Acta Math. Hungar.66 (1995), 275-288.

A LATTICE CONSTRUCTIONAND CONGRUENCE-PRESERTvING

EXTENSIONS

G. GRATZERl (Winnipeg) and E. T. SCHMIDT2 (Budapest)

1. Introduction

To find a simple proof of the congruence lattice characterization theoremof finite lattices, H. Lakser and the first author (see (1]) introduced a specialtype of finite partial lattices: a meet-senlilattice in which any two elementswith a common upper bound have a join. If M is such a partial lattice,then the ideal lattice ofM is a congruence-preserving extension of M; that

every congruence of M has exactly one extension to the ideal lattice.In [2], we introduced the llalne chopped lattice for such lattices,

no longer necessarily finite. Of course, if lit is no longer we cannoteXlpe(~t the ideal lattice to a congruence-preserving extension. It is naturalto consider, instead, finitely ideals; unfortunately, they do not, ingeneral, form a lattice. In 2 we introduce Condition (FG) underwhich the finitely generated ideals form a lattice.

Given two lattices A and B, sharing the sublattice C == An B, we obtainthe lattice M (A, B) by amalgamation. If C is a principal ideal of both Aand B, then M(A, B) is a chopped lattice.

In Section 3, we introduce (see Definition 3) a set of sufficient conditionsM(A, B) is a chopped lattice. If A and B the conditions

of Definition 3, we shall call A, B a chopped pair. Theorem 1 states that ifA, B is a chopped pair, then M(A, B) is a chopped lattice. The concept ofa chopped pair does not seem enough to compute with In Section

we introduce two stronger versions: sharp and full chopped pairs.In Section 5 we investigate finitely generated ideals in M(A., B) for a

chopped pair A, B. For a sharp chopped pair A and B, ifCAn B satisfiesthe Ascending Chain Condition, then we obtain Condition (FG) (whichguarantees that the finitely generated ideals form a lattice) for M(A, B).

In Section 6 we investigate modular lattices. If A, B is a sharp choppedpair and both A and B are modular, then M(A, B) satisfies Condition (FG)

1 The research of the first author was supported by the NSERC of Canada.2 The research of the second author was supported by the National Foun-

dation for Scientific Research, under Grant No. 1903.

© 1995 Akademiai Kiado, Budapest

276 G. GRATZER and E. T. SCHMIDT

(Theorem 3). If A, B is a full chopped pair, then it is enough to assume thatone of them is modular to obtain the same conclusion (Theorem 4).

In Section 7 we deal with the problem whether every lattice has a propercongruence-preserving extension. We apply Theorem 4 to prove that if thereexists a nontrivial distributive interval in a lattice, then it has a proper con­gruence-preserving extension.

A modular example of a congruence-preserving extension is outlined inSection 6.

1.1. Notation. We refer the reader to (1] for the basic concepts andnotation.

In a lattice L, (x, y]L denotes the interval in L, and (alL the principalideal generated by a. If there is no confusion, the subscript is dropped.

If L is a sublattice of K, then we call K an extension of L. If L has azero, and it is also the zero of K, then K is {OJ-extension of L.

2. Chopped lattices

A chopped lattice M is a lattice L with zero, 0, and unit, 1, with the unitremoved: M = L - {I}; on M, 0 is a nullary operation, A is an operation,and V is a partial operation. Equivalently, a chopped lattice M is a meet­semilattice with zero, 0, in which any two elements having an upper boundhave a join. M will be regarded as a partial algebra (M; A, V, 0).

We shall use the concept of extension for chopped lattices; observe that,by definition, an extension of a chopped lattice isa {OJ-extension.

An ideal I of M is a subset of M containing 0 with the following twoproperties for x, y EM:

x E I and y ~ x imply that y E I.If x, y E I and x V y exists, then x V y E 1.

For H ~ M, there is a smallest ideal (H] of M containing H. If an idealI can be represented in the form (H] for some finite set H, then the ideal Iis called finitely generated. In particular, for a EM, we let (a] = ({a}] bethe principal ideal generated by a in M, that is,

(a] ={x I x E M and x ~ a}.

