describe the new results, conditions (T,) and (Tv) have to be introduced.

Let K be a finite lattice, p E K , and J E K . We call ( p , J) a pair iff p 6 J, (JI 2 2, and p < V J . A pair ( p , J) is minimal iff whenever ( p , J’) is also a pair and for every a E J’ there exists a b E J with a< b, then J c J‘. (Minimal pairs are the most econo- mical way of describing the join-structure of a lattice.) A finite lattice K satisfies the condition (Tv) iff a rank function r can be defined on K (with integer values) such that if (p, J ) is a minimal pair of K and q J , thus r ( p ) < r (q ) . The condition (T,) is the dual of (Tv).

THEOREM. (i) K is sharply transferable.

(ii) K satisfies (TA), (T,), and (W). (iii) K can be embedded in F(3) .

For a finite lattice K the following conditims are equivalent:

This result was discovered in many parts. H. S. Gaskill introduced the conditions (T,) and (T,) in his Ph. D. Thesis, see also H. S. Gaskill [1972], and he proved that (ii) implies (i) and that (i) implies (T,) and (Tv). I suggested a constructionforlattices failing (T,) to show that they are not sharply transferable; in a related vein, C. R. Platt came up with a construction that applies to all finite lattices. This is the construction given in H. s. Gaskill, G. Grlitaer, and C. R. Platt [1975]; as a corollary of Platt’s construction one obtains one of the results on transferability mentioned above. Meanwhile, H. S. Gaskill and C. R. Platt [1975] proved that sharp transferability and (W) are together equivalent to enibeddability in F(3) . Finally, I proved that sharp transferability implies (W). Thus the equivalence of (i)-(iii) is established.

H. Lakser observed that the result of K. A. Baker and A. W. Hales can be sharpened : every finite projective lattice is sharply transferable.

All distributive sharply transferable and all distributive transferable lattices have been described by T. Tan [1976].

If all lattices considered are restricted to an equational class H we get the concepts of H-transferable and sharply H-transferable lattices.

According to H. S. Gaskill [1972a] (see also E. Nelson [1974]) every finite distributive lattice is sharply D-transferable.

G. Grlitzer and T. Tan proved that if K is D-transferable, then for every a C K , (a] and {b I b t a} are finite ; moreover, if K is sharply D-transferable, then K does not contain &2x&o as a sublattice. It is natural to conjecture that the linear sums of t,ype u) of finite distributive lattices are the infinite sharply D-transferable lattices.

For a finite lattice K , the embeddability of K into a lattice L can be expressed as a first-order property. Thus K is transferable iff this first-order property is preserved

Coiiclutling Rimarks 315

when passing from I ( L ) to L. The general problem of investigating and classifying such properties is discussed in G. Griitzer [1970]. For some interesting contributions to this problem, see K. A. Baker and A. W. Hales [1974]. An important first-order property of this type is investigated in W. Poguntke and I. Rival [1974].

Many problems proposed in FC are still open. As Piet Hein [1969] writes: Problems worthy-of attack-prove their worth-by hitting back.

The following is an account of the work done in relation to problems in FC; it is not claimed to be comprehensive.

3. Solved in J. Sichler [1972]. 9. Solved in C. R. Platt [1976] and D. Kelly and I. Rival [1975a].

12. and 13. A negative answer is provided for the case K =M in A. P. Huhn [1975].

14. and 15. These are discussed above. 16. n =Z and 3 are done in R. Padmanabhan [1972]. 18. An affirmative solution is given in R. W. Quackenbush [1974]. 21. and 23. Some relevant references are given in the Further Topics and Refer-

29. Solved in R. Padmanabhan [1974a]. 30. See the references given to Problem 11.3. 31. According to J. D. Monk this can be solved affirmatively under the Generalized

32. An affirmative solution was found independently by B. Jbnsson (unpublished)

33. References to various solutions are given in the Further Topics and References

34. The question raised in the problem was answered in the negative by T. Tan.

35. See V. Vilhelm [1955], J. Jakubik [1975], and F. Sik [a]. 39. Solved by M. E. Adams [1974] and R. Balbes [1972], independently. 47. A contribntion to this problem was made in H. Lakser [1973a]. The problem

48. Solved in M. E. Adams and D. Kelly [a]. 51. Answered in the negative in G . Griitzer, B. Jhsson , and H. Lakser [1973]. 54. Solved in T. Katrifibk [1972]. 55. Solutions are offered in T. Katrihbk [1973a] and H. A. Priestley [1974]. 56. A description is given in A. Urquhart [1973]. 57. See 31. E. Adams [1973a], T. Katrihbk [c], and H. A. Priestley [1975]. 63. This is verified in M. E. Adams [1976]. (The result of M. E. Adams was announ-

ced in Notices Amtr. Math. Soc. 20 (1973): A-560. A new proof is announced by A. Wr6nski [1976].)

The problem is still open for K =L.

ences of Chapter I.

Continuum Hypothesis.

and K. D. Magill [1970].

of Chapter 11.

The problem remains unsolved.

is solved in M. E. Adams and D. Kelly [b].

65. Solved by T. Katriilhk [1973c]. 66. The case n = 1 is considercd in T. Katrihhk [1966] and [1968]. I hope I shall be able to keep the reader informed about developments related to

the problems proposed in this book through the journal Algebra Universalis.

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TABLE OF NOTATION

EXPLANATION

riienibership sign inclusion proper inclusion union intersection empty set (void set) set difference Cartesian product

ordered pair

equivalent notations for pEA2, 1, 192

set of all subsets of a set X

A x A

APPEARS ON PAGE(S)

1 1 6 6

10 10

162 162

2 2 2 4

V 4

SYMBOL APPEARS ON PAGE(S)

15, 16 17 17 17 17 2 2

268 268 172 61 2

5 8 7

111 111 268 89 21

113 19

19,42 18,80 18,42

26,192 49,112

24 21

151

Table of Notation 363

SYMBOL APPEARS ON PAGE( S)

L' 23 L x K 22 11(L, I i E I ) 23 rIa(Li I i E I ) 23 1 A * B 265 ( B ; A, v, I , 0, 1) 48 BIZ 95

SPECIAL LATTICES A N D ALGEBRAS

119 48 92 97 16 44 33 32 43 33

204 11,59

219,235,238 11,59

7 118

CLASSES OF LATTICES A N D OPERATORS

69 32 32 32

235,236 235,238 235,236

44 258 230 241 229 235 255

SYMBOL APPEAHS ON PAGE(S)

hlod (C) 227 P 229 PR 234 ps 230 pu 231 S 229 Si 233

SUBSETS AND FAMILIES OF SUBSETS O F LATTICES

(a1 17 [a 1 19 [a, bl 17 (a, b1 17 [a, 6 ) 17 (a , b ) 17 Cen( L) 156 Comp(A) 292 D(L) 113

(HI 17 [HI 17

[HI 19 [HI, 89 I" 67,91 J ( L ) 61 M ( L ) 63

65,100 r (a) 61,64,100 T ( I ) 100

49,112 W )

