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POLYNOMIALS IN FINITE GEOMETRY

Peter Sziklai

20 February, 2008

Foreword iii

-1 Special foreword

Before the real Foreword, let me say some words about this book. It is ready inthe sense that this is the topic that I wanted to write about and this is how Iwanted to write about it. Naturally, this book is not finished in the sense that (i)it is about an active field of mathematics, which changes rapidly; (ii) if you haveworked on a book for years then it is hard to stop it: every day you can have anew idea how to slightly improve it; (iii) no editor/referees have seen it yet, hencethere may (must!) be typos, inaccuracies, missing citations, line overflow errors...left. It would be nice to increase the number of figures as well.

However, I think that in the current state of this volume it is possible to decideabout its values and shortcomings. Although I understand that it can be evaluatedby this current version, in the spirit of (i) and (ii) above I will keep an updatedversion on the secret, unlinked webpagehttp://www.cs.elte.hu/˜sziklai/poly.html , (where I will enlist all thechanges performed, too), just for pleasure. When a decision of publication isreached, the editors of the publisher can decide between this and a possibly up-dated version.

0 Foreword

A most efficient way of investigating combinatorially defined pointsets of projectivespaces over finite fields is associating polynomials to them. This technique was firstused by Jamison and Bruen, then, followed by several people, became a standardmethod; nowadays, the contours of a growing theory can be seen already.

The polynomials we use should reflect the combinatorial properties of the pointset,then we have to be equipped with enough means to handle our polynomials and getan algebraic description about them; finally, we have to translate the informationgained back to the original, geometric language.

The first investigations in this field examined the coefficients of the polynomials,and this idea proved to be very efficient. Then the derivatives of the polynomialscame into the play and solving (differential) equations over finite fields; a thirdbranch of results considered the polynomials as algebraic curves. The idea of asso-ciating algebraic curves to pointsets goes back to Segre, recently a bunch of newapplications have shown the strength of this method. Finally, due to Gacs’s recentresults, dimension arguments on polynomial spaces have become fruitful.

This book is an attempt to collect and classify the most interesting ways of ap-plying polynomials in finite geometry.

We focus on combinatorially defined (point)sets of projective geometries. Theyare defined by their intersection numbers with lines (or other subspaces) typically,like arcs, blocking sets, nuclei, caps, ovoids, etc. Here we do not want to give a

iv

complete account on them, however some of the best results will be presented.In order to show a wider scope of different applications, some other fields aregently touched like group theory, graph theory, etc.This book is divided into chapters. The first one (Background) contains selectedtools for a finite geometer, from the basic facts to some theory of polynomials overfinite fields; proofs are only provided when they are short or interesting for us. Weprovide slightly more information than the essential background for the secondchapter. After the basic facts (Section 5) the most useful representations of affineand projective spaces are presented (6), then we introduce our main tool, the Redeipolynomial associated to pointsets (7,8). The coefficients of Redei polynomials areelementary symmetric polynomials themselves, what we need to know about themand other invariants of subsets of fields or spaces is collected in Section 9. Themultivariate polynomials associated to pointsets can be considered as algebraicvarieties, so we can use some basic parts of algebraic geometry (10). Finally, inSection 11 some background needed for stability results is presented.The second (and main) chapter contains several results of finite Galois geometry,where polynomials play a main role. We start with general results on intersectionnumbers of planar pointsets (12,13). Then we turn to special cases as arcs, max-imal arcs, unitals, semiovals, untouchable sets. (14-17). The next highlight is thetopic of directions (18,19) and blocking sets (20-22). Section 23 shows the newresultant method for stability results. Affine blocking sets and nuclei are consid-ered in sections 24,25. Section 26 introduces an interesting mixed representation.Stability theorems for flocks, some basic facts for spreads and a nice result onovoids is proved in Sections 27,28. Finally some non-geometrical applications arecollected in Section 29.The last chapter contains hints and solutions for the exercises.

Contents

-1 Special foreword . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii

0 Foreword . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii

1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

2 Acknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

3 Definitions, basic notation . . . . . . . . . . . . . . . . . . . . . . . 3

1 Background 5

5 Finite fields and polynomials . . . . . . . . . . . . . . . . . . . . . 5

5.1 Some basic facts . . . . . . . . . . . . . . . . . . . . . . . . 5

5.2 Self-dual normal bases . . . . . . . . . . . . . . . . . . . . . 6

5.3 Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

5.4 Differentiating polynomials . . . . . . . . . . . . . . . . . . 8

5.5 Polynomials vanishing at many points, Alon’s Combinato-rial Nullstellensatz and the Ball-Serra refinement . . . . . . 9

5.6 Additive, linearized polynomials . . . . . . . . . . . . . . . 16

5.7 The random-like behaviour . . . . . . . . . . . . . . . . . . 17

5.8 Lacunary polynomials . . . . . . . . . . . . . . . . . . . . . 19

6 Representations of affine and projective geometries . . . . . . . . . 22

6.1 Subspaces, subgeometries . . . . . . . . . . . . . . . . . . . 24

6.2 Transformations . . . . . . . . . . . . . . . . . . . . . . . . 25

7 The Redei polynomial and its derivatives . . . . . . . . . . . . . . 26

7.1 The Redei polynomial . . . . . . . . . . . . . . . . . . . . . 26

7.2 Differentiation in general . . . . . . . . . . . . . . . . . . . 30

7.3 Hasse derivatives of the Redei polynomial . . . . . . . . . . 32

7.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

8 Univariate representations . . . . . . . . . . . . . . . . . . . . . . . 35

v

vi Contents

8.1 The affine polynomial and its derivatives . . . . . . . . . . . 36

8.2 The projective polynomial . . . . . . . . . . . . . . . . . . . 37

8.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

9 Symmetric polynomials and invariants of subsets . . . . . . . . . . 39

9.1 The Newton formulae . . . . . . . . . . . . . . . . . . . . . 39

9.2 The invariants uf , vf and wf . . . . . . . . . . . . . . . . . 43

9.3 Arbitrary subsets . . . . . . . . . . . . . . . . . . . . . . . . 47

9.4 The generalized formulae . . . . . . . . . . . . . . . . . . . 51

9.5 Resultants . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

10 Elements of algebraic geometry . . . . . . . . . . . . . . . . . . . . 61

10.1 Conditions implying linear (or low-degree) components . . . 65

11 Finding the missing factors, removing the surplus factors . . . . . . 68

2 Polynomials in geometry 71

12 Lines meeting a pointset . . . . . . . . . . . . . . . . . . . . . . . . 71

12.1 Prescribing the intersection numbers . . . . . . . . . . . . . 71

12.2 A general result . . . . . . . . . . . . . . . . . . . . . . . . . 72

12.3 Another immediate corollary . . . . . . . . . . . . . . . . . 73

13 Sets with constant intersection numbers mod p . . . . . . . . . . . 74

13.1 Sets with intersection numbers 0 mod r . . . . . . . . . . . 76

13.2 Small and large super-Vandermonde sets . . . . . . . . . . . 80

14 Arcs, Segre . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

15 Maximal arcs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

15.1 Existence: hyperovals . . . . . . . . . . . . . . . . . . . . . 89

15.2 Non-existence . . . . . . . . . . . . . . . . . . . . . . . . . . 93

15.3 Embeddability . . . . . . . . . . . . . . . . . . . . . . . . . 96

16 Unitals, semiovals . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98

17 Sets without tangents: where Segre meets Redei . . . . . . . . . . . 102

17.1 The proof using Segre’s method . . . . . . . . . . . . . . . . 103

17.2 Untouchable sets in Galois planes of even order . . . . . . . 109

18 Directions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110

18.1 When D is (contained in) a subgroup of GF(q)∗ . . . . . . . 112

18.2 When D is (contained in) a subgroup of q+1√

1 ⊂ GF(q2)∗ . 115

18.3 Sets determining ≤ q+12 directions: affine linear pointsets . 115

18.4 Small blocking sets of Redei type are linear . . . . . . . . . 116

Contents vii

19 Over prime fields: the Gacs method . . . . . . . . . . . . . . . . . . 122

19.1 Permutation polynomials . . . . . . . . . . . . . . . . . . . 124

19.2 Directions in the plane . . . . . . . . . . . . . . . . . . . . . 125

19.3 Linear combinations of three permutation polynomials . . . 131

20 Projective linear pointsets . . . . . . . . . . . . . . . . . . . . . . . 134

21 Blocking sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136

21.1 One curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140

21.2 Three new curves . . . . . . . . . . . . . . . . . . . . . . . . 144

21.3 Three old curves . . . . . . . . . . . . . . . . . . . . . . . . 147

21.4 Small blocking sets . . . . . . . . . . . . . . . . . . . . . . . 149

22 Multiple blocking sets . . . . . . . . . . . . . . . . . . . . . . . . . 154

22.1 A t (mod p) result . . . . . . . . . . . . . . . . . . . . . . . 155

23 Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162

23.1 Stability theorems for small blocking sets and the Szonyi-Weiner method . . . . . . . . . . . . . . . . . . . . . . . . . 164

23.2 Stability theorems for blocking sets: the prime case . . . . . 167

23.3 Stability theorems for blocking sets: the non-prime case . . 170

23.4 Stability theorem for sets of even type . . . . . . . . . . . . 176

23.5 Number of lines meeting a (q + 2)-set . . . . . . . . . . . . 179

24 Jamison, Brouwer-Schrijver, affine blocking sets . . . . . . . . . . . 180

24.1 Applications of the punctured Nullstellensatz . . . . . . . . 184

24.2 Redei polynomials in three indeterminates . . . . . . . . . . 189

24.3 Redei polynomials in many indeterminates . . . . . . . . . 191

24.4 Affine blocking sets . . . . . . . . . . . . . . . . . . . . . . . 195

25 Nuclei and affine blocking sets . . . . . . . . . . . . . . . . . . . . . 198

25.1 Lower nuclei . . . . . . . . . . . . . . . . . . . . . . . . . . 201

25.2 Internal nuclei . . . . . . . . . . . . . . . . . . . . . . . . . 203

25.3 Higher dimensions . . . . . . . . . . . . . . . . . . . . . . . 203

26 Mixed representations . . . . . . . . . . . . . . . . . . . . . . . . . 205

27 Flocks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207

27.1 Partial flocks of the quadratic cone in PG(3, q) . . . . . . . 208

27.2 Partial flocks of cones of higher degree . . . . . . . . . . . . 211

28 Spreads and ovoids . . . . . . . . . . . . . . . . . . . . . . . . . . . 214

28.1 Spreads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214

28.2 Ovoids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215

viii Contents

29 Non-geometrical applications . . . . . . . . . . . . . . . . . . . . . 21929.1 Normal factorizations of Abelian groups . . . . . . . . . . . 21929.2 The Paley graph . . . . . . . . . . . . . . . . . . . . . . . . 22029.3 Representing systems . . . . . . . . . . . . . . . . . . . . . 22129.4 Wielandt’s visibility theorem . . . . . . . . . . . . . . . . . 22229.5 Burnside’s theorem . . . . . . . . . . . . . . . . . . . . . . . 22229.6 Jaeger’s conjecture . . . . . . . . . . . . . . . . . . . . . . . 22329.7 Graphs containing p-regular subgraphs . . . . . . . . . . . . 22529.8 Blokhuis’ proof of Bollobas’ theorem . . . . . . . . . . . . . 22529.9 Alon-Furedi . . . . . . . . . . . . . . . . . . . . . . . . . . . 22629.10 Chevalley-Warning . . . . . . . . . . . . . . . . . . . . . . . 22629.11 Cauchy-Davenport . . . . . . . . . . . . . . . . . . . . . . . 22729.12 Another application . . . . . . . . . . . . . . . . . . . . . . 228

3 Hints and short solutions to the exercises 231

1 Of Chapter I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2312 Of Chapter II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235

4 Glossary of concepts 241

5 Notation 243

Index 244

References 245

1. Introduction 1

1 Introduction

This one is not intended to be an introductory textbook about finite geometries,finite fields nor polynomials. There are very good books of these kinds available,e.g. Ball-Weiner [23] for a smooth and fascinating introduction to the concepts offinite geometries, the three volumes of Hirschfeld and Hirshfeld-Thas [77, 78, 79]as handbooks and Lidl-Niederreiter [90] for finite fields. Still, the interested reader,even with a little background, may find all the definitions and basic informationhere (the Glossary of concepts at the end of the volume can help) to enjoy thisinterdisciplinary field in the overlap of geometry, combinatorics and algebra.

We would like to use a common terminology. Parts and reformulations of resultsof some interest for non-geometers will be emphasized.

Bruen and Fisher called the polynomial technique as “the Jamison method” andsummarized it in performing three steps: (1) Rephrase the theorem to be provedas a relationship involving sets of points in a(n affine) space. (2) Formulate thetheorem in terms of polynomials over a finite field. (3) Calculate. (Obviously, step3 carries most of the difficulties in general.) In some sense it is still “the method”,we will show several ways how to perform steps 1-3.

We have to mention the book of Laszlo Redei [103] from 1970, which inspired anew series of results on blocking sets and directions in the nineties.

There are a few survey papers on polynomial methods, by Blokhuis [35, 36], Szonyi[122], Ball [7], I included here some of their material. The form of a book allowsto show the algebraic background in more details; also to insert the most recentresults.

The typical theories in this book have the following character. Define a class of(point)sets (of a geometry) in a combinatorial way (which, in general, means re-strictions on the intersections with subspaces); examine its numerical parameters(usually the spectrum of sizes in the class); find the minimal/maximal values of thespectrum; characterize the extremal entities of the class; show that the extremal(or other interesting) ones are “stable” in the sense that there are no entities ofthe class being “close” to the extremal ones.

There are some fundamental concepts and ideas that we feel worth to put intolight all along this treatise:

• an algebraic curve (or envelope) containing all the “interesting” lines of apointset;

• the Lemma of tangents with its generalizations and analogues;

• resultants as sophisticated tools improving the simple use of Bezout’s theo-rem;

• examination of (lacunary) coefficients of polynomials;

2 Introduction

• considering (sub)spaces generated by polynomials.

Apology. There are some theorems being proved more than once in this book,these multiple proofs were included in order to show how the different ideas andtechniques can lead to the same result; sometimes they even serve as touchstonesof a new method.

Also, this book contains a few proofs being longer than the usual ones. Theywere included to show some elaborate series of ideas that lead to practical use ofpolynomial techniques.

There are exercises, like a hundred and twenty. One type of them is for the neces-sary technicalities where no further idea is needed (so boring but useful), anothertype is where an argument analogous to the preceding ones can solve the question(so good to practice), and a third type which needs minor new ideas (so the in-teresting ones). We provide short solutions for a great part of them at the end ofthe volume.

There are side comments on the margins. They might help reading, the edi-tors/referees are kindly asked to decide whether they should be (i) left as theyPlease!

are, or (ii) included into the normal text, or (iii) the number of them can beincreased (I would be happy to add more).

2 Acknowledgements

Most of this book is about the work of Simeon Ball, Aart Blokhuis, Andras Gacs,Tamas Szonyi and Zsuzsa Weiner. They, together with the author, contributed inroughly one half of the references; also their results (and sometimes even their textswith minor modifications: thanks to them for letting me do so) form an importantpart of this volume. Not least, I always enjoyed their warm, joyful, inspirating andsupporting company in various situations in the last some years. I hope that theselection and the choice of the topics covered in this book makes them happy aswell, and that being the “topic” of a book is at least as funny as writing it. Mostof them have seen a preliminary version of it, I am grateful for all the suggestionsthey made.

Above all I am deeply indebted to Tamas Szonyi, from whom most of my knowledgeand most of my enthusiasm for finite geometries I have learned.

It was Gabor Korchmaros who suggested (and not only to me) some seven yearsago to write a book like this. Probably any of us, or maybe someone else, couldhave done it. Of course this one is about the way how I can see the topic.

Last but not least I am grateful to all the colleagues and friends who helped mein any sense in the last some years: researchers of Ghent (and among them mymultiple coauthor Leo Storme), Potenza, Naples, Barcelona, Eindhoven and, ofcourse, Budapest.

3. Definitions, basic notation 3

3 Definitions, basic notation

We will not be very strict and consistent in the notation (but at least we’ll try tobe). However, here we give a short description of the typical notation we are goingto use.If not specified differently, q = ph is a prime power, p is a prime, and we work in theDesarguesian projective (or affine) space PG(n, q) (AG(n, q), resp.), each space co-ordinatized by the finite (Galois) field GF(q). The n-dimensional vectorspace overGF(q) will be denoted by V(n, q) or simply by GF(q)n. When discussing PG(n, q)and the related V(n+ 1, q) together then for a subspace dimension will be meant rank=dim+1

projectively while vector space dimension will be called rank. A field, which is notnecessarily finite will be denoted by F.In general capital letters X,Y, Z, T, ... will denote independent variables, whilex, y, z, t, ... will be elements of GF(q). A pair or triple of variables or elements inany pair of brackets can be meant homogeneously, hopefully it will be always clearfrom the context and the actual setting.We write X or V = (X,Y, Z, ..., T ) meaning as many variables as needed; Vq =(Xq, Y q, Zq, ...). As over a finite field of order q for each x ∈ GF(q) xq = x holds,two different polynomials, f and g, in one or more variables, can have coincidingvalues “everywhere” over GF(q). In this case we ought to write f ≡ g, as forunivariate polynomials f(X), g(X) it means that f ≡ g (mod Xq−X) in the ringGF(q)[X]. However, as in the literature f ≡ g is used in the sense ‘f and g areequal as polynomials’, we will use it in the same sense; though simply f = g andf(X) = g(X) may denote the same.Throughout this book we mostly use the usual representation of PG(n, q).This means that the points have homogeneous coordinates (x, y, z, ..., t) wherex, y, z, ..., t are elements of GF(q). The hyperplane [a, b, c, ..., d] of the space haveequation aX + bY + cZ + ...+ dT = 0.When PG(n, q) is considered as AG(n, q) plus the hyperplane at infinity, then wewill use the notation H∞ for that (‘ideal’) hyperplane. If n = 2 then H∞ is calledthe line at infinity `∞.According to the standard terminology, a line meeting a pointset in one point willbe called a tangent and a line intersecting it in r points is an r-secant (or a line oflength r).This book is about combinatorially defined (point)sets of (mainly projective oraffine) finite geometries. They are defined by their intersection numbers with lines(or other subspaces) typically. The most important definitions and basic informa-tion are collected in the Glossary of concepts at the end of this book. These are:blocking sets, arcs, nuclei, spreads, sets of even type, etc.Warning. In this book a curve is allowed to have multiple components, so in factthe curves considered here are called cycles in a different terminology.

4 Introduction

Chapter 1

Background

5 Finite fields and polynomials

5.1 Some basic facts

Here the basic facts about finite fields are collected. For more see [90].

For any prime p and any positive integer h there exists a unique finite field (orGalois field) GF(q) of size q = ph. The prime p is the characteristic of it, meaninga + a + ... + a = 0 for any a ∈ GF(q) whenever the number of a’s in the sum is(divisible by) p. The additive group of GF(q) is elementary abelian, i.e. (Zp,+)h

while the non-zero elements form a cyclic multiplicative group GF(q)∗ ' Zq−1, anygenerating element (often denoted by ω) of it is called a primitive element of thefield.

For any a∈ GF(q) aq = a holds, so the field elements are precisely the roots of so if a 6= 0 thenaq−1 = 1and Xq−1 − 1is the rootpolynomial ofGF(q)∗

Xq −X. (Lucas’ theorem implies, see below, that) we have (a+ b)p = ap + bp forany a, b ∈ GF(q), so x 7→ xp is a field automorphism. GF(q) has a (unique) subfieldGF(pt) for each t|h; GF(q) is an h

t -dimensional vectorspace over its subfield GF(pt).The (Frobenius-) automorphisms of GF(q) are x 7→ xp

i

for i = 0, 1, ..., h − 1,forming the complete, cyclic automorphism group of order h. Hence x 7→ xp

t

fixesthe subfield GF(pgcd(t,h)) pointwise (and all the subfields setwise!); equivalently,(Xpt −X)|(Xq −X) iff t|h.

One can see that for any k not divisible by (q − 1),∑a∈GF(q) a

k = 0. From this, write GF(q)∗ =

{ω0, ω1, ..., ωq−2}and considerPq−2i=0 ω

ik =

ωk(q−1)−1ωk−1

.

if f : GF(q) → GF(q) is a bijective function then∑x∈GF(q) f(x)k = 0 for all

k = 1, ..., q − 2. See also Dickson’s theorem.

We will frequently use Lucas’ theorem when calculating binomial coefficients(nk

)in finite characteristic, so “modulo p”: let n = n0 + n1p + n2p

2 + ... + ntpt, k =

5

6 I. Background

k0 + k1p + k2p2 + ... + ktp

t, with 0 ≤ ni, ki ≤ p − 1, then(nk

)≡(n0k0

)(n1k1

)...(ntkt

)(mod p). In particular, in most cases we are interested in those values of k when(nk

)is non-zero in GF(q), so modulo p. By Lucas’ theorem, they are precisely the

elements ofMn

Mn = {k = k0 + k1p+ k2p2 + ...+ ktp

t : 0 ≤ ki ≤ ni}.

Exercise 5.1. Prove Lucas’ theorem!We define the trace and norm functions on GF(qn) as Trqn→q(X) = X+Xq+Xq2 +...+Xqn−1

and Normqn→q(X) = XXqXq2 ...Xqn−1, so the sum and the product of

all conjugates of the argument. Both maps GF(qn) onto GF(q), the trace functionis GF(q)-linear while the norm function is multiplicative.

Exercise 5.2. Show that Tr and Norm are in some sense unique, i.e. any GF(q)-linear function mapping GF(qn) onto GF(q) can be written in the form Trqn→q(aX)with a suitable a ∈ GF(qn) and any multiplicative function mapping GF(qn) ontoGF(q) can be written in the form Normqn→q(Xa) with a suitable integer a. (SeeExercise 5.27 and also Section 8.1 for Tr).

5.2 Self-dual normal bases

GF(qn) is a vector-space over its subfield GF(q), so several bases can be chosen.There are some “natural choices” like normal bases of form {ω, ωq, ωq2 , ..., ωqn−1},the conjugates of a certain primitive element. Another “nice requirement” is self-duality (with respect to the inner product 〈a, b〉 = Trqn→q(ab)), i.e. the innerproduct of two basis elements bi and bj is Trqn→q(bibj) = 0 if i 6= j and 1 if i = j.The following theorem answers the question whether self-dual normal bases exist.

Result 5.3. [89] For a prime power q and n > 1 a self-dual normal basis of GF(qn)over GF(q) exists if and only if either n is odd, or n ≡ 2 (mod 4) and q is even.

5.3 Polynomials

Here we summarize some properties of polynomials over finite fields. Givena field F, a polynomial f(X1, X2, ..., Xk) is a finite sum of monomial termsai1i2...ikX

i11 X

i22 · · ·Xik

k , where each Xi is a free variable, ai1i2...ik , the coefficient ofthe term, is an element of F. The (total) degree of a monomial is i1 + i2 + ...+ ikif the coefficient is nonzero and −∞ otherwise. The (total) degree of f , denotedby deg f or f◦, is the maximum of the degrees of its terms. These polynomialsform the ring F[X1, X2, ..., Xk]. A polynomial is homogeneous if all terms havethe same total degree. If f is not homogeneous then one can homogenize it, i.e.transform it to the following homogeneous form: Zdeg f · f(X1

Z ,X2Z , ...,

XkZ ), which

is a polynomial again (Z is an additional free variable).

5. Finite fields and polynomials 7

Given f(X1, ..., Xn) =∑ai1...inX

i11 · · ·Xin

n ∈ F[X1, ..., Xn], and the ele-ments x1, ..., xn ∈ F then one may substitute them into f : f(x1, ..., xn) =∑ai1...inx

i11 · · ·xinn ∈ F; (x1, ..., xn) is a root of f if f(x1, ..., xn) = 0.

A polynomial f may be written as a product of other polynomials, if not (exceptin a trivial way) then f is irreducible. If we consider f over F, the algebraicclosure of F, and it still cannot be written as a product of polynomials over Fthen f is absolutely irreducible. E.g. X2 + 1 ∈ GF(3)[X] is irreducible but notabsolutely irreducible, it splits to (X + i)(X − i) over GF(3) where i2 = −1.X2 + Y 2 + 1 ∈ GF(3)[X,Y ] is absolutely irreducible. Over the algebraic closureevery univariate polynomial splits into linear factors.In particular, x is a root of f(X) (of multiplicity m, see below) if f(X) can bewritten as f(X) = (X − x)m · g(X) for some polynomial g(X), m ≥ 1.Over a field any polynomial can be written as a product of irreducible polynomials(factors) in an essentially unique way (so apart from constants and rearrangement).Let f : GF(q) → GF(q) be a function. Then it can be represented by the linearcombination

∀x ∈ GF(q) f(x) =∑

a∈GF(q)

f(a)µa(x),

where this is Lagrangeinterpolation

µa(X) = 1− (X − a)q−1

is the characteristic function of the set {a}. In other terms it means that anyfunction can be given as a polynomial of degree ≤ q − 1. As both the number offunctions GF(q) → GF(q) and polynomials in GF(q)[X] of degree ≤ q−1 is qq, this both a vec-

torspace ofdim = q overGF(q)

representation is unique.Let now f ∈ GF(q)[X]. Then f , as a function, can be represented by a polynomialf of degree at most q−1, this is called the reduced form of f . The degree of f will multiplicity of

a root maychange whenreducing f

be called the reduced degree of f .

Proposition 5.4. For any (reduced) polynomial f(X) = cq−1Xq−1 + ...+ c0,∑

x∈GF(q)

xkf(x) = −cq−1−k0 ,

where k = t(q − 1) + k0, 0 ≤ k0 ≤ q − 2. In particular,∑

x∈GF(q)

f(x) = −cq−1.

Proof:∑x x

kf(x) =∑x

∑q−1i=0 cix

i+k =∑q−1i=0 ci

∑x x

i+k = −cq−1−k0 .

Exercise 5.5. If f is bijective (permutation polynomial) then the reduced degree offk is at most q − 2 for k = 1, ..., q − 2.

We note that (i) if p 6∣∣∣ |{t : f(t) = 0}| then the converse is true; see also Exer-

cise 9.13, whichis Dickson’s

8 I. Background

(ii) it is enough to assume it for the values k 6≡ 0 (mod p).

Exercise 5.6. If q = 2h then the quadratic equation aX2 + bX + c = 0 has nosolution iff Tr2h→2

(acb2

)= 1.

Let’s examine GF(q)[X] as a vector space over GF(q).

Exercise 5.7. Gacs [67] For any subspace V of GF(q)[X], dim(V ) = |{deg(f) : f ∈V }|.

Exercise 5.8. [56] Consider the space Sλ spanned as Sλ = 〈(X − a)λ : a ∈ GF(q)〉.Write λ =

∑h−1i=0 αip

i, where 0 ≤ αi < p and let Mλ = {∑h−1i=0 βip

i : 0 ≤ βi ≤ αi}.q = ph

Then Sλ has a basis {Xr : r ∈Mλ}.

In several situations we will be interested in the zeros of (uni- or multivariate)polynomials. Let a = (a1, a2, ..., an) be in GF(q)n. We shall refer to a as a pointin the n-dimensional vector space V(n, q) or affine space AG(n, q). Consider an fin GF(q)[X1, ..., Xn], f =

∑αi1,i2,...,inX

i11 · · ·Xin

n .

We want to define the multiplicity of f at a. It is easy if a = 0 = (0, 0, ..., 0). Letm be the largest integer such that for every 0 ≤ i1, ..., in, i1 + ...+ in < m we haveαi1,i2,...,in = 0. Then we say that f has a zero at 0 with multiplicity m.multaf

def= m

For general a one can consider the suitable “translate” of f , i.e. fa(Y1, ..., Yn) =f(Y1 +a1, Y2 +a2, ..., Yn+an), and we say that f has a zero at a with multiplicitym if and only if fa has a zero at 0 with multiplicity m.m ≥ 1 ⇔

f(a)=fa(0)=0

5.4 Differentiating polynomials

Given a polynomial f(X) =∑ni=0 aiX

i, one can define its derivative ∂Xf = f ′X =f ′ in the following way: f ′(X) =

∑ni=0 iaiX

i−1. Note that if the characteristic pdivides i then the term iaiX

i−1 vanishes; in particular deg f ′ < deg f − 1 mayoccur. Multiple differentiation is denoted by ∂iXf or f (i) or f ′′, f ′′′ etc. If a is aroot of f with multiplicity m then a will be a root of f ′ with multiplicity at leastm− 1, and of multiplicity at least m iff p|m. Also if k ≤ p then a is root of f withmultiplicity at least k iff f (i)(a) = 0 for i = 0, 1, ..., k − 1.

We will use the differential operator ∇ = (∂X , ∂Y , ∂Z) (when we have three vari-ables) and maybe ∇i = (∂iX , ∂

iY , ∂

iZ) and probably ∇iH = (Hi

X ,HiY ,Hi

Z), whereHi stands for the i-th Hasse-derivation operator (see 7.3). The only propertieswe need are that Hj Xk =

(kj

)Xk−j if k ≥ j (otherwise 0); Hj is a linear op-

erator; Hj(fg) =∑ji=0Hif Hj−ig; a is root of f with multiplicity at least k iff

Hif(a) = 0 for i = 0, 1, ..., k − 1; and finally HiHj =(i+ji

)Hi+j .


We might use the following differential equation:

V · ∇F = X∂XF + Y ∂Y F + Z∂ZF = 0,

where F = F (X,Y, Z) is a homogeneous polynomial in three variables, of totaldegree n. Let F (X,Y, Z, λ) = F (λX, λY, λZ) = λnF (X,Y, Z), then

nλn−1F (X,Y, Z) = (∂λF )(X,Y, Z, λ) = (X∂XF + Y ∂Y F + Z∂ZF )(λX, λY, λZ).

It means that if we consider V · ∇F = 0 as a polynomial equation then(∂λF )(X,Y, Z, λ) = 0 identically, which holds if and only if p divides n = deg(F ).If we consider our equation as (V · ∇F )(x, y, z) = 0 for all (x, y, z) ∈ GF(q)3, anddeg(F ) is not divisible by p, then the condition is that F (x, y, z) = 0 for everychoice of (x, y, z), i.e. F ∈ 〈Y qZ − Y Zq, ZqX − ZXq, XqY −XY q〉, see later.

5.5 Polynomials vanishing at many points, Alon’s CombinatorialNullstellensatz and the Ball-Serra refinement

We will quite often get into a situation when our polynomials have many roots,sometimes they vanish almost everywhere in their domain. It was Bruen and laterAlon who started to explore this situation; here we show some results of this kind.An incredible number of wonderful applications were found later, we will see someof the geometrical ones and in the last section some more non-geometrical onesas well. Recently Ball and Serra achieved a new improvement on Alon’s Nullstel-lensatz, (which promises new applications, in fact they found some already), weshow that as well. Therefore this Section is essentially based on [1] and [22].

Exercise 5.9. Let S be a subset of GF(q)2 and f ∈ GF(q)[X,Y ], such that f(aY +b, Y ) = 0, for all (a, b) ∈ S. If |S| > deg(f) then f(X,Y ) ≡ 0.

Exercise 5.10. • Show that G(x1, ..., xn) = 0 for all (x1, ..., xn) ∈ GF(q)n if andonly if G is of form g1(X1, ..., Xn)·(Xq

1−X1)+...+gn(X1, ..., Xn)·(Xqn−Xn).

• Similarly show that G(x1, ..., xn) = 0 for all (x1, ..., xn) ∈ (GF(q)∗)n if andonly if G ∈ 〈(Xq−1

1 − 1), ..., (Xq−1n − 1)〉.

• Show that all (x1, ..., xn) ∈ GF(q)n are t-fold zeros of G(X1, ..., Xn) if andonly if G is an element of the ideal Jt

Jt = Jt(X1, ..., Xn) = 〈(Xq1−X1)i1(X

q2−X2)i2 ...(Xq

n−Xn)in : i1+i2+...+in = t〉.

• Similarly show that all (x1, ..., xn) ∈ (GF(q)∗)n are t-fold zeros ofG(X1, ..., Xn) iff G ∈ 〈(Xq−1

1 − 1)i1(Xq−12 − 1)i2 ...(Xq−1

n − 1)in : i1 +i2 + ...+ in = t〉.

Theorem 5.11. Let f ∈ GF(q)[X1, ..., Xn] satisfy f(0, 0, ..., 0) 6= 0 and f(a) = 0 forall a 6= 0. Then deg(f) ≥ n(q − 1).

10 I. Background

Proof Let f(0) = γ 6= 0 and g(X1, ..., Xn) =∏ni=1(1 − Xq−1

i ). Then 1γ f − g

vanishes everywhere hence it is in J1.

Exercise 5.12. Let t be a positive integer, and let the polynomial f ∈GF(q)[X1, ..., Xn] satisfy f(0, 0, ..., 0) 6= 0 and multaf ≥ t for all a 6= 0. Thena generalization

deg(f) ≥ (n+ t− 1)(q − 1).

Exercise 5.13. Give an example when in Exercise 5.12 deg(f) = (n+ t− 1)(q− 1)holds, for n+ t− 1 ≤ q + 1.

Exercise 5.14. Suppose that the curves C1, ..., Cm cover AG(2, q) \ {(0, 0)} but nei-ther of them contains the origin. Then

∑i deg(Ci) ≥ 2(q − 1).

Theorem 5.15. Let A,B ⊆ GF(q)∗. Let f(X,Y ) ∈ GF(q)[X,Y ] satisfy (i) f(0, 0) 6=anothergeneralization 0 and (ii) f(a, b) = 0 whenever a ∈ A or b ∈ B. Then f can be written as f = g+h,

where h ∈ J1(X,Y ), deg(g) ≤ deg(f) and deg(f) ≥ |A|+ |B|.

Proof: Let’s reduce f modulo (Xq−X) and modulo (Y q−Y ), this gives f = g+hwith degX(g) ≤ q−1, degY (g) ≤ q−1, h ∈ J1(X,Y ), deg(h) ≤ deg(f) and clearlydeg(f) ≥ deg(g). Write g(X,Y ) =

∑ci(Y )Xi.

As g(x, b) = 0 for any x ∈ GF(q), b ∈ B and degX(g),degY (g) ≤ q − 1, it followsthat g(X, b) = 0, so each ci(b) = 0. Hence

∏b∈B(Y − b) divides g, and similarly∏

a∈A(X − a) divides g.

Note that it gives another proof of Theorem 5.11 for n = 2, with A = B = GF(q)∗.In applications we often need the “homogeneous version” of Exercise 5.10. Notethat e.g. XqY − XY q = Y (Xq − X) − X(Y q − Y ). If f(X,Y ), a homogeneouspolynomial in X and Y , of total degree d, vanishes everywhere on GF(q)2 thenf ∈ 〈(Xq −X), (Y q − Y )〉, so f(X,Y ) = f1(X,Y )(Xq −X) + f2(X,Y )(Y q − Y ),where f1 and f2 can be chosen to be homogeneous polynomials of total degreed− q. Now the terms of low degree must disappear, so Xf1 + Y f2 = 0. Hence Ydivides f1; let f1(X,Y ) = Y g(X,Y ), then

f(X,Y ) = g(X,Y )(XqY −XY q).

Alon’s Nullstellensatz

Hilbert’s Nullstellensatz is the fundamental theorem stating that if F is an alge-braically closed field, and f ; g1, ..., gm are polynomials in F[X1, ..., Xn], where fvanishes over all common zeros of g1, ..., gm, then there is an integer k and polyno-mials h1, ..., hm in F[X1, ..., Xn] so that fk =

∑mi=1 higi. In the special case m = n,

where each gi is a univariate polynomial of the form∏s∈Si(Xi − s), a stronger

conclusion holds, by Alon’s result [2].


Theorem 5.16. Let F be an arbitrary field, and let f = f(X1, ..., Xn) be apolynomial in F[X1, ..., Xn]. Let S1, ..., Sn be nonempty subsets of F and definegi(Xi) =

∏s∈Si(Xi − s). If f vanishes over all the common zeros of g1, ..., gn

(that is; if f(s1, ..., sn) = 0 for all si ∈ Si), then f ∈ 〈g1, ..., gn〉 and there arepolynomials h1, ..., hn ∈ F[X1, ..., Xn] satisfying deg(hi) ≤ deg(f)−deg(gi) so that

f =n∑i=1

higi.

Moreover, if f ; g1, ..., gn lie in R[X1, ..., Xn] for some subring R of F then thereare polynomials hi ∈ R[X1, ..., Xn] as above.

As a consequence of the above one can prove the following:

Theorem 5.17. Let F be an arbitrary field, and let f = f(X1, ..., Xn) be a poly-nomial in F[X1, ..., Xn]. Suppose that deg(f) =

∑ni=1 ti, where each ti is a non-

negative integer, and suppose the coefficient of∏ni=1X

tii in f is nonzero. Then, if

S1, ..., Sn are subsets of F with |Si| > ti, there are s1 ∈ S1, s2 ∈ S2, ..., sn ∈ Sn sothat f(s1, ..., sn) 6= 0.

To prove Theorem 5.16 we need the following lemma.

Lemma 5.18. Let P = P (X1, ..., Xn) be a polynomial in n variables over an ar-bitrary field F. Suppose that the degree of P as a polynomial in Xi is at most tifor 1 ≤ i ≤ n, and let Si ⊂ F be a set of at least ti + 1 distinct members of F. IfP (X1, ..., Xn) = 0 for all n-tuples (x1, ..., xn) ∈ S1 × ...× Sn then P ≡ 0.

Proof: We apply induction on n. For n = 1, the lemma is simply the assertionthat a non-zero polynomial of degree t1 in one variable can have at most t1 distinctzeros. Assuming that the lemma holds for n−1, we prove it for n (n ≥ 2). Given apolynomial P = P (X1, ..., Xn) and sets Si satisfying the hypotheses of the lemma,let us write P as a polynomial in Xn; that is,

P =tn∑i=0

Pi(X1, ..., Xn−1)Xin,

where each Pi is a polynomial withXj-degree bounded by tj . For each fixed (n−1)-tuple (x1, ..., xn−1) ∈ S1×...×Sn−1 , the univariate polynomial P (x1, ..., xn−1, Xn)vanishes for all xn ∈ Sn, and is thus identically 0. Thus Pi(x1, ..., xn−1) = 0 forall (x1, ..., xn−1) ∈ S1 × ...× Sn−1. Hence, by the induction hypothesis, Pi ≡ 0 forall i, implying that P ≡ 0.

Proof of Theorem 5.16 Define ti = |Si| − 1 for all i. By assumption,

f(x1, ..., xn) = 0 ∀(x1, ..., xn) ∈ S1 × ...× Sn (1)

12 I. Background

For each i, 1 ≤ i ≤ n, let

gi(Xi) =∏s∈Si

(Xi − s) = Xti+1i −

ti∑j=0

gijXji .

Observe that, if xi ∈ Si then gi(xi) = 0; that is, xti+1i =

∑tij=0 gijx

ji . (2)

Let f be the polynomial obtained by writing f as a linear combination of monomi-als and replacing, repeatedly, each occurrence of Xui

i (1 ≤ i ≤ n), where ui > ti, bya linear combination of smaller powers of Xi, using the relations (2). The resultingpolynomial f is of degree at most ti in Xi, for each 1 ≤ i ≤ n, and is obtained fromf by subtracting from it products of the form higi, where the degree of each polyno-mial hi ∈ F[X1, ..., Xn] does not exceed deg(f)−deg(gi) (and where the coefficientsof each hi are in the smallest ring containing all coefficients of f and g1, ..., gn.)Moreover, f(x1, ..., xn) = f(x1, ..., xn), for all (x1, ..., xn) ∈ S1× ...×Sn, since therelations (2) hold for these values of x1, ..., xn. Therefore, by (1), f(x1, ..., xn) = 0for every n-tuple (x1, ..., xn) ∈ S1× ...×Sn and hence, by the Lemma, f ≡ 0. Thisimplies that f =

∑ni=1 higi, and completes the proof.

Proof of Theorem 5.17 We may assume that |Si| = ti + 1 for all i. Suppose thestatement is false, and define gi(Xi) =

∏s∈Si(Xi− s). By Theorem 5.16 there are

polynomials h1, ..., hn ∈ F[X1, ..., Xn] satisfying deg(hj) ≤∑ni=1 ti − deg(gj) so

that f =∑ni=1 higi. By assumption, the coefficient of

∏ni=1X

tii on the left hand

side is nonzero, and hence so is the coefficient of this monomial on the right handside. However, the degree of higi = hi

∏s∈Si(Xi − s) is at most deg(f), and if

there are any monomials of degree deg(f) in it they are divisible by Xti+1i . It

follows that the coefficient of∏ni=1X

tii on the right hand side is zero, and this

contradiction completes the proof.

For nice applications see Section 29.Now we present the punctured version of Alon’s Nullstellensatz by Ball and Serra[22], which states that if f vanishes at nearly all, but not all, of the common zerosof some polynomials g1(X1), ..., gn(Xn) then every I-residue of f , where the idealI = 〈g1, ..., gn〉, has a large degree. As a consequence we prove a converse of thecorollary to Alon’s Nullstellensatz. The corollary to Alon’s Nullstellensatz statesthat if f has a term of maximum degree Xr1

1 ...Xrnn , where ri = |Si|− ti and ti > 0

for all i, then a gridD1×...×Dn containing the points of the grid S1×...×Sn wheref does not vanish, satisfies |Di| ≥ ti. The converse, which will follow as a corollaryto the punctured version of Alon’s Nullstellensatz, states that if D1 × ... ×Dn isa grid containing the points of the grid S1 × ... × Sn where f does not vanish,then f has a term Xr1

1 ...Xrnn , where ri satisfies |Si| − 1 ≥ ri ≥ |Si| − |Di| for all

i. Furthermore, Ball and Serra extend Alon’s Nullstellensatz to functions which


have multiple zeros at the common zeros of g1, g2, ..., gn and prove a puncturedversion of this generalised version.The following corollary is slightly more general than Theorem 5.17. Note thatunder the hypothesis there is always at least one point of the grid where f doesnot vanish.

Corollary 5.19. If f ∈ F[X1, X2, ..., Xn] has a term of maximum degree Xr11 ...Xrn

n ,where ri = |Si| − ti and ti ≥ 1 for all i, then a grid which contains the points ofS1 × ...× Sn where f does not vanish, has size at least t1 × ...× tn.

Proof: Suppose that there is a grid M1 × ... ×Mn, where |Mj | < tj for some j,containing all the points S1 × ...× Sn where f does not vanish. Let

ej(Xj) =∏

mj∈Mj

(Xj −mj).

The polynomial fej is zero at all points of S1×...×Sn and has a term of maximumdegree Xr1

1 ...Xrnn X

|Mj |j . Note that rj + |Mj | < |Sj | and ri < |Si| for i 6= j. By

Theorem 5.16 the polynomial fej =∑ni=1 gihi for some polynomials hi. The terms

of maximum degree in fej have degree in Xi at least |Si| for some i, which is acontradiction.

Punctured Combinatorial Nullstellensatz

In Alon’s Combinatorial Nullstellensatz, Theorem 5.16, the function f was as-sumed to have zeros at all points of the grid S1 × S2 × ... × Sn. In the case thatthere is a point in S1 × S2 × ... × Sn where f does not vanish a slightly differentconclusion holds. The following can be thought of as a punctured version of Alon’sCombinatorial Nullstellensatz.Let F be a field and let f be a polynomial in F[X1, X2, ..., Xn]. For i = 1, ..., n, letDi and Si be finite non-empty subsets of F, where Di ⊂ Si, and define

gi(Xi) =∏si∈Si

(Xi − si), and li(Xi) =∏di∈Di

(Xi − di).

Theorem 5.20. If f vanishes over all the common zeros of g1, g2, ..., gn except atleast one element of D1 × D2 × ... × Dn, where it is not zero, then there arepolynomials h1, h2, ..., hn ∈ F[X1, X2, ..., Xn] satisfying deg(hi) ≤ deg(f)−deg(gi)and a non-zero polynomial w, whose degree in Xi is less than |Si| and whose totaldegree is at most deg f , with the property that

f =n∑i=1

higi + w,

14 I. Background

and

w = un∏i=1

gili,

for some non-zero polynomial u.

Note that the theorem gives the lower bound deg f ≥∑ni=1(|Si| − |Di|).

Proof: We can write

f =n∑i=1

gihi + w,

for some polynomials hi of degree at most the degree of f minus the degree of gi,and a polynomial w, where the degree of w in Xi is less than the degree of gi andthe overall degree of w is at most the degree of f . For each i the polynomial flihas zeros on all common zeros of gi, by assumption, and hence so does wli. ByAlon’s Nullstellensatz there are polynomials vi with the property that

wli =n∑i=1

givi.

However the degree of Xj in wli, for j 6= i, is less than the degree of gj(Xj) and sowli = givi. Thus gi divides wli. Note that li divides gi, so this divisibility impliesgi/li divides w. Hence

w = un∏i=1

gili

for some polynomial u and u is not zero since 0 6= f(d1, d2, ..., dn) =w(d1, d2, ..., dn) for some di ∈ Di.

The following corollary is a converse of the corollary to Alon’s Nullstellensatz,Corollary 5.19.

Exercise 5.21. If D1 × ... × Dn is a grid containing all the points of the gridS1×...×Sn where f does not vanish, then f has a term Xr1

1 ...Xrnn , where |Si|−1 ≥

ri ≥ |Si| − |Di|.

Combinatorial Nullstellensatzen with multiplicity

In this section we shall consider polynomials that have zeros of multiplicity, seethe end of Section 5.3. The following proof of Theorem 5.22 is based on the proofof Theorem 1.3 in [53].Let T be the set of all non-decreasing sequences of length t on the set {1, 2, ..., n}.For any τ ∈ T , let τ(i) denote the i-th element in the sequence τ .


Let F be a field and let f be a polynomial in F[X1, X2, ..., Xn]. Suppose thatS1, S2, ..., Sn are arbitrary non-empty finite subsets of F and define

gi(Xi) =∏si∈Si

(Xi − si).

Theorem 5.22. If f has a zero of multiplicity t at all the common zerosof g1, g2, ..., gn then there are polynomials hτ in F[X1, X2, ..., Xn], satisfyingdeg(hτ ) ≤ deg(f)−

∑i∈τ deg(gi), such that

f =∑τ∈T

gτ(1)...gτ(t)hτ .

Proof: This can be proved by double induction on n and t. If n = 1 and f hasa zero of degree t for all si ∈ Si then f = g(X1)th(X1) for some polynomial h.If t = 1 then the theorem is Alon’s Nullstellensatz, Theorem 5.16. For the detailssee [22].

Theorem 5.22 has the following corollary.

Exercise 5.23. Let F be a field and let f be a polynomial in F[X1, X2, ..., Xn] andsuppose that f has a term Xr1

1 Xr22 ...Xrn

n of maximal degree. If S1, S2, ..., Sn arenonempty subsets of F with the property that for all τ ∈ T , the set of non-decreasingsequences of length t on {1, 2, ..., n}, there exists an i such that∑

i∈τ|Si| > ri,

then there is a point a = (a1, a2, ..., an), with ai ∈ Si, where f has a zero ofmultiplicity at most t− 1.Note that this statement with t = 1 is the original corollary to Alon’s Nullstellen-satz that has proven so useful. Specifically, if ri < |Si| for all i then there is a point(a1, a2, ..., an), with ai ∈ Si, where f does not vanish. When t = 2 the hypothesisis that ri < |Si| for all i except possibly one, when i = 1 say, for which r1 < 2|S1|.The conclusion is that there is a point (a1, a2, ..., an), with ai ∈ Si, where f has asimple zero (of multiplicity one) or does not vanish. When t = 3 the hypothesis isthat ri < |Si| for all i except possibly two values, 1 and 2 say, for which r1 < 3|S1|and r2 < 2|S2|.The following is a version of Theorem 5.20 for functions that have many zeroswith multiplicity.Let F be a field and let f be a polynomial in F[X1, X2, ..., Xn]. For i = 1, ..., n, letDi and Si be finite non-empty subsets of F, where Di ⊂ Si, and define

gi(Xi) =∏si∈Si

(Xi − si), and li(Xi) =∏di∈Di

(Xi − di).

16 I. Background

Theorem 5.24. If f has a zero of multiplicity at least t at all the common zerosof g1, g2, ..., gn, except at at least one point of D1 ×D2 × ... ×Dn where it has azero of multiplicity less than t, then there are polynomials hτ in F[X1, X2, ..., Xn],satisfying deg(hi) ≤ deg(f)−

∑i∈τ deg(gi), and a non-zero polynomial u satisfying

deg(u) ≤ deg(f)−∑ni=1(deg(gi)− deg(li)), such that

f =∑τ∈T

gτ(1)...gτ(t)hτ + un∏i=1

gili.

Moreover, if there is a point of D1 ×D2 × ...×Dn where f is non-zero, then forany j,

deg(f) ≥ (t− 1)(|Sj | − |Dj |) +n∑i=1

(|Si| − |Di|).

5.6 Additive, linearized polynomials

A polynomial f ∈ GF(q)[X] is additive if f(U+V ) = f(U)+f(V ) or, alternatively,f(x+ y) = f(x) + f(y) for all x, y ∈ GF(q). It is GF(pe)-linear if it is additive andfor all λ ∈ GF(pe) ⊆ GF(q), f(λX) = λf(X) or f(λx) = λf(x) for each x ∈ GF(q).If all the exponents in f are powers of pe then f is obviously GF(pe)-linear.

We remark that GF(p)-linearity and additivity are equivalent.

Note that the polynomial X6 +X4 +X2 +X ∈ GF(3)[X] is additive (GF(3)-linear)but the exponents are not powers of 3. This cannot occur if the degree is at mostq − 1.

Theorem 5.25. A polynomial f ∈ GF(q)[X] of degree at most q−1 is GF(pe)-linearif and only if all of its exponents are powers of pe.

Proof: The “if” part is obvious. Consider GF(q) as an he -dimensional vectorspace

over GF(pe), then f is a GF(pe)-linear transformation of it. The number of such

transformations is (pe)(he )

2

. The polynomials of degree at most q − 1 and withall exponents being powers of pe are among these transformations, the number ofthese polynomials is q

he = (pe)(

he )

2

.

Exercise 5.26. (Segre-Bartocci) A linear map f : GF(q) → GF(q) is nonsingular ifand only if for the coefficients of its unique polynomial form f(X) =

∑h−1i=0 aiX

pi ,

the matrix A =

a0 a1 ... ah−1aph−1 ap0 ... aph−2

ap2

h−2 ap2

h−1 ... ap2

h−3

.

.

.

aph−1

1 aph−1

2 ... aph−1

0

is nonsingular.


Note that the determinant of the Segre-Bartocci matrix is always an element ofGF(p).

Exercise 5.27. A polynomial f ∈ GF(q)[X] of degree ≤ q − 1 maps to the subfieldGF(pe) and is GF(pe)-linear if and only if it is of the form f(X) = Trq→pe(aX)with some a ∈ GF(q).

Theorem 5.28. Fundamental Theorem of Additive Polynomials. Let f(X) ∈ F[X] separable: splitsinto deg(f) dis-tinct linear fac-tors

be a separable polynomial with roots {x1, ..., xm}. Then f(X) is additive if andonly if S = {x1, ..., xm} is a(n additive) subgroup of (F,+).

Proof: The “additive ⇒ subgroup” direction is obvious. To show the other, con-sider

f(X) =m∏i=1

(X − xi).

If y ∈ S then f(X+s) = f(X). Now for any y put g(X) = f(X+y)−f(X)−f(y).Here deg g < deg f = m and g(xi) = 0 for i = 1, ...,m, hence g(X) = 0 and f isadditive.

5.7 The random-like behaviour

In this section we prove a lemma, which is a generalization of a result of Szonyi.It is interesting in itself, and it is used in several applications as well. In fact thislemma is a consequence of the character sum version of Weil’s estimate. In orderto formulate it, we need

Definition 5.29. Let f1(X), ..., fm(X) ∈ GF(q)[X] be given polynomials. We saythat their system is d–power independent, if no partial product fs1i1 f

s2i2...f

sjij

(1 ≤j ≤ m; 1 ≤ i1 < i2 < ... < ij ≤ m; 1 ≤ s1, s2, ..., sj ≤ d− 1) can be written as aconstant multiple of a d-th power of a polynomial.

Equivalently, one may say that if any product fs1i1 fs2i2...f

sjij

is a constant multiple ofa d-th power of a polynomial, then this product is ‘trivial’, i.e. for all the exponentsd|si, i = 1, ..., j. Now

Lemma 5.30. Let f1, ..., fm ∈ GF(q)[X] be a set of d–power independent poly-nomials, where d|(q − 1), d,m ≥ 2. Denote by N the number of solutions{x ∈ GF(q) : fi(x) is a d-th power in GF(q) for all i = 1, ...,m}. Then |N− q

dm | ≤√q∑mi=1 deg fi.

Note that this lemma implies that, under some natural conditions, one can solvea system of equations

χd(fi(X)) = δi (i = 1, ...,m),

18 I. Background

where the δi-s are d-th complex roots of unity, and χd is a multiplicative characterof order d. So ‘the d-th power behaviour’ can be prescribed if the polynomials are‘independent’.It can be interpreted as ‘being a d-th power’ is like a random event of probability1d .Some words about the condition d|(q− 1): if d and q− 1 are co-primes, then everyelement is a d-th power in GF(q). If g.c.d.(d, q − 1) = d1 and we write d = d1d2

and g.c.d.(d2, q − 1) =g.c.d.(d1, d2) = 1, then the lemma can be applied with d1,as d-th and d1-th powers are the same in this case.We remark that Szonyi [126] proved this lemma for d = 2. [114] contains a generalbound; the proof below is a modified version of a lemma for linear polynomials inBabai, Gal and Wigderson [5].We need the character sum version of Weil’s estimate:

Result 5.31. ([90], Thm. 5.41) Let f(X) be a polynomial over GF(q) and r thenumber of distinct roots of f in its splitting field. If χe is a multiplicative character(of order e) of GF(q) and f(X) 6= cg(X)e, then

|∑

x∈GF(q)

χe(f(x)) | ≤ (r − 1)√q.

Proof of Lemma 5.30: First note that we use the definition χ(x) = χd(x) = xq−1d .

Let {ε0 = 1, ε1, ε2, ..., εd−1} be the set of d-th complex roots of unity. Let h(Z) =Zd−1Z−1 = 1 + Z + ... + Zd−1; then h(1) = d, h(εj) = 0 for j = 1, ..., d − 1 andh(0) = 1. Define H(x) =

∏mi=1 h(χ(fi(x))).

If x is a solution then H(x) = dm, if x is a root of some fi then H(x) = 0 orH(x) = dm−1. In the remaining cases H(x) = 0. Hence, if N denotes the numberof solutions and D :=

∑mi=1 deg fi then the sum S =

∑x∈GF(q)H(x) satisfies

Ndm ≤ S ≤ Ndm +Ddm−1. (∗)

H(x) is a product of sums of d terms each. Let’s expand the product to the sum ofdm terms. Let Ψ denote the set of the dm functions ψ : {1, ...,m} → {0, ..., d− 1},which will serve to index this sum. Now

S =∑

x∈GF(q)

∑ψ∈Ψ

m∏i=1

(χ(fi(x)))ψ(i) =∑

x∈GF(q)

∑ψ∈Ψ

χ(fψ(x)),

where χ(fψ(x) =∏mi=1 fi(x)

ψ(i). Let ψ0(i)def= 0 for all i and Ψ∗ = Ψ \ {ψ0}. Afterso ∀x fψ0

(x)=1

switching the order of summation and separating the term corresponding to ψ0,this (“main”) term will be q. For the “error term” R = S − q we have

|R| ≤∑ψ∈Ψ∗

∣∣∣ ∑x∈GF(q)

χ(fψ(x))∣∣∣.


By our assumption for the d-th power independence of {f1, ..., fm} and as each ψ(i)is at most d− 1, we can use Theorem 5.31 (note that fψ have at most D distinctroots in the splitting field), so the inner sum has absolute value ≤ (D−1)

√q. Hence

|R| ≤ dm(D − 1)√q and from (∗) we get |N − q

dm | ≤ (D − 1)√q + D/d ≤ D

√q

(as if D/d >√q then Lemma 5.30 clearly holds).

The bounds in Result 5.31 can be strengthened when f(X) and χ are quadratic.

Exercise 5.32. Let q be an odd prime power, f(X) = aX2 + bX + c ∈ GF(q)[X]with a 6= 0, and let χ denote the quadratic character GF(q) → {1,−1, 0}. Then∑

x∈GF(q)

χ(ax2 + bx+ c) ={−χ(a), if b2 − 4ac 6= 0;(q − 1)χ(a), if b2 − 4ac = 0.

5.8 Lacunary polynomials

A polynomial is called lacunary if an interval is missing from its terms, i.e. someconsecutive coefficients happen to be zero. A polynomial of GF(q)[X] is fully re-ducible if it splits to linear factors over GF(q). These two requirements proved tobe hard to satisfy simultaneously, there are a few results stating that fully re-ducible lacunary polynomials (the extent of lacunarity must be specified) shouldhave some particular feature.The theory of lacunary polynomials is treated in Redei’s book [103], where thefollowing two problems are considered: and a third

problem as atool for Problem2Problem 5.33. Problem 1 of Redei Let d|q − 1, d > 1. Determine the fully

reducible polynomials f(X) = Xq−1d + g(X) for which deg g ≤ q−1

d2 , X 6 |g, and fhas no multiple factors.

Problem 5.34. Problem 2 of Redei Determine the fully reducible polynomialsf(X) = Xq + g(X), f(X) ∈ GF(q)[X] \ GF(q)[Xp], for which deg g ≤ q+1

2 .

In geometric applications a modified version of Problem 2 will be crucial, where f it is meaningfulif only q 6= pis allowed to be a polynomial of Xp:

Problem 5.35. Problem 2’ Determine the fully reducible polynomials f(X) =Xq + g(X), f(X) ∈ GF(q)[X], for which deg g ≤ q+1

2 .

Problem 1 is rather easy:

Theorem 5.36. (Redei, [103]) If d > 2 then the only solutions of Problem 1 are theEuler-binoms X

q−1d − α, where α = u

q−1d for a u ∈ GF(q)∗. If d = 2 and q ≡ 1

(mod 4) then there are other solutions as well, namely the four polynomials so if −1 is asquare

20 I. Background(X

q−14 ± 1

)(X

q−14 ± γ

),

where γ2 = −1.

Let’s consider Problem 2. The conditions in it do not seem natural. The followinglemma ([103], Satz 18) shows that the degree of g is always at least q+1

2 unlessf(X) = Xq −X. The degree of a polynomial h is denoted by h◦.

Lemma 5.37. Let s be a power of p with 1 ≤ s < q and suppose that

Xq/s + g(X) ∈ GF(q)[X] \ GF(q)[Xp]

is fully reducible over GF(q). Then either s = 1 and g(X) = −X orin particular s=1

gives g◦ ≥ q+12

g◦ ≥ q + s

s(s+ 1).

Proof: A zero of Xq/s + g of multiplicity k is a zero of (Xq/s + g)′ = g′ ofmultiplicity at least k − 1 and a zero of Xq −X with multiplicity 1. Hence

Xq/s + g | ((Xq/s + g)s − (Xq −X))g′ = (gs +X)g′.

By assumption g′ 6= 0 so either gs = −X and hence s = 1 or the right hand sideis non-zero and q/s ≤ sg◦ + g◦ − 1.

At this stage one can solve Problem 2 in the q = p prime case.

Theorem 5.38. [103] If q = p 6= 2 is a prime then the solutions of Problem 2 areprecisely the four polynomials

f(X) = (X + a)((X + a)

p−12 − σ

)((X + a)

p−12 − στ

), σ = ±1, τ = 0, 1.

It means that either f has one single and p−12 double roots, or it has one root of

multiplicity p+12 and p−1

2 other single roots.Proof: As the translation x 7→ x+a does not change the problem, we can assumethat in g(X) = a0X

p+12 + a1X

p−12 + ... + a p+1

2the coefficient a1 is zero. Fromthis translation

changes a p+12

as well becausefrom Xp we getan extra ap = a

the proof of Lemma 5.37 (s = 1) one can see that f(X)|(g(X) + X)g′(X); astheir degrees are equal, one is constant times the other. Comparing the leadingcoefficients we get

a20

2f(X) =

a20

2(Xp + g(X)) = (g(X) +X)g′(X). (∗)

In particular


(i) g(X) +X and g′(X) are fully reducible;(ii) g(X) +X divides Xp −X hence its roots are all single.

In (∗) one can observe that the coefficients of Xp−1, ..., Xp+32 are all zero. From

this, comparing coefficients we get that a2 = a3 = ... = a p+32

= 0, so g(X) =a0 6= 0, a1 = 0

a0Xp+12 + a p−1

2X + a p+1

2. Now g′(X) = a0

2 Xp−12 + a p−1

2will be reducible only

if a p−12

= ±a02 , 0. Now using full reducibility and comparing the coefficients the

statement of the theorem follows.

The following generalization of Lemma 5.37 is easy:

Exercise 5.39. [32] Suppose that f(X) = Xqg(X) + h(X) ∈ GF(q)[X] (q = pn, pprime) is fully reducible over GF(q), (g, h) = 1. Then either f(X) = a(Xq −X);f(X) = aXq + b = (aX + b)q; or q = p and max(g◦, h◦) ≥ p+1

2 ; or n ≥ 2 andmax(g◦, h◦) ≥ p[n+1

2 ].

The best (currently available) result of this kind is the following (up to my knowl-edge):

Theorem 5.40. [39] Suppose that f(X) = Xqg(X) + h(X) ∈ GF(q)[X] (q = pn, pprime) is fully reducible over GF(q), (g, h) = 1. Let k = max(g◦, h◦) < q. Let e bemaximal such that f is a pe-th power. Then we have exactly one of the followingcases:

(1) e = n and k = 0;

(2) e ≥ 2n/3 and k ≥ pe;

(3) 2n/3 > e > n/2 and k ≥ pn−e/2 − 32pn−e;

(4) e = n/2 and k = pe and f(X) = aTrq→√q(bX + c) + d or f(X) =

aNormq→√q(bX + c) + d for suitable constants a, b, c, d, e;

(5) e = n/2 and k ≥ pe⌈

14 +

√(pe + 1)/2

⌉;

(6) n/2 > e > n/3 and k ≥ pn/2+e/2 − pn−e − pe/2, or, if 3e = n+ 1 and p ≤ 3then k ≥ pe(pe + 1)/2;

(7) n/3 ≥ e > 0 and k ≥ ped(pn−e + 1)/(pe + 1)e;

(8) e = 0 and k ≥ (q + 1)/2;

(9) e = 0, k = 1 and f(X) = a(Xq −X).

22 I. Background

The following lemma will be required, for example to prove Theorem 18.14.

Lemma 5.41. Let s be a power of p with 1 ≤ s < q and suppose that

f = Xq/s + g ∈ GF(q)[X] \ GF(q)[Xp]

is fully reducible over GF(q). If 3(g′)◦ < 2q/s− sg◦ then

f ∈ 〈1, X,Xs, Xs2 , ..., Xq/s〉GF(q).

Proof: We have the same divisibility as in Lemma 5.37. Let m(X) be defined by

(Xq/s + g)m = (X + gs)g′. (∗)

Nowm◦ = sg◦−q/s+(g′)◦. Differentiating (∗), we get another equation; combiningthem we conclude to

(Xq/s + g)(g′′m− g′m′) = (m− 1)(g′)2.

The degree of the right hand side is at most m◦+2g′◦ ≤ sg◦− q/s+3(g′)◦ < q/s,so g′′m− g′m′ = 0, implying (m− 1)g′ = 0 and hence m = 1. It follows that g′ isconstant and from the differential equation (∗) it is an easy task to deduce that fis in the required span.

6 Representations of affine and projective geometries

In order to profit from the Desarguesian structure of an affine or projective geom-etry, a few ways of representation or coordinatization is used. In each of them thefollowing questions should be answered:

• How can the subspaces be seen or handled?

• How can one calculate the intersection or span of two subspaces?

• How can the subgeometries and other “nice subsets”, like quadrics, Hermi-tian surfaces, etc. be seen or handled?

• How can transformations (collineations, polarities) be represented?

• How can general pointsets be handled?

The standard representation of AG(n, q) is just GF(q)n = V(n, q), the n-standardrepresentationof AG(n, q) dimensional vectorspace over GF(q), where the k-subspaces of AG(n, q) are rep-

resented by the k-dimensional affine subspaces (i.e. translates of linear subspaces)of GF(q)n, for k = 0, 1, ..., n, and incidence is the natural one.

6. Representations of affine and projective geometries 23

In the standard homogeneous representation of PG(n, q) the k-subspaces ofstandardhomogeneousrepresentationof PG(n, q)

PG(n, q) are represented by the (k+1)-dimensional linear subspaces of GF(q)n+1 =V(n + 1, q), for k = 0, 1, ..., n, and incidence is the natural one. In particular,points can be represented by nonzero vectors (v1, ..., vn+1) and another vector(w1, ..., wn+1) represents the same point if 〈v〉 = 〈w〉, i.e. there is a λ ∈ GF(q) forwhich wi = λvi for i = 1, ..., n+ 1.

When coordinatizing, an inner product should be chosen and used, in these tworepresentations the easiest is the standard inner product.

Hence a hyperplane, i.e. a 1-codimensional subspace can be represented by any(nonzero) vector [v1, ..., vn+1] being orthogonal to it(s points), i.e. for any pointu of the hyperplane

∑uivi = 0; and another vector [w1, ..., wn+1] represents the

same hyperplane if 〈v〉 = 〈w〉.

Exercise 6.1. Calculate the number of k-dimensional subspaces of V(n, q), AG(n, q)and PG(n, q). You may want to use the notation

[ab

]q

= (qa−1)(qa−1−1)...(qa−b+1−1)(qb−1)(qb−1−1)...(q−1)

q-binomials

and θi =[i+11

]q.

As AG(n, q) can be imagined as an n-dimensional vectorspace over GF(q), and sois GF(qn), it is quite natural to identify them. The incidence structure of AG(n, q) affine

big fieldrepresentationwill be recognised in the following way: three points A,B,C are collinear if for

the corresponding elements a, b, c of GF(qn) there exists a λ ∈GF(q) such thatc = (1 − λ)a + λb or equivalently λ(a − b) = a − c. Taking (q − 1)th powers killsthe coefficient λ ∈ GF(q) and we get (a− b)q−1 = (a− c)q−1. That is why we canidentify the directions with the θn−1-th roots of unity in GF(qn), which are the(q − 1)-st powers, and then one can say that A and B determine the direction(a− b)q−1.

In this and in the next representation, when an inner product is needed, the usualone is 〈x, y〉 = Trqn→q(xy) (or Trqn+1→q(xy)); we will see that this is the mostnatural one.

As PG(n, q) can be imagined as an (n + 1)-dimensional vectorspace over GF(q)‘modulo’ multiplication by non-zero scalars, it is quite natural to identify it with projective

big fieldrepresentationGF(qn+1)∗/GF(q)∗, i.e. the elements x and y of GF(qn+1)∗ represent the same

point of PG(n, q) if and only if x = λy for some λ ∈ GF(q). Note that in thiscase xq−1 = yq−1 as λq−1 = 1 for all λ ∈ GF(q)∗; this is again the common trickto eliminate factors from a subfield. As usually the hyperplanes are representedby the same set as the points (referring to the geometrically self-dual structure),we will make it so. The incidence structure of PG(n, q) will be recognized in thefollowing way: the point A will be incident with the hyperplane H iff for thecorresponding elements a and h of GF(qn+1) we have Trqn+1→q(ah) = 0. For thisthe most convenient way is the use of a trace-orthogonal normal base (if exists, asfor example in the planar case), see Section 5.2.

24 I. Background

Suppose that ω is a primitive element of GF(qn+1), i.e.

GF(qn+1)∗ = {1 = ω0, ω1, ω2, ..., ωqn+1−2}.

Then for the subfield GF(q)∗ = {1 = ω0, ωθn , ω2θn , ..., ω(q−2)θn}. Now one cancyclic projectiverepresentation

It is commonto imaginePG(n, q) as thehyperplane atinfinity, i.e. setof directions ofAG(n + 1, q).

see what ‘modulo GF(q)∗’ means: PG(n, q) is represented as the factor group ofthese two cyclic groups; for calculations the convenient representation is either{ω0, ω1, ω2, ..., ωθn−1} or {ω0, ωq−1, ω2(q−1), ..., ω(θn−1)(q−1)}, so points (and thehyperplanes) are represented by the θn-th roots of unity in GF(qn+1), which arethe (q − 1)-st powers. In the first one the map ωi 7→ ωi+1 (mod θ)n, in the sec-ond one the map ωi(q−1) 7→ ω(i+1)(q−1) (mod qn+1 − 1) gives a cyclic (“Singer-”)automorphism, being regular on both the points and the hyperplanes. This repre-sentation is very useful if the cyclic structure, or some very regular substructureof the space is examined.