Id M denotes the lattice of ideals of M. 0 bviously, Id M is a lattice.Idfg M, the finitely generated ideals of M, form a join-sublattice of Id M.

By identifying a E M with (a], we regard Id M an extension of M.

DEFINITION 1. A chopped lattice M satisfies Condition (FG) if everyfinitely generated ideal is a finite union of principal ideals.

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A LATTICE CONSTRUCTION AND CONGRUENCE·PRESERVING EXTENSIONS 277

U ... U (an]., J:::::. (bI ]

deed, ifI:::::.

(FG), Idfg M is a sut)lalttl(~e

thenInJ:::::. U«ai A 11 < '< ·1< '<••. := ~ := 'n, .:= J :=

tolloWlnf! two len(lmas.

a chopped every congruence ret('l,r;eo~n

of M to IdM.

PROOF. Let e be a congruence of M; define a relation e on Id M asfollows:

I= J 0)

(a] n I is prlIllClp;a.l.

i =. j (mode),

IEnt"'!O~''l;rL!l that if a E

I:::::. (al]U U (an],

~eIleri:LLt;~U by {a A aI, , a Ait has a b MnJ .

.................4.~" ..."...I... of E> to JE S, I=. J .),=. J A (a] (mod .). By the statement previous

is a E J suc~ that (a] A J_ :::::. (b]; obviously, a =. b (mod 9).I=. J E». • 9._1, J E S with J J e). By the assumption on

as

Indeed,

IS an "'......".........."".Lv.u

congruencesso every

Therefore,

is a Ci

there is a dj Icongruences hold for •.

J (mod.), eM has a unique extension to

1'nlln"futnn' is true:

(modS).

•join of n+ mThus 8 = 9,o

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278 G. GRATZER a.nd E. T. SCHMIDT

LEMMA 4. Let M be a chopped lattice satisfying Condition (FG). ThenIdfg M is a congruence-preserving extension of M.

In fact, a congruence-preserving {O}-extension.

3. Chopped pairs

Let A and B be lattice, let C = A n B f:. 0. Then we can form theamalgamation M =M(A,B) of A and B over C. It is well-known that onM we can define a partial ordering:

DEFINITION 2. The partial ordering ~M is defined on M as follows:(1) For x, yEA, let x ~M y iff x ~A y.(2) For x, y E B, let x ~M y iff x ~B y.(3) For x E A and y E B, let x ~M 11 iff there exists acE C such that

x ~A c and c ~B y; and symmetrically, for 3( E Band 11 E A.

The subscripts of ~ will be dropped whenever there is no danger ofconfusion.

We shall use the following notation: M(A, B) = AU B is the posetobtained by amalgamating A and B over C. In A we form the ideal fAgener-ated by C; we set CA. = fA - C; symmetrically, we define fB and CBoNote that the ideal CM generated by C in M is the disjoint union of C, CA,and CB.

Sometimes, the poset M(A,B) is a chopped lattice. The next definitionformulates some natural conditions under which this is the case.

DEFINITION 3. A pair of lattices A and B is called a chopped pair iffthe following conditions are satisfied:

(1) The lattices A and B have a common zero, O.(2) Let C denote the lattice An B. Then C has a largest element i.(3) For x E CM,. there is a smallest x E C satisfying x ~ X.(4) For x E M(A, B), there is a largest!. E C satisfying!. ~ x.(5) For x E CA and y E CB, the two elements: x V fj (formed in A) and

x V y (formed in B) are comparable (in M(A, B».(6) For x E A - Band y E B - A, the two elements: x Ay (formed in A)

and!.A y (formed in B) are comparable (in M(A,B». -

THEOREM 1. Let A, B be a chopped pair. Then M(A, B) is a choppedlattice.