CONGRUENCES, POLYNOMIALS, AND QUOTIENTS

20 20 74 73 74

139,142 85

116 24 20

364 Table of Notation

SYMBOL APPEARS ON PAGE(S1 SYMBOL APPEARS ON PAGE(S)

131 268 268 269 22

287 267

27 129 129 129 129 129 1 30 130 130

131

269 90 4

90 1

102 102 103

MISCELLANEOUS

274 274 274 269 269 269 269

15,16 180 180 235 100

108,184 31

72,78,202 203 48 48 51

176 243 243 227 243 204

INDEX

A-connected, 187 a-disjoint, 121 Abian, S., 55, 316 Absorption identity, 4 Accessible element

Adanis, M. E., 54, G8, 123, 124, 128, 306, 306, 307, 308, 309, 315, 316 Affine geometry, 22 1 Aigner, M., 9, 316 Alexander, J. W., 108, 316 Alexander’s Theorem, 105 Algebra, 52

join-, 122

Boolean, 48 generalized Boolean, 67 Heyting, 125 injective, 124 partial, 45 Post, 126 projective, 124 Stone, 112

subdirectly irreducible, 116 (See also Universal algebra)

sub-, 48, 11 1

Algebraic closure Space, 184 Algebraic function, 73

Algebraic lattice, 79 Algebraically closed,

relativcAly, 189 Alternating scquence of perspectivities, 164 Altwegg, M., 62, 316 Ainalganiated V-formation, 252

Amalgamation l ’roprty, 253

Boolean, 73

strongly, 252

Strong, 253 Weak, 263

Arnemiya, I., 224, 316, 317

Anderson, F. W., 121, 317 Anderson, L. W., 312, 317 Andima, S. J., 312, 317 Antichain, 2

Antisymmetry, 1 AntoniuR, R., 55, 317 Aregkin, G. Ja., 111, 317 Arguesian identity, 199 Arguesian lattice, 199 Arithmetical ring, 124 Artin, E., 209, 317 Artmann, B., 224, 317 Ascending Chain Condition, 12 Associativity, 4 Associative identity, 4 Associative lattice,

Atom, 49

Atomic Boolean polynomial, 70 Atomic lattice, 189

relatively, 297 .4tomistic lattice, 179 Augmented ideal lattice, 18 Automorphism, 16 Avann, S. P., 124, 218, 317 Axiom,

maximum sized, 11 8

weakly, 250

dual, 49

Pasch, 202 Steinitz-MacLane Exchange, 178

Babai, L., 54, 317 Bacsich, P., 261, 318 Baer, R., 209, 318 Baker, K. A., 61, 53, 55, 56, 124, 188, 223, 241, 242, 244, 245, 246, 250, 260 261, 305, 313, 314, 315, 318 Bakker, J. W. cle, 56, 318

366 Index

I3albt.s, lt., 50, 53, 120, 121, 123, 124, 125, 126, 305, 315, 318, 319 I h e , 108

Basis. 47 stth-, 108

equational, 243 hoinogeneous, 224 irretlundant equational, 243

13astcrfieJtl, J. G., 222, 319 Beazer, R., 125, 126, 319 Behrens, E. A., 124, 319 Bclnap, N. D., 126, 319 Berginan, G. &I., 124, 219, 252, 319 13erinti11, J., 122, 127, 303, 319 I3et \vctsiiness rrlation, 52 13i-iiiii(liit~ly eoitiI)lt~ii~c~iitc.tl l a t t ic*c., 297 Binary bracketing, 14 Rinary operations, 4. 52 13innry polynot nial, 27 Hinary relation, 1

Rirkhoff, Q., 3. 45. 51. ti4, 79, 115, 117, 121, 122. 123, 138. 140, 146, 154, 155, l6X, 172, 1 7 8 , 179, 189, 220, 229, 304, 306, 3 11,

Uirklioff-i'vlrngrr Theorern, 154 Blair, R. L., 121, 320 I3oc>k, 8, 192

triviaJ, 192 13luriit~ntal, L., 320 13oolean algebra, 48

generalized, 6 7 rigid, 126

transitive extension o f , X

312, 313, 319, 320

Boolean algebraic f i i i 1 c . t ion, 7 3 Hooltari lattice. 48

gczncralized, 77 R-generated by a chain, 92 R-generatetl by a distributive lattice. 97

atomic, 70 Hooleaii pol y noniial, 69

Boolean ring, 78 Boolean space, 104

generalized, 109 Rooleanization, 10,; Bound,

greatest lower, 2 least upper, 2 lower, 2 upper, 2

Bouncletl, k-, 244

I3ountled posct, 47 I3rackct ing,

binary, 14 right, 14

I%racketing function, 15 I<rcadth, 242 Hrown, A. U., 56, 320

Bruns, G., 53, 123, 125, 320, 321 13uchi, J. I t . , 97, 122, 321 lhlttafuoeo, E.. 120, 321 Hyrd, I<. D., 122, 321 Byrnr, L., 53. 321

Hrowii, T. J. , 320

C'-relation. 2x9 g-rrducetl f r w produet, 2x9 C'ainpbell, A. I ) . , 45, 321 C'ttnfiel(1, E. 12.. 309 Canonical form, 284 Ceiionical representation. 280 Csrathi.otlory, C., 313, 321 Carroll, L.. 321 Cartesian pro(1ric.t. 23 Cell, 220 Centre of a bountlctl lattice, 136 Chain, 2

length of, 2 tnaxitnal, 68 strongly inaxinial, 96

Chain Conclitioii, Ascending, 12 Descending, 50 lnfinite Jordan- Holdtar, 17s Jordan-Holder, 172

n1-, 307 Chain group, 22 1 Chang, C. C., 123, 126, 321 Chcn, C. C., 54, 5 6 , 113, 119, 120. 115, 289, 293, 297, 303, 304, 321, 322 Church, R., 121, 322 Cibulskis, J. M., 322 Circuit, 190 Class,

congruence, 20 equational, 32 implicational, 235 locally finite rquatiorial, 234, 255 regular equational, 303

Index 367

Class, trivial equational, 32

Clopen set, 104 CloscYl,

Closed clement, 5 2 Closed set. 107, 184 Closure, 108 Closurc tiiap, 52

algebraic. 184

relati\ ely algebraicall>-, 189

Closurt. sl’ac’, 184

Cobountlary operator, 222 Cofinal. 258 Cohvn, 1. S.. 124, 322 Colin, 1’. JI., 124, 322 Cohopfian lattice, 309 Collinear, 203 Comer, S . I)., 236, 2ti0, 261, 262, 922 Cottitiioii l<c~fiticttic~tit J’ropt‘rt:,-, 2 7 ; Co I 111 I iu t a t ive c 1 iagru I i 1. :1:3 Cottitiiutati\ i ty , 4 Cotnttirttati~-t~ iclcsritity, 4 Cot1ipH.c.t t~letiicnt. 79

Cotlipact space, 108

Compactly generated lattice, 79 Conipactness Theoreiii, 235 Comparability relation, 9 Co t i iparable cot tipleilien t s, 292 Comparable elements, 2

CoI1lpat‘t set, 108

lot‘ally, 109

Complement, 47

pseudo-, 48 cotiiparable, 292

relativr., 47 relative pseudo-, 51

C o I lip1 e men t a r y , 2 8 8 Cotiiplementation,

latticr with, 297 transitive, 297

bi-uniqrielj-, 297 left-, 190 relativelj, 48 sectionally, 83, 18.5 uniquely, 293

Cotnpletiiented lattice, 48

Cotnplete lnfiriito Distribrrti\ ( * I t l c . t i t ity. 97 Coniplete latticcb, 21

conditionall:, , 26

25 GrHtzer

Corit1)lete sublattice, 15ti CotnI)lctely frcely generated,

Cornplt~tely nicct-irreduciblr, 1 1 7 Coruplctcly priiric d u d ideal, 23s Completion, 12:%

RlacNeille, 98 Complexity, 56 Coriiponent sitbset, 287 C‘ontlition.