6.1 Subspaces, subgeometries

It is vital to be able to describe subspaces and subgeometries in affine and projec-tive spaces. In the standard representations they are quite obvious however.In the affine big field representation the points x of a hyperplane are Trqn→q(ax)+b = 0, where a ∈ GF(qn)∗ and b ∈ GF(q). A suitable linear combination of k hy-perplane polynomials gives an equation

∑n−k+1j=0 αjX

qj +β = 0 whose zeros corre-spond to an (n−k+1)-dimensonal subspace, the intersection of the k hyperplanes.In particular, lines are given by Xq − αX + β = 0, and for a line joining x1 andx2 we have α = (x1 − x2)q−1. The non-zero (q − 1)-th powers are θn−1-th rootsand β = αx1 −

x1 = αx2 − x2 of unity in GF(qn), so we see the one-to-one correspondence between the θn−1-throots of unity in GF(qn) and the θn−1 direction of lines in AG(n, q).In the projective representation we have a slightly more complicated situation.Given a set of indeterminates {Xi : i = 0, ..., n} the hyperplanes of PG(n, q) (i.e.subspaces of V(n+ 1, q) of rank n) are given by linear homogeneous equations ofthe form

n∑i=0

ciXi = 0, (∗)

where (c0, c1, ..., cn) is a point of PG(n, q). The points of PG(n, q) are subspaces ofrank 1 in V(n+1, q) which in GF(qn+1) are given by the sets of zeros of equations ofthe form Xq = uX where uq

n+qn−1+...+q+1 = 1. This is a necessary and sufficientcondition on u for the polynomial Xq − uX to divide Xqn+1 − X and hence besplitting poly of

GF(qn+1) a polynomial that splits completely into distinct linear factors over GF(qn+1).Hence we got again that it makes sense to refer to the points of PG(n, q) as(qn+ qn−1 + ...+ q+1)-st roots of unity in GF(qn+1). In GF(qn+1) the polynomialobserve the link

with the cyclicrepresentation!

Trqn+1→q(γqi

X) = γqi

X + γqi+1Xq + ...+ γq

i+n−1Xqn−1

+ γqi+n

Xqn

6. Representations of affine and projective geometries 25

splits completely into distinct linear factors over GF(qn+1), has degree qn andis linear over GF(q). Hence we choose γ = ω to be a fixed primitive elementof GF(qn+1) and consider the hyperplane Xi = 0 of PG(n, q) as the equationTrqn+1→q(ωq

i

X) = 0, over GF(qn+1), and in general the hyperplane (*) as theequation

Trqn+1→q

((n∑i=0

ciωqi

︸︷︷︸c

) X)

= 0.

In other words, the points x of a hyperplane are Trqn+1→q(cx) =

0 = cxn∑i=0

cqi−1xq

i−1. Writing z = xq−1 and a = cq−1 we get∑ni=0 a

(qi−1)/(q−1)z(qi−1)/(q−1).

A suitable linear combination of k hyperplane polynomials gives an equation∑n−k+1j=0 αjZ

θj + β = 0 whose zeros correspond to an (n − k + 1)-dimensionalsubspace, the intersection of the k hyperplanes. In particular lines are given byZq+1 − αZ + β = 0 (where there exist relations between α and β depending onthe direction), and for a line joining z1 and z2 (viewed as (q − 1)-th powers) wehave α = (zq+1

1 − zq+12 )/(z1 − z2).

For example, the lines of PG(3, q) represented in GF(q4) are obtained by lookingat the set of zeros of polynomials whose zeros are zeros of two such hyperplanepolynomials and we conclude that these have the form L(Z) := Zq

2+ cZq + eZ

for some c and e in GF(q4). These polynomials must have q2 distinct zeros inGF(q4) and hence divide Zq

4 − Z. The polynomial Lq2 − cq

2Lq − (eq

2 − cq2+q)L

(mod Zq4−Z) has degree q and q2 zeros and is therefore identically zero. Equating

coefficients gives the following necessary and sufficient conditions that

cq+1 = eq − eq2+q+1 and eq

3+q2+q+1 = 1. (∗∗)

6.2 Transformations

As is known, the automorphism group of PG(n, q) and AG(n, q) are PΓL(n+ 1, q)and AΓL(n, q), respectively. Here we describe them in the non-standard represen-tations as well.

Theorem 6.2. Any automorphism ϕ : AG(n, q) → AG(n, q), ϕ : x 7→ ϕ(x), wherethe point x is represented in the standard way, can be written in the form ϕ(x) =Mxσ + v, where M is a non-singular linear map , v ∈ V(n, q) gives a translation i.e.

M ∈ GL(n, q)and σ : GF(q) → GF(q) is a field automorphism.Frobenius-automorphism,possibly =idTheorem 6.3. Any automorphism ϕ : PG(n, q) → PG(n, q), ϕ : x 7→ ϕ(x), where

the point x is represented with homogeneous coordinates, can be written in the formϕ(x) = Mxσ, where M is a non-singular linear map and σ : GF(q) → GF(q) is a i.e. M ∈ GL(n+

1, q)field automorphism.Frobenius-automorphism,possibly =id

26 I. Background

So we have to understand what a linear transformation, a translation and a fieldautomorphism means in the big field representation r : V(n, q) → GF(qn).Translation is easy: adding the vector v to each point of V(n, q) is the very sameas adding its representative r(v) to each element of GF(qn).Field automorphism is also easy: the automorphism σ of GF(q) is just among theautomorphisms of GF(qn), and it “commutes with r”, i.e. it is the same to use itin the big field representation.The interesting one is the linear map. Let its matrix be M = (mij), so Mx =y, yi =

∑n−1j=0 mijxj . On the other hand, a polynomial f(X) = c0X + c1X

q +

c2Xq2+...+cn−1X

qn−1, with c0, ..., cq−1 ∈ GF(qn) defines a map GF(qn) → GF(qn),

which is linear over GF(q). One can check that the number of matrices is equalto the number of these GF(q)-linear polynomials (i.e. qn

2). See Section 5.6 about

polynomials being linear over a subfield.Given M , the corresponding f(X) depends on the base chosen in GF(qn). If abase of the form {ω, ωq, ωq2 , ..., ωqn−1} was chosen, then to find fM we haver(Mx) =

∑n−1k=0 ykω

qk =∑n−1k=0

∑n−1j=0 mkjxjω

qk =∑n−1j=0 xj

∑n−1k=0 mkjω

qk and

fM (r(x)) =∑n−1i=0 ci

∑n−1j=0 xjω

qj+i =∑n−1j=0 xj

∑n−1i=0 ciω

qj+i so to find the coeffi-cients c0, ..., cn−1 we have to solve the linear equation

ωq0ωq1

ωq2... ωqn−1

ωq1ωq2

ωq3... ωq0

ωq2ωq3

ωq4... ωq1

ωqn−1ωq0

ωq1... ωqn−2

c0c1c2

cn−1

=

P

k mk0ωqkPk mk1ωqkPk mk2ωqk

Pk mk,n−1ωqk

= M>

ωq0

ωq1

ωq2

...

ωqn−1

The assumption that M is a non-singular linear map is equivalent to the following:fM (X) has no root in GF(qn)∗.

7 The Redei polynomial and its derivatives

7.1 The Redei polynomial

Generally speaking, a Redei polynomial is just a (usually multivariate) polynomialwhich splits into linear factors. We use the name Redei polynomial to emphasizethat these are not only fully reducible polynomials, but each linear factor cor-responds to a geometric object, usually a point or a hyperplane of an affine orprojective space.Let S be a pointset of PG(n, q), S = {Pi = (ai, bi, ..., di) : i = 1, ..., |S|}.The (Redei-)factor corresponding to a point Pi = (ai, bi, ..., di) is PiV = aiX +biY + ... + diT . This is simply the equation of hyperplanes passing through Pi.When we decide to examine our pointset with polynomials, and if there is no

7. The Redei polynomial and its derivatives 27

special, distinguished point in S, it is quite natural to use symmetric polynomialsof the Redei-factors. The most popular one of these symmetric polynomials is theRedei-polynomial, which is the product of the Redei-factors, and the *-polynomial,which is the (q − 1)-th power sum of them. I have not found

any good namefor it yet

Definition 7.1. The Redei-polynomial of the pointset S is defined as follows:

RS(X,Y, ..., T ) = R(X,Y, ..., T ) :=|S|∏i=1

(aiX + biY + ...+ diT ) =|S|∏i=1

Pi ·V.

The points (x, y, ..., t) of R, i.e. the roots R(x, y, ..., t) = 0, correspond to hyper-planes (with the same (n+1)-tuple of coordinates) of the space. The multiplicity of the fundamen-

tal propertyof Redeipolynomials

a point (x, y, ..., t) on R is m if and only if the corresponding hyperplane [x, y, ..., t]intersects S in m points exactly.

Given two pointsets S1 and S2, for their intersection

RS1∩S2(X,Y, ..., T ) = gcd(RS1(X,Y, ..., T ) , RS2(X,Y, ..., T )

)holds, while for their union, if we allow multiple points or if S1 ∩ S2 = ∅, we have

RS1∪S2(X,Y, ..., T ) = RS1(X,Y, ..., T ) · RS2(X,Y, ..., T ).

Definition 7.2. The *-polynomial of S is

GS(X,Y, ..., T ) = G(X,Y, ..., T ) :=|S|∑i=1

(aiX + biY + ...+ diT )q−1.

If a hyperplane [x, y, ..., t] intersects S in m points then the corresponding m termswill vanish, hence G(x, y, ..., t) = |S|−m (in other words, all m-secant hyperplanes modulo the char-

acteristicwill be solutions of G(X,Y, ..., T )− |S|+m = 0).The advantage of the *-polynomial (compared to the Redei-polynomial) is that itis of lower degree if |S| ≥ q. The disadvantage is that while the Redei-polynomialcontains the complete information of the pointset (S can be reconstructed fromit), the *-polynomial of two different pointsets may coincide. This is a hard taskin general to classify all the pointsets belonging to one given *-polynomial.The *-polynomial of the intersection of two pointsets does not seem to be easyto calculate; the *-polynomial of the union of two pointsets is the sum of their obviously

*-polynomials.The next question is what happens if we transform S. Let M ∈ GL(n+ 1, q) be alinear transformation. Then

RM(S)(V) =|S|∏i=1

(MPi) ·V =|S|∏i=1

Pi · (M>V) = RS(M>V).

28 I. Background

For a field automorphism σ, Rσ(S)(V) = (RS)(σ)(V), which is the polynomial RS

but all coefficients are changed for their image under σ.Similarly GM(S)(V) = GS(M>V) and Gσ(S)(V) = (GS)(σ)(V).The following statement establishes a further connection between the Redei poly-nomial and the *-polynomial.

Lemma 7.3. (Gacs) For any set S,

RS · (GS − |S|) = (Xq −X)∂XRS + (Y q − Y )∂YRS + ...+ (T q − T )∂TRS .

In particular, RS(GS − |S|) is zero for every substitution [x, y, ..., t].

Proof: Trivial induction on |S|.

We remark that

∂XG(X,Y, ..., T ) = −|S|∑i=1

ai(aiX + biY + ...+ diT )q−2.

Note that αq−2 = 1α for all 0 6= α ∈ GF(q). Compare this to the derivative of the

Redei polynomial, see below.

Next we shall deal with Redei-polynomials in the planar case n = 2. This caseis already complicated enough, it has some historical reason, and there are manystrong results based on algebraic curves coming from this planar case. Most ofthe properties of “Redei-surfaces” in higher dimensions can be proved in a verysimilar way, but it is much more difficult to gain useful information from them.

Let S be a pointset of PG(2, q). Let LX = [1, 0, 0] be the line {(0, y, z) : y, z ∈GF(q), (y, z) 6= (0, 0)}; LY = [0, 1, 0] and LZ = [0, 0, 1]. Let NX = |S ∩ LX | andNY , NZ are defined similarly. Let S = {Pi = (ai, bi, ci) : i = 1, ..., |S|}.

Definition 7.4. The Redei-polynomial of S is defined as follows:

R(X,Y, Z) =coefficient-polynomials

|S|∏i=1

(aiX+biY +ciZ) =|S|∏i=1

Pi ·V=r0(Y,Z)X |S|+r1(Y,Z)X |S|−1 + ...+r|S|(Y, Z).

For each j = 0, ..., |S|, rj(Y, Z) is a homogeneous polynomial in two variables,either of total degree j precisely, or (for example when 0 ≤ j ≤ NX − 1) rjis identically zero. If R(X,Y, Z) is considered for a fixed (Y, Z) = (y, z) as apolynomial of X, then we write Ry,z(X) (or just R(X, y, z)). We will say that R isa curve in the dual plane, the points of which correspond to lines (with the sametriple of coordinates) of the original plane. The multiplicity of a point (x, y, z)the fundamen-

tal propertyof Redeipolynomials

on R is m if and only if the corresponding line [x, y, z] intersects S in m pointsexactly.


Remark 7.5. Note that if r = 1, i.e. [x, y, z] is a tangent line at some (at, bt, ct) ∈S, then R is smooth at (x, y, z) and its tangent at (x, y, z) coincides with the onlylinear factor containing (x, y, z), which is atX + btY + ctZ.

Exercise 7.6. Let S be the pointset of the parabola X2 − Y Z in PG(2, q). Provethat GS(X,Y, Z) = Xq−1 if q is even and GS(X,Y, Z) = (X2 − 4Y Z)

q−12 if q is

odd. What is the geometrical meaning of it?

Exercise 7.7. Let S be the pointset of the parabola X2 − Y Z in PG(2, q). Provethat

RS(X,Y, Z) = Y∏

t∈GF(q)

(tX + t2Y + Z) = Y (Zq + Y q−1Z − C q−12Yq−12 Z

q+12 −

−C q−32X2Y

q−32 Z

q−12 − C q−5

2X4Y

q−52 Z

q−32 − ...− C1X

q−3Y Z2 − C0Xq−1Z),

where Ck = 1k+1

(2kk

)are the famous Catalan numbers.

Remark. If there exists a line skew to S then w.l.o.g. we can suppose that LX∩S =∅ and all ai = 1. If now the lines through (0, 0, 1) are not interesting for somereason, we can substitute Z = 1 and now R is of form

R(X,Y ) =|S|∏i=1

(X + biY + ci) = X |S| + r1(Y )X |S|−1 + ...+ r|S|(Y ).

This is the affine Redei polynomial. Its coefficient-polynomials are rj(Y ) =σj({biY + ci : i = 1, ..., |S|}), elementary symmetric polynomials of the linearterms biY + ci, each belonging to an ‘affine’ point (bi, ci). In fact, substitutingy ∈ GF(q), biy+ ci just defines the point (1, 0, biy+ ci), which is the projection of(1, bi, ci) ∈ S from the center ‘at infinity’ (0,−1, y) to the line (axis) [0, 1, 0].One may ask what happens with the affine Redei polynomial if the pointset{(bi, ci)} is translated by the vector (u, v) or enlarged by λ. Then RS+(u,v)(X,Y ) =RS(X + uY + v, Y ) and RλS(X,Y ) = λ|S|RS( 1

λX,Y ).

Exercise 7.8. Suppose that S = {(x, f(x)) : x ∈ GF(q)} ⊂ AG(2, q), i.e. the graphof a function Y = f(X), where f(X) = anX

n + an−1Xn−1 + ...+ a0. Express the

Redei polynomial of S, or some coefficient if it, from the ai-s.

How can one “see” the points of S from its Redei polynomial?

1. They can be seen from the factors of R(X,Y, Z);

2. Let {1, ω, ω2} be a base of GF(q3) over GF(q). Then if bω + cω2 is a root ofR(X,ω, ω2) then (−1, b, c) is a point of S. (You can’t see the points (0, b, c)this way.)

30 I. Background

3. If S is an affine pointset then, using the affine Redei polynomial R(X,Y )and the GF(q2)-representation of AG(2, q), with ω ∈ GF(q2) \ GF(q), nowR(X,ω) has roots exactly a+ bω where (−a, b) are the points of S.

7.2 Differentiation in general

Here we want to introduce some general way of “differentiation”. Give each pointPi the weight µ(Pi) = µi for i = 1, ..., |S|. Define the curve

R′µ(X,Y, Z) =|S|∑i=1

µiR(X,Y, Z)

aiX + biY + ciZ. (∗)

If ∀µi = ai then R′µ(X,Y, Z) = ∂XR(X,Y, Z), and similarly, ∀µi = bi means ∂YRand ∀µi = ci means ∂ZR.

Theorem 7.9. Suppose that [x, y, z] is an m-secant with S ∩ [x, y, z] ={Pti(ati , bti , cti) : i = 1, ...,m}.

(a) If m ≥ 2 then R′µ(x, y, z) = 0. Moreover, (x, y, z) is a point of the curve R′µof multiplicity at least m− 1.

(b) (x, y, z) is a point of the curve R′µ of multiplicity at least m if and only iffor all the Ptj ∈ S ∩ [x, y, z] we have µtj = 0.

(c) Let [x, y, z] be an m-secant with [x, y, z] ∩ [1, 0, 0] 6∈ S. Consider the line[0,−z, y] of the dual plane. If it intersects R′µ(X,Y, Z) at (x, y, z) with in-tersection multiplicity ≥ m then

∑mj=1

µtjatj

= 0.

Proof: (a) Suppose w.l.o.g. that (x, y, z) = (0, 0, 1) (so every ctj = 0).Substituting Z = 1 we have R′µ(X,Y, 1). In the sum (∗) each term offor help

see Section 10 ∑i 6∈{t1,...,tm} µi

R(X,Y,1)aiX+biY+ci

will contain m linear factors through (0, 0, 1), so, afterexpanding it, there is no term with (total) degree less than m (in X and Y ).

Consider the other terms contained in

∑i∈{t1,...,tm}

µiR(X,Y, 1)aiX + biY

=R(X,Y, 1)

RS∩[0,0,1](X,Y, 1)

m∑j=1

µtjRS∩[0,0,1](X,Y, 1)atjX + btjY

.

Here R(X,Y,1)RS∩[0,0,1](X,Y,1)

is non-zero in (0, 0, 1). Each term RS∩[0,0,1](X,Y,1)atjX+btjY

contains at

least m− 1 linear factors through (0, 0, 1), so, after expanding it, there is no termwith (total) degree less than (m − 1) (in X and Y ). So R′µ(X,Y, 1) cannot havesuch a term either.


(b) As RS∩[0,0,1] ′

µ (X,Y, 1) is a homogeneous polynomial in X and Y , of totaldegree (m − 1), (0, 0, 1) is of multiplicity exactly (m − 1) on R(X,Y, 1), unlessRS∩[0,0,1] ′

µ (X,Y, Z) happens to vanish identically.

Consider the polynomials RS∩[0,0,1](X,Y,1)atjX+btjY

. They are m homogeneous polynomials

in X and Y , of total degree (m − 1). Form an m ×m matrix M from the coeffi-cients. If we suppose that atj = 1 for all Ptj ∈ S ∩ [0, 0, 1] then the coefficient of

Xm−1−kY k in RS∩[0,0,1](X,Y,1)atjX+btjY

, so mjk is σk({bt1 , ..., btm} \ {btj}) for j = 1, ...,m

and k = 0, ...,m − 1. So M is the elementary symmetric matrix (see in Section 9on symmetric polynomials) and |detM | =

∏i<j(bti − btj ), so if the points are all

distinct then detM 6= 0. Hence the only way of RS∩[0,0,1] ′

µ (X,Y, 1) = 0 is when∀j µtj = 0.In order to prove (c), consider the line [0,−z, y] in the dual plane. To calculate itsintersection multiplicity with R′µ(X,Y, Z) at (x, y, z) we have to look at R′µ(X, y, z)and find out the multiplicity of the root X = x. As before, for each term of∑i 6∈{t1,...,tm} µi

R(X,y,z)aiX+biy+ciz

this multiplicity is m, while for the other terms wehave ∑

i∈{t1,...,tm}

µiR(X, y, z)

aiX + biy + cz=

R(X, y, z)RS∩[x,y,z](X, y, z)

m∑j=1

µtjRS∩[x,y,z](X, y, z)atjX + btjy + ctjz

.

Here R(X,y,z)RS∩[x,y,z](X,y,z)

is non-zero at X = x. Now RS∩[x,y,z](X, y, z) =∏mj=1 atjX+

btjy + ctjz.

Each term RS∩[x,y,z](X,y,z)atjX+btj y+ctj z

is of (X-)degree at most m − 1. We do know that the

degree of∑mj=1 µtj

RS∩[x,y,z](X,y,z)atjX+btj y+ctj z

is at least (m− 1) (or it is identically zero), asthe intersection multiplicity is at least m−1. So if we want intersection multiplicity≥ m then it must vanish, in particular its leading coefficient

(m∏j=1

atj )m∑j=1

µtjatj

= 0.

Remark. If ∀µi = ai, i.e. we have the partial derivative w.r.t X, then eachµtjatj

are

equal to 1. The multiplicity in question remains (at least) m if and only if on thecorresponding m-secant [x, y, z] the number of “affine” points (i.e. points differentfrom (0,−z, y)) is divisible by the characteristic p.

In particular, we may consider the case when all µ(P) = 1.Consider

R1 =∑b∈B

R(X,Y, Z)b1X + b2Y + b3Z

= σ|B|−1({b1X + b2Y + b3Z : b ∈ B}).

32 I. Background

For any ≥ 2-secant [x, y, z] we have R1(x, y, z) = 0. It does not have a lin-ear component if |B| < 2q and B is minimal, as it would mean that all thelines through a point are ≥ 2-secants. Somehow this is the “prototype” of “allthe derivatives” of R. E.g. if we coordinatize s.t. each b1 is either 1 or 0, then∂1XR =

∑b∈B\LX

R(X,Y,Z)b1X+b2Y+b3Z

, which is a bit weaker in the sense that it containsthe linear factors corresponding to pencils centered at the points in B ∩ LX .Substituting a tangent line [x, y, z], with B ∩ [x, y, z] = {a}, into R1 we getR1(x, y, z) =

∏b∈B\{a}(b1x + b2y + b3z), which is non-zero. It means that R1

contains precisely the ≥ 2-secants of B. In fact an m-secant will be a singularpoint of R1, with multiplicity at least m− 1.A similar argument shows, that in general, if

Rt = σ|B|−t({b1X + b2Y + b3Z : b ∈ B}),

then for any ≥ t + 1-secant [x, y, z] we have Rt(x, y, z) = 0. It does not have alinear component if |B| < (t + 1)q and B is minimal, as it would mean that allthe lines through a point are ≥ t + 1-secants. Somehow this is the “prototype”of “all the t-th derivatives” of R. E.g. if we coordinatize s.t. each b1 is either1 or 0, then ∂tXR = σ|B|−t({b1X + b2Y + b3Z : b ∈ B \ LX}), which is a bitweaker in the sense that it contains the linear factors corresponding to pencilscentered at the points in B ∩ LX . Substituting a t-secant line [x, y, z] into Rt weget Rt(x, y, z) =

∏b∈B\[x,y,z](b1x+b2y+b3z), which is non-zero. It means that R0

contains all the ≥ t+ 1-secants but none of the t-secants of B. In fact Rt containsall the m-secants with multiplicity at least m− t.

7.3 Hasse derivatives of the Redei polynomial

The next theorem is about Hasse derivatives of R(X,Y, Z). (For its properties seeSection 5.4.)

Theorem 7.10. (1) Suppose [x, y, z] is an r-secant line of S with [x, y, z] ∩ S ={(asl , bsl , csl) : l = 1, ..., r}. Then (Hi

XHjYH

r−i−jZ R)(x, y, z) =

R(x, y, z) ·∑

m1<m2<...<mimi+1<...<mi+jmi+j+1<...<mr

{m1,...,mr}={1,2,...,r}

asm1asm2

...asmi bsmi+1...bsmi+j csmi+j+1

...csmr ,

where R(x, y, z) =∏l 6∈{s1,...,sr}(alx + bly + clz), a non-zero element, inde-

pendent from i and j.

(2) From this we also have∑0≤i+j≤r

(HiXH

jYH

r−i−jZ R)(x, y, z)XiY jZr−i−j = R(x, y, z)

r∏l=1

(aslX+bslY+cslZ),

constant times the Redei polynomial belonging to [x, y, z] ∩ S.


(3) If [x, y, z] is a (≥ r + 1)-secant, then (HiXH

jYH

r−i−jZ R)(x, y, z) = 0.

(4) If for all the derivatives (HiXH

jYH

r−i−jZ R)(x, y, z) = 0 then [x, y, z] is not

an r-secant.

(5) Moreover, [x, y, z] is a (≥ r+1)-secant iff for all i1, i2, i3, 0 ≤ i1 + i2 + i3 ≤ rthe derivatives (Hi1

XHi2Y H

i3Z R)(x, y, z) = 0.

(6) The polynomial∑0≤i+j≤r

(HiXH

jYH

r−i−jZ R)(X,Y, Z)XiY jZr−i−j

vanishes for each [x, y, z] (≥ r)-secant lines.

(7) In particular, when [x, y, z] is a tangent line to S with [x, y, z] ∩ S ={(at, bt, ct)}, then

(∇R)(x, y, z) = ( (∂XR)(x, y, z), (∂YR)(x, y, z), (∂ZR)(x, y, z) ) = (at, bt, ct).

If [x, y, z] is a (≥ 2)-secant, then (∇R)(x, y, z) = 0. Moreover, [x, y, z] is a(≥ 2)-secant iff (∇R)(x, y, z) = 0.

Proof: (1) comes from the definition of Hasse derivation and from aslx+ bsly +cslz = 0; l = 1, ..., r. In general (Hi1

XHi2Y H

i3Z R)(X,Y, Z) =∑

m1<m2<...<mi1mi1+1<...<mi1+i2

mi1+i2+1<...<mi1+i2+i3|{m1,...,mi1+i2+i3

}|=i1+i2+i3

am1am2 ...ami1 bmi1+1 ...bmi1+i2cmi1+i2+1 ...cmi1+i2+i3

∏i 6∈{m1,...,mi1+i2+i3}

(aiX+biY+ciZ).

(2) follows from (1). For (3) observe that after the “r-th derivation” of R stillremains a term asix + bsiy + csiz = 0 in each of the products. Suppose that forsome r-secant line [x, y, z] all the r-th derivatives are zero, then from (2) we getthat

∏rl=1(aslX + bslY + cslZ) is the zero polynomial, a nonsense, so (4) holds.

Now (5) and (7) are proved as well. For (6) one has to realise that if [x, y, z] is anr-secant, still

∏rl=1(aslx+ bsly + cslz) = 0.

Or: in the case of a tangent line

∇R =|S|∑j=1

∇(Pj ·V)∏i 6=j

Pi ·V =|S|∑j=1

Pj

∏i 6=j

(Pi ·V).

7.4 Examples

In this section we compute the polynomials (curves) for some well-known pointsets. Thecomputations are also used to illustrate several results and ideas of this book. In the cases

34 I. Background

when the pointset is a blocking set, we use the notation f = (f1, f2, f3) and g = (g1, g2, g3)for certain curves (see Section 21.2) satisfying R = f · (X, Y, Z) = g · (Y qZ−Y Zq, ZqX−ZXq, XqY −XY q).

Example 7.11.

The line [a, b, c]. Its Redei polynomial is a(Y qZ−Y Zq)+b(ZqX−ZXq)+c(XqY −XY q).

Example 7.12.

A conic. For the Redei polynomial of the parabola X2 − Y Z see Exercise 7.7.

Example 7.13.

The projective triangle. Let q be odd, B = {(1, 0, 0); (0, 1, 0); (0, 0, 1)} ∪ {(a2, 0, 1);(1,−a2, 0); (0, 1, a2) : a ∈ GF(q)∗}. Then

R(X, Y, Z) = XY Z((−Z)q−12 −X

q−12 )(X

q−12 − Y

q−12 )((−Y )

q−12 − Z

q−12 ) =

(Xq −X)Y Z[Zq−12 − (−Y )

q−12 ] + (Y q − Y )XZ[(−X)

q−12 − Z

q−12 ]+

(Zq − Z)XY [(−Y )q−12 − (−X)

q−12 ] =

(XqY−XY q)Z[Zq−12 −(−Y )

q−12 −(−X)

q−12 ]+(Y qZ−Y Zq)X[(−X)

q−12 −Z

q−12 −(−Y )

q−12 ]+

(ZqX − ZXq)Y [(−Y )q−12 − Z

q−12 − (−X)

q−12 ].

Note that g = 0 iff [x, y, z] = [α2, 1, 0] or [1, 0,−α2] or [0,−α2, 1], so for the 2-secants.

Example 7.14.

The sporadic almost-Redei blocking set. The affine plane of order 3 can be embedded intoPG(2, 7) as the points of inflexion of a non-singular cubic. The 12 lines of this plane covereach point of PG(2, 7), so in the dual plane they form a blocking set of size 12 = 3(7+1)/2,but its maximal line-intersection is only 4 = (12− 7)− 1. A characterization of it can befound in [69].

A representation of this blocking set is the following: its affine part is U ={(x,−x6 + 3x3 + 1, 1) : x ∈ GF(7)} ∪ {(0,−1, 1)}, the infinite part is D ={(1, 0, 0), (1, 1, 0), (1, 2, 0), (1, 4, 0)}. Now

R(X, Y, Z) = X10Y 2−X10Z2−X7Y 5−2X7Y 3Z2 +3X7Y Z4−X4Y 8 +X4Z8 +XY 11 +XY 9Z2 + 4XY 7Z4 + XY 3Z8

from which we get f =

(X3Y 2−X3Z2−Y 5−2Y 3Z2 +3Y Z4, −X4Y +XY 4 +XY 2Z2 +4XZ4, X4Z +XY 3Z)

and g = (XY 2Z, X3Z + 2Y 3Z, X3Y − Y 4 + 3Z4).

Note that g(x, y, z) = (0, 0, 0) iff (x, y, z) ∈ {[1, 0, 0]; [1, 2, 0]; [1, 4, 0]} and all three are2-secants.

Example 7.15.

8. Univariate representations 35

The Baer-subplane.

In PG(2, q) the standard embedding of a Baer subplane is {(a, b, 1) : a, b ∈ GF(√

q)} ∪{(m, 1, 0) : m ∈ GF(

√q)} ∪ {(1, 0, 0)}.

Now its Redei polynomial is

R(X, Y, Z) =Y

a,b∈GF(√

q)

(aX + bY + Z)Y

a∈GF(√

q)

(aX + Y )X =

(Xq −X)[Y Z√

q − Y√

qZ] + (Y q − Y )[X√

qZ −XZ√

q] + (Zq − Z)[XY√

q −X√

qY ] =

(Y qZ − Y Zq)X√

q + (ZqX − ZXq)Y√

q + (XqY −XY q)Z√

q.

Here the equation of the blocking set is

X√

q(c√

qb− cb√

q) + Y√

q(a√

qb− ab√

q) + Z√

q(a√

qc− ac√

q) = 0.

Example 7.16.

In PG(2, q3) let U = {(x, x + xq + xq2, 1) : x ∈ GF(q3)}, and D the directions determined

by them, |D| = q2 + 1.

Now for B = U ∪D we get

f = ( Xq2Z + Xq2−qY qZ −Xq2−qY Zq + Y q2

Z − Y Zq2, Xq2

Z + Xq2−q+1Zq + XZq2,

−Xq2+1 −Xq2Y −Xq2−q+1Y q −XY q2

)

and g = ( −Xq2, Xq2

+ Xq2−qY q + Y q2, Xq2−qZq + Zq2

).

Example 7.17.

In PG(2, q3) let U = {(x, xq, 1) : x ∈ GF(q3)}, and D the directions determined by them,D = {(1, aq−1, 0) : a ∈ GF(q3)∗}, |D| = q2 + q + 1.

Then, after the linear transformation (1, aq−1, 0) 7→ (1− aq−1, aq−1 − β, 0), where β is a(q − 1)-st power, we have

rNZ = (X − βY )q2+q+1 − (X − Y )q2+q+1.

Example 7.18.

The Hermitian curve. In PG(2, q), the Hermitian curve, which is a unital, {(x, y, z) :x√

q+1 + y√

q+1 + z√

q+1 = 0} in PG(2, q) has the following Redei polynomial:

R(X, Y, Z) = (X√

q+1 + Y√

q+1 + Z√

q+1)q−√q+1 −Xq√

q+1 − Y q√

q+1 − Zq√

q+1.

8 Univariate representations

Here we describe the analogue of the Redei polynomial for the big field represen-tations.

36 I. Background

8.1 The affine polynomial and its derivatives

After the identification AG(n, q) ↔ GF(qn), described in Section 6, for a subsetS ⊂ AG(n, q) one can define the root polynomialroot polynomial

BS(X) = B(X) =∏s∈S

(X − s) =∑k

(−1)kσkX |S|−k;

and the direction polynomialdirectionpolynomial

F (T,X) =∏s∈S

(T − (X − s)q−1) =∑k

(−1)kσkT |S|−k.

Here σk and σk denote the k-th elementary symmetric polynomial of the set S and{(X − s)q−1 : s ∈ S}, respectively. The roots of B are just the points of S whileF (x, t) = 0 iff the direction t is determined by x and a point of S, or if x ∈ S andt = 0.If F (T, x) is viewed as a polynomial in T , its zeros are the θn−1-th roots of unity,moreover, (x− s1)q−1 = (x− s2)q−1 if and only if x, s1 and s2 are collinear.In the special case when S = Lk is a k-dimensional affine subspace, one may thinkthat BLk will have a special shape.We know that all the field automorphisms of GF(qn) are Frobenius-automorphismsx 7→ xq

m

for i = 0, 1, ..., n − 1, and each of them induces a linear trans-formation of AG(n, q). Any linear combination of them, with coefficients fromGF(qn), can be written as a polynomial over GF(qn), of degree at most qn−1.These are called linearized polynomials. Each linearized polynomial f(X) induceslinearized

polynomials a linear transformation x 7→ f(x) of AG(n, q). What’s more, the converse isalso true: all linear transformations of AG(n, q) arise this way. Namely, distinctlinearized polynomials yield distinct transformations as their difference has de-gree ≤ qn−1 so cannot vanish everywhere unless they were equal. Finally, boththe number of n × n matrices over GF(q) and linearized polynomials of formc0X + c1X

q + c2Xq2 + ... + cn−1X

qn−1, ci ∈ GF(qn) is (qn)n. This is the same

argument as in Section 5.6.Here we show (what we did already in 6.1) that

Proposition 8.1. (i) The root polynomial of a k-dimensional subspace of AG(n, q)containing the origin, is a linearized polynomial of degree qk;(ii) the root polynomial of a k-dimensional subspace of AG(n, q) is a linearizedpolynomial of degree qk plus a constant term.

Proof: (i) For k = 0 the statement is obvious, B0 = X. Any one-dimensionallinear subspace is of form {aλ : λ ∈ GF(q)} = aGF(q) for some a ∈ GF(qn)∗.The root polynomial of the one-dimensional subspace corresponding to GF(q) isBGF(q) = Xq −X, so the root polynomial of aGF(q) is BGF(q)(a−1X), which is alinearized polynomial of degree q.

8. Univariate representations 37

Now if L is a subspace of dimension k > 1 then let K be a subspace of L ofdimension k− 1. By induction, the root polynomial BK is a linearized polynomialinducing a linear map with kernel K. The image of L is one-dimensional, BK(L) =M . It is easy to check that BL(X) = BM (BK(X)).

(ii) Any subspace is a translate of a linear subspace; consider any subspace v + Lwith v ∈ GF(qn), its root polynomial is BL(X − v). As BL is a linear map, wehave BL(X − v) = BL(X)−BL(v).

A more compact form of the equation of a hyperplane is Trqn→q(aX) + b = 0,where a, b ∈ GF(qn), a 6= 0. Indeed, for the root polynomial

BH(X) =n−1∑i=0

ciXqi + b

of the hyperplaneH, the polynomial cn−1BqH−c

q+1n−1(X

qn−X)−cqn−2BH has degreeat most qn−2 and vanishes in all the qn−1 points of H, hence it is identically zero.Equating the coefficients of Xqi for 0 ≤ i ≤ n− 2 implies the trace function form Here is the

proof ofExercise 5.2!above.

Now we examine the derivative(s) of the affine root polynomial (written up witha slight modification). Let S ⊂ GF(qn) and consider the root and direction poly-nomials of S[−1] = {1/s : s ∈ S}:

B(X) =∏s∈S

(1− sX) =∑k

(−1)kσkXk;

F (T,X) =∏s∈S

(1− (1− sX)q−1T ) =∑k

(−1)kσkT k.

For the characteristic function χ of S[−1] we have |S|−χ(X) =∑s∈S(1−sX)q

n−1.Then, as B′(X) = B(X)

∑s∈S

11−sX , we have (X −Xqn)B′ = B(|S| −

∑s∈S(1−

sX)qn−1) = Bχ, after derivation B′ + (X −Xqn)B′′ = B′χ+Bχ′, so B′ ≡ (Bχ)′

and (as Bχ ≡ 0) we have BB′ ≡ B2χ′.

8.2 The projective polynomial

After the identification PG(n, q) ↔ GF(qn+1)/GF(q)∗, described in Section 6, fora subset S ⊂ PG(n, q) one can define C0(X) =

∏s∈S Tr(sX) =

= X |S|(∏s∈S

s)∏s∈S

((Xq−1)θn−1(sq−1)θn−1 +...+(Xq−1)θ1(sq−1)θ1 +Xq−1sq−1+1

).

38 I. Background

So it is enough to consider C(X) =∏s∈S

((n−1∑i=0

(Xq−1)θi(sq−1)θi) + 1)

=

∏β∈S[q−1]

((n−1∑i=0

Y θiβθi) + 1)

=∏

β−1∈S[q−1]

((n−1∑i=0

Y θiβθn−1−θi) + βθn−1

), where Y =

Xq−1.

Here the direction polynomial is

F (U,X) =∏s∈S

(1− sX(1− sX)q−1U).

For a value x, the linear factors of F (U, x) have zeros of the form u−1 = sx(1 −sx)q−1 which satisfies 1+v+vq+1+vq

2+q+1+...+vθn−1 = 0 for v = u−1. Moreover,s1x(1 − s1x)q−1 = s2x(1 − s2x)q−1 if and only if 1/x, s1 and s2 are collinear, sothere is a one-to-one correspondence between the zeros v and the θn−1 directionof lines through 1/x.

8.3 Examples

This section is about the polynomials for some well-known pointsets, like in a formerpart. We leave them as exercises.

Affine sets

For an affine line we have the equation Xq − c0X + b = 0 as we have seen already.

Exercise 8.2. Calculate the polynomials of the following affine pointsets.

For an ellipse, i.e. the q + 1 points of a conic contained in AG(2, q).

An affine Baer-subplane: A = AG(2,√

q) = {(a, b) : a, b ∈ GF(√

q)} ! {a + bω : a, b ∈GF(

√q)} ⊂ GF(q2) .

Projective sets

Exercise 8.3. Calculate the polynomials of the following projective pointsets.

A conic.

The projective triangle., see Section 7.4.

A Baer-subplane PG(2,√

q) of PG(2, q) in “standard position”.

The cyclic embedding of the Baer-subplane: GF(q3) = 〈w〉, B = 〈wq−√q+1〉.The cyclic (q −√

q + 1)-arc

The Hermitian curve.

9. Symmetric polynomials and invariants of subsets 39

9 Symmetric polynomials and invariants of subsets

9.1 The Newton formulae

In this section we recall some classical results on symmetric polynomials. For moreinformation and the proofs of the results mentioned here, we refer to [135].The multivariate polynomial f(X1, ..., Xt) is symmetric, if f(X1, ..., Xt) =f(Xπ(1), ..., Xπ(t)) for any permutation π of the indices 1, ..., t. Symmetric polyno-mials form a (sub)ring (or submodule over F) of F[X1, ..., Xt]. The most famousparticular types of symmetric polynomials are the following two:

Definition 9.1. The k-th elementary symmetric polynomial of the variablesX1, ..., Xt is defined as

σk(X1, ..., Xt) =∑

{i1,...,ik}⊆{1,...,t}

Xi1Xi2 · · ·Xik .

σ0 is defined to be 1 and for j > t σj = 0, identically.

Given a (multi)set A = {a1, a2, ..., at} from any field, it is uniquely determined byits elementary symmetric polynomials, as

t∑i=0

σi(A)Xt−i =t∏

j=1

(X + aj).

Definition 9.2. The k-th power sum of the variables X1, ..., Xt is defined as

πk(X1, ..., Xt) :=t∑i=1

Xki .

The power sums determine the (multi)set a “bit less” than the elementary sym-metric polynomials. For any fixed s we have

s∑i=0

(s

i

)πi(A)Xs−i =

t∑j=1

(X + aj)s

but in general it is not enough to gain back the set {a1, ..., at}. Note also thatin the previous formula the binomial coefficient may vanish, and in this case it“hides” πi as well.One may feel that if a (multi)set of field elements is interesting in some sense thenits elementary symmetric polynomials or its power sums can be interesting as well.E.g.

A = GF(q): σj(A) = πj(A) ={

0, if j = 1, 2, ..., q − 2, q ;−1, if j = q − 1. πq−1(GF(q)) =Q

α∈GF(q)∗α = −1

is Wilson’s theo-rem

40 I. Background

If A is an additive subgroup of GF(q) of size pk: σj(A) = 0 whenever p 6 |j < q − 1holds. Also πj(A) = 0 for j = 1, ..., pk − 2, pk.If A is a multiplicative subgroup of GF(q) of size d|(q − 1): σj(A) = πj(A) = 0 forj = 1, ..., d− 1.

The fundamental theorem of symmetric polynomials: Every symmetric polynomialcan be expressed as a polynomial in the elementary symmetric polynomials.

There are several proofs for this theorem, the most well-known one uses inductionfor the exponent-series of the terms after ordering the terms in a lexicographicway; one may formulate the proof in the language of initial ideals (see [72]), orthere is another one using Galois-theory, etc.According to the fundamental theorem, also the power sums can be expressedin terms of the elementary symmetric polynomials. The Newton formulae areequations with which one can find successively the relations in question. Essentiallythere are two types of them:

kσk = π1σk−1 − π2σk−2 + ...+ (−1)i−1πiσk−i + ...+ (−1)k−1πkσ0 (N1)

andπt+k − πt+k−1σ1 + ...+ (−1)iπt+k−iσi + ...+ (−1)tπkσt = 0. (N2)

In the former case 1 ≤ k ≤ t, in the latter k ≥ 0 arbitrary. Note that if we defineσi = 0 for any i < 0 or i > t, and, for a fixed k ≥ 0, π0 = k, then the followingequation generalizes the previous two:

k∑i=0

(−1)iπiσk−i = 0. (N3)

Exercise 9.3. Prove the Newton identities by differentiating

B(X) =∏s∈S

(1 + sX) =|S|∑i=0

σiXi.

Symmetric polynomials play an important role when we use Redei polynomials, ase.g. expanding the affine Redei polynomial

∏i(X+aiY + bi) by X, the coefficient

polynomials will be of the form σk({aiY + bi : i}).Magic. (Gacs) Let f(X) =

∑q−1i=0 ciX

i be a polynomial from GF(q)[X]. Considerthe following (affine Redei-) polynomial of its graph:∏

a∈GF(q)

(X − aY − f(a));


then its homogeneous part of degree q − 1 is precisely the homogenized form off , i.e.

∑q−1i=0 ciX

iY q−1−i.

Proof: Let γk be the coefficient of Xq−1−kY k in the Redei-polynomial, we proveby induction that γk = cq−1−k. Let GF(q) = {a1, ..., aq} and bi = f(ai).We know that γ0 = −

∑bi = cq−1 and it can be called

σk,1(a1,...,aq ; b1,...,bq)

γk = (−1)k+1∑

i1<...<ik

ai1 · · · aik∑

i 6={i1,...,ik}

bi

if 1 ≤ k ≤ q − 1. Hence γk = (−1)k+1σk(a1, ..., aq)∑qi=1 bi + γk−1, where γk−1 is

defined as γk−1, with bi replaced by aibi.Let f(X) = (c0 + cq−1)X+

∑q−1i=2 ci−1X

i; then f(ai) = aibi. By induction we mayassume that γk−1 is the coefficient of Xq−k in f(X). Hence we obtain γk = γk−1 =cq−k−1 for k < q − 1; finally γq−1 = −cq−1 + γq−2 = c0.

Corollary 9.4. Expanding the Redei-polynomial

∏a∈GF(q)

(X − aY − f(a)) =q∑

k=0

rk(Y )Xq−k,

for k = 1, ..., q − 2 we get degY (rk) ≤ k − 1; equality holds iff cq−k 6= 0.

Proof: The first part is easy: rk(Y ) = (−1)kσk({(aY + f(a)) : a ∈ GF(q)}) sothe leading coefficient is just σk(GF(q)) = 0. The second part, the case of equalityfollows from the previous statement.

Given S = {x1, x2, ..., xn}, the following determinant is called the Vandermonde-determinant of S:

V dM(x1, x2, ..., xn) =

∣∣∣∣∣∣∣∣∣1 1 ... 1x1 x2 ... xn

x21 x2

2 ... x2n

...

xn−11 xn−1

2 ... xn−1n

∣∣∣∣∣∣∣∣∣ =∏i<j

(xi − xj)

The P -adic (P = pe, p prime) Moore-determinant of S is

MRDP (x1, ..., xn) =

∣∣∣∣∣∣∣∣∣∣

x1 x2 ... xn

xP1 xP

2 ... xPn

xP2

1 xP2

2 ... xP2n

.

..

xPn−1

1 xPn−1

2 ... xPn−1n

∣∣∣∣∣∣∣∣∣∣=

∏(λ1,...,λn)∈PG(n−1,P )

n∑i=1

λixi

42 I. Background

Note that this formula gives the value of the determinant up to a non-zero con-stant only, one way to find the exact value is choosing each homogeneous vector usually we ask

det = 0 or not(λ1, λ2, ..., λn) in such a way that its first non-zero coordinate is 1.The elementary symmetric determinant of S is∣∣∣∣∣∣∣∣∣

1 1 ... 1σ1(S \ {x1}) σ1(S \ {x2}) ... σ1(S \ {xn})σ2(S \ {x1}) σ2(S \ {x2}) ... σ2(S \ {xn})

...σn−1(S \ {x1}) σn−1(S \ {x2}) ... σn−1(S \ {xn})

∣∣∣∣∣∣∣∣∣ =∏i<j

(xi − xj)

Exercise 9.5. Give a unified proof for the determinant formulae! (Hint: considerx1, ..., xn as free variables.)

Exercise 9.6.

(a) Prove that V dM(x1, x2, ..., xn) 6= 0 iff {x1, ..., xn} are pairwise distinct.

(b) Prove that MRDpe(x1, x2, ..., xn) 6= 0 iff {x1, ..., xn} are independent overGF(pe).

(c) Prove the following version of Wilson’s theorem:∏y∈〈x1,...,xn〉GF(pe)

y 6=0

y = (−1)nMRDpe(x1, ..., xn)pe−1.

Exercise 9.7. Prove the following general form of the elementary symmetric de-terminant: Given S = {x1, x2, ..., xn, xn+1, ..., xm},∣∣∣∣∣∣∣∣∣

1 1 ... 1σ1(S \ {x1}) σ1(S \ {x2}) ... σ1(S \ {xn})σ2(S \ {x1}) σ2(S \ {x2}) ... σ2(S \ {xn})

...σn−1(S \ {x1}) σn−1(S \ {x2}) ... σn−1(S \ {xn})

∣∣∣∣∣∣∣∣∣ =∏

1≤i<j≤n

(xi − xj),

so somehow the elements xn+1, ..., xm “do not count”.

Exercise 9.8. Prove that∏λ∈GF(q)n

(T + λ1X1 + λ2X2 + ...+ λn−1Xn−1 + λn) =

=

∣∣∣∣∣∣∣∣∣∣∣∣∣

T X1 X2 ... Xn−1 1T q Xq

1 Xq2 ... Xq

n−1 1

T q2Xq2

1 Xq2

2 ... Xq2

n−1 1

.

..

..

.

T qn Xqn

1 Xqn

2 ... Xqn

n−1 1

∣∣∣∣∣∣∣∣∣∣∣∣∣/∣∣∣∣∣∣∣∣∣∣

X1 X2 ... Xn−1 1Xq

1 Xq2 ... Xq

n−1 1

Xq2

1 Xq2

2 ... Xq2

n−1 1

...

Xqn−1

1 Xqn−1

2 ... Xqn−1

n−1 1

∣∣∣∣∣∣∣∣∣∣.


9.2 The invariants uf , vf and wf

In this section, which is extracted from [135] mostly, we consider a fixed polynomialf(X) =

∑ni=0 ciX

i ∈ GF(q)[X] of degree n. Put sk = σk({f(x) : x ∈ GF(q)}), so∏x∈GF(q)(X + f(x)) =

∑qk=0 skX

q−k.

• Let u = uf be the smallest positive integer k with sk 6= 0 if such k exist, u

otherwise set u = ∞.

• Let v = vf denote the number of distinct values of f (so v ≥ 2 if we assume v

that f is not a constant function). If v = q then f is called a permutationpolynomial (PP).

• Let w = wf be the smallest positive integer with pk = πk({f(x) : x ∈ w

GF(q)}) =∑x∈GF(q) f(x)k 6= 0 if such k exists, otherwise set w = ∞.

One can find another description of u and w in Proposition 9.9.

Firstly we study the relations among u, v, w, n, q. Then we shall investigate thenumber Mf of c ∈ GF(q) such that f(X)− cX is a PP. Mf

With any family S of q elements of GF(q), counted possibly with multiplicity, wemay associate the same invariants u = uS , v = vS , w = wS as above; obviouslythe order of the values is irrelevant. However, n measures the “complexity” of thisset (ordered in a certain sense) somehow. If g(X) is a permutation polynomialthen f(g(X)) has the same invariants u, v and w as f ; and the same is true if f isreplaced by af(X) + b with a, b ∈ GF(q), a 6= 0 (and b = 0 if v = 1).

The similar invariants can be defined for a (multi)set S of cardinality 6= q. Notethat if |S| < q then the multiset S ∪ (q− |S|){0} of size q has the same invariantsu and w as S, while v, the number of distinct elements of S, increases by one iff0 6∈ S.

If |S| > q and v = q then for S′ = S \ GF(q) we have uS′ = uS and wS′ = wS ifany of these values is less than q − 1.

Proposition 9.9. (a) q− u is the degree of Xq −∏a∈GF(q)(X − f(a)). This poly-

nomial is equal to c ∈ GF(q) iff f(X) = c is a constant polynomial. Hencev ≥ 2 implies 1 ≤ u < q and v = 1 implies u = q or u = ∞ according tof = 0 or f = c 6= 0.

(b) q−1−w = deg(∑

a∈GF(q)(X−f(a))q−1)

= deg((∏a∈GF(q)(X−f(a)) )′

).

(c)∑a∈GF(q) a

kf(a) = −cq−1−k for 0 ≤ k < q − 1 and∑a∈GF(q) a

q−1f(a) =−(c0 + cq−1). Hence n = q − 1 iff u = 1 iff w = 1. For arbitrary f we havethat f has reduced degree ≤ q − l− 2 iff

∑a∈GF(q) a

kf(a) = 0 for 0 ≤ k ≤ l.(We will see it later again.)

44 I. Background

(d) From (c) it is clear that w is the smallest positive integer k such that thereduction of f(X)k mod Xq − X has degree q − 1, if such k exists, andotherwise w = ∞.

Note that if v ≥ 2 then (c) implies that one can change the order of the valuessuch that the polynomial will have degree q− 1 or q− 2 (in fact one transpositionis enough); while the invariants u, v, w remain the same.

Lemma 9.10. (a) If 1 ≤ k ≤ q and k < u+ w then pk = (−1)k−1ksk.(b) If ksk 6= 0 for some 1 ≤ k ≤ q − 1 then w is the smallest k with this property,otherwise w = ∞.

For (a) use the Newton identity, for (b) see (d) below.

Here we simply enlist some of the relations known for u, v, w, n, q, most of themare easy to prove (an asterisk denotes some of the non-obvious ones):

Proposition 9.11.

(a) v = 1 ⇔ n = 0 or n = −∞, u = q in the first and u = ∞ in the lattercase;

(b) if n ≥ 1 then v ≥ q/n and hence v ≥ [ q−1n + 1]. v = q/n ⇒ w = ∞ or

n = 1. v = [ q−1n + 1] implies n|q − 1 or w = ∞. If g.c.d.(n, p) = 1 then

the case n|q−1 can occur only. If n < p then f is of form a(X− b)n+ cwith a 6= 0.

(c) n = q − 1 ⇔ u = 1 ⇔ w = 1;

(d) If w < ∞ then w < q (since f(a)q = f(a)); and p 6 |w (since pkp =(pk)p). Compare to the section on Vandermonde sets, 9.3;

(e) w <∞⇒ w < v;

(f) w ≥ q−1n , with equality iff n|q − 1;

(g) w = ∞ ⇔ for each a ∈ GF(q) the number na of preimages of a isdivisible by p, see Exercise 9.17.

(h) w = ∞⇒ n ≥ p, v ≤ q/p (if n ≥ 1);

(i) if f(X) = Xp −X then w = ∞, n = p, v = q/p;

(j) if f(X) = Xn then w = v − 1;

(k) n|q − 1 ⇒ v ≤ q − 1;

(l) w <∞⇒ u ≤ w, with equality iff p 6 |u;(m) w = ∞⇒ p|u or u = ∞;

(n) (∗) w <∞⇒ w < 2(q − v) or v = q;


(o) (∗) u ≥ q−1n , with equality iff n|q − 1;

(p) 2 ≤ v < q ⇒ u+ v ≤ q;

(q) v < q ⇒ v < q − q−1n (if n ≥ 1);

(r) v < q ⇒ w ≤ 2q/3− 1 if w <∞, (this bound is probably not optimal);

(s) v < q ⇒ v + w ≤ 4q/3 − 1 if w < ∞, (this bound is probably notoptimal);

(t) v < q ⇒ v + w ≤ q if v > q − p, v > 1.

The proof of (e) is obvious from the first equality of Proposition 9.9 (b): if g(X) =∑a∈GF(q)(X − f(a))q−1 then for every b, which is not in the range of f(X), we

have f(b) = 0. For another proof (and a generalization) see Exercise 9.14.

Proposition 9.12. The following properties are equivalent if 1 ≤ n < q:

(a) v = q, i.e. f is a permutation polynomial;

(b) u = q − 1;

(c) u > q − q/n;

(d) u > q − v;

(e) v > q − q−1n ;

(f) w = q − 1;

(g) 2q/3− 1 < w <∞;

(h) q − q+1n < w <∞;

(i) q − u ≤ w <∞;

(j) u > q−12 and w <∞.

Exercise 9.13. Prove that if wf ≥ q−1 then either (i) f is a permutation polynomialand

∑x f(x)q−1 = −1; or (ii)

∑x f(x)q−1 = 0 and f assumes every value 0 mod p

times.A general form of Proposition 9.11 (e) is the following:

Exercise 9.14. (P. Das) Let f(X1, ..., Xn) be a polynomial, its value set (range) isV (f) = {f(α1, ..., αn) : (α1, ..., αn) ∈ GF(q)n}. Define wf as the smallest positiveinteger k such that

∑(α1,...,αn)∈GF(q)n

f(α1, ..., αn)k 6= 0. Prove (using a Vander-

monde determinant) that |V (f)| ≥ wf + 1.

We defined Mf as the number of elements in C = {c ∈ GF(q) : f(X) − cX is a Mf

permutation polynomial}. An alternative definition can be given with the help ofthe following

46 I. Background

Proposition 9.15. Let D = Df = { f(a)−f(b)a−b : a 6= b ∈ GF(q)}, the set of difference

quotients (or directions), determined by (the graph of) f . Then −C = GF(q) \D.

Later we will use the notation N = Nf = |D| for the number of directions deter-mined by f . The obvious connection is

Mf +Nf = q.

For more on directions we refer to Section 18. Note that Mf is “a bit less invariant”than u, v, w are; in general, if f is changed for a · f(g(X)) + b ·X + c, Mf remainsthe same only if g(X) is a linear polynomial (and not an arbitrary permutation,as in the earlier cases, but there the term bX was not allowed).

Let nk denote the reduced degree of f(X)k. Here we summarize some propertiesnk

of Mf :

1. If f(X) + cX is a permutation polynomial then for 1 ≤ k < q − 1, in thenotation

∏a∈GF(q)(X−aY −f(a)) =

∑qk=0 rk(Y )Xq−k (see Magic), we have

rk(c) = 0. So if rk(Y ) 6= 0 then degY (rk) ≥Mf .

2. n1 < q −Mf .

3. If Mf < p then nk < q −Mf + k − 1 for all k ≥ 1.

4. If Mf + 1 < p and n1 < q −Mf − 1 then nk < q −Mf + k − 2 for all k ≥ 1.

5. If f(X) = Xn and n|q then Xn − cX is a PP iff an − ca 6= 0 if a 6= 0.Hence (if n > 1) q −Mf is the number of (n − 1)-st powers in GF(q)∗, soq −Mf = q−1

g.c.d.(n−1,q−1) . In particular, for n = q/p we have Mf = q − q−1p−1 ,

while n =√q yields Mf = q −√q − 1.

6. Mf ≤ 1 if 1 < n4 < p = q.

7. Mf ≤ n− 1 if n 6 |q and n4 ≤ q.

8. Mf ≤ q − 1 with equality iff f has reduced degree ≤ 1.

9. Mf ≤ q − q−1n−1 if 1 < n < q.

10. If Mf < q − 1 then Mf ≤ q −√q − 1. See Section 21.

11. (Lovasz-Schrijver) If q = p thenMf ≤ p−32 . IfMf = p−3

2 then f(X) = Xp+12 ,

up to a linear transform; see Theorem 19.1 as well.


12. (Redei) If Mf >q−32 then

(i) if f(X)−cX is not a PP then the number of preimages of each valuef(a)− ca is divisible by p;

(ii) Mf ≥ q − q−1p−1 and p|(Mf + 1);

(iii) if n < q then f ′(X) = f ′(0).

(iv) in fact much more is known: in this case, after assuming f(0) = 0, fis “linear over GF(pe)” for some e|h (so GF(pe) is a subfield of GF(q = ph));for details see [43] and Theorem 18.14.

In Section 9.4 we shall see that Mf is somewhat related to double symmetricpolynomials and double power sums as well.

9.3 Arbitrary subsets

Now we consider the more general case when the number of elements is not nec-essarily q.

Definition 9.16. Let 1 < t < q. We say that T = {y1, ..., yt} ⊆ GF(q) is aVandermonde-set, if πk =

∑i yki = 0 for all 1 ≤ k ≤ t− 2.

Vandermonde-sets were first defined and studied in [70]. This part is a generaliza-tion from [120].

Here we do not allow multiple elements in T . Observe that the power sums do notchange if the zero element is added to (or removed from) T . Note that in generalthe Vandermonde property is invariant under the transformations y → ay + b(a 6= 0) if and only if p|t; if p 6 |t then a “constant term” tbk occurs in the powersums. (It may help in some situations: we can “translate” T to a set with π1 = 0if needed.)

If p|t then a t-set cannot have more than t − 2 zero power sums (so in this caseVandermondeness means w = wT = t−1). This is an easy consequence of the factthat a Vandermonde-determinant of distinct elements cannot be zero: consider theproduct

1 1 ... 1y1 y2 ... yty21 y22 ... y2t...

yt−11 y

t−12 ... y

t−1t

111

.

.

.1

,

it cannot result in the zero vector.

Note that we have already seen that for a set of q elements the value w can be ∞.Recall that in that case the set has to contain multiple elements, in fact we hadw = ∞ ⇔ all the multiplicities are divisible by the characteristic p.

48 I. Background

Exercise 9.17. Prove that if 1 < t ≤ q, |T | = t and multiplicities are allowedthen wT = ∞ ⇔ all the multiplicities of the elements of T are divisible by thecharacteristic p.

The proof above, with slight modifications, shows that in general a t-set cannothave more than t−1 zero power sums (so for a Vandermonde-set wT is either t−1t − 1

or t). If the zero element does not occur in T then consider the producty1 y2 ... yty21 y22 ... y2t...

yt−11 y

t−12 ... y

t−1t

yt1 yt2 ... ytt

111

.

.

.1

,

it cannot result in the zero vector as the determinant is still non-zero. If 0 ∈ Tthen remove it and we are again in the zero-free situation.If for a set T of cardinality t we have that πk(T ) = 0 for k = 1, ..., t− 1, so wT = tthen such a set can be called a super-Vandermonde set. Note that the zero elementis never contained in a super-Vandermonde set (removing it, for the other t − 1elements all the first t − 1 power sums would be zero, which is impossible). Thesame argument gives the first examples of super-Vandermonde sets:

Example 9.18. If T is a Vandermonde set, containing the zero element, then T \{0}is a super-Vandermonde set. In particular, if T is a Vandermonde set and |T | = tis divisible by the characteristic p, then for any a ∈ T , the translate T − a is aVandermonde set, containing the zero element.

In the next proposition we characterize the Vandermonde-property.

Proposition 9.19. Let T = {y1, ..., yt} ⊆ GF(q). The following are equivalent

(i) T is a Vandermonde set, i.e. wT = t− 1;

(ii) the polynomial f(Y ) =∏ti=1(Y −yi) is of the form Y t

′g(Y )p+aY +b (where

0 ≤ t′ ≤ p− 1, t′ ≡ t mod p);

(iii) for the polynomial χ(Y ) = −∑ti=1(Y − yi)q−1, tY q−1 + χ(Y ) has degree

q − t; moreover

(iv) for some Q = ps, t < Q, the polynomial tY Q−1 −∑ti=1(Y − yi)Q−1 has

degree Q− t.

Proof: The coefficients of χ are the power sums of the set T , so (i) and (iii)are clearly equivalent. (i) ⇔(iv) is similar. The equivalence of (i) and (ii) is aneasy consequence of the Newton formulae relating power sums and elementarysymmetric polynomials.


Note that for the function χ in (iii), t + χ(Y ) is the characteristic function oft+χ(Y ) = χ(Y )if p|t

T , that is it is 1 on T and 0 everywhere else. (i) means that a Vandermonde setis equivalent to a fully reducible polynomial of form gp(Y ) + Y t + cY . (In theimportant case when p|t we have gp(Y ) + Y .)And now we characterize the super-Vandermonde-property.

Proposition 9.20. Let T = {y1, ..., yt} ⊆ GF(q). The following are equivalent

(i) T is a super-Vandermonde set, i.e. wT = t;

(ii) the polynomial f(Y ) = Πti=1(Y − yi) is of the form Y t

′g(Y )p + c (where

0 ≤ t′ ≤ p− 1, t′ ≡ t mod p);

(iii) for the polynomial χ(Y ) = −∑ti=1(Y − yi)q−1, tY q−1 + χ(Y ) has degree

q − t− 1; moreover

(iv) for some Q = ps, t < Q, the polynomial tY Q−1 −∑ti=1(Y − yi)Q−1 has

degree Q− t− 1.

Proof: Very similar to the Vandermonde one above.

Here come some really motivating examples for Vandermonde sets.

Example 9.21. Let q be a prime power.

(i) Any additive subgroup of GF(q) is a Vandermonde set.

(ii) Any multiplicative subgroup of GF(q) is a (super-)Vandermonde set.

(iii) For q even, consider the points of AG(2, q) as elements of GF(q2). Any q-setcorresponding to the affine part of a hyperoval with two infinite points is aVandermonde set in GF(q2).

(iv) Let q be odd and consider the points of AG(2, q) as elements of GF(q2) and aq+1-set A = {a1, ..., aq+1} in it, intersecting every line in at most two points(i.e. an oval or (q+ 1)-arc). Suppose that it is in a normalized position, i.e.∑ai = 0. Then A is a super-Vandermonde set in GF(q2).

Proof: (i) Suppose T is an additive subgroup of size t in GF(q). We want to provethat Proposition 9.19 (ii) is satisfied, that is f(Y ) = Πy∈T (Y − y) has only termsof degree divisible by p, except for the term Y . By Theorem 5.25 if we prove thatf is additive, hence GF(p)-linear, then this implies that f has only terms of degreea power of p.Consider the polynomial in two variables F (X,Y ) = f(X) + f(Y ) − f(X + Y ).First of all note that it has full degree at most t and that the coefficient of Xt andY t is zero. Considering F as a polynomial in X, we have

F (X,Y ) = r1(Y )Xt−1 + r2(Y )Xt−2 + · · ·+ rt(Y ),

50 I. Background

where ri(Y ) (i = 1, . . . t) is a polynomial in Y of degree at most i (and deg(rt) ≤t − 1). Now F (X, y) ≡ 0 for any y ∈ T (as a polynomial of X), so all ri-s haveat least t roots. Since their degree is smaller than this number, they are zeroidentically, so we have F (X,Y ) ≡ 0, hence f is additive.

(ii) Suppose T is a multiplicative subgroup of size t in GF(q). Then the polynomialf(Y ) = Πt

i=1(Y − yi) is of the form Y t − 1 so Proposition 9.20 (ii) is satisfied, weare done.

(iii) Let {x1, ..., xq} ⊆ GF(q2) correspond to the affine part of the hyperoval Hand ε1 and ε2 be (q+1)-st roots of unity corresponding to the two infinite points.Consider the polynomial χ(X) =

∑qi=1(X − xi)q−1. For any point x out of the

hyperoval every line through x meets H in an even number of points, and since(x − xi)q−1 represents the slope of the line joining the affine points x and xi,we have that χ(x) = ε1 + ε2 for any x /∈ {x1, ..., xq}. There are q2 − q differentchoices for such an x, while the degree of χ is at most q − 2, so χ(X) ≡ ε1 + ε2identically (that is, all coefficients of χ are zero except for the constant term), soby Proposition 9.19 (iv), we are done.

(iv) A short proof is that by Segre’s theorem 14.9 such a pointset is a conic if q isodd, so affine equivalent to the “unit circle” {α ∈ GF(q2) : aq+1 = 1}, which is amultiplicative subgroup.

Exercise 9.22. Given an additive subgroup G ≤ (GF(q),+), with baseG = 〈g1, ..., gt〉GF(P ), write up its root polynomial

∏g∈G(Y − g) as in Proposition

9.19 (ii).

For a multiplicative subgroup H = 〈α〉 ≤ (GF(q)∗, ·), |H| = t, its root polynomialis∏h∈H(Y − h) = Y t − 1.

Note that Proposition 9.19 (iv) implies that if T ⊆ GF(q1) ≤ GF(q2) then T is aVandermonde-set in GF(q1) if and only if it is a Vandermonde-set in GF(q2).

There are other interesting examples as well.

Example 9.23. Let q = qt−10 , then in GF(q) T = {1} ∪ {ωqi0 : i = 0, ..., t − 2} for

some element ω ∈ GF(q)∗ satisfying Trq→q0(ωk) = −1 for all k = 1, ..., t− 1.

As:t−2∑i=0

(ωqi0)k =

t−2∑i=0

(ωk)qi0 = Trq→q0(ω

k) = −1.

Note that such ω exists for several triples (t, q0, q), here I enlist some values; ”-”means that such ω does not exist, while ”x” means that the only element with theproperty above is 1 ∈ GF(qt−1

0 ), see Exercise 9.24 :


q0 t 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 183 x x x - x - x x4 x x x x x x ?5 - x - x - ?7 - - - x - ?8 x x x x x ?9 x - x - x - x ?

Exercise 9.24. Prove that if g.c.d.(q0, t) 6= 1 then the only element with theproperty above is 1 ∈ GF(qt−1

0 ).

Exercise 9.25. Let T = {y1, ..., yt} be a super-Vandermonde set. Prove that(y1y2− 1)(y1

y3− 1)· · ·(y1yt− 1)

= t.

Note that it is easy check the previous condition for a multiplicative subgroup;also that any element of T can play the role of y1, so in fact we have t conditions.For applications see for instance Section 13, Proposition 13.9 and its consequence,Theorem 13.12.

9.4 The generalized formulae

Following Gacs in this section, we define two functions in the independent variablesa1, ..., at and b1, ..., bt over any field K, where t is a positive integer. Let k and lbe two non-negative integers, (k, l) 6= (0, 0), k, l ≤ t.

Definition 9.26. The elementary symmetric polynomial of order (k, l) is defined as

σk,l(a1, ..., at; b1, ..., bt) =∑

{i1,...,ik}; {j1,...,jl}⊆{1,...,t}{i1,...,ik}∩{j1,...,jl}=∅

ai1ai2 · · · aik · bj1bj2 · · · bjl .