PROOF. There are two claims to verify.Claim 1: M(A, B) is a meet-semilattice. Let x, y E M(A,B). We have

to find u = infM(A,B){x, V}. We shall distinguish several cases.Case 1.1: x, yEA. Let u =x Ay be formed in A. Obviously in M(A, B),

u ~ x and u ~ y. Now let 'Ii E M(A, B) be a common lower bound of x andy. There are two subcases to consider.

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A LATTICE CONSTRUCTION AND CONGRUENCE.PRESERVING EXTENSIONS

Case v E A. By Definition v is a common bound of xand 11 A, v 'U.

Case LIb: v E By Definition 2.3, are elements Cx and cyin C such that v ~B Cx ~A x and v ~B cy ~A y. Then Cx 1\ cy E C, andv Cft A cy u. So indeed, u = infM (A,B){X,1I}.

Case 1.2: y E B. Proceed as Case 1.1.x E 11 E B. In of n'l""Ci'U"",Uc> cases, we can assume

tha.t x E A - B lIE B - A. by 2.3, any commonbound must be in CM, we can replace x by x A i and y by y Ai. So againreferring to cases, we can assume xE CA and 11 E Nowtake a common bound v of x 11.

Now we claim of the common bounds vE A,one, x A Indeed, x A 11 is a lower bound. If t Ethent 11 in M(A, B), hence by Definition 2.3, is acEt ~A C y. Obviously, C ~ y, and sot x 1\ y, as claimed.

Now we claim that of the common lower bounas v E B, there is a largestone, ~ A y' To proceed as previous paragraph.

Finally, by Definition 3.6, x 1\ 11 and ~ A 11 are comparable, hence

inf~~)li/'l E B, lIa:~~t;::::~A ~~3.completes the proof of Claim 1.

Claim InM(A, B), any two elements, x and having a common upperbound, v, have a join. Let x, yE .M(A, B), v be an upper boundx andy. We have to find u = SUPM(A,B){X,y}. We shall distinguishcases.

Case x, 11 E A. Form u =x Vy A. We have to show that if t isany upper bound of x and y in M (A, B), then u ~ t.

Case 2.1a: tEA. This case is obvious.2.1b: t E B. By Definition

x Cft t 11 ~A cy ~B t.again, by 2.3, U ~M(A,B) t, COlnp,Letllng

Case 2.2: x,y E B. Proceed as in Case2.3: x E 11 E B. In view of Cases 2.1-2.2, we can assume

that x E A - Band y E B - A. Without generality, we can assumethat t E It that 11 E we distinguish two subcases.

2.3a: x E If tEA is an bound of xy, x VV1i ~ t. Similarly, if t E B is an of x and 11, then x VlI ~ t. ByDefinition 3.5, the elements x V fi and x V yare comparable, hence,

SUP{x,lI} = V xVy}.

Case2.3b: x ¢ case, no upper bound of x issup{x, y} =x Vfi Tn'l"'fnOn

Case 2.4: x E B 11 E A. as Case 2.3.This completes the proof of Claim 2 and lemma. 0

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280 G. GRATZER and E. T. SCHMIDT

4. Some examples and special cases

It is easy to give examples that the last two strange conditions of Defi­nition 3 do not follow from the others. Here is one: let A = B be the directproduct of the two element chain {O, I} with the three element chain {O, a, I}.The elements are of the form (x, y), where x E {O, I} and y E {O, a, I}. Wemake A and B disjoint (we shall denote (x, y) E A by (x, Y)A' and the samefor B), then we identify elements as follows:

(0,0)A with (0,0)B;

(l,O)A with (O,l)B;(O,l)A with (l,O)B;(l,l)A with (l,l)B'So C = {(O, O)A' (1, O)A, (0, l)A' {I, l)A} is a four-element Boolean lattice.

It is easy to see that Definitions 3.1-3.4 hold, but both Definitions 3.5 and 3.6fail. Indeed, let x = (a,O)A E CA and Y = (a,O)B E CB. Then x = (l,O)Aand y = {l,O)B = (O,l)B' Hence,

xVy= (a,l)A andxVy= (a,l)B'

and these two elements are not comparable.If A, B is a chopped pair, then we know that in M(A, B) any pair of

elements with a common upper bound has a join. To perform computationswe need more; we must have a formula for the join we can work with.