Asct.nding Chain, 12 1)cscending Chain, 50 I nfinitc. Jorclan-Holder Chain, 178 Jordan-Holt lw Chain, 172 Lower Covering, 162 nt-chain, 307 Alac.Lane, 178 i\Ittl’ccv type, 2.20 Upper Covering, 1 ti2

lattice, 44

Conditionally caomplete lattice, 26 Congruence class, 20 Congruence distributive, 2 2 i Congruence 14:xtension Property, 120 Congritenc~e kernel of a homomorpliistn, 22 Congruence lattice, 21 Congruence tuodular, 220 Congrucnce relation, 20, 57, 81, 1 1 1

distributive, 149 fully invariant 228 icleal kernel of, 22, 133 invariant, 234 join-, 25, 109 ,loin-represeiitublc, 133 taeet-representable, 135 neutral, 149 order-representable, 138 principal, 74 representable, 136 separable, 155 standard, 149

Cougrrience relation generated b y H, 73 Congruance relations,

p m n u t i n g , 67, 149 C o tigruence Substitution Property, 122 ~‘ollllected,

A - . 187 trianglv, 297. 301

Con t i w tion, Galois, 51

368 Index

Constant Imp), eventually, 300

join-, 122 upper, 189

Continuous lattice, 189

Continuous map, 108 Convex, 17 Convex sublattice generated by a subset, 17 Coordinatization Theorem of Projective Geometry, 208 Cornish, W. H., 123, 126, 322 Cotlar, M., 123, 322 Cover, 10

lower i-, 268 proper lower i-, 268 proper upper i-, 268 upper i - , 268

Lower, 162 Upper, 162

Covering Condition,

Covering relation, 10 Crapo, H. H., 3, 222, 223, 322, 323 Crawley, P., 121, 123, 158, 159, 218, 220, 303, 304, 305, 323 Croisot, R., 179, 220, 221, 323 Curry, H. B., 126, 323 Curzio, M., 54, 323 Cycle,

?&-, 294

Daigneault, A., 261, 323 Davey, B. A., 55, 123, 126, 306, 324 Davey-Sands Theorem, 307 Davis, A. C., 55, 324 Day, A., 126, 126, 159, 219, 220, 256, 260, 261, 262, 306, 309, 324 Day, G. W., 98, 123, 324 Dean, R. A., 55, 110, 284, 286, 281, 288, 303, 304, 305, 324, 326 Decomposition,

linear, 288 Dedekind, R., 39, 121, 124, 325 Demarr, R., 55, 325 De Morgan’s Identities, 48 Dense element, 11 3 Dense lattice, 119 Dense set, 43, 11 3 Dssargues’ Theorein, 205 I)cscending, 12

Dtwwiding Chain Condition, 50 Devid6, V., 55, 325 Diagrani, 10, 33

coininutativt,, 33 optimal, 10 planar, 10

Diamond, 59 Diamond associated with a trianglc, 299 Diamond, A. H., 53, 325 Difference,

Dilworth, R. P., 31, 81, 121, 123, 130, 131, 154, 158, 170, 188, 191, 214, 218, 220, 222, 223, 272, 286, 287, 289, 293, 303, 304, 325, 326 Dinlension of a subspacc, 185 Dimension of a poset,

Direct join representation, 218 Direct power, 23 Direct product, 23 Directed set, 122, 189 Directly indecomposable lattice, 153 Disconnected space,

totally, 104, 109 Disjoint,

a-, 121 Disjunction of equations,

universal, 243 Disinantlable, 51 Distance function, 194 Distributive,

eongruence, 221 0-, 126

symmetric, 72

order, 54

Distributive congruence relation, 149 Distributive eleitient, 138

Distributive ideal, 146 Distributive Identity,

Complete Infinite, 97 Join Infinite, 90 Meet Infinite, 90

dually, 138

Distributive inequality, 29 Distributive join-settiilltttic.~, 99 Distributive lattice, 30

Distributive lattice R-genrrated by FI

chain, 92 Distributive product,

freo { O , 1}-, 106

0-, 126

Index 369

Di\ ision ring, 204 Dohoux, M., 312, 326 Doubly irreiliiriblc, 49 DoGling, T. A, , 222, 326 Drake, I). A., 222, 326 DragkoviEova, H ., 326 Dual, 2 Dill31 atolll, 49 Dual ideal, 19, 42, 100

coinpletely prime, 238 generated by H, 19 primp, 19, 100 principal, 19 prol’e’. 19

Dritdity l’rinciple, 3, 0 Dually clistributivc elrnlent, 138 Dually standard t~letncxnt, 138 Ihhrc4-.Jacotin, M. L., 179, 221, 326 Thinran, D. G., 326 Dunn, J. M., 126, 326 Duthie, W’. D., 56, 326 Dwinger, I%., 53, 123, 125, 126, 326

Etlrlberg, M., 222, 326 Edge geometry, 187 Edge lattice, 187 Ehrenfeucht, A., 223, 228, 3 % Eklof, 1’. C., 261, 327 Elenient,

closed, 52 wiiipect, 79 dcnse, 113 llistributive, 138 tloubly irreducible, 49 iliially distributive, 138 tli~ally standard, 138 iilrntity, 2 incoiiiparable. 2 join-accessible, 122 join-inaccessible, 122 join-irreducible, 49 tnaxiinal, 12 n,rc.t-irrccluciblr, 49 n(~lItra1, 138 pthrspective, 180 st a d a r i l , 13X unit, 2 zcro, 2

Elementarily rquivalent, 128 T~:inbetltling, 16, 43 “5’

Enibedding l’ropcrty, 253 Ihdomorphislli, 1 ti, 293, 295, 297 lhgclking, K., 124, 327 Epstein, G., 126, 327 Equational basis, 243

irredundant, 243 Equational class, 3 2

congruence modular, 220 locdly finite, 234, 255 rc>gular, 303 trivial, 32

Equational class generated by a lattice,234 Equations,

Equivalencw lattice, 192 Equivalence relation, 20 Equivalent, 235

eleinentarily, 128 Ertliis, P., 313, 327 Jhler, L., 221 Evans, T., 298, 299, 908, 327 Eventually constant map, 300 Exchange Axiom,