In other words, σk,l is the coefficient of Xt−l−kY k in the Redei polynomial∏ti=1(X + aiY + bi); or the coefficient of XkY l in

∏ti=1(aiX + biY + 1). It shows

again that σk,l’s determine the set {a1, ..., at; b1, ..., bt}.For k = l = 0, we define σ0,0(a1, ..., at; b1, ..., bt) := 1 identically.

Definition 9.27. The power sum of order (k, l) is defined to be

πk,l(a1, ..., at; b1, ..., bt) :=t∑i=1

aki bli.

Evaluating π0,l or πk,0, 00 may occur, which is defined to be 1.

52 I. Background

Note that for k = 0, σ0,l(a1, ..., at; b1, ..., bt) = σl(b1, ..., bt), the l-th elementarysymmetric polynomial of the bi-s, and σk,0(a1, ..., at; b1, ..., bt) = σk(a1, ..., at) sim-ilarly.

Taking l = 0 or k = 0 in the second definition, we get the corresponding powersum of the ais or bis, respectively.

The functions in the previous two definitions are somewhat ‘parallelly symmetric’;they do not change value if we take the same permutation of the first t and second tvariables simultaneously. A direct analogue of the fundamental theorem mentionedin the previous section does not seem to be true for such functions, but the Newtonformulae can be generalized:

Theorem 9.28. For fixed k and l, define π0,0 := k + l and σa,b = 0 for any a > t,b > t or a+ b > t. The following holds:

k∑r=0

l∑s=0

(−1)r+s(r + s

r

)πr,sσk−r,l−s = 0. (NG)

Proof: It is easy to see that on the left hand side we have a homogeneous poly-nomial of degree k + l with monomials ari b

siai1 · · · aik−r · bj1 · · · bjl−s , where all

indices are different and 0 ≤ r ≤ k, 0 ≤ s ≤ l.

If r and s are both positive, then we can get such a term from three summands:πr,sσk−r,l−s, πr−1,sσk+1−r,l−s and πr,s−1σk−r,l+1−s. Hence the coefficient of sucha monomial is (−1)r+s

(r+sr

)+ (−1)r+s−1(

(r+s−1r−1

)+(r+s−1r

)) = 0. (Note that if

the monomial in question exists, then necessarily k + 1− r ≤ t, l + 1− s ≤ t andk + l + 1 − r − s ≤ t, hence σr,s−1 and σr−1,s are not zero by definition. Similarremarks will hold for the next two cases.)

If r = 0, s > 1, then there are two summands giving the monomial in question:π0,sσk,l−s and π0,s−1σk,l+1−s, so the coefficient is (−1)s

(s0

)+ (−1)s−1

(s−10

)= 0

again. The s = 0, r > 1 case is similar.

The {r, s} = {0} and {r, s} = {0, 1} cases are the same, so what is left is to showthat the coefficient of a monomial of type ai1 · · · aik · bj1 · · · bjl is also zero. We getthis term k + l times from π0,0σk,l, −k times from π1,0σk−1,l and −l times fromπ0,1σk,l−1, so its coefficient is certainly zero (here the binomial coefficients are all1).

Remark 9.29. Taking l = 0, the theorem coincides with (N3).

Remark 9.30. There is a slightly more general formula:

k∑r=0

l∑s=0

f(r, s)πr,sσk−r,l−s = 0, (NG∗)


were π0,0 is defined to be 1 and the function f(r, s) is defined as follows. Let a andb be two fixed field elements, f(0, 0) = ak + bl, and for (r, s) 6= (0, 0),

f(r, s) = (−1)r+s((

r + s− 1s

)a+

(r + s− 1

r

)b

). (NG∗∗)

Proof: Looking through the proof of 9.28, it is easy to see that the functionf(r, s) has to satisfy the following equations: f(0, 0) + kf(1, 0) + lf(0, 1) = 0;f(r, 0) + f(r + 1, 0) = 0 for all r ≥ 1; f(0, s) + f(0, s + 1) = 0 for all s ≥ 1; andf(r, s)+f(r−1, s)+f(r, s−1) = 0 for all r ≥ 1, s ≥ 1. Now defining f(1, 0) = −a,f(0, 1) = −b, it can be seen by induction on k + l that the general solution forsuch a function is (NG∗∗).

(NG∗) seems to be much stronger, but (NG) has turned out to be sufficient forall applications found so far. In many applications, the set U = {(ai, bi) : i =1, ..., t} will be the graph (or almost a graph) of a polynomial. For the graph ofa polynomial, the double power sum πk,l becomes

∑x x

kf(x)l, the meaning ofwhich can be understood through Proposition 5.4.

For a demonstration of the strength of using Redei polynomials together withthese formulae see 15.10.

The invariant Wf

For any polynomial f , let

Wf =min{k+ l :∑

x∈GF(q)

xkf(x)l 6= 0}=min{k+ l : πk,l(a1, ..., aq; f(a1), ..., f(aq)) 6=0}

(where {ai : i} = GF(q)). Here k and l are non-negative integers; for k = 0 orl = 0, x0 or f(x)0 is defined to be the polynomial 1. By the remarks at the endof the previous section, Wf is the smallest k + l, for which xkf(x)l has reduceddegree p− 1. Wf is very similar to wf above. The second and third statements ofthe next lemma can be implicitly found in [91].

Lemma 9.31. (Gacs [66]) (i) For any of the transformations mentioned after Propo-sition 9.15 for Mf , the value of Wf remains unchanged;(ii) For any polynomial f , Wf +Nf ≥ p+ 1; Nf : number

of determineddirections(iii) For any non-linear polynomial, Wf ≤ p−1

2 with equality if and only if f(x) isaffinely equivalent to x2 or x

p+12 ;

(iv) For any polynomial of degree between 2 and p+12 , Wf ≤ p−1

3 , unless it isaffinely equivalent to x2 or x

p+12 .

54 I. Background

A polynomial of the form xkf(x)l will be called a double power of f , the degree ofxkf(x)l is defined to be k + l.

We will need the following result of A. Biro:

Theorem 9.32. (A. Biro [28]) Suppose the graph of f ∈ GF(p)[X] is contained inthe union of two intersecting lines. Then Wf = p−1

2 or Wf = p−13 or Wf ≤ p−1

4 .There is an example with p−1

4 > Wf >p−15 .

Exercise 9.33. Prove the equalities or compute the value of the symmetricfunctions of the interesting sets below:σk(GF(q)) = 0 if k = 1, ..., q − 2 or q and = −1 if k = q − 1;

πk =∑

t∈GF(q)

tk = 0 if (q − 1) 6 |k and = −1 if (q − 1)|k;∑{t1,...,tk;tk+1,...,tk+l}⊆GF(q)

all distinct

t1t2...tkt2k+1t

2k+2...t

2k+l =?;

∑S⊆GF(q)|S|=k

σj(S) =?

9.5 Resultants

In this section the classical notion of resultant is reconsidered. The author doesnot know who found the generalization considered below. It appears in a theoremand its corollary by Green and Eisenbud [72], and probably already in Coolidge[58]; also in Szonyi [125] where it was first applied in finite geometry. The methodwas refined later in Weiner [137].

As is known from classical algebra, given two polynomials f(X) = a0Xk+a1x

k−1+... + ak−1X + ak and g(X) = b0X

l + b1xl−1 + ... + bl−1X + bl, not both a0 and

b0 being zero, then their greatest common divisor has degree at least one iff thedeterminant of the following matrix of size (l + k)× (l + k)

R0(f, g)=R0 =

a0 0 0 ... 0 b0 0 ... 0a1 a0 0 ... 0 b1 b0 ... 0a2 a1 a0 ... 0 b2 b1 ... 0

.

.

.. . .

.

.

.. . .

ak ak−1 ak−2 ... bk bk−1 ... 0

0 ak ak−1 .... . . ...

. . . a0 bl−1

. . . a1 bl bl−1

.

.

.0 bl

.

.

.. . .

ak−1 bl bl−10 ... ... ak 0 ... 0 bl


is zero. This matrix is sometimes called the Sylvester matrix Syl(f, g) of f and g;the determinant of it is the resultant of f and g and is equal to al0b

k0

∏i,j(αi−βj) =

bk0∏j f(βj) = al0

∏i g(αi), where {αi : i = 1, ..., k} and {βj : j = 1, ..., l} are the

roots of f and g, resp., in the algebraic closure. Note also that the resultant is ahomogeneous polynomial of degree l in the variables a0, ..., ak and of degree k inb0, ..., bl.We will frequently need the following

Theorem 9.34. Let f(X) = a0Xk +a1x

k−1 + ...+ak−1X +ak and g(X) = b0Xl+

b1xl−1 + ...+ bl−1X + bl, not both a0 and b0 being zero. If their greatest common

divisor has degree m then the next matrix is non-singular, moreover, their greatestcommon divisor has degree at least m + 1 iff the determinant of the followingmatrix of size (k + l − 2m)× (k + l − 2m)

Rm(f, g) = Rm =

a0 0 0 ... 0 b0 0 ... 0a1 a0 0 ... 0 b1 b0 ... 0a2 a1 a0 ... 0 b2 b1 ... 0

.

.

.. . .

. . .

al−m−1 al−m−2 al−m−3 ... a0 bl−m−1 bl−m−2 ... b0 0 0al−m al−m−1 al−m−2 ... a1 bl−m bl−m−1 ... b0 0

. . ....

. . ....

.

.

.. . . b0

.

.

....

.

.

.ak+l−2m−1 ... ... ak−m bk+l−2m−1 bk+l−2m−2 ... bl−m

l−m

k−m

is zero. (ai and bj are considered to be zero if they are out of the range 0 ≤ i ≤k, 0 ≤ j ≤ l.)

Note that when the greatest common divisor of the two polynomials is supposedto be large then the matrix Rm is small.Proof: Let r(X) be the greatest common divisor of f and g. Denote by c thequotient f/r and let d = g/r. We can suppose that r is monic (of degree m), so

c(X) = a0Xk−m + c1X

k−m−1 + c2Xk−m−2 + ...+ ck−m

andd(X) = b0X

l−m + d1Xl−m−1 + d2X

l−m−2 + ...+ dl−m.

For these polynomials we have r = f/c = g/d. In other words, this means thatfd− gc = 0 or equivalently

f(d− b0Xl−m)− g(c− a0X

k−m) = a0Xk−mg − b0X

l−mf, (∗)

56 I. Background

and this polynomial equation can be interpreted as a system of linear equationsfor the coefficients d1, ..., dl−m, c1, ..., ck−m.

More precisely, compute the coefficient of Xk+l−m−1, Xk+l−m−2, ..., Xm in (∗), ityields the system of linear equations

Rm(d1, ..., dl−m,−c1, ...,−ck−m)T =

(a0b1 − b0a1, a0b2 − b0a2, ..., a0bk+l−2m − b0ak+l−2m)T .

It is important to note that the solutions of the system of linear equations cor-respond to solutions of the polynomial equation, if we know in advance thatthe greatest common divisor r has degree at least m. Indeed, if (d1, ..., dl−m,c1, ..., ck−m) is a solution of our system of linear equations, then the polynomialsc(X) = a0X

k−m + c1Xk−m−1 + ... + ck−m and d(X) = b0X

l−m + d1Xl−m−1 +

... + dl−m will have the property that fd − gc has degree less than m. Since thispolynomial is divisible by r, it must be identically zero.

Now the polynomial equation (and hence the system of linear equations) has aunique solution if r has degree exactly m. To see this, first observe that f/r =c, g/r = d is a solution of the polynomial equation of degree exactly k −m, thismeans that there is at least one solution. (Remember that r was supposed to bemonic.) Suppose that there are two solutions of the equations, yielding d′, d′′ andc′, c′′. Then from fd′−gc′ = 0 and fd′′−gc′′ = 0 we get f(d′− d′′)−g(c′− c′′) = 0.Now the degree of c′− c′′ is less than k−m; so we get a contradiction after dividingby r: in c(d′− d′′) = d(c′− c′′) we have that c and d are relatively primes, so e.g. ddivides (d′−d′′), but the latter has smaller degree so d′ = d′′, and similarly c′ = c′′.Therefore the solution is unique if deg(r) = m, det Rm 6= 0, and the solution canbe obtained by Cramer’s rule.Cramer’s rule

If the degree of r is strictly bigger, say m′ > m, then detRm is zero, since thepolynomial equation fd − gc = 0 cannot have a unique solution. Indeed, we candetermine d, c of degree l −m′ and k −m′ resp., and then multiply both by thesame arbitrary monic polynomial of degree m′ −m.

Note that if a0 = b0 = 0 then both det Rm and the right hand side of the equation(∗) is zero.

In the applications the coefficients aj and bj are polynomials in Y (or homoge-neous polynomials in (Y,Z) say), which satisfy deg(aj),deg(bj) ≤ j. Then, by theremark on Cramer’s rule above, the coefficients of d and c will be quotients of twopolynomials, where the denominator is detRm(Y ). Multiply each coefficient of dand c by detRm(Y ) to obtain the polynomials dm(X,Y ) and cm(X,Y ). Then thefollowing holds.

Result 9.35. (Szonyi [125]) Suppose that f(X) =∑ai(Y )Xn−i and g(X) =∑

bi(Y )Xn−1−i satisfy deg(ai), deg(bi) ≤ i and not both a0 and b0 are zero. Then


the determinant of Rm(Y ) in Theorem 9.34 has degree at most (l − m)(k − m)deg det Rm(Y )

as a polynomial in Y . The total degree of the polynomial cm(X,Y ) is at most(l−m+1)(k−m), and the total degree of dm(X,Y ) is at most (l−m)(k−m+1).

Proof: To estimate the degree of the determinant Rm(Y ), observe that the degreeof the (i, j)-th entry (i, j = 1, ..., k + l− 2m) is at most i− j if we are on the left-hand side, and it is at most i − j + l − m if we are on the right-hand side ofRm(Y ). So each term in the expression of the determinant has Y -degree at most(l −m)(k −m).The solutions d1, ..., dl−m,−c1, ...,−ck−m can be obtained using Cramer’s rule;d0 = b0, c0 = a0. If we put the right hand side of the equation in place of the j-thcolumn, then the degree will be bigger than that of Rm by j or j− l+m accordingas j is in the left or right part of the matrix, from which the second part of theresult follows.

Note that the elements of Rm(Y ) depend on ai(Y ), i ≤ k − m, and on bj(Y ),j ≤ l −m. This is also true for the polynomials am(X,Y ) and cm(X,Y ). Hencein Result 9.35 it is enough to assume that deg(ai) ≤ i, when i ≤ k − m, anddeg(bj) ≤ j, when j ≤ l −m.How can these observations be applied? Consider two homogeneous polynomialsf(X,Y, Z) and g(X,Y, Z), f(a, b, c) = 0 iff the line [a, b, c] has some (good) prop-erty Pf , and g(a, b, c) = 0 iff the line [a, b, c] has some (good) property Pg. We wantto find/count the infinite points (0,−c, b) through which there pass > m affinelines being good in both sense, i.e. for which deg gcd(f(X, b, c), g(X, b, c)) > mfor a certain value m. From Theorem 9.34 it follows that these are all roots ofRm(Y, Z) = Rm(f(X,Y, Z), g(X,Y, Z)). See Theorem 12.1 for a nice illustration.The polynomials c, d, cm(X,Y ), dm(X,Y ) seem to appear in the reasonings aboveas by-products; however at some point, e.g. in stability results, they play a crucialrole. The idea will be the following. Like before, we are given f(X,Y, Z) andg(X,Y, Z), where g is responsible for some “bad” property of lines (and f for somegood property). So we are interested in f

gcd(f,g) but this is somewhat meaningless.

Well, then we can calculate f(X,b,c)gcd(f(X,b,c),g(X,b,c)) , but it does not say anything about

the change of the situation when we alter (b, c). Suppose that there is a so-called“typical index” m on the line [1, 0, 0], meaning that for “most of” the points(0, b, c), the degree of gcd(f(X, b, c), g(X, b, c)) is exactly m. Then, like in Theorem9.34, construct Rm(Y,Z) = Rm(f(X,Y, Z), g(X,Y, Z)) and the other polynomialsc(X,Y, Z), cm(X,Y, Z). We have estimates on their degree, and we do know thatfor all typical points (so for most of the points of [1, 0, 0]) f(X,b,c)

gcd(f(X,b,c),g(X,b,c)) =c(X, b, c). So c(X,Y, Z) is zero for most of the (a, b, c)-s we are interested in. Theonly problem is that we gained c(X,Y, Z) from Cramer’s rule, so it is a fractionwith denominatorR(Y,Z). Hence if we prefer polynomials / algebraic curves (as weusually do), we have to consider c(X,Y, Z)Rm(Y, Z) = cm(X,Y, Z) instead. Note

58 I. Background

that in principle we know it precisely, it can be calculated from the coefficientpolynomials ai(Y,Z) and bj(Y, Z) as above.For illustrations see the proof of Theorem 15.17 in [137].

We want to refine the statement that if the greatest common divisor has degreebigger than m then det Rm is zero; we will show that as the degree gets bigger,the determinant becomes “more and more zero”.More precisely, we know that if for the fix value Y = y the degree of the greatestcommon divisor of f(X, y) and g(X, y) is bigger than m, then the determinant ofRm(y) is 0; hence (Y −y) is a factor of det Rm(Y ). The refinement is the following

Theorem 9.36. For Y = y, if deg(gcd(f(X, y), g(X, y))) = m thendet(Rm(f(X, y), g(X, y))) is non-zero. If deg(gcd(f(X, y), g(X, y))) ≥ m + t then(Y − y)t|det(Rm(f(X,Y ), g(X,Y ))).

The homogeneous version is:

Theorem 9.37. Let f(X,Y, Z)=∑ai(Y, Z)Xk−i and g(X,Y, Z)=

∑bi(Y, Z)X l−i

be homogeneous polynomials in three variables and assume that not both a0 and b0are zero. For (Y, Z) = (y, z), if deg(gcd(f(X, y, z), g(X, y, z))) = m ≥ 0 then

det(Rm(f(X, y, z), g(X, y, z)))

is non-zero. If deg(gcd(f(X, y, z), g(X, y, z))) ≥ m+ t then

(zY − yZ)t|det(Rm(f(X,Y, Z), g(X,Y, Z))).

For the proof the next lemma is needed.

Lemma 9.38. Let M(Y, Z) be an n × n matrix with entries being polynomialsdepending on Y and Z and s ≥ 1. Assume that for the fix value y and z, (zY −yZ)s

divides all (n− 1)× (n− 1) subdeterminants of M(Y, Z), then (zY − yZ)s+1 is afactor of detM(Y, Z).

Proof: Let M∗(Y, Z) be the transpose of the matrix obtained by replacing eachelement mij of M(Y,Z) by (−1)i+j times the corresponding (n− 1)× (n− 1) sub-determinant of M(Y, Z). Note that M∗(Y,Z)M(Y,Z) = (detM(Y, Z))I, where Iis the n×n identity matrix; hence detM∗(Y,Z) = (detM(Y,Z))n−1. By assump-tion the elements of M∗(Y, Z) are divisible by (zY − yZ)s, therefore (zY − yZ)sndivides detM∗(Y, Z) = (detM(Y,Z))n−1 and so the lemma follows.

Proof of Theorem 9.37 As before, we want to determine the coefficients of thepolynomials d(X) and c(X), for that f(X, y, z)d(X) − g(X, y, z)c(X) = 0 holds,


deg c = k − m and deg d = l − m. Again this equation can be interpreted as asystem of linear equations with matrix Rm(y, z), for the coefficients of c and d.

Since the degree of the greatest common divisor of f(X, y, z) and g(X, y, z) ism+ t, the polynomials d and c are the products c(X) = f(X,y,z)

gcd(f(X,y,z),g(X,y,z)) (Xt+

u1Xt−1 + · · ·+ut) and d(X) = f(X,y,z)

gcd(f(X,y,z),g(X,y,z)) (Xt+u1X

t−1 + · · ·+ut), whereevery ui can be chosen freely. This means that the (k + l − 2m) × (k + l − 2m)matrix Rm(y, z) has rank k+l−2m−t. Hence the (k+l−2m−t+1)×(k+l−2m−t+1) subdeterminants of detRm(y, z) are all 0. As det Rm(Y,Z) is a homogeneouspolynomial, (zY − yZ) divides all (k + l − 2m − t + 1) × (k + l − 2m − t + 1)subdeterminants of det Rm(Y, Z). The result follows by applying Lemma 9.38 ttimes.

Later we will see how Theorem 9.37 can be used and this is one of the observationsthat lead to improvement on several results.

Let’s say some words about the connections with Bezout’s theorem 10.6. Apply-ing Theorem 9.37 with m = 0 for the homogeneous polynomials f(X,Y, Z) andg(X,Y, Z), and using the notation my,z = deg gcd(f(X, y, z), g(X, y, z)), i.e. the“common intersection multiplicities” with the line [0,−z, y], we have that the num-ber of common points of the two curves (even if counted with these multiplicities)is at most Bezout∑

(y,z)

my,z ≤ deg det R0(f, g) = kl.

In fact here the common intersection multiplicities with horizontal lines, i.e. linesthrough (1, 0, 0) are added up.

It becomes more interesting when we fix a “typical multiplicity” m and we want toestimate the number of (y, z)-s for which f(X, y, z) and g(X, y, z) has more than mcommon roots. If this “excess” is ty,z, i.e. m+ty,z = deg gcd(f(X, y, z), g(X, y, z)),then Theorem 9.37 gives

excess sum

Theorem 9.39.

∑(y,z)ty,z≥0

ty,z ≤ deg det Rm(f, g) = (deg(f)−m)(deg(g)−m).

As an application let’s consider the case when we are interested in g.c.d.(Xq −X, f(X)), so the distinct factors of f . If we look for more than m factors, we need

60 I. Background

det

1 0 ... 0 a0 0 ... 00 1 ... 0 a1 a0 ... 0

.

.

.. . .

. . .

0 0 ... 1 ak−m−1 ak−m−2 ... a0 0 00 0 ... 0 ak−m ak−m−1 ... a0 0

. . ....

. . ....

.

.

.. . . a0

−1...

.

.

....

0 −1 ... 0 aq+k−2m−1 aq+k−2m−2 ... ak−m

= 0

where the -1’s do not appear in the lower left corner if q+ k− 2m− 1 < q− 1 i.e.k/2 < m. In this case the determinant is just the determinant of the lower rightsubmatrix. In any case, if the coefficients ai are polynomials of Y -degree at mosti, then the degree of the determinant above is at most (q −m)(k −m).The second very important example is when we work on g.c.d.(f(X), f ′(X)), i.e.we look for the multiple factors of f . If we want more than m multiple factors, weneed the determinant of0BBBBBBBBBBBBBBBBBBB@

a0 ... 0 ka0 0 ... 0a1 ... 0 (k − 1)a1 ka0 ... 0

.

.

.. . .

. . .

ak−m−2 ... a0 (m+ 2)ak−m−2 (m+ 3)ak−m−3 ... ka0 0 0ak−m−1 ... a1 (m+ 1)ak−m−1 (m+ 2)ak−m−2 ... ka0 0

. . .. . .

.

.

.

.

.

.. . . ka0

.

.

....

.

.

.a2k−2m−2 ... ak−m (2m+ 2− k)a2k−2m−2 (2m+ 3− k)a2k−2m−3 ... (m+ 1)ak−m−1

1CCCCCCCCCCCCCCCCCCCAto be zero. Again, if the coefficients ai are polynomials of degree at most i, then

the degree of this determinant is at most (k −m)(k −m− 1).

Additive/linearized polynomials, P -adic resultant

Let f(X), g(X) be monic separable GF(pe)-linear polynomials (i.e. each of themsplits into distinct linear factors), and let S = {x1, ..., xn} and U = {t1, ..., tm} beP = pe

bases for the GF(pe)-vectorspaces of their roots (see Theorem 5.28), respectively.The P -adic resultant of them is defined as

RP (f, g) =MRD(x1, ..., xn, t1, ..., tm)

MRD(x1, ..., xn)MRD(t1, ..., tm).

Theorem 9.40. (1) RP (f, g)pe−1 = R( f(X)

X , g(X)X );Example:

pe = 3,n = m = 1,x1 = a, t1 = b

f(X)=X3−a2Xg(X)=X3 − b2X

R( f(X)X

,g(X)X

) =

(a2−b2)2

RP (f, g)=b2−a2

(2) RP (f, g) = MRD(f(t1),...,f(tm))MRD(t1,...,tm) = (−1)nmMRD(g(x1),...,g(xn))

MRD(x1,...,xn) ;

(3) RP (f, g) = 0 ⇔ the GF(pe)-vectorspaces 〈x1, ..., xn〉 and 〈t1, ..., tm〉 have anon-trivial intersection.

10. Elements of algebraic geometry 61

Proof: Easy (using the Moore determinant formula).

10 Elements of algebraic geometry

A typical method in Galois geometry is to associate an algebraic curve to a pointsetof the plane or the space. The classical example of this method is Segre’s theoryof complete arcs or the theory of Redei polynomials.The usual idea is that investigating a pointset in PG(2, q) we are trying to collectall the interesting (“deviant”, “irregular”) lines in an envelope (so a curve in thedual plane, or a dual curve), where the meaning of “interesting” depends on thespecial properties of the pointset. E.g. in the case of arcs Segre introduced a dualcurve containing all the tangent lines. He proved the existence of that curve usingMenelaos’ theorem only; in some other cases we need the explicite form of a curve.For finding something similar to it we often write up the Redei-polynomial of thepointset and we consider the Redei polynomial, or another polynomial derivedfrom it, as an algebraic curve.For using a very little from the enormous theory of algebraic geometry we collecta few results here concerning algebraic curves. Some parts of this section are takenfrom Szonyi [128]. We use a slightly simplified terminology, the interested readercan find several thousands pages of introductory books on algebraic geometry, wemention [64], [107] or the recent excellent Hirschfeld-Korchmaros-Torres [81].An algebraic plane curve defined over a field F is just a polynomial f(X,Y ) intwo variables or a homogeneous polynomial f(X,Y, Z) in three variables, withcoefficients from F. For a constant λ ∈ F, λf is considered to be the same curve.The irreducible components of the curve f correspond to the irreducible factorsof the polynomial f . We may quite often switch back and forth between bivariate minor abuse of

notation(‘affine’) polynomials and their homogenized 3-variate (‘projective’) forms. Definedover F, the curve can be viewed over any field G containing F. The points of f overG are just the pairs (x, y) or homogeneous triples (x, y, z) for which f(x, y) = 0 orf(x, y, z) = 0, resp., where x, y, z ∈ G (so (x, y) ∈ AG(2,G) or (x, y, z) ∈ PG(2,G)).Obviously, the same f can have more zeros (points) over a larger field.

Exercise 10.1. Let F = GF(q). How many different curves (projectively inequiva-lent homogeneous polynomials) of degree n are there?

Exercise 10.2. Let S be a set of points in AG(2, q). Then there exists a two-variablepolynomial of degree at most

√2|S| − 1, so that it vanishes at every point of S.

One can write the affine curve f(X,Y ) in the form

f(X,Y ) = f0 + f1(X,Y ) + f2(X,Y ) + ...+ fn(X,Y ),

62 I. Background

where fi is a homogeneous polynomial in two variables of total degree i. Nowthe point (0, 0) has multiplicity k on f(X,Y ) if k is maximal such that f can bewritten as

f(X,Y ) = fk(X,Y ) + fk+1(X,Y ) + ...+ fn(X,Y ),

so f0 = f1 = ... = fk−1 = 0. Multiplicity of another point P (a, b) can bedefined translating it to the origin, i.e. we can examine f(X + a, Y + b) instead.The tangents of f(X,Y ) at (0, 0) are the k linear factors of fk(X,Y ) above,they may well be defined over the algebraic closure F; tangents at anotherpoint can be defined translating it to the origin. A point is simple if it hasmultiplicity 1. At a simple (projective) point (x, y, z) the (unique) tangent lineis [(∂Xf)(x, y, z), (∂Y f)(x, y, z), (∂Zf)(x, y, z)]. Equvalently, (x, y, z) is a singularpoint iff ((∂Xf)(x, y, z), (∂Y f)(x, y, z), (∂Zf)(x, y, z)) = (0, 0, 0).

As the following result shows, a curve can have only a restricted number of pointswith multiplicity more than one.

Result 10.3. [64, p.118] Let F be a projective plane curve of degree n withoutmultiple components. Then for the multiplicities mP of its points

∑P∈F

mP (mP−1)2 ≤(

n2

)holds.

One can define the intersection multiplicity (denoted by I(f ∩ g;P ) or I(P ; f ∩ g))of two curves f and g at a point P (x, y, z) of the plane. The direct definition israther complicated in the general case; here is a list of some easy facts which willbe enough for our purposes:

(a) If at least one of f and g does not go through P then obviously I(f∩g;P ) = 0.

(b) If P is a simple point on both f and g and the tangents at P are distinctthen I(f ∩ g;P ) = 1. In particular, if f 6= g are linear, meeting at P , thenI(f ∩ g;P ) = 1.

(c) P is not on a common component of f and g ⇔ I(f ∩ g;P ) <∞.

(d) I(f ∩ g;P ) = I(g ∩ f ;P ).

(e) I(f ∩ g;P ) = I(f ∩ g + hf ;P ).

(f) If g is of degree 1 (i.e. a line) then I(f ∩ g;P ) can be calculated easily: onecan express one of the variables from g(X,Y, Z) = 0 and substitute it into f ;then the resulting homogeneous polynomial in two variables will vanish at P ,the multiplicity of this root will be the intersection multiplicity. Note that, asa univariate polynomial of degree n has at most n roots (and the same holdsfor homogeneous polynomials in two variables), this intersection multiplicity


is at most the degree of f , except when after the substitution f vanishesidentically, which means that g was a factor of f . In this case (and generallywhen f and g have a common factor through P ) we say I(f ∩ g;P ) = ∞.

(g) Note that if g is linear and it happens to be a tangent of f at P thenI(f ∩ g;P ) ≥ 2.

(h) I(f ∩ (g1g2);P ) = I(f ∩ g1;P ) + I(f ∩ g2;P ).

(b’) A more general form of (b) is: I(f ∩ g;P ) ≥ mP (f)mP (g), where mP is themultiplicity of P on the curve in question.

We note that there is a unique intersection multiplicity satisfying these properties(and being invariant under projective transformations), so they can be used todefine intersection multiplicity. Also note that these properties can be used toactually compute the intersection multiplicity of two curves. Let us briefly recallhow this works for the origin. Since any point can be transformed to the origin,this approach works for other points, too.We wish to compute I(P ;F ∩ G), where P = (0, 0). Let f(X) = F (X, 0),g(X) = G(X, 0) and d(X) = g.c.d.(f(X), g(X)). Then we can write d(X) =a(X)f(X) + b(X)g(X) by the Euclidean algorithm. Put f1(X) = f(X)/d(X),g1(X) = g(X)/d(X). Then f1(X)a(X) + g1(X)b(X) = 1, which means that(

−f1(X) g1(X)b(X) a(X)

)(G(X,Y )F (X,Y )

)=(R(X,Y )U(X,Y )

)is an invertible linear transformation. Hence the ideals (F,G) and (H,U) are thesame, so I(P ;F ∩G) = I(P ;H∩U). Since Y divides R(X,Y ) = −f1(X)G(X,Y )+g1(X)F (X,Y ), so we can use property (h) of the intersection multiplicity. Thisprocess can be continued and after some steps we can recognize that a certainintersection multiplicity is 0, using property (a).As an illustration the following useful lemma can be proved, which can be foundin [107], Lemma 9.2, see also p. 87 in [107].

Lemma 10.4. Let P be the origin.

(1) If I(P ;F ∩ Y ) = r, I(P ;G ∩ Y ) = s and r ≥ s, then I(P ;F ∩G) ≥ s.

(2) If r > s and P is a simple point of F , then I(P ;F ∩G) = s.

Proof: We only prove (1). In determining the intersection multiplicity with theapproach indicated above, then we haveXs|d(X). So when we compute I(P ;H∩U)with H = Y H1, we get I(P ;Y H1 ∩U) = I(P ;Y ∩U) + ..., where ... ≥ 0. Now putU = U1(X) + Y U2(X,Y ). Then I(P ;Y ∩U) = I(P ;Y ∩U1). Since the terms of Uthat do not contain Y come from similar terms of F and G, and d(X) is divisibleby Xs, Xs divides U1(X). But then I(P ;Y ∩ U1(X)) ≥ I(P ;Y ∩Xs) = s. (2) is

64 I. Background

left as an exercise.

Exercise 10.5. Prove case (2) above!

In the applications we need bounds on the common points of two curves, andbounds on the number of points on a single curve.

Theorem 10.6. (Bezout) If f and g has no common component then the num-ber of their common points, counted with intersection multiplicities, is at mostdeg(f) deg(g). Over the algebraic closure F always equality holds.

The proof of Bezout’s theorem is based on resultants, compare to Section 9.5.Now let F = GF(q). How many points can a curve f ∈ GF(q)[X,Y, Z] of degree nhave over GF(q)? Denote this number by Nq = Nq(f).

Theorem 10.7. Hasse-Weil, Serre For an absolutely irreducible non-singular alge-braic curve f ∈ GF(q)[X,Y, Z] of degree n we have

|Nq(f)− (q + 1)| ≤ gb2√qc ≤ (n− 1)(n− 2)√q.

In the theorem g denotes the genus of f , we do not define it here. Note that Nqcounts the points of f with multiplicities, also that for GF(q) ⊂ GF(q1) we haveNq(f) ≤ Nq1(f).It happens that some absolutely irreducible component of f cannot be defined overGF(q) (but it still has some GF(q)-rational points, i.e. points in PG(2, q)). Thenthe following bound can be used:

Exercise 10.8. Prove that for an absolutely irreducible algebraic curvef ∈ GF(q)[X,Y, Z] of degree n, that cannot be defined over GF(q), we haveNq(f) ≤ n2.

In some cases the Hasse-Weil bound can be changed for a better one, for examplewhen q = p is a prime number.

Theorem 10.9. Stohr-Voloch [111] For an irreducible algebraic curve f ∈GF(q)[X,Y, Z], q = ph, of degree n with Nq rational points over GF(q) we have(i) if n > 1, not every point is an inflexion and p 6= 2 then Nq ≤ 1

2n(q + n− 1);(ii) if n > 2, not every point is an inflexion and p = 2 then Nq ≤ 1

2n(q + 2n− 4);(iii) if q = p and n > 2 then Np ≤ 2n(n− 2) + 2

5np;(iv) if q = p, 3 ≤ n ≤ 1

2p and f has s double points then Np ≤ 25n(5(n−2)+p)−4s.

In (i) the condition is automatically satisfied if q = p is a prime.


Exercise 10.10. Let f(X) be a polynomial of degree at most 4√q, (q odd), which

assumes square elements of GF(q) only. Prove that f = g2 for a suitable polynomialg(X).

10.1 Conditions implying linear (or low-degree) components

In the applications it is typical that after associating a curve to a certain set (inprinciple), the possible linear components of the curve have a very special meaningfor the original problem. Quite often it more or less solves the problem if one canprove that the curve splits into linear factors, or at least contains a linear factor.Here some propositions ensuring the existence of linear factors are gathered. Theusual statement below considers the number of points (in PG(2, q)) of a curve.We will use two numbers: for a curve C, defined by the homogeneous polynomialf(X,Y, Z), Mq denotes the number of solutions (i.e. points (x, y, z) ∈ PG(2, q))for f(x, y, z) = 0, while Nq counts each solution with its multiplicity on C. Mq ≤ Nq

Exercise 10.11. Barlotti-bound Prove that if a curve of order n has no linearfactors over GF(q) then Nq ≤ (n− 1)q + n.

Exercise 10.12. Prove that a curve of degree n defined over GF(q), without linearcomponents, has always Nq ≤ (n − 1)q + n

2 points in PG(2, q). Hint: let k bemaximal such that every tangent of the curve contains at least k points of the thecurve (counting without multiplicity, in PG(2, q)). Show that (i) Nq ≤ (n−1)q+k;(ii) Nq ≤ (n− 1)q + (n− k).

Conjecture 10.13. We conjecture that a curve of degree n defined over GF(q),without linear components, has always Nq ≤ (n− 1)q + 1 points in PG(2, q).

For n = 1, 2,√q + 1, q − 1 it would be sharp as the curves X2 − Y Z,X

√q+1 +

Y√q+1 +Z

√q+1 and αXq−1 +βY q−1− (α+β)Zq−1 (where α, β, α+β 6= 0) show.

It can be called the Lunelli-Sce bound for curves, see Section 25.Note that the conjecture is true

(i) if there exists a line skew to the curve and (q, n) = 1, see Corollary 25.6;

(ii) if n ≤ √q+ 1 then q+ 1 + (n− 1)(n− 2)

√q ≤ nq− q+ 1 proves it by Weil’s

bound Theorem 10.7;

(iii) if the curve has a singular point in PG(2, q);

(iv) if n ≥ q + 2.

The statement (ii) can be proved by induction: if C has more points then it cannotbe irreducible, so it splits to the irreducible components C1 ∪ C2 ∪ ... ∪ Ck withdegrees n1, ..., nk; if each Ci had ≤ (ni−1)q+1 points then in total C would have≤∑ki=1(niq− q+ 1) = nq− k(q− 1) < nq− q+ 1 points. So at least one of them,

66 I. Background

Cj say, has more than njq− q+ 1 points. By Exercise 10.8 Cj can be defined overGF(q) and Weil does its job again.For (iii) the Barlotti-bound, recounted looking around from a singular point, willwork.

The truth of the conjecture would also mean that the counterexamples for theLunelli-Sce cenjecture are not pointsets of curves, see Section 25, Corollary 25.6.

Let Cn be a curve of degree n defined over GF(q), with at least qn affine points(counted with multiplicity). (In fact it’s enough to have qn− δ points, see below).We want to prove that it has a linear component, then repeat everything for n−1and (n − 1)q. The next proposition, which may be omitted in the final version,deals with curves of high degree as well.

Proposition 10.14. A curve of degree n defined over GF(q), without linear compo-nents, has always Nq ≤ nq points in AG(2, q). More precisely, if tq < n ≤ (t+ 1)qthen Nq ≤ max{tq2, (q + 1)(n− t− 1)}.tq2 < nq;

(t+1)q−n+(t+1) ≥ 0

(For t = 0 it gives Nq ≤ (n− 1)q + n− 1.)

The key idea is that if you find a line intersecting Cn in at least n+ 1 points thenit is a (linear) component by Bezout. If not then Cn is a (k, n)-arc and one canestimate its cardinality.

Let tq < n ≤ (t+ 1)q and suppose that Cn is a curve with

Nq > max{tq2, (q + 1)(n− t− 1)}

points. Then Nq > tq2, so there is a point P of maximal multiplicity m ≥ t + 1.Consider a tangent line ` through P , then I(Cn ∩ `, P ) ≥ m + 1, so there are atmost n−m− 1 further points on ` \P . As there exist at least m tangents throughP , we have

Nq ≤ m+ (q + 1)(n−m)−m = (q + 1)(n−m) ≤ (q + 1)(n− t− 1),

which is a contradiction.

The following lemma is a generalization of a result by Szonyi. For d = 1 it can befound in Sziklai [113], which is a variant of a lemma by Szonyi [124].

Lemma 10.15. Let Cn, 1 ≤ d < n be a curve of order n defined over GF(q), notcontaining a component defined over GF(q) of degree ≤ d. Denote by N the numberof points of Cn in PG(2, q). Choose a constant 1

d+1 + d(d−1)√q

(d+1)(q+1) ≤ α. Assume thatn ≤ α

√q − 1

α + 1. Then N ≤ n(q + 1)α.


It works also with α > 1d+1 + 1+d(d−1)

√q

(d+1)q , n < α√q− d+ 2, N < αnq. Here α = 1

d

can be written (that is often needed) when d ≤ 6√q.

Proof: Suppose first that Cn is absolutely irreducible. Then Weil’s theorem ([136],[77]) gives N ≤ q+1+(n−1)(n−2)

√q ≤ n(q+1)α. (The latter inequality, being

quadratic in n, has to be checked for n = d+ 1 and n = α√q − 1

α + 1 only.)If Cn is not absolutely irreducible, then it can be written as Cn = Di1 ∪ ... ∪ Dis ,where Dij is an absolutely irreducible component of order ij , so

∑sj=1 ij = n. If

Dij can not be defined over GF(q), then it has at most Nij ≤ (ij)2 ≤ ij(q + 1)αpoints in PG(2, q) (see Ex. 10.8). If Dij is defined over GF(q), then the Weil-boundimplies again that Nij ≤ ij(q + 1)α. Hence

N =s∑j=1

Nij ≤s∑j=1

ij(q + 1)α = n(q + 1)α.

For applications see Section 11, Theorem 23.1, Theorem 23.2 and Theorem 27.8.

Exercise 10.16. Prove that if q = p is prime and α > 25 then in the theorem above

n ≤ ( 12α−

15 )p+ 2 is enough for N ≤ n(p+ 1)α.

In the applications quite often happens that we do not know the number of pointsof a certain curve but we know the numbers of intersections with a pencil of lines(each intersection counted with intersection multiplicity).

Lemma 10.17. Let C, 1 ≤ d < n be a curve defined by the homogeneous polynomialF ∈ GF(q)[X,Y, Z], deg(F ) = n. Suppose that C does not contain a componentdefined over GF(q) of degree ≤ d. Assume that C does not contain (multiple com-ponents nor) components with vanishing partial derivative with respect to X. Let`1, ..., `q+1 be the lines of the pencil centered at P (1, 0, 0). Denote by N0 the num-ber of intersection points of C and the lines {`1, ..., `q+1}, each point on `j countedwith the intersection multiplicity of `j and C (but we do not count P even if it ison C). Choose a constant 1

d+1 + d(d−1)√q

(d+1)(q+1) ≤ α. Assume that n ≤ α√q − 1

α + 1.Then N0 ≤ n(q + 1)α+ n(n− 1).

Proof: The only difference here, compared to the situation of Lemma 10.15, isthat N0 can be bigger than N , the number of points of C. As before we can countcomponentwise. The difference N0 − N comes from points where the tangent ofan irreducible component Gij of the curve is horizontal, that is the intersectionpoints of Gij and ∂XGij . If ∂XGij is not zero and degGij = ij then by Bezout’stheorem ij(ij − 1) is an upper bound for the number of points in Gij ∩ ∂XGij . SoN0 ≤ N +

∑j ij(ij − 1) ≤ n(q + 1)α+ n(n− 1) .

Exercise 10.18. (De Beule - Gacs) Let I consist of all the linear combinations of themonomials Xp+1, Y p+1, Zp+1, XpY,XY p, Y pZ, Y Zp, ZpX,ZXp. Suppose that fis a homogeneous quadratic polynomial for which f

p+12 is in I. Then f is reducible.

68 I. Background

11 Finding the missing factors, removing the surplus

factors

Here we treat a very general situation, with several applications in future.Given A = {a1, ..., aq−ε} ⊂ GF(q), all distinct, let F (X) =

∏q−εi=1 (X − ai) be their

root polynomial. We would like to find the “missing elements” {aq−ε+1, ..., aq} =GF(q) \ A, or, equivalently, G∗(X) =

∏qi=q−ε+1(X − ai). Obviously, G∗(X) =

Xq−XF (X) , so F (X)G∗(X) = Xq−X. Expanding this, and introducing the elementary

symmetric polynomials

σj = σj(A), σ∗k = σk(GF(q) \A),

we get Xq −X =(Xq−ε−σ1X

q−ε−1 +σ2Xq−ε−2−...±σq−ε−1X ∓σq−ε)(Xε−σ∗1Xε−1 +σ∗2X

ε−2−...± σ∗ε−1X ∓ σ∗ε ),from which σ∗j can be calculated recursively from the σk-s, as the coefficient ofXq−j , j = 1, ..., q − 2 is 0 = σ∗j + σ∗j−1σ1 + ...+ σ∗1σj−1 + σj ; for example

σ∗1 = −σ1; σ∗2 = σ21 − σ2; σ∗3 = −σ3

1 + 2σ1σ2 − σ3; etc. (1)

Note that we do not need to use all the coefficients/equations, it is enough to doit for j = 1, ..., ε. (The further equations can be used as consequences of the factthat the ai-s are pairwise distinct, see e.g. Theorem 25.7 and Lemma 25.8.)The moral of it is that the coefficients of G∗(X) can be determined from thecoefficients of F (X) in a “nice way”.

* * *

Let now B = {b1, ..., bq+ε} ⊃ GF(q) be a multiset of elements of GF(q), and letF (X) =

∏q+εi=1 (X−bi) be their root polynomial. We would like to find the “surplus

elements” {bk1 , ..., bkε} = B \ GF(q), or, equivalently, G(X) =∏εi=1(X − bki).

Obviously, G(X) = F (X)Xq−X , so F (X) = (Xq − X)G(X). Suppose that ε ≤ q − 2.

Expanding this equation and introducing the elementary symmetric polynomials

σj = σj(B), σk = σk(B \ GF(q)),

we get Xq+ε − σ1Xq+ε−1 + σ2X

q+ε−2 − ...± σε−1Xq+1 ∓ σεX

q ± ......± σq+ε−1X ∓ σq+ε = (Xq −X)(Xε − σ1X

ε−1 + σ2Xε−2 − ...± σε−1X ∓ σε) =

= Xq+ε−σ1Xq+ε−1+σ2X

q+ε−2−...±σε−1Xq+1∓σεXq+ terms of lower degree.

From this σj can be calculated even more easily then in the previous case:if ε ≥ q− 1 thenit’s slightly morecomplicated

σk = σk for all k = 1, ..., ε. (2)

11. Finding the missing factors, removing the surplus factors 69

* * *

In both case suppose now that instead of the “elements” {ai} or {bj} we have(for example) linear polynomials ciY + di and a set S ⊆ GF(q) such that for eachy ∈ S the set Ay = {ciy + di : i} consists of pairwise distinct elements of GF(q),or, similarly, the multiset By = {ciy+ di : i} contains GF(q). Then the σk-s in thereasonings above become polynomials in Y , with degY (σk) ≤ k. Now one cannotspeak about polynomials σ∗k(Y ) (or σk(Y ), resp.) as there is no guarantee that themissing values (or the surplus values) for different y-s can be found on ε lines. Sofirst we define σ∗k(y) (or σk(y), resp.), meaning the coefficient of Xε−k in Gy(X) orG∗y(X), so the elementary symmetric function of the missing (or surplus) elementswhen substituting Y = y ∈ S. However, the equations for the σ∗k-s or σk-s are stillvalid. So one may define the polynomials analogously to (1):

σ∗1(Y ) def= −σ1(Y ); σ∗2(Y ) def= σ21(Y )− σ2(Y );

σ∗3(Y ) def= −σ31(Y ) + 2σ1(Y )σ2(Y )− σ3(Y ); etc.

or analogously to (2):

σk(Y ) def= σk(Y ) for all k = 1, ..., ε

with the help of them. Note that from the defining equations it is obvious that

degY σ∗k(Y ) ≤ k and degY σk(Y ) ≤ k.

Now we can define the algebraic curve

G∗(X,Y ) def= Xε − σ∗1(Y )Xε−1 + σ∗2(Y )Xε−2 −...± σ∗ε−1(Y )X ∓ σ∗ε (Y )

or in the other case

G(X,Y ) def= Xε − σ1(Y )Xε−1 + σ2(Y )Xε−2 − ...± σε−1(Y )X ∓ σε(Y ).

As before, for each y ∈ S we have that the roots of G(X, y) are just the missing(or the surplus) elements of Ay or By, resp. Our aim is to factorize G∗(X,Y ) orG(X,Y ) into linear factors X − (αiY + βi). To do so, observe that G∗(X,Y ) hasmany points in GF(q)×GF(q): for any y ∈ S we have ε solutions of G∗(X, y) = 0,i.e. the ε missing values after substituting Y = y in the linear polynomials ciY +di,so after determining the sets Ay. So G∗(X,Y ) has at least ε|S| points.A similar reasoning is valid for G(X,Y ). If it splits into irreducible componentsG = G1G2 · · ·Gr, with degGi = degX Gi = εi,

∑εi = ε, then for any y ∈ S,

the line Y = y intersects Gi in εi points, counted with intersection multiplicity.So the number of points on Gi is at least εi|S| − εi(εi − 1), where the secondterm stands for the intersection points of Gi and ∂XGi, where the intersectionmultiplicity with the line Y = y is higher than the multiplicity of that point onGi. So, unless some Gi has zero partial derivative w.r.t. X, we have that G has atleast

∑εi|S| − εi(εi − 1) ≥ ε|S| − ε(ε− 1) points.

70 I. Background

Now we can use Lemma 10.15, Lemma 10.17 (or any similar result) repeatedly,with d = 1, it will factorize G(X,Y ) into linear factors of the form X− (αjY +βj)if degG(X,Y ), which is at most ε in our case, is small enough, i.e. if ε <

√q and

|S| > max{ ε−1+ 4√q√q , 1

2}·(q+1) in the first and |S| > max{ ε−1+ 4√q√q , 1

2}·(q+1)+ε−1in the second case.It means, that one can add ε linear polynomials αiY +βi in the first case such thatfor any y ∈ S, the values {ciy + di} ∪ {αjy + βj} = GF(q). In the second case wehave a weaker corollary: for any y ∈ S, the values {ciy+di} \ {αjy+βj} = GF(q),which means that adding the new lines αjY + βj “with multiplicity= −1” thenS × GF(q) is covered exactly once. (What we do not know in general, that theselines were among the given q + ε lines, so whether we could remove them.)Finally, these lines (or similar objects), covering S × GF(q) usually have someconcrete meaning when applying this technique; this book contains several suchapplications, see Theorem 27.4, Exercise 27.6, etc.The arguments above are easy to modify when we change some of the conditions,for example when ai or bi is allowed to be some low degree (but non-linear) poly-nomial of Y .

Exercise 11.1. Use the second (“surplus”) case above to prove that a blockingset B ⊂ PG(2, q) with |B ∩ AG(2, q)| = q + 1 affine points always contains a(naffine) point that is unnecessary (i.e. it can be deleted without violating the blockingproperty).

Exercise 11.2. Let fi(T ), i = 1, ..., q − ε be polynomials of degree at most d, andsuppose that their graphs {(t, fi(t)) : t ∈ GF(q)} are pairwise distinct. Prove thatif ε < ... then one can find fq−ε+1(T ), ..., fq(T ), each of degree at most d such thatthe graphs of these q polynomials partition the affine plane.

Exercise 11.3. Let fi(T ), i = 1, ..., q − ε be polynomials, each from a subspace Uof GF(q)[T ] with 1 ∈ U , and suppose that their graphs {(t, fi(t)) : t ∈ GF(q)} arepairwise distinct. Prove that if ε < ... then one can find fq−ε+1(T ), ..., fq(T ), eachfrom U , such that the graphs of these q polynomials partition the affine plane.

Chapter 2

Polynomials in geometry

12 Lines meeting a pointset

12.1 Prescribing the intersection numbers

Suppose that a function m : L → N is given, where L is the set of lines of PG(2, q).The problem is to find conditions, necessary and/or sufficient, under which we canfind a pointset S such that |S ∩ `| = m(`) for all ` ∈ L.Note that one can pose the similar question for any hypergraph.If A denotes the incidence matrix of the plane, m = (m(`1),m(`2), ...,m(`q2+q+1))is the weight-vector, then the problem is reduced to finding a (“characteristicvector”) v such that Av = m. As A is non-singular, we have v = A−1m. Nowone can turn the question around: for which m will v be of the required type, forexample a non-negative, integer or 0-1 vector?It is easy to compute that

A−1 =1qAT − 1

q(q + 1)J.

We also know that the eigenvalues ofA are q+1 (with multiplicity 1 and eigenvector(1, 1, ..., 1));

√q and −√q (their multiplicities depend on the order of points and

lines we choose).

In most cases m is not given, we know some of its properties only. It means thata certain set M of weight-vectors is given (for example, the set of all vectors witheach coordinate from a small fixed set of integers, say {0, 1, 2}); and we want toknow some property (for example, the possible Hamming-weight, i.e. the numberof nonzero coordinates) of (0-1) vectors v satisfying Av ∈M .

71

72 II. Polynomials in geometry

12.2 A general result

Metsch conjectured [95] and Weiner [130] proved the following

Theorem 12.1. Let B ⊆ PG(2, q) be a pointset of size b on r lines of a pencilcentered at P 6∈ B. Then the number of lines meeting B in at least one point is atmost

b(q + 1)− (r(b− r + 1)− 1).

It can be reformulated as: the points of B “determine” at least r(b − r + 1) − 1lines, where a line containing t ≥ 2 points of B is counted with multiplicity t− 1.First of all observe that there are point sets for which the given bound is sharp.Assume that r − 1 is the order of a subplane π in PG(2, q) and let B be a propersubset of π containing r collinear points (i.e. a subline), hence it blocks all the linesof π. So the number of lines meeting B is ((r−1)2+(r−1)+1)+|B|(q+1−r), wherethe first part is the number of lines of π, the second is the number of tangentsthrough the points of B. The number of lines meeting B through a point P ofπ \B is r and so for such a point the bound in the conjecture is sharp.Note that this statement is stronger than the following well known result of Jami-son [84] and Brouwer and Schrijver [49].

Theorem 12.2. (Jamison, Brouwer and Schrijver) A blocking set in AG(2, q) con-tains at least 2q − 1 points.

Proof: Assume to the contrary that there is a blocking set B in AG(2, q) so that|B| ≤ 2q − 2. Embed AG(2, q) in PG(2, q) and let P be and ideal point. Now thevalue r in Theorem 12.1 above is q and so the total number of lines meeting B isat most 1 + qq + (|B| − q)(q + 1− q) ≤ q2 + q − 1; which is a contradiction, sinceB blocks all the q2 + q affine lines.

Note that there are blocking sets of size 2q − 1, e.g. the points of two intersectingand many other

lines.

Exercise 12.3. Prove Jamison’s theorem using the Combinatorial Nullstellensatz(Theorems 5.16, 5.17).

Exercise 12.4. [123] If the smallest blocking set of AG(2, q) has size r then inPG(2, q) the smallest blocking set, which contains two different minimal blockingsets, is of size r + 2, and vice versa.Note that it implies a kind of “stability” of blocking sets of size ≤ 2q.There are blocking sets of size less than 2q − 1 in some non-Desarguesian affineplanes of order q, see [54]; which shows that a purely combinatorial proof cannotbe given for neither Theorem 12.2 nor Theorem 12.1. To prove Theorem 12.1 thefollowing lemma is essential.

12. Lines meeting a pointset 73

Exercise 12.5. (Blokhuis and Brouwer, [44]) Let B be a blocking set in PG(2, q)and let P be an essential point of B. Then there are at least 2q+ 1− |B| tangentsthrough P .

Lemma 12.6. Let `∞ = [1, 0, 0] be called now the line at infinity in PG(2, q) and letS be a point set in PG(2, q)\`∞. Assume that through the ideal point (0,−z, y) therepass t (affine) lines meeting S. Denote by nt+h the number of ideal points throughthat there pass exactly t+ h lines meeting S. Then

∑q−th=1 hnt+h ≤ (|S| − t)(q− t).

Proof: For the points of S write {(ai, bi, 1)} and consider the three-variableRedei polynomial of S, that is R(X,Y, Z) =

∏|S|i=1(X + aiY − biZ) =∑|S|

j=0 rj(Y, Z)X |S|−j . Recall that deg rj = j. It follows from the fundamentalproperty of the Redei polynomial, that degX gcd(R(X, y, z), Xq −X) = t.

For the polynomials R and Xq − X construct the matrix Rt(Y,Z) introducedin Section 9.5. By Theorem 9.39, for the excess sum in question

∑q−th=1 hnt+h ≤

deg(detRk(Y, Z)) ≤ (|S| − t)(q − t).

Proof of Theorem 12.1: For the line at infinity `∞ choose an m-secant ofB, m > 0, passing through P . Note that now the line at infinity meetsB. Again denote by n(r−1)+h the number of ideal points through that therepass exactly (r − 1) + h affine lines meeting B. Let us sum up the numberof affine lines meeting B through the ideal points, in total we get at mostqm+

((q + 1−m)(r − 1) +

∑q−(r−1)h=1 hn(r−1)+h

); where the first part corresponds

to the points of `∞∩B, the second to the points of `∞ \B. The result follows fromLemma 12.6 immediately as 1+qm+(q+1−m)(r−1)+(b−m−(r−1))(q−(r−1)) =b(q + 1)− (r(b− r + 1)− 1).

It has a nice (combinatorial) corollary in the 3-space:

Corollary 12.7. [95] Suppose B is a set of b ≥ q + 1 points in PG(3, q). Then Bmeets at most q2 + q + 1 + bq2 lines, with equality if and only if B is a planarblocking set.

Exercise 12.8. Formulate and prove the analogue of the corollary in the plane(where a combinatorial proof can be given easily).

12.3 Another immediate corollary

An m-fold blocking set in AG(2, q) is a set of points intersecting each line in atleast m points. Blokhuis ([31]) showed that an m-fold blocking set S in AG(2, q),where (m, q) = 1, has at least (m+ 1)q − 1 points. Later Ball ([6]) extended this


result to arbitrary m, he showed that if e(m) is the maximal exponent such thatpe(m)|m, then |S| ≥ (m+ 1)q − pe(m), see Corollary 24.25.In this subsection we show that Theorem 9.39 immediately implies a lower boundon the size of an m-fold blocking set in AG(2, q). In general, except when m|q,this result is weaker than Ball’s result.

Corollary 12.9. (Szonyi-Weiner [130]) The size of an m-fold blocking set in AG(2, q)is at least (m+ 1)q −m.

Proof: Assume to the contrary that there exists an affine m-fold (not necessarilyminimal) blocking set B of size (m + 1)q − m − 1. Let ` be an (m + k)-secant,k ≥ 0, where |B| − (m + k) 6= qm. Such a line ` can be chosen, since countingthe points of B on the lines through a point of B and on the lines through anaffine point not in B shows that the intersection numbers of B with lines takeat least two different values. Change the coordinate system, so that ` is the idealline and (∞) ∈ B. Now B contains at least m points from each line, except fromthe ‘old’ line at infinity that is skew to B. Denote this line by `′ and by (y`′) theideal point of it in this new coordinate system. Let U = B \ ` = {(ai, bi)}i andconsider the Redei polynomial of U , that is R(X,Y ) =

∏|B|−(m+k)i=1 (X + aiY −

bi) =∑|B|−(m+k)j=0 rj(Y )X |B|−(m+k)−j . By the properties of the Redei polynomial,

degX gcd(R(X, y`′), (Xq − X)m) = m(q − 1) and for any (y) ∈ ` \ (B ∪ (y`′)),degX gcd(R(X, y), (Xq − X)m) = mq. For the polynomial R and (Xq − X)m

and for the value s = max(degXR, qm) − m(q − 1), construct the matrix Rsintroduced in Section 9.5. By Result 9.34, the determinant of this matrix is notzero. Furthermore, similarly as it is in the proof of Lemma 12.6, one can show thatdeg(detRs) ≤ m(q −m − k − 1). This is a contradiction, since by Theorem 9.39,m(q −m− k) ≤ deg(detRs).

13 Sets with constant intersection numbers mod p

In [34, 40] the following is proved:

Proposition 13.1. Let S be a pointset in AG(2, q), suppose that every line intersectsS in 1 (mod p) points, or is completely disjoint from S. Then |S| ≤ q − p+ 1.

Proof: Counting the points of S on the lines through some fixed point s ∈ S wehave |S| ≡ 1 (mod p). After the AG(2, q) ↔ GF(q2) identification define

f(X) =∑s∈S

(X − s)q−1,

it is not identically zero as the coefficient of Xq−1 is 1. Note that for x ∈ GF(q2)the value of (x − s)q−1 depends on the direction of the line joining x and s. If

13. Sets with constant intersection numbers mod p 75

x ∈ S then every direction will occur with multiplicity divisible by p, hence all thepoints of S are roots of f , which is of degree q − 1. The biggest value ≡ 1 mod pbelow q is q − p+ 1.

There are examples of sets like in the statement above, e.g. some (1 mod p)collinear points, or a projective subplane of order <

√q completely contained in

AG(2, q).

Exercise 13.2. Prove that in Proposition 13.1 it does not make any difference if Sis allowed to be a multiset.

We remark that the projective case is totally different: there are very big pointsetsin PG(2, q) with 1 (mod p) -secants only, e.g. the plane itself, or a unital, etc.

Exercise 13.3. (Blokhuis) Prove the following generalization: Let S be a pointsetin AG(n, q), n ≥ 1, and suppose that every hyperplane intersects S in 1 (mod p)points, or is completely disjoint from S. Then |S| ≤ q − p+ 1.

Exercise 13.4. Let S be a pointset in AG(2, q) and suppose that for some1 ≤ r ≤ p − 1, every line intersects S in r (mod p) points, or is completelydisjoint from S. Then |S| ≤ q − p + r. Why does this statement become not veryinteresting after proving if?

The situation is different if we consider the projective plane.

Theorem 13.5. Given a pointset S = {(ai, bi, ci) : i = 1, ..., s} = {(ai, bi, 1) : i =1, ..., s1} ∪ {(aj , bj , 0) : j = s1 + 1, ..., s} ⊆ PG(2, q), the following are equivalent:

(i) S intersects each line in r mod p points for some fixed r;

(ii) G(X,Y, Z) =∑|S|i=1(aiX + biY + ciZ)q−1 ≡ 0;

(iii) for all 0 ≤ k+ l ≤ q−1,(k+lk

)6= 0 mod p, we have

∑|S|i=1 a

q−1−k−li bki c

li = 0. here 00 = 1

(iv) for all 0 ≤ k + l ≤ q − 2,(k+lk

)6= 0 mod p, we have

∑s1i=1 a

ki bli = 0

and for all 0 ≤ m ≤ q − 1,∑si=1 a

q−1−mi bmi = 0.

Proof: Note that if each line intersects S in r mod p points then |S| ≡ r (mod p). count |S| from apoint not in SSo let r be defined by |S| ≡ r (mod p). If each line [x, y, z] intersects S in r mod

p points then (mod p) |S| − r ≡ 0 terms (aix+ biy+ ciz)q−1 will be 1 in G(x, y, z)hence G(x, y, z) = 0. As degG ≤ q − 1 we have (i)⇒(ii). One can turn it around:if G(x, y, z) = 0 then the number of terms (aix+ biy + ciz)q−1 with nonzero (i.e.=1) value should be zero mod p, so (ii)⇒(i).


For the rest consider the coefficient of Xq−1−k−lY kZl in G (for 0 ≤ k+ l ≤ q−1),it is (

q − 1k + l

)(k + l

k

) |S|∑i=1

aq−1−k−li bki c

li = 0

if (ii) holds and vice versa. Finally (iii)⇔(iv) is obvious.

Many interesting pointsets (small blocking sets, unitals, maximal arcs, evenpointsets, in particular (0, 2, t)-arcs and hyperovals) have constant modulo p in-tersection numbers with lines; we may give the name (generalized) Vandermondeset to such sets. (We recall here Section 9.3.)

Take the “affine” part of a Vandermonde-set, i.e. points with ci 6= 0 and supposethe rest doesnot count in thepower sum that all its points are written as (ai, bi, 1). After the AG(2, q) ↔ GF(q2) identifi-

cation this point becomes ai + biω for some generator ω of GF(q2). Substituting(1, ω, Z) into G we get

0 = G(1, ω, Z) =∑

(ai,bi,1)∈S

((ai + biω) + Z)q−1 +∑

(aj ,bj ,0)∈S

(aj + bjω)q−1 =

q−2∑k=0

±Zq−1−k∑

(ai,bi,1)∈S

(ai + biω)k +∑

(ai,bi,ci)∈S

(ai + biω)q−1

which means that the affine part of a (generalized) Vandermonde set, consideredas a set in GF(q2), has power sums equal to zero for exponents 1, ..., q − 2. (Thelast, constant term is just G(1, ω, 0) = 0.)

13.1 Sets with intersection numbers 0 mod r

Here we continue the examination of sets of PG(2, q) intersecting every line in aconstant number of points mod p. This section is based on [15]. For the moregeneral k-dimensional case see Section 26. The proofs in both sections are similarand are streamlined and then generalised versions of the proof in [14].

Theorem 13.6. Let 1 < r < q = ph. A pointset S ⊂ PG(2, q) which is incident with0 mod r points of every line has |S| ≥ (r− 1)q+ (p− 1)r points and r must divideq.

Proof: Let us first see that r divides q. By counting the points of S on linesthrough a point not in S we have that |S| = 0 mod r. By counting points of S onlines through a point in S we have |S| = 1 + (−1)(q + 1) mod r and combiningthese two equalities we see that q = 0 mod r.


Assuming |S| < r(q + 1) (for if not the theorem is proved) there is an externalline to S, so we can view S as a subset of GF(q2) ' AG(2, q) and consider thepolynomial

R(X,Y ) =∏b∈S

(X + (Y − b)q−1) =|S|∑j=0

σj(Y )X |S|−j .

For all y, b and c ∈ GF(q2) the corresponding points of AG(2, q) are collinear if andonly if (y − b)q−1 = (y − c)q−1 and each factor X + (y − b)q−1 of R(X, y) dividesXq+1 − 1 whenever y 6= b.For y ∈ S we have

R(X, y) = X(Xq+1 − 1)r−1g1(X)r,

and for y 6∈ SR(X, y) = g2(X)r.

In both cases σj(y) = 0 for 0 < j < q and r does not divide j. The degree of σj isat most j(q−1) and there are q2 elements in GF(q2), hence σj ≡ 0 when 0 < j < qand r does not divide j. So

R(X,Y ) = X |S|+σrX |S|−r+σ2rX|S|−2r+...+σqX |S|−q+σq+1X

|S|−q−1+...+σ|S|.

For all y ∈ GF(q2) we have

∂R

∂Y(X, y) =

(∑b∈S

−(y − b)q−2

X + (y − b)q−1

)R(X, y).

In all terms the denominator is a divisor of Xq+1 − 1 so multiplying this equalityby Xq+1 − 1 we get an equality of polynomials and we see that

R(X, y) | (Xq+1 − 1)∂R

∂Y(X, y),

or even betterR(X, y)Gy(X) = (Xq+1 − 1)

∂R

∂Y(X, y) =

(Xq+1 − 1)(σ′rX|S|−r + σ′2rX

|S|−2r + ...+ σ′qX|S|−q + σ′q+1X

|S|−q−1 + ...). (∗)

Here G = Gy is a polynomial in X of degree at most q+1−r. The term of highestdegree on the right-hand side of (∗) that has degree not 1 mod r is of degree |S|and has coefficient σ′q+1, where ′ is differentiation with respect to Y .First examine y 6∈ S. As R(X, y) is an r-th power, any non-constant term in G,with degree not 1 mod r would give a term on the right-hand side of degree > |S|and not 1 mod r, but such a term does not exist. Hence every term in G has degree1 mod r except for the constant term which has coefficient σ′q+1.


For any natural number κ and i = 1, . . . , r−2 the coefficient of the term of degree|S| − i(q + 1)− κr (which is not 0 or 1 mod r) on the right-hand side of (∗) is

−σ′i(q+1)+kr + σ′(i+1)(q+1)+κr

and must be zero. However if (r − 1)(q + 1) + κr > |S| then σ(r−1)(q+1)+κr ≡ 0and we have σ′i(q+1)+κr = 0 for all i = 1, . . . , r− 2. Now consider the coefficient ofthe term of degree |S|−κr. On the right hand side of (∗) this has coefficient −σ′κr(since σ′q+1+κr = 0). The only term of degree zero mod r in G is the constantterm which is σ′q+1. The coefficient of the term of degree |S| − κr in R(X, y) isσκr. Hence

σκrσ′q+1 = −σ′κr for all y 6∈ S. (∗∗)

If y ∈ S then σq+1(y) = 1 and if y 6∈ S then σq+1(y) = 0. Let

f(Y ) =∏y∈S

(Y − y).

Then fσq+1 = (Y q2 − Y )g(Y ) for some g ∈ GF(q2)[Y ] of degree at most |S| − 1

(the degree of σq+1 is at most q2 − 1). Differentiate and substitute for a y ∈ Sand we have f ′(y) = −g(y). Since the degree of f ′ and g are less than |S| we haveg ≡ −f ′. Now differentiate and substitute for a y 6∈ S and we get σ′q+1f = f ′.Thus for y 6∈ S we have σκrf ′/f = −σ′κr and so (fσκr)′(y) = 0. The polynomial(fσκr)′ has degree at most κr(q − 1) + |S| − 2, which is less than q2 − |S| ifκr ≤ q − 2r. So from now on let |S| = (r − 1)q + κr. The polynomial (fσκr)′ ≡ 0and so fσκr is a p-th power. Hence fp−1 divides σκr.If κ ≤ p − 2 then (p − 1)(r − 1)q + κr(p − 1) > κr(q − 1) and so σκr ≡ 0.However the polynomial whose terms are the terms of highest degree in R(X,Y )is (X + Y q−1)|S| which has a term X(r−1)qY κr(q−1) since

(|S|κr

)= 1. Thus σκr has

a term Y κr(q−1) which is a contradiction. Therefore κ ≥ p− 1.