DEFINITION 4. A chopped pair of lattices, A and B, is called sharp iff

x V Y =xV y,

for x E CA and y E CB, and

x 1\ 1!.. =~ 1\ y,

for x E A - Band y E B - A.

There are many equivalent forms of these conditions; for instance, thefirst is equivalent to

x Vy E C,

for x E CA and y E CB; or to

x Vy =xVy.

Observe that if A and B form a sharp chopped pair, then in M(A, B),we have x 1\ y E C, for x E CA and y E CB; and x V y E C, for x E CA andy ECB·

Two important examples of chopped pairs follow in which C is largestand smallest possible:

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A LATTICE CONSTRUCTION AND CONGRUENCE-PRESERVING EXTENSIONS 281

EXAMPLE 1. C· = (i] is a principal ideal of both A and B.We considered this special case for finite lattices in a previous paper [2].

In this case, CA == CB == 0; for every x E M(A, B), ~ =x 1\ i; and for everyx E C == CM, x == x. The conditions of Definition 3 and Definition 4 aretrivially satisfied - in fact,

x V Y =x V y = x V y and x A '!l. =~ 1\ Y = x 1\ Y 1\ i.

EXAMPLE 2. C ={O, i}.In this case, again, the conditions of Definition 3 are trivially satisfied ­

in fact,x V y =x V y == i and x A'!l. =~ A Y == O.

In these two examples, the conditions of Definition 3 and Definition 4hold in a much stronger form.

We name the first example:

DEFINITION 5. A chopped pair of lattices, A and B, is called full ifC =(i]A =(i]B·

5. Finitely generated ideals

In this section, we shall investigate conditions under which M(A, B)satisfies Condition (FG). The following two lemmas are easy to verify, butthey are crucial to our investigations. First some definitions.

DEFINITION 6. Let A, B be a chopped pair, C =An B. Let a E A - Cand b E B - C. We define the elements:

ao= a,

bo=b,

b1= bo V ao 1\ i

al= ao V b1 A i

b2=b1 Val Ai bVal 1\ i)

a2= a2 V b2 A i (= a V bl A i)

bn+1= bn Va;:Ai (= b V a;:Ai)

an+l= an V bn+1 1\ i (= a V bn+1 A i)

(formed in B),

(formed in A),

(formed in B),

(formed in A),

(formed in B),

(formed in A),

See Figure 1 - the white filled elements are in A (and maybe in C); theshaded elements are in B (and maybe in C), and the black filled elementsare in C.

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282 Go GRATZER a.nd Eo To SCHMIDT

e i

1

5. Let A B a choppedfollowing inequalities hold:(1) a == ao ~ al ~ a2 ~ ... (in A),

b bo ~ bI ~ b2 ~ ••• (in B),

M(A,B), the

and<---.'<~ A = U2 A 1, = ...

bn+l ;

and an+2, it follows== It is now

Therefore,

- ... ,

= b V an 1\ i

definition of--..--:-==an+2 A i, so

allPROOF. Let an

bn+2 ;::;: b V an+l

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A LATTICE CONSTRUCTION AND CONGRUENCE-PRESERVING EXTENSIONS 283

Finally,an A. i ~ bn+1 A. i ~ an+l A. i ~ bn+2 A. i,

and bn+1 A. i =bn+2 A. i; therefore,

an A. i = bn+1 A. i = an+1 A. i = bn+2 A. i, ... ,

so sequence (3) also terminates. Conversely, if sequence (3) terminates,then sequences (1) and (2) terminate by the definitions of an+l and bn+1in Definition 6. 0

LEMMA 6. Let A and B be a sharp chopped pair; let a E A - C, b EE B - C. The ideal (a,b] of M(A, B) generated by {a,b} can be described asfollows:

(a, b] =U((an]A I n < w) U U((bn]B I n < w) .

This is not a finitely generated ideal if, and only if, none of the sequences ofLemma 5 terminate. If (a, b] is a finitely generated ideal, then (a, b] = (an] UU (bn] for some n < w.