Existence Theorem for Free Products, 110 Exchange lattice, 179 Kxteneion Property,

Congruence, 120

universal disjunction of, 243

Steinitz-MacLane, 178

Faber, V., 223, 226, 327 Factor lattice, 21 Fajtlowicz, S., 223, 226, 256, 327 Field of sets, 65 Filippov. N. D., 54, 327 Filter, 19

prime, 19 princ+ipaI, 19 ultra, 19

Finch, 1’. D., 313, 327 Finite dimensional lattice, 183 I‘initr rquational class,

Finite length, 2 Finite lattice,

locally, 157 Finite partition, 193 Finite V-formation, 255 Finitely generated V-formation, 255 Finitely presented lattice, 298 Finitely subdirectly irreducible, 235

locally, 234, 255

370 Index

Finkbeiner, D. T., 158, 169, 220, 327 First Isomorphism Theorem, 26, 151, 160 Fishburn. P. C., 51, 63, 327 Fixed-Point Theorem, 55 Flat, 184 Fleischer, I., 55, 327 Frattini sublattice, 54 Frayne, T., 260, 327 Free K-product, 105, 267 Free K (0, 1)-product, 106 Free lattice,

generated by P, 33 on m generators, 33 over K generated by %, 43 over K generated by P, 32

g-reduced, 289

Existence Theorem for, 110 Structure Theorem for, 272

Free product,

Free Products,

Free projective plane, 216 Free variable, 235 Free {O,l}-distributive product, 106 Free (0,l)-product, 106 Freely generated,

lattice completely, 44 Freese, R., 55, 124, 178, 219, 220, 222, 250, 261, 305, 309, 327, 328 Freidman, P. A., 124, 328 Fremlin, D. H., 313, 328 Freudenthal, H., 312 Fried, E., 159, 224, 260, 261, 304, 313, 328, 329 Friedman, H., 14, 329 Frink, O., 49, 118, 121, 122, 125, 208, 211, 220, 311, 312, 329 Frontera-MarquBs, B., 223, 329 Frucht, R., 306, 329 Fryer, K. D., 224, 329 Fuchs, L., 124, 313, 329 Fujiwara, S., 180, 223, 329 Fujiware, T., 220, 329 Fully invariant congruence relation, 228 Fully invariant subdirect representation, 299 Fully ordered set, 2 Funayama, N., 40, 42, 79, 91, 92, 123, 158, 330 Func tion,

algebraic, 73

Boolean algebraic, 73 bracketing, 15 distance, 194 height, 172 Mobius, 191 polynomial, 27

Gaifman, H., 121, 330 Gallai, T., 121 Galois connection, 5 1 Galvin, F., 286, 288, 306, 330 Ganter, B., 218, 330 Gaskill, H. S., 55, 305, 313, 314, 330 Gelfand, I. M., 219, 330 Generalized Boolean algebra, 67 Generalized Boolean lattice, 77

R-generated by a distributive lattice. 87

Generalized Boolean spare, 109 Generated,

lattice completely freely, 44 Generated by a lattice,

equational class, 234 Generated by H,

congruence relation, 73 Generated by a subset,

convex sublattice, 1 7 Generating set, 1 7 Geometric lattice, 179 Geometry, 184

affine, 221 Coordinatization Tlieoreni of Projrrtive, 208 edge, 187 n-stufige, 221 planar. 221 projective, 204

GhouilB-Houri, A., 9, 330 Gilbert, E. N., 121, 330 Gillman, L., 312, 331 Gilmore, P. C., 9, 331 Gladstien, K., 313, 331 Glivenko, V., 49, 91, 125, 331 Gluhov, M. M., 47, 331 Graded poset, 178 Graph, 187 Grayev, M., 219, 333 Greatest lower bound, 2 Greene, C., 191, 222, 333 Gumm, H. P., 284

Index 371

Hajek, O . , 333 Htijnal, A., 313, 333 Hales, A. W., 55, 66, 121, 196, 305, 313, 314, 315, 333, 334 Half-open interval, 17 Hall, NI., 31, 214, 224, 334 Hall, Ph., 191, 334 Halmos, P. R., 53, 124, 259, 334 llalperin, I., 161, 224, 334 Hmf, W., 123, 334 Hansel, G., 121, 334 Harper, L. H., 188, 334 Harrison, M. A., 53, 334 Hartinanis, J., 223, 334 Hashirnoto, J., 65, 67, 68, 78, 85, 86, 88, 121, 140, 146, 149, 151, 157, 159, 354, 335 Haustlorff space, 108 Havas, G., 54, 5 7 , 336 I-Lavel, H., 218, 335 Hecht, T., 335 Hetlrlin, Z., 306, 335 Height function, 172 Hein, P., 315, 335 Hell, P., 306, 335 Hereditary, 44, ti1 Herrmann, C., 47, 158, 219, 220, 224, 245, 260, 261, 335 Keyting algebra, 126 Higgs, D., 25, 336 Hoffman, A. J., 9, 336 Honieomorphisn~, 108 Homogeneous basis, 224 Homomorphic image, 2 2

Honiolnorphisin, 16, 43, 45, 48, 111, 295 maximal, 46

congruence kernel of, 22 ideal kernel of, 22 join-, 16 meet-, 16

{ O , l } - , 47 { O h , 47

H 01 mi I lorphist 11 Theore i n , 2 2 Hong, D. X., 250, 260, 261, 262, 299, 308, 336 Hopfian lattice, 298 Horn, A., 50, 53, 55 , 121, 123, 124, 125, 126, 305, 336 I+uang, S., 14, 396 Huguet, I)., 14, 336 Huhn, A. P., 219, 220, 252, 315, 336

Huntington, E. V., 53, 304, 336 Hutchinson, G., 219, 261, 336

i-cover, lower, 268 proper lower, 268 proper upper, 268 upper, 268

Ideal, 17, 42, 80 completely prime dual, 238 distributive, 146 dual, 19, 42, 100 generated by H, 17 iliaxima1 prime, 67 minimal prime, 67, 115 neutral, 146 normal, 98 prime, 17, 100 prime dual, 19 principal, 1 7 principal dual, 19 proper, 1 7 proper dual, 19 standard, 146

of a congruence relation, 22, 133 of a homomorphism, 22

Ideal lattice, 18 augmented, 18

Ideal of order TI,

standard, 159 Idempotency, 4 Idempotent identity, 4 Identity, 28

ldeal kernel,

absorption, 4 arguesian, 199 associative, 4 commutative, 4 Complete Infinite Distributive, 97 De Morgan’s, 48 idempotent, 4 Join Infinite:Distributive, 90 lattice, 28 Meet Infinite Distributive, 90 semiassociative, 15 shearing, 161 Stone, 112