Corollary 13.7. A code of dimension 3 whose weights and length have a commondivisor r and whose dual minimum distance is at least 3 has length at least (r −1)q + (p− 1)r.

A maximal arc in a projective plane is a set of points S with the property thatevery line is incident with 0 or r points of S, see Section 15. Apart from the trivialexamples of a point, an affine plane and the whole plane, that is where r = 1, qor q + 1 respectively, there are examples known for every r dividing q for q even,see e.g. Denniston [59].

Corollary 13.8. There are no non-trivial maximal arcs in PG(2, q) when q is odd.


Proof: A maximal arc has (r − 1)q + r points, see Exercise 14.1.

After Theorem 13.5 it is quite natural to return to Vandermonde-like properties(see also Section 9.3). To end this section and to kind of prepare for the next one,here is an application of the Vandermonde property from [70]:

|K| = q + t, ∀`line |`∩K| = 0, 2or tProposition 13.9. Let q be even and suppose that K is a (q+t, t)-arc of type (0, 2, t)

having t points on the line at infinity, {(y1), (y2), ..., (yt)}, say. If the y-axis is alsoa t-secant, then the set {y1, y1, ..., yt} is a Vandermonde-set.

Proof: First of all note that (∞) cannot be in K, since it would contradict theeasy observation that through any point of K there is one t-secant and q 2-secants.Let U = {(ai, bi) : i = 1, ..., q} be the affine part of K. Suppose a1 = ... =at = 0 and consider the affine Redei polynomial R(X,Y ) =

∏qi=1(X + aiY +

bi) =∑qj=0 rj(Y )Xq−j , of U . According to the fundamental property of the Redei

polynomial and to the properties of K, for a fixed Y = y 6= yi, the polynomialR(X, y) is a square (each affine line through (y) intersects K in an even numberof points, hence each root of R(X, y) has even multiplicity), while for Y = yi(i = 1, ..., t), R(X, yi) = Xq − X. This means that the coefficient polynomialrq−1(Y ) is 1 on the set T = {y1, ..., yt} and zero elsewhere (note that since thecharacteristic is two, we have 1 = −1). This is also true about the polynomialχ(Y ) =

∑ti=1(Y − yi)q−1, so since these polynomials have degree at most q − 1,

they have to be the same. On the other hand (using that a1 = ... = at = 0), theY -degree of R and hence also of rq−1 is at most q − t, so using Proposition 9.19(iii), we are done.

Exercise 13.10. Let B ⊂ PG(2, q) be a pointset, |B| = q+k, with every intersectionnumber being 1 mod p and suppose that B∩`∞ = k. Prove that B\`∞ ⊂ AG(2, q),considered as a subset of GF(q), is a Vandermonde-set.We note that much more is true if k ≤ q+1

2 : such a set, which is a blocking setof Redei type, is always a (translate of a) subspace of GF(q2), considered as avectorspace over a suitable subfield (hence an additive subgroup of GF(q2)); seeTheorem 18.14.

Exercise 13.11. Let B ⊂ PG(2, q) be a pointset, |B| = q+k, with every intersectionnumber being 1 mod p. Let `1 and `2 be two lines such that |`1∩B| = |`2∩B| = kand B ∩ `1 ∩ `2 = ∅. W.l.o.g. let `1 = [1, 0, 0] the X-axis and `2 = [0, 0, 1] the lineat infinity. Prove that `1 ∩B and `2 ∩B are Vandermonde sets!

Theorem 13.12. [70] The t-secants of a (q+t, t)-arc of type (0, 2, t) are concurrent.

Proof: Suppose to the contrary that (after transformation) the line at infinityand the two axes are t-secants.


Let T = {(y1), ..., (yt)} be the intersection of the arc and the line at infinity.According to Proposition 13.9, T is a Vandermonde-set. On the other hand, theaffine transformation switching the two affine coordinates (that is (u, v) → (v, u))(which extends to the ideal line as (y) → (1/y)) interchanges the two axes, whilethe set T is replaced by T ′ = {1/y1, ..., 1/yt} (note that (0) and (∞) cannot be inT ). It follows that again by Proposition 13.9, T ′ is also a Vandermonde set. So wehave that

∑i yki = 0 for any 1 ≤ k ≤ t− 2 and for any q − t+ 1 ≤ k ≤ q − 2.

Now consider the polynomial χ(Y ) =∑ti=1(Y −yi)q−1. According to Propositions

13.9 and 9.19, it has degree q − t and it has q − t different roots (the complementof T ). On the other hand, since

∑i yki = 0 for any q − t + 1 ≤ k ≤ q − 2, it is

divisible by Y t, which means that 0 is a root of multiplicity at least t. This is acontradiction.

13.2 Small and large super-Vandermonde sets

If in Proposition 9.20(ii) we write Y tf( 1Y ) then we get a polynomial of degree t

and its roots are { 1y : y ∈ T}. Hence a super-Vandermonde set is equivalent to a

fully reducible polynomial of form gp(Y ) + Y t, t > p · deg g.Let’s explore this situation. Firstly, if q = p is a prime then the only possibility isf(Y ) = Y t+ c, i.e. a transform of the multiplicative group {y : yt = 1}, if it exists(so iff t|q − 1).If f(Y ) = gp(Y ) + Y t is a fully reducible polynomial without multiple roots thenwe can write it as Y q − Y = f(Y )h(Y ). Now we may use the trick I have learntfrom Gacs: differentiating this equation one gets

−1 = tY t−1h(Y ) + f(Y )h′(Y ).

Substituting a root y1 of f we get h(y1) = −1tyt−1

1= − 1

t yq−t1 . Suppose that t > q

2 ,

then h(Y ) = − 1tY

q−t holds for more values than its degree hence it is a polynomialidentity implying a contradiction unless q− t = 1. As t = q

2 is impossible (it wouldimply p = 2 and f would be a power), we have that either t = q − 1 (and thenh(Y ) = Y so f(Y ) = Y q−1 − 1) or t ≤ q−1

2 .For describing small and large super-Vandermonde sets we need to examine thecoefficients of the original equation Y q−Y = f(Y )h(Y ) carefully. What does smalland large mean? We know that any additive subgroup of GF(q) forms a Vander-monde set, so removing the zero element from it one gets a super-Vandermondeset. The smallest and largest non-trivial additive subgroups are of cardinality pand q/p, respectively. Note that the super-Vandermonde set, derived from an addi-tive subgroup of size p, is a transform of a multiplicative subgroup. This motivatesthat, for our purposes small and large will mean “of size < p” and “of size > q/p”,resp.


Theorem 13.13. [120] Suppose that T ⊂ GF(q) is a super-Vandermonde set of size|T | < p. Then T is a (transform of a) multiplicative subgroup.

Proof: Since t < p the polynomial f(Y ) is of the form f(Y ) = Y t−b0. As f(Y ) isa fully reducible polynomial without multiple roots, it implies that b0 has preciselyt distinct t-th roots, t|q − 1 and T is a coset of a multiplicative subgroup.

Theorem 13.14. [120] Suppose that T ⊂ GF(q) is a super-Vandermonde set of size|T | > q/p. Then T is a (transform of a) multiplicative subgroup.

Proof: Let us write Y q − Y = f(Y )h(Y ), where f(Y ) = Y t + bmpYmp +

b(m−1)pY(m−1)p+...+bpY p+b0 and h(Y ) = Y q−t+aq−t−1Y

q−t−1+...+a2Y2+a1Y .

Consider the coefficient of Y 1, Y 2, ..., Y q in this equation. We getY 1 : −1 = a1b0

Y j : aj = 0 if 2 ≤ j ≤ t and j 6= 1 (mod p)Y j : aj = 0 if t+ 1 ≤ j ≤ 2t and j 6= 1, t+ 1 (mod p)Y j : aj = 0 if 2t+ 1 ≤ j ≤ 3t and j 6= 1, t+ 1, 2t+ 1 (mod p) and so on, generallyY j : aj = 0 if kt+ 1 ≤ j ≤ (k + 1)t and j 6= 1, t+ 1, ..., kt+ 1 (mod p).Y p+1 : ap+1b0 + a1bp = 0Y 2p+1 : a2p+1b0 + ap+1bp + a1b2p = 0generallyY kp+1 : akp+1b0 + a(k−1)p+1bp + ...+ ap+1b(k−1)p + a1bkp = 0, for k = 1, 2, ...,m.Y t+1 : a1 + b0at+1 + bpat−p+1 + b2pat−2p+1 + ...+ bmpat−mp+1 = 0The indices of coefficients a are of the form t − kp + 1. Since t − kp + 1 < t andt− kp+ 1 6= 1 (mod p) (because t 6= 0 (mod p) is true) these coefficients are 0.So the equation is of the formY t+1 : a1 + b0at+1 = 0.Y 2t+1 : at+1 + b0a2t+1 + bpa2t−p+1 + ...+ bmpa2t−mp+1 = 0The indices j of coefficients aj are t < j < 2t. These coefficients are 0 if j 6= 1, t+1(mod p). It means 2t+ 1 6= 1 (mod p) so 2t 6= 0 (mod p) which means p 6= 2. Theother condition 2t+ 1 6= t+ 1 (mod p) is satisfied by any t. HenceY 2t+1 : at+1 + b0a2t+1 = 0 if p 6= 2.SimilarlyY 3t+1 : a2t+1 + b0a3t+1 + bpa3t−p+1 + ...+ bmpa3t−mp+1 = 0.The indices are between 2t and 3t here. The coefficients are 0 if 3t+1 6= 1, t+1, 2t+1(mod p). It gives only one new condition: 3t + 1 6= 1 (mod p) so 3t 6= 0 (mod p)which means p 6= 3. The two other conditions has occurred earlier: p 6= 2 and t 6= 0(mod p).


Y 3t+1 : a2t+1 + b0a3t+1 = 0 if p 6= 2, 3.GenerallyY lt+1 : a(l−1)t+1 +b0alt+1 +bpalt−p+1 + ...+bmpalt−mp+1 = 0, for l = 1, 2, ..., n−1.The indices are of the form t− kp+ 1 and they are between (l− 1)t and lt. Hencethe coefficients a are 0 if lt+ 1 6= 1, t+ 1, ..., (l− 1)t+ 1 (mod p). It gives (l− 1)tconditions:lt+ 1 6= 1 (mod p) so p 6= l;lt+ 1 6= t+ 1 (mod p) so p 6= (l − 1);lt+ 1 6= 2t+ 1 (mod p) so p 6= (l − 2);and so onlt+ 1 6= (l − 2)t+ 1 (mod p) so p 6= 2; finallylt+ 1 6= (l − 1)t+ 1 (mod p) so t 6= 0, which is true.Hence generally we getY lt+1 : a(l−1)t+1 + b0alt+1 = 0 if p 6= 1, 2, ..., l.In particular, substituting l = n− 1 into this equation we getY (n−1)t+1 : a(n−2)t+1 + b0a(n−1)t+1 = 0 if p 6= 1, 2, ..., n− 1.The greatest index of a coefficient a can be q − t− 1.(n− 1)t < q − 1 and nt ≥ q − 1 because of the definition of n.It means that (n− 1)t ≥ q − t− 1 so (n− 1)t+ 1 ≥ q − t.It implies that a(n−1)t+1 (which occurred in the previous equation) does not exist.So we have two possibilities:Case 1. (n− 1)t+ 1 = q − t, so t = q−1

n and the equation is of the form

Y (n−1)t+1 : a(n−2)t+1 + b0 = 0. (Hence we can write 1 instead of a(n−1)t+1.)Case 2. (n− 1)t+ 1 > q − t, so the equation is of the formY (n−1)t+1 : a(n−2)t+1 = 0 if p 6= 1, 2, ..., n − 1. We will now prove that it leads toa contradiction.Substituting a(n−2)t+1 = 0 into the equation

Y (n−2)t+1 : a(n−3)t+1 + b0a(n−2)t+1 = 0, we get a(n−3)t+1 = 0.We can substitute this again into the equationY (n−3)t+1 : a(n−4)t+1 + b0a(n−3)t+1 = 0, and we get a(n−4)t+1 = 0.Substituting this in a decreasing order we getY t+1 : a1 + b0at+1 = 0 so a1 = 0.Hence −1 = a1b0, so a1 6= 0, Case 2 implied a contradiction. It means that Case 1will occur, so t = q−1

n if p 6= 1, 2, ..., n−1. In other words t|q−1 if p 6= 1, 2, ..., n−1.Hereafter, we can write 1 instead of aj if j = (n−1)t+1, and 0 if j > (n−1)t+1.Y (n−1)t+1 : a(n−2)t+1 + b0 = 0 so a(n−2)t+1 = −b0.


Substituting this into the equationY (n−2)t+1 : a(n−3)t+1 + b0a(n−2)t+1 = 0, we get

Y (n−2)t+1 : a(n−3)t+1 + b0b0 = 0 so a(n−3)t+1 = b02.

Substituting this in a decreasing order we getY lt+1 : a(l−1)t+1 = (−b0)n−l for l = n− 1, n− 2, ..., 1. FinallyY t+1 : a1 = (−b0)n−1.Substituting this into −1 = a1b0, we get −1 = (−b0)n−1b0 so 1 = (−b0)n.We are going to examine the equation that belongs to Y lt+kp+1. First we write upY (n−1)t+p+1 : a(n−2)t+p+1 + bp + b2pa(n−1)t+p+1 + ... = 0We have already seen that the coefficients a occurring in this equation are 0,because these are the same as in the equation of Y (n−1)t+1. SoY (n−1)t+p+1 : a(n−2)t+p+1 + bp = 0.SimilarlyY (n−1)t+2p+1 : a(n−2)t+2p+1 + b2p = 0, generally

Y (n−1)t+kp+1 : a(n−2)t+kp+1 + bkp = 0 for k = 1, 2, ...,m.On the other handY lt+p+1 : a(l−1)t+p+1 + b0alt+p+1 + bpalt+1 = 0, for l = 1, 2, ..., n− 1.Generally we getY lt+kp+1 : a(l−1)t+kp+1 + b0alt+kp+1 + bpalt+(k−1)p+1 + ... + bkpalt+1 = 0 for l =1, 2, ..., n− 1 and k = 1, 2, ...,m.In particular, if l = 1 the equation is of the formY t+kp+1 : akp+1 + b0at+kp+1 + bpat+(k−1)p+1 + ...+ bkpat+1 = 0.

Lemma 13.15. bp = b2p = ... = bmp = 0.

Proof: We prove it by mathematical induction.Step 1. First we prove that bp = 0. Consider the equationY (n−2)t+p+1 : a(n−3)t+p+1 + b0a(n−2)t+p+1 + bpa(n−2)t+1 = 0. (∗)We have seen thatY (n−1)t+p+1 : a(n−2)t+p+1 + bp = 0 so a(n−2)t+p+1 = −bp and

Y (n−1)t+1 : a(n−2)t+1 = −b0.Substituting these into the equation (∗), we getY (n−2)t+p+1 : a(n−3)t+p+1− b0bp− b0bp− b0 = 0 so a(n−3)t+p+1 = 2b0bp. Generallywe can writeY lt+p+1 : a(l−1)t+p+1 + b0alt+p+1 + bpalt+1 = 0 for l = n− 1, n− 2, ..., 1. (∗∗)SubstitutingY (l+1)t+p+1 : alt+p+1 = (−1)n−l−1(n− l − 1)bn−l−2

0 bp and


Y (l+1)t+1 : alt+1 = (−b0)n−l−1 into the equation (∗∗), we get

Y lt+p+1 : a(l−1)t+p+1 = (−1)n−l(n− l)bn−l−10 bp for l = n− 1, n− 2, ..., 1. If l = 0

it means

Y p+1 : ap+1b0 + a1bp = 0. (∗ ∗ ∗)Substituting

Y t+p+1 : ap+1 = (−1)n−1(n− 1)bn−20 bp and

Y t+1 : a1 = (−b0)n−1 into (∗ ∗ ∗), we get

Y p+1 : (−1)n−1nbn−10 bp = 0.

In this equation −1 6= 0 (mod p), n 6= 0 (mod p) and b0 6= 0 (mod p) (from theequation a1b0 = −1). It means that bp = 0.

Step 2. Suppose bp = b2p = ... = b(s−1)p = 0. We show that bsp = 0. Consider

Y (n−2)t+sp+1 : a(n−3)t+sp+1 + b0a(n−2)t+sp+1 + bspa(n−2)t+1 = 0. (?)

We have seen that

Y (n−1)t+sp+1 : a(n−2)t+sp+1 + bsp = 0 so a(n−2)t+sp+1 = −bsp and

Y (n−1)t+1 : a(n−2)t+1 = −b0.Substituting these into the equation (?), we get

Y (n−2)t+sp+1 : a(n−3)t+sp+1 − b0bsp − b0bsp = 0 so a(n−3)t+sp+1 = 2b0bsp.

Generally we can write

Y lt+sp+1 : a(l−1)t+sp+1 + b0alt+sp+1 + bspalt+1 = 0 (??) for l = n− 1, n− 2, ..., 1.

Substituting

Y (l+1)t+sp+1 : alt+sp+1 = (−1)n−l−1(n− l − 1)bn−l−20 bsp and

Y (l+1)t+1 : alt+1 = (−b0)n−l−1 into the equation (??), we get

Y lt+sp+1 : a(l−1)t+sp+1 = (−1)n−l(n− l)bn−l−10 bsp for l = n− 1, n− 2, ..., 1.

If l = 0 it means

Y sp+1 : asp+1b0 + a1bsp = 0. (? ? ?)

Substituting

Y t+sp+1 : asp+1 = (−1)n−1(n− 1)bn−20 bsp and

Y t+1 : a1 = (−b0)n−1 into (? ? ?), we get

Y sp+1 : (−1)n−1nbn−10 bsp = 0.

In this equation −1 6= 0 (mod p), n 6= 0 (mod p) and b 6= 0 (mod p) (from theequation a1b0 = −1). It means that bsp = 0.

So we have got bp = b2p = ... = bmp = 0. It means that f(Y ) is of the form

f(Y ) = Y t + b0, and that t|q − 1 so t = q−1n and (−b0)n = 1. Hence

14. Arcs, Segre 85

f(Y ) = Yq−1n + b0, where (−b0)n = 1 if p 6= 1, 2, ..., n− 1 (mod p). So the roots of

f(Y ) are the elements of a coset of a multiplicative subgroup of order t.

14 Arcs, Segre

A (k, n)-arc in a projective plane is a set of k points such that each line intersectsit in at most n points. It is complete if it cannot be extended to a (k + 1, n)-arc.

Considering (k, n)-arcs, Barlotti [24] showed that for 1 < n < q+1, k ≤ qn−q+nand equality can only hold when n divides q.

Exercise 14.1. Verify Barlotti’s bound. Observe that in case of equality each lineintersects the (k, n)-arc in either 0 or n points and so n|q.

In this section we consider the case n = 2, so simply k-arcs. Note that, by Exercise14.1, (q + 2)-arcs exist only if q is a power of 2. So if q is odd then arcs are ofsize ≤ q + 1. In some sense Segre’s theory was a starting point of modern Galoisgeometry.

Exercise 14.2. Let the lines `1, `2 and `3 form a triangle in PG(2, q). Prove thatthere exists a set {m1, ...,mq−1} of lines (mi 6= `j), covering the non-vertex pointsof the triangle if and only if q is even.

The Segre curve is not used in general as a polynomial in three variables, butas an object with general, local or global, algebraic-geometric properties. Strictlyspeaking it is not obvious why to include the “Segre method” to this book. Idecided to treat it here as well because of (i) historical reasons; (ii) it inspireddirectly or indirectly the outburst of the polynomial methods; (iii) there are resultswhere Segre-like arguments can be used together with other polynomial methods,see e.g. Section 17.

A generalization of the theorem of Menelaos is the following.

Let T be an arbitrary triangle in PG(2, q). The points on the sides of T maybe identified by coordinates as follows. Take T as fundamental triangle (i.e. thevertices of T are A1(1, 0, 0), A2(0, 1, 0) and A3(0, 0, 1)), and place the unit point(1, 1, 1) arbitrarily (not on a side of T of course). Points on the side A2A3 are ofform (0, 1, c), points on the side A3A1 are of form (a, 0, 1), and points on the sideA1A2 are of form (1, b, 0). These coefficients will be called the “coordinates” ofthe points. Note that these three points are collinear (so they are contained in an this is Menelaos

algebraic curve of degree one) precisely when abc = −1.

More generally, we have the following:


Theorem 14.3. A necessary and sufficient condition that 3n points, distributed inthree n-tuples on the three sides of T (some possibly coinciding with each other,but none with any vertex of T ) are contained in an algebraic curve C of degree n,is that the product of their “coordinates” is

n∏i=1

ai

n∏i=1

bi

n∏i=1

ci = (−1)n (∗).

If a point P appears m times in an n-tuple then we require that at P the inter-section multiplicity of C and the corresponding line is precisely m. (It also impliesthat C does not contain any side of the triangle.)Proof: We are looking for a curve of form

f(X,Y, Z) =∑

i+j+k=n

αijkXiY jZk.

Let {Ps = (1, bs, 0) : s = 1, ..., n} be the given (multi)set of points on A1A2.Then F (X,Y, 0) must factor to d3(b1X − Y )(b2X − Y ) · · · (bnX − Y ); similarlyF (0, Y, Z) = d1(c1Y − Z)(c2Y − Z) · · · (cnY − Z) and F (X, 0, Z) = d2(a1Z −X)(a2Z − X) · · · (anZ − X), where d1, d2 and d3 are suitable non-zero con-stants. These equations determine all the coefficients αij0, αi0k, α0,jk. In par-ticular, αn00 = d3b1b2 · · · bn = d2(−1)n, α0n0 = d1c1c2 · · · cn = d3(−1)n andα00n = d2a1a2 · · · an = d1(−1)n. The product of these equations gives exactly (∗).On the other hand, if (∗) holds, let’s fix one of the constants, say d1 = 1, then d2, d3

and all the coefficients αij0, αi0k, α0,jk are determined. The remaining coefficients“n−1

2

”free

coefficients αijk, i, j, k 6= 0 can be chosen arbitrarily.

As it is used frequently, we formulate the dual version of the above, which is thena generalization of the theorem of Ceva.Let T be an arbitrary triangle in PG(2, q). The lines through the vertices of Tmay be identified by coordinates as follows. Take T as fundamental triangle (i.e.the vertices of T are A1(1, 0, 0), A2(0, 1, 0) and A3(0, 0, 1)), and place the unitpoint (1, 1, 1) arbitrarily (not on a side of T of course). Lines through the verticesof T other than the sides will have an equation of the form X2 = −bX3, X3 =−cX1 and X1 = −aX2, with a, b, c ∈ GF(q)∗. This coefficient will be called the“coordinate” of the line. Note that these three lines, i.e. [0, 1, b], [c, 0, 1] and [1, a, 0]are concurrent (so they are contained in a pencil, i.e. an algebraic envelope ofdegree one) precisely when abc = −1. More generally, we have the following dualversion:

Result 14.4. A necessary and sufficient condition that 3n lines, distributed in threeCeva: n = 1

n-tuples through the three vertices of T (some possibly coinciding with each other,but none with any side of T ) are contained in an algebraic envelope C of degree ncontaining no side of T , is that the product of their “coordinates” is (−1)n.

14. Arcs, Segre 87

We are especially interested in the following extension :

Result 14.5. (Segre) Let us consider any number k (≥ 3) of lines L = {L1, ..., Lk}in PG(2, q) no three of which are concurrent (that is a dual k-arc), and on eachline Li a set Gi of n points, not necessarily distinct, but each on one line of Lonly. In order that there exists an algebraic curve C of degree n, containing eachGi but not containing any line of L it is necessary and sufficient that each of the(k3

)triplets of sets Gi satisfies the condition in the previous Result 14.3.

Finally, we remark that if k > n, then the curve C is unique, for if C∞ and C∈both satisfy the requirements above, the same holds for any linear combination ofthem. Now form a linear combination C that contains an extra point P on one ofthe lines Li. Since C intersects the line Li in too many points it will contain theline Li as a factor, and since it introduces more intersections with the other linesas well, all lines of L will factor into C but C only has degree n and k > n.Let us also underline that the curve does not contain any point which is on twolines of the dual k-arc A.The most general form of this theorem we are aware of is the following.

Result 14.6. [134] Let us consider any number k (≥ 3) of lines L = {L1, ..., Lk}in PG(2, q), and on each line Li a set Gi of n points, not necessarily distinct, buteach on one line of L only. Suppose that for each J ⊆ {1, 2, ..., k} of the followingtype there is a curve CJ of degree n intersecting the lines Lj : j ∈ J at the givenpoints Gj with the required multiplicities:(a) {Lj : j ∈ J} is a triangle;(b) {Lj : j ∈ J} are all concurrent.Then there exists an algebraic curve C of degree n, intersecting each Li in the givenpoints Gi with the required multiplicities. In this case, if k > n then the curve C isunique.

Exercise 14.7. Prove unicity, so the last sentence of the result.Note that the result does not hold any more if we require (b) for a bounded numberof concurrent lines of L only. But if the hypothesis holds for any t concurrent linesof L, t ≤ n+ 2, then it holds in general.Note also that if we know that a curve C of degree n contains some pointsPs = (as, bs, cs), s = 1, 2, ... then each f(as, bs, cs) = 0 gives a homogeneous linear This argument is

useful for dis-tinct points Psonly.

equation for the coefficients αijk, i + j + k = n. If the curve is unique then thesystem of equations has a unique solution (in the projective sense, so up to aconstant multiplier). The matrix of the system of linear equations contains entrieslike aisb

jscks .

There are many applications of the results of this section. See Section 17 on setswithout tangents, Section 16 on unitals, etc.We begin with an easy application of Ceva’s theorem.


Proposition 14.8. (Bichara, Korchmaros) Let K ⊆ PG(2, q) be a set of q+2 points,and suppose that A1, A2, A3 ∈ K are such that any line meeting {A1, A2, A3}intersects K in exactly two points. Then q is even.

Proof: Choose A1, A2, A3 as the base points of a coordinate frame.Construct a (q− 1)× 3-as matrix M , whose rows are indexed by K\{A0, A1, A2},whose columns are indexed by A0, A1, A2. The element ofM determined by (P,Ai)will be the coordinate λi(P ) of the line PAi. Ceva’s theorem gives that the productof the three elements in one row of M will be 1. Therefore the product of allelements of M will also be 1. On the other hand, in each column there are q − 1pairwise different nonzero element of GF(q). This means that in every column wesee each nonzero element of GF(q). By Wilson’s theorem their product is (−1),whence 1 = (−1)3 follows. This implies that q is even.

Of course the most celebrated application is Segre’s theorem on arcs.

Theorem 14.9. Let S ⊂ PG(2, q) be a (q+1)-arc, i.e. a set of q+1 points, no threeof them being collinear. If q is odd then S is a conic.

Proof: ([23]) Suppose w.l.o.g. that S contains the points (1, 0, 0), (0, 1, 0) and(0, 0, 1). Let the tangents at these points be Z = cY , X = aZ and Y = bX,respectively. Let {Si | i = 1, 2, . . . , q − 2} be the other q − 2 points of S.Let (1, bi, 0) be the point that is the intersection of the line (0, 0, 1), Si with theline Z = 0. Let (0, 1, ci) be the point that is the intersection of the line (0, 0, 1), Siwith the line X = 0 and let (ai, 0, 1) be the point that is the intersection of theline (0, 1, 0), Si with the line Y = 0. Since S is an oval, the set

{ai | i = 1, 2, . . . q − 2} ∪ {a}

contains every non-zero element of GF(q) and similarly for the bi’s and the ci’s.The lines Y = biX, X = aiZ and Z = ciY have a point in common (namely Si),hence aibici = 1.The product of all the non-zero elements of GF(q) is −1. Therefore

−1 =

(q−2∏i=1

ai

)a =

(q−2∏i=1

bi

)b =

(q−2∏i=1

ci

)c,

(q−2∏i=1

aibici

)abc = −1

hence abc = −1.Note that if −1 = 1, so when q is even then we got a(n algebraic) proof for the(combinatorial) fact, that the tangents of a (q+1)-arc are concurrent (and so their

15. Maximal arcs 89

intersection point, called the nucleus, can be added to the arc, which results in a(q + 2)-arc or hyperoval).Let f(X,Y, Z) := −cXY + acY Z + XZ. The conic defined from f = 0 containsthe points (1, 0, 0), (0, 1, 0) and (0, 0, 1) and it follows from the fact that abc = −1that it has tangents Y = cX, X = bZ and Z = aY at these points.We have to show that f (the conic defined by f) contains another arbitrary pointS of S and then we have finished. Let {P,Q,R} = {(1, 0, 0), (0, 1, 0), (0, 0, 1)}. Letf1 be the quadratic form that defines the conic f1 = 0 that contains the points P ,Q and S and whose tangents coincide with the tangents of S at those points. Inthe same way define f2 for the points P , R and S.Let tP be the tangent at the point P . The forms f , f1 and f2 modulo tP have adouble zero at P , so we can multiply by a scalar so that the three forms coincideat all the points of tP . The forms f1 and f2 modulo tS share a double point at Sand coincide at tS ∩ tP , so they coincide at all the points of tS . In the same waythe forms f and f1 coincide at all the points of tQ and the forms f and f2 coincideat all the points of tR. But then f1 and f2 also coincide at tQ ∩ tR. So in all f1and f2 also coincide on two lines and point not on these lines. Hence the quadraticform f1 − f2 is zeros on two lines and another point and is therefore identicallyzero. So f1 = f2 and in the same way f = f1 = f2. Hence the conic defined by theequation f = 0 contains S.

15 Maximal arcs

After Barlotti’s bound (Exercise 14.1), (k, n) arcs of size k = qn − q + n arecalled maximal. In the Desarguesian projective plane of order q, Denniston [59]constructed maximal arcs for every divisor of q, when q is even. Ball, Blokhuis andMazzocca [13] proved that for q odd, there is no maximal arc, see Theorem 15.12.Their proof was simplified in Ball-Blokhuis [14] and then generalized in [15], as wehave seen in Theorem 13.6.

15.1 Existence: hyperovals

Here we give some information about the case n = 2 (hence q = 2h), i.e. abouthyperovals. For other even values of n, Denniston, Thas and Mathon constructedexamples.Consider a hyperoval Hf of PG(2, q), with oval polynomial f(X), i.e. Hf = or o-polynomial

{(1, 0, 0), (0, 1, 0)} ∪ {(x, f(x), 1) : x ∈ GF(q)}; we require f(0) = 0, f(1) = 1.

Exercise 15.1. Hf is a hyperoval if and only if f(X) is a permutation polynomialand Fs(X) = f(X+s)−f(s)

X is a permutation polynomial for all s ∈ GF(q), with N.B.:Fs(0) = f′(s)


Fs(0) = 0.

Exercise 15.2. Prove that an oval polynomial is of formq > 2 even

f(X) = a2X2 + a4X

4 + ...+ aq−2Xq−2.

If f(X) = Xr is a monomial then everything becomes easier.It does notmean that wecan characterizemonomialhyperovals...

Exercise 15.3. f(X) = Xr is an oval polynomial if and only if (r, q − 1) = 1,(r − 1, q − 1) = 1 and f(X+1)−f(X)

X is a permutation polynomial.

Exercise 15.4. In particular, f(X) = X2k is an oval polynomial of PG(2, 2h) ifand only if (k, h) = 1.

It may happen that the graph of f has the property that the group of translationsof AG(2, q) acts transitively on it, i.e. for any c ∈ GF(q), f(X) + f(c) = f(X + c).It is equivalent

to saying that fis additive. In this case Hf is called a translation oval. Then, from Exercise 15.1 we have that

Hf is a translation oval if and only if f(X) is an additive permutation polynomial,and also f(X)/X is a permutation polynomial.

Recall Theorem 5.25, saying that every additive function GF(q) → GF(q) is alinearized polynomial a0X + a1X

p + a2Xp2 + ... + ah−1X

ph−1. See also Exerciseq = ph

5.26.

Exercise 15.5. The map f(X) =∑h−1i=0 aiX

pi is an additive permutation of GF(q)such that f(X)/X is also a permutation, if and only if for the Segre-Bartocci-like

matrix A(λ) = det

λ a1 ... ah−1

aph−1 λp ... aph−2

ap2

h−2 ap2

h−1 ... ap2

h−3...ap

h−1

1 aph−1

2 ... λph−1

= λ2h−1 − 1 holds.

Proposition 15.6. (Payne, Hirschfeld) The oval polynomial of a translation oval isof form f(X) = X2k (and (k, h) = 1 of course).

Proof: [97] We know that the oval polynomial (being additive) is of the formf(X) = a2X

2 + a4X4 + a8X

8 + ...+ aq/2Xq/2 and, by Exercise 15.1 now f(X)/X

is a permutation polynomial. We have to prove that only one ai can be nonzero.

In Exercise 15.5 degλA(λ) = 2h − 1. It is easy to see that the coefficient ofλt, 0 < t < 2h − 1 can be calculated as follows. Let t =

∑hi=1 ti2

i−1, ti ∈ {0, 1};and let tj1 , ..., tjr be precisely the binary digits of t being equal to zero. Let Mt bethe r × r principal minor of A(λ) obtained by selecting the rows and columns ofindices j1, ..., jr and then substituting λ = 0. Then detMt is the coefficient of λt.

15. Maximal arcs 91

Hence we need that each k×k principal minor of A =

0 a1 ... ah−1

aph−1 0 ... aph−2

ap2

h−2 ap2

h−1 ... ap2

h−3...ap

h−1

1 aph−1

2 ... 0

,

0 < k < 2h−1 − 1 is singular.Now for 2 ≤ k < h let {i1, ..., ik} ⊂ {1, ..., h}. By induction on k one can provethat ai2−i1 · ai3−i2 · · · aik−ik−1 · ai1−ik = 0, where the indices are modulo h.The coefficient of λ0 is clearly detA = 1. Expanding detA and using the obser-vation of the previous paragraph on cyclic product, we see that at most (henceprecisely) one ai is nonzero.

Exercise 15.7. If 1 < r, r1, r2, r3 < q− 1, rr1 ≡ 1 (mod q− 1), (r− 1)(r2 − 1) ≡ 1(mod q − 1), r + r3 ≡ 1 (mod q − 1), then the hyperovals HXr , HXr1 , HXr2 andHXr3 are projectively equivalent.

Exercise 15.8. HXr is regular (conic + nucleus) iff r = 2, q/2 or q − 2.We remark that in PG(2, 2h) non-regular hyperovals exist iff h ≥ 4. As for h = 5and for h ≥ 7 there exist translation hyperovals; for n = 4 Lunelli and Sce foundHf with

f(X) = (a2X7 + e7X6 + e3X5 + e13X4 + eX3 + e7X2 + eX)2, where e4 = e+ 1;

finally for h = 6 we have

f(X) = X62 +X30 +X24

+e21(X60+X58+X54+X52+X46+X44+X40+X38+X34+X16+X14+X10+X8+X4)

+e42(X50 +X48 +X36 +X32 +X26 +X20 +X18 +X12 +X6), where e6 = e+ 1.

For h = 1, 2, 3 it is easy to check that every hyperoval is regular.

Proposition 15.9. Segre In PG(2, 2h) f(X) = X6 is an oval polynomial iff h isodd.

Proof: By Exercise 15.3 we have to check whether (X+1)6+1X = X5 +X3 +X is

bijective in the case when h is odd. Suppose that for s 6= t s5 +s3 +s = t5 +t3 +t,from this, dividing out by (s + t) and introducing u = s2 + t2, v = st we get thequadratic u2 + u(v + 1) + v2 + v + 1 = 0. Dividing by (v + 1)2 and introducingw = u

v+1 we get (w + 1v+1 )2 + (w + 1

v+1 ) + 1 = 0. By Exercise 5.6 there is nosolution of Y 2 + Y + 1 if Tr2h→2(1) = h = 1, so when h is odd.


We demonstrate the strength of using the Redei polynomial together with thegeneralised Newton formulae (NG) from Section 9.4 by giving a short proof for atheorem of Glynn.It is easy to see that a set of q + 2 points is a hyperoval if and only if it meetsevery line in an even number of points. (Through a point of the set in questionevery line contains at least one more point, but 1 + (q + 1) · 1 = q + 2, so everyline has to meet it in 0 or 2 points.) For even more on even sets (sets of even type)see e.g. Section 23.4. Glynn gave an equivalent description of oval polynomials asfollows.

Theorem 15.10. (Glynn) A polynomial f(x) is an oval polynomial if and only if(k+ll

)∑xkf(x)l = 0 for every 1 ≤ k+ l ≤ q− 1, k+ l odd, {k, l} 6= {0, q− 1}, and∑

f(x)q−1 = 1.

Proof: (Gacs) First suppose H = {(t, f(t)) : t} ∪ {(0), (∞)} is a hyperoval inPG(2, q) and let U = {(t, f(t)) : t ∈ GF(q)} be its affine part. Consider the affineRedei polynomial

R(X,Y ) =∏

t∈GF(q)

X + tY − f(t) =q∑i=0

ri(Y )Xq−i

of U . H is a hyperoval precisely if every line meets it in an even number of points.For lines with direction y 6= 0 this is equivalent to the condition that R(X, y) is asquare. Since q is even, and deg(rj) ≤ j, this is equivalent to rj(Y ) = 0 identicallyfor all 1 ≤ j ≤ q − 3, j odd, and rq−1(Y ) = Y q−1 − 1 (since R(X, 0) = Xq −X).Considering the coefficients of the rjs, we get σk,l = 0 for all 1 ≤ k + l ≤ q − 1,σk,l =P

t1,...,slall distinct

t1· · ·tkf(s1)· · ·f(sl) k + l odd, except for k = q − 1 or l = q − 1, when it is −1 (note that since thecharacteristic is 2, we have −1 = 1). This implies (k + l)σk,l = 0 for all k and l,except for σ0,q−1 = σq−1,0 = 1.From this, by induction on k + l and using (NG), we get the desired condition.

On the other hand, suppose f(x) is a polynomial satisfying our conditions and letH = {(t, f(t)) : t} ∪ {(0), (∞)}.Again induction on k + l and (NG) give that σk,l = 0 for all 1 ≤ k + l ≤ q − 1,k + l odd, except for k = q − 1 or l = q − 1, when it is −1. This implies that forany fixed y 6= 0, R(X, y) is a square, so all lines with slope not in {0,∞} meetH in an even number of points. This is also true about vertical lines, since f is apolynomial and (∞) is in H , so what is left is to show that horizontal lines meetH in one affine point, i.e. f is bijective.Note that the bijectivity of f is equivalent to R(X, 0) =

∏t(X + f(t)) = Xq +X.

Now R(X, 0) = Xq + σ0,1Xq−1 + ...+ σ0,q = Xq +X + g(X)2 (since σ0,k = 0 for

odd k, except for k = q − 1, when it is 1). But H ′(X, 0) = 1, so R(X, 0) cannothave a multiple root, i.e. R(X, 0) = Xq −X.

15. Maximal arcs 93

Denniston-arcs

Still in GF(q), q = 2h, choose β which makes T 2+βT+1 an irreducible polynomial.Consider the quadratic curves

Fλ(X,Y, Z) = X2 + βXY + Y 2 + λZ2 λ ∈ GF(q).

For λ = 0 it defines the point (0, 0, 1) only, while for other values these are irre-ducible conics contained in AG(2, q). The nucleus of each Fλ (λ 6= 0) is (0, 0, 1).

As the additive group of GF(q) is elementary abelian, it has subgroups of everysize 2k, 0 ≤ k ≤ h.

Exercise 15.11. (Denniston [59]) Let H be a multiplicative subgroup of GF(q) ofsize n = 2k, 0 ≤ k ≤ h. Define

SH =⋃λ∈H

( the points of Fλ ).

Prove that SH is a (qn− q + n, n)-arc (so a maximal arc).

15.2 Non-existence

We saw in Section 13.1 a more general statement and its proof. Here we show theproof by Ball and Blokhuis from [14] of the following result:

Theorem 15.12. Ball, Blokhuis, Mazzocca [13] In PG(2, q), with q odd and n < q,maximal (k, n)-arcs do not exist.

First we can assume that B is a maximal (k, n)-arc contained in AG(2, q) (externallines always exist if n < q), with |B| = k = (n − 1)q + n. As usual, we representAG(2, q) by GF(q2) so B ⊂ GF(q2).

For simplicity assume that 0 6∈ B and let B[−1] = {b−1 : b ∈ B}. Define thepolynomial

B(T ) =∏b∈B

(1− bT ) =∞∑i=0

(−1)iσiT i,

where σi is the i-th elementary symmetric polynomial of the set B, so σi = 0 ifi > k. Also define the polynomial

F (S, T ) =∏b∈B

(1− (1− bT )q−1S) =∞∑i=0

(−1)iσiSi,

where σi is the i-th elementary symmetric polynomial of the set of polynomials{(1− bT )q−1 : b ∈ B}, so degT σi ≤ i(q − 1).


Lemma 15.13. (i) If t0 ∈ GF(q2) \ B[−1], then F (S, t0) is an n-th power.

(ii) If t0 ∈ B[−1], then F (S, t0) = (1− Sq+1)n−1.

Proof: (i) When t0 = 0, the polynomial f(s, 0) =∏b∈B(1 − S) = (1 − S)k and

n|k.When t0 6= 0, then t−1

0 is a point not contained in the arc, whence every linethrough t−1

0 contains a number of points of B that is either 0 or n. In the multiset{(t−1

0 − b)q−1 : b ∈ B}, every element therefore occurs with multiplicity n, so thatin F (S, t0) every factor occurs exactly n times.

(ii) Every line passing through the point t−10 contains exactly n − 1 other points

of B; hence the multiset (t−10 − b)q−1 : b ∈ B consists of every (q + 1)-st root of

unity repeated n− 1 times, together with the element 0. This gives

F (S, t0) =∏b∈B

(1− (t−10 − b)q−1tq−1

0 S) = (1− tq2−1

0 Sq+1)n−1 = (1− Sq+1)n−1.

Corollary 15.14. (i) The functions σi are identically zero for 0 < i < q unless n|i.(ii) The polynomial B divides σn, which is not identically zero; the degree of σn/Bis at most q − 2n.

Proof: From the form of F in both cases of the lemma, it follows that σi(t0) = 0for all t0 ∈ GF(q2) with 0 < i < q unless n|i; since the degree of σi is at mosti(q− 1) < q2, these functions are identically zero. Therefore the first coefficient ofF that is not necessarily identically zero is σn.

It is immediate that σn(0) =(kn

). With q = ph, n = pe,

k = (n− 1)q + n = (p− 1)ph+e−1 + ...+ (p− 1)ph + pe;

hence by Lucas’ theorem(kn

)≡ 1 (mod p), so is not identically zero. On the

other hand, the coefficient of Sn in (1 − Sq+1)n−1 is zero; so σn(t0) = 0 fort0 ∈ B[−1]. Equivalently, B divides σn. Hence the degree of σn/B is at mostn(q − 1)− nq + q − n = q − 2n.

Corollary 15.15.

(i) F (S, T ) = 1 +k∑i=1

(−1)iσinSin +k∑i=1

(−1)iσi(q+1)Si(q+1) (mod T − T q

2).

(ii) BF (S, T ) = B +Bk∑i=1

(−1)iσinSin (mod T − T q2).

15. Maximal arcs 95

Proof: By the previous corollary, the first coefficient of F that is not necessarilyidentically zero is σn. Since, for all t0 ∈ GF(q2), the function σj vanishes unless n|jor (q+1)|j, it follows that T −T q2 |σj . If n does not divide j, then σj still vanishesfor t0 ∈ GF(q2) \ B[−1], and since B|σn, the divisibility relation T − T q

2 |σnσjfollows.

Proof of Theorem 15.12: We shall show that σ2n is a p-th power. This, together

with (a) B|σn and (b) σn is not identically zero, leads to a contradiction for p 6= 2.

Calculating the derivative of B(T ) gives

B′(T ) =∑b∈B

−b1− bT

B(T ) = −(∑b∈B

∞∑i=0

bi+1T i)B(T ).

Since bq2

= b, it follows that

(T − T q2)(∑b∈B

∞∑i=0

bi+1T i) =∑b∈B

q2−1∑i=0

biT i =∑b∈B

(1− bT )q2−1.

Now, −∑b∈B(1− bT )q

2−1 takes the value 1 for T = t0 ∈ B[−1] and is zero for allother elements of GF(q2). Since σq+1 takes the same values and they are both ofdegree q2 − 1 it follows that σq+1 = −

∑b∈B(1− bT )q

2−1. This gives

(T − T q2)B′ = σq+1B.

Differentiating it and multiplying by B gives that

BB′ ≡ B2σ′q+1 (mod T − T q2).

However, differentiating F (S, T ) with respect to T , it follows that

∂TF (S, T ) = (∑b∈B

−b(1− bT )q−2S

1− (1− bT )q−1S)F (S, T ) =

k∑i=0

(−1)iσ′iSi.

Multiplying this by (1 − Sq+1) and putting T = t0 ∈ GF(q2), the expression inbrackets becomes a polynomial in S. Hence

F (S, t0)|(1− Sq+1)∂TF (S, t0).

Define the quotient of this division to be Rt0(S). It follows that

Rt0(S) = −σ′n(t0)Sn + Rt0(S)S2n + σ′q+1(t0)Sq+1,


where Rt0(S) is an n-th power (considered as a function of S). Abusing notationwe define the polynomial R(S, T ) with the property that for t0 ∈ GF(q2) R(S, t0) =Rt0(S). Then we have that

FR = (1− Sq+1)∂TF (mod T − T q2).

Multiplying it by B,

( q−q/n+1∑i=1

(−1)iBσinSin)R ≡ (1− Sq+1)B∂TF (mod T − T q

2). (∗)

Equating the coefficients of Sq+1+n we get

−σ′q+1+nB +Bσ′n ≡ −σ′q+1Bσn = −B′σn (mod T − T q2).

HenceBσ′q+1+n = (Bσn)′ (mod T − T q

2).

Equating successively the coefficients of Si(q+1)+n in (∗) for 1 < i < n− 1 gives

Bσ′i(q+1)+n ≡ Bσ′(i−1)(q+1)+n ≡ (Bσn)′ (mod T − T q2).

This implies that Bp−1 divides σn, which gives a contradiction for p 6= 2, since thedegree of σn is at most n(q − 1) and it is not identically zero.

15.3 Embeddability

There are several improvements on Barlotti’s bound, when n is not a divisor ofq. Lunelli and Sce [94] showed that k ≤ (n− 1)q + n− 3 and if q is large enoughcompared to n, then k ≤ (n− 1)q + 8n/13.Let K be an arbitrary (qn − q + n − ε, n)-arc in PG(2, q). When n divides q, anatural question is whether K is incomplete if ε is small enough, that is whetherit can be completed to a maximal arc. For ε = 1, this was shown by Thas [132].Ball, Blokhuis [12] showed it for ε < n/2 when q/n > 3, for ε < 0.476n whenn = q/3 and for ε < 0.381n when n = q/2. This result was improved by Hadnagyand Szonyi [73], namely they proved it for ε ≤ 2n/3, if q/n is large enough.When n = 2, (k, n)-arcs are simply called k-arcs and maximal arcs (of size k =q+ 2) are called hyperovals. In this case Segre [106] showed that a (q+ 2− ε)-arc,ε ≤ √

q, can be extended to a hyperoval. When 4 < q is a square, this resultis sharp, since there are complete (q + 1 − √

q)-arcs, see Boros and Szonyi [48],Fisher, Hirschfeld and Thas [63] and Kestenband [85]. The only known example of(q + 1−√q)-arcs comes from the cyclic representation of the plane: observe that

15. Maximal arcs 97

q2 + q + 1 = (q +√q + 1)(q − √

q + 1) and take every (q +√q + 1)-th point in

the Singer-cycle (i.e. a point orbit of a subgroup of size (q −√q + 1) in the cyclicrepresentation), it takes some calculations to prove that they form a complete arc.It is nice to remark that if you take every (q −√q + 1)-th point instead then youget the (q +

√q + 1) points of a Baer subplane PG(2,

√q).

Note that these examples exist for odd square q-s as well, and it is conjectured thatalso in the odd case they are the second largest complete arcs, however the bestknown bound is q− 1

2

√q+ 5 by Hirschfeld and Korchmaros. (For a more detailed

description of the known results see e.g. [82].) We remark that Hirschfeld andKorchmaros were also able to prove the stability of the second largest completearc in the even square case:

Result 15.16. [80] A complete arc of PG(2, q), q = 22e, e ≥ 3 has size either q+2,or q −√q + 1, or at most q − 2

√q + 6.

The proofs of the Segre or the Hirschfeld-Korchmaros theorems are based on theprofound examination of the properties of an algebraic curve associated to the arc.

Theorem 15.17. (Weiner) Assume that K is a (qpe−q+pe−ε, pe)-arc in PG(2, q),q = ph, p prime. Suppose also that ε ≤ √

q/4 and pe <√q. Then K can be

embedded in a maximal arc.

After some hesitation we omit its (nice but lengthy) proof. For e = 1 see Szonyi[125], his method was improved by Weiner, who proved it for general e in [137].For large n = p, the results of Ball-Blokhuis and Hadnagy-Szonyi turn out to bebetter, while for small p Theorem 15.17 is stronger. This and Theorem 15.12 ofBall, Blokhuis and Mazzocca on the non-existence of maximal arcs for p > 2, yieldan upper bound on the size of a (k, pe)-arc.The following result and proof by Weiner contains some of the key ideas of theproof of Theorem 15.17; it is similar to Segre’s idea as it associates an(other)algebraic curve (envelope) to the arc. The adventage is that this proof seems tobe easier and there is a nice way to generalize it to (k, pe)-arcs of PG(2, q) to getTheorem 15.17. We will use Theorem 9.37.

Theorem 15.18. Any arc in PG(2, q), q even, of size greater than q −√q + 1 canbe embedded in a hyperoval.

Proof: ([137]) Assume to the contrary, that K is not a hyperoval, hence |K| =q+ 2− ε, where 1 ≤ ε ≤ √

q. If there is a point, not in the arc, through that therepass only 0- and 1-secants, then adding this point to the arc we still get an arc.Repeating this process until there is no more such a point, we obtain an arc K.We show that K is a hyperoval.First of all observe that there are exactly ε 1-secants passing through a point ofK, hence the total number of 1-secants is ε(q + 2− ε) 6= 0.


Define the index of a point as the number of 1-secants through it. Let s be theindex of a point P on some 0-secant `, we show that s is at most ε.Let ` be the line X = 0 (so [1, 0, 0]) and R(X,Y, Z) be the Redei polynomialof K. For a fixed (0,−z, y) ∈ `, consider a root x of R(X, y, z), its multiplicityis 1 or 2 according as the line [x, y, z] is a 1- or a 2-secant. If x is a root ofmultiplicity 2 then its multiplicity as a root of ∂XR(X, y, z) is still at least 2, soR(X, y, z) is

fully reducible deg(g.c.d.(R(X, y, z), ∂XR(X, y, z))) is exactly (q+2−ε)−(the number of 1-secantsthrough the point (0,−z, y)).For some s now write up the resultant R(Y,Z) = Rs(R(X,Y, Z), ∂XR(X,Y, Z))as in Section 9.5, its (homogeneous) degree is s(s− 1). We have to give an upperbound for the points on ` with index less than s: if ty,z denotes the number of1-secants through (0,−z, y) then∑

(0,−z,y)∈`

min{s− ty,z, 0} ≤ s(s− 1).

So counting the 1-secants through the points of ` we get at least (q+1)s−s(s−1),that is at most ε(q + 2− ε); from which s ≤ ε or s ≥ q + 2− ε follows. Note thatthere can be no point with index at least q+ 2− ε, since such a point would haveindex q + 2 − ε exactly and it could have been added to K. Furthermore, sincethrough each point outside K there passes at least one 0-secant, the argumentabove implies that the index of any point is at most ε.Hence the 1-secants form a dual (ε(q + 2 − ε), ε)-arc and so when ε ≤ √

q, thecontradiction follows by Barlotti’s bound 14.1.

In general, probably√q is not the right order of magnitude for ε, except for the

case n = pe = 2. However if we want a bound for ε not depending on pe, the bestorder of magnitude we can get is

√q.

Combining Theorem 15.17 with the result of Ball, Blokhuis and Mazzocca Theorem15.12 on non-existence of maximal arcs, the next corollary follows:

Corollary 15.19. A (k, pe)-arc in PG(2, q), q = ph, p > 2 prime, has size less thanqpe − q + pe − 1

4

√q.

16 Unitals, semiovals

A unital is a 2− (q3 + 1, q + 1, 1)-design. We use the same name for a pointset Uof PG(2, q2) if it, together with its (q + 1)-secants, forms a unital in the previoussense. It means that we have |U | = q3 + 1 points, and every line intersects U ineither 1 or q+1 points, the size and this intersection property is equivalent to the

16. Unitals, semiovals 99

definition. It is also obvious by elementary counting that there is a unique tangentline through each point of a unital. Hence unitals are minimal blocking sets (ofmaximum possible size, in fact) and semiovals as well.The absolute points of a unitary polarity of PG(2, q2) form a unital, it is calledthe classical unital or a Hermitian curve. This curve is projectively equivalent to

Xq+11 +Xq+1

2 +Xq+13 = 0.

Note that for a Hermitian curve, the tangent lines of collinear points are concur-rent, i.e. they meet in the polar point of that line. We will see below that thisproperty characterizes Hermitian curves, also giving us an opportunity to show anapplication of Result 14.6 and the Segre technique.Various characterizations of a Hermitian curve have been given. Lefevre-Percsyand independently Faina and Korchmaros proved the following:

Result 16.1. Let H be a unital in PG(2, q2). If every (q + 1)-secant of H meets Hin a Baer subline then H is a Hermitian curve.

Hirschfeld, Storme, Thas and Voloch gave a characterization in terms of algebraiccurves:

Result 16.2. In PG(2, q2), q 6= 2, any algebraic curve of degree (q+1), without linearcomponents, and with at least q3+1 points in PG(2, q2) must be a Hermitian curve.

Thas [133] proved the following

Theorem 16.3. Let H be a unital in PG(2, q2). If the tangents of H at collinearpoints of H are concurrent, then H is a Hermitian curve.

The next lemma was the first step in Thas’ proof:

Lemma 16.4. Let H be a unital in PG(2, q2) with the property that the tangents ofH at collinear points of H are concurrent. Let P1, P2, P3 be non-collinear pointsof H and let Li be the tangent of H at Pi. Choose the coordinates in such a waythat L1 ∩ L2 = (0, 0, 1), L2 ∩ L3 = (1, 0, 0), L1 ∩ L3 = (0, 1, 0). If P1 = (0, b, 1),P2 = (1, 0, c) and P3 = (a, 1, 0) then (abc)q+1 = 1.

The proof is a nice example of the Segre-like arguments, see the first part ofTheorem 14.9. We leave it as an exercise.

Exercise 16.5. Prove the lemma of Thas above!In his original proof Thas showed, using this lemma, that every (q + 1)-secantmeets H in a Baer subline, then by Result 16.1 H is classical. Here we follow theproof of [134], which uses Result 16.2 instead.Proof of Theorem 16.3, sketch. Let L1, L2 and L3 be non-concurrent tangentsof H at the tangent points P1, P2 and P3. By the lemma and Segre’s Theorem


14.3, there is an algebraic curve C′ intersecting each Li at Pi with intersectionmultiplicity q + 1. Next, let Q 6∈ H and let M1,M2, ...,Mq+1 the tangents of Hthrough Q. Then the tangent points Q1, Q2, ..., Qq+1 are collinear, let ` be the linecontaining them. The curve of degree q + 1, consisting of ` counted q + 1 times,intersects Mi at Qi with intersection multiplicity q + 1. Now from Result 14.6 itfollows that there exists a curve C of degree q+1 intersecting each tangent of H atits tangent point with intersection multiplicity q+1. As C contains all points of H,it has at least q3+1 points. If C contains a linear component L, then L contains allthe q3 +1 points of H, a contradiction. Now by Result 16.2 of Hirschfeld, Storme,Thas and Voloch, if q 6= 2 then C is a Hermitian curve. As |C| = |H| = q3 + 1,C = H and H is a Hermitian curve. The case of q = 2 is easy to check.

A semioval is a set of points such that there is a unique tangent at each point.A semioval is regular if all non-tangent lines intersect it in either 0 or a constantnumber a of points.A semioval is called almost oval if there are exactly q − 1 2-secants through eachof its points. This means that an almost oval semioval C has q + a points andthrough each point of C there is a unique tangent and a unique (a + 1)-secant(a > 1). For a = 1 these sets were called seminuclear sets by Blokhuis and Bruenin [45].In [34] Blokhuis characterizes almost oval semiovals.

Theorem 16.6. (Blokhuis) Let C be an almost oval semioval in PG(2, q), |C| =q+a. Then either q = a+2 and C consists of the symmetric difference of two lines,with one further point removed from each line, or q = 2a+3 and C is projectivelyequivalent to the projective triangle, that is the set defined in Example 18.2.

The proof of this theorem is based on two propositions, the first one can be provedusing the Redei polynomial.

Proposition 16.7. Let ` be an (a+ 1)-secant of an almost oval semioval C, |C| =q + a. Then the tangents of C at the different points of ` ∩ C are concurrent.

Proof: Choose coordinates so that ` = `∞ and let U = C \ `. Let U = {(ai, bi) :i = 1, . . . , q − 1} and consider the affine Redei polynomial R(X,Y ) of U . For an(m) ∈ ` ∩ C we have that R(X,m)|(Xq − X). Since degX(R) = q − 1, for these(m)’s the quotient (Xq −X)/R(X,m) can be computed; it has to be X − r1(m).This means that the line Y = mX + r1(m) is the tangent of C at (m). Sincedeg(r1) = 1, r1(m) = am + b, so the tangents are the lines Y = mX + am + b.These lines pass through (−a, b).

The following generalization of Proposition 16.7 can be proved in exactly the sameway as the original one.

16. Unitals, semiovals 101

Proposition 16.8. Let U be a set of q − 1 points in AG(2, q), which determines atmost q−1 directions. Then there is a unique point w such that U ∪{w} determinesthe same directions as U .

Proof: Take an infinite point (m) not determined by U . Then there is a uniqueline tm with slope m which does not meet U . These lines pass through a point w;this can be proved exactly as in the proof of Proposition 16.7. (The details are leftas an exercise.)

This result can be generalized further, namely for sets of size less than q− 1. Thiswill be discussed in Theorem 23.1.Not much is known about semiovals, Thas proved lower and upper bounds fortheir size (and nonexistence results for their higher dimensional analogues). Forregular semiovals Blokhuis and Szonyi [40] showed that a has to divide q − 1 andthe points not on the semiovals are on 0 or a tangents. Thus in this case there isa dual (regular) semioval. For regular semiovals in PG(2, q) more can be said.

Proposition 16.9. [40] Let S be a regular semioval in PG(2, q). Then S is a unital,or the tangents at collinear points of S are concurrent.

Proof: (Hint:) Try to copy the proof of the previous two propositions.

We have seen above that unitals with the property that tangents at collinear pointsare concurrent were characterized by Thas as classical unitals (i. e. Hermitiancurves).

Using the results of Blokhuis and Szonyi, and applying an approach similar toThas’ above, Gacs in [68] could give a nice proof of Segre-type for the following.

Theorem 16.10. In PG(2, q) any regular semioval is either an oval or a unital.

To end this section we show an easy non-existence result.

Theorem 16.11. For q odd there exist no semioval of size q + 2, i.e. a set S ⊂PG(2, q) with the property that through each point of S there pass exactly onetangent.

Proof: Observe that through any point of S there pass exactly one 3-secant, andthe tangent and 3-secant lines together cover the points of the plane, as q + 2 isodd, hence counting the points of S on the lines through any point P 6∈ S wehave to find an odd number of odd secants. Therefore they form a dual blockingset of size q + 2 + q+2

3 . It is minimal and does not contain a dual line (pencil),


hence by Szonyi’s Corollary 21.21, every dual line intersects it in 1 mod p points.But there are intersections of size 2 (with pencils centered at the points of S),contradiction.

17 Sets without tangents: where Segre meets Redei

This section is based on [42], [41] and [130]. Sets without tangents were introducedby Blokhuis, Seress and Wilbrink [42], who called them untouchable sets. As a setof size ≤ q+1 always has a 1-secant (through each point of it), an untouchable sethas at least q+2 points. For q odd, Blokhuis, Seress and Wilbrink proved that thesize of a set without tangents in PG(2, q) is at least q+ 1

4

√2q+ 2. For q even, the

plane PG(2, q) always has a hyperoval, which is an untouchable set of minimumcardinality; here the question is to find the size of the second smallest untouchableset. When all lines intersect a set in an even number of points, then such a setis a codeword in the dual code of the plane. For codewords of small weight, seeChapter 6 of the book [3] by Assmus and Key. Note that the cardinality q+4 canbe realized in the planes of order 4, 8 and 16, but probably in no other planes.All known examples have at least q +

√q points, see Korchmaros, Mazzocca [88].

They constructed q+ t-sets of type (0, 2, t) for t ≥ √q. More examples and a proof

that the t-secants are concurrent for such sets can be found in Gacs, Weiner [70],see Theorem 13.12.

As mentioned in [42], there are untouchable sets of size 2q, and this is best possiblefor q = 3, 5. Smaller examples were also constructed, the best one for odd q’s hassize 2q − q/p, where q = pn. For more constructions in the case q odd, and forplanes of small order, we refer to the original paper [42]. For q even see someexamples from [41] at the end of this section.

The method introduced in the paper by Blokhuis, Seress and Wilbrink is to as-sociate a curve to an untouchable set, whose points correspond to the lines inter-secting the set in at least 3 points. This is a nice application of Segre’s Lemmaof tangents. Their bound follows from estimating the number of singular points ofthis curve. For q even, the construction of the curve still works, but typically wehave no information on the number of its singular points. In the next section webriefly recall the construction of these curves.

The Segre-curve, here containing exactly the > 2-secants of a pointset is, intu-itively, the Redei-curve minus the points corresponding to 2-secants (if there areno tangents). The result we are going to prove here is the following.

Theorem 17.1. Let S be a set of points in PG(2, q), without tangents. Then

(a) if q is odd then |S| > q + 2 +√

18q;

17. Sets without tangents: where Segre meets Redei 103

(b) if q is even and S has an odd line then |S| > q + 2 +√

16q.

In Section 17.2 we show an improvement by the Szonyi-Weiner resultant method.In the case when q is even they proved that a set without tangents, having an oddline, is of size > q + 3b√qc − 7 if q > 16. Their proof is very short, and elegant,but it is based on other strong results of their stability method.

17.1 The proof using Segre’s method

Curves associated to untouchable sets using Segre’s method

Let S be a set without tangents in PG(2, q), 2|q, |S| = q + 2 + ε. Throughout thissection we suppose that q > 6ε2 + 5ε + 1, at some points we only need that ε isless than

√q/6. Since |S| is close to q+ 2, and the set has no tangents, almost all

lines intersect it in 0 or 2 points. We wish to show that the lines intersecting S inmore than 2 points are contained in an algebraic envelope of degree ε. For this wewill need the generalization of the theorem of Ceva, see Theorems 14.5,14.3.Let’s remember that if k > n, then the envelope C of Segre is unique. Let us alsounderline that the envelope does not contain any 2-secant of the k-arc A.

Lemma 17.2. Let T be any triangle contained in S, (S is a set without tangentsand |S| = q+2+ ε), with the property that the sides of this triangle are 2-secants.

(a) Suppose that q is even. Consider the i-secants through the vertices of T ,each with multiplicity (i− 2). Then these lines are contained in an algebraicenvelope of degree ε.

(b) Suppose that q is odd. Consider the i-secants through the vertices of T , eachwith multiplicity 2(i − 2). Then these lines are contained in an algebraicenvelope of degree 2ε.

Proof: Looking at S from a vertex of T we see that the sum of the multiplicitiesof the i-secants through this vertex is just ε. Consider a (q − 1 + ε) × 3 matrixwhose rows are indexed by the points of S \T , and the columns are indexed by thevertices A0, A1, A2 of T . As the notation suggests, choose T as the fundamentaltriangle. Put the “coordinate” of the line PAi in position (P,Ai) of this matrix.Then the product of the coordinates in each row will be −1. In each column wesee the coordinates of the i-secants (i − 1) times each, and every other elementof GF(q) once. Since the product of the elements of GF(q) is −1, the product ofthe remaining elements is (−1)q+ε. This simply means that the i-secants withmultiplicity (i − 2) satisfy the condition in Result 14.3 if q is even (i.e. −1 = 1),but they do not satisfy it when q is odd. However, by the argument above, thei-secants with multiplicity 2(i− 2) will satisfy the required condition, we pay forit by taking square, i.e. the resulting curve will be of degree 2ε.


Note that when ε is small then most of the lines will be 2-secants or skew to S.We want to collect, as usual when using Segre-curves, the “rare”, “irregular” linesin one curve (envelope).

Lemma 17.3. If 6ε2 + 5ε+ 1 < q, then there exists an algebraic envelope of degreeε in the even and 2ε in the odd case, containing all the i-secants of S with i > 2,but containing no 2-secants of S.

Proof: On the points of S we form a graph Γ as follows: vertices of Γ are thepoints of S, two points are adjacent if the line joining them is an i-secant withi > 2. Since the number of i-secants through a point of S is at most ε, there areat least q + 1 − ε lines through the points which are 2-secants. Hence the degreeof a point is at most 2ε. One can show that there is a set A ⊂ S of at least3ε independent points, for which Result 14.5 can be applied, and one can gluetogether the locally obtained curves if (2ε+ 1)(3ε+ 1) > q.

The proof in the q = odd case

We have a curve C of degree 2ε with the property that it intersects each pencilof S in any i-secant line ` of it with multiplicity 2(i − 2). As for each of the ipoints (pencils) on `, C has a multiple intersection in `, it follows that ` cannotbe a simple “point” of C. We are going to use Result 10.3, stating that a curvewithout multiple components can only have a very limited number of multiplepoints: for a projective plane curve F of degree n without multiple components∑P∈F

mP (mP−1)2 ≤

(n2

)holds.

So we get rid of the multiple components: write C = E · D2, where E no longercontains multiple components. The essential observation is that E is of positive(even) degree as otherwise D would be a curve of degree ε with the property thatit intersects each pencil of S in every one of its i-secant lines ` with multiplicity(i − 2), but we have seen that such a curve does not exist. As C fully intersectsevery pencil P ∈ S (since

∑`3P 2(i` − 2) = 2ε), the same holds for E . It follows

from the fact that the intersection numbers are still even (we subtracted an evennumber from 2(i− 2)) and if an the intersection number is positive for one pencilP containing ` then ` ∈ E so the intersection number is positive (and ≥ 2) for alli` points on P . As a consequence, ` is a multiple point of E .

For ` ∈ E let µ(P ) denote the multiplicity of ` on E and for a pencil P ∈ S letµ(`, P ) be the intersection multiplicity of E and the pencil P at `. Then µ(`, P ) ≥µ(`), where “>” holds if and only if the pencil P is a “tangent” of E at `. Also thenumber of pencils P ∈ S that are tangents at ` is at most µ(`). Let the degree ofE be 2m; since E fully intersects each pencil P ∈ S, we get∑

P∈S

∑`3P

µ(`, P ) = 2m(q + 2 + ε).


Let’s change the order of summation and use the following estimate:∑P∈`

µ(`, P ) ≤ µ(`)(ε− µ(`)) + 2mµ(`) ≤ 3εµ(P ). (∗)

This estimate is obtained as follows: if the pencil P is a tangent (here wemean tangent in the algebraic sense, so this neither implies or is implied by|E ∩ the pencil P | = 1) of E at `, then we know no better than that µ(`, P ) ≤ 2m,and this happens at most µ(`) times. For the remaining points of ` we haveµ(`, P ) = µ(`). Finally we used 2m ≤ 2ε. Using Result 10.3 we obtain∑

`∈E

µ(`)(µ(`)− 1) ≤ 2m(2m− 1). (∗∗)

We now combine (∗) and (∗∗), where in (∗∗) we restrict ourselves to those ` ∈ Ethat are on a point of S. For those `, we have that µ(`) ≥ 2, so certainly

∑` µ(`) ≤

2m(2m− 1). Combining this with (∗) we get

(2m− 1)3ε ≥ q + 2 + ε.

Since 2m ≤ 2ε, we get q ≤ 6ε2 − 4ε− 2 < 6ε2 + 5ε+ 1, a contradiction.

The proof in the q = even case: Segre and Redei hand in hand

Here the idea is to combine the above method with a Redei polynomial approachintroduced for (k, n)-arcs in [125]. Besides the Segre-curve of the previous sec-tions, another curve will be associated to the untouchable set, with many similarproperties as the Segre-curve, implying that they have many points in common.