PROOF. Let R = U«an]A In < w) U U«bn]B In < w). If we know thatR is an ideal of M(A, B), then it is straightforward to verify that R is theideal of M(A,B) generated by {a,b}, and the rest follows from Lemma 5.

So we verify that R is an ideal of M(A, B).Firstly, let x E Rand y ~ x in M(A, B). Without loss of generality

we can assume that x ~ an for some nand y ~ x. If YEA, then y ~ an;therefore y ~ an in A, and so y E R. If y E B, then y ~ an, and so y ~ an A." i ~ bn . This implies that y ~ bn in B, therefore y E R; completing theproof of y E R.

Secondly, let x, y E R, and let x and y have a common upper bound zin M(A, B). Without loss of generality we can assume that z E A. We wantto show that x V y E R. We shall distinguish several cases.

Case 1: x, yEA.Case 1.1: x ~ an and y ~ am for some nand m.In this case, as in all

the subsequent cases, we can assume without loss of generality that n =m.Then x V y ~ an, so x V y E R.

Case 1.2: x ~ an and y ~ bn. Since yEA and bn E B, the conditiony ~ bn implies that y ~ i. Hence, y ~ bn " i ~ an, and so x Vy ~ an, yieldingxVYER.

Case 1.3: x ~ bn and y ~ an' Proceed as in Case 1.2.Case 1.4: x ~ bn and y ~ bn • As in Case 1.2, we can verify that x ~ an

and y ~ an, so Case 1.1 completes this case.Case 2: x E A, y E B. Observe that y ~ i since y ~ Z, Y E Band Z E A.

Case 2.1: x ~ an and y ~ an' So x V y = X VY ~ an, hence x V y E R.Case 2.2: x ~ an and y ~ bn. Since y ~ i, it follows that y ~ bn " i, so

y ~ an; hence x V y ~ an, yielding x V y E R.

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284 G. GRATZER and E. T. SCHMIDT

an' Using the argument of Case 2.2, wewhich is symmetric to Case 1.1. Hence

Case 2.3: x ~ bn and y ·~an' Proceed as in Case 2.2.Case 2.4: x ~ bn and y ~ On. Then as in 2.2, x ~ an and y ~ an,

so we can proceed as in Case 1.Case 3: x E B, yEA. This is symmetric to Case 2.Case 4: x, yE B.

Case 4.1: x S an and""V'" ............ that x ~ bn';;' and yx Vy E R.

Case 4.2: x ~ an and y ~ On' Again, x E B and x ~ an imply thatx ~ 011.+1, which is to Case 1.1.

Case 4.3: x andy ~ an' Proceed as CaseCase 4.4: x and y ~ On' This is symmetric to Case 1.1. 0

Observe that this lemma fully describes all finitely generated ideals, sincea finitely generated ideal M(A, B) is obviously one- or two-generated.

Now we prove:

THEOREM 2. Let A and B form a sharp chopped pair,and let C =Ann B. us assume that C satisfies the Ascending Chain Condition. ThenM(A, B) satisfies condition (FG), and Idfg M(A, B) is a congruence-preserv­ing extension of M(A, B) (in fact, a congruence-preserving {OJ-extension).

PROOF. If C the Ascending Chain Condition, then sequence(3) of Lemma 5 must By Lemma 5, sequences (1) and (2)termulate, and so the statement of the Theorem follows from Lemma 6.

the statement congruence-preserving extension fol-lows from Lemma 4. 0

For full chopped pairs, Definition 6, Lemma and Lemma 6 take on amuch simpler form:

DEFINITION A, B be a full chopped pair, C = A n B. Let a EE A - and 0 E B - C. Then we define the elements:

ao= a,b,boY A

al= ao V (01 A

V A i)a2= al V (02 A i)

oV (al 1\

a V (01 1\ i)),

bn+1=bn V (an 1\ i) (= bv (an 1\ i»,an+l= an V 1\ i) (= a V (bn+1 1\ i»,

Figure 2 - elements are A (and maybe in C); theshaded elements are in B (and maybe in C), and the black filled elementsareActa. Ma.tkematica H1LngaricQ, ee, HJSS