Jdentity element, 2 Image,

homomorphic, 22

372

Image,

Iuiplication, 234 Implicational class, 235 Implies, 235 Inaccessible element,

join-, 122 Incomparable elements, 2 Indecornposable lattice,

directly, 153 Independent set, 167 Inequality, 28

distributive, 29 lattice, 28 modular, 29 triangle, 194

iiiaxinial honiomorphic, 46

Infimuiii, 2 Infinite Distributive Identity,

Complete. 97 Join, 90 Meet, 90

Infinite Jordan-Holder C h i n Conditioi Injective, 124, 255 Injective structure, 128 Interval, 1 7

half-open, 17 open, 1 7 prime, 94

Intrinsic topology, 31 2 Invariant congruence relation, 234

fully, 228, 299 Inverse map, 25 Iqbalunissa, 140, 150, 158, 169, 337 Irreducible,

completely meet-, 11 7 doubly, 49 finitely subdirectly, 235 join-, 49 meet-, 49 subdirectly, 116

Irredundant equational basis, 243 Irredundant representation, 62 Isomorphic, 15, 16 Isomorphism, 15, 16, 43 Isomorphism Theorem, 162

First, 26, 151, 160 Second, 116, 151

Isotone map, 16 Ivanov, S. G., 159, 337 Iwaniura, T., 122, 337

Index

Jakubik, J., 54, 123, 158, 159, 315, 337 Jameson, G., 312, 337 Janowitz, M. F., 126, 157, 159, 337, 338 Jauch, J. M., 313, 338 Jensen, Chr. U., 124, 338 Jerison, M., 312, 338 Johnson. J., 260, 338 Join, 4 Join-accessible element, 122 Join-congruence relation, 25, 109 Join-continuous lattice, 122 Join-homoinorphisin, 1 G Join-inaccessible element, 122 Join Infinite Distributivc Identity, 90 Join-irreducible, 49 Join-representable congruence relation, 135 Join representation,

Join-semilattice, 7

.

direct, 218

distributive, 99 n, 178 modular, 109

Join table, 9 Jbnsson, B., 120, 123, 124, 126, 128, 158, 171, 194, 197, 198, 199, 201, 205, 211, 212, 217, 220, 223, 224, 232, 233, 234, 237, 242. 246, 260, 256, 257, 260, 261, 262, 263, 272, 273, 280, 285, 286, 288, 303, 304, 305, 306, 309, 315, 338, 339 Jordan-Holder Chain Condition, 172

Infinite, 178

k-bounded, 244 K-product,

free, 105, 267 K-transferable lattice, 314

sharply, 314 Kakutani, S., 312 Kalman, J. A., 8, 339 Kalmbach, G., 125, 339 Kannan, V., 126, 339 KantoroviE, L. V., 312, 339 Keplansky, I., 54, 312, 339 Kappos, D. A., 122, 340 KatBtov, M., 124, 340 KatrihBk, T., 126, 128, 315, 340 Kelly, D., 17, 51, 54, 56, 123, 124, 128, 171, 304, 305, 308, 315, 341 Kelly, L. M., 222, 341 Kelly-Rival Theorem, 56

Index 373

Kernel of a cwngrucnce relation,

Kcmiel of u Iioinornorphistll, ideal, 22, 133

congruencr, 22 itleal, 22

liiefer, J. E., 288, 305, 341 Kitiiura, N., 55, 341 Kintlerinann, M., 219, 341 Kinugawa, S., 121, 140, 146, 341 Klass, hI. J., 222, 341 Kleitman, D., 121, 222, 341 Knaster, B., 55, 341 Kogalovskii, S. It., 55, 230, 341

Kolibiar, M., 53, 54, 120, 342 Kotnutu, A., 122, 342 Kontlraiov, G. F., 159, 342 Kontorovifi, P. G., 159, 342 Kostinsky, A., 303, 305, 342 Krein, &I. G., 312 Kuwra, T., 305 KuroL, A. G., 163, 219, 342 Kurog-Ore Theorem, 163

~ ~ ) i ~ , K . nf., 54, 55, 56, 178, 341, 342

Latlzianska, Z., 306, 342 Lakser, H., 51, 54, 55. 81, 84, 86, 120, 123, 124, 125, 127, 178, 207, 213, 223, 234, 235, 256, 257, 258, 260, 261, 272, 274, 275, 276, 282, 299, 305, 306, 308, 3 13, 314, 3 15, 342. 343 Latiipe, W. A., 223, 309 31 1, 343 Latnbek, J., 306, 343 Larson, R. E., 312, 343 Lattice, 3, 5

algebraic, 79 arguesian, 199 atomic, 189 atotnistic, 179 augmented ideal, 18 bi-uniqucly cotnpletuentetl, 297 Roolean, 48 cohopfian, 309 c3ornpactly generated, 79 rarnplernentetl, 48 coinpletr, 24 coiidi~ionally cotnpl(~tc, 26 c*ongruenc*e, 2 I continuous, 189 tlense, 119 direclly indeeoinposablr, 153

distnantlable, 51 distributive, 30 edge, 187 cquivalence, 192 rxchange, 179 factor, 21 finite dimensional, 183 finitely presented, 298 free, 33 generalized Boolean, 77 geometric, 179 hopfian, 298 itleal, 18 join-continuous, 122 K- transferable, 3 14 lef t-complemented, 190 locally finite, 157 locally modular, 218 M-symmetric, 176 m-universal, 304 tnatroid, 179 inodular, 30 modular partial 224 rdistributive modular, 219, 252 partial, 40 partition, 192 planar, 51 pseudocomplemented, 48, 11 1 quotient, 21 R-generated Boolean, 97 relatively atomic, 297 relatively complemented, 48 sectionally complemented, 85, 135 setniinodular, 172 sharply K-transferable, 314 sharply transferable, 57, 31 3 simple, 153 splitting, 239 Stone, 112 strongly representable, 312

topological, 3 12 transferable, 56, 313 uniquely complemented, 293 upper continuous, 189 WA, 260 weak partial, 41 weakly associative, 250 weakly modular, 134 0-distributive, 126

sub-, 16

374 Index

Lattice Completely freely generated, 44 Lattice identitiy, 28 Lattice inequality, 28 Lattice polynomial, 27

Lattice R-generated by a chain,

Lattice with complementation, 297 Lattice with pseudocomplementation, 11 1 Lattices,

Least upper bound, 2 Lee, K. B., 125, 128, 343 Left-complemented lattice, 190 Length,

Length of a chain, 2 Lesieur, L., 179, 221, 343 Line, 185, 202

n-ary, 27

distributive, 92

elementary equivalence of, 128

finite, 2

perspective with respect to, 205 refracted, 217

Linear decomposition, 288 Linear subepace, 203 Linearity, 1 Linearly ordered set, 2 Livgic, A. H., 219, 343 Locally compact space, 109 Locally finite equational class, 234, 256 Locally finite lattice, 157 Locally modular lattice, 218 LOB, J., 231, 236, 343 Lower bound, 2

greatest, 2 Lower Covering Condition, 162 Lower i-cover, 268

Lowig, H. F. J., 53, 343 Lozier, F. W., 126, 343 Lubell, D., 222, 343 Lunnon, F., 121, 343 Lyndon, R. C., 223, 262, 261, 343

proper, 268

m-chain condition, 307 m-universal, 304 M-symmetric lattice, 176 Mackey, G. W., 313, 344 MacLane, S., 178, 179, 189, 220, 344 MacLane Condition, 178 MacNeille, H. M.,‘87, 98, 123, 344 MacNeille completion, 98