Finally, we can show that the two curves have a common linear component andderive a contradiction from this fact, if the size of the set is not far from q+2. Thissecond method does not work when the set is contained in the dual of the codegenerated by the lines of PG(2, q), that is when each line intersects our untouchableset in an even number of points.

Now another curve will be associated to the untouchable set. This method relieson the paper [125]. Again, the geometric statements will be proved and we willuse the algebraic results of Section 9.5.

Consider a line ` which will be the line at infinity. Let U = S \ ` = {(ai, bi) : i =1, . . . , |U |} and write the Redei polynomial as

R(X,Y ) =|U |∏i=1

(X + aiY − bi) =|U |∑j=0

rj(Y )X |U |−j .


Note that degY (rj) ≤ j. First let us try to interpret in terms of R(X, y) the factthat through (y) there pass exactly s odd lines. (Here an odd line is an i-secantwith i odd. Note that i ≥ 3 automatically.) From now on the index of a point willbe the number of odd lines passing through it.

Lemma 17.4. The point (y) /∈ S has index s if and only if the greatest commondivisor of R(X, y) and ∂XR(X, y) has degree exactly |U | − s.

Proof: Since the characteristic is 2, for (y) /∈ S the polynomial R(X, y) has rootswith even multiplicity, and roots corresponding to the odd lines. If we consider∂XR(X, y), then the roots of even multiplicity remain with at least this multiplicityand the multiplicity of the other roots decreases by 1. In other words, the greatestcommon divisor of R(X, y) and ∂XR(X, y) has degree exactly |U | − s.

Bounds for the number of odd lines is also needed in the proof.

Lemma 17.5. The total number of odd lines is at most ε|S|/3 and at least q + 1.

Proof: As we have seen in the previous part, the number of i-secants, with i ≥ 3,through any point of S is at most ε. Therefore, the total number of odd lines isat most ε|S|/3. For the lower bound note that when |S| is even then the numberof odd lines passing through any point is even, hence the odd secants form a dualeven set. When |S| is odd, the number of odd lines passing through a point is odd,so they form a dual blocking set. It is well-known that a blocking set of PG(2, q)Look around

from a point notin the blockingset!

is of size at least q + 1.

The main idea of the proof is that for a fixed Y = y the polynomial

b(X, y) = R(X, y)/ gcd(R(X, y), ∂XR(X, y))

can be expressed using the coefficients r1(y), ... of R(X,Y ). Choose a parameter sand consider the polynomial b = bs of degree s in X, which is obtained by solvingthe system of linear equations in the proof of Theorem 9.34 (s is not fixed yet!).We will try to choose a typical index s for the line at infinity; that is an s forwhich through most of the points there pass exactly s odd lines. The next lemmasays that such a typical index exists.

Proposition 17.6. Assume that ` intersects S in 0 or 2 points. Then there is aunique s = s(`) such that s ≤ ε and at least q − ε

√q points on ` have index at

most s and at least 56q − ε

√q of them have index exactly s.

Proof: Since there are at most ε|S|/3 ≤ εq/2 odd lines, the average index of apoint is at most ε/2. Hence there are at least q+1

2 points on the line at infinitywith index at most ε. Go down from ε and stop at the largest s ≤ ε, where there


are points of index s. Then the determinant of R = Rs in Theorem 9.34 is notidentically zero as a polynomial in Y , and we know the following from Theorems9.34 and 9.35:

(A) There are at most s(s− 1) points for which detR = 0.(B) The points whose index is less than s are among these points, and detR 6= 0for points of index s.

Hence there are at least q+12 − s(s − 1) points that have index exactly s. Using

the same trick, the first t > s for which there can be points of index t must satisfyt(t − 1) ≥ q+1

2 , since for the points of index at most s the determinant of Rt inTheorem 9.34 must vanish. This implies t ≥

√q/2. Since the average index is at

most ε|S|/(3(q + 1)), there are at most ε|S|√

2/(3√q) points having index larger

than s. This is certainly at most√

2ε(√q + 1)/3 ≤ ε

√q, since ε ≤

√q/6. We

already saw in (A) that there are at most s(s− 1) ≤ q/6 points of index less thans, so the result follows.

Proposition 17.7. Let s be the typical index for the line ` at infinity. Then thereis a curve c`(X,Y, Z) with the following properties:

(1) degX(c`) ≤ s, deg(c`) ≤ s2;

(2) c`(x, y, 1) = 0 implies that [y,−1, x] is an odd line, if the point (1, y, 0) hasindex s.

Proof: Denote the typical index by s. By Lemma 17.4 this means that the g.c.d.of R(X,Y ) and ∂XR(X,Y ) has degree degX R− s. Consider c` as a homogeneouspolynomial constructed in the proof of Theorem 9.34. By Result 9.35, the polyno-mial c(X) detR = c` has X-degree s and total degree at most s2. If a fixed point(1, y, 0) of the line at infinity has index s, then the odd lines through it correspondto the roots of c`s(X, y). This also shows (2).

The two curves have a common component

In this section we would like to choose the line at infinity to be a 2-secant of S, sothat the typical index of it s > 0 holds. Next we show that such 2-secants exist.

Lemma 17.8. Assume that there is a line intersecting S in an odd number of points.Then there exists a 2-secant of S, such that the typical index for it is not zero.

Proof: The envelope C(X0, X1, X2) obtained by Segre’s method has degree lessthan

√q/6 and it contains all the odd secants of S, hence by Bezout’s theorem


any pencil with vertex having index at least√q/2 must be a component of C. So

there are at most√q/6 points with index at least

√q/2.

Choose a 2-secant ` that contains no point with index at least√q/2. (Such a 2-

secant exists, since through a point of S there pass at least (q+1−ε) 2-secants andthere are only

√q/6 points with index at least

√q/2.) Assume that the typical

index on ` is zero. Note that there are no points of ` \ S with index greater thanthe typical index of `. (Since by the proof of Proposition 17.6 such a point wouldhave index at least

√q/2.) Hence each point of ` \ S has index zero and the odd

lines intersect ` in the points of S ∩ `. So there can be at most 2ε odd lines intotal, which contradicts our lower bound for the number of odd lines, see Lemma17.5.

Let the line at infinity be a 2-secant of S and assume that for the typical indexs > 0 holds. Then the curve C(X0, X1, X2) obtained by Segre’s method has X2-degree ε (which is its total degree). SubstituteX0 = Y ,X1 = −1, andX2 = X in Cto get the curve g(X,Y ). According to the results in the second section, this curvecontains all odd lines. Similarly, let us denote by c(X,Y ) the curve constructed inthe previous section.

Lemma 17.9. The curves c and g have a common component.

Proof: The curve c has at least ( 56q − ε

√q)s points, which correspond to odd

lines. These points are also points of the curve g. If the two curves had no commoncomponent, then Bezout’s theorem would give that

s(56q − ε

√q) ≤ s2ε,

which is impossible, since s ≤ ε.

Now we are able to prove our main theorem.

Theorem 17.10. The size of an untouchable set in PG(2, q), q even, having oddlines, is at least q + 1 +

√q/6.

Proof: Assume to the contrary, that there exists an untouchable set in PG(2, q),having odd lines, with size q + 2 + ε, where ε <

√q/6− 1. Note that then 6ε2 +

5ε + 1 < q. Consider a common (absolutely irreducible) component of c and gguaranteed by the previous lemma. Let the X-degree of this component be u,which is its total degree. Then the component has at least ( 5

6q − ε√q)u points.

Using the Weil bound one immediately sees that this component has to be linear.But that implies the existence of a point with at least ( 5

6q−ε√q) odd lines through

it. This is a contradiction, since such a set would have at least 3( 56q−ε

√q) points.


17.2 Untouchable sets in Galois planes of even order

Here we improve Theorem 17.1 in the case when q is even. The proof is based onTheorem 23.22.First we show that a not too large untouchable set is very close to be a set of eventype, hence we can apply Theorem 23.22 to such sets.

Lemma 17.11. Let U be a set without tangents in PG(2, q), q even. Assume thatU has q + 2 + ε points, then the number of odd-secants is at most ε|U |/3.

Proof: Through any point of U there pass at most ε odd-secants. An odd-secantcontains at least 3 points of U , therefore the total number of odd-secants is atmost ε|U |/3.

Theorem 17.12. (Szonyi-Weiner [130]) The size of an untouchable set U inPG(2, q), 16 < q even, having odd lines, is larger than q + 3b√qc − 7.

Proof: Assume that |U | ≤ q + 3b√qc − 7. By Lemma 17.11, the number of odd-secants, δ of U is less than (b√qc + 1)(q + 1 − b√qc). By Theorem 23.22, wecan construct a set of even type from U by modifying d δ

q+1e ≥ 1 points. If P isa modified point, then through P there pass at least q + 1 − (d δ

q+1e − 1) odd-secants of U . But then counting the points of U on the lines through P we getthat |U | > 2(q − b√qc), when P ∈ U , and |U | > 3(q − b√qc), when P 6∈ U ; whichis a contradiction.

Concluding remarks

In this section first we give examples for small sets without tangents but havingodd lines. First of all note that the union of two non-disjoint hyperovals H1 andH2 is always such a set of size 2(q + 2)− |H1 ∩H2|. In [86] it was showed that inPG(2, 16) there exist two hyperovals (a Lunelli-Sce and a regular one) intersectingeach other in 8 points; hence there is a non-even type, untouchable set of size 28in PG(2, 16). By the construction method introduced in [70], the above set canbe used to obtain non-even type, untouchable sets in PG(2, 16h), (h ≥ 2), of size74 · 16h − 16h−1 and of size 7

4 · 16h.

For untouchable sets in planes of even order we were lucky in the sense that bothSegre’s method and Redei’s method worked. The Redei method can be generalizedto the following situation: let S be a set having only 0-secants and i-secants, wherei ≥ p, q = ph. Then the size of S is at least qp − q + p, and in case of equalityS is a maximal arc. By Theorem 15.12 of Ball, Blokhuis and Mazzocca [13], suchan arc does not exist if p is odd. Even in that case, one can show that |S| hasto be at least qp − q + p + ε with ε ≥ 4

√q/2, if S has an i-secant for which i is


not divisible by p. For this, one has to copy more or less the entire proof in [125].The reason why the result is weaker than for our case p = 2, is that the methodbased on Segre’s lemma of tangents does not seem to work in this case. Similarly,instead of i ≥ p one can suppose that i ≥ pr and the situation is more or less thesame in this case, too.

When pr > 4√q/2, then a better bound can be obtained by simple counting argu-

ments:

Exercise 17.13. Let S be a set of size nq+ n− q+ ε which intersects every line ineither 0 or at least n points for some n = pr. Then ε ≥ n.

18 Directions

The theory of directions determined by a pointset is one of the most prominentones in finite geometry. There are several applications of it, within geometry (e.g.blocking sets of Redei type), in group theory (Hajos-Redei type results), for per-mutations, etc. One can see the original question from at least four points of view.

Consider a pointset U = {Pi(ai, bi) : i = 1, ..., |U |} ⊂ AG(2, q). One can say thatU determines the direction/infinite point (m) ∈ `∞, where m ∈ GF(q) ∪ {∞} ifthere are points Pi and Pj such that m = bj−bi

aj−ai . The set of determined directions1st form

will be denoted by D, its cardinality by N = |D|. When |U | > q then D = `∞ (i.e.N = q+1) by the pigeon hole principle, as through any (m) ∈ `∞ there are q affinelines, so at least one of them will contain at least two points of U , determining(m); this shows that only |U | ≤ q is interesting. The most examined case is when|U | = q, we are going to discuss it in details, but some results concerning smallerU can be found in this book, like Theorem 23.1, etc. The general feeling is that if|U | is close to q then it can be extended to a q-set determining the same directions,while small pointsets U can behave “as they want”. Note that a random pointsetof size at least

√2q log(q) determines every direction with high probability.

Suppose U is a set of q points which does not determine all directions. Thenapplying an affine transformation, we can achieve that the direction of verticallines is not determined. But this means that our point set in question can beconsidered as the graph of a function f(X), over the underlying field. Since over2nd form

a finite field every function can be represented by a polynomial, we may assumef ∈ GF(q)[X]. (Note that lines correspond to linear polynomials.) We will say thata polynomial f determines a direction, if its graph determines it, which means thatthe set of determined directions is Df = D = { f(x)−f(y)

x−y : x 6= y}. This was theoriginal question Redei considered, he tried to determine the number of differencequotients a polynomial can have.

It is easy to verify that D = {c ∈ GF(q) : f(X)− cX is not bijective }, so a third3rd form

formulation of our problem is to look for polynomials, for which there are many

18. Directions 111

c-s such that f(X)− cX is a permutation. Compare with the definition of Mf inSection 9.In his book [103] Redei proved deep theorems about fully reducible, lacunarypolynomials. As an application of his algebraic results, he considered the problemof determining the possible values N = Nf = |Df | can take, that is, how manydifferent directions a point set can determine. For an arbitrary prime-power q, heproved that for Nf < q+1

2 , Nf cannot take all values, it has to be an element ofthe union of some intervals.For q = p prime, he proved much more, namely that unless U is a line (and henceN = 1), N is at least p+1

2 . Later Megyesi realised that the case p+12 is impossible,

so their result can be formulated as follows:

Theorem 18.1. (Redei and Megyesi [103]) If a set of p points in AG(2, p) is not aline, then it determines at least p+3

2 directions.

This bound is sharp, as the (unique) Example 18.2, constructed by Megyesi, shows,see below. Unicity is proved by Lovasz and Schrijver, see Theorem 19.1. For theproof of Theorem 18.1, which is a classical lacunary polynomial argument, see thefirst part of Theorem 18.14.Now we present a series of examples constructed by Megyesi [103] for arbitraryprime-powers.

Example 18.2. For a d|q − 1, 1 < d < q − 1, let G be the multiplicative subgroupof GF(q) of order d. Set U = {(x, 0) : x ∈ G} ∪ {(0, x) : x /∈ G}, that is put G andthe complement of G on the x and y axes, respectively. An easy calculation showsthat the number of determined directions is q + 1− d.

Perhaps it is useful to give Example 18.2 in the graph form as well. In general, apolynomial giving the example on two lines with N = q + 1 − d is XχG, whereχG is the characteristic function of G, the multiplicative subgroup of order d. Apolynomial having N = q+3

2 is f(X) = Xq+12 .

We remark that the Megyesi example belonging to N = q+32 , together with the

determined directions as points at infinity, is called the projective triangle. It is aset of size 3

2 (q + 1) in PG(2, q).

After the publication of Redei’s book, it turned out that the problem of deter-mining the possible values of N and looking for sets determining a small numberof directions has an important application: a certain type of blocking sets (called 4th form

blocking sets of Redei type) in PG(2, q) of size q + N is equivalent to a set of qpoints in AG(2, q) determining N directions.

Exercise 18.3. Show that for a blocking set B ⊂ PG(2, q) and a line `, |B \ `| ≥ qholds, and in case of equality B\`, considered as a pointset of AG(2, q) = PG(2, q)\`, determines (as determined directions) the “necessary” points of B ∩ `.


In the general case today everything is known for N < q+32 : Ball, Blokhuis,

Brouwer, Storme and Szonyi [43], and then the beautiful improved proof of Ball[9] determined all possible values, see Theorem 18.14.

18.1 When D is (contained in) a subgroup of GF(q)∗

Given D ⊆ `∞ = GF(q) ∪ {∞}, let FD denote the set of all polynomials f deter-mining directions from D, i.e. for which Df ⊆ D.

Lemma 18.4. [56] Let D be a non-trivial subgroup of the multiplicative groupGF(q)∗. Then the polynomials in FD, with the operation of composition, form apermutation group of the ground set GF(q).

Proof: As (0) 6∈ D, each f ∈ FD is a permutation. Let f, g ∈ FD, thenf(g(x))−f(g(y))

x−y = f(g(x))−f(g(y))g(x)−g(y)

g(x)−g(y)x−y , so the product of two elements of D,

hence also in D.

Exercise 18.5. Prove that the permutation group FD is3/2-transitive:transitive andall stabilizershave orbits ofequal size.3/2-trans. ⇒primitive orFrobeniusgroup.

3/2-transitive on GF(q), moreover, even the subgroup F linD = {f(X) = aX + b :

a ∈ D, b ∈ GF(q)} ⊆ FD is 3/2-transitive on GF(q).

Exercise 18.6. [56] Prove that if D ⊆ GF(pe)∗ for some subfield GF(pe) < GF(q)then FD = F lin

D , i.e. the only functions determining directions from D are thelinear ones with slope from D.

As the multiplicative group GF(q)∗ is cyclic, its subgroups are exactly of the formD = {x : xd = 1} for some d|(q− 1). We are going to prove the following theoremof Carlitz, McConnel, Bruen and Levinger [56]:

Theorem 18.7. If Df ⊆ D = {x : xd = 1} for some d|(q− 1) then f is of the formf(X) = a+ bXpj , where a ∈ GF(q), b ∈ D and m = q−1

d divides (pj − 1).

Proof: ([56]) Let V denote the vectorspace of functions GF(q) → GF(q), or, equiv-alently, the space of polynomials in GF(q)[X] of degree ≤ q − 1. The element g ofFD acts on f ∈ V as fg(X) = f(g(X)).This proof will go in four steps.

The first is about the linear space of homomorphisms V → V , which commutewith elements of the permutation group FD; that is let

HomFD (V, V ) = { Φ : V → V | Φ( f(g(X) ) = Φ(f)(g(X)) for all f ∈ V, g ∈ FD }.

Define also the monomial functions ϕi(X) = Xid for i = 0, 1, ...,m = q−1d . Finally

recall µa(X) = 1 − (X − a)q−1, the characteristic function of the set {a}, seeSection 5.3.

18. Directions 113

Step 1. If Φ ∈ HomFD (V, V ) then Φ(µ0) can be written as a linear combi-nation of the monomials ϕi, i = 0, ...,m. Conversely, given any linear com-bination

∑mi=0 λiϕi, there is a unique element Φ ∈ HomFD (V, V ) such that

Φ(µ0) =∑mi=0 λiϕi.

To prove this we need the following

Exercise 18.8. The dimension of HomFD (V, V ) over GF(q) is just the number ofthe orbits of (FD)0 (the stabilizer of 0) on GF(q), i.e. m+ 1.

For a ∈ GF(q) define the translation Ta(X) = X − a. Then µTa0 = µa. As anyf ∈ V has the form f(X) =

∑a∈GF(q) f(a)µa(X) =

∑a∈GF(q) f(a)µTa0 (X), for any

Φ ∈ HomFD (V, V ) and f ∈ V we have

Φ(f) =∑

a∈GF(q)

f(a)Φ(µTa0 ) =∑

a∈GF(q)

f(a)Φ(µ0)Ta ,

hence Φ is uniquely determined by the image of µ0.To finish Step 1 we have to prove that Φ(µ0) is some linear combination of themonomials ϕi, i = 0, ...,m. Let Φ(µ0)(X) =

∑q−1i=0 λiX

i and f(X) = bX ∈ F linD

(so b ∈ D). Then µf0 (X) = µ0(bX) = µ0(X), so

Φ(µ0)(X) = Φ(µf0 )(X) = Φ(µ0)f (X) =q−1∑i=0

λi(bX)i.

Thus∑q−1i=0 λi(1 − bi)Xi = 0, hence all the coefficients λi(1 − bi) are zero for

i = 0, 1, ..., q − 1. As b was an arbitrary element of D, it means that λi = 0 unlessd|i and we are done.

Step 2. We prove that for each k = 0, 1, ...,m, Uk = 〈(X − a)kd : a ∈ GF(q)〉 is anFD-module (i.e. for any g ∈ FD, f ∈ Uk also fg ∈ Uk).To see this observe that if Φ is in HomFD (V, V ) then Φ(V ) is an FD-module. ByStep 1 there exists a Φ in HomFD (V, V ) for which Φ(µ0)(X) = Xkd. Further

Φ(µa)(X) = Φ(µTa0 )(X) = (Φ(µ0))Ta(X) = (X − a)kd.

Step 3. Recall Exercise 5.8. There we defined Sλ = 〈(X − a)λ : a ∈ GF(q)〉 λ =Pn−1i=0 αip

i

0 ≤ αi < pand Mλ = {∑n−1i=0 βip

i : 0 ≤ βi ≤ αi}. Since d|(q − 1) we have (d, p) = 1 and1, (d − 1) ∈ Md. In Step 2 we have seen that Sd is an FD-module. Thus, for any εr(X) = Xr

g ∈ FD and r ∈ Md, εgr(X) = εr(g(X)) = g(X)r is a polynomial in Sd. We aregoing to use this.

Lemma 18.9. Let f ∈ FD. Then f(X) = a+ bXt, where a = f(0), b = f(1)− f(0)and td ≡ d (mod q − 1).


Proof: Let a = f(0). Since the translation Ta is in FD, Ψ(X) = Ta(f(X)) =f(X) − a is also in FD and Ψ(0) = 0. As 1, (d − 1) ∈ Md, the functions εΨ1 = Ψand εΨ1 = Ψd−1 (after reduction) are in Sd. Now, Ψ ∈ FD implies that Ψ(x)

x =Ψ(x)−Ψ(0)

x ∈ D for x 6= 0. From the definition of D, it follows that Xd = Ψ(X)d =Ψd−1 afterreduction modXq −X Ψ(X)Ψ(X)d−1. But Xd can be a product of two polynomials of degree ≤ d only

if Ψ(X) = bXt and Ψ(X)d−1 = b′Xd−t, where bb′ = 1 and 0 < t < d. ThenΨd(X) = bdXtd = Xd, if and only if td ≡ d (mod q − 1) and bd = 1. Further,b = Ψ(1) = f(1)− a = f(1)− f(0).

Step 4. We now show that the only possible choices for t above are t = pj . W.l.o.g.we may assume that f ∈ FD has the form f(X) = Xt. FD is a group containingthe translations Ta. Thus for any fixed α 6= 0, h(X) = f(Tα(X)) = (X − α)t isin FD. By Step 3, h(X) = a + bXt′ = (X − α)t, where a = h(0) = (−α)t andb = h(1)− h(0). We have the equation (X −α)t− (a+ bXt′) = 0 as it is of degree< q − 1. Hence t = t′, b = 1 and (X − α)t = Xt + (−α)t. Since α was arbitrary,this shows that f(x + α) = (x + α)t = xt + αt for all x, α ∈ GF(q), so f is anautomorphism of GF(q), hence t = pj and by Step 3, d(pj − 1) ≡ 0 (mod q− 1).

Exercise 18.10. Prove that if D = {x ∈ GF(q) : xd = λ} for some fixed λ for whichλq−1d = 1, then FD consists of the functions f(X) = a+ bXpj where q−1

d |(pj − 1),a ∈ GF(q) and b ∈ D.

The theorem above has some further applications in geometry. In Section 18.4we will see that pointsets of size q determining at most q+1

2 directions can beclassified: they are GF(pe)-linear pointsets in AG(2, q), where GF(pe) is a subfieldof GF(q). It also means that the set D of determined directions is a GF(pe)-linearpointset of (the line at infinity ') PG(1, q), generated by h

e + 1 points. (For theq = ph

definition of generation see Section 20.) So one can use that we are in the situationthat {(xd) : x ∈ GF(q)} ⊇ 〈(a1), ..., (am+1)〉GF(pe) for some a1, ..., am+1 ∈ GF(q).

One can also try to alter the proof above a little bit to prove the following

Conjecture 18.11. If Df ⊆ D = {x : xd = 1} ∪ {0} for some d|(q− 1) then f is ofthe form f(X) = a+ bXpj , where a ∈ GF(q), b ∈ D and m = q−1

d divides (pj − 1).

It follows from Theorem 18.1 of Redei and Megyesi that if q = p is a prime thenthe Conjecture holds and f is a linear function.

Exercise 18.12. Suppose that a set of disjoint parabolas {Y = aiX2 + bi : i =

1, ..., q} partition AG(2, q). Then (i) the bi-s are all distinct; (ii) writing ai = f(bi),the directions determined by the function f are minus non-squares and possibly (0);finally (iii) if q = p is a prime and we do not allow a line as a degenerate parabolathen all ai-s are equal and they are the vertical translates of the same parabola.

18. Directions 115

18.2 When D is (contained in) a subgroup of q+1√

1 ⊂ GF(q2)∗

Well, it almost never happens, if we mean proper subgroups, see Theorem 29.5. Forthis being meaningful we have to put our pointset U into the affine plane AG(2, q)which is identified with GF(q2) and the directions with the (q+1)-th roots of unity.Theorem 29.5 says that, given 1 < d < q + 1, if U determines directions from onecoset of the multiplicative subgroup of order d, then U is a line.

Aart Blokhuis has recently asked the opposite question: is it possible that U doesnot determine any direction of the multiplicative subgroup of order d (unless U is aline)? Such (positive or negative) results would give rise several applications. Onemay feel that such sets U probably do not exist. For d = q+1

2 Blokhuis’ theorem in[29], so Theorem 29.5 shows this. The next possible case d = q+1

3 is already openas far as I know.

18.3 Sets determining ≤ q+12

directions: affine linear pointsets

Linear pointsets has gained an important role in the theory of blocking sets. Firstwe give the definition of affine linear pointsets:

Definition 18.13. A pointset S ⊆ AG(n, q) is called GF(pe)-linear if

(i) GF(pe) is a subfield of GF(q), and so e|h

(ii) there is an affine space AG(n′, q) containing AG(n, q) such that S is a one-to-one projection of a subgeometry AG(t, pe) ⊂ AG(n′, q) from a suitable subspace(“vertex”) V onto PG(n, q).

Algebraically this means that if we suppose that S contains the origin and has size|S| = (pe)t, then one can choose t points (vectors) v1, ..., vt of AG(n, q) such that Sis the vectorspace spanned by them over GF(pe), i.e. S = {

∑ti=1 λivi : λ1, ..., λt ∈

GF(pe)}.

Suppose that we have an affine pointset S with the suspect that it is GF(pe)-linearin the affine sense. W.l.o.g. suppose that the origin is in S. Then S is GF(pe)-lineariff (i) (a1, b1), (a2, b2) ∈ S implies (a1 + a2, b1 + b2) ∈ S and (ii) (ca1, cb1) ∈ S forall c ∈ GF(pe). Now changing the representation to the GF(q2)-one, we concludethat S is GF(pe)-linear iff substituting Y = ω ∈ GF(q2) \ GF(q) into its affineRedei polynomial R(X,Y ), R(X,ω) contains terms with exponents being powersof pe only. As R was defined over GF(q), it is equivalent to saying that all theX-exponents of R(X,Y ) are powers of pe.

Everything is similar in AG(n, q) for bigger n. But it gets much harder if S isnon-affine, see Section 20.


18.4 Small blocking sets of Redei type are linear

Here we present a short proof of the following theorem of Blokhuis, Ball, Brouwer,Storme and Szonyi [43], which was refined and turned to its current beautiful formby Ball [9]:

Theorem 18.14. Let |U | = q be a pointset in AG(2, q), q = ph, p prime, and letN be the number of directions determined by U . Let s = pe be maximal such thatevery line intersects U in a multiple of s points. Then one of the following holds:(i) s = 1 and q+3

2 ≤ N ≤ q + 1;

(ii) GF(s) is a subfield of GF(q) and qs + 1 ≤ N ≤ q−1

s−1 ;(iii) s = q and N = 1.Moreover, if s ≥ 3 then U is a GF(s)-linear pointset.

Proof: Before proving the lower bounds, note first that the upper bounds aretrivial. Let P be a point of U . If s > 1 then the points of U lie on lines incidentwith P and with a direction in D. Therefore

N(s− 1) + 1 ≤ q.

The degree of a polynomial g is denoted by g◦. We will require Lemma 5.37 byRedei on lacunary polynomials, and Lemma 5.41 as well.Let U = {(ai, bi) : i = 1, ..., q} be our set determining the directions D. Wecan assume that (∞) 6∈ D (so one may think about U as a graph of a functionf(X) : GF(q) → GF(q)), N = |D| and s = pe is maximal such that every lineintersects U in a multiple of s points. The (affine) Redei polynomial isσj(Y ) =

σj({aiY − bi :i = 1, ..., q})

R(X,Y ) =q∏i=1

(X + aiY − bi) =q∑j=0

σj(Y )Xq−j .

The polynomial in one variable R(X, y) has a repeated factor if and only if (y) ∈ D.Hence for (y) 6∈ D

R(X, y) = Xq −X,

and σj(y) = 0 for j = q and j = 1, 2, ..., q − 2.The overall degree of R(X,Y ) is q and (so) the degree of σj(Y ) is at most j, andthe coefficient of Y j in σj(Y ) is σj({ai}) = σj(GF(q)). Hence for j = 1, 2, ..., q− 2the degree of σj(Y ) is at most j − 1 and moreover, from the previous paragraph,non-determined

directions has at least q−N distinct zeros. So the polynomial σj(Y ) = 0 whenever j ≤ q−N .Therefore

R(X,Y ) = Xq +N−1∑j=0

σq−j(Y )Xj .

Let y ∈ D. The line c = yX − Y is incident with exactly k points of U if anddetermineddirections

18. Directions 117

only if (X + c) is a factor of R(X, y) with multiplicity exactly k. By assumptionthis k is always a multiple of s, hence R(X, y) is an s-th power. We can writeR(X, y)1/s = Xq/s + g(X) for some g(X) = gy(X); since s was chosen to bemaximal there exists a y ∈ D such that g′y(X) 6= 0. It follows from Lemma 5.37that, for any y ∈ D such that g′(X) 6= 0 we have g◦ ≥ q+s

s(s+1) . Let N0 be maximal N0

such that σq−N0 is not identically 0. We have shown that so σ1=...=σq−(N0+1)=0

soσk,l({a1, .., aq ;b1, .., bq}) = 0∀k+l ≤ q−(N0+1)

q + s

s(s+ 1)≤ N0 ≤ N − 1.

This completes the proof of Redei and Megyesi for the case s = 1. In the remainder see Thm 18.1

of the proof s ≥ 2.

If j 6= 1 and s does not divide j then σj(y) = 0 for all y ∈ GF(q) so σj(Y ) = 0.Therefore

R(X,Y ) = Xq +N0/s∑j=0

σq−js(Y )Xjs + σq−1(Y )X.

In what follows the Hasse derivatives are taken with respect to Y . The k-th Hassederivative of R(X,Y ) is Hk

Y (R(X,Y )) = Hassederivatives

=∑

(s1,s2,...,sk)

as1as2 ...ask∏t6=si

(X+atY−bt) =∑

(s1,s2,...,sk)

(as1

X + as1Y − bs1)...(

askX + askY − bsk

)R(X,Y ).

(1)For all (ai, bi) and y ∈ D

X + aiy − bi | R(X, y)− (Xq −X) = X +N0/s∑j=0

σq−js(y)Xjs.

Note that σq−1(y) = 0 for y ∈ D as R(X, y) is an s-th power.

If we evaluate the equation (1) for y ∈ D and multiply by the polynomial abovek times then we get an equality of polynomials in X the right-hand side of whichcontains a factor R(X, y) which divides the left-hand side. That is

R(X, y) | (X +N0/s∑j=0

σq−js(y)Xjs)kHkY (R)(X, y). (2)

The polynomial

HkY (R)(X, y) =

N0∑j=0

HkY (σq−j)(y)Xj

has degree at most N0 so the right-hand side of (2) has degree at most (k+ 1)N0.


The polynomial X +∑N0/sj=0 σq−js(y)Xjs is not zero as this would imply s = 1.

If (k + 1)N0 ≤ q − 1 then the left hand side of (2) has degree larger than theright-hand side and we conclude that

HkY (σq−js)(y) = 0 for all 0 ≤ j ≤ N0/s.

Let us assume that N ≤ q/s.

Now N0 ≤ N − 1 ≤ q/s − 1 and (k + 1)N0 ≤ sN0 ≤ q − s < q − 1 wheneverk ≤ s− 1. Hence for k ≤ s− 1 the k-th Hasse derivatives of the polynomials σq−jsare zero when evaluated at y ∈ D. We shall show that they are in fact identicallyzero when js = N0.It follows immediately from the definition of Hasse derivatives that Hs−1

Y (f) is ans-th power for any polynomial f . A zero of an s-th power is a zero of multiplicityat least s. Hence there are at least sN zeros of the polynomial Hs−1

Y (σq−N0).However the degree of this polynomial is less than q − N0 < sN0 < sN . HenceHs−1Y (σq−N0)(Y ) ≡ 0. But Hs−2(σq−N0) is then an s-th power, it also has at least

sN zeros and is therefore identically zero. We continue by induction. Assume thatHs−i(σq−N0) ≡ 0 for i = 1, 2, ..., j − 1. Then Hs−j(σq−N0) is an s-th power andif j < s then it has at least sN zeros and is therefore identically zero. Finally forj = s we have that σq−N0 itself is an s-th power. However σq−N0 is zero for ally 6∈ D and so has at least s(q −N) zeros which is far greater than its degree. Butσq−N0 6≡ 0 by assumption, a contradiction.Hence N ≥ q/s+ 1, the lower bound is proved.

Now we are going to prove that GF(s) is a subfield. Without loss of generality wecan assume that (0, 0) ∈ U .Note that

HkY (R)(X, y)′ = Hk

Y (σq−1)(y),

where f ′ is the derivative with respect to X, and note that the polynomial

σq−1(Y ) =q∏i=1

(aiY − bi).

Moreover, y is a zero of σq−1 exactly k times if and only if the line joining y to(0, 0) contains exactly k points, other than (0, 0), of U .Let D∞ ⊆ D be such that for all y ∈ D∞ the point (y) is joined to (0, 0) by ans-secant. Then Hk

Y (σq−1)(y) = 0 for all k = 0, 1, ..., s− 2 and Hs−1Y (σq−1)(y) 6= 0.

Hence Hs−1Y (R)(X, y) = 0.

Counting points of U on lines through the origin

|D \D∞| ≤ q − 1−N(s− 1)s

≤ q

s− 1.

18. Directions 119

Hence |D∞| ≥ N − qs2 + 1.

LetQ(X), a polynomial of degree at most sN0−q, be the quotient in the divisibility

R(X, y) | (X +N0/s∑j=0

σq−js(y)Xjs)s−1Hs−1Y (R)(X, y),

where y ∈ D∞ and so Hs−1Y (R)(X, y) 6= 0. Differentiating with respect to X we

getR(X, y)Q′(X) =

(X+N0/s∑j=0

σq−js(y)Xjs)s−1Hs−1Y (σq−1)(y)−(X+

N0/s∑j=0

σq−js(y)Xjs)s−2Hs−1Y (R)(X, y).

The right-hand side of this equation has degree at most (s − 1)N0 < q while ifQ′ 6= 0 then the left-hand side has degree at least q. We conclude Q′ = 0 and

Hs−1Y (R)(X, y) = Hs−1

Y (σq−1)(y)(X +N0/s∑j=0

σq−js(y)Xjs).

Extracting s-th roots and incorporating any constant multiple in the quotient Qwe have

R1/sQ1/s = X +N0/s∑j=0

σq−js(y)Xjs.

Differentiating again with respect to X we see R1/s(Q1/s)′ + (R1/s)′Q1/s = 1.Now ((R1/s)′Q1/s)◦ ≤ N0/s − 1 + N0 − q/s < q/s and the degree of R1/s isq/s so (Q1/s)′ = 0. Therefore Q1/s and (R1/s)′ are constant and hence for someγ ∈ GF(q)

R(X, y) = γ(X +N0/s∑j=0

σq−js(y)Xjs)s.

Now it is easy to check that since

R(X, y) = Xq +N0/s∑j=0

σq−js(y)Xjs

we haveR(X, y) ∈ 〈1, Xs, Xs2 , ..., Xq/s, Xq〉GF(q).

In particular this implies that q is a power of s and equivalently that GF(s) is asubfield of GF(q).


If j 6= si for some i then σj(y) = 0 for all y ∈ (GF(q) \D) ∪D∞. By comparingthe degree of σj with the number of zeros of σj we see that σj ≡ 0 wheneverj − 1 ≤ q −N +N − q/s2 + 1.

Let y ∈ D\D∞ be such that R(X, y) ∈ GF(q)[Xs]\GF(q)[Xsp]. Then R(X, y)1/s =Xq/s + g(X) and by Lemma 5.37 we know that g◦ ≥ q+s

s(s+1) . However by theprevious paragraph if g◦ 6= q/s2 then g◦ < q/s3 which contradicts this. Henceg◦ = q/s2 and (g′)◦ < q/s3. If s > 2 then Lemma 5.41 implies that

R(X, y) ∈ 〈Xs, Xs2 , ..., Xq/s, Xq〉GF(q).

In the final part of the proof we show that U is a GF(s)-linear set.

Let Dt ⊆ D be the subset of elements of D for which y ∈ Dt implies

R(X, y) ∈ GF(q)[Xt] \ GF(q)[Xtp].

Counting points of U on lines incident with a point of U gives

|⋃t>s

Dt| ≤q − 1−N(s− 1)

t− s≤ q − s2

s(t− s).

If s 6= si for some i then σq−j(y) = 0 for all y ∈ GF(q) \ (⋃t>sDt). By degrees

σq−j ≡ 0 for j ≥ q−s2s(t−s) .

Let t > s be minimal such that Dt 6= ∅ and let y ∈ Dt. Write

R(X, y) = Xq + g(X) + h(X)t

where g(X) has non-zero terms of degree sj for some j. We have that

(ht)◦ <q − s2

s(s− t).

If t is not a power of s then si < t < si+1 for some i. Extracting t-th roots andfollowing the proof of Lemma 5.37 we get

R1/t|(X + g + ht)h′.

For 1 ≤ j ≤ i we have that

Xq/sj + g1/sj + ht/sj

= 0 (mod R1/t).

The polynomial g1/sj has degree at most q/sj+1 and has terms of degrees that arepowers of s. We can reduce g in the divisibility modulo R1/t to a polynomial in

〈X,Xs, Xs2 , ..., Xq/si+1, ht/s, ..., ht/s

i

〉.

18. Directions 121

Note that

th◦ ≤ q − s2

s(t− s)≤ q

si+1.

So the right-hand side of the divisibility has degree less than q/t after the reductionof g and is therefore zero. However h′ 6= 0 and so

ht ∈ 〈X,Xs, Xs2 , ..., Xq/si+1, ht/s, ..., ht/s

i

〉

which is impossible since t is not a power of s.

If t = si for some i > 1 then we go through the same argument re-placing h′ by (g1/t + h)′ and concluding that some linear combination ofX,Xs, Xs2 , ..., Xq/si+1

, h, hs, ..., hsi−1

is zero. It follows from this that

h ∈ 〈X,Xs, Xs2 , ..., Xq/si〉.

Hence for all y ∈ GF(q) we have R(X, y) ∈ 〈X,Xs, Xs2 , ..., Xq〉.

Let ω ∈ GF(q2) \ GF(q) and map AG(2, q) to GF(q2) by (a, b) 7→ −aω + b. Theimages of the points of U under this map are zeros of

R(X,ω) =q∏i=1

(X + aiω − bi) ∈ 〈X,Xs, Xs2 , ..., Xq〉GF(q2).

Now R(X1 + X2, ω) = R(X1, ω) + R(X2, ω) and R(αX,ω) = αR(X,ω) for allα ∈ GF(s).

We will show a more geometric proof for this theorem, see Corollary 21.27.

This result has an analogue for arbitrary dimensional k-blocking sets of Redeitype:

Theorem 18.15. Storme, Sziklai [108] Let U ⊆ AG(n, q), |U | = qk, and let D ⊆ H∞be the set of directions determined by U . If D ≤ q+3

2 qk−1+qk−2+qk−3+...+q2+q,then U is a GF(pe)-linear set for some subfield GF(pe) of GF(q).

Note that this theorem is sharp, as the cone (cylinder), with base the q affinepoints of a projective triangle in a plane Π 6⊂ H∞, and with a (k− 2)-dimensionalvertex subspace contained in H∞ \ Π, has qk points and determines q+3

2 qk−1 +qk−2 + qk−3 + ...+ q2 + q + 1 directions. For a strong improvement of this resultconsult Theorems 24.14 and 24.19.


19 Over prime fields: the Gacs method

For the q = p prime case, as we have seen in Theorem 18.1, U ⊂ AG(2, p) deter-mines either 1 direction (i.e. U is an affine line), or at least p+3

2 directions. It waspossible to characterize equality here already in 1981:

Theorem 19.1. (Lovasz and Schrijver [91]) For every prime p > 2, up to affinetransformation there is a unique set of p points in AG(2, p) determining p+3

2 di-rections.

Choosing d = q−12 in the construction of Megyesi we get an example withN = q+3

2 ,so this should be the unique set in the theorem above. Note that the analogue overR, i.e. the union of three halflines {(a, 0, 1), (1,−a, 0), (0, 1, a) : a ≥ 0} ⊂ PG(2,R)is still a blocking set. It is also worth to write the (affine part of the) Megyesiexample as the graph of the function X

q+12 . The analogue of it over R is the

absolute value function |X| with determined directions D = {(m) : −1 ≤ m ≤ 1}.Proof: W.l.o.g. we can assume that (0, 0), (1, 0) and (0, 1) are in our affinepointset. There are several ways available to prove this theorem. One of themis to write up the Redei polynomial R(X,Y ) =

∏pi=1(X + aiY − bi) for the points

{(ai, bi) : i} = U of our pointset. Then substituting a determined direction y, onecan see that we are in the situation of Theorem 5.38. Considering the possiblemultiplicities of roots (i.e. the distribution of the points of U on the affine linesthrough (y) ), observing that each point of U is contained in a collinear triple of Uif p ≥ 7 (for p = 3, 5 everything is easy), and using some elementary geometricalarguments we get the result.Note that here Theorem 5.38 has done most of the job.

Exercise 19.2. Make the geometric argument precise in the proof above.

Also after p+32 , there is a big gap in the possible values of N , the number of

determined directions. The main result here is the following:

Theorem 19.3. Gacs [67] For every prime p, besides lines and the example char-acterized by Lovasz and Schrijver, any set of p points in AG(2, p) determines atleast [2p−1

3 ] + 1 directions.

One may recall Megyesi’s construction 18.2, there N = q + 1− d where d|q − 1 isthe size of a multiplicative subgroup of GF(q). So after q+1− q−1

2 = q+32 one gets

q + 1 − q−13 = 2 q−1

3 + 2(whenever 3|q − 1), so for primes the bound in Theorem19.3 is one less than a possible sharp result.

The examples of Megyesi are all contained in the union of two lines. This propertywas characterized by Szonyi. For primes the result can be formulated as follows:

19. Over prime fields: the Gacs method 123

Theorem 19.4. (Szonyi [121]) Suppose U is a set of p points in AG(2, p), p prime,which is contained in the union of two lines and denote by N the number ofdetermined directions. If 1 < N < p − 1, then N = p + 1 − d for a d|p − 1, andafter linear transformation, U = {(x, 0) : x ∈ K} ∪ {(0, x) : x /∈ K} , where K isthe union of some cosets of the multiplicative subgroup of GF(p), of order d.

For another description of pointsets covered by two lines see Theorem 9.32 of Birofrom [28] as well, which also shows the limits of the method we are using.

From now on our underlying field will be GF(p), where p is a prime. For anypolynomial f , N = Nf will denote the number of directions f determines. Asit was pointed out in the section on symmetric polynomials, if f and g are affinetransforms of each other (that is f(X) = ag(bX+c)+dX+e, where a, b, c, d, e ∈ F,a, b 6= 0), or if f and g are bijective and f−1 = g, then Nf = Ng and Wf = Wg.

Our next duty is to recall the parameter Wf on polynomials over finite fields(introduced in Section 9.4), which has a close relationship with N as Wf +Nf ≥p + 1 (see Lemma 9.31), and which will help us to take the advantages of thealgebraic terminology:

Wf = min{k + l :∑

x∈GF(p)

xkf(x)l 6= 0}.

Here k and l are non-negative integers; for k = 0 or l = 0,X0 or f(X)0 is defined tobe the polynomial 1. Remember that Wf is the smallest k+ l, for which Xkf(X)l

has reduced degree p− 1.

Throughout this section the result of Exercise 5.7 will be used without mentioningit; i.e. for any subspace V of GF(q)[X], dim(V ) = |{deg(f) : f ∈ V }|.Theorem 19.3 will be proved, in a slightly more general form, as Theorem 19.6.We will follow the arguments of Ball-Gacs-Sziklai [17] in the following sections.

There are some natural ways to generalize Theorem 19.3. One is to consider theproblem in AG(2, q) for any prime-power q. After the classification result in [43, 9](see Theorem 18.14) for the case N < q+3

2 , one should ask, whether Theorem 19.3is true in general and is there a gap in the possible values of N after q+3

2 . A proofis known for the case q = p2 only.

Proposition 19.5. (Gacs-Lovasz-Szonyi) For any prime p, in AG(2, p2) the onlypoint set determining p2+3

2 directions is the one coming from Example 18.2. Thereare no examples for p2+3

2 < N < p2+p2 + 1.

This result is sharp, there is a construction by Polverino, Szonyi and Weiner [102]for a pointset determining q+

√q

2 + 1 directions, whenever q is a square.


19.1 Permutation polynomials

Let p be a prime. We say that f ∈ GF(p)[X] is a permutation polynomial if themapping x 7→ f(x) is a permutation of GF(p), and by Mf we denoted the numberMf

of elements a ∈ GF(p) for which f(X) + cX is a permutation polynomial. Theconnection with the directions determined by f(X) is straightforward. Namely,the set of parallel lines defined by the equation cX1 + X2 = a, where a runsthrough the elements of GF(p), all contain exactly one point of the graph of f ifand only if f(X) + cX is a permutation polynomial. Thus the direction of theselines is not a direction determined by f . Hence Mf +Nf = p.

Therefore the Redei-Megyesi result 18.1 says that if Mf ≥ (p− 1)/2 then f(X) =cX+d for some c, d ∈ GF(p); in other words, the graph of f , {(x, f(x)) : x ∈ GF(p)}is a line.

Similarly Theorem 19.3 says that if Mf ≥ 2dp−16 e + 1, then (f(X) − (cX +

d))(f(X)− (bX + e)) = 0 for some b, c, d, e ∈ GF(p); in other words, the graph off is contained in the union of two lines. The new method, invented by Gacs wasvery fruitful, it became the key technique of other results as well, see below.

In the Megyesi examples 18.2, G is a multiplicative subgroup of GF(p) of size dand f(X) = XχG(X), where χG(X) is the characteristic function of G. Here if1 < d < q − 1 then Mf = d− 1.

In Theorem 19.4 Szonyi proved that if the graph of f is contained in the union oftwo lines and Mf ≥ 2, then the graph of f is affinely equivalent to a generalisedexample of Megyesi detailed above. In the generalised Megyesi example G canbe replaced by a union of cosets of a multiplicative subgroup of GF(p). In thegeneralised example the value of Mf is again d− 1 for some divisor d of p− 1.

Thus, the above results imply that, either Mf ≤ 2dp−16 e, f is affinely equivalent

to Xp+12 , or f is linear.

In Theorem 19.6 and on, we will follow [17] when proving a version of the “planar”Gacs result. Then in Theorem 19.15 a “3-dimensional” version will be stated. Weshall prove that if there are more than (2dp−1

6 e+ 1)(p+ 2dp−16 e)/2 ≈ 2p2/9 pairs

(c, d) ∈ GF(p)2 with the property that x 7→ f(x) + cg(x) + dx is a permutation ofGF(p) then there are elements a, b, e ∈ GF(p) such that f(x) + ag(x) + bx+ e = 0,for all x ∈ GF(p); in other words the graph of (f, g), {(x, f(x), g(x)) | x ∈ GF(p)},is contained in a plane.

The situation is analogous to the planar case. The set of parallel planes defined bythe equation dX1 +X2 +cX3 = a, where a runs through the elements of GF(p), allcontain exactly one point of the graph of (f, g) if and only if f(X) + cg(X) + dXis a permutation polynomial.


19.2 Directions in the plane

In [67] it was proved that if Mf > (p− 1)/4 and Wf ≥ 2dp−16 e+ 2 then the graph

of f is contained in the union of two lines.

Letπk(Y ) =

∑x∈GF(p)

(f(x) + xY )k.

It’s a simple matter to check (see Section 5.1), that if x 7→ f(x)+ax is a permuta-tion then πk(a) = 0 for all 0 < k < p− 1. Since the polynomial πk(Y ) has degreeat most k − 1 (the coefficient of Y k is

∑x∈GF(p) x

k = 0) it is identically zero forall 0 ≤ k − 1 < Mf , unless Mf = p− 1 in which case f is linear. Hence if f is notlinear then Wf − 1 ≥Mf .

The original proof of Theorem 19.3 [67] showed that if Mf ≥ 2dp−16 e+1, then the

graph of f is contained in the union of two lines.

In this section we shall prove the following, slightly stronger theorem.

Theorem 19.6. [17] If Mf ≥ (p− 1)/6 and Wf ≥ 2dp−16 e+ 2 then the graph of f

is contained in the union of two lines.

The values Wf and Mf are invariant under affine transformations and inversion.Replacing f by its inverse is the transformation which switches coordinates, inother words if we switch coordinates then the graph of f , {(x, f(x)) | x ∈ GF(p)},becomes the graph of f−1. Let E(f) denote the set of all polynomials that can beobtained from f by applying affine transformations and inversions.

Let (f i)◦ be the degree of the polynomial f i modulo Xp − X. Unless statedotherwise all equations are to be read modulo Xp −X.

Note that for any polynomial g of degree less than p the sum −∑x∈GF(p) g(x) is

equal to the coefficient of Xp−1 of g.

Lemma 19.7. If 3 ≤ f◦ ≤ (p− 1)/2 then Wf ≤ (p+ 1)/3.

Proof: Write p− 1 = af◦ + b with 0 ≤ b < f◦. The degree of f(X)aXb is p− 1,so we have Wf ≤ a+ b.

If f◦ = 3 then a+ b ≤ (p− 2)/3 + 1 = (p+ 1)/3.

If (p+ 1)/3 ≤ f◦ ≤ (p− 1)/2 then a+ b = 2 + p− 1− 2f◦ ≤ (p+ 1)/3.

If (p+1)/4 ≤ f◦ ≤ (p−1)/3 then a+b = 3+p−1−3f◦ ≤ 3+p−1−3(p+1)/4 =(p+ 1)/4 + 1 ≤ (p+ 1)/3 for p ≥ 11.

If 4 ≤ f◦ ≤ (p+ 1)/4 then a+ b ≤ (p− b− 1)/f◦ + b ≤ p/f◦ + (bf◦ − b− 1)/f◦ ≤p/f◦ + f◦ − 2. This is at most (p + 1)/3 if and only if the quadratic inequality3(f◦)2 − (p+ 7)f◦ + 3p ≤ 0 is satisfied. For p ≥ 20, the inequality is satisfied forboth f◦ = 4 and f◦ = (p+1)/4, so it holds for all values between 4 and (p+1)/4.


For p < 20 a case by case analysis suffices to show that a+ b ≤ (p+ 1)/3.

Note that for f◦ = 2 we have Wf = (p− 1)/2 and Mf = 0 and for f◦ = 1 we haveWf = p− 1 and Mf = p− 1.

Lemma 19.8. If f◦ = (p+1)/2 then either Wf ≤ (p+5)/4 or f is affinely equivalentto X

p+12 .

Proof: After applying a suitable affine transformation we can suppose thatf(X) = X

p+12 + g(X) where g◦ ≤ (p− 3)/2.

If g◦ ≤ 1 then by applying another linear transformation we can subtract g fromf and hence f is affinely equivalent to X

p+12 .

Suppose g◦ ≥ 2. Write (p − 3)/2 = ag◦ + b with 0 ≤ b < g◦ and consider thepolynomial

f(X)a+1Xb =a+1∑i=0

(a+ 1i

)Xi p+1

2 +bg(X)a+1−i.

We claim that the only term in the sum that has a term of degree Xp−1 (moduloXp − X) is g(X)aX

p+12 +b. Let r(X) = g(X)a+1−iXi p+1

2 +b (modulo Xp − X), atypical term in the sum (note that all the binomial coefficients are non-zero). Ifi is even then r(X) = g(X)a+1−iXb+i, which has degree (a + 1 − i)g◦ + b + i =(p−3)/2+g◦− (g◦−1)i < p−1. If i 6= 1 is odd then r(X) = g(X)a+1−iX

p−12 +i+b,

which has degree (a+ 1− i)g◦ + (p− 1)/2 + i+ b = p− 2 + g◦ − (g◦ − 1)i < p− 1.Hence f(X)a+1Xb has degree p− 1 which implies Wf ≤ a+ 1 + b.Finally, note that a + b ≤ (p − 3)/(2g◦) + g◦ − 1, which is at most (p + 1)/4 if2 ≤ g◦ ≤ (p−3)/4. If g◦ > (p−3)/4 then a = 1 and b = (p−3)/2−g◦ < (p−3)/4and so a+ b < (p+ 1)/4.

Let s = d(p− 1)/6e.We will assume from now on that Wf ≥ 2s+ 2. By the definition of Wf the sum∑

x∈GF(p)

xkf(x)

has no term of degree Xp−1, for all k = 0, 1, . . . , 2s, and therefore the degree of fis at most p− 2s− 2. By Lemma 19.7 and Lemma 19.8 the degree of f is at least(p+ 3)/2.

Lemma 19.9. There is polynomial in h ∈ E(f) with one of the following properties.Either

(i) for all i such that 1 ≤ i ≤ 2s, (hi)◦ ≤ h◦ + i− 1 and (h2)◦ = h◦ + 1, or


(ii) for all i such that 1 ≤ i ≤ 2s, (hi)◦ ≤ (h2)◦ + i− 2 and (h3)◦ = (h2)◦ + 1,

and h has no root in GF(p).

Proof: Letd(f) = max{(f i)◦ − i | 1 ≤ i ≤ 2s}

and let d = d(f1) be maximal over all polynomials in E(f). The fact that f◦ ≥(p+ 3)/2 implies that d ≥ (p+ 1)/2.

Let π(Y ) = πp−1−d(Y ). The coefficient of Y p−1−d−j in π(Y ) is(p−1−dj

)∑x∈GF(p) x

p−1−d−jf j which, by the definition of d, is non-zero for at leastone j where 1 ≤ j ≤ 2s. Hence π(Y ) 6≡ 0.

If for all a such that f(X) + aX is a permutation polynomial we have π(a) =π′(a) = π′′(a) = 0 then (Y − a)3 divides π(Y ) and since Mf ≥ (p − 1)/6 thedegree of π, π◦ = p− 1− d ≥ 3Mf ≥ (p− 1)/2 which isn’t the case.

Since 0 < p−1−d < p−1 we have already seen that π(a) = 0, so either π′(a) 6= 0or π′′(a) 6= 0 for some a.

Let f2 be the inverse of the function f(X) + aX.

If0 6= π′(a) = −(d+ 1)

∑x∈GF(p)

x(f + ax)p−2−d

then∑z∈GF(p) f2(z)z

p−2−d 6= 0 and so f◦2 ≥ d + 1. By the maximality of d,f◦2 = d + 1 and so (f i2)

◦ − i ≤ f◦2 − 1. If (f22 )◦ ≤ f◦2 then let f3 = f2 + cX where

c is chosen so that (f23 )◦ ≥ f◦3 + 1 and f3 is not a permutation polynomial. Note

that f23 = f2

2 + 2cXf2 + c2X2.

If0 6= π′′(a) = (d+ 1)(d+ 2)

∑x∈GF(p)

x2(f + ax)p−3−d

then∑z∈GF(p)(f2(z))

2zp−3−d 6= 0 and so (f22 )◦ ≥ d + 2. By the maximality of d,

(f22 )◦ = d+ 2 and so (f i2)

◦ − i ≤ (f22 )◦ − 2. If (f3

2 )◦ ≤ (f22 )◦ then let f3 = f2 + cX

where c is chosen so that (f33 )◦ ≥ (f2

3 )◦+1 and f3 is not a permutation polynomial.

Finally, let e be an element not in the image of f3 and let f4 = f3 − e. Then f4has no root in GF(p).

Lemma 19.10. There is a polynomial in h ∈ E(f) for which there exist polynomialsF , G and H, where H◦ − 2 = F ◦ − 1 = G◦ = r ≤ s− 2, (F,G) = 1 and

Fh+Gh2 = H.


Note that this implies that h satisfies the conditions of Lemma 19.9 (i).Proof: Let h be a polynomial satisfying the conditions of Lemma 19.9. SinceWh ≥ 2s+ 2 we have (hi)◦ ≤ p− 2s− 3 + i.Define subspaces of the vector space of polynomials of maximum degree p− 1

ψj = {Fh+Gh2 | F ◦ ≤ j, G◦ ≤ j − 1},

where j ≤ s− 1. If there are polynomials F and G such that Fh+Gh2 = 0 thensince h has no root F +Gh = 0 which is impossible since (hG)◦ is at least 3s andat most 5s− 3 < p− 1. Thus the dimension of ψj is 2j + 1.Since Wh ≥ 2s+ 1 and 2(j + 1) ≤ 2s, the sum over GF(p) of the evaluation of theproduct of any two elements of ψj is zero, hence the sum of the degrees of any twoelements of ψs−1 is not equal to p−1. The maximum degree of any element of ψs−1

is p− s−3 and so only half of the degrees in the interval [s+2, . . . , p−1− (s+2)]can occur. But dimψs−1 = 2s − 1 > (p − 1 − (s + 2) − (s + 1))/2 and so there isan element H of degree at most s+ 1 in ψs−1.Let H be of minimal degree, so (F,G) = 1.If h satisfies case (i) of Lemma 19.9 then (h2)◦ = h◦ + 1 and r = G◦ = F ◦ − 1.Moreover Fh2 +Gh3 = Hh and (h3)◦ ≤ h◦ + 2 implies H◦ ≤ r + 2.If h satisfies case (ii) of Lemma 19.9 then (h3)◦ = (h2)◦ + 1 ≥ h◦ + 2 and (h4)◦ ≤(h2)◦ + 2. Let F ◦ = r+ 1 and so G◦ ≤ r. The equation Fh3 +Gh4 = Hh2 impliesH◦ ≤ r+ 2. If G◦ ≤ r− 1 then Fh2 +Gh3 = Hh implies r+ 2 + h◦ ≥ H◦ + h◦ =r + 1 + (h2)◦ and so (h2)◦ = h◦ + 1. But then Fh+Gh2 = H implies G◦ = r.Either way we have r = G◦ = F ◦ − 1 ≥ H◦ − 2.Let r1 = h+aX and F1 = F−2aXG, G1 = G and H1 = H−a2X2G+aXF . ThenF1r1 + G1r

21 = H1 and we can choose a so that H1 has degree r + 2. Now when

we look at ψr+1 for r1 we find F1, G1 and H1 as required. Note that (F,G) = 1implies (F1, G1) = 1.

We wish to prove r = 0. So let us assume r ≥ 1 and define i to be such that(i− 2)r + 1 ≤ s < (i− 1)r + 1 for r ≥ 2 and i = s for r = 1. Note that r ≤ s− 2implies i ≥ 3 and that s + r − 1 ≤ 2s − i if i = 3 or i = s and also if both i ≥ 4and r ≥ 2, since r ≤ (s− 1)/2 and i ≤ (s− 1)/2.

Lemma 19.11. There is a polynomial h ∈ E(f) and a polynomial G, where G◦ =r ≤ s−2, such that for all j = 2, . . . , i, there is an Fj and an Hj with the propertythat (Fj , G) = 1,

Fjh+Gj−1hj = Hj ,

F ◦j ≤ (j − 1)(r + 1), H◦j ≤ (j − 1)r + j and H◦

i = (i− 1)r + i.

Proof: Let r1 satisfy the conditions of Lemma 19.10. We start by proving thatthere is an h ∈ E(f) for which (hi−1)◦ ≥ h◦ + i− 2.


If (ri−11 )◦ ≤ r◦1 + i − 3 then let h = r1 + aX. Choose a so that hi−1 =∑i−1

j=0

(i−1j

)(aX)i−j−1rj1 has degree at least h◦ + i − 2 while at the same time

the degree of F − 2aXG is r + 1 and the degree of H − a2X2G+ aXF is r + 2.We will prove the lemma by induction. Lemma 19.10 implies that for j = 2 wecan take F2 = F and H2 = H.Define Fj = −(Fj−1F +Hj−1G) and Hj = −HFj−1. It can be checked by induc-tion, multiplying by Gh and using Gh2 = H − Fh, that

Fjh+Gj−1hj = Hj .

The degrees satisfy F ◦j ≤ (j − 1)(r + 1) and H◦j ≤ (j − 1)r + j and (Fj , G) = 1,

since (F,G) = 1 by Lemma 19.10 and (Fj−1, G) = 1 by induction.Now (hi−1)◦ ≥ h◦+ i− 2 and the equation Fi−1h+Gi−2hi−1 = Hi−1 implies thatF ◦i−1 ≥ (i− 2)(r + 1) and so F ◦i−1 = (i− 2)(r + 1). Finally Hi = −HFi−1 impliesH◦i = (i− 1)r + i.

Let h satisfy the conditions of Lemma 19.11. Note that this implies that h satisfiesthe conditions of Lemma 19.10 and Lemma 19.9 (i). Define

φj = {Ah+Bhi | A◦ ≤ j, B◦ ≤ j + 1− i}.

Note that Hi ∈ φ(i−1)r+i−1 and that (i− 1)r + i− 1 ≤ s+ r + i− 2 ≤ 2s− 1.

Lemma 19.12. For j ≤ 2s − 1 all polynomials of φj have degree at least H◦i and

those of degree at most p− 2− h◦ are multiples of Hi.

Proof: If Ah+Bhi = 0 then, since h has no root in GF(p), A+Bhi−1 = 0. Thedegree of Bhi−1 is at most p− 4 and at least (p+ 3)/2 and so A = B = 0. Thusthe dimension of φj is 2j + 3− i.Suppose that φj contains a polynomial C of degree n but no polynomial of degreen + 1. Then φj+1 contains a polynomial of degree n + 1, X · C for example, anda polynomial of degree one more than the maximum degree of an element of φj .However dimφj+1 = dimφj+2, so n is unique. Moreover, the polynomials of degreen+ 1 in φj+1 are multiples of a polynomial of degree n in φj .Since j ≤ 2s−1, φj contains no element of degree p−1−h◦. Now Hi ∈ φ(i−1)r+i−1

and is a polynomial of degree less than p − 1 − h◦. It is not a multiple of anypolynomial in φj for j < (i−1)r+i−1, since if it were there would be a non-constantpolynomialK and polynomials A and B with the property that (KA)h+(KB)hi ∈φ(i−1)r+i−1, with (KA)◦ ≤ (i−1)r+ i−1 and (KB)◦ ≤ (i−1)r, which would be aconstant multiple ofHi. This is not possible since (Fi, G) = 1. Thus all polynomialsin φj of degree at most p−2−h◦ are multiples of Hi and in particular have degreeat least H◦

i .


The following lemma contradicts the previous one which implies that our assump-tion that r ≥ 1 was incorrect.

Lemma 19.13. There is a non-zero polynomial of degree less than H◦i in φj for

some j ≤ 2s− 2.

Proof: Suppose r ≥ 2 and so i ≤ s. Let

∆ = {Ah+B2h2 + . . .+Bi−1h

i−1 + Chi | A◦ ≤ s− 1, B◦j ≤ r − 1, C◦ ≤ s− i}.

Since Wh ≥ 2s + 1 the sum of the degrees of any two elements of ∆ is not equalto p − 1. The maximum degree of any element of ∆ is p − s − 3 and so onlyhalf of the degrees in the interval [s + 2, . . . , p − 1 − (s + 2)] can occur, in otherwords at most b(p − 4 − 2s)/2c ≤ 2s − 2 of the degrees in this interval occur. Ifdim∆ = (i− 2)r + 2s− i+ 1 then there is a polynomial

E = Ah+B2h2 + . . .+Bi−1h

i−1 + Chi

in ∆ of degree at most s+2− ((i−2)r+2s− i+1− (2s−2)) = s− (i−2)r+ i−1.If dim∆ < (i− 2)r + 2s− i+ 1 then E = 0 ∈ ∆ non-trivially. Either way there isa polynomial E ∈ ∆ with not all A,Bj , C zero where E◦ ≤ s− (i− 2)r + i− 1.Substituting Gj−1hj = Hj − hFj we have

Gi−2E = Gi−2Ah+ CGi−2hi +i−1∑j=2

BjGi−1−j(Hj − hFj)

and rearranging

Gi−2E −i−1∑j=2

BjGi−1−jHj = (Gi−2A−

i−1∑j=2

BjFjGi−1−j)h+ CGi−2hi.

Checking the degrees on the right-hand side we see that the left-hand side is apolynomial in φj for some j ≤ 2s− 2.The degree of the left-hand side is at most max{s+ i− 1, ir − r + i− 2} which isless than H◦

i = (i− 1)r + i.If r = 1 then take i = s and define ∆ as above. There is a polynomial E in ∆ ofdegree at most s+1 and the degree of Gi−2E is at most 2s−1 which is the degreeof Hs. If we have equality then by Lemma 19.12 the polynomial

(Gs−2A−s−1∑j=2

BjFjGs−1−j)h+ CGs−2hs

is a constant multiple of Fsh+Gs−1hs which implies CGs−2 is a constant multipleof Gs−1 which it is not since one has degree s− 2 and the other s− 1.


We can now prove Theorem 19.6.Proof: By the previous lemmas there exist polynomials h ∈ E(f) and F of degree1 and H of degree 2 such that h2 + Fh = H. Thus (h + F/2)2 = H + F 2/4. Allvalues of H + F 2/4 are squares and so H + F 2/4 = (aX + b)2. Hence (h+ F/2−aX − b)(h+ F/2 + aX + b) = 0 and the graph of h (and so the graph of f too) iscontained in the union of two lines.

19.3 Linear combinations of three permutation polynomials

Let M(f, g) be the number of pairs (a, b) ∈ GF(p)2 for which f(X) + ag(X) + bXis a permutation polynomial. Let

W (f, g) = min{k + l +m |∑

x∈GF(p)

xkf(x)lg(x)m 6= 0}.

Let again s = dp−16 e. Before we prove the main result of this section we need the

following lemma

Lemma 19.14. [17] If M(f, g) > (2s+ 1)(p+ 2s)/2 then W (f, g) ≥ 2s+ 2 or thereare elements c, d, e ∈ GF(p) such that f(x) + cg(x) + dx+ e = 0 for all x ∈ GF(p).

Proof: Let πk(Y, Z) =∑x∈GF(p)(f(x) + g(x)Y + xZ)k.

By Section 5.1, if f(X)+ag(X)+bX is a permutation polynomial then πk(a, b) = 0for all 0 < k < p− 1. Write

πk =∏

σj(Y, Z),

where each σj is absolutely irreducible. Then∑σ◦j = π◦k ≤ k.

Let Nj be the number of solutions of σj(a, b) = 0 in GF(p) for which f(X) +ag(X) + bX is a permutation polynomial.If λσj ∈ GF(p)[Y, Z], for some λ in an extension of GF(p), and σ◦j ≥ 2 then byTheorem 10.9(i) Nj ≤ σ◦j (p+ σ◦j − 1)/2.Suppose σ◦j = 1 and there are at least (p+ 1)/2 pairs (a, b) for which σj(a, b) = 0and f(X) + ag(X) + bX is a permutation polynomial. Let σj = αY + βZ + γ.If α 6= 0 then there are (p + 1)/2 elements b ∈ GF(p) with the property thatαf(X) − (βb + γ)g(X) + bαX = αf(X) − γg(X) + b(αX − β) is a permutationpolynomial. By Redei and Megyesi’s theorem mentioned in the introduction, thisimplies that αf(X)− γg(X) is linear and hence there are elements c, d, e ∈ GF(p)such that f(x) + cg(x) + dx + e = 0 for all x ∈ GF(p). If α = 0 then there are(p+ 1)/2 elements a ∈ GF(p) with the property that βf(X)− γX + aβg(X) is apermutation polynomial. The set of p points {(βf(x) − γx, βg(x)) | x ∈ GF(p)}


may not be the graph of a function but it is a set of p points that does notdetermine at least (p+ 1)/2 directions. Thus it is affinely equivalent to a graph ofa function that does not determine at least (p − 1)/2 directions and so by Redeiand Megyesi’s theorem, it is a line. Hence, there are elements c, d and e with theproperty that c(βf(x)−γx)+dβg(x)+ e = 0 for all x ∈ GF(p). Thus, either thereare elements c, d, e ∈ GF(p) such that f(x) + cg(x) + dx+ e = 0 for all x ∈ GF(p)or Nj ≤ (p− 1)/2.Suppose λσj 6∈ GF(p)[Y, Z] for any λ in any extension of GF(p). The polynomialsσj =

∑αnmY

nZm and σj =∑αpnmY

nZm have at most (σ◦j )2 zeros in common

by Bezout’s theorem. However if (y, z) ∈ GF(p)2 and σj(y, z) = 0 then σj(y, z) = 0.Hence

Nj ≤ (σ◦j )2 ≤ σ◦j (p+ σ◦j − 1)/2,

whenever σ◦j ≤ (p− 1)/2.Thus if πk 6≡ 0 and k ≤ (p−1)/2 then N(πk), the number of solutions of πk(y, z) =0 in GF(p) for which f(X) + ag(X) + bX is a permutation polynomial, satisfies

N(πk) ≤∑

Nj ≤∑

σ◦j (p+ σ◦j − 1)/2 ≤ k(p− 1)/2 +12

∑(σ◦j )

2

≤ k(p− 1)/2 +12(∑

σ◦j )2 = (k(p− 1) + k2)/2.