A LATTICE CONSTRUCTION AND CONGRUENCE-PRESERVING EXTENSIONS 285.i

Fig,2

LEMMA 7. Let A and B be a full chopped pair. Then in M(A, B), thefollowing inequalities hold:

(4) a =ao ~ a1 ~ a2 ~ (in A),

(5) b =bo ~ bI ~ b2 ~ (in B),and

(6) . <b . < '<b' < . < (' C)ao 1\ ~ == 1 1\ ~ == aI 1\ ~ == 2 1\ ~ == a2 1\ ~ == ••• zn •

If, for some n, an == an+1, then (4) terminates at n, and (5) terminates atn +1; and symmetrically, for (5). If (6) does not terminate, neither do (4)and (5).

The proof of this lemma is a simplified version of the proof of Lemma 5.Lemma 6 remains valid for full chopped pairs; in this case, the sequences anand bn will be the ones defined in Definition 7.

6. Modular lattices

By inspecting Figure 1, we can see that if A and B are modular, then alot of elements must collapse. In fact, we have the following result:

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286 G, GRATZER and K T.. SCHMIDT

1)

3. A and B form a sharp CfU.»f)f)lea us assumethat both A and B are modular. M(A, Q"{<f1QlfI'~Q condition (FG),and Idfg M (A, B) is a congruence-preserving extension M( A, B) (in fact,a congruence-preserving {O}-extension).

PROOF. Let A and B modular. The equations (see

:= aoA A i):= «0 Ai,

hold M(A, B). By the ......'-" ............<.10....

a1 A i. Soof A, the two eq111ataOlls that

:= 61 Ai.

By the modularity of B, a similar argument yields that b2 i\. i a1 andso on. So the sequence (3) has only one or two members; it terminates.By 5, the sequences (1) and (2) terminate. So the statement of theTheorem follows from 6.

Finally, the statement congruence-preserving foI-Lemma 0

We can prove a "".,. .....'''' ......n .. statement full chopped pairs.

8. Let A, B a full chopped pair.. If A is a modular lattice,

b] := U (b1].

PROOF. As in Theorem 3, the modularity of A implies that b1 /\ i := al Ai.Hence b2 := b1 V (al ./\ i) := b1 V (b1 /\ i) := hI, and a2 := a1 V (b2 A i) := al VV (b1 A i) := al V (a1 A i) := al' So statement of the follows fromLl'='.U.I..l.l.lQ! 6. 0

So now we can a form of 3 for chopped

THEOREM 4. A, B be a fvJl chopped pair. If A isa modular lattice,M(A, B) satisfies condition (FG).

7. Congruence-preserving extensions

In we raised the following question:

Is it true every lattice with more one element hasa proper congruence-preserving extension?

proved in [2] the finite case is true. result isby the following tn€!or~~m:

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A LATTICE CONSTRUCTION AND CONGRUENCE-PRESERVING EXTENSIONS 287

THEOREM 5. Let L be a lattice with zero, 0. If there exists an elemento > °in L such that the interval [0,0] is distributive, then L has a propercongruence-preserving extension K.

PROOF. To prove this result, we need a construction due to the secondauthor. Let M 3 denote the five-element modular nondistributive lattice onthe set {O, a, b, c, 1}, and let D be a bounded distributive lattice. Let

M3 [D] ={(x, y, z) E D3 I x A Y =x A z =Y A z} .

Then Ms[D] is a modular lattice; it contains Ms as a {O, 1}~sublattice (onthe set { (O, 0, 0), (1, 0,0), {O, 1, O}, {O, 0, 1}, {1, 1, 1}} ), and each prime intervalof this Ms contains (in M3 [D]) a copy of D; for instance, the interval[{O, 0, O}, {1, 0, O}] can be described as {(d, 0, 0) IdE D}. If we identifyD with { (d, 0, 0) IdE D}, we find that the lattice Ms[D] is a congruence­preserving {O}-extension of D.