Maeda, F., 20, 159, 180, 189, 220, 221, 224, 344 Maeda, S., 159, 176, 189, 220, 221, 344 Magill, K. D., 315, 344 Makkai, M., 54, 258, 261, 344, 345 Mal’cev, A. I., 158, 235, 345 Mal’cev type condition, 220 Mandelker, M., 126, 345 Map,

closure, 52 continuous, 108 eventually constant, 300 inverse, 25 open, 108

Maranda, J.-M., 128, 345 Matching between subsets, 188 Matroid, 221 Matroid lattice, 179 Maximal chain, 68

Maximal element, 12 Maximal homomorphic image, 46 Maximal prime ideal, 67 Maximum sized antichain, 11 8 McKenzie, R. N., 8, 32, 52, 53, 56, 126, 127, 220, 225, 238, 239, 241, 242, 246, 246, 262, 260, 261, 262, 306, 306, 346 McKenzie-Kostinsky Theorem, 307 MoKinsey, J. C. C., 63, 345 McLaughlin, J. E., 123, 158, 304, 346 Meet, 4 Meet-homomorphisni, 16 Meet Infinite Distributive Identity, 90 Meet-irreducible, 49

completely, 11 7 Meet-polynomial, 69 Meet-representable congruence relation, 135 Meet-representation, 194 Meet-seniilattice, 7

Meet table, 9 Mean, R. A., 122, 345 Menger, K., 154, 220, 345 Michler, G., 124, 345 ,

Milner, E. C., 313, 346 Minimal, 12 Minimal pair, 314 Minimal polynomial, 275 Minimal prime ideal, 67, 115 Minimal representation, 275, 346

strongly, 96

relatively pseudoconiplemented, 61

Index 375

Mirsky, L., 222 Mitchke, A., 219, 346 Mittelstadt, P., 313, 346 Mobius, -4. F., 191, 346 MoEulskii, E. N., 219, 346 Motlel, 235 Mod u 1 ar ,

congruence, 220 scnii-, 172

Modular inequality, 29 Modular join-semilattice, 109 Motlular 1at tire, 30

locally, 218 n-distributive, 219, 252 weakly, 134

Modular pair. 176 ?vlodiilar partial lattice 224 Monk, G. S., 224, 346 Monk, J. D., 126, 127, 315, 346 Monoid, 293 Monotone map, 16 Monteiro, A., 53, 346 Moore, J. T., 346 bIord, A. C., 260, 346 Morley, M. D., 123, 346 Mostowski, A., 92, 97, 346 Murata, K., 220, 346 Murskii, V. L., 252, 261, 346 Murty, P. V. R.. 125, 347 Musti, R., 120, 347 Mutually rigid, 296 Mycielski, J., 223, 226, 347

ti-ary lattice polynomial, 27 n-ary operation, 52 pi-cycle, 294 ii-dktributive modular lattice, 21 9, 252 ji-partition, 223 Nachbin, L., 65, 122, 347 Nakano, H., 312, 347 Nakayama, T., 79, 347 Nation, J. B., 55, 178, 220, 225, 305, 347 Nelson, E., 303, 314, 347 Nelson, 0. T., Jr., 242, 347 Ncinitz, W. C., 126, 347 Nrrode, A., 106, 347 Neuniann, B. FT., 228, 260, 347 Neuniann, H., 260, 347 Nruiiiann, J. von, 90, 121, 156, 166, 168, 208, 220, 223, 224, 304, 313, 347, 348

Neutral congruence relation, 149 Neutral element, 138 Neutral ideal, 146 Ng, Kung-fu, 312, 348 Noether, E., 124, 348 Nondegenerete, 204 Normal ideal, 98 Normal sequence of perspectivities, 164 Nullary operation, 52

Oehinke, R. H., 55, 110, 348 Ogasawara, T., 297, 304, 348 O h , Ph., 128, 305, 306, 348 Open interval, 17

Open map, 108 Open set, 5 Operation,

half-, 17

binary, 4, 52 n-ary, 52 nullary, 52 partial, 40 unary, 52

Operator, 260 coboundary, 222

Optinial diagram, 10 Order dimension, 54 Order o f t , 110 Order-preserving map, 1 G Order -representable congruence relation, 1 38 Ordered set,

linearly, 2 partially, 1

partial, 1 quasi-, 8

fully, 2

Ordering relation,

Ore,O., 51, 120, 124, 138, 139, 145, 163, 172, 193, 200, 201, 218, 219, 220, 223, 348 Ore Theorem, 225

Padmanabhan, R., 8, 32, 52, 53, 220, 260, 261, 297, 304, 315, 348, 349 Pair, 314

minitnal, 314 niodular, 176

Papangelou, F., 122, 349 I’apert, D., 55, 110, 349 Pappus’ Theorem, 2 18 Partial algebra, 45

376 Index

l’artial lattice, 40 modular, 224 weak, 41

Partial operation, 40 Partial ordering relation, 1 Partial plane, 214 Partial transversal, 222 Partially ordered set, 1

I’artially ordered topologicd vector space, 312 Partition, 192

finite, 193

(See also Poset)

~ t - , 223 Partition lattice, 192 Pasch Axiom, 202 Pelczar, A., 55, 349 Pentagon, 59 Peressini, A. L., 312, 349 Perkins, P., 261, 349 Permuting congruence relations, 67, 149 Perniutti, R., 55, 349 Perspective, 129

down, 129 up, 129 weakly, 130

Perspective elements, 180 Perspective with respect to a line, 205 Perspective with respect to a point, 205 Perspectivities,

alternating sequence of, 164 normal sequence of, 164

Pierce, R. S., 123, 255, 349 Pinkser, A. G., 312, 349 Piron, C., 313, 349 Pitcher, E., 52, 349 Pixley, A. F., 124, 349 Planar iliagram, 10 Planar geometry, 221 Planar lattice, 51 Plane, 207

projective, 214 partial, 214 free projective, 216

Polynoniial, 27, 267 atomic Boolean, 70 binary, 27 Boolean, 69 lattice, 27 meet-, 69 minimal, 275 n-ary lattice, 27 rank of, 27, 269 unary, 27

Polynomial function, 27 Ponomarev, V. A., 219, 350 Poset, 1

boundeil, 47 dual, 2 graded, 178 of finite length, 2 sub-, 2 unordered, 2 width of, 2