By hypothesis πk ≡ 0 or

(2s+ 1)(p+ 2s)/2 < Nk ≤ (k(p− 1) + k2)/2,

which gives k ≥ 2s+ 2. Now

πk(Y, Z) =k∑l=0

k−l∑m=0

(k

l

)(k − l

m

) ∑x∈GF(p)

xk−l−mf(x)lg(x)m

Y mZk−l−m,

and so W (f, g) ≥ 2s+ 2.

Theorem 19.15. [17] If M(f, g) > (2s + 1)(p + 2s)/2 then there are elementsc, d, e ∈ GF(p) such that f(X) + cg(X) + dX + e = 0.

Proof: If p = 3 and M(f, g) > (2s + 1)(p + 2s)/2 = 15/2 then there is a c suchthat f(X) + cg(X) + bX is a permutation polynomial for all b ∈ GF(p), which canonly occur if there is a constant e such that f(X) + cg(X) + e = 0.So suppose p ≥ 5 and that there are no elements c, d, e ∈ GF(p) with the propertythat f(X) + cg(X) + dX + e = 0.Clearly Wf+ag ≥W (f, g) for all a ∈ GF(p) and W (f, g) ≥ 2s+2 by Lemma 19.14.


There is an a1 ∈ GF(p) with the property that

Mf+a1g ≥M(f, g)/p ≥ (p− 1)/6.

By Theorem 19.6 there are constants c, d, c′, d′ ∈ GF(p) with the property that

(f + a1g + cX + d)(f + a1g + c′X + d′) = 0

so the graph of (f, g), the set of points {(x, f(x), g(x)) | x ∈ GF(p)}, is containedin the union of two planes.By Redei and Megyesi’s theorem, since we have assumed that the graph of f+a1gis not a line, Mf+a1g ≤ (p − 1)/2 and so there is an a2 6= a1 with the propertythat

Mf+a2g ≥ (M(f, g)− (p− 1)/2)/(p− 1) ≥ (p− 1)/6.

Thus the graph of (f, g) is contained in the union of two other planes, differentfrom the ones before. The intersection of the two planes with the two planes isfour lines and so the graph of (f, g) is contained in the union of four lines.Similarly, since (M(f, g) − (p − 1))/(p − 2) ≥ (p − 1)/6 and (M(f, g) − 3(p −1)/2)/(p − 3) ≥ (p − 1)/6, there is an a3 and an a4 with that property thatMf+a3g ≥ (p−1)/6 and Mf+a4g ≥ (p−1)/6 and so the graph of (f, g) is containedin two other distinct pairs of planes. The four lines span three different pairs ofplanes and so the graph of (f, g) is contained in the union of two lines and hencea plane, which is a contradiction.

There is an example when q is an odd prime (power) congruent to 1 modulo 3with M(f, g) = 2(q − 1)2/9 − 1 where the graph of (f, g) is not contained in aplane, which shows that the bound is the right order of magnitude.Let E = {e ∈ GF(q) | e(q−1)/3 = 1} ∪ {0}. Then the set S ={(e, 0, 0), (0, e, 0), (0, 0, e) | e ∈ E} is a set of q points. If π, the plane defined by

X1 + aX2 + bX3 = c,

is incident with (e, 0, 0) for some e ∈ E then c ∈ E. Likewise if it is incident with(0, e, 0) for some e ∈ E then a/c ∈ E and if it is incident with (0, 0, e) for somee ∈ E then b/c ∈ E.If π is incident with two points of S then either a ∈ E, b ∈ E or a/b ∈ E. Thusif a, b and a/b are not elements of E then π and all the planes parallel to π areincident with exactly one point of S. There are 2(q − 1)2/9 such sets of parallellines.If we make a change of coordinates so that {x1 = x | x ∈ GF(q)} is one such set ofparallel planes then there are functions f and g for which S = {(x, f(x), g(x)) | x ∈GF(q)}. Each other set of parallel lines with the above property corresponds toa pair (a, b) such that f(X) + ag(X) + bX is a permutation polynomial. Thus


M(f, g) = 2(q − 1)2/9 − 1. Explicitly the functions f and g can be defined byf(x) = χR(x)x and g(x) = χεH(x)x, where χH is the characteristic function ofH = {t3 | t ∈ GF(p)} and ε is a primitive third root of unity.

20 Projective linear pointsets

In Section 18.3 we defined affine linear pointsets. The general geometric definitionin the projective case is the following:

Definition 20.1. A pointset B ⊆ PG(n, q) is called GF(pe)-linear if

(i) GF(pe) is a subfield of GF(q), and

(ii) there is a projective space PG(n′, q) containing PG(n, q) such that B is aprojection of a subgeometry PG(t, pe) ⊂ PG(n′, q) from a suitable subspace(“vertex” or “center”) V to PG(n, q).

Note that here dimV = n′−n−1 and the projection is not necessarily one-to-one.Projective linear pointsets are more complicated then affine ones. One way of thealgebraic description is that B is GF(pe)-linear iff one can choose t + 2 points(vectors) v0, v1, ..., vt, vt+1 of PG(n, q) such that B is the “span” of them overGF(pe), i.e.

(1) their homogeneous coordinates are chosen in such a way that v0 = v1 + v2 +...+ vt+1;

(2) v1, ..., vt+1 are independent over GF(pe);

(3) B=〈v0, v1, ..., vt+1〉GF(pe) ={t+1∑i=1

λivi : λ1, ..., λt+1 ∈ GF(pe), (λ1, ..., λt+1) 6=(0, 0, ..., 0)}.

(As usual, for any µ ∈ GF(q)∗ and any vector u, the point µu represents the samepoint). So we get (pe)t+1−1

pe−1 points, possibly counted with multiplicities.In this case some points may well coincide. Let’s examine the structure of mul-tiple points! For any point u ∈ B consider the (homogeneous) t + 1-tuplesLu = {(λ1, ..., λt+1)} ⊂ PG(t, pe) defining it with u =

∑t+1i=1 λivi. Obviously any

Lu is a projective subspace of PG(t, pe) (so all the multiplicities are of the form(pe)i+1−1pe−1 for i ∈ {0, 1, ..., t}), and {Lu : u ∈ B} is a partition of PG(t, pe).

Exercise 20.2. Prove that, because of (2) above, most of the subspaces Lu are infact points of PG(t, pe).Define the matrix

U = (vT1 , vT2 , ..., v

Tt+1),

20. Projective linear pointsets 135

then B = 〈v0, v1, ..., vt+1〉GF(pe) is the image of PG(t, pe) = {(λ1, ..., λt+1) 6=(0, 0, ..., 0) : ∀λi ∈ GF(pe)} under the map λ 7→ UλT .

Two linear combinations P =∑t+1i=1 λivi and Q =

∑t+1i=1 µivi define coinciding

points if there exists an α ∈ GF(q) such that P = αQ, so∑t+1i=1(λi − αµi)vi = 0.

It can happen if the vi-s are dependent over GF(q) (as usually they are). Considerthe ((n+ 1)× (t+ 1)) matrix U defined above. Let W be the projective subspaceof PG(t, q) consisting of nonzero vectors w for which UwT = 0. (It may be theempty set if n ≥ t.) If W = ∅ then all the points of B are distinct.

Note that, counting without multiplicities, the number of points of B satisfies

|B| ≤ (pe)t+1 − 1pe − 1

;

and the author conjectures that if GF(pe) is the “maximum subfield of linearity”then

(pe)t + (pe)t−1 + 1 ≤ |B| ≤ (pe)t+1 − 1pe − 1

holds as well. Intuitively it would mean that we cannot lose more than a (t− 2)-dimensional subspace, collapsing into one point.

Theorem 20.3. The Redei polynomial of B is

R(X,Y, ..., T ) =∏

(λ1,...,λt+1)∈PG(t,pe)

(t+1∑i=1

λivi)V =∑

π∈Sym({1,2,...,t+1})

sgn(π)t+1∏i=1

(viV)(pe)π(i)

=

= det

v1V v2V ... vt+1V

(v1V)pe

(v2V)pe

... (vt+1V)pe

(v1V)p2e

(v2V)p2e

... (vt+1V)p2e

...(v1V)p

te

(v2V)pte

... (vt+1V)pte

= MRDpe(v1V, v2V, ..., vt+1V).

Proof: To prove this first observe that both R and the determinant is of degree1 + pe + p2e + ...+ pte. Hence it is enough to prove that each factor (

∑t+1i=1 λivi)V

of R appears in the determinant as well.

W.l.o.g. suppose that λ1 6= 0. Multiply the first column of the de-terminant by the constant λ1, then successively add λ2·(the second col-umn), ..., λt+1·(the (t + 1)-th column) to the first column, this pro-cess does not change the determinant essentially. Now the first column is( (

∑t+1i=1 λivi)V, ((

∑t+1i=1 λivi)V)p

e

, ..., ((∑t+1i=1 λivi)V)p

te

)>, so each en-try in it is divisible by the factor (

∑t+1i=1 λivi)V, hence the same holds for the


determinant as well.

Suppose that a a hyperplane x = [x, y, ..., t] contains precisely one point P =∑t+1i=1 λivi point of B (λi ∈ GF(pe)). It means that

0 = Px =t+1∑i=1

λivix.

Take the (pe)j-th power of this equation, it is∑t+1i=1 λi(vix)

pje = 0, meaning thatlinear combination of the columns of the determinant above (after substitutingV = x), with the same λi-s, result in the zero vector, hence the value of thedeterminant is zero.

Exercise 20.4. Prove the other direction, i.e. if the determinant is zero for somesubstitution V = x = (x, y, ..., t) then there is a point P =

∑t+1i=1 λivi ∈ B on the

hyperplane [x, y, ..., t].

Intuitively, if a hyperplane [x, y, ..., t] contains (pe)k+1−1pe−1 points of B, as the inter-

section is a linear set generated by some u0, u1, ..., uk+1 ∈ B, then it means thatthere are k + 1 independent equations for the columns of the determinant above(i.e. k+1 independent vectors λ = (λ1, ..., λt+1), each coming from a uj , expressedfrom the vi-s), hence the rank of the matrix is (t+ 1)− (k + 1).

Exercise 20.5. Prove that it implies that (x, y, ..., t) is a point of R(X,Y, ..., T ) withmultiplicity (pe)k+1−1

pe−1 precisely (as it has to be).

If for example pte = q then B is a blocking set with respect to hyperplanes, sincefor any hyperplane [x, y, ..., t] the Redei polynomial R(x, y, ..., t) vanishes: the firstand the last rows of the determinant above are identical after the substitution.Note that now R(X,Y, ..., T ) may contain multiple factors. (Removing all but onecopies of a multiple point the blocking property remains intact.)

21 Blocking sets

A blocking set with respect to k-dimensional subspaces is a pointset meeting everyk-subspace. As a blocking set plus a point is still a blocking set, we are interestedin minimal ones (with respect to set-theoretical inclusion) only. Note that in a(projective) plane the only interesting case is k = 1.In any projective plane of order q the smallest blocking set is a line (of size q+1).In PG(2, q) there exist minimal blocking sets of size ∼ 3

2q; the projective triangleof size 3(q + 1)/2 if q is odd and the projective triad (which is a linear pointsetin fact) of size 3q/2 + 1 if q is even. In general, in PG(n, q) it is easy to construct

21. Blocking sets 137

a blocking set with respect to k-dimensional subspaces; it is straightforward toprove that the smallest example is a subspace of dimension n − k (so consistingof qn−k+1−1

q−1 ∼ qn−k points), this example is called trivial. Another easy one is acone, with a planar blocking set as a base and an (n−k−2)-dimensional subspaceas vertex; if the base was of size ∼ 3

2q then the blocking set will be of size ∼ 32qn−k

roughly. A blocking set with respect to k-dimensional subspaces of PG(n, q) is saidto be small if it is smaller than 3

2 (qn−k + 1), in particular in the plane it meansthat |B| < 3(q + 1)/2.

A most interesting question of the theory of blocking sets is to classify the smallones. A natural construction (blocking the k-subspaces of PG(n, q)) is a subgeom-etry PG(h(n− k)/e, pe), if it exists (recall q = ph, so 1 ≤ e ≤ h and e|h).It is one of the earliest results concerning blocking sets, due to Bruen [51], that anontrivial blocking set of a projective plane of order q is of size ≥ q+

√q+ 1, and

equality holds if and only if it is a Baer subplane (i.e. a subgeometry of order√q).

It is easy to see that the projection of a blocking set, w.r.t. k-subspaces, from avertex V onto an r-dimensional subspace of PG(n, q), is again a blocking set, w.r.t.the (k + r− n)-dimensional subspaces of PG(r, q) (where dim(V ) = n− r− 1 andV is disjoint from the blocking set).

A blocking set of PG(r, q), which is a projection of a subgeometry of PG(n, q), iscalled linear. (Note that the trivial blocking sets are linear as well.) Linear blockingsets were defined by Lunardon, and they were first studied by Lunardon, Politoand Polverino [93], [98].

Conjecture 21.1. The Linearity Conjecture. In PG(n, q) every small blocking set,with respect to k-dimensional subspaces, is linear.

There are some cases of the Conjecture that are proved already.

Theorem 21.2. For q = ph, every small minimal non-trivial blocking set w.r.t.k-dimensional subspaces is linear, if

(a) n = 2, k = 1 (so we are in the plane) and

(i) (Blokhuis [33]) h = 1 (i.e. there is no small non-trivial blocking set atall);

(ii) (Szonyi [123]) h = 2 (the only non-trivial example is a Baer subplanewith p2 + p+ 1 points);

(iii) (Polverino [99]) h = 3 (there are two examples, one with p3 + p2 + 1and another with p3 + p2 + p+ 1 points);

(iv) (Blokhuis, Ball, Brouwer, Storme, Szonyi [43], Ball [9]) if p > 2 andthere exists a line ` intersecting B in |B ∩ `| = |B| − q points (so ablocking set of Redei type);


(b) for general k:

(i) (Szonyi and Weiner) [129] if h(n−k) ≤ n, p > 2 and B is not containedin an (h(n− k)− 1)-dimensional subspace;

(ii) (Storme-Weiner [110] (for k = n− 1), Bokler and Weiner [138]) h = 2,q ≥ 16;

(iii) (Storme-Sziklai [108]) if p > 2 and there exists a hyperplane H inter-secting B in |B ∩ H| = |B| − qn−k points (so a blocking set of Redeitype).

There is an even more general version of the Conjecture. A t-fold blocking set w.r.t.k-subspaces is a pointset which intersects each k-subspace in at least t points.Multiple points may be allowed as well.

Conjecture 21.3. The Linearity Conjecture for multiple blocking sets: In PG(n, q)any t-fold blocking set B, with respect to k-dimensional subspaces, is the unionof some (not necessarily disjoint) linear pointsets B1, ..., Bs, where Bi is a ti-foldblocking set w.r.t. k-dimensional subspaces and t1 + ...+ ts = t; provided that t and|B| are small enough (t ≤ T (n, q, k) and |B| ≤ S(n, q, k) for two suitable functionsT and S).

Note that there exists a ( 4√q+1)-fold blocking set in PG(2, q), constructed by Ball,

Blokhuis and Lavrauw [19], which is not the union of smaller blocking sets. (Thismultiple blocking set is a linear pointset.)

First we study 1-fold (planar) blocking sets.

As an appetizer, we present here Blokhuis’ theorem, which was a real breakthroughat 1994. It was conjectured by Jane di Paola in the late 1960’s.

Theorem 21.4. (Blokhuis [33]) In PG(2, p), p prime, the size of a non-trivial block-ing set is at least 3(p+ 1)/2.

Proof 1: Let B = U ∪D, U = B ∩AG(2, p) = {(ai, bi) : i = 1, ..., p+ s} with s ≥ 1and D = B \ AG(2, p) = {(∞), (y1), ..., (yk−s)}, so |B| = p + 1 + k, and supposethat B is a minimal non-trivial blocking set. Consider the affine Redei polynomial

R(X,Y ) =p+s∏i=1

(X + aiY + bi) =p+s∑j=0

rj(Y )Xp+s−j

of U , where degY rj(Y ) ≤ j and r0(Y ) = 1. If (y) 6∈ B then the affine lines of slopey should be blocked by U . It means that R(X, y) vanishes for all X = x ∈ GF(p)(y) 6∈ D

hence R(X, y) = (Xp −X)t(X) for some t(X) of degree s. Consequently, R(X, y)does not contain terms of exponent s + 2, s + 3, ..., p − 1, so the polynomialsrs+1(Y ), ..., rp−2(Y ) vanish for all y 6∈ D, so for p − k + s values. It means that


rs+1(Y ) = ... = rp−k+s−1(Y ) = 0 identically, so the terms Xk+1, ..., Xp−1 are allmissing even from R(X,Y ).Now consider a point at infinity from D, i.e. some yr. As R(X, yr) still do not(yr) ∈ D

contain the terms Xk+1, ..., Xp−1, we have

R(X, yr) = Xpg(X) + h(X),deg(g) = s,deg(h) ≤ k.

It is almost the situation of Exercise 5.39, excepting that g(X) and h(X) may lacunary,isn’t it?have a common factor. Dividing out the common factors we get Xpg1(X)+h1(X),

which is still totally reducible, deg(g1) ≤ s,deg(h1) ≤ k. As s ≤ k, Exercise 5.39gives k ≥ p+1

2 , so |B| = p+ k + 1 ≥ 32 (p+ 1). The only cases we have to exclude

are (i) Xpg1(X) + h1(X) = (aX + b)p and (ii) Xp −X | Xpg1(X) + h1(X). Bothare impossible as (i) would imply that B contains a whole line, while (ii) wouldmean that Xp − X | R(X, yr) so the point (yr) could be deleted without loss ofthe blocking property, contradicting the minimality of B.

Proof 2: Let B = {(ai, bi) : i = 1, ..., p+ k} ∪ {(∞)}. The affine Redei polynomial

R(X,Y ) =p+k∏i=1

(X + aiY + bi)

vanishes for all X = x, Y = y, x, y ∈ GF(p). Hence R(X,Y ) is in the ideal〈(Xp − X), (Y p − Y )〉 and we can write R(X,Y ) = (Xp − X)F (X,Y ) + (Y p −Y )G(X,Y ), where f and g are of total degree k. Let R0 denote the part of R thatis homogeneous of total degree p+k, and let F0 and G0 denote the parts of F and so R0(X, Y ) =Q

(X + aiY )G resp. that are homogeneous of total degree k. Now R0 = XpF0 + Y pG0. As theequation is homogeneous let’s put Y = 1, and define h(X) = R0(X, 1), f(X) =F0(X, 1) and g(X) = G0(X, 1).So f(X) =

∏i(X + ai) = Xpf(X) + g(X) is a fully reducible polynomial with

f◦, g◦ ≤ k. Using Exercise 5.39 after dividing out the common factors of f and gwe can finish the argument as in Proof 1.

These proofs seem to be very similar, the first treats the case s ≥ 1, the second thecase s = 0. The little “philosophical” difference led to different trains of thought,as we will see later. This theorem somehow became a probe of new ideas examiningblocking sets, at least two more proofs will be presented, see Theorem 21.4 andCorollary 21.22.

One can formulate the chase for minimal (nontrivial) blocking sets (in PG(2, q))in an algebraic way as follows. Consider the polynomial ring GF(q)[X,Y, Z] andits subset GF(q)[X,Y, Z]hom, the homogeneous polynomials of any degree. Thefully reducible polynomials form the multiplicative sub-semigroups R and Rhom


in them. Let GF(q)[X,Y, Z]0 and GF(q)[X,Y, Z]hom,0 denote the sets (ideals) ofpolynomials vanishing everywhere in GF(q)× GF(q)× GF(q).Both GF(q)[X,Y, Z]0 and GF(q)[X,Y, Z]hom,0, as ideals, can be generated bythree polynomials from R (and Rhom, resp.), for example GF(q)[X,Y, Z]0 =〈(Xq−X); (Y q−Y ); (Zq−Z)〉 and GF(q)[X,Y, Z]hom,0 = 〈(Y qZ−Y Zq); (ZqX−ZXq); (XqY − XY q)〉. Note also that for any a, b, c ∈ GF(q) the polynomiala(Y qZ − Y Zq) + b(ZqX − ZXq) + c(XqY − XY q) is still totally reducible. (Itis the Redei polynomial of the pointset consisting of the points of the line [a, b, c].)The blocking set problem is now equivalent to finding minimal polynomials, w.r.t.divisibility, as partial order, in

(Rhom ∩ GF(q)[X,Y, Z]hom,0).

The trivial blocking sets, as we have seen, correspond to the minimal polynomialsa(Y qZ − Y Zq) + b(ZqX − ZXq) + c(XqY −XY q).

21.1 One curve

So let |B| = q+ k be our blocking set. We often suppose that |B| < 2q. Recall theRedei polynomial of B:

R(X,Y, Z) =∏

(ai,bi,ci)∈B

(aiX + biY + ciZ) =q+k∑j=0

rj(Y, Z)Xq+k−j .

Definition 21.5. ([47], [123]) Let C be the curve of degree k defined by

f(X,Y, Z) = r0(Y,Z)Xk + r1(Y, Z)Xk−1 + . . .+ rk(Y, Z).

Note that as deg(rj) = j (or rj = 0), the polynomial f(X,Y, Z) is homogeneousof degree k indeed.

Lemma 21.6. If the line LX [1, 0, 0] contains the points {(0, bij , cij ) : j = 1, ..., NX}then

rNX (Y, Z) = (∏as 6=0

as)NX∏j=1

(bijY + cijZ) | R(X,Y, Z);

rNX (Y, Z) | f(X,Y, Z);

so f can be written in the form f = rNX f , where f(X,Y, Z) is a homogeneouspolynomial of total degree = X-degree = k −NX . In particular, if LX is a Redeiline then f = rNX . One can write R(X,Y, Z) = rNX (Y, Z)R(X,Y, Z) as well.

Proof: obvious from the definitions: rNX |ri ∀i. Indeed, rNX contains the X-freefactors of R; NX is the smallest index j for which rj is not identically zero. As, by


definition, rj is gained from R =∏

(aiX + biY + ciZ) by adding up all the partialproducts consisting of all but j (biY + ciZ) factors and j non-zero ai factors,each of these products will contain all the factors of rNX , so rNX |rj ∀j.

Note that the curve rNX consists of NX lines on the dual plane, all passing through[1, 0, 0].On the other hand if k < q then

f = HqXR =

∑{s1,s2,...,sq}

as1as2 ...asq∏

j 6∈{s1,s2,...,sq}

(ajX + bjY + cjZ).

Obviously it is enough to sum for subsets {Ps1 ,Ps2 , ...,Psq} ⊆ B \ LX .If one coordinatizes B such that each ai is either 0 or 1, then

f = rNX f = rNX∑

J⊆{1,2,...,q+k}|J|=k−NXai 6=0 ∀i∈J

∏j∈J

(X + bjY + cjZ).

AlsorNX = Hq+k−NX

X R = Hk−NXX f.

The next proposition summarizes some important properties of the Redei polyno-mial and of this curve.

Theorem 21.7. ([123])

(1.1) For a fixed (y, z) (where (0,−z, y) 6∈ B), the element x is an r-fold root ofRy,z(X) = R(X, y, z) if and only if the line with equation xX+yY +zZ = 0intersects B in exactly r points.

(1.2) Suppose Ry,z(X) = 0, i.e. (0,−z, y) ∈ B. Then the element x is an (r − 1)-fold root of R(X, y, z) if and only if the line with equation xX+yY +zZ = 0intersects B in exactly r points.

(2.1) For a fixed (0,−z, y) 6∈ B the polynomial (Xq − X) divides Ry,z(X).Moreover, if k < q − 1 then Ry,z(X) == (Xq − X)f(X, y, z) for every(0,−z, y) 6∈ B; and f(X, y, z) splits into linear factors over GF(q) for thesefixed (y, z)’s.

(2.2) If the line [0,−z, y] (where (0,−z, y) 6∈ B) meets f(X,Y, Z) at (x, y, z) withmultiplicity m, then the line with equation xX + yY + zZ = 0 meets B inexactly m+ 1 points.

This theorem shows that the curve f has a lot of GF(q)-rational points and helpsus to translate geometric properties of B into properties of f .


Proof: (1.1) and (1.2) are straightforward from the definition of the Redei poly-nomial. The multiplicity of a root X = x is the number linear factors in theproduct defining R(X,Y, Z) that vanish at (x, y, z), which is just the number ofpoints of B lying on the line [x, y, z]. The first part of (2.1) follows from (1.1) andthe well-known fact that

∏x∈GF(q)(X −x) = Xq −X. The rest of (2.1) is obvious.

To prove (2.2) note that if the intersection multiplicity is m, then x is an (m+1)-fold root of Ry,z(X). Now the assertion follows from (1.1).

The facts given in Theorem 21.7 will be used frequently without further reference.The next lemma shows that the linear components of f (or the curve C defined byf = 0) correspond to points of B which are not essential.

Lemma 21.8. ([123])

(1.1) If a point P (a, b, c) ∈ B \ LX is not essential, then aX + bY + cZ dividesf(X,Y, Z) (as polynomials in three variables).

(1.2) Conversely, if NX < q + 2 − k and aX + bY + cZ divides f(X,Y, Z), then(a, b, c) ∈ B \ LX and (a, b, c) is not essential.

(2.1) If a point P (0, b, c) ∈ B∩LX is not essential, then Xq−X divides R(X,−c, b)(as polynomials in three variables).

(2.2) Conversely, if Xq−X divides R(X,−c, b), then (0, b, c) cannot be an essentialpoint of B.

Proof: (1.1): Take a point Q(0,−z0, y0) 6∈ B. For this Q(0,−z0, y0) there are atleast two points of B on the line PQ, hence (aX + by0 + cz0) divides f(X, y0, z0).In other words, the line L : aX + bY + cZ and C have a common point for(Y, Z) = (y0, z0). This happens for q + 1 − NX values of (y0, z0), so Bezout’stheorem implies that L is a component of C.(1.2): Conversely, if aX+bY +cZ divides f(X,Y, Z), then for every Q(0,−z0, y0) 6∈B the line through Q and (a, b, c) intersects B in at least two points. If (a, b, c) /∈ B,then |B| ≥ 2(q+ 1−NX) +NX . Putting |B| = q+ k gives a contradiction. Hence(a, b, c) ∈ B. Since every line through (0,−z0, y0) 6∈ B, contains at least two pointsof B, the point (a, b, c) cannot be essential.(2.1) and (2.2) can be proved in a similar way as (1.1) and (1.2).

If the line [1, 0, 0] is a tangent, or if B is a small blocking set, then the previouslemma simply says that there are no linear components of f if |B| < 2q. Note thatalso in Segre’s theory there is a lemma corresponding to this one (see [77], Lemmas10.3.2 and 10.4.), and it plays an important role in proving the incompleteness ofarcs.


Recall also a lower bound on the number of GF(q)-rational points of certain com-ponents of f , see Blokhuis, Pellikaan, Szonyi [47].

Lemma 21.9. ([47]) (1) The sum of the intersection multiplicities I(P, f ∩ `P )over all GF(q)-rational points of f is at least deg(f)(q+ 1)− deg(f)NX , where `Pdenotes the line through P and (1, 0, 0) (the “horizontal line”). If g is a componentof f , then the corresponding sum for g is at least deg(g)(q+1)−deg(g)(NX), whereg0 = g.c.d.(g, rNX ) and g = g0g.(2) Let g(X,Y, Z) be a component of f(X,Y, Z) and suppose that it has neithermultiple components nor components with zero partial derivative w.r.t. X. Thenthe number of GF(q)-rational points of g is at least

deg(g)(q + 1)− deg(g)(NX + deg(g)− 1)

Proof: Let g = g0g, where g0 contains the product of some linear components(hence g0|rNX ) and g has no linear component; s = deg(g), s = deg(g). First notethat the linear components of rNX all go through (1, 0, 0) while f does not. Forany fixed (Y, Z) = (y, z), for which (0,−z, y) 6∈ B, the polynomial f(X, y, z) is theproduct of linear factors over GF(q), hence the same is true for every divisor g off . So the number of points, counted with the intersection multiplicity of g and thehorizontal line at that point, is at least s(q + 1−NX) + deg(g0)(q + 1). To countthe number of points without this multiplicity we have to subtract the number ofintersections of g and g′X (see [47]); Bezout’s theorem then gives the result. Notealso that in this counting the common points of g and g′X are counted once ifthe intersection multiplicity I(P ; g ∩ `P ) is not divisible by p, and the points withintersection multiplicity divisible by p are not counted at all. Hence we have atleast s(q + 1−NX) + (s− s)(q + 1)− s(s− 1) points of g.

These elementary observations already yield interesting results on blocking sets.We mention without a proof that Lemma 21.9, combined with the Weil-estimate onthe number of rational points of a curve gives the result of Bruen |B| ≥ q+

√q+1.

We repeat a lemma of Blokhuis and Brouwer.

Proposition 21.10. ([44]) There are at most k2−k+1 lines that meet B in at leasttwo points.

Proof: Exercise 12.5, with |B| = q + k, gives that the total number of tangentsis at least (q + k)(q + 1− k), which means that there are at most k2 − k + 1 linesintersecting B in at least two points.

Now we are ready to prove Blokhuis’ theorem 21.4 in the prime case.

Theorem 21.11. (Blokhuis [33]) In PG(2, p), p prime, the size of a non-trivialblocking set is at least 3(p+ 1)/2.


Proof: Take a component g = g (of degree s) of f . Since p is prime, it cannothave zero partial derivative with respect to X. Therefore it has at least s(p+ 1)−s(NX + s− 1) points by Lemma 21.9. On the other hand, again since p is prime,it cannot be non-classical with respect to lines. Therefore, by the Stohr-Volochtheorem 10.9, it has at most s(p+ s− 1)/2 GF(p)-rational points. This implies

s(p+ 1)− s(NX + s− 1) ≤ s(p+ s− 1)/2,

which means s ≥ (p+ 5− 2NX)/3. In particular, if |B| < p+ 1 + 2(p+ 3)/3 andwe choose LX to be a tangent, then the curve f must be (absolutely) irreducible.Now Lemma 21.9 can be applied to f itself and it says that f has at least (k −1)(p + 1) − (k − 1)(k − 1) points. On the other hand, the previous lemma showsthat it can have at most k2 − k + 1 points over GF(p). Solving the inequalitypk − k(k − 1) ≤ k2 − k + 1 implies k ≥ (p+ 2)/2.

Let us repeat the interesting observation that for |B| < p + 1 + 2(p + 3)/3 thecurve f itself must be irreducible. A second proof of a slightly weaker bound willbe given in Proposition 21.20. In case of a prime-power q the same argument showsthat for |B| ≤ p + (p + 3)/3 each component of f whose partial derivative withrespect to X is not identically zero must be non-classical with respect to lines.However, the existence of such non-classical components can be excluded by usinga pair or triple of curves, as we shall see soon.

21.2 Three new curves

In this subsection we introduce three nice curves. We use the notation V =(X,Y, Z);Vq = (Xq, Y q, Zq) and Ψ = V × Vq = ((Y qZ − Y Zq), (ZqX −ZXq), (XqY − XY q)). Let B be a minimal blocking set of PG(2, q). SinceR(X,Y, Z) vanishes for all homogeneous (x, y, z) ∈ GF(q) × GF(q) × GF(q), wecan write it as

R(X,Y, Z) = Ψ · g = det(V,Vq,g) =

(Y qZ−Y Zq) g1(X,Y, Z)+(ZqX−ZXq) g2(X,Y, Z)+(XqY −XY q) g3(X,Y, Z),

where g1, g2, g3 are homogeneous polynomials of degree k − 1 in three variablesand g = (g1, g2, g3). Note that g is not determined uniquely, it can be changedby g′ = g + g0V for any homogeneous polynomial g0 ∈ GF(q)[X,Y, Z] of totaldegree k − 2, if k < q; and for g′ = g + g0V + g00Vq for arbitrary homogeneouspolynomials g0 of degree k − 2 and g00 of degree k − q − 1.Why is this a most natural setting? For example observe that if B is the line[a, b, c] then R = (a, b, c) ·Ψ.Now one can define f = V× g. Then f · (Vq −V) = f1(Xq −X) + f2(Y q − Y ) +f3(Zq −Z) = R, and f1, f2, f3 are homogeneous polynomials of degree k. If k < qthen, by this “decomposition” of R, f is determined uniquely. Conversely, if for


some g′ also f = V×g′ holds then g′ = g+gV for some homogeneous polynomialg of degree k − 2.We also remark that, as V · (V × g) = 0, we have Vf = 0. For another proof see21.16.If k ≥ q then f is not necessarily unique in the decompositon of R. But if wechoose f = V × g for some g then 21.16 remains valid (otherwise it may happenthat V · f is not the zero polynomial).The following lemma summarizes some fundamental properties of g.

Proposition 21.12. (1.1) If a point P (a, b, c) ∈ B is not essential, then there existsan equivalent g′ = g + g0V (or g′ = g + g0V + g00Vq) of g such that aX +bY + cZ divides g′i(X,Y, Z), i = 1, 2, 3 (as polynomials in three variables).

(1.2) Conversely, if NX < q+2−k and aX+bY +cZ divides each gi(X,Y, Z), i =1, 2, 3, then (a, b, c) ∈ B and (a, b, c) is not essential.

(2) If B is minimal then g1, g2 and g3 have no common factor.

Proof: (1.1) In this case R0 = R/(aX + bY + cZ) still vanishes everywhere, so itcan be written in the form R0 = g0Ψ, so Ψ · (g0(aX + bY + cZ)− g) = 0.(1.2) Now aX + bY + cZ divides R as well, so (a, b, c) ∈ B. Deleting it, theRedei polynomial of the new pointset is (Y qZ − Y Zq) g1(X,Y,Z)

aX+bY+cZ + (ZqX −ZXq) g2(X,Y,Z)

aX+bY+cZ + (XqY −XY q) g3(X,Y,Z)aX+bY+cZ , so it remains a blocking set.

(2) Such a factor would divide R as well, which splits into linear factors. Then fora linear factor see (1.2).

We want to “evaluate” R along a line [a, b, c] of the dual plane (so we examine thelines through (a, b, c) of the original plane). We use the notation

φ(X,Y, Z)∣∣∣[a,b,c]

= φ(bZ − cY, cX − aZ, aY − bX).

In general f(X,Y, Z)∣∣∣[a,b,c]

= f(−bY − cZ, aY, aZ) = f(bX,−aX − cZ, bZ) =

f(cX, cY,−aX − bY ), where e.g. f(−bY − cZ, aY, aZ) can be used if a 6= 0 etc.

Theorem 21.13. (1)

R∣∣∣[a,b,c]

=(ag1

∣∣∣[a,b,c]

+ bg2

∣∣∣[a,b,c]

+ cg3

∣∣∣[a,b,c]

)((ZY q − ZqY )

∣∣∣[a,b,c]

)(the last factor should be changed for (XZq −XqZ)

∣∣∣[a,b,c]

if a = 0 and b 6= 0

and for (Y Xq−Y qX)∣∣∣[a,b,c]

if a = b = 0 and c 6= 0, “normally” these factors

are identical when restricting to aX + bY + cZ = 0).


(2) (a, b, c) ∈ B if and only if ag1∣∣∣[a,b,c]

+bg2∣∣∣[a,b,c]

+cg3∣∣∣[a,b,c]

= 0. It means that

if one considers this equation as an equation in the variables a, b, c then thepoints of B are exactly the solutions of it.

(3) If (a, b, c) ∈ B then consider

R

aX + bY + cZ

∣∣∣∣∣[a,b,c]

=

(a(g1

∣∣∣[a,b,c]

+ bg2

∣∣∣[a,b,c]

+ cg3

∣∣∣[a,b,c]

)(aX + bY + cZ)

∣∣∣[a,b,c]

((ZY q−ZqY )

∣∣∣[a,b,c]

)

(4) Suppose (a, b, c) 6∈ B then if a line [x, y, z] through (a, b, c) is an r-secant ofB, then (x, y, z) is a root of ag1

∣∣∣[a,b,c]

+bg2∣∣∣[a,b,c]

+cg3∣∣∣[a,b,c]

with multiplicity

r − 1.

Proof: Easy calculations. For instance to prove (2) we simply need

R∣∣∣[a,b,c]

= ag1

∣∣∣[a,b,c]

+ bg2

∣∣∣[a,b,c]

+ cg3

∣∣∣[a,b,c]

= 0.

See Example 7.4 for showing the use of (2) above: there we get that the equationof the “canonical” Baer subplane is

Ga,b,c(X,Y, Z) = X√q(c

√qb− cb

√q)+Y

√q(a

√qb−ab

√q)+Z

√q(a

√qc−ac

√q) = 0,

meaning that the Baer subplane is just {(a, b, c) ∈ PG(2, q) : Ga,b,c(X,Y, Z) ≡ 0}.The map [x, y, z] 7→ [g1(x, y, z), g2(x, y, z), g3(x, y, z)], acting on the lines, is aremarkable one.

Proposition 21.14. Let [x, y, z] be a tangent line to B at the point (at, bt, ct) ∈ B.Then [g1(x, y, z), g2(x, y, z), g3(x, y, z)] is also a line through (at, bt, ct), differentfrom [x, y, z].

If [x, y, z] is a secant line then [g1(x, y, z), g2(x, y, z), g3(x, y, z)] is either [x, y, z]or meaningless (i.e. [0, 0, 0]).

Obviously, if g(x, y, z) = [0, 0, 0] then [x, y, z] is a ≥ 2-secant as the ≤ 1-st deriva-tives are 0.

Proof: Recall Theorem 7.10, here we have

(at, bt, ct) = ( (∂XR)(x, y, z), (∂YR)(x, y, z), (∂ZR)(x, y, z) ) =

( −yg3(x, y, z)+ zg2(x, y, z), xg3(x, y, z)− zg1(x, y, z), yg1(x, y, z)−xg2(x, y, z) ).

Now the scalar product with (g1(x, y, z), g2(x, y, z), g3(x, y, z)) vanishes.


or:

(at, bt, ct)·g(x, y, z) = (∇R)(x, y, z)·g(x, y, z) = ((x, y, z)×g(x, y, z))·g(x, y, z) = 0.

Here (x, y, z) 6= g(x, y, z) as their cross product is (at, bt, ct).If [x, y, z] is a secant line then there are more than one components of R goingthrough (x, y, z) (see Theorem 7.10) hence

0 = (∇R)(x, y, z) = (x, y, z)× g(x, y, z).

or: we can use Theorem 21.13.We also calculate ∇R for future use: ∇R = Ψ(∇ ◦ g) + Vq × g.The following is true as well. We will see (Theorem 21.19, Corollary 21.23) thatif B is small, then R can be written in the form

Xq(Y gXY + ZgXZ) + Y q(XgY X + ZgY Z) + Zq(XgZX + Y gZY ),

where the polynomials gXY , ... contain only exponents divisible by p. Surprisinglyor not, gXY = −gY X , ..., as 0 = Xf1 +Y f2 +Zf3 = XY (gXY +gY X)+Y Z(gY Z +gZY ) + ZX(gZX + gXZ). One can take ∂X∂Y , ∂Y ∂Z , ∂Z∂X of both sides.Now g = (gY Z , gZX , gXY ), which is a kind of “natural choice” for g.Or: gXY = ∂Y f1 = −∂Xf2,..., or similarly (if p 6= 2) g = ∇×f and f = V×(∇×f);g = ∇1

R × (∇qHR).

21.3 Three old curves

In this section we will present the method using three old algebraic curves. Asan application we show Szonyi’s result [123] that blocking sets of size less than3(q + 1)/2 intersect every line in 1 modulo p points. This immediately impliesBlokhuis’ theorem for blocking sets in PG(2, p).Let now B be a minimal blocking set of PG(2, q). Since R(X,Y, Z) vanishes for all(x, y, z) ∈ GF(q)× GF(q)× GF(q), we can write it as

R(X,Y, Z)=(Xq−X)f1(X,Y, Z)+(Y q−Y )f2(X,Y, Z)+(Zq−Z)f3(X,Y, Z)=W·f ,

where f = (f1, f2, f3) and deg(fi) ≤ k as polynomials in three variables. Note thatf1 here is the same as the polynomial f defined in Definition 21.5 and examinedin Section 21.1; while f2 and f3 behave very similarly.

Proposition 21.15. (Lovasz, Szonyi) Let [x, y, z] be a tangent line to B at the point(at, bt, ct) ∈ B. Then

f(x, y, z) = (f1(x, y, z), f2(x, y, z), f3(x, y, z)) = (at, bt, ct)

as homogeneous triples.


Proof: Recall Theorem 7.10, here we have

((∂XR)(x, y, z), (∂YR)(x, y, z), (∂ZR)(x, y, z)) = −(f1(x, y, z), f2(x, y, z), f3(x, y, z))

Or:

(at, bt, ct) = (∇R)(x, y, z) = (∇(W ·f))(x, y, z) = ((∇◦W)f +(∇◦f)W)(x, y, z) =

−I f(x, y, z) + 0 = −f(x, y, z).

Lemma 21.16. If k < q − 1 then V · f = Xf1 + Y f2 + Zf3 = 0. (Hence R = Vqfas well).

Proof: If [x, y, z] is a tangent then by 21.15 V · f vanishes, and by the end ofTheorem 7.10 it also vanishes if [x, y, z] is at least a 2-secant. As the degree is lessthan q we are done. Or: it is just Theorem 7.10 (6) with r = 1.

Note also that f = ∇qHR; and −f = (∇ ◦ f)V. Moreover, in general −∇i−1H f =

(∇iH ◦ f)V. It follows from the derivation of f ·V = 0.From Theorem 7.10 (5) one can see that each of the curves f1, f2, f3 go through thepoint (x, y, z) of the dual plane corresponding to a secant line [x, y, z]. Where arethe other (extra) points of e.g. f1? They are exactly the points of rNX of Lemma21.6, so points on factors corresponding to points with ai = 0.If one fixes (Y, Z) = (y, z) then R(X, y, z) is divisible by (Xq−X). If R(X, y, z) 6=0, so if (0, y, z) 6∈ B ∩ LX then for an (x, y, z) ∈ GF(q)× GF(q)× GF(q) if the linewith equation xX+yY +zZ = 0 intersects B in at least two points (cf. Proposition21.7 (2.2)) then f1(x, y, z) = 0. One can repeat the same reasoning for f2, f3 andthis immediately gives the following lemma:

Lemma 21.17. ([123]) The curves fi have almost the same set of GF(q)-rationalpoints. The exceptional points correspond to lines intersecting LX , LY or LZ in apoint of B.

Proof: Since this observation is crucial, a direct proof is also included. Considerthe Redei polynomial R(X,Y, Z). For an element (x, y, z) ∈ GF(q) × GF(q) ×GF(q) we get −f1(x, y, z) = ∂XR(x, y, z) and similarly −f2(x, y, z) = ∂YR(x, y, z).Since R is a product of linear factors and R(x, y, z) = 0, ∂XR(x, y, z) = 0 if andonly if there are two linear factors vanishing at (x, y, z), or if R(X, y, z) = 0 (i.e.(0,−z, y) ∈ B). The similar statement holds for ∂YR, hence the two derivatives arezero for the same values (x, y, z), except in the cases described in the statement.

Lemma 21.18. ([123]) If k < q−1 then the polynomials f1, f2 and f3 cannot have acommon factor. Moreover, e.g. f1 and f2 have a common factor g iff (0, 0, 1) ∈ Band g = Z.


Proof: Such a common factor must divide R(X,Y, Z), hence it must be divisibleby aiX + biY + ciZ for some i. Lemma 21.8 (2) gives (N = 1, k ≤ q− 2) that thepoint (ai, bi, ci) can be deleted, a contradiction.Suppose that g is a common factor of f1 and f2, then from Xf1 + Y f2 +Zf3 = 0we have g|Zf3.

Therefore, (f1, f2, f3) is a triple of polynomials (curves) having no common factor(component), but they pass through almost the same set of GF(q)-rational points.Using Bezout’s theorem it immediately gives Lemma 21.10 back.

21.4 Small blocking sets

Lemmas 21.17 and 21.18 can also be used to show that all the components off have identically zero partial derivative with respect to X. Note that for anycomponent h of f the total degree of h is the same as its degree in X.

Theorem 21.19. ([123]) If k ≤ (q+1)/2 and g(X,Y, Z) is an irreducible polynomialthat divides f1(X,Y, Z), then g′X = 0.

Proof: Suppose to the contrary that g is a component of f1 with nonzero partialX-derivative, denote its degree by deg(g) = s. By Lemma 21.9 the number ofGF(q)-rational points on g is at least s(q + 2 − NX − s). Since these points arealso on f2, Bezout’s theorem gives s(q + 2 − NX − s) ≤ sk, since by Lemma21.18, if f2 and g has a common component (i.e. g itself) then it cannot be acomponent of f3 and one can use Bezout for g and f3 instead. This immediatelyimplies q + 2 ≤ k +NX + s and from NX + s ≤ k it follows that k ≥ (q + 2)/2, acontradiction.

Note that it implies that all the X-exponents appearing in f1 are divisible by p(as rNX does not involve X); and a similar statement holds for the Y -exponentsof f2 and for the Z-exponents of f3. Let’s define e, the (algebraic) exponent of B,as the greatest integer such that f1 ∈ GF(q)[Xpe , Y, Z], f2 ∈ GF(q)[X,Y p

e

, Z] andf3 ∈ GF(q)[X,Y, Zp

e

]. By the Theorem e ≥ 1.

Proposition 21.20. If q = p is a prime and |B| < p+ 2p+4−NX3 , then the curve f1

is irreducible (and similarly for f2, f3).

Proof: Suppose to the contrary that e.g. f1 is not irreducible, and let g be acomponent of f1 of degree at most (k−NX)/2. The proof of Theorem 21.19 givesp+ 2 ≤ k +NX + deg(g) ≤ 3k/2 +NX/2, that is 2(p+2)−NX

3 ≤ k.

The following corollary, due to Szonyi, generalizes the similar result of Redei onblocking sets of Redei type.


Corollary 21.21. ([123]) If B is a blocking set of size less than 3(q + 1)/2, theneach line intersects it in 1 modulo p points.

Proof: Take a line ` and coordinatise such that ` ∩ LX ∩ B = ∅. If ` = [x, y, z]then rNX (y, z) 6= 0. Since all the components of f1 contain only terms of exponent(in X) divisible by p, for any fixed (Y, Z) = (y, z) the polynomial f1(X, y, z) =rNX (y, z)f1(X, y, z) itself is the p-th power of a polynomial. This means that atthe point P (x, y, z) the “horizontal line” (i.e. through P and (1, 0, 0)) intersectsf1(X,Y, Z) with multiplicity divisible by p (and the same is true for f1), so byTheorem 21.7 the line [x, y, z] intersects B in 1 modulo p points.

Note that now we have |B| ≡ 1 (mod p). Of course, this theorem also impliesBlokhuis’ theorem in the prime case.

Corollary 21.22. (Blokhuis [33]) If q = p is a prime, then |B| ≥ 3(q+ 1)/2 for thesize of a non-trivial blocking set.

Corollary 21.23. If B is a blocking set of size less than 3(q + 1)/2,then the X-exponents in f1, the Y -exponents in f2 and the Z-exponents in f3 are 0 (mod p)e; moreover all the exponents appearing inR(X,Y, Z), f1, f2, f3; rNX (Y, Z), rNY (X,Z), rNZ (X,Y ), are 0 or 1 (mod p)e.

Proof: The first statement is just Theorem 21.19. From this the similar statementfollows for fi: the X-exponents in f1, the Y -exponents in f2 and the Z-exponentsin f3 are 0 (mod p)e.Consider a term aXαpe+1Y βZγ of Xf1 in the identity Xf1 + Y f2 + Zf3 ≡ 0. Itshould be cancelled by Y f2 and Zf3, which means that it should appear in eitherone or both of them as well with some coefficient. It cannot appear in both of them,as it would imply exponents like Xαpe+1Y β

′pe+1Zγ′pe+1, but the exponents must

add up to k+1, which is 2 mod pe, a contradiction. So this term is cancelled by itsnegative, for example contained in Y f2, then it looks like −aXαpe+1Y β

′pe+1Zγ ,where the exponents, again, add up to k + 1, which is 2 mod pe, hence γ ≡ 0(mod pe), so the original term of f1 was of form aXαpeY β

′pe+1Zγ′pe .

For rNX (Y, Z), rNY (X,Z) and rNZ (X,Y ) recall that they are also homogeneouspolynomials of total degree 1 (mod pe) and for instance f1 = rNX f1 and deg f1 =degX f1, so in f1 the terms of maximal X-degree have 0 or 1 mod pe exponents(as terms of f1), on the other hand they together form rNXX

k−NX .Finally R = Xqf1 + Y qf2 + Zqf3 so R has also 0 or 1 mod pe exponents only.

Note that in fi other exponents can occur as well. Comparing the exponents onecan find Y ∂Y f1 + Z∂Z f1 = X∂X f1 + Y ∂Y f1 + Z∂Z f1 = 0 as well.


The (geometric) exponent eP of the point P can be defined as the largest integerfor which each line through P intersects B in 1 mod peP point. It can be proved(e.g. [39]) that the minimum of the (geometric) exponents of the points in B isequal to e defined above.

Theorem 21.24. [118] Let B be a blocking set with exponent e. If for a certain line|` ∩B| = pe + 1 then GF(pe) is a subfield of GF(q) and ` ∩B is GF(pe)-linear.

Proof: Choose the frame such that ` = LX and (0, 0, 1); (0, 1, 0); (0, 1, 1) ∈ `∩B.Consider f = f1, now rNX (Y, Z) is a homogeneous polynomial of (total) degree pe+1, with exponents 0, 1, pe or pe+1, so of form αY p

e+1 +βY Zpe

+γY pe

Z+δZpe+1.

As rNX (0, 1) = rNX (1, 0) = rNX (1,−1) = 0 we have rNX = Y pe

Z − Y Zpe

=∏(a,b)∈PG(1,pe)(aY + bZ).

Now we can disclose one of our main goal: to get as close as we can to the proofof the conjecture that every small blocking set is linear.By the following proposition, a blocking set with exponent e has a lot of (pe + 1)-secants (so “nice substructures”). Similar arguments can be found in [37].

Proposition 21.25. Let P be any point of B with exponent eP .

(1) (Blokhuis) There are at least (q − k + 1)/peP + 1 secant lines through P .

(2) Through P there are at most 2(k − 1)/peP − 1 long secant lines, i.e. linescontaining more than peP +1 points of B (so at least q/peP −3(k−1)/peP +2 (peP + 1)-secants).

(3) There are at most 4k− 2peP − 4 points Q ∈ B \ {P} such that PQ is a longsecant.

(4) There are at least q − 3k + 2pe + 4 points in B with (point-)exponent e.

Proof: (1) was proved by Blokhuis using lacunary polynomials. To prove (2)denote by s the number of (peP +1)-secants through P and let r be the number of(≥ 2peP + 1)-secants through P . Now speP + 2rpeP + 1 ≤ q+ k. From (1) s+ r ≥(q−k+1)/peP +1, so q/peP−(k−1)/peP +r+1 ≤ s+2r ≤ q/peP +(k−1)/peP hencer ≤ 2(k−1)/peP −1 and s ≥ q/peP −(k−1)/peP +1−r ≥ q/peP −3(k−1)/peP +2.For proving (3) subtract the number of points on (peP +1)-secants through P from|B|, it is ≤ q + k − (q/peP − 3(k − 1)/peP + 2)peP − 1 = 4k − 2peP − 4. There isat least one point P ∈ B for which eP = e. On the pe + 1-secants through it (by(2)) we find at least 1 + pe(q/pe − 3(k − 1)/pe + 2) points, each of exponent e, itproves (4).

Recall that there are at least q+1−k tangent lines through P , so at most k secants.We also know from Szonyi [123] that q/pe+1 ≤ k ≤ q/pe+q/p2e+2q/p3e+ ... Now


“almost all” line-intersections of B are GF(pe)-linear (in fact they are isomorphicto PG(1, pe) in the non-tangent case).

Corollary 21.26. [118] For the exponent e of the blocking set, e|h (where q = ph).

Proof: By Proposition 21.25 B has a lot of short secants. By Theorem 21.24 theseintersections are all isomorphic to PG(1, pe), so GF(pe) is a subfield of GF(ph) =GF(q).

Now we can give a very short proof for Theorem 18.14 in the case when pe > 13.

Corollary 21.27. [118] Small blocking sets of Redei type, with pe > 13, are linear.

Proof: Suppose LZ is the Redei-line, O = (0, 0, 1) ∈ B, eO = e, and take anyP ∈ B \LZ , with eP = e, and any α ∈ GF(pe). Claim: αP (affine point!) is also inB. If OP is a short secant then it is obvious.Consider the short secants through P , there are at least q/pe − 3(k − 1)/pe + 2.Most of them, at least q/pe − 3(k − 1)/pe + 2 − 4k−2pe−4

pe−1 ≥ qpe −

7k−3k/pe

pe−1 , say{ì : i ∈ I}, contain at least two points Q1, Q2 ∈ B, such that OQ1 and OQ2 areshort secants.For any of them, say ì, take αì, it contains αP . If all of {αì : i ∈ I} were longsecants then they would contain at least 2pe( qpe −

7k−3k/pe

pe−1 ) > q + k points of B,contradiction if pe > 13. Say α` is a short secant, then αP ∈ B ∩ α` and eαP = eas well.Let U0 be the set of affine points of B with exponent e. Now we have that U0 isinvariant for magnifications from any center in U0 and with any scale α ∈ GF(pe),so it forms a vectorspace over GF(pe). As its size is q− 3k+ 2pe + 4 ≤ |U0| ≤ q wehave |U0| = q and it contains all the affine points of B.

Consequences

The bounds for the sizes of small blocking sets are now the following.

Corollary 21.28. Let B be a minimal blocking set of PG(2, q), q = ph, of size|B| < 3(q + 1)/2. Then there exists an integer e, called the exponent of B, suchthat1 ≤ e|h,and

q + 1 + ped q/pe+1

pe+1 e ≤ |B| ≤ 1+(pe+1)(q+1)−√

(1+(pe+1)(q+1))2−4(pe+1)(q2+q+1)

2 .

If |B| lies in the interval belonging to e and pe 6= 4 then each line intersects B in1 modulo pe points. Most of the secants are (pe + 1)-secants, they intersect B in apointset isomorphic to PG(1, pe).


These bounds are due to Blokhuis, Polverino and Szonyi, see [99, 123], and asymp-totically they give q + q

pe −qp2e + q

p3e − ... ≤ |B| ≤ q + qpe + q

p2e + 2 qp3e + ... . Note

that for q = p2s and q = p3s, where s is a prime, the lower bound is sharp:|B| ≥ q + q/ps + 1 and |B| ≥ q + q/p2s + 1, resp.The 1 mod pe property was established by Szonyi; our Theorem 21.26 shows thatonly a very few of the intervals of Szonyi, Blokhuis, Polverino contain values fromthe spectrum of blocking sets, i.e. only those with e|h. The linearity of short secantsis Theorem 21.24, on their number see Proposition 21.25.Let S(q) denote the set of possible sizes of small minimal blocking sets in PG(2, q).

Corollary 21.29. Let B be a minimal blocking set of PG(n, q), q = ph, with respectto k-dimensional subspaces, of size |B| < 3

2 (qn−k + 1), and of size |B| <√

2qn−k

if p = 2. Then

• |B| ∈ S(qn−k);

• if p > 2 then ((|B| − 1)(qn−k)n−2 + 1) ∈ S((qn−k)n−1).

If p > 2 then there exists an integer e, called the exponent of B, such that

1 ≤ e|h,

for which every subspace that intersects B, intersects it in 1 modulo pe points. Also|B| lies in an interval belonging to some e′ ≤ e, e′|h. Most of the k-dimensionalsubspaces intersecting B in more than one point, intersect it in (pe + 1) pointsprecisely, and each of these (pe + 1)-sets is a collinear pointset isomorphic toPG(1, pe).

Most of this was proved by Szonyi and Weiner in [129]. Consider the line deter-mined by any two points in a (pe + 1)-secant k-subspace, this line should containpe + 1 points. Then the technique of [129] can be used to derive a planar minimalblocking set (in a plane of order qn−k) with the same exponent e: firstly embedPG(n, q) into PG(n, qn−k) where the original blocking set B becomes a blocking setw.r.t. hyperplanes, then choose an (n− 3)-dimensional subspace Π ⊂ PG(n, qn−k)not meeting any of the secant lines of B and project B from Π onto a planePG(2, qn−k) to obtain a planar minimal blocking set, for which Theorem 21.24and Proposition 21.25 can be applied, implying e|h(n− k).Now in PG(n + 1, q) ⊇ PG(n, q) build a cone B∗ with base B and vertex V ∈PG(n+ 1, q) \PG(n, q); then B∗ will be a (small, minimal) blocking set in PG(n+1, q) w.r.t. k-dimensional subspaces. The argument above gives e|h(n+ 1− k), soe | g.c.d.(h(n− k), h(n+ 1− k)) = h.

Exercise 21.30. Let B ⊂ PG(2, q) be a double blocking set of size 2q + k. Then

R(X,Y, Z) = W F(X,Y, Z) WT ,

where F(X,Y, Z) = (fij(X,Y, Z))3×3.


22 Multiple blocking sets

In this section we give a generalization of the Redei-polynomial approach of block-ing sets. Some of the statements below (and much more) can be found in [37], someothers in [62].Let B = {(ai, bi, ci)} be a t-fold blocking set in PG(2, q), having possibly weighted(i.e. multiple) points, |B| = tq + k.We use the very same Redei polynomial as we did before.

R(X,Y, Z) =|B|∏i=1

(aiX + biY + ciZ)

=|B|∑j=0

X |B|−jrj(Y, Z)

=∑

0≤j1,j2,j3≤tj1+j2+j3=t

fj1j2j3(X,Y, Z)(Xq −X)j1(Y q − Y )j2(Zq − Z)j3 , (∗)

where deg(rj) ≤ j, j = 0, . . . , |B|, and each fj1j2j3 , if present, is a homogeneouspolynomial of degree k, for 0 ≤ j1, j2, j3 ≤ t, j1 + j2 + j3 = t. The latter equationabove follows from the fact that this Redei polynomial is zero t times everywherein PG(2, q).The polynomials fj1j2j3 satisfy a (low degree) polynomial equation:

Lemma 22.1. If k + t < q then∑0≤j1,j2,j3≤tj1+j2+j3=t

Xj1Y j2Zj3fj1j2j3(X,Y, Z) = 0

andR(X,Y, Z) =

∑0≤j1,j2,j3≤tj1+j2+j3=t

fj1j2j3(X,Y, Z)Xj1qY j2qZj3q.

Proof: For the second equation observe that in (∗) the left hand side is homoge-neous of degree tq+ k so all the terms on the right hand side with degree 6= tq+ kmust vanish.As a special case, consider the lowest degree terms of (∗), they are of degree k+ t,and they must disappear after summing them, this proves the first equation.

As a corollary note that fj1j2j3 = Hj1qX Hj2q

Y Hj3qZ R as polynomials, and also for

every x, y, z ∈ GF(q) we have fj1j2j3(x, y, z) = −(Hj1XH

j2Y H

j3Z R)(x, y, z).

22. Multiple blocking sets 155

We would like to understand what the (GF(q)-rational points of the) curves fj1j2j3mean. For ft00 it is obvious: (x, y, z) is a point of it if and only if the line [x, y, z]is either an (≥ t + 1)-secant (this is the typical case), or if [x, y, z] intersects theline [1, 0, 0] in a point of B. Similar statements hold for f0t0 and f00t.

Proposition 22.2. If the line [x, y, z] is an (≥ t+1)-secant then fj1j2j3(x, y, z) = 0for all 0 ≤ j1, j2, j3 ≤ t, j1 + j2 + j3 = t.

Proof: Fix 0 ≤ j1, j2, j3 ≤ t, j1 + j2 + j3 = t and some (≥ t + 1)-secant line[x, y, z]. Recall Theorem 7.10 (3) stating that

(Hj1XH

j2Y H

j3Z R)(x, y, z) = 0 and

also fj1j2j3(x, y, z) = −(Hj1XH

j2Y H

j3Z R)(x, y, z).

22.1 A t (mod p) result

Let B be a minimal weighted t-fold blocking set in PG(2, q), with |B| = tq+ t+ k,where 2t + k < q + 2. From now on we are going to use a “more affine” point ofview, still based on [37] and [62].Assume that the line l∞ is an m-secant to B. Consider PG(2, q) as the affineplane AG(2, q) with l∞ as the line at infinity. Assume that B ∩ l∞ = D ={(∞), . . . , (∞), (y1), . . . , (ym−s)}, where (∞) is a point of weight s ofB (1 ≤ s ≤ t),where some of the other points of D might be multiple points of B, and thatU = B \D = {(ai, bi) : i = 1, . . . , tq + t+ k −m}, where U is a multiset when Bhas affine multiple points.Now we redefine the Redei polynomial associated to the t-fold blocking set B,(with (∞) deleted).

Definition 22.3 (The Redei polynomial of the set B).

R(X,Y ) =m−s∏j=1

(Y − yj)tq+t+k−m∏

i=1

(X + aiY − bi)

=m−s∏j=1

(Y − yj)tq+t+k−m∑

i=0

Xtq+t+k−m−iri(Y ) (2.1)

= (Xq −X)tf0(X,Y ) + (Xq −X)t−1(Y q − Y )f1(X,Y )+ · · ·+ (Y q − Y )tft(X,Y ), (2.2)

where deg(ri) ≤ i, i = 0, . . . , tq + t+ k −m, and deg(fi) ≤ k + t− s, i = 0, . . . , t.

Choose a point (b,m), b 6= bj , m 6= yj . Consider R(X,m) = (Xq −X)tf0(X,m).By the properties of the Redei polynomial, the line Y = mX + b intersects U inmore than t points iff X = b is a root of R(X,m) with multiplicity ≥ t+1 iff (b,m)


is a point of the algebraic curve f0(X,Y ). Considering R(b, Y ) = (Y q−Y )tft(b, Y )instead, we get that the line Y = mX + b intersects U in more than t points iff(b,m) is a point of the algebraic curve ft(X,Y ).

Therefore, these two algebraic curves f0 and ft have essentially the same set ofGF(q)-rational points.

If m = yj or b = bj , and the line Y = mX + b intersects U in more than t points,then still f0(b,m) = ft(b,m) = 0. Because of the above mentioned divisibilities,f0(b,m) = 0 or ft(b,m) = 0 does not imply that Y = mX+b intersects U in morethan t points.

In this section we will assume that there is no line contained in B. As the followingtheorems will show, this is no restriction when 2t+ k < q + 2.

Theorem 22.4. Let B be a minimal weighted t-fold blocking set of PG(2, q), with|B| = tq + t+ k, where 2t+ k < q + 2, containing a line `. Then B is the sum ofthe line ` and the minimal weighted (t− 1)-fold blocking set B∗, obtained from Bby reducing the weight of every point P of ` by one.

Proof: Since ` ⊆ B, |` ∩B| ≥ q + 1.

If |`∩B| ≥ q+ t, then after reducing the weight of every point of ` by one, a newweighted set B∗ is obtained which still intersects every line in at least t−1 points.Since B is a minimal weighted t-fold blocking set, also B∗ is a minimal weighted(t− 1)-fold blocking set.

Assume now that q+1 ≤ |B∩`| < q+t. Reduce again the weight of every point on `by one, and add a minimal number of simple points P1, . . . , of ` back so that againa weighted (t−1)-fold blocking set B∗ is obtained, hence |B∗∩`| = t−1. We need toadd at most t−1 points P1, . . . to achieve this, hence |B∗| ≤ tq+t+k−(q+1)+t−1 =(t− 1)q+2(t− 1)+ k. A particular feature of a point Pi, i = 1, . . ., is that the line` is the only (t− 1)-secant to B∗ passing through Pi.

Finally, we show that through Pi, there pass at least two (t−1)-secants, hence theabove case cannot occur. Now we choose our coordinate system in such a way that(∞) ∈ B, Pi is an affine point (a, b), and (`∞ ∩ `) 6∈ B∗ and (∞) has multiplicitys. Suppose that |`∞ ∩B∗| = m and write up the Redei polynomial. Since B∗ is a(t− 1)-fold blocking set, we get that:

R(X,Y ) =m−s∏j=1

(Y − yj)tq+t+k−(m−s)∏

i=1

(X + aiY − bi)

= (Xq −X)t−1f0(X,Y ) + (Xq −X)t−2(Y q − Y )f1(X,Y )+ · · ·+ (Y q − Y )t−1ft−1(X,Y ), (2.3)

where deg(fi) ≤ |B∗| − q(t− 1)− s = 2(t− 1) + k − 1, i = 0, . . . , t− 1.


The argument before this theorem shows that if a line Y = mX + b intersects B∗

in more than (t−1) points, then (b,m) is a point of the curve f0. Each line except` through the point Pi(a, b) intersects B∗ in at least t points. These lines are pointsof the line X+aY − b in the dual plane. Hence X+aY − b intersects f0 in at leastq − 1 points (we do not see the vertical line here). Since degf0 < q − 1, Bezout’stheorem implies that the line X + aY − b is a component of f0. Suppose that ` isthe line ` = Y +m′X + b′. Then f0(b′,m′) = 0 and since `∩ `∞ 6∈ B∗, ` intersectsB∗ in at least t points. This is a contradiction, hence q + 1 ≤ |B ∩ `| < q + t doesnot occur.

As the next example shows, the above theorem is sharp.

Example 22.5. Let S be the set of points lying on the lines of a dual hyperoval inPG(2, q), q even. Then S is a ( q2 +1)-fold blocking set of size ( q2 +1)q+( q2 +1) (eachpoint in S has multiplicity one). Note that now t = q

2 +1, k = 0 and 2t+k = q+2.If we delete a line of S, then the resulting point set is not a q

2 -fold blocking set.

Remark 22.6. Theorem 22.4 has several important applications.

(1) It first of all shows that when characterizing minimal weighted t-fold blockingsets of size tq + t + k, where 2t + k < q + 2, in PG(2, q), it is possible to assumethat they do not contain any lines.

(2) Moreover, also when proving the t (mod p) result for a minimal weighted t-foldblocking set B, |B| = tq+ t+k, where 2t+k < q+2, it is possible to assume thatthere are no lines contained in B. For, if there is a line ` contained in B, Theorem22.4 shows that it is possible to reduce the weight of every point of ` by one inorder to obtain a new minimal weighted (t−1)-fold blocking set B∗. Proving the t(mod p) result for B is now reduced to proving the (t− 1) (mod p) result for B∗.

(3) We are also now able to characterize weighted minimal t-fold blocking setsof size tq + t and to exclude the existence of minimal t-fold blocking sets of sizetq + t+ 1.

Theorem 22.7. A weighted t-fold blocking set B in PG(2, q), of size |B| = tq + t,where 2t < q + 2, is a sum of t lines.

There does not exist a weighted minimal t-fold blocking set B in PG(2, q) of size|B| = tq + t+ 1, 2t+ 1 < q + 2.

Proof: Suppose that tq + t ≤ |B| ≤ tq + t + 1. Then counting the incidences ofthe points of B with the lines through a point R not in B, we have that through Rall the lines are t-secants if |B| = tq+ t and there is exactly one (t+1)-secant andq t-secants if |B| = tq+ t+1. Counting the incidences of the points of B with thelines through a point R′ ∈ B, we get that R′ lies on at least one line ` completelycontained in B, when |B| = tq + t and R′ lies on at least one line ` completelycontained in B or R′ lies on q+1 (t+1)-secants, when |B| = tq+ t+1. This latter


case means that R′ is not minimal, hence we can assume that each point of B lieson at least one line completely contained in B.Now the t points of any t-secant (which must exist) and Theorem 22.4 show thatB contains the sum of t lines, which is a t-fold blocking set already, of size tq+ t.

Remark 22.8. One can observe now that a weighted t-fold blocking set in PG(2, q),of size tq+ t, where 2t < q+2, intersects every line in t (mod p) points; also thatthrough any point of it there pass at least q + 1− t t-secants.

By the theorem above, from now on we can (and will) assume that k ≥ 2 for theminimal blocking sets considered.