Now let D =[0, a], and let A =M3 [D]. Then A has a spanning Ms;let i = (a,O,O). Let B = L, and define i = a in B. Then An B = (i],and A, B.. form a full chopped pair in which A is modular. So we can formthe chopped lattice M(A, B). Obviously, M(A, B) is a proper congruence­preser.ving {OJ-extension of L. By Theorem 4, (FG) holds for M(A, B).Therefore, by Lemma 4, Idfg M(A, B) is a congruence-preserving {O}-exten­sion of M(A, B). We conclude that Idfg M(A, B) is a proper congruence-pre­serving {O}-extension of L. 0

The following result is a generalization of Theorem 5.

THEOREM 6. Let L be a lattice. If there exist a nontrivial distributiveinterval in L, then L has a proper congruence-preserving extension K.

PROOF. Let [a,,8] be a nontrivial distributive interval in L. Let us formthe lattice B = [a) in L. Obviously, B satisfies the conditions of Theorem 5;therefore, B has a congruence-preserving {OJ-extension Kl • Clearly, B is anideal of K I and a dual ideal of L; hence we can glue Land Kl over B; letK be the resulting lattice.

Let e be a congruence relation on L. Let eB be the restriction of eto B. Since K l is a congruence-preserving extension of B, there is a uniqueextension • of eB to K 1. It is easy to see that e = e u • is the uniqueextension of e to K. Hence K is a congruence-preserving extension of L.Obviously, it is a proper extension. 0

8. A modular example

It is easy to give examples of classes of lattices that have proper con­gruence-preserving extensions that have nothing to do with distributivity.

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288 G. GRATZER and E. T. SCHMIDT: A LATTICE CONSTRUCTION ...

For instance, every simple lattice with more than one element has a propersimple extension; this is obviously a proper congruence-preserving extension.

In this section we outline a modular example with no proper distributivesublattice.

Let C be a continuous geometry with zero, 0, and unit, 1. Then C hasthe following properties:

(1) For a < b, the interval [a,.b] is isomorphic to C.(2) C is a simple lattice.Let I be a nonprincipal ideal of C and F a nonprincipal dual ideal of C

satisfying In F == 0. Let L be the sublattice I U F. The congruence latticeof L is the three element chain.

We choose in Ca spanning Ma == {o < a,. b, e < 1}. The interval [0, a] isisomorphic to C. Therefore, we find in [0, a] a copy 10. of I and a copy Fa ofF. The projectivities in the spanning Ma define the ideals and dual ideals,Ib' Ie, Pb, Fe in the intervals [0, b] and [0, c]. Similarly, we obtain the idealT: and dual ideal F: in [a, 1], II: and Fb in [b, 1], 16 and F~ in [e,1].

Let I be the ideal of C generated by the three "small" ideals, la, Ib'Ie. Similarly, the three dual ideals Fi:, Fb,Fi generate a dual ideal F. Weconsider the sublattice

It is easy to see thatK is a sublattice of C, and it is a congruence-preservingextension of the sublattice L ~ [0, a].

Referenees

G. Gra.tzer, General Lattice Theory, Academic Press, New York, N. BirkhauserVerlag, Basel, Akademie Verlag, Berlin, 1978.

[2] G. Gra.tzerand E. T. Schmidt,The Strong Independence Theorem for automorphismgroups and congruence lattices of finite lattices. Submitted to Beitriige zurAlgebm und Geometrie.

(Received April

DEPARTMENT OF MATHEMATICSUNIVERSITY OF MANITOBAWINNIPEG,. MAN. R3T 2N2CANADA

DEPARTMENT OF MATHEMATICSTRANSPORT ENGINEERING FACULTYTE(::B:N!lCj~L UNIVERSITY OF BUDAPEST,.tt1l\r..,-'1l\"I'·"' ..... RKP. 91111 BUDAPESTHUNGARY

Acta Mathematica Hv,'ngarica 66, 1995

1993; revised February 8, J994)