Post, E. L., 126, 261, 350 Post algebra, 126 Potts, D. H., 261, 350 Power,

direct, 33 ultra, 235

Pregeometry, 185 Prenowitz, W., 221, 350 Preserving map,

Priestley, H. A., 123, 315, 350 Prime dual ideal, 19, 100

completely, 238 Prime filter, 19 Prime ideal, 17, 100

maximal, 67 minimal, 67, 115

Prime interval, 94 Prime over I, 231 Prime product, 231 Prime quotient, 130 Primitive set, 135 Principal congruence relation, 74 Principal dual ideal, 19

order-, 16

Platt, C. K.., 54, 57, 58, 272, 309, 313, 314, 315, 349, 350 Poguntke, W., 55, 261, 315, 350 Point, 202 Duality, 3, 6

perspective with respect to, 206 Polin, V., 309

Principal filter, 19 Principal ideal, 17 Principle,

Product, ,!?-reduced free, 289

Index 377

Product, Cartrsian, 23 tlirrct, 23 free K-, 105, 267 free K {0,1}-, 106 free {0,1}-, 100 frpe {0,1}- distributivc, 106 prime, 231 reduced, 23 1 ultra, 231

Existence Tlicorein for Frer. 110 Structure Theoretii for Free. 272

Subdirect, 11 7

Produc te,

Product Representation Throrrtii,

Product topology, 108 I’rojective quotients, 129

Projective algebra, 124 Projective elements, 180 Projective Geometry, 204

Coordinatization Theorem of, 208 Projective plane, 214

free, 216 nontlegenerate, 204

Projective space, 202 nondegenerate, 204

Projectivity Property, 158 Proper dual itltd, 19 Proper ideal, 17 Proper lower i-cover, 268 Proper quotient, 129 Proper sublattic.cl, 68 Proper upper &cover, 268 Property,

weakly, 130

Atiialgamation, 253 Coininon Refinement, 277 Congruence Extension, 120 Congruence Substitution, 122 Embedding, 253 Projectivity, 158 Strong Amalgarnat ion, 253 Substitution, 20 Weak Amalganiation. 259

relative, 51

lattice with, 11 1

Pseudocomplement, 48

Pseudoconiplrmen tat ion.

I’seudocomplementetl lattice, 48, 111

I’seudoco tnplemen ted meet-seinilttt tice.

l’seudoconipleinente~ semilattice, 48 PutllBk, P., 225, 311, 312, 350 l’ultr, A., 306, 350

rrlatively, 51

Quackenbush, It. W., 53, 54, 123, 124, 262, 312, 313, 315, 350, 351 Quasi-ordering relation, 8 Quotient, 129

pririie, 129 proper, 129 sub-, 129

Quoticmt latticae, 21

R-generated Boolean lattice, 97 R-generated distributive lattice, 92 R-generated sublattice, 87, 97 Rajagopalan, M., 126, 351 Itaney, G. N., 123, 361 Rank of countable Boolean algebra, 96 Rank of polynomial, 27, 269 Hank of subspace, 185 Itao, V. V. R., 125, 351 Rasiowa, H., 126, 351 Reduced free product,

Reduced product, 231 Redunclant representation, 62 Refinement Property,

Common, 277 Reflexivity, 1 Refracted line, 21 7 Regular equational class, 303 Reichel, H. C., 123, 351 Relation, 1

betweenness, 52 binary, 1

comparability, 9 congruence, 20, 57, 81, 111

distributive congruence, 149 equivalence, 20 ideal kernel of a congruence, 22, 133 join-congruence, 25, 109 join-representable congruence, 1 35 inept-representable congruence, 135 neutral congruence, 149 order-representable, 138

g-, 289

C-, 289

covering, 10 0

378 Index

Relation perniuting congruence, 67, 149 partial ordering, 1 principal congruence, 74 quasi-ordering, 8 representable congruence, 135 separable congruence, 155 standard congruence, 149

congruence, 73 Relation generated by H,

Relative complement, 47 Relative pseudocomplement, 51 Relative sublattice, 40 ltelatively algebraically closed subfield, 189 Relatively atoniic lattice, 29 7 Relatively complemented lattice, 48 Rclatively pseudocomplemented meet-semilattice, 51 Itennie, B. C., 312, 351 Representable congruence relation, 135

join-, 135 meet-, 135 order-, 138

Representable lattice, strongly, 3 12

Representation, 194 canonical, 280 direct join, 218 irredundant, 62 meet-, 194 minimal, 275 redundant, 62 subdirect, 11 7 type 1, 198 type 2, 197 type, 3, 194

Representation Theorem, Subdirect Product, 11 7

Representative, 204 Retract, 259 Reznikoff, T., 1C1, 351 Khodes, J. B., 55, 351 Itibenboim, P., 53, 351 RieEan, J., 53, 351 Rieger, L,, 65, 99, 124, 351 Hiesz, F., 312 Right bracketing, 14 Rigid, 126, 295

mutually, 296

Ring, arithmetical, 124 Boolean, 78 division, 204

€Ling of sets, 62 Ringel, C. M., 219, 351 Rival, I., 17, 51, 54, 55, 68, 121, 171, 218!

224,263,315, 352 Roberts, I?. S., 51,53,352 Robinson, A., 258, 352 Rolf, H. L., 280,352 Rosenberg, I., 223, 352 Rosenbloom, P. C., 126, 352 Rosicky, J., 312,352 Rota,G.-C., 3, 188, 191, 222,223, 352 Rousseau, G., 126, 352 R.-Salines, B., 55, 353 Hudeanu, S., 53, 124, 353

Sachs, D., 54,196,223,353 SaliK, V. N., 297, 304, 353 Sampei, Y., 123,353 Sands, B., 55,305,309,353 Sasaki,U., 180,221,223,297,304, 353 Sauer, G., 219, 353 Schein, B. M., 55,353 Schmidt, E. T., 20, 55, 57, 68, 74, 78, 81, 84, 85, 86, 97, 98, 113, 115, 119, 122, 125, 127, 130, 134, 139, 140, 142, 145, 147, 149, 155, 157, 158, 159, 160, 171, 220, 223, 225, 311,354 Schiitzenberger, N., 198,205,261, 354 Schwan, W., 170,354 Scott, D. S., 260, 354 Second Isomorphism Theorem, 116, 151 Sectionally Complemented lattice, 85, 135 Seibert, W., 219,354 Semiassociative identity, 15 Semigroup,

Semilettice, 7 standard, 260

distributive join-, 99 join-, 7 meet-, 7 modular join-, 109 pseudocomplemented, 48 relatively pseudocomplemented meet-, 51 sharply transferable, 57 transferable, 56

Semimodular lattice, 172

Index 379

s:untl~llce, 235

Separable congruence relation, 166 Seqncmce of prrspectivities,

uni\ ersal, 249

alternating, 164 nortnal, 164

sewin, L. N., 54,364 Sharp, H., 121, 354 Sharply K-transferable lattice, 314 Sharply transferable lattice, 57, 31 3 Sharply transferable semilattice, 57 Shearing identity, 161 Sholantler, M., 5 3 , 354 Sichlw, J. , 223, 277, 295, 297, 299, 302, 303, 306, 307, 308, 309, 316, 364 Sik, P., 315,354 Sikorski, It., 121, 123. 126, 355 Silcock, H. L., 124, 356 Siinplr lattic-r, 153 Sioson. 1”. M., 53. 356 Sltala, H., 250, 313, 355 Skrletal elerneiit, 112 Skcleton, 112 Skolein, T., 125 Skornjakov. L. A., 224,365 Smiley, M. F., 52. 355 Sinith, F. A., 158, 355

Sorkin, Ju. I., 52, 280, 287, 355 Sorkin’s Theoreni, 306 Sbs, V. T., 224, 319, 355 Spanned by X.