In this section, from now on, we assume that B does not contain any line, hence|B| = tq+t+k ≥ tq+t+2, and we suppose that |B| < tq+(q+3)/2. (Note that sincek ≥ 2, we still have 2t+k < q+2.) Furthermore, we choose our coordinate systemso that `∞ is a t-secant and the point (∞) in B has multiplicity s, where 1 ≤ s ≤ t.

Lemma 22.9. The polynomial∏t−sj=1(Y − yj) divides f0(X,Y ).

Proof: By (2.1),

R(X,Y ) =tq+k∑i=0

ri(Y ) ·t−s∏j=1

(Y − yj)

Xtq+k−i.

So every coefficient polynomial of a term Xtq+k−i is divisible by∏t−sj=1(Y −yj). By

(2.2), the high degree part∏t−sj=1(Y −yj) ·Xtq+k+ · · ·+ rk(Y ) ·

∏t−sj=1(Y −yj) ·Xtq

must be equal to Xtqf0(X,Y ), when one compares the X-degrees of the twoexpressions (2.1) and (2.2) for R(X,Y ). So

∏t−sj=1(Y − yj) divides f0(X,Y ).

If X = 0 intersects U in the, possible weighted, points (0, bj), j = 1, . . . , z, then asimilar argument shows that

∏bj

(X − bj) divides ft(X,Y ), where the product istaken over the values bj , according to their weights.

Theorem 22.10. Let B be a minimal weighted t-fold blocking set of PG(2, q), with|B| = tq+ t+ k < tq+ (q+ 3)/2 and k ≥ 2. Then every point of B lies on at leastq + 1− k − t different t-secants.

Proof: Let P = (a, b) ∈ U and suppose (∞) ∈ B, |l∞ ∩ B| = t. Assume that Plies on more than k + t different lines sharing at least t + 1 points with B. Then


more than k of those lines intersect l∞ in a point not belonging to B. Each of theselatter lines defines a point of f0(X,Y )/

∏t−sj=1(Y − yj). More precisely, they define

intersection points, in the dual plane, of the algebraic curve f0(X,Y )/∏t−sj=1(Y −

yj) = 0 with the line X + aY − b = 0. The polynomial f0(X,Y )/∏t−sj=1(Y − yj)

has at most degree k, so by Bezout’s theorem, the linear term X + aY − b is afactor of f0(X,Y )/

∏t−sj=1(Y − yj).

Consider a line through P with slope m 6= yj , m 6= ∞, so that we can use thearguments above.

Plugging Y = m into R(X,Y ), we get

R(X,m) =t−s∏j=1

(m− yj)tq+k∏i=1

(X + aim− bi) = (Xq −X)tf0(X,m).

The fact that X+aY −b is a linear factor of f0 means geometrically that the linesthrough P with slope m 6= yj , m 6= ∞, intersect U in at least t+ 1 points.

Assume that a line ` through P with slope m = yj or m = ∞ is a t-secant. Thenreplace the line at infinity l∞ by a new line at infinity l∗∞ not intersecting ` in apoint of B, and containing exactly t points of B.

In the affine plane with l∗∞ as line at infinity, we use the same notations, but incombination with the symbol * to make the distinction with the notations usedin the original affine plane. The theorem of Bezout will now lead to the desiredcontradiction.

Through P , there pass at least q + 1 − t lines that intersect B in at least t + 1points. Whenever a line contains at least t + 1 affine points of B, it defines apoint of f∗0 (see the reasoning at the beginning of this section). This implies thatif the coordinates of P in the new coordinate system are equal to (a∗, b∗), thenthe line X∗ + a∗Y ∗ − b∗ = 0 meets the curve f∗0 in at least q + 1 − t − t points.We subtracted t from q + 1 − t since t of those lines through P could contain apoint of B belonging to l∗∞. Then such a line through P , not containing a pointat infinity of B, and containing at least t+ 1 affine points (a∗i , b

∗i ) defines a point

of the polynomial f∗0 (X∗, Y ∗)/∏t−sj=1(Y

∗ − y∗j ). This latter polynomial has degreeat most k < (q + 3)/2− t.

Then, if q + 1− 2t ≥ (q + 3)/2− t > k ≥ deg(f∗0 ), Bezout’s theorem implies thatthis linear component X∗ + a∗Y ∗ − b∗ is a component of f∗0 , so also the line `through P intersects U∗ in at least t + 1 points, since its point at infinity in thenew coordinate system does not lie in B, and so the observations made before thistheorem apply. Note that the condition q + 1 − 2t ≥ (q + 3)/2 − t implies thatt ≤ (q − 1)/2.

We have proved that all lines through P ∈ B are at least (t + 1)-secants to B;this contradicts the minimality of B. So a point of B lies on at most k + t linessharing more than t points with B.


Corollary 22.11. Let B be a weighted t-fold blocking set of PG(2, q), with |B| =tq + t+ k < tq + (q + 3)/2 and 2t < q + 2. Assume that P is an essential point ofB. Then there are at least q + 1− k − t different t-secants through P .

Proof: Delete the non-essential points of B one-by-one until a minimal t-foldblocking set B′ is obtained. By Theorem 22.10 and Remark 22.8, there will be atleast q+1− (|B′|− tq) different t-secants of B′ through P . Now if we add back thepoints of B \B′, then through P , we will see at least q+ 1− (|B′| − tq)− |B \B′|t-secants to B.

Let B be a minimal weighted t-fold blocking set of size tq + t+ k, where t+ k <(q + 3)/2 and k ≥ 2. Recall the definition of the Redei polynomial from thebeginning of this section, the following lemma is straightforward.

Lemma 22.12. If a line Y = mX + b intersects B ∩U in more than t points, thenf0(b,m) = · · · = ft(b,m) = 0.

We will also need

Lemma 22.13. The algebraic curve f0(X,Y ) = 0 does not have linear componentsdepending on the variable X.

Proof: Such a linear component depending on X should have the formX+aY − b = 0. The proof of Theorem 22.10 then shows that the point P = (a, b)is a non-essential point of B; which contradicts the minimality of B.

Lemma 22.14. If B is minimal, then the polynomials f0, . . . , ft cannot have acommon divisor different from Y − yj.

Proof: Such a polynomial would divide R(X,Y ); so would be linear. This canonly be of the form Y − yj .

We now come to the main theorem of this section: the proof of the t (mod p)result.


Theorem 22.15. Let B be a minimal weighted t-fold blocking set in PG(2, q), q = ph,p prime, h ≥ 1, with |B| = tq + t + k, t + k < (q + 3)/2, k ≥ 2. Then every lineintersects B in t (mod p) points.

Proof: By Remark 22.6, it is possible to assume that B does not contain anylines. We will assume that the line at infinity intersects B in t points.Let h(X,Y ) be an absolutely irreducible component of f0(X,Y )/

∏t−sj=1(Y −yj) of

degree larger than one. Similar arguments as in the case t = 1 imply that ∂Xh ≡ 0.If Y = m 6= yi, we obtain R(X,m) = (Xq − X)tf0(X,m), having t (mod p)solutions since f0(X,m) is a p-th power. So every line Y = mX+b, not containinga point of B at infinity, intersects B in t (mod p) points.For every line ` of which we are not yet sure that it intersects B in t (mod p)points, it is possible to find a new line at infinity intersecting B in t points andintersecting ` in a point not belonging to B. Repeating the previous argumentsnow shows that also ` intersects B in t (mod p) points.

The next corollary follows from Theorem 22.10 and Remark 22.8.

Corollary 22.16. Let B be a weighted t-fold blocking set in PG(2, q), q = ph, pprime, h ≥ 1, with |B| = tq + t + k, t + k < (q + 3)/2, 2t < q + 2. Assume thatall the points of B on the line ` are essential. Then ` intersects B in t (mod p)points.

When each line intersects B in t (mod q) points, then the characterization of Bis immediate.

Proposition 22.17. Let B be a minimal weighted t-fold blocking set in PG(2, q) ofsize tq+ t+ k, where t+ k < (q+ 3)/2, k ≥ 2. Assume that each line intersects Bin t (mod q) points. Then B is a sum of t (not necessarily different) lines.

Proof: Let ` be a line of PG(2, q) not contained in B. Let P ∈ ` \ B. Since allthe lines, different from `, through P contain at least t points of B, ` contains atmost t+ k points of B.Every point R of B lies on at least one line containing more than t points of B, soon a line ` containing at least t+ q points of B. Since t+ k < t+ q, the precedingparagraph implies that ` is contained in B. By Theorem 22.4, B is the sum of thisline ` and a (t− 1)-fold blocking set B∗ intersecting every line in (t− 1) (mod q)points. Repeating the above argument shows that B is a sum of t lines.

These results are enough to give a determine a lower bound on the size of a minimalweighted t-fold blocking set B in PG(2, q), q = ph, p prime, h ≥ 1.


We again assume that B does not contain any lines, for it is trivially possible toconstruct a minimal weighted t-fold blocking set in PG(2, q) by taking a sum B oft lines. Then |B| = t(q + 1).

Theorem 22.18. Let B be a minimal weighted t-fold blocking set in PG(2, q), q = ph,p prime, h ≥ 1, with |B| = tq + t+ k, t+ k < (q + 3)/2, containing no lines.Assume that h(X,Y ) is a component of f0, which can be written as h(X,Y ) =g(Xpe , Y ) with g′X 6≡ 0. Then k ≥ q+pe

pe+1 − t+ 1.

We omit the details now, the interested reader may see Proposition 3.6 in [37] forall the ideas needed.

23 Stability

We start with a result of [113], which is a generalization of the main result of [124].Let D be a set of directions in AG(2, q). A set U ⊂ AG(2, q) is called a D-set if Udetermines precisely the directions belonging to D.

Theorem 23.1. Let U be a D-set of AG(2, q) consisting of q − ε points, whereε ≤ α

√q and |D| < (q+ 1)(1−α), 1/2 ≤ α ≤ 1. Then U is incomplete, i.e. it can

be extended to a D-set Y with |Y | = q.

Proof: In the proof we follow the method of Section 11.Let U = {(ai, bi) : i = 1, ..., q − ε}, suppose (∞) ∈ D.If (y) 6∈ D then the values aiy − bi are all distinct elements of GF(q). Define theRedei polynomial of U as

R(X,Y ) =q−ε∏i=1

(X + aiY − bi) =q−ε∑j=0

rj(Y )Xq−ε−j .

Then deg(rj) ≤ j. Let Ry(X) = R(X, y), then Ry|(Xq − X) if and only if theelements in A(y) = {−aiy + bi : i = 1, ..., q − ε} are pairwise distinct, that is,when (y) 6∈ D. (We define A(Y ), a set of linear polynomials, in the similar way.)In this case let σj = σj(A(y)) be the j-th elementary symmetric polynomial of theelements in A(y), and σ∗j = σ∗j (A(y)) = σj(GF(q) \ A(y)) be the j-th elementarysymmetric polynomial of the remaining elements. Note that σj = (−1)jrj , and byinduction, using the recursive formula

σ∗j = −σj −j−1∑k=1

σkσ∗j−k,

hence in principle σ∗j can be expressed by the σi-s, so by the ri-s (i = 0, ..., j),which means that σ∗j can be considered as a polynomial in Y . Now the recursiveformula shows that degY (σ∗j ) ≤ j as well. (See also Section 25 on nuclei.)

23. Stability 163

Define the polynomial f(X,Y ) = Xε − σ∗1Xε−1 + σ∗2Xε−2 −+...+ (−1)εσ∗ε . Here

f is of total degree ε and if (y0) 6∈ D, then R(X, y0)f(X, y0) = Xq −X. For suchy0-s, we have

f(X, y0) =∏

β∈GF(q)\Ay0

(X − β),

so the curve F defined by f(X,Y ) = 0 has precisely ε distinct simple points (x, y0).So F has at least

N ≥ (q + 1− |D|)ε > (q + 1)αε

simple points in PG(2, q).Now, using Lemma 10.15 with the same α, we have that F has a linear componentX+aY − b over GF(q). Then −ay+ b 6∈ Ay if (y) 6∈ D. Let U∗ = U ∪{(a, b)}, thenR∗(X, y) = R(X, y)(X + ay − b) divides Xq − X for all (y) 6∈ D, as Xq − X =R(X, y)f(X, y) and (X + aY − b)|f(X,Y ). Hence U∗ does not determine anydirections not in D, so U∗ is also a D-set. Repeating this procedure we end upwith a D-set consisting of q points.

Comparing this to Theorem 18.14 one can see that if U ⊂ AG(2, q) determinesN ≤ q+1

2 directions and U is of size q − ε, with ε small then still we know thestructure of U .The analogue of Theorem 23.1 is the following version of the results of [69, 47]:

Theorem 23.2. Let U ⊂ AG(2, q) be a pointset of size |U | = q+ε, where ε ≤ α√q−1

and 12 + 1

4√q ≤ α ≤ 1. Suppose that there are more than α(q + 1) points on `∞,

through which every affine line contains at least one point of U ; let the complement nuclei of U∪P∞for some P∞∈∞of this pointset on `∞ be called D. Then one can find ε points of U , such that

deleting them the remaining q points will still block all the affine lines through thepoints of `∞ \D.

Proof: Let U = {(ai, bi) : i = 1, ..., q + ε}, suppose (∞) ∈ D. Define the Redeipolynomial of U as

R(X,Y ) =q+ε∏i=1

(X + aiY − bi) =q+ε∑j=0

rj(Y )Xq+ε−j .

Then deg(rj) ≤ j. Let Ry(X) = R(X, y), then (Xq−X)|Ry if and only if GF(q) ⊂A(y) = {−aiy + bi : i = 1, ..., q + ε} for the multiset A(y), that is, when (y) 6∈ D.Similarly, let A(Y ) = {−aiY + bi : i = 1, ..., q + ε}, a set of linear polynomials. Inthis case let σj = σj(A(y)) be the j-th elementary symmetric polynomial of theelements in A(y), and σj = σj(A(y)) = σj(A(y) \ GF(q)) be the j-th elementarysymmetric polynomial of the “extra” elements. Note that σj = (−1)jrj , and likein Section 11, we have σj = σj , and we can define

σj(Y ) def= σj(Y ) = (−1)jrj(Y ).


Define the polynomial f(X,Y ) = Xε − σ1Xε−1 + σ2X

ε−2 −+...+ (−1)εσε. Heref is of total degree ε and if (y0) 6∈ D then R(X, y0) = (Xq −X)f(X, y0). For suchy0-s, we have

f(X, y0) =∏

β∈Ay0\GF(q)

(X − β),

so the curve F defined by f(X,Y ) = 0 has precisely ε distinct simple points (x, y0).So F has at least

N ≥ (q + 1− |D|)ε > (q + 1)αε

simple points in PG(2, q).Now, using Lemma 10.15 with the same α, we have that F has a linear componentX + aY − b over GF(q). Then −ay + b has multiplicity at least two in A(y) if(y) 6∈ D. Now (Xq −X)(X + ay− b) divides R(X, y) for all (y) 6∈ D, as R(X, y) =(Xq − X)f(X, y) and (X + aY − b)|f(X,Y ). Suppose that the point (a, b) 6∈ U .Then counting the points of U on the lines connecting (a, b) to the points of `∞\D,we find at least 2|`∞ \D| ≥ q + 1 + ε points (at least 2 on each), a contradiction.Hence (a, b) ∈ U , and one can delete (a, b) from U . Repeating this procedure weend up with a set consisting of q points and still not determining any direction in`∞ \D.

Usually it is difficult to prove that, when one finds the “surplus” element(s), thenthey can be removed, i.e. they were there in the original set. Here the “meaning”of a non-essential point (i.e. each line through it is an ≥ 2-secant) helped.

Exercise 23.3. Prove that if S1, S2 ⊆ PG(2, q) are two pointsets, with characteristicvectors vS1 ,vS2 and weight (or line-intersection) vectors mS1 = AvS1 ,mS2 =see Section 12

AvS2 , thenA = the inci-dence matrix ofthe plane ||mS1 −mS2 ||2 = ||vS1 − vS2 ||+ q · ||vS1 4 vS2 ||

where ||x|| =√∑

x2i and 4 denotes symmetric difference.

Note that it means that the (euclidean) distance of the line-intersection vectors ofany two pointsets is at least

√q.

The next sections are taken from Szonyi-Weiner [130], which is a collection ofwonderful stability results, most (but not all) of them based on the new resultantmethod of the authors.

23.1 Stability theorems for small blocking sets and the Szonyi-Weiner method

If we delete a few (say, ε) points from a blocking set, then we get a point setintersecting almost all except of at most a few (εq) lines. A point set which is“close” to be a blocking set is a point set that intersects almost all lines. A stability

23. Stability 165

type question asks whether all point sets having only a few 0-secants can beobtained from a blocking set by deleting a few points of it. Of course, if we onlyhad a few 0-secants, then by adding one-one point to each of these lines (hence intotal still only a few points), we obtain a blocking set. So we have to be carefulby choosing sensible bounds in such stability theorems.The next result is about the stability of the smallest blocking sets, that is, aboutthe lines.

Result 23.4. (Erdos and Lovasz, [61]) A point set of size q + 1 in PG(2, q), withless than q

√q − q skew lines always contains at least q −√q points from a line.

Note that this result is sharp, since if we delete√q points from a Baer subplane,

then the resulting set will have q√q − q skew lines exactly and it has at most√

q + 1 collinear points.By the result of Blokhuis and Brouwer (Exercise 12.5) we know that through anessential point of a blocking set B there pass at least 2q + 1− |B| tangents. Thisresult helps to estimate the number of 0-secants we get by deleting an essentialpoint from a small blocking set. So if we delete ε essential points from a smallblocking set then we get at least ε q2 skew lines. The main result of this section isthe following theorem.

Theorem 23.5. Let B be a point set in PG(2, q), q ≥ 81, of size less than 32 (q+1).

Denote the number of 0-secants of B by δ, and assume that it is at most 13q√q,

when |B| < q +√q and at most q2

3(|B|−q) otherwise.

Then B can be obtained from a blocking set by deleting at most 2δq points of it.

Note that the assumptions in this theorem are chosen such that δ ≤ 13q√q and

δ ≤ q2

3(|B|−q) will both hold whenever |B| < 32 (q + 1).

When the size of B is large, the theorem above is immediate.

Lemma 23.6. Assume that for the size of B, 54q ≤ |B| < 3

2 (q + 1) holds andδ ≤ q2

3(|B|−q) , q ≥ 81. Then B can be obtained from a blocking set by deleting atmost 2δ

q points of it.

Proof: Since any two lines intersect, we can add at most δ2 points to B so that we

get a blocking set B′. This blocking set will have size at most |B|+ q2

6(|B|−q) . Thisfunction takes its maximum at |B| = 5

4q when |B| is in the interval [54q,32q + 1]

(and q ≥ 81). Hence |B′| ≤ 2312q and so by Result 12.5, through each of its essential

point there pass at least 112q tangents. Since δ ≤ 4

3q, this shows that indeed wecould have added at most 4

3q/112q = 16 points to B to obtain a blocking set of

size at most 32q + 17. Again by Result 12.5, through each of its essential point

there pass at least 12q − 16 tangents, hence again Result 12.5 implies that indeed


we needed less than 5 points (here we use again that q ≥ 81). Repeating thisargument once more, we see that we need at most 2 (≤ 2δ

q ) points.

As the above lemma shows, Theorem 23.5 is weak when the size of B is in theupper part of the given interval. The aim of the next two sections is to improveon this result, when |B| is relatively large.To prove Theorem 23.5 we need the following two lemmas.

Lemma 23.7. Let S be a point set of size less than 2q in PG(2, q), q ≥ 81, andassume that the number of external lines δ of S is less than (q2 − q)/2.

(1) Denote by s the number of external lines of B passing through a point P .Then (2q + 1− |S| − s)s ≤ δ.

(2) If |S| ≤ 54q and δ ≤ 1

3

√qq, then through any point there are at most

δ2q+1−|S| + 1

2 or at least 2q + 1− |S| − δ2q+1−|S| −

12 external lines to S.

Proof: Let `∞ be a line intersecting S in k > 0 points and let (∞) ∈ S. Fur-thermore, assume that |S| − k 6= q. This can be done, otherwise every line thatintersects S would intersect it in k points. Since |S| < 2q, counting the pointsof S through a point in S, we get that k = 2. Hence S is a hyperoval, but thiscontradicts our assumption on δ.Assume that there is an ideal point different from (∞) through that there passt affine lines intersecting S \ `∞ in at least 1 point. Denote by nt+h, the numberof ideal points different from (∞) through that there pass (t + h) affine linesintersecting S \ `∞. Hence by Lemma 12.6,

∑q−th=1 hnt+h ≤ (|S| − k − t)(q − t).

Suppose that P is a point of `∞ \ S and assume that through P there pass q − taffine lines not intersecting S \ `∞. Denote by r(q−t)−h, the number of ideal pointsdifferent from (∞) through that there pass (q − t)− h affine skew lines. Throughthese points there pass t+h affine lines intersecting S\`∞, hence

∑q−th=1 hr(q−t)−h =∑q−t

h=1 hnt+h and from above this is at most (|S|−k− t)(q− t). Hence by countingthe number of skew lines through the points of `∞ \ S, we get a lower bound onδ:

(q + 1− k)(q − t)−q−t∑h=1

hr(q−t)−h ≤ δ. (2.4)

Using∑q−th=1 hr(q−t)−h =

∑q−th=1 hnt+h ≤ (|S| − k − t)(q − t) and substituting

s = (q − t), we get the first part of the lemma:

(2q + 1− |S| − s)s ≤ δ. (2.5)

To prove the second part of the lemma, we estimate the discriminant of (2.5) (frombelow) by (2q + 1 − |S|) − ( 2δ

2q+1−|S| + 1) (here we use the fact that δ ≤ 13q√q).

23. Stability 167

Hence s ≤ δ2q+1−|S| + 1

2 or s ≥ 2q + 1− |S| − δ2q+1−|S| −

12 .

Lemma 23.8. Assume that |B| ≤ 54q. Let N be the set of points through that there

are at least 2q + 1 − |B| − δ2q+1−|B| −

12 external lines to B. Then B ∪ N is a

blocking set.

Proof: Assume to the contrary that there exists a line ` external to B, so thatthrough each point of ` there pass less than 2q + 1− |B| − δ

2q+1−|B| −12 external

lines of B. Then by Lemma 23.7 (2), through each of these points there pass atmost δ

2q+1−|B| + 12 external lines (including `). Counting the external lines of B

through the points of `, we get an upper bound on δ.

δ ≤ 1 + (q + 1)(δ

2q + 1− |B|− 1

2) = δ +

δ(|B| − q)2q + 1− |B|

− q + 12

+ 1 (2.6)

This is a contradiction, since δ ≤ q2

3(|B|−q) . Hence on each of the external lines,there is at least one point through that there pass at least 2q+1−|B|− δ

2q+1−|B|−12

external lines.

Proof of Theorem 23.5: By Lemma 23.6, we may assume that |B| ≤ 54q. We

construct the point set B′, by adding the points through that there pass at least2q + 1 − |B| − δ

2q+1−|B| −12 external lines to B. Since δ ≤ 1

3q√q and |B| ≤ 5

4q,there pass at least 11

36q skew lines through such points. Counting the external linesthrough these points, we see at least 11

36q + ( 1136q − 1) + (11

36q − 2) + . . . 0-secants,which shows that there were less than d 1

4qe such points (here we use that q ≥ 81).Hence |B′| < 3q/2. By Lemma 23.8, B′ is a blocking set. Let ε be the minimumnumber of points we need to add to B in order to obtain a blocking set B∗.(|B∗| ≤ |B′| < 3q/2). By Result 12.5, through each essential point of this blockingset (such are the points of B∗ \ B) there pass at least 2q + 1 − |B| − ε ≥ q/2tangents, i.e. external lines to B. Hence in total B has at least εq/2 external lines,which shows that ε ≤ 2δ/q.

23.2 Stability theorems for blocking sets: the prime case

In this section we will improve on the stability theorem of the previous section,when the order of the plane is prime. Blokhuis ([33]) proved that a blocking setwith less than 3

2 (q+1) points must contain a line. We are going to show that if Bis a point set with |B| ≤ 3

2 (q+ 1)− ... that has at most ε(q+ 1) 0-secants, then itcontains a huge part of a line. Here ε can even be cq, where c is a small constant.The proof is motivated by [33].


Lemma 23.9. Let B be a point set in PG(2, q), |B| < 32 (q + 1). Assume that there

are at most ε(q+ 1) skew lines to B. Then the total number τ of 1-secants of B isat least (q + 1)(2q − |B| − 2ε). Hence there is a point P of B so that there are atleast 2

3 (2q − |B| − 2ε) 1-secants through P .

Proof: Take a 0-secant ` of B. If there is no such line then B is a blocking set and(by Result 12.5) through any essential point of B there pass at least 1

2 (q + 1)− 1tangents. Let the points of ` be denoted by P1, . . . , Pq+1 and let νi be the numberof 0-secants, τi be the number of tangents through Pi. Looking at B from Pi onegets that q − (νi + τi) ≤ (|B| − τi)/2, which implies that 2νi + τi ≥ 2q − |B|.Summing over all i we get that (q + 1)(2q − |B|) ≤ 2ε(q + 1) + τ , from whichτ ≥ (q + 1)(2q − |B| − 2ε) follows. On the other hand, if we add up the numberof tangents at the points of B, we get τ , so there is a point which has at least theaverage number of tangents.

Theorem 23.10. Let ∆ be the integer part of√

2ε(q + 1) − 1. Let B be a set ofpoints of PG(2, q), q = p prime, that has at most ε(q + 1) 0-secants for someε < 1

4 (q − 6). Suppose that |B| < 32 (q + 1−∆). Then there is a line that contains

at least q − 2ε points of B.

Proof: Choose the coordinate system in such a way that (∞) is a point of Bwith at least 2

3 (2q− |B| − 2ε) tangents, one of them being the line at infinity. LetU = {(ai, bi) : i = 1, . . . , |B|−1} be the affine part of B. The 0-secants of B can bewritten as Y = mjX + bj , j = 1, . . .. Consider the polynomial a(x, y) of smallestdegree ∆, which vanishes at the points (bj ,mj), j = 1, . . . , ε(q + 1). By Exercise10.2, ∆ ≤

√2ε(q + 1)− 1. Now write up the polynomial

R(X,Y ) =

(∏i

(X + aiY − bi)

)a(X,Y ).

The first product is the Redei polynomial of U . This polynomial R vanishes forevery (x, y), hence it can be written as

R(X,Y ) = (Xq −X)f(X,Y ) + (Y q − Y )g(X,Y ),

where deg(f),deg(g) ≤ |B| − 1 − q + ∆. As in Blokhuis [33], consider the termsof highest degree of this equation and substitute Y = 1 in it. Then we get apolynomial equation

r∗(X) =(∏

(X + ai))a∗(X) = Xqf∗(X) + g∗(X),

where Xq - g∗(X). We may suppose that f∗ and g∗ are coprime, since otherwisewe could divide by their greatest common divisor and obtain an equation of the

23. Stability 169

same type with smaller degrees. Denote by s the maximum of the degrees of f∗

and g∗ after this division. Now we can continue copying Blokhuis’ proof. The rootsof r∗(X) in GF(q) are also roots of Xf∗(X) + g∗(X). The multiple roots of r∗(X)in GF(q) are also roots of Xq(f∗(X))′+(g∗(X))′. The roots not in GF(q) are rootsof a∗(x). Hence

r∗(X)|(Xf∗(X) + g∗(X))((f∗(X))′g∗(X)− (g∗(X))′f∗(X))a∗(X). (2.7)

If the polynomials on the right hand side of (2.7) are non-zero, then comparingthe degrees gives q + s ≤ s + 1 + 2s − 2 + ∆, that is s ≥ (q + 1 − ∆)/2. Sinces ≤ |B| − 1− q + ∆, it gives that |B| ≥ 3

2 (q + 1)− 32∆; which is a contradiction.

The third term on the right hand side of (2.7) cannot be the zero polynomial, sincethe terms of highest degree of a(X,Y ) formed a homogeneous polynomial and so(Y − 1) cannot be a factor of it.If the first term on the right hand side of (2.7) is the zero polynomial then r∗(X)is divisible by (Xq −X). Since a∗(X) has degree at most ∆, the remaining q −∆factors of (Xq −X) must come from the product

∏(X + ai). Geometrically this

would imply that through the point (∞) there passed at most ∆ + 1 tangents,which contradicts the choice of (∞). (Here we use that ∆ + 1 < 2

3 (2q− |B| − 2ε).)If the second term is zero, then, since f∗ and g∗ are coprime, f∗(X)|(f∗(X))′ andsimilarly g∗(X)|(g∗(X))′. Hence (f∗(X))′ = (g∗(X))′ = 0. For q = p prime, itimplies that either |B| ≥ 2q+1−∆ (which is not possible by our upper bound on|B|) or aXq + b divides r∗(X). Since aXq + b = (aX+ b)q, and at most ∆ of thesefactors can come from a∗(X), it implies that there is a line ` (through (∞)) thatcontains at least q+1−∆ points of B. Finally, assume that |`∩B| = q+1−k, k ≤ ∆.Then the 0-secants pass through the k missing points of `. Since |B| ≤ 3

2q+1− 32∆,

the number of 0-secants is at most k(q − ( 32q + 1− 3

2∆− q − 1 + k)) ≤ 12k(q + 1).

Hence k ≤ 2ε.

Note that there is no restriction on how small ε can be. Chosing ε to be slightlyless than 1

q+1 one gets Theorem 21.4 of Blokhuis.

Remark 23.11. (1) The condition ε < 14 (q − 6) is not really necessary. Indeed,

fix the size of B. Determine the largest ε so that |B| < 32 (q + 1 − ∆). Then

ε < 14 (q − 6) will automatically hold if |B| > q/2 (a more exact bound comes

from |B| < 32 (q + 1 − ∆)). On the other hand, if a set B has size cq for some

c < 1 then the number of 0-secants is at least (1− c)q(q+ 1). This can be seen bycounting 0-secants through points of a fixed 0-secant. Hence our theorem gives anon-trivial bound only if (1− c)q(q+1) < ε(q+1), where |B| = cq = 3

2 (q+1−∆).Roughly speaking this gives that for a fixed c < 1, the value of ε has to be smallerthan (1 − 2

3c)2q/2, which gives the equation 4c2 + 6c − 9 = 0 for the critical

value of c. The positive root of the equation is (−3 +√

45)/4. Since it is biggerthan 1

2 , this shows that ε = (q − 6)/4 is never reached. Let us underline that


the result gives a non-trivial stability theorem also for sets of size cq, c < 1 (butc > (−3 +

√45)/4 ≈ 0.927).

(2) However, even when |B| = q + 1, one cannot expect only ε missing points.Indeed, if we take an affinely regular δ-gon K, inscribed in an ellipse of AG(2, q),and the set of q + 1 − δ directions N not determined by K, then the number of0-secants od B = K ∪N is exactly δ(q − δ), so when δ ≥ √

q, then this is smallerthan δq. For example, if δ = (q+1)/2, then δ(q−δ) is roughly qδ/2. The bound onε in Theorem 23.10 does not allow this value, but it shows that the combinatorialargument at the end of the proof cannot really be improved.

23.3 Stability theorems for blocking sets: the non-prime case

As mentioned before, in this section we are going to improve on Theorem 23.5,when the size of our point set is relatively large. The proof is guided by the ideas ofthe paper [123], where it is shown that each line intersects a small minimal blockingset in 1 mod p points. In case of an almost blocking set B, a very similar argumentto those in [123] will show that B contains a point set intersecting almost everyline in 1 mod p points and so its size cannot be too large. Finally, using resultants(see Section 9.5), we show that we can add a few points to B so that we obtain ablocking set.

Theorem 23.12. Let B be a point set in PG(2, q), q = ph, h > 1, q ≥ 81, andsuppose that the number of 0-secants, δ, of B is at most 1

1000pq. Assume that|B| < 3

2 (q+1)− (p+3)2

√2δ. Then B can be obtained from a blocking set by deleting

at most 2δq points of it.

Note that when |B| > q+ 10003

qp , then this theorem is an improvement on Theorem

23.5.To prove the theorem our main aim is first to show that B can be embedded in ablocking set of size less than 3

2 (q+1), and then the result will follow immediately.For the points (av, bv, cv) of B, consider the three-variable Redei polynomial.

RB(X,Y, Z) =|B|∏v=1

(cvX + avY − bvZ) =|B|∑j=0

rj(Y,Z)X |B|−j . (2.8)

Note that rj(Y, Z) is a homogeneous polynomial of degree j. As in the case ofthe two variable Redei polynomial, this polynomial also encodes the intersectionmultiplicities of B and the lines:

For a fixed (z, y, 0) ∈ `, the element x ∈ GF(q) is an r-fold root of R(X, y, z) ifand only if the line with equation zY = yX + xZ intersects U in exactly r points.

Hence the line wY = uX + vZ is a 0-secant of B, if and only if R(u, v, w) 6= 0.The following lemma is similar to Exercise 10.2.

23. Stability 171

Lemma 23.13. There exists a three-variable homogeneous polynomial a(X,Y, Z)over GF(q), of degree at most

√2δ−1, such that for each skew line wY = uX+vZ

to B, where w, u, v ∈ GF(q), a(u, v, w) = 0.

Delete each linear component (ciX + aiY − biZ) of a and add the correspond-ing projective point (ai, bi, ci), if it was not in B, to B. We will denote the newpolynomial by a1 and the new set by B1. Hence |B1| < 3(q + 1)/2− p+1

2

√2δ − 1.

First of all, note that if B1 has at most one 0-secant, then by putting one pointon this line, we obtain a blocking set of size less than 3

2 (q + 1) containing B. Sofrom now on we assume that B1 has at least two 0-secants, and we choose our newcoordinate system in such a way, that the lines Z = 0 and X = 0 are skew to B1.Note that, in this new coordinate system, B1 is an affine point set. Let a∗(X,Y, Z)be the polynomial a1(X,Y, Z) in this new coordinate system. From now on we willsubstitute Z = 1, hence we will consider the two variable (affine) Redei polynomialRB1(X,Y ) and the two variable polynomial a∗(X,Y ). By the construction of a∗,the polynomial RB1a

∗ vanishes for all (x, y) ∈ GF(q) × GF(q). If there is a factord(X,Y ) of a∗(X,Y ), so that (RB1

a∗

d )(x, y) = 0 for every (x, y) pair, then we deletethis factor from a∗. We repeat this process until there is such factor.

Remark 23.14. Hence we obtained a polynomial a(X,Y ), such that t := dega =dega − (|B1| − |B|) ≤

√2δ − 1, deg(RB1a) < 3(q + 1)/2 − p+1

2

√2δ − 1, and

a(X,Y ) has no linear component. Furthermore, RB1(x, y)a(x, y) = 0 for everypair (x, y) ∈ GF(q)× GF(q), and a is minimal in the sense that this property willnot hold if we delete any factor of a.

Hence, we can write RB1a as:

RB1(X,Y )a(X,Y ) = (Xq −X)f(X,Y ) + (Y q − Y )g(X,Y ), (2.9)

where deg(f), deg(g) ≤ |B1| − q+deg(a) = |B1| − q + t < (q + 1)/2− p+12

√2δ.

Denote by C(X,Y ) the product of the common factors (with multiplicity) off(X,Y ) and g(X,Y ).

RB1(X,Y )a(X,Y ) = C(X,Y )((Xq −X)f(X,Y ) + (Y q − Y )g(X,Y )), (2.10)

where f and g have no common factors. Furthermore, deg(f), deg(g) and deg(C)are at most |B1| − q+deg(a) = |B1| − q + t < (q + 1)/2 − p+1

2

√2δ. By the

minimality of a, the polynomials C and a cannot have a common factor, henceC(X,Y )|RB1(X,Y ).

Corollary 23.15. All the factors of C(X,Y ) have multiplicity one and are of form(X + akY − bk). Furthermore, the points (ak, bk, 1) corresponding to the linearfactors of C(X,Y ) are in B1.


Let C denote the set of these points, hence |C| =degXC. From above, C ⊂ B1.Denote the point set B1\C by B2. Note that B2 is an affine point set and constructthe Redei polynomial RB2(X,Y ) of B2. Then:

RB2(X,Y )a(X,Y ) = (Xq −X)f(X,Y ) + (Y q − Y )g(X,Y ), (2.11)

As the next lemma shows, through each point of C, there pass a few 1-secants only.

Lemma 23.16. If (X+akY −bk) is a factor of C(X,Y ) then the number of 1-secantsof B1 through (ak, bk, 1) (∈ B1) is at most t+ 1.

Proof: We have seen that (X + akY − bk) cannot divide a(X,Y ). Hence byBezout’s theorem, a and (X + akY − bk) have at most t common points.For any value y, x = bk − aky is an at least 2-fold root of the right hand side ofequation (2.10), and so this also holds for the left hand side. Hence from above,there are at least q− t values y, such that x = bk−aky is an at least 2-fold root ofRB1(X, y). For these values, the line Y = yX + x (through (ak, bk, 1)) intersectsB1 in at least two points. Now the lemma follows, since the lines through (ak, bk, 1)are either of type Y = yX + x or vertical.

Lemma 23.17. For any (x, y) ∈ GF(q)×GF(q), if f(x, y) = 0, then g(x, y)a(x, y) =0.

Proof: Suppose that for a fixed Y = y, x is a root of f(X, y). Then, by equation(2.11), the intersection multiplicity of RB2(X,Y )a(X,Y ) and the line Y = y in(x, y) is at least two. Now assume that a(x, y) 6= 0, then by the properties of Redeipolynomials we have that the line Y = yX+x intersects B2 in at least two points.Hence the intersection multiplicity of the line X = x and RB2(X,Y ) in (x, y) isalso at least two, and so by equation (2.11), g(x, y) = 0.

Lemma 23.18. Let ri(X,Y ) be irreducible polynomials over the algebraic closureof GF(q) dividing f . If ∂ri

∂X 6= 0 then∑i degXri ≤

qp .

Proof: Let GF(q) denote the algebraic closure of GF(q), and assume thatdegXri = si. Then for any i, the sum of the intersection multiplicities I(P, ri∩ `P )over GF(q)×GF(q) is exactly siq, where `P denotes the horizontal line through P .First we show that the above sum over GF(q)×GF(q) is less than siq−sipt. Assumeto the contrary that the sum of these intersection multiplicities is at least siq−sipt.To get the number of points of ri over GF(q) × GF(q) without multiplicities, wehave to subtract the number of intersections of ri and ∂ri

∂X . Hence by Bezout’stheorem, we get that ri has at least siq − sipt− si(si − 1) GF(q)-rational points.By Lemma 23.17, these points are also on ga. Now we show that ri 6 |ga. Since f

23. Stability 173

and g have no common factors, if ri|ga, then ri|a. Hence ri also divides the righthand side of equation (2.11) and so it divides (Y q − Y )g(X,Y ). Again as f and ghave no common factors, ri|(Y q−Y ) and so ri = Y −m, m ∈ GF(q); but this is notpossible since ri|a and a has no linear component. Hence we can apply Bezout’stheorem on the polynomials ri and ga:

siq − sipt− si(si − 1) ≤ si(q + 1

2− (p− 1)

2

√2δ).

After simplifying the inequality, we get that (q + 1)/2− p+12

√2δ ≤ si; which is a

contradiction since si ≤ degxf < (q + 1)/2− p+12

√2δ.

Hence the sum of the intersection multiplicities I(P, ri∩`P ) over (GF(q)\GF(q))×GF(q) is at least sipt. For any fixed Y = y, RB2(X, y) splits into linear factors,hence f(X, y) can have at most t roots that are not from GF(q). So the sum of theintersection multiplicities I(P, f ∩ `P ) over (GF(q) \ GF(q))× GF(q) is at most tq,from which pt

∑i si ≤ tq follows.

Hence for a fixed Y = y, f(X, y) is almost a p-th power.

f(X, y) = (wy(X))puy(X), (2.12)

where degXuy ≤ q/p. Using equations (2.11) and (2.12), we get that for anyy ∈ GF(q),

RB2(X, y)a(X, y) = (Xq −X)f(X, y) = (Xq −X)(wy(X))puy(X). (2.13)

As the next lemma shows, this equation helps to bound the size of B2.

Lemma 23.19. The size of B2 is less than q + 7 qp .

Proof: We will estimate the sum S of the intersection multiplicities I(P, f ∩ rP ),where rP are the horizontal lines through those points P ∈ GF(q)× GF(q), wherethese intersection multiplicities are at least p.For a fixed value y ∈ GF(q), f(X, y) = (wy(X))puy(X) splits into linear factorsover GF(q), hence the sum S of the above intersection multiplicities is at leastq(degX(f)−maxydeg(uy)). By Lemma 23.18 and by equation (2.13), this is atleast q(|B2|+ t− q − q/p).Now we will give an upper bound for S. First of all note that

I(P, f ∩ rP ) = I(P,RB2 ∩ rP ) + I(P, a ∩ rP )− 1,

for all points P . By Bezout’s theorem, the sum of the intersection multiplicitiesI(P, a ∩ rP ) over all points P ∈ GF(q) × GF(q) is at most qdega = qt. To give an


upper bound for the sum of the intersection multiplicities I(P,RB2 ∩ rP ) (whenfor each P , I(P, f ∩ rP ) ≥ p), we will distinguish between the points P accordingas I(P,RB2 ∩ rP ) ≥ (p+ 2)/2 and I(P,RB2 ∩ rP ) < (p+ 2)/2.By the fundamental properties of the Redei polynomial, the first case considersthose points P (x, y) that correspond to lines Y = yX+x intersecting B2 in at least(p+2)/2 points. Hence for these points P , the sum of the intersection multiplicitiesI(P,RB2 ∩ rP ) is exactly the number of incident point-line pairs, where the pointlies in B2 and the line is a ≥ (p+ 2)/2-secant. The number of ≥ (p+ 2)/2-secantsthrough a point of B2 is at most 2(|B2|−1)/p, hence the number of these incidentpoint-line pairs is at most 2|B2|(|B2| − 1)/p.Now we bound the sum of the intersection multiplicities I(P,RB2 ∩ rP ) over thepoints P , for that I(P,RB2 ∩ rP ) < (p + 2)/2 (and I(P, f ∩ rP ) ≥ p). For thesepoints, I(P, a∩rP ) ≥ p/2 and so I(P, a∩ f) ≥ p/2. Hence by Bezout’s theorem thenumber of such points P is at most (degfdega)/(p/2). For each of these points P ,I(P,RB2 ∩ rP ) ≤ (p+ 1)/2, hence the sum of these intersection multiplicities is atmost p+1

p degfdega = p+1p (|B2| − q)t. Hence:

q(|B2|+ t− q − q

p) ≤ S ≤ 2|B2|(|B2| − 1)

p+p+ 1p

(|B2| − q)t+ qt.

Subtract q(|B2|+ t− q − qp ) from both sides of the above inequality.

0 ≤ 2|B2|(|B2| − 1)p

+p+ 1p

(|B2| − q)t− q(|B2| − q − q

p)

When q = ph, h > 2, (using that |B2| ≤ 32q), this inequality yields immediately

that |B2| < q + 7 qp .

So now we only consider the case q = p2. For |B2|, the right hand side of theabove inequality is a quadratic expression which is positive when |B2| = q andnegative when |B2| = q + 7 qp (the lemma is only interesting if q + 7 qp <

32q (since

|B2| ≤ 32q and so we may suppose that p > 13). Divide the above inequality by

2p , so the coefficient of the quadratic term is 1 and the constant term is ((p +1)qt+ q2p+ q2)/2. It follows from the previous argument that one of the roots ofthis quadratic expression is less than q + 7 qp <

32q; hence the other is larger than

((p+1)qt+ q2p+ q2)/3q. This is larger than 32q, when p > 13 and since |B2| ≤ 3

2q,we get that |B2| < q + 7 qp .

Lemma 23.20. The number of 0-secants of B2, δ′, is at most 140q

√pq.

Proof: Originally B had at most 11000pq 0-secants. To obtain B2 we added some

points and deleted the points of C. By Lemma 23.16, after deleting the points

of C, we may get at most |C|(t + 1) ≤ q2

√2

1000pq new 0-secants. Hence δ′ ≤

23. Stability 175

q2

√2

1000pq + 11000pq < q

√1

2000pq + q 11000

√pq.

Lemma 23.21. The number of 0-secants of B2 through a point is either at mostδ′

2q+1−|B2| + 150p or at least 2q + 1− |B2| − ( δ′

2q+1−|B2| + 150p).

Proof: By Lemma 23.7, if through a point P there pass s external lines of B2,then (2q + 1− |B2| − s)s ≤ δ′. Estimate the discriminant (from below) by

(2q + 1− |B2|)− (2δ′

2q + 1− |B2|+

250p). (2.14)

Note that, 2q + 1 − |B2| ≥ 0.5q, and so we get that s ≤ δ′

2q+1−|B2| + 150p or

s ≥ 2q + 1− |B2| − ( δ′

2q+1−|B2| + 150p).

Proof of Theorem 23.12. First we show that B can be embedded in a blocking setof size less than 3

2 (q + 1).Construct the point set B1 (see the beginning of this section). As we have alreadynoticed there, if B1 has only at most one 0-secant then adding one point to B1,we get a blocking set of size less than 3

2 (q + 1). Otherwise construct the point setB2.By Lemma 23.19, B2 has less than q+7 qp points. Hence if the number of 0-secantsδ′ of B2 (which is at least the number of 0-secants of B) is at most 1

21pq, thenTheorem 23.5 finishes our proof.Finally, we show that δ′ is indeed smaller than 1

21pq. Assume to the contrarythat δ′ > 1

21pq. By the construction of B2, all the 0-secants of B2 that are notskew to B pass through one of the points of C. There are at least δ′ − 1

1000pqsuch 0-secants, hence there exists a point P ∈ C through that there pass at leastδ′

|C| −pq

1000|C| 0-secants of B2. We show that this contradicts Lemma 23.21. To this,note that |B2| ≥ 39

40q. Otherwise counting the number of 0-secants through thepoints of a skew line to B2, we get more than 1

40q2 ≥ 1

40q√pq skew lines; which is

a contradiction by Lemma 23.20.First we show that δ′

|C| −pq

1000|C| >δ′

2q+1−|B2| + 150p. We may assume, that 2q +

1 − |B2| ≥ dq, where 4140 ≥ d > 1

2 . Recall that B1 = B2 ∪ C and |B1| ≤ 32q, so

|C| ≤ (d − 12 )q. Hence it is enough to show that δ′

(d− 12 )q

− p1000(d− 1

2 )> δ′

dq + 150p.

Now δ′ > 121pq; hence we need to see that 1

21 ( 1/2

d(d− 12 )

) − 11000(d− 1

2 )> 1

50 . Afterdifferentiating the left hand side of the above inequality, one can see that it isdecreasing in the interval (1

2 ,4140 ]. So for d = 41

40 , it takes its minimum, which islarger than 1

50 .To get a contradiction with Lemma 23.21, now we only have to show that thenumber of 0-secants of B2 through any point of C is less than 2q + 1 − |B2| −


( δ′

2q+1−|B2| + 150p). By Lemma 23.16, there are at most t + 1 <

√2pq1000 0-secants

through any point of C. Using Lemma 23.20 and that |B2| ≤ 32q, we get the

t+ 1 < 2q + 1− |B2| − ( δ′

2q+1−|B2| + 150p).

Hence by Theorem 23.5, we can add at most 221p < p points to B2, so that we

obtain a blocking set. Note that we constructed B2 from B by adding at mostdega <

√2δ points to B and deleting some points. Hence in total we can add at

most√

2δ + p points to B in order to get a blocking set. This blocking set willhave size less than 3

2 (q + 1) (here we assume that δ > 1, see the beginning of thissection). Hence similarly as at the end of the proof of Theorem 23.5, we get thatε ≤ 2δ

q .

23.4 Stability theorem for sets of even type

A set of even type U is a point set intersecting each line in even number of points.By counting the points of S on the lines through a point of S and on the linesthrough a point not in S, one can see immediately that q must be even. Hencefrom now on in this section we will assume that 2|q. The smallest sets of eventype are the hyperovals, they have q + 2 points. All other examples known tothe authors have at least q +

√q points; see Korchmaros, Mazzocca [88]. They

constructed (q + t)-sets of type (0, 2, t) for t ≥ √q. More examples and a proof

that the t-secants are concurrent for such sets can be found in Gacs, Weiner [70].It is easy to see that the symmetric difference of any two sets of even type is a setof even type.

A set of almost even type is a point set having only a few odd-secants. If we delete apoint from a set of even type U then the new point set will have q+1 odd-secants,and of course this will be also the case when we add a point to U . If we modify εpoints then we will get at least ε(q+1− (ε−1)) and at most ε(q+1) odd-secants.In this section we are going to prove the following stability theorem.

Theorem 23.22. Assume that the point set H in PG(2, q), 16 < q even, has δodd-secants, where δ < (b√qc + 1)(q + 1 − b√qc). Then there exists a unique setH′ of even type, such that |(H ∪H′) \ (H ∩H′)| = d δ

q+1e.

Remark 23.23. Note that a complete arc of size q−√q+1 has (√q+1)(q+1−√q)

odd-secants, which shows that Theorem 23.22 is sharp, when q is a square. (Sincethe smallest sets of even type are hyperovals.) For the existence of such arcs see[48], [63] and [85].

Let `∞ be the line at infinity intersecting the point set M in even number ofpoints, furthermore let M\ `∞ = {(av, bv)}v and M∩ `∞ = {(yi)}i. Consider the

23. Stability 177

following polynomial:

g(X,Y ) =|M\`∞|∑v=1

(X+avY−bv)q−1+∑

yi∈M∩`∞

(Y−yi)q−1+|M| =q−1∑i=0

ri(Y )Xq−1−i,

(2.15)Note that degri ≤ i.

Lemma 23.24. Assume that the line at infinity contains even number of points ofM. Through a point (y) there pass s odd-secants of M if and only if the degree ofthe greatest common divisor of g(X, y) and Xq −X is q − s.

Proof: To prove this lemma we only have to show that x is a root of g(X, y) ifand only if the line Y = yX + x intersects M in even number of points. To this,for the (x, y) pairs, one has to count the parity of zero and non-zero terms.

Remark 23.25. Assume that the line at infinity is an even-secant and suppose alsothat there is an ideal point, different from (∞), with s odd-secants. Let nh denotethe number of ideal points different from (∞), through that there pass s − h odd-secants of the point set M. Then Lemma 23.24 and Corollary 12.6 imply that∑s−1h=1 hnh ≤ s(s− 1).

Lemma 23.26. Let M be a point set in PG(2, q), 16 < q even, having less than(b√qc+1)(q+1−b√qc) odd-secants. Then the number of odd-secants through anypoint is at most b√qc+ 1 or at least q − b√qc.

Proof: Pick a point P with s odd-secants and let `∞ be an even-secant of Mthrough P . (If there was no even-secant, then the lemma follows immediately.) ByRemark 23.25, counting the number of odd-secants through `∞ \ (∞), we get that:

qs− s(s− 1) ≤ δ.

This is a quadratic inequality for s, where the discriminant is larger then q + 1−2(b√qc+ 2). Hence s < b√qc+ 2 or s > q + 1− (b√qc+ 2).

Proposition 23.27. Let M be a point set in PG(2, q), 16 < q even, having less than(b√qc+1)(q+1−b√qc) odd-secants. Assume that through each point there pass atmost 3b√qc odd-secants. Then the total number of odd-secants δ of M is at mostb√qcq − q + 2b√qc+ 1.

Proof: Assume to the contrary that δ > b√qcq − q + 2b√qc + 1. Pick a pointP and let `∞ be an even-secant of M through P . Assume that there are s odd-secants through P . If there is a point Q on this even-secant through which there


pass at least s odd-secants, then choose the coordinate system so that Q is (∞).Then by Remark 23.25, counting the number of odd-secants through `, we get alower bound on δ:

(q + 1)s− s(s− 1) ≤ δ.

Since δ < (b√qc + 1)(q + 1 − b√qc), from the above inequality we get that s <b√qc + 1 (hence s ≤ b√qc) or s > q + 1 − b√qc, but by the assumption of theproposition the latter case cannot occur.Now we show that through each point there are at most b√qc odd-secants. Theabove argument and Lemma 23.26 show that on each even-secant there is at mostone point through that there pass b√qc + 1 odd-secants and through the rest ofthe points there are at most b√qc of them. Assume that there is a point R withb√qc+1 odd-secants. Since δ > b√qc+1, we can find an odd-secant ` not throughR. From above, the number of odd-secants through the intersection point of aneven-secant on R and ` is at most b√qc. So counting the odd-secants through thepoints of `, we get at most (q − b√qc)(b√qc − 1) + (b√qc+ 1)b√qc+ 1, which isa contradiction, so there was no point with b√qc+ 1 odd-secants.This means that the odd-secants form a dual b√qc-arc, hence δ ≤ (b√qc − 1)(q+1) + 1, which is a contradiction again; whence the proof follows.

Proof of Theorem 23.22 By Lemma 23.26, through each point there pass eitherat most b√qc + 1 or at least q − b√qc odd-secants. Consider the points throughthat there pass at least q−b√qc odd-secants. If such a point was in H then deleteit, otherwise add it to H. Denote this new set by H′. Note that the number ofmodified points is less than 2b√qc, hence through each point there pass at most3b√qc odd-secants of H′. Denote by δ′, the number of odd-secants of H′. Notethat when we modify (delete or add) a point then each odd-secant through thatpoint will become an even-secant and vice-versa. Hence, if through a point therepassed s ≥ q− b√qc odd-secants of H, then at least s− 3b√qc of them will be aneven secant of H′. So through this point there will pass at most q+1−(s−3b√qc)new odd-secants. Since, when q > 16, s ≥ q+ 1− (s− 3b√qc), the number of oddsecants δ′ of H′ is at most δ. By Proposition 23.27, δ′ ≤ b√qcq − q + 2b√qc + 1.Our first aim is to show that H′ is a set of even type.Let P be an arbitrary point with s odd-secants, and let `∞ be an even-secantthrough P . Assume that there is a point on `∞ with at least s odd-secant throughit. Then as in Proposition 23.27, counting the number of odd-secants through `,we get a lower bound on δ′:

(q + 1)s− s(s− 1) ≤ δ′.

This is a quadratic inequality for s, where the discriminant is larger then (q+ 2−2 δ

′+qq+1 ). Hence s < δ′+q

q+1 or s > q + 2 − δ′+qq+1 , but by the construction of H′, the

latter case cannot occur.

23. Stability 179

Now we show that there is no point through that there pass at least δ′+qq+1 odd-

secants. On the contrary, assume that T is a point with δ′+qq+1 ≤ s odd-secants. We

choose our coordinate system, so that the ideal line is an even-secant through Tand T 6= (∞). Then from above, through each ideal point, there pass less thans(≥ δ′+q

q+1 ) odd-secants. First we show that there exists an ideal point through thatthere pass exactly (s − 1) odd-secants. Otherwise by Remark 23.25, 2(q − 1) ≤s(s− 1); but this is a contradiction since by Lemma 23.26, s ≤ b√qc+ 1. Let (∞)be a point with (s− 1) odd-secants. Then as before, we can give a lower bound onthe total number of odd-secants of H′:

(s− 1) + qs− s(s− 1) ≤ δ′

Bounding the discriminant (from below) by (q+2−2 δ′+qq+1 ), it follows that s < δ′+q

q+1

or s > q + 2 − δ′+qq+1 . This is a contradiction, since by assumption, the latter case

cannot occur and the first case contradicts our choice for T .Hence through each point there pass less than δ′+q

q+1 odd-secants. Assume that ` isan odd-secant of H′. Then summing up the odd-secants through the points of `we get that δ′ < (q + 1) δ

′−1q+1 + 1, which is a contradiction. So H′ is a set of even

type.

To finish our proof we only have to show that |(H ∪H′) \ (H ∩H′)| = d δq+1e. As

we saw at the beginning of this proof, the number ε of modified points is smallerthan 2b√qc. On one hand, if we construct H from the set H′ of even type, thenwe see that δ ≥ ε(q + 1 − (ε − 1)). Solving the quadratic inequality we get thatε < b√qc+ 1 or ε > q+ 1− b√qc, but from above this latter case cannot happen.On the other hand, δ ≤ ε(q+ 1). From this and the previous inequality (and fromε ≤ b√qc), we get that δ

q+1 ≤ ε ≤ δq+1 + b√qc(b√qc−1)

q+1 . Hence the theorem follows.

For applications of Theorem 23.22 see the next section and also Section 17.2.

23.5 Number of lines meeting a (q + 2)-set

Stability of hyperovals was already studied by Blokhuis and Bruen. A hyperoval isnot only nice in the sense that it intersects each line in even number of points, butit is also extremal in the sense that considering point sets of size q+2, a hyperovalintersects the least number of lines, that is

(q+22

). The next result shows that a

(q + 2)-set intersecting a bit more than(q+22

)lines can always be constructed by

modifying a hyperoval a little bit.

Result 23.28. (Blokhuis, Bruen [44]) Let H be a point set in PG(2, q), q even,of size q + 2. Assume that the number of lines meeting H in at least 1 point is(q+22

)+ ν, where ν ≤ q

2 . Then H is a hyperoval or there exist two points P and Q,so that (H \ P ) ∪Q is a hyperoval.


Now we will try to improve on this result. Hence we will consider (q + 2)-setsintersecting a bit more than

(q+22

)+ q

2 lines.

Theorem 23.29. Let H be a point set in PG(2, q), 16 < q even, of size q + 2.Assume that the number of lines meeting H in at least 1 point is

(q+22

)+ ν, where

ν < 14 (b√qc+ 1)(q+ 1−b√qc). Then there exists a set H∗ of even type, such that

|(H ∪H∗) \ (H ∩H∗)| = d 4νq+1e.

Proof: First we show that almost all lines meeting H intersect it in 2 points. Letli denote the number of lines intersecting H in i points. We will do the standardcounting arguments to get a lower bound on l2. That is:

q+1∑i=1

li =(q + 2

2

)+ ν (2.16)

q+1∑i=1

ili = (q + 2)(q + 1) (2.17)

q+1∑i=2

i(i− 1)li = (q + 2)(q + 1) (2.18)

Calculating ((2.17)-(2.16)), and then subtracting it twice from (2.18), we get

q+1∑i=2

(i− 1)li =(q + 2

2

)− ν (2.19)

q+1∑i=3

(i− 2)(i− 1)li = 2ν (2.20)

Hence from (2.20), we have that∑q+1i=3 (i − 1)li ≤ 2ν and so using (2.19), we

have that l2 ≥(q+22

)− 3ν. Hence the total number of odd-secants is at most(

q+22

)+ ν − (

(q+22

)− 3ν) = 4ν. So Theorem 23.22 finishes our proof.

24 Jamison, Brouwer-Schrijver, affine blocking sets

Here we reconstruct the short proof of Brouwer and Schrijver [49] for the theoremon (1-)blocking sets in AG(n, q). It was proved independently and in a more generalform by Jamison [84].

Theorem 24.1. In AG(n, q) a blocking set B with respect to hyperplanes has at least|B| ≥ n(q − 1) + 1 points.

24. Jamison, Brouwer-Schrijver, affine blocking sets 181

Note that the bound is sharp: n concurrent lines form such a blocking set (butthere are several other examples as well, no classification of the minimal blockingsets can be hoped for).Proof: We can assume that the origin 0 = (0, 0, ..., 0) is in B, let B′ = B\{0}. B′blocks all hyperplanes not through 0, hence, as any such hyperplane has equationof form w1X1 + w2X2 + ...+ wnXn = 1 for some w1, ..., wn, not all zero, we havethat the polynomial

F (Y1, ..., Yn) =∏

b∈B′(b1Y1 + b2Y2 + ...+ bnYn − 1)

vanishes at every (w1, ..., wn) 6= 0. It means that for each i we have (Y q−1i −

1)|F (Y1, ..., Yn), so∏ni=1(Y

q−1i − 1)|F (Y1, ..., Yn). As F (0) 6= 0, its degree, so |B′|,

is at least n(q − 1).

After removing the origin from B (w.l.o.g. it was in there), the remaining pointsblock all hyperplanes not through 0. After dualisation, as Jamison did, one can nonzero hyper-

planesformulate Theorem 24.1 like:

Theorem 24.2. In AG(n, q) if a set B of nonzero hyperplanes cover AG(n, q) \ {0}then it has at least |B| ≥ n(q − 1) hyperplanes.

In fact Jamison proved the following more general result:

Theorem 24.3. Any covering of AG(n, q)\{0} with k-dimensional affine subspacesnot through the origin, contains at least qn−k − 1 + k(q − 1) subspaces; where0 ≤ k < n. Moreover, this bound is always sharp.

Proof: Suppose to the contrary that S is a family of nonzero k-subspaces coveringAG(n, q) \ {0} and |S| < qn−k − 1 + k(q − 1). We represent AG(n, q) as GF(qn).Let p(X) ∈ GF(qn)[X] be the product of all root polynomials pL(X) as L rangesthough the k-subspaces of S. As p(x) = 0 for all x ∈ GF(qn)∗, we have

p(X) = (Xqn−1 − 1)t(X) (1)

for some t(X). Let aj be the coefficient of the term of degree j(qn − 1) in p(X)and bj the coefficient of the term of degree j(qn − 1) in t(X). From (1) we obtainthe recursive relations

a0 = −b0and aj = bj−1 − bj for j > 0. (2)Since 0 is not covered, 0 6= p(0) = a0 so b0 6= 0. We want to show that aj = 0for j ≥ 1. From this and (2), bj = b0 6= 0 for all j, which is a contradiction as apolynomial can have finitely many nonzero terms only.


By |S| < qn−k − 1 + k(q − 1) the degree of p(X) is strictly less than

qk(qn−k − 1 + k(q − 1)) ≤ (k + 1)qn − (k + 1)qk < (k + 1)(qn − 1);

whence aj = 0 for all j ≥ k + 1. We are going to prove aj = 0 for 1 ≤ j ≤ k.As p(X) is the product of root polynomials (each a linearized polynomial plus aconstant term), we deduce that if aj 6= 0 then

j(q − 1) = c0 + c1q + ...+ ckqk, (3)

where each ci ≥ 0 and

c0 + ...+ ck ≤ |S| < qn−k − 1 + k(q − 1). (4)

We want to prove that such an expansion (3), (4) cannot hold. Suppose that it≥ is for constantfactors fromsome rootpolynomial

exists, choose one with∑i ci minimal. Then it is easy to see that

ci ≤ q − 1 for 0 ≤ i ≤ k − 1. (5)

As all terms in (3) are nonnegative,

ck ≤ jqn−k − 1, (6)

since otherwise jqn−kqk > j(qn − 1).Suppose j = 1. If both in (5) and (6) equalities hold then

∑i ci = k(q−1)+qn−k−1,

contradicting (4). So one of the inequalities must be strict. Consequently,

k∑i=0

ciqi < (q − 1) + (q − 1)q + ...+ (q − 1)qk−1 + (qn−k − 1)qk = qn − 1,

contradicting (3).Suppose j > 1. Let m be the least integer such that j ≤ qm. By induction on j itis easy to see that j ≤ qj−1, hence 0 < m ≤ j−1 ≤ k−1. Considering (3) moduloqm we obtain

qm − j ≡ c0 + c1q + ...+ cm−1qm−1 (mod qm).

As m − 1 ≤ k − 1 we can apply (5) here and conclude that the sum on the righthand side lies between 0 and qm − 1, so it is not only a congruence but actuallyan equality. Thus we may write

j(qn − 1) = qm − j + cmqm + ...+ ckq

k.

If ci = q − 1 when m ≤ i ≤ k − 1 and ck = jqn−k − 1, then

cm + ...+ ck = (k −m)(q − 1) + jqn−k − 1 ≥ (k −m)(q − 1) + qn−k − 1 +mqn−k


> (k −m)(q − 1) + qn−k − 1 +m(q − 1) = qn−k − 1 + k(q − 1).

This contradicts (4), so either one of the inequalities in (5) must be strict for somei between m and k − 1, or inequality (6) must be strict. In either case

c0+c1q+ ...+ckqk < qm−j+(q−1)qm+ ...+(q−1)qk−1+(jqn−k−1)qk = jqn−j,

and this contradicts (3). Therefore the representation (3) is impossible if |S| ≤qn−k − 1 + k(q − 1). Consequently, aj = 0 for all j > 0.

Note that even if |S| = qn−k − 1 + k(q − 1), then aj is nonetheless 0 for all j > 1.Of course, since a covering does exist in this case, a1 may be nonzero. Thereforea1 is the critical coefficient.

For k = n− 1 Bruen [53] generalized it in the following way for t-covers (i.e. wheneach point is covered at least t times):

Theorem 24.4. Any t-covering of AG(n, q) \ {0} with hyperplanes not through theorigin, contains at least (n+ t− 1)(q − 1) hyperplanes. Dually:

Any t-fold blocking set of AG(n, q) (with respect to the hyperplanes) contains atleast (n+ t− 1)(q − 1) + 1 points.

Proof: Let B be a set of t-covering hyperplanes, defined by the linear equationsfi = 0, i = 1, ..., |B|. Then

∏|B|i=1 fi is a polynomial not vanishing at the origin but

vanishing t times elsewhere. Now Exercise 5.12 finishes the proof. For the dualstatement suppose w.l.o.g. that the origin is contained in the blocking set, and usethe standard duality in PG(n, q) containing AG(n, q). (x0, x1, ..., xn) 7→

[x0, x1, ..., xn]

We mention without proof that Bruen generalized Theorem 24.3 for t-coverings: which is similarto Jamison’s

Theorem 24.5. Any t-covering of AG(n, q)\{0} with k-dimensional affine subspacesnot through the origin, contains at least

tqn−k − 1 +k−1∑i=0

bi

subspaces, where qk − t =∑k−1i=0 biq

i, 0 ≤ bi < q, kt ≤ qk.

Exercise 24.6. [53] Let π be a dual translation plane of order q2, with (dual) kernel affine plane

of order q. Then any blocking set of π contains at least q2 + 2(q − 1) points.

See also Theorem 27.1.


24.1 Applications of the punctured Nullstellensatz

In this section, which is based on [22], we show applications to the puncturedNullstellensatz by Ball and Serra (see Section 5.5).

Consider two lines `1 and `2 of a projective plane over a field F and finite non-intersecting subsets of points Si of ì. Let A be a set of points with the propertythat every line joining a point of S1 to a point of S2 is incident with a point of A.If we asked ourselves how small can A be then obviously we could simply chooseA to be the smaller of the Si and clearly we can do no better. If, however, weimpose the restriction that one of the lines joining a point P1 of S1 to a pointP2 of S2 is not incident with any point of A then it is not so obvious how smallcan A can be. We will see that at least |S1|+ |S2| − 2 points are needed, which isclearly an attainable bound, for example take A to be (S1 ∪ S2) \ {P1, P2}. Thenext theorem provides the bound to arbitrary dimension and to sets that have notjust one point incident with the lines joining a point of S1 to a point of S2, but afixed number t of points.

Let F be an arbitrary field.

Theorem 24.7. Let `1, `2, ..., `n be n concurrent lines spanning PG(n,F) and letSi be a subset of points of ì for each i not containing the point of intersectionof the lines ì. Let A be a set of points with the property that every hyperplane〈s1, s2, ..., sn〉, where si ∈ Si is incident with at least t points of A except thosehyperplanes 〈d1, d2, ..., dn〉, where di is an element of a subset Di of Si for alli, which may be incident with less than t points of A. If there is a hyperplane〈d1, d2, ..., dn〉, where di is an element of a subset Di for all i, which is incidentwith no point of A, then for all j

|A| ≥ (t− 1)(|Sj | − |Dj |) +n∑i=1

(|Si| − |Di|).

Proof: Let H be a hyperplane that meets the lines ì in a point of Si but isnot incident with any point of A. Apply a collineation of PG(n,F) that takes`1, `2, ..., `n to the axes of AG(n,F), the affine space obtained from PG(n,F) byremoving the hyperplane H, and takes the point H ∩ ì to the point 〈ei〉, where

ei = (0, ..., 0,i1, 0, ..., 0).

We can then assume that A is a subset of AG(n,F), the affine space obtained fromPG(n,F) by removing the hyperplane H. The hyperplane H is defined by theequation Xn+1 = 0.

Let 0 ∈ Ti be the subset of F with the property that s−1 ∈ Ti \ {0} if and only if〈sei + en+1〉 is a point of Si. Note that the line ì, after applying the collineation,is 〈ei, en+1〉 and |Ti| = |Si|. Let 0 ∈ Ei be the subset of F with the property that


d−1 ∈ Ei \ {0} if and only if 〈dei + en+1〉 is a point of Di. Define

f(X1, X2, ..., Xn) =∏a∈A

((n∑i=1

aiXi

)− 1

).