Solovay, It. M., 355

subspace, 184 spans, 176, 1x4 Spec.truiir of a lattice, 54 Speed, T. P., 123, 124, 356 Spencer, J. H., 126, 3.56 Sperner, E., 188, 356 Sperner’s Lemma, 188 Splitting, 239 Splitting lattice, 239 Splitting Theorem, 272, 303 Standard congruence relation, 149 Standard element, 138

Standard ideal, 146 Standard ideal of order n, 159 Standard semigroup, 260 Stanley, R. P., 121. 356 Steiner, A. K., 312, 356

dually, 138

Stcintz-MacLane Exchange Axiom, 178 Steven, D., 121,356 Stone, M. H., 63, 64, 65, 78, 99, 103, 104, 356 Stone algebra, 112 Stone identity, 112 Stone lattice, 112 Stone space, 100,106 Strong Amalgamation Property, 253 Strongly amalgamated, 252 Strongly maximal chain, 96 Strongly representable lattice, 312 Structure Theorem of Free Products, 272 Subalgebra, 48, 11 1

Subbase, 108 Subdirect Product Represent at ion Thro- rein, 11 7 Subdirect representation, 11 7

fully invariant, 299 Subdirectly irreducible, 116

finitely, 235 Sublattice, 16

convex, 1 7 complete, 156 Frattini, 54 generated by a subset, 1 7 proper, 68 R-generated, 87,97 relative, 40

Submodule, 204 Subposet, 2 Subquotient, 129 Subset,

generated by a subset, 110

P911-9 47

component, 287 convex, 1 7 generating, 1 7 hereditary, 44, 61

Subspace, 184 rank of, 185 dimension of, 185 linear, 203

Subspace spanned by X, 184 Substitution Property, 20

Supreinurn, 2 Swamy, U. M., 125,356 Symmetric difference, 72

Congruence, 122

380 II ldRX

Sy t ninetric lattice, N-, 176

To-space, 108 TI-space, 108 T2-space, 108 Table,

join, 9 meet, 9

Takeuchi, K., 68,219,356 Tainari, ?., 14, 356 Tamaschke, O., 220, 356 Tainura, S., 120, 356 Tan, S. K., 54,356 Tan, T., 127, 314, 315, 357 Tanaka, T., 159, 357 Tarski, A., 53,55,92, 97,98, 230,261, 357 Thickness of subset, 189 Thomason, S. K., 223,357 Thrall, R. M., 357 Tihonov’s Theorem, 109 Tolkien, J. R. R., 357 Topological lattice, 3 12 Topological space, 107 Topological vector space,

partially ordered, 3 12 Topology,

intrinsic, 312 product, 108

Totally disconnected space, 104, 109 Traczyk, T., 126,357 Transcue, W. R., 52,357 Transferable lattice, 56, 313 H-, 314 sharply, 57, 313 sharply K-, 314

sharply, 57 Transferable semilattice, 56,

Transitive complementation, 297 Transitive extension of binary relation, 8 Transitivity, 1 Transposition, 196 Transversal,

Tree, 200 Triangle, 205, 297, 299

Triangle connected, 297, 301 Triangle inequality, 194 Triple, 11 3, 258

partial, 222

diamond associated with, 299

Trivial block, 192 Trivial equational class, 32 Trotter, W. T., Jr., 357 Troy, L. A., 122,357 T h l a , J., 225 Tutte, W. T., 222, 357 Tverberg, H., 121, 358 Type of algebra, 110 Type 1 representation, 198 Type 2 representation, 197 Type 3 representation, 194

UDE, 243 Ultrafilter, 19 Ultra product, 231 Ultra power, 235 Unary operation, 52 Unary polynomial, 27 Unimodal, 188 Unit, 2 Uniquely complemented lattice, 293 Universal, m-, 304

Universal algebra, 52 type of, 110

Universal disjunction of equations, 243 Universal sentence, 249 Unordered poset, 2 Upper bound, 2

Upper continuous lattice, 189 Upper Covering Condition, 162 Upper i-cover, 268

Urquhart, A., 127,315,358

least, 2

proper, 268

V-formation, 252 finite, 255 finitely genrated, 255

Vaidyanathaswamy, R., 312,358 Varadarajan, V. S., 313, 358 Variable,

Variety,

Varlet, J., 56, 115,125, 126, 358 Vaught, R. L., 123,358 Veblen, O., 208. 358 Vector space,

free, 235

(See Equational class)

partially ordered topological, 312

Index 381

Villii~liii, V., 315, 358 Vopt.iiktt, P., 306, 358 Vulih, R. Z., 312, 358

WA lattice, 250 Waerclrn, 13. L. van tlrr, 190, 358 Wdlttce, A. D., 312 \VaIltiian, H., 312, 359 \Vang, Shih-chiang, 158, 359 Ward, &I., 54, 57, 218, 304, 359 Watcrman, A. G., 303, 359 \Vmk A ~nalgamation Property, 253 Weak partial Iatticp, 41 1Veakly associati\re lattice, 250 Weakly moclulttr lattice, 134 Wrakly perspective quotients, 130 Weakly projective quotients, 130 Weisner, L., 191, 359 Werner. H., 124, 359 Whalry, T. I]., 54, 359

221, 224, 241, 242, 250, 260, 263, 299, 303, 359, 360 Wilson, R. J., 222, 360 Wilson. It. M., 222, 360 Wolk, R., 57, 124, 170, 171,312, 360 Wolk, E. S., 55, 360 Wonenburger, M. J., 360

Wrbnski, A., 315, 360

Yasuhara, M., 258, 261, 361 Young, W. H., 208,361

Zacher, G., 55, 361 Zappa, G., 220,361 Zero, 2 0-distributive lattice, 126 { 0)-honiomorphism, 47 { 0,i }-distxibutive product,

{O,l }-hoinotiiorphisrii, 47

\\'Tong, Yau-chuen, 31 2, 360

free, 106

Whitinan, Ph. AT., 31, 55, 194, 196, 223, 269, {O,l}-product, free, 106

Whitney, H., 179, 359 free K, 106 Width. 2 {O,l}-sublattice, 47 Wilcox, L. R., 176, 177. 190, 220, 359 Zilber, J., 54 Wille, H., 53, 54, 124, 135, 136, 218, 219, Zorn's Lemma, 63

272, 274, 282, 285, 286, 303, 305, 359

General Lattice Theory George Gratzer

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