The affine hyperplanes∑ni=1 tiXi = 1, where ti ∈ Ti are not all zero, are the affine

hyperplanes spanned by points s1, s2, ..., sn, where si ∈ Si. By hypothesis thereare t points of A incident with these hyperplanes, unless ti ∈ Ei for all i, and sof has a zero of multiplicity t at (t1, t2, ..., tn), unless ti ∈ Ei for all i.However 0 ∈ Ei for all i and F (0, 0, ..., 0) = (−1)|A|, so there is an element ofT1 × T2 × ...× Tn where f does not vanish. Theorem 5.24 implies that for all j

|A| = deg(f) ≥ (t−1)(|Tj |−|Ej |)+n∑i=1

(|Ti|−|Ei|) = (t−1)(|Sj |−|Dj |)+n∑i=1

(|Si|−|Di|).

The above theorem also holds for any multi-set A.The condition that there is a hyperplane that is not incident with a point of A isessential. If we do not impose this condition then for t < n any collection of t ofthe sets Si would be appropriate for A.For t = 1 the bound is tight: take A =

⋃ni=1(Si \Di).

Theorem 24.7 has some corollaries. The following theorem, which, after Theorem24.7, can be formulated as an exercise, is due to Bruen [53]. It was initially provenfor t = 1 by Jamison (Theorem 24.1); see also the independent proof found byBrouwer and Schrijver [49].It is nice to prove Theorem 24.4 again, using Theorem 24.7:

Exercise 24.8. If every hyperplane of AG(n, q) is incident with at least t points ofa set of points A, then A has at least (n+ t− 1)(q − 1) + 1 points.The bound in this theorem can be improved slightly in many cases when t ≤ q aswas proven in [31]. In Theorem 24.11 we shall investigate when this improvementapplies to the more general Theorem 24.10 below.Firstly let us look at a consequence of Theorem 24.7 for projections.

Theorem 24.9. If there are m − 1 points x1, x2, ..., xm−1 of PG(n,F) that projectm collinear points S1 onto m collinear points S2 then there is a further point xmwhich also projects S1 onto S2.

Proof: Suppose that there is no such point xm which also projects S1 onto S2.There are m lines `1, ..., `m that join a point of S1 to a point of S2 but are notincident with any of the points x1, x2, ..., xm−1. The points of S1 and S2 are all


contained in the same plane and so any two lines `i and `j are incident. If theyare not all incident with a common point xm then we can choose m − 2 pointsy1, ..., ym−2 such that yi is incident with `i but is not incident with `m and ym−2 isthe intersection of the lines `m−2 and `m−1. (We may have to relabel the lines toensure that `m is not incident with the intersection of the lines `m−2 and `m−1.)The set A = {x1, x2, ..., xm−1, y1, ..., ym−2} has the property that every line, except`m, that joins a point of S1 to a point of S2 is incident with a point of A, whichcontradicts Theorem 24.7.

The following theorem is almost the dual of Theorem 24.7. The proof is shorteras here we do not need an initial transformation.

Theorem 24.10. Let A be a set of hyperplanes of AG(n,F) and let Di be a non-empty proper subset of Si, a finite subset of F. If every point (s1, s2, ..., sn), wheresi ∈ Si, is incident with at least t hyperplanes of A except at least one point ofD1 ×D2 × ...×Dn, which is incident with no hyperplane of A, then for all j

|A| ≥ (t− 1)(|Sj | − |Dj |) +n∑i=1

(|Si| − |Di|).

Proof: Define

f(X1, X2, ..., Xn) =∏((

n∑i=1

aiXi

)− an+1

),

where each factor in the product corresponds to a hyperplane, defined by theequation

∑ni=1 aiXi = an+1, in A. By hypothesis the polynomial f has a zero of

multiplicity t at all the points of S1 × S2 × ... × Sn except at least one point ofD1 ×D2 × ...×Dn where it is not zero. By Theorem 5.24 the bound follows.

If Si = GF(q) and Di = {0} then Theorem 24.10 implies that a set of hyperplanesA with the property that every non-origin point of AG(n, q) is incident with atleast t hyperplanes of A is a set of at least (n+ t− 1)(q − 1) hyperplanes, whichdualising gives Bruens theorem (Exercise 24.8) again.

We end this section by proving the following theorem which is similar to Theorem24.10 but in which there are translations of the set of hyperplanes of AG(n, q), notincident with the origin, which also cover most, but not all, of the points of thegrid S1 × ...× Sn.

For any λ ∈ Fn and A, a finite subset of Fn, define A+ λ = {a+ λ|a ∈ A}.In the following a punctured grid is a set of points (S1× ...×Sn) \ (D1× ...×Dn),where Di is a proper non-empty subset of some subset Si ⊆ F.


Theorem 24.11. Let Λ be a set of vectors of Fn with the property that λ ∈ Λ ifand only if there is a finite punctured grid Gλ , punctured at the origin and whosedimensions do not depend on λ, with the property that every point of Gλ is incidentwith at least t hyperplanes defined by equations of the form

b1X1 + ...+ bnXn = 1,

for some b ∈ A+ λ. Let m be minimal such that for all λ ∈ Λ∏s∈S1\D1

(X − s) = 1 +Xmrλ(X)

for some polynomial rλ, where Gλ is the punctured grid (S1× ...×Sn)\ (D1× ...×Dn). If there are non-negative integers j and k with the property that either

k ≤ j ≤ min{t− 1,m− 1, |{λ1|(λ1, ..., λn) ∈ Λ ∩ (−A)}| − 1}

ork + 1 ≤ j ≤ min{t− 1,m− 1, |{λ1|(λ1, ..., λn) ∈ Λ}| − 1},

and (∑ni=1(|Si| − |Di|) + (t− 1)(|S1| − |D1|) + k

j

)6= 0

then

|A| ≥n∑i=1

(|Si| − |Di|) + (t− 1)(|S1| − |D1|) + k + 1.

Proof: Suppose that |A| =∑ni=1(|Si| − |Di|) + (t− 1)(|S1| − |D1|) + k. Let

fλ(X1, X2, ..., Xn) =∏a∈A

((n∑i=1

(ai + λi)Xi

)− 1

).

The degree of fλ is |A| − 1 + ε, where ε = 0 if λ = −a for some a ∈ A and ε = 1 ifnot. By hypothesis the polynomial f has a zero of multiplicity t at all the pointsof S1 × S2 × ... × Sn except at least one point of D1 ×D2 × ... ×Dn where it isnon-zero. Let

gi(Xi) =∏si∈Si

(Xi − si) and li(Xi) =∏di∈Di

(Xi − di).

By Theorem 5.24

fλ(X1, ..., Xn) =∑τ∈T

gτ(1)...gτ(t)hτ + u1

n∏i=1

gili,


for some polynomial u1 of degree at most (t−1)(|S1|−|D1|)+k−1+ε. Since thereis a point, the origin, of D1 ×D2 × ...×Dn where fλ is non-zero, the polynomialu1 is non-zero at this point. The polynomial in one variable

fλ(X, 0, ..., 0) = gt1h+ u2g1l1,

for some polynomial h and polynomial u2 of degree at most deg u1.By hypothesis fλ has a zero of multiplicity t at all the points (s1, 0, ..., 0), wheres1 ∈ S1 \D1. Therefore fλ(X, 0, ..., 0) is divisible by(

g1l1

)t= (1 +Xmrλ(X))t.

Let A1 be the multiset where a1 appears as an element of A1 the number of timesit appears as the first coordinate of an element of A. Thus

fλ(X, 0, ..., 0) = gt1h+ u2g1l1,

can be written as∏a1∈A1

((a1 + λ1)X − 1) = (1 +Xmrλ)t(u3 + hlt1),

for some polynomial u3 of degree at most deg u2 − (t − 1)(|S1| − |D1|), which isat most k − 1 + ε. Since 0 ∈ D1 we can write hlt1 = Xth2 for some polynomialh2. Thus, the coefficient of Xj in the right hand side of the equation above is zerofor all j for which k + ε ≤ j ≤ min{m − 1, t − 1}. On the left-hand side of thisequation the coefficient of Xj is a polynomial in λ1 of degree at most j where theterm λj1, if it appears in the polynomial, has coefficient

(|A|j

).

If the number of λ1 which appear as first coordinate in the vectors in Λ is morethan j, then the coefficient of Xj , which is a polynomial in λ1 of degree at mostj, must be identically zero, and so the binomial coefficient is zero.

The following corollary is a slight generalisation of a result of Blokhuis (Theorem2.2) from [31], see Section 25 as well for this and the next one.

Corollary 24.12. A set A of points of AG(n, q) with the property that every hyper-plane is incident with at least t points of A, has size at least

(n+ t− 1)(q − 1) + k + 1,

provided that there exists a j with the property that k ≤ j ≤ min{t− 1, q− 2} and(−n− t+ k + 1

j

)6= 0.


Proof: Apply Theorem 24.11 with Λ = −A, Si = GF(q) and Di = {0} for alli = 1, 2, ..., n. Note that for all i,∏

s∈Si\Di

(X − s) = Xq−1 − 1,

so m = q−1 and that |{λ1|(λ1, ..., λn) ∈ Λ∩(−A)}| = |{−a1|(a1, ..., an) ∈ A}| = q.We conclude that if there are non-negative integers j and k with the property thatk ≤ j ≤ min{t− 1, q − 2} and(

(n+ t− 1)(q − 1) + k

j

)=(−n− t+ k + 1

j

)6= 0,

then |A| ≥ (n+ t− 1)(q − 1) + k + 1.

The following corollary is from Blokhuis [31] for (t, q) = 1 and [8] in general.

Exercise 24.13. A set of points of AG(2, q) with the property that every line isincident with at least t points of A has size at least (t+ 1)q − (t, q).

24.2 Redei polynomials in three indeterminates

The next two sections come from Ball-Lavrauw [21].In most of the applications of Redei polynomials we have a set of points withcertain intersection properties with hyperplanes. Now we shall look at how to useRedei polynomials in situations where we have a set of points that have certainintersection properties with smaller dimensional subspaces. The obvious place tostart is in three dimensional affine or projective space, so let A be a pointset ofAG(3, q) and define

R(T, S, U) =∏

(x,y,z)∈A

(T + xS + yU + z).

We define polynomials σj in two indeterminates, of total degree at most j, bywriting

R(T, S, U) =|A|∑j=0

σj(S,U)T |A|−j .

Let a and b be any elements of GF(q) and consider the factors of the polynomial

R(T, aU + b, U) =∏

(x,y,z)∈A

(T + (ax+ y)U + bx+ z).

The point (x, y, z) is incident with the line defined by the hyperplanes aX+Y = αand bX + Z = β if and only if ax + y = α and bx + z = β if and only if the


corresponding factor in R(T, aU + b, U) is T + αU + β. Note that for a fixeda and b these q2 lines are parallel. Identifying the affine point (x, y, z) with theprojective point (x, y, z, 1) then the line defined by the (hyper)planes aX+Y = αand bX + Z = β is incident with point (1,−a,−b, 0). Indeed, every affine lineincident with (1,−a,−b, 0) can be defined by the intersection of two planes of theform aX + Y = α and bX + Z = β, for some α and β, and this line is incidentwith k points of A if and only if T +αU + β occurs as a factor of R(T, aU + b, U)with multiplicity k. Let us fix a situation to see how we can make use of this.Let A be a set of q2 points that does not determine every direction, so we canassume there is a point (1,−a,−b, 0) whose corresponding parallel class of q2 linesare each incident with exactly one point of A. Each factor in R(T, aU + b, U) isdistinct and we have

R(T, aU+b, U) =∏

α,β∈GF(q)

(T+αU+β) = T q2−((Uq−U)q−1+1)T q+(Uq−U)q−1T.

However, we also have that

R(T, aU + b, U) =q2∑j=0

σj(aU + b, U)T |A|−j

and hence for 1 ≤ j ≤ q2 − q − 1, σj(aU + b, U) ≡ 0.So we have that σj(S,U) ≡ 0 mod S − aU − b from which it follows that

S − aU − b | σj(S,U).

If there are more pairs (a, b) than the degree of σj for which this holds thenσj(S,U) ≡ 0. So in our situation let N be the number of directions (1,−a,−b, 0)not determined by two points of A. Then

R(T, S, U) = T q2+q2−N∑j=0

σq2−j(S,U)T j .

When we evaluate both S = s and U = u we are looking at the intersectionproperties of A with planes. Indeed, a factor T +α has multiplicity k in R(T, s, u)if and only if the plane sX +uY +Z = α is incident with k points of A. However,R(T, s, u) = T q

2+ g(T ), where deg g ≤ q2−N , and moreover factorises into linear

factors in GF(q)[T ]. In [20] Ball and Lavrauw prove the following

Theorem 24.14. If A is a set of q2 points of AG(3, q) that does not determine atleast peq directions for some e then every plane meets A in 0 mod pe+1 points.

By Theorem 18.15 of Storme and Sziklai we know that if the number of directionsdetermined is less than q(q+ 3)/2 then the set A is GF(s)-linear for some subfield


GF(s) of GF(q). In the case when q is prime this means that A is a plane. Theorem24.14 says that in the case when q is prime and we have a set of points thatdetermine less than q2 + 2 directions then every plane contains 0 mod q points.It would be interesting to know if this implies that A is a cone. Some immediatecorollaries of Theorem 24.14 relate to ovoids of the generalised quadrangles T2(O)or T ∗2 (O).

Corollary 24.15. [20] Let O be an ovoid of the generalised quadrangle T ∗2 (O). Theplanes of AG(3, q) are incident with zero mod p points of O.

Corollary 24.16. [20] Let O be an ovoid of the generalised quadrangle T2(O) con-taining the point (1). The planes of AG(3, q) are incident with zero mod p pointsof O.

In the case when O is a conic the generalised quadrangle T2(O) is isomorphic toQ(4, q). Since we can choose any point of Q(4, q) to be the point (1), the corollarysays that every elliptic quadric meets an ovoid of Q(4, q) in 1 mod p points ornot at all. (See also Theorem 28.2 in Section 28.2) In fact one can eliminate thepossibility that an elliptic quadric and an ovoid are disjoint but that does involveslightly more work, see [18]. A short counting argument leads to the followingtheorem.

Theorem 24.17. When q is prime an ovoid of Q(4, q) is an elliptic quadric.

Although the Redei polynomial considered in [18] is the same as we have usedhere, they used a more roundabout argument involving the Klein correspondenceto deduce the fact that for nearly all j the identity σj(aU,U − a) ≡ 0 holds for alla ∈ GF(q). Now we see that from σj(aU+b, U) ≡ 0, one only needs to substitute Uwith U − a and put b = a2 to obtain the same equivalence. The previous theoremcan be pushed a little further. De Beule and Metsch recently proved the following.

Theorem 24.18. When q is prime a set of q2 + 2 points that blocks every line ofQ(4, q) is an elliptic quadric together with a point.

24.3 Redei polynomials in many indeterminates

Let us prove the natural generalisation of Theorem 24.14 to AG(n, q) by usingRedei polynomials in n variables.

Theorem 24.19. If A is a set of qn−1 points of AG(n, q) that does not determineat least peq directions for some e then every hyperplane meets A in 0 mod pe+1

points.

Proof: Let A ⊂ AG(n, q) = PG(n, q) \ π∞ where π∞ is defined by the equationX0 = 0. Let D be the set of directions determined by A, i.e.

D = {P ∈ π∞|P ∈ 〈Q,R〉, Q,R ∈ A}.


Define the Redei polynomial as

R(T,X) =∏

(a1,...,an)∈A

(T +

n∑i=1

aiXi

),

where X = (X1, ..., Xn), and define the polynomials σj(X) by writing

R(T,X) =qn−1∑j=0

T qn−1−jσj(X).

Note that the total degree of σj is at most j. Let P = (0, y1, y2, ..., yn) ∈ π∞ \Dbe a direction not determined by A and since P must have a non-zero coordinate,we may assume that ym = 1, for some m. Now consider the Redei polynomialR(T,X) modulo

∑ni=1 yiXi. We get

R(T,X) ≡∏

(a1,...,an)∈A

(T +

n∑i=1

(ai − amyi)Xi

)mod

n∑i=1

yiXi.

For all γ ∈ GF(q)n−1, the line defined by the n− 1 equations Xi − yiXm = γiX0,where i 6= m, contains the point P , which is a direction not determined by A, andso contains exactly one point of A. Hence there is exactly one (1, a1, ..., an) ∈ Asuch that ai − amyi = γi for all i 6= m and we have

R(T,X) ≡∏

γ∈GF(q)n−1

(T +

n∑i=1i6=m

γiXi

)mod

n∑i=1

yiXi.

The above polynomial is linear over GF(q) in T and so

R(T,X) ≡ T qn−1

+n−2∑k=0

σqn−1−qk(X)T qk

+ σqn−1(X) modn∑i=1

yiXi

and we conclude that σj(X) ≡ 0 mod∑ni=1 yiXi, whenever j 6= qn−1 and

j 6= qn−1 − qk for some k. Hence for each P = (0, y1, y2, ..., yn) ∈ π∞ \Dn∑i=1

yiXi | σj(X)

in these cases and so σj(X) ≡ 0 whenever deg(σj) ≤ j < |π∞ \D|. By hypothesis|π∞ \D| ≥ peq and so we can write

R(T,X) = T qn−1

+qn−1∑j=peq

σj(X)T qn−1−j .


Now let x ∈ GF(q)n be any vector in n coordinates and let d be maximal suchthat R(T, x) ∈ GF(q)[T p

d

] \ GF(q)[T pd+1

]. We wish to prove that d ≥ e+ 1. WriteR(T, x) = S(T )p

d

, so S ∈ GF(q)[T ]\GF(q)[T p] and importantly ∂TS 6= 0. Moreover

S(T ) = T qn−1/pd + S1(T ),

where deg(S1) ≤ qn−1/pd − qpe−d. Since S(T ) is the product of linear factors

S(T ) | (T q − T )∂TS.

The degree of the right-hand side of this divisibility is less than q + deg(S1) ≤q+ qn−1/pd− qpe−d. If d ≤ e, the degree of the left-hand side, which is the degreeof S and equal to qn−1/pd, will be greater than the degree of right-hand side andwe conclude that the right-hand side of the divisibility must be zero. However thisimplies that ∂TS is zero, a contradiction, hence d ≥ e+ 1.We have shown that for all x ∈ GF(q)n,

R(T, x) =∏

(1,a1,...,an)∈A

(T +

n∑i=1

aixi

)∈ GF(q)[T p

e+1]

and so all its factors T +α occur with multiplicity a multiple of pe+1. Hence thereare a multiple of pe+1 points of A satisfying

∑ni=1 aixi = α, or, in other words,

there are 0 modulo pe+1 points of A on the hyperplane defined by the equation∑ni=1 xiXi = α. This concludes the proof.

Now the following improvement to Theorem 24.19 is given with a shorter andsimpler proof from [11]. The main difference here is the use of GF(q) × GF(qn−1)as a model for AG(n, q) in place of the natural GF(q)n.

Theorem 24.20. Let q = ph and 1 ≤ pe < qn−2. If there are more than pe(q −1) directions not determined by a set S of qn−1 points in AG(n, q) then everyhyperplane meets S in 0 modulo pe+1 points.

In [11] sets of points that reach or almost reach the bound (for some special valuesof e) are constructed.If f is a function in n − 1 variables over the finite field GF(q) then the set ofpoints S = {(x, f(x))|x ∈ GF(q)n−1} is a set of qn−1 points of AG(n, q) and theset of directions determined by the function f is defined to be the set of directionsdetermined by S. We call S the graph of f , for obvious reasons. Note that for anyset S of qn−1 points which does not determine all the directions, by applying anaffine transformation so that the projective point (0, ..., 0, 1) is a non-determineddirection, we can construct a function whose graph is the set S.Proof: Let us consider the points of AG(n, q) as a set of elements of GF(q) ×GF(qn−1) and S a set of qn−1 points. We can assume, after making an affine


transformation if necessary, that the hyperplane with first coordinate zero does notcontain exactly qn−2 points of S. Then the set of directions not determined by Sconsists of projective points (1,m) where m runs through some set N ⊆ GF(qn−1).Let x = (x1, x2) and y = (y1, y2) be two points of AG(n, q) with the property thatx1 6= y1 and there is an element µ ∈ GF(qn−1) such that x1µ − x2 = y1n − y2.Then, (x1 − y1)µ = x2 − y2 and the projective point (x1 − y1, x2 − y2) = (1, µ).Thus, (1, µ) is the direction determined by the points x and y.For all m ∈ N the projective point (1,m) is not determined by S and so the set{x1m−x2 | x = (x1, x2) ∈ S} consists of distinct elements, and hence all elements,of GF(qn−1).Define a polynomial r in two variables and polynomials σj , in one variable ofdegree at most j(q − 1), by

r(X,Y ) =∏x∈S

(X − (x1Y − x2)q−1) =qn−1∑j=0

σj(Y )Xqn−1−j .

Now r has the property that for all m ∈ N ,

r(X,m) =∏x∈S

(X − (x1m− x2)q−1) =∏

λ∈GF(qn−1)

(X − λq−1) =

= X(X(qn−1−1)/(q−1) − 1)q−1 = Xqn−1 +Xqn−1−(qn−1−1)/(q−1) + ...+X.

Thus, for all m ∈ N and 1 ≤ j < (qn−1− 1)/(q− 1) we have σj(m) = 0. However,|N | > pe(q − 1) ≥ deg(σj) for all j ≤ pe and since a polynomial has at most asmany zeros as its degree, σj is identically zero for all 1 ≤ j ≤ pe.Let cj be coefficient of the term of degree j(q− 1) in the polynomial σj(Y ). Then

∏x∈S

(X − xq−11 ) =

qn−1∑j=0

cjXqn−1−j = Xqn−1

+ cpe+1Xqn−1−pe−1 + ...+ cpn−1 .

Now x1 ∈ GF(q) and so xq−11 = 1 unless x1 = 0. If we write t for the number of

points of x ∈ S with first coordinate zero then

∏x∈S

(X − xq−11 ) = Xt(X − 1)q

n−1−t =qn−1∑j=0

(−tj

)(−1)jXqn−1−j .

Comparing the above two equations we see that(−tpr

)= 0 mod p for all 0 ≤ r ≤ e

and so t = 0 mod pe + 1, by Lucas’ Theorem. Thus the number of points of Sincident with the hyperplane with first coordinate zero is 0 modulo pe+1. However,this hyperplane was chosen arbitrarily among the hyperplanes that do not containexactly qn−2 points of S. Hence, every hyperplane is incident with 0 modulo pe+1

points of S.


Corollary 24.21. Let q = ph and 1 ≤ peqt < qn−2, where e, t ∈ Z. If there are morethan peqt(q − 1) directions not determined by a set S of qn−1 points in AG(n, q)then every (n−1−s)-dimensional subspace, where 0 ≤ s ≤ t, meets S in 0 modulope+1qt−s points.

Proof: Let Σn−1−s be a (n−1−s)-dimensional subspace which does not containqn−2−s points of S.

We first prove that there is a hyperplane containing Σn−1−s which does not containexactly qn−2 points of S. So we assume s ≥ 1 since otherwise there is nothing toprove. If every subspace of dimension n−s which contains Σn−1−s contains exactlyqn−s−1 points of S then

|S| = qn−1 = n+ (qn−s−1 − n)(qs+1 − 1)/(q − 1),

where n is the number of points of S contained in Σn−s−1. This implies thatn = qn−1 which by assumption it is not. Therefore we can find a chain of subspacesΣn−s−1 ⊂ Σn−s ⊂ ... ⊂ Σn−1 with the property that |S \ Σn−1| 6= qn−2. ThusΣn−1 meets the subspace at infinity in a subspace consisting entirely of determineddirections.

There are (qn − qn−1)/(qn−s − qn−s−1) = qs subspaces of dimension n − s thatcontain Σn−1−s and are not contained in Σn−1 and so one of them meets the sub-space at infinity in a subspace that contains at least peqt−s(q−1) non-determineddirections. Applying Theorem 24.20 to this subspace, the corollary follows.

24.4 Affine blocking sets

This part is from Ball [8].

The following proposition will be analysed later to provide lower bounds on thesizes of multiple blocking sets with respect to hyperplanes in AG(n, q). For themoment we satisfy ourselves with proving that for t-fold blocking sets of AG(n, q),which have at most (n+t−1)q−n points, a set of linear equations has a nontrivialsolution. The degree of a polynomial f will be denoted f◦.

Proposition 24.22. Let S, a set of (t + n − 1)(q − 1) + k − 1 points in AG(n, q),be a t-fold blocking set with respect to hyperplanes. Then there exist σi(q−1) withσ0 = 1 such that

bj/(q−1)c∑i=0

(nq − q − n+ k − t− i(q − 1)

j − i(q − 1)

)σi(q−1) = 0 (mod p)

for all j such that k − 1 ≤ j < t.


Proof: Let S, a set of (t + n − 1)(q − 1) + k − 1 points in AG(n, q), be a t-foldblocking set with respect to hyperplanes and assume 0 ∈ S. Define

F (X1, ..., Xn) =∏s∈S

(s1X1 + ...+ snXn + 1),

then F ◦ = |S| − 1 = (t + n − 1)(q − 1) + k − 2. Write F (X) = W (X) + V (X),where V ∈ Jt and all terms divisible by Xc1q

1 Xc2q2 · · ·Xcnq

n with c1 + ... + cn = tappear only in V . Note that W (0) 6= 0 as F (0) 6= 0. By hypothesis we have thatXt

1F ∈ Jt and hence that Xt1W ∈ Jt. Moreover, considering the possible terms

that can appear in Xt1W it follows that Xt

1W = (Xq1 −X1)W1 where W1 ∈ Jt−1.

Continuing for X2, ..., Xn we see that we can write

W (X) = U(X)n∏i=1

(Xq−1i − 1),

where U◦ ≤ (t− 1)(q − 1) + k − 2 and U(0) 6= 0.On every hyperplane X1 = c there exist at least t points of S and therefore(Xq−1

1 − 1)t divides

F (X1, 0, ..., 0) = (Xq−11 − 1)U(X1, 0, ..., 0) + (Xq

1 −X1)tV (X1) =∏s∈S

(s1X1 + 1),

where V (X1, 0, ..., 0) = (Xq1 −X1)tV (X1). Let S be a multi-subset of the multiset

S[1] := {s1|(s1, ..., sn) ∈ S} of size |S|− tq such that the multi-set S[1] \ S containseach element of GF(q) repeated exactly t times. Define the symmetric functionsσj by ∏

a∈S

(aX + 1) =nq−q−n+k−t∑

j=0

σjXj .

Put X1 = X, s1 = a and F (X, 0, ..., 0) = (Xq−1 − 1)tf(X), let V (X, 0, ..., 0) =v(X) and U(X, 0, ..., 0) = (Xq−1 − 1)t−1u(X) and rewrite the above as

f(X) = u(X) +Xtv(X) =∑

σjXj ,

where u◦ ≤ k−2. The coefficient of Xj implies that σj = 0 whenever k−1 ≤ j < t.For all λ ∈ GF(q) the set Sλ = {(s1 + λ, s2, ..., sn)|(s1, s2, ..., sn) ∈ S} is a t-foldblocking set with respect to hyperplanes of AG(n, q) and we can make a shift ofthe other n − 1 coordinates to ensure that 0 is in Sλ. Let σ(λ)

j denote the j-th

symmetric function of the set Sλ = {a + λ|a ∈ S}, then it follows that σ(λ)j = 0

whenever k − 1 ≤ j < t. Now

σ(λ)j

∑(a1 + λ)(a2 + λ) · · · (aj + λ) =

j∑r=0

(|S| − r

j − r

)σrλ

j−r,


where the first sum extends over all possible combinations of a1+λ, a2+λ, ..., aj+λin Sλ. Hence for k − 1 ≤ j < t

k−1∑r=0

(nq − q − n+ k − t− r

j − r

)σrλ

j−r = 0 (mod p).

This is zero when reduced modulo λq − λ since it is valid for all λ ∈ GF(q).Therefore

bj/(q−1)c∑i=0

(nq − q − n+ k − t− i(q − 1)− r

j − i(q − 1)− r

)σi(q−1)+r = 0 (mod p)

where 0 ≤ r < q − 1. We are specifically interested in getting equations involvingσ0, since this is non-zero and so we conclude that for k − 1 ≤ j < t

bj/(q−1)c∑i=0

(nq − q − n+ k − t− i(q − 1)

j − i(q − 1)

)σi(q−1) = 0 (mod p).

Theorem 24.23. For t < q a t-fold blocking set with respect to hyperplanes inAG(n, q) has at least (t + n − 1)(q − 1) + k points provided there exists a j suchthat k − 1 ≤ j < t and the binomial coefficient

(k−n−t

j

)6= 0 (mod p).

Proof: In fact this is an almost trivial consequence of Proposition 24.22. Assumethat there is a t-fold blocking set with (t+n−1)(q−1)+k−1 points. Since t < q wehave that j < q − 1 and there is only one term in the sum. Lucas’ theorem allowsus to ignore the terms in the top of the binomial coefficient divisible by q, and wededuce that

(k−n−tj

)σ0 =

(k−n−t

j

)= 0 (mod p), which gives a contradiction if a j

in the condition of the theorem exists.

Corollary 24.24. For t < q a t-fold blocking set with respect to hyperplanes inAG(n, q) has at least (t+ n− 1)q − n+ 1 points provided

(−nt−1

)6= 0 (mod p).

Corollary 24.25. For t < q a t-fold blocking set with respect to hyperplanes inAG(n, q) has at least (t+ 1)q − pe(t) points (where e(t) is maximal such that pe(t)

divides t).

Proof: Let k = t − pe(t) + 1 and j = t − 1, write t = t0pe(t) where p 6 |t0. Then(−pe(t)−1

t−1

)σ0 =

( −2pe(t)+pe(t)−1(t0−1)pe(t)+pe(t)−1

)=( −2t0−1

)(mod p), which is non-zero.

Now we are well prepared for the next section on nuclei and affine blocking setsas well.


25 Nuclei and affine blocking sets

The theory of nuclei shows a nice application of the GF(q2)-representation.

Definition 25.1. P 6∈ B is a t-fold nucleus of B ⊂ PG(2, q) if all the lines throughP intersect B in at least t points.

A 1-fold nucleus is called nucleus simply. The motivation of this notion is thatthe nucleus of an oval in a plane of even order is a nucleus in this sense as well.Obviously, if B has a nucleus then |B| ≥ t(q + 1). The set of nuclei is sometimesN(B)

denoted by N(B).It is also interesting to dualise this concept. Given L, a set of lines in PG(2, q), aline ` 6∈ L is a t-fold nuclear line of it if the lines of L cover each point of ` at leastt times. It means that if you consider L as a totally reducible algebraic curve, then` intersects it in each point (of `) with multiplicity at least t.

Theorem 25.2. If B ⊂ PG(2, q), |B| = q+1 and it is not a line then it has at mostq − 1 nuclei.

Proof 1: There exists a line disjoint from B so w.l.o.g. B ∈ AG(2, q) and no infinitepoint can be a nucleus. Using the GF(q2)-representation of the affine plane, definef(X) =

∑b∈B(X − b)q−1. If x is a nucleus then the set of (x − b)q−1 is exactly

the complete set of (q+ 1)-st roots of unity, so their sum is 0, hence x is a root off , which is of degree q − 1.

Proof 2: Consider the following polynomial defined by B:

F (X,Y, Z) =∑

(b1,b2,b3)∈B

(b1X + b2Y + b3Z)q−1;

we will prove that if (a1, a2, a3) is a nucleus of B then (a1X + a2Y +a3Z)|F (X,Y, Z). Then it follows that∏

(a1,a2,a3)∈N(B)

(a1X + a2Y + a3Z) | F (X,Y, Z),

which is of degree q − 1.Take a line [v1, v2, v3] meeting B in precisely one point. Then F (v1, v2, v3) = 0 asof the q + 1 terms of the sum one is zero and the other q are all 1.Now w.l.o.g. assume that a1 6= 0 and write F (X,Y, Z) = (a1X + a2Y +a3Z)g(X,Y, Z) + G(Y, Z) with G(Y, Z) homogeneous of degree q − 1 or identi-cally zero. Since [u, v, w] with u = −a2v+a3w

a1is a line through (a1, a2, a3) and

hence contains a unique point of B, so F (u, v, w) = 0 and a1u + a2v + a3w = 0.Hence G(v, w) = 0 for all v, w ∈ GF(q). So G cannot have degree q − 1 hence it is

25. Nuclei and affine blocking sets 199

identically zero.

Proof 3: This proof uses the Segre-trick:

Lemma 25.3. (Segre, Korchmaros) Let B ⊂ PG(2, q) a pointset of size q + 1 andA1, A2, A3 three non-collinear nuclei of B. Write Bi = AjAk∩B, where {i, j, k} ={1, 2, 3}. Then the three points B1, B2, B3 ∈ B are collinear.

Proof of the lemma Let A1, A2, A3 be the base points of the coordinate frame,let B1 = (0, 1, b1), B2 = (b2, 0, 1), B3 = (1, b3, 0), where b1b2b3 6= 0. Apply theSegre-trick for the lines joining Ai to to points of B \ {B1, B2, B3}, so build amatrix of size (q − 2) × 3, with rows indexed by the points of B \ {B1, B2, B3}and with columns indexed by A1, A2 and A3. The entry in position (P,Ai) isthe “coordinate” of the line PAi, so if the equation of the line is X3 = λ1X2

or X1 = λ2X3 or X2 = λ3X1 then this λi will appear in the matrix. By Ceva’stheorem, the product of the three elements in any row is 1. On the other hand,in each column all but one elements of GF(q)∗ appears, the missing ones are justthe coordinates of the lines AiBi. As the product of the elements of GF(q)∗ is −1by Wilson’s theorem, the (product of all the entries) times b1b2b3 is on the onehand = b1b2b3, on the other hand = (−1)3. So b1b2b3 = −1, hence by Menelaos’theorem B1, B2 and B3 are collinear.

Now we are ready for Proof 3 (which is due to Blokhuis and Mazzocca): Againwe can restrict ourselves to AG(2, q). Define a map f : N(B) → GF(q)∗ in thefollowing way: take an arbitrary P ∈ N(B) and put f(P ) = 1. If Pi ∈ N(B),let Bi = PPi ∩ B (which is unique). Now Bi = λiP + (1 − λi)Pi for a suitableλi ∈ GF(q)∗ \ {1}. Then put f(Pi) = λi

λi−1 . Denote the affine coordinates likeP (p1, p2), Pi(p1

i , p2i ) and Bi(b1i , b

2i ). It is easy to check that in AG(3, q) the points

(p1, p2, f(P )), (p1i , p

2i , f(Pi)) and (b1i , b

2i , 0) are collinear. lift to

3-space!

The Segre-Korchmaros lemma implies that if B′ ∈ B is on the line PiPj thenBi, Bj , B

′ and hence (p1i , p

2i , f(Pi)), (p1

j , p2j , f(Pj)) and (b′1, b′2, 0) are collinear. In

particular f is injective, so |N(B)| ≤ q − 1.

A more general form of it is the following

Theorem 25.4. (Blokhuis) If B ⊂ AG(2, q), |B| = t(q+ 1) + k− 1, k < q, then thenumber of its t-fold nuclei is at most k(q−1), supposing that

(t+k−1k

)6≡ 0 (mod p)

holds.

Proof: As in the previous proof, after identifying AG(2, q) and GF(q2), define thepolynomial

F (X,T ) =∏b∈B

(T − (X − b)q−1).


(Note that∑

(X − b)q−1, used in the previous proof, is just the coefficient ofT t(q+1)+k−2.) For a t-fold nucleus x, the multiset {(x − b)q−1 : b ∈ B} containsevery (q + 1)-st root of unity at least t times, so the polynomial F (x, T ) in onevariable is divisible by (T q+1 − 1)t for each t-fold nucleus x.

Let σj(X) denote the j-th elementary symmetric polynomial of the terms (X −b)q−1. As a polynomial of X, its degree is ≤ j(q − 1), with equality if

(|B|j

)is

nonzero in GF(q); for j = k this is the condition in the statement. As F (X,T ) =∑|B|j=0(−1)jσj(X)T |B|−j , in case of X = x being a t-fold nucleus, we have

F (x, T ) = (T q+1 − 1)t(T k−1 + ...).

On the right hand side the coefficient of T t(q+1)−1 is zero as k < q, on the left handside this coefficient is (−1)kσk(x). So every t-fold nucleus x is a root of σk(X),which is of degree k(q − 1).

A nice refinement and re-frasing is the following

Theorem 25.5. (Ball) If B is a t-fold blocking set of AG(2, q), e(t) is the maximalexponent for which pe(t)|t then |B| ≥ (t+ 1)q − pe(t).

For e(t) = 0, i.e. when g.c.d.(t, q) = 1, it was proved by Blokhuis. In particular,Jamison

when t = 1 it is Jamison’s result 12.2.

Proof: First let’s see the connection between nuclei and blocking sets. If B is at-fold blocking set then all the points in its complement are t-fold nuclei of B.

Suppose to the contrary that B is a blocking set of size (t + 1)q − pe(t) − 1 =t(q + 1) + (q − t− pe(t))− 1, so by the previous theorem (k = q − t− pe(t)) it hasat most (q − t− pe(t))(q − 1) nuclei, which is less then the size of the complementof B, so B cannot be a t-fold blocking set provided

(t+k−1k

)6= 0 modulo p. Write

t = pe(t)c where p 6 |c. Using Lucas’ theorem and calculating modulo p we have(t(q + 1) + k − 1t(q + 1)− 1

)=(tq + q − pe(t) − 1

tq + t− 1

)=

(q − pe(t) − 1cpe(t) − 1

)=(q − 2pe(t) + pe(t) − 1

(c− 1)pe(t) + pe(t) − 1

)=(q/pe(t) − 2c− 1

)(pe(t) − 1pe(t) − 1

),

which is non-zero mod p as, using base p, each digit of q/pe(t)− 2 is (p− 1) exceptthe last one which is (p− 2), but the last digit of c− 1 cannot be p− 1 as p 6 |c.

Corollary 25.6. If K is a (k, n)-arc of AG(2, q), and e is the maximal exponentsuch that pe|n then |K| ≤ (n− 1)q + pe.


Proof: Let B be the complement of K in AG(2, q). It is a t = (q−n)-fold blockingset and the previous theorem gives the bound.

Lunelli and Sce conjectured that the size of a (k, n)-arc K in PG(2, q) is k ≤(n − 1)q + 1. It was disproved it by Hill and Mason, their counterexample is thedisjoint union of some Baer subplanes; but if g.c.d.(n, q)=1 and there exists a lineskew to K then the previous Corollary shows that the Lunelli-Sce bound is true.

25.1 Lower nuclei

Now suppose that B ⊂ AG(2, q), |B| = t(q+1)−k+1, and we want to investigateits t-fold lower nuclei, so the points from which B looks like a (|B|, t)-arc. Theresult is quite similar to the case of nuclei:

Theorem 25.7. If B ⊂ AG(2, q), |B| = t(q + 1) − k + 1, then the number of itst-fold nuclei is at most k(q − 1), supposing that

(t(q+1)k

)6≡ 0 (mod p) holds.

As we did before, let’s identify AG(2, q) and GF(q2), so B ⊂ GF(q2). Let E bethe multiset of (q + 1)-st roots of unity in GF(q2), each with multiplicity t. LetB(X) = {(X − b1)q−1, (X − b2)q−1, ..., (X − b|B|)q−1} where B = {b1, ..., b|B|}.Then the point represented by x ∈ GF(q2) is a t-fold (lower) nucleus of B if andonly if it is not in B and B(x) ⊆ E as a multiset.

First we prove two lemmas.

Lemma 25.8. Let S ⊂ E, |S| = t(q + 1)− r, σi = σi(S), σ∗i = σi(E \ S). Then

r∑i=0

σ∗i σr+1−i = 0.

Proof: (Xq+1 − 1)t =∏a∈E(X − a) =

∏b∈S(X − b)

∏β∈(E\S)(X − β) =∑t(q+1)−r

i=0 (−1)iσiXt(q+1)−r−i∑rj=0(−1)jσ∗jX

r−j . The coefficient of Xt(q+1)−u inthis expression, u = 1, 2, ..., q − 2, is

r∑i=0

σ∗i σu−i = 0. (u)

From (u), (u = 1, ..., r), σ∗u can be expressed as

σ∗u = −u−1∑i=0

σ∗i σu−i. (u∗)


Now the next, (r+ 1)-st equality gives the desired relation among the elementarysymmetric polynomials, as no more σ∗i can appear (note that the upper limit is rinstead of r + 1):

r∑i=0

σ∗i σr+1−i. (r + 1)

We remark that the lemma can be proved in a shorter way, but the equationsabove will be needed; this way one can see that (r + 1) is the first (i.e. minimaldegree) non-trivial condition that can be proved.

Lemma 25.9. For j ≥ 1∑jl=0(−1)l

(a+l−1l

)(aj−l)

= 0 holds.

Proof: Since (X + 1)−a =∑∞l=0

(a+l−1l

)(−1)lX l and (X + 1)a =

∑∞l=0

(al

)X l, it

follows that

1 = (X + 1)−a(X + 1)a =∞∑j=0

Xj∞∑l=0

(−1)l(a+ l − 1

l

)(a

j − l

),

which immediately gives the lemma.

Now we are ready for theProof of Theorem 25.7: Let

G(X) =k−1∑i=0

σ∗i (B(X))σk−i(B(X)).

If x is a t–fold nucleus then Lemma 25.8 applies, that is

G(x) =k−1∑i=0

σ∗i (B(x))σk−i(B(x)) = 0.

We state that G(X) has degree k(q−1). All the nuclei are roots of it. We determinethe coefficient of Xk(q−1). Since k−1 < q, all the equations (u) hold for u = 1, ..., k.Let zi denote the coefficient of Xi(q−1) in

σ∗i ((X − b1)q−1, (X − b2)q−1, ..., (X − b|B|)q−1).

Then for i = 1, ..., k − 1

zi = −i−1∑j=0

zj

(a

i− j

).

By Lemma 25.9 for i = 1, ..., k − 1

zi = (−1)i+1

(a+ i− 1

i

).


Then for our polynomial (k):

“zk” = −k−1∑j=0

zj

(a

k − j

)= −

k−1∑j=0

(−1)j+1

(a+ j − 1

j

)(a

k − j

)from which

zk = (−1)k−1

(t(q + 1)

k

)6= 0

by the assumption of the Theorem. So the polynomial is of degree k(q−1) exactly,it is not the zero polynomial, hence it has at most so many roots.

25.2 Internal nuclei

A point of B not contained in a collinear triplet is usually called an internalnucleus. The set of internal nuclei is denoted by IN(B). Further result on internalnuclei can be found e.g. in [139], we only state a nice result of Wettl here. Theproof is left to the reader as an exercise.

Theorem 25.10. (Wettl) Let q be odd, B be a set of q+ 1 points in PG(2, q). ThenIN(B) is contained in a conic.

Proof: (Hint: Show that there is a unique tangent at each P ∈ IN(B). ApplySegre’s lemma for a triangle A1, A2, A3 ∈ IN(B)).

25.3 Higher dimensions

Now we generalize the previous results to higher dimensions. Here we identifyAG(n, q) with GF(qn).

Definition 25.11. P 6∈ B is a t-fold nucleus of B ⊂ PG(n, q) if all the lines throughP intersect B in at least t points.

A 1-fold nucleus is called nucleus simply. Obviously, if B has a nucleus then |B| ≥tθn−1.

Theorem 25.12. If B ⊂ AG(n, q), |B| = θn−1 and it is not a hyperplane then ithas at most q − 1 nuclei.

Exercise 25.13. Modify Proof 1 and Proof 2 of Theorem 25.2 as needed!

A more general form of it is the following


Theorem 25.14. Let B ⊂ PG(n, q), |B| = t(q + 1) + k − 1, k < q, suppose thatthere are exactly i points of it on the ideal hyperplane PG(n, q) \ AG(n, q). Thenthe number of its t-fold nuclei contained in AG(n, q) is at most (k + r)(q − 1) forany r ≥ 0 such that the binomial coefficient

(t+k−i−1k+r


It is a generalisation of Blokhuis’ [31]. For r = 0 this is in [112].

Exercise 25.15. Modify the proof of Theorem 25.4 as needed!From 25.12, so in the case t = k = 1 we have that there are q − 1 nuclei only.However, if B intersects every hyperplane of PG(n, q), Blokhuis and Mazzoccaproved the following.

Theorem 25.16. If B of size θn−1 intersects every hyperplane of PG(n, q) then ithas at most qn−1 − qn−2 nuclei; moreover there exist sets attaining this bound.

In the extremal case when r = qn−2 − 1 we have that qn−2|θn−1 − i or we wouldhave been able to find a smaller r for which the binomial coefficient was non-zero.It follows then that i ≡ θn−3 (mod qn−2) or in words that every hyperplane meetsB in θn−3 (mod qn−2) points. From this it can be deduced that the classificationof θn−1-sets with qn−1 − qn−2 nuclei is equivalent to the classification in the caseof (q + 1)-sets in PG(2, q) having q − 1 nuclei, see ... .As in the planar case we have the following consequence for affine blocking sets.

Theorem 25.17. (Ball) If B is a t-fold blocking set of AG(n, q), e(T ) is the maximalexponent for which pe(t)|t then |B| ≥ (t+ 1)q − pe(t).

Exercise 25.18. Modify the proof of Theorem 25.5 as needed!

For e(t) = 0, i.e. when g.c.d.(t, q) = 1, it was proved by Blokhuis. In particular,when t = 1 it is Jamison’s result.

Lower nuclei

Now suppose that B ⊂ AG(n, q), |B| = tθn−1 − k + 1, and we want to investigateits t-fold lower nuclei, so the points from which B looks like a (|B|, t)-cap. Theresult is quite similar to the case of nuclei:

Theorem 25.19. Let B ⊂ PG(n, q), |B| = t(q + 1) − k + 1, k < q, suppose thatthere are exactly i points of it on the ideal hyperplane PG(n, q) \ AG(n, q). Thenthe number of its t-fold lower nuclei contained in AG(n, q) is at most (k+r)(q−1)for any r > 0 such that the binomial coefficient

(t−ik+r


Exercise 25.20. Modify the proof of Theorem 25.7 as needed!

26. Mixed representations 205

26 Mixed representations

This part is from [15]. It is a continuation of Section 13.1, where the planar casewas arranged. Here we see the high dimensional generalization, and what makesit work is the following “mixed representation” of the space.Let k ≥ 4. If we view GF(q)k−3 × GF(q2) as the (k − 1)-dimensional vector spaceover GF(q) then the points of AG(k−1, q) can be viewed as a = (1, a1, a2, . . . , ak−2)where ak−2 ∈ GF(q2) and the other ai are elements of GF(q). In the quotient spaceof w0 = (1, 0, . . . , 0, y0), w1 = (0, 1, 0, . . . , 0, y1),. . . , wk−3 = (0, . . . , 0, 1, yk−3) thepoint a is given by (0, . . . , 0, y0−ak−2 +

∑k−3i=1 aiyi). As in Section 13.1, the points

a and b quotient to the the same point in PG(1, q) if and only if (y0 − ak−2 +∑k−3i=1 aiyi)

q−1 = (y0− bk−2 +∑k−3i=1 biyi)

q−1 if and only if 〈w0, w1, . . . , wk−3, a〉 =〈w0, w1, . . . , wk−3, b〉. Note that y0 − ak−2 +

∑k−3i=1 aiyi = 0 if and only if a ∈

〈w0, w1, . . . , wk−3〉.

Theorem 26.1. A set of points S in PG(k − 1, q) which is incident with 0 mod rpoints of every hyperplane has at least (r − 1)q + (p − 1)r points, where 1 < r <q = ph and k ≥ 4.

Proof: Either there is a co-dimension 2 subspace that is incident with no pointsof S or |S| ≥ q2 + q + 1 (from which the theorem follows) or every co-dimension3 subspace is incident with a point of S (which if were the case either |S| > q3 orevery co-dimension 4 subspace is incident with a point of S.)Counting points of S on hyperplanes containing this co-dimension 2 subspaceeither |S| ≥ r(q + 1) (and hence the thereom is proved) or there is a hyperplaneincident with no points of S. Moreover |S| = 0 mod r.Let s be the greatest common divisor of the non-trivial intersections that S haswith the co-dimension 2 subspaces. If s > 1 then take a co-dimension 2 subspaceincident with ms points of S and by induction each hyperplane containing thisco-dimension 2 subspace contains at least (s − 1)q + (p − 1)s points of S and so|S| ≥ ((s−1)q+(p−1)s−ms)(q+1)+ms > (r+1)q. Thus there is a co-dimension2 subspace incident with just one point of S. Counting points of S on hyperplanescontaining this co-dimension 2 subspace we see that |S| = 1 + (q+ 1)(−1) mod r.Combining this with |S| = 0 mod r we have that q = 0 mod r.We can view S as a subset of GF(q)k−3 × GF(q2) ' AG(k − 1, q) and consider thepolynomial

R(X,Y ) =∏a∈S

(X + (Y0 − ak−2 +k−3∑i=1

aiYi)q−1) =|S|∑j=0

σj(Y )X |S|−j .

For all y = (y0, y1, . . . , yk−3) ∈ GF(q2)k−2 the points w0 = (1, 0, . . . , 0, y0), w1 =(0, 1, 0, . . . , 0, y1),. . . , wk−3 = (0, . . . , 0, 1, yk−3), a and b span the same hyperplaneif and only if (y0 − ak−2 +

∑k−3i=1 aiyi)

q−1 = (y0 − bk−2 +∑k−3i=1 biyi)

q−1 6= 0.


Suppose that W = 〈w0, w1, . . . , wk−3〉 is a (k−3)-dimensional subspace is incidentwith t points of S. By hypothesis every (k − 2)-dimensional subspace contains amultiple of r points of S and so

R(X, y) = Xt(Xq+1 − 1)r−t0g(X)r,

where t0 = t mod r.For all y ∈ GF(q2)k−2 the polynomial σj(y) = 0 whenever 0 < j < q and r doesnot divide j. However σj has degree at most j(q − 1) and so, for 0 < j < q and rdoes not divide j, the polynomial σj ≡ 0.Let |S| = (r − 1)q + κr.For future reference note that if t = 0 then σq+1(y) = 0. If t = 1 then σq+1(y) = 1and σκr(y) = 0 since the degree of g in this case is at most κ− 1.Let ′ be differentiation with respect to any one of the variables Yi. As in the proofof the planar case, Theorem 13.6, we have

R(X, y) | (Xq+1 − 1)∂R

∂Y(X, y).

If |W ∩ S| = 0 then R(X, y) is an r-th power and in exactly the same way as inthe proof of Theorem 13.6 we have

σkrσ′q+1 = −σ′kr.

Let

f(Y ) =∏a∈S

(Y0 − ak−2 +k−3∑i=1

aiYi).

If |W ∩ S| > 0 then f(y) = 0 and if |W ∩ S| = 0 then σq+1(y) = 0, thereforefσq+1 = 0 for all y ∈ GF(q2)k−2. So there are polynomials ri for i = 1, . . . , k − 2such that

fσq+1 =k−2∑i=1

(Y q2

i − Yi)ri(Y ),

and the degree of ri is at most deg f+deg σq+1 − q2 ≤ |S| − 1.Now choose any (y1, . . . , yk−3) ∈ GF(q2)k−3 and b ∈ S and set y0 = bk−2 −∑k−3i=1 biyi. Then b ∈W so |W ∩S| ≥ 1. Differentiate fσq+1 with respect to Yi and

evaluate at y. Since |W ∩ S| ≥ 1 it follows that f(y) = 0 and so σq+1f′ = −ri.

If |W ∩ S| > 1 then f ′(y) = 0 and hence ri(y) = 0 = −f ′(y). If |W ∩ S| = 1then σq+1 = 1 and ri(y) = −f ′(y). Thus the polynomial ri + f ′ has at least|S|q2(k−3) zeros but is a polynomial of degree at most |S| − 1 over GF(q2) ink− 3 indeterminates and is either zero or has at most (|S| − 1)q2(k−3) zeros. Thusri = −f ′ and

fσq+1 = −k−2∑i=1

(Y q2

i − Yi)∂f

∂Yi.

27. Flocks 207

Now again differentiate fσq+1 but now evaluate for y ∈ GF(q2)k−2 where |W∩S| =0. Then σq+1(y) = 0 and

σ′q+1 = f ′/f.

Combining this with σκrσ′q+1 = −σ′κr we have that (fσκr)′ = 0.

Returning to the case where |W ∩ S| = 1 we have that σκr(y) = f(y) = 0 and so(fσκr)′ = 0. And in the case |W ∩S| = t > 1 the polynomial f has a zero of degreet and so again (fσκr)′ = 0. The polynomial (fσκr)′ is zero for all y ∈ GF(q2)k−2

but is a polynomial of degree at most |S|+κr(q−1) < q2. Thus (fσκr)′ ≡ 0. Thisholds for which ever indeterminate Yi we choose to differentiate with respect toand so we conclude that fσκr ∈ GF(q2)[Y p0 , . . . , Y

pk−3]. Hence fp−1 divides σκr.

If κ ≤ p− 2 then (p− 1)(r− 1)q+ κr(p− 1) > κr(q− 1) and so σκr ≡ 0. Howeverthe polynomial whose terms are the terms of highest degree in R(X,Y0, 0, . . . , 0)is (X + Y q−1

0 )|S| which has a term X(r−1)qYκr(q−1)0 since

(|S|κr

)= 1. Thus σκr has

a term Y κr(q−1) which is a contradiction. Therefore κ ≥ p− 1.

Corollary 26.2. A code whose weights and length have a common divisor r andwhose dual minimum distance is at least 3 has length at least (r − 1)q + (p− 1)r.

27 Flocks

Theorem 27.1. (Fisher and Thas 1979, Orr 1973) Given an elliptic quadric inPG(3, q) and a set of q − 1 pairwise disjoint conics partitioning all but two of itspoints, then the q − 1 planes of those conics must contain a common line (thatmisses the quadric).

Proof: Use stereographic projection from one of the uncovered points, mappingall the other points onto the affine plane AG(2, q) in such a way that the otheruncovered point is mapped to the origin. All the conics become pairwise disjointcircles of the affine plane (or the inversive plane over GF(q)). One can imaginethem as conics in the plane having two conjugate imaginary points at infinity incommon. Now the theorem can be translated saying that a family of q−1 disjointcircles covers AG(2, q) \ {0} if and only if all circles are centered at the origin.

Now we identify AG(2, q) \ {0} and GF(q2) \ {0} as usual; the points are exactlythe roots of Zq

2−1 − 1 = 0.

The j-th circle is the set {z : (z − aj)(z − aj)q = rj}, where aj ∈ GF(q2), rj ∈GF(q)∗. Set cj(Z) = (Z − aj)(Z − aj)q − rj , then the theorem says

∏q−1j=1 cj(Z) =

Zq2−1 − 1 if and only if all aj = 0.

Exercise 27.2. Prove that this “translation” is correct!


Now∏q−1j=1 cj(Z) =

∏q−1j=1(Z

q+1−ajZq)+ terms of degree at most (q+1)(q−2)+1.Equating terms of degree greater than (q + 1)(q − 2) + 1 one sees that the claimis∏q−1j=1 Z

q(Z − aj) = Zq2−1 if and only if all aj = 0, i.e.

∏q−1j=1(Z − aj) = Zq−1 if

and only if all aj = 0.

Exercise 27.3. Generalize the theorem in the following way: If q − 1 norm sets{z ∈ GF(qn) : Normqn→q(z − aj) = rj}, where aj ∈ GF(qn), rj ∈ GF(q)∗, partitionthe points of GF(qn)∗ = AG(n, q) \ {0}, then all aj must be 0.We are going to use norm sets in Section 29.12.

27.1 Partial flocks of the quadratic cone in PG(3, q)

A flock of the quadratic cone of PG(3, q) is a partition of the points of the conedifferent from the vertex into q irreducible conics. Associated with flocks are someelation generalised quadrangles of order (q2, q), line spreads of PG(3, q) and, whenq is even, families of ovals in PG(2, q), called herds. In [109] Storme and Thasremark that this idea can be applied to partial flocks, obtaining a correspondencebetween partial flocks of order k and (k + 2)-arcs of PG(2, q), and constructingherds of (k + 2)-arcs. Using this correspondence, they can prove that, for q > 2even, a partial flock of size > q −√q − 1 if q is a square and > q −

√2√q if q is a

nonsquare, is extendable to a unique flock.Applying this last result, Storme and Thas could give new and shorter proofs ofsome known theorems, e.g., they can show directly that if the planes of the flockhave a common point, then the flock is linear (this originally was proved by Thasrelying on a theorem by D. G. Glynn on inversive planes, and is false if q is odd).Here we prove the following

Theorem 27.4. Assume that the planes Ei, i = 1, ..., q − ε intersect the quadraticcone C ⊂ PG(3, q) in disjoint irreducible conics. If ε < 1

4 (1− 1q+1 )

√q then one can

find additional ε planes (in a unique way), which extend the set {Ei} to a flock.

Proof: Let C be the quadratic cone C = {(1, t, t2, z) : t, z ∈ GF(q)}∪{(0, 0, 1, z) :z ∈ GF(q)} ∪ {(0, 0, 0, 1)} and C∗ = C \ {(0, 0, 0, 1)}. Suppose that the planes Eiintersect C∗ in disjoint conics, and Ei has the equation X4 = aiX1 + biX2 + ciX3,for i = 1, 2, ..., q − ε.Define fi(T ) = ai + biT + ciT

2, then Ei ∩ C∗ = {(1, t, t2, fi(t)) : t ∈ GF(q)} ∪{(0, 0, 1, ci)}. Let σk(T ) = σk({fi(T ) : i = 1, ..., q− ε}) denote the i-th elementarysymmetric polynomial of the polynomials fi, then degT (σk) ≤ 2k. As for any fixedT = t ∈ GF(q) the values fi(t) are all distinct, we would like to find

Xq −X∏i(X − fi(t))

,

27. Flocks 209

the roots of which are the missing values GF(q) \ {fi(t) : i = 1, ..., q − ε}.We are going to use the technique of Section 11. In order to do so, we definethe elementary symmetric polynomials σ∗j (t) of the “missing elements” with thefollowing formula:Xq −X =(Xq−ε−σ1(t)Xq−ε−1+σ2(t)Xq−ε−2−...±σq−ε(t)

)(Xε−σ∗1(t)Xε−1+σ∗2(t)Xε−2−...±σ∗ε (t)

);

from which σ∗j (t) can be calculated recursively from the σk(t)-s, as the coefficientof Xq−j , j = 1, ..., q− 2 is 0 = σ∗j (t) + σ∗j−1(t)σ1(t) + ...+ σ∗1(t)σj−1(t) + σj(t); forexample

σ∗1(t) = −σ1(t); σ∗2(t) = σ1(t)2 − σ2(t); σ∗3(t) = −σ1(t)3 + 2σ1(t)σ2(t)− σ3(t);

etc. Note that we do not need to use all the coefficients/equations, it is enough todo it for j = 1, ..., ε.Using the same formulae, obtained from the coefficients of Xq−j , j = 1, ..., ε, onecan define the polynomials

σ∗1(T )=−σ1(T );σ∗2(T )=σ1(T )2− σ2(T );σ∗3(T )=−σ1(T )3+2σ1(T )σ2(T )− σ3(T );(1)

up to σ∗ε . Note that degT (σ∗j ) ≤ 2j. From the definition(Xq−ε−σ1(T )Xq−ε−1+...±σq−ε(T )

)(Xε−σ∗1(T )Xε−1+σ∗2(T )Xε−2−...±σ∗ε (T )

)is a polynomial, which is Xq − X for any substitution T = t ∈ GF(q), so it isXq −X + (T q − T )(...). Now define

G(X,T ) = Xε − σ∗1(T )Xε−1 + σ∗2(T )Xε−2 − ...± σ∗ε (T ), (2)

from the recursive formulae it is a polynomial in X and T , of total degree ≤ 2εand X-degree ε.For any T = t ∈ GF(q) the polynomial G(X, t) has ε roots in GF(q) (i.e. themissing elements GF(q) \ {fi(t) : i = 1, ..., q − ε}), so the algebraic curve G(X,T )has at least N ≥ εq distinct points in GF(q) × GF(q). Suppose that G has nocomponent (defined over GF(q)) of degree ≤ 2. Let’s apply the Lemma with d = 2and α < 1

2 (1− 1q+1 ); (as 2ε ≤ α

√q) we have

εq ≤ N ≤ 2ε(q + 1)α,

which is false, so G = G1G2, where G1 is an irreducible factor over GF(q) of degreeat most 2. If degX G1 = 2 then degX G2 = ε − 2, which means that G1 has atmost q + 1 and G2 has at most (ε− 2)q distinct points in GF(q)×GF(q) (at mostε− 2 for each T = t ∈ GF(q)), contradiction (as G has at least εq).


Both G1 and G2, expanded by the powers of X, are of leading coefficient 1. So G1

is of the form G1(X,T ) = X−fq−ε+1(T ), where fq−ε+1(T ) = aq−ε+1 +bq−ε+1T +cq−ε+1T

2. Let the plane Eq−ε+1 be defined by X4 = aq−ε+1X1 + bq−ε+1X2 +cq−ε+1X3.The plane Eq−ε+1 intersects C∗ in {(1, t, t2, fq−ε+1(t)) : t ∈ GF(q)} ∪{(0, 0, 1, cq−ε+1)}. Now we prove that for any t ∈ GF(q) the points{(1, t, t2, fi(t)) : i = 1, ..., q − ε} and (1, t, t2, fq−ε+1(t)), in other words, thevalues f1(t), ..., fq−ε(t); fq−ε+1(t) are all distinct. But this is obvious from(Xq−ε − σ1(t)Xq−ε−1 + σ2(t)Xq−ε−2 − ...± σq−ε(t)

)(X − fq−ε+1(t)

)| Xq −X.

Now one can repeat all this above and get fq−ε+2, ..., fq, so we have

G(X,T ) =q∏

qε+1

(X − fi(T ))

and the values fi(t), i = 1, ..., q are all distinct for any t ∈ GF(q). The only remain-ing case is “t = ∞”: we have to check whether the intersection points Ei ∩ C∗on the plane at infinity X1 = 0, i.e. the values c1, ..., cq−ε︸︷︷︸

Γ

; cq−ε+1, ..., cq︸︷︷︸Γ∗

are all

distinct (for Γ we know it). (Note that if q planes partition the affine part of C∗

then this might be false for the infinite part of C∗.) From (1), considering theleading coefficients in each defining equality, we have

σ1(Γ∗) =−σ1(Γ); σ2(Γ∗) = σ1(Γ)2−σ2(Γ); σ3(Γ∗) =−σ1(Γ)3+2σ1(Γ)σ2(Γ)−σ3(Γ);

etc., so

Xq − X =(Xq−ε − σ1(Γ)Xq−ε−1 + σ2(Γ)Xq−ε−2 − ...σq−ε(Γ)

)(Xq−ε −

σ1(Γ∗)Xq−ε−1+ σ2(Γ∗)Xq−ε−2− ...σq−ε(Γ∗)), which completes the proof.

Exercise 27.5. Prove that if q = p is a prime then in Theorem 27.4 the conditionε < 1

4

√q can be changed for the weaker ε < 1

40p + 1 (so the result is muchstronger).

Exercise 27.6. Assume that the planes Ei, i = 1, ..., q + ε intersect the quadraticcone C ⊂ PG(3, q) in disjoint irreducible conics that cover the cone minus itsvertex. If ε < 1

4 (1 − 1q+1 )

√q then one can find ε planes (in a unique way), such

that if you remove the points of the irreducible conics, in which these ε planesintersect C, from the multiset of the original cover then every point of C (exceptthe vertex) will be covered precisely once.

Exercise 27.7. Let P be the parabola Y = X2 in AG(2, q). Suppose that the secantlines Y = aiX + bi, i = 1, ..., q−3

2 and the tangent Y = a0X + b0 = 0 meet P inq − 2 distinct points. Find the formula of the missing secant Y = αX + β (i.e.express α, β with the ai, bi-s).

* * *

27. Flocks 211

27.2 Partial flocks of cones of higher degree

Using the method above one can prove a more general theorem on flocks of cylin-ders with base curve (1, T, T d). This is from [116].

Theorem 27.8. For 2 ≤ d ≤ 6√q consider the cone {(1, t, td, z) : t, z ∈

GF(q)} ∪ {(0, 0, 1, z) : z ∈ GF(q)} ∪ {(0, 0, 0, 1)} = C ⊂ PG(3, q) and letC∗ = C \ {(0, 0, 0, 1)}. Assume that the planes Ei, i = 1, ..., q− ε, Ei 63 (0, 0, 0, 1),intersect C∗ in pairwise disjoint curves. If ε < b 1

d2√qc then one can find addi-

tional ε planes (in a unique way), which extend the set {Ei} to a flock, (i.e. qplanes partitioning C∗).

The proof (see below) starts like in the quadratic case. Using elementary symmetricpolynomials we find an algebraic curve G(X,Y ), which “contains” the missingplanes in some sense. The difficulties are (i) to show that G splits into ε factors,and (ii) to show that each of these factors corresponds to a missing plane. For(i) we use our Lemma 10.15. For (ii) we have to show that most of the possibleterms of such a factor do not occur, which needs a linear algebra argument ona determinant with entries being elementary symmetric polynomials; this matrixmay be well-known but the author could not find a reference for it.

Proof of Theorem 27.8. Suppose that the plane Ei has the equation X4 = aiX1 +biX2 + ciX3, for i = 1, 2, ..., q − ε.

Define fi(T ) = ai + biT + ciTd, then Ei ∩ C = {(1, t, td, fi(t)) : t ∈ GF(q)} ∪

{(0, 0, 1, ci)}. Let σk(T ) = σk({fi(T ) : i = 1, ..., q−ε}) denote the k-th elementarysymmetric polynomial of the polynomials fi, then degT (σk) ≤ dk.

We proceed as in the quadratic case and so we define the polynomials

σ∗1(T )=−σ1(T ); σ∗2(T )= σ1(T )2−σ2(T ); σ∗3(T )=−σ1(T )3+2σ1(T )σ2(T )−σ3(T ); ...(1∗)

up to σ∗ε . Note that degT (σ∗j ) ≤ dj. From the definition(Xq−ε−σ1(T )Xq−ε−1+...±σq−ε(T )

)(Xε−σ∗1(T )Xε−1+σ∗2(T )Xε−2−...±σ∗ε (T )

)is a polynomial, which is Xq −X for any substitution T = t ∈ GF(q), so it is ofthe form Xq −X + (T q − T )(...). Now define

G(X,T ) = Xε − σ∗1(T )Xε−1 + σ∗2(T )Xε−2 − ...± σ∗ε (T ), (2)

from the recursive formulae it is a polynomial in X and T , of total degree ≤ dεand X-degree ε.

For any T = t ∈ GF(q) the polynomial G(X, t) has ε roots in GF(q) (i.e. themissing elements GF(q) \ {fi(t) : i = 1, ..., q − ε}), so the algebraic curve G(X,T )has at least N ≥ εq distinct points in GF(q) × GF(q). Suppose that G has no


component (defined over GF(q)) of degree ≤ d. Let’s apply the Lemma with asuitable 1

d+1 + 1+d(d−1)√q

(d+1)q ≤ α < 1d , n = degG ≤ dε ≤ 1

d

√q − d+ 3

2 , we have

εq ≤ N ≤ dεqα < εq,

which is false, so G = H1G1, where H1 is an irreducible factor over GF(q) of degreeat most d. If degX H1 = dX ≥ 2 then degX G1 = ε − dX , which means that H1

has at most q + 1 + (dX − 1)(dX − 2)√q and G1 has at most (ε − dX)q distinct

points in GF(q)×GF(q) (at most ε−dX for each T = t ∈ GF(q)), so in total G has

εq ≤ N ≤ (ε− dX + 1)q + 1 + (dX − 1)(dX − 2)√q,

a contradiction if 2 ≤ dX ≤ √q + 1, so degX H1 = 1.

One can suppose w.l.o.g. that both H1 and G1, expanded by the powers of X, areof leading coefficient 1. So H1 is of the form H1(X,T ) = X − fq−ε+1(T ), where

fq−ε+1(T ) = aq−ε+1 + bq−ε+1T + cq−ε+1Td + δq−ε+1(T ),

where δq−ε+1(T ) is an “error polynomial” with terms of degree between 2 and d−1.At the end of the proof we will show that δq−ε+1 and other error polynomials arezero.

Now one can repeat everything for G1, which has at least (ε− 1)q distinct pointsin GF(q) × GF(q) (as H1 has exactly q and H1G1 has at least εq). The similarreasoning gives G1 = H2G2, where H2(X,T ) = X − fq−ε+2(T ) with fq−ε+2(T ) =aq−ε+2 + bq−ε+2T + cq−ε+2T

d + δq−ε+2(T ). Going on we get fq−ε+3, ..., fq (wherefor j = q − ε + 1, ..., q we have fj(T ) = aj + bjT + cjT

d + δj(T ), where δj(T )contains terms of degree between 2 and (d− 1) only). Hence

G(X,T ) =q∏

q−ε+1

(X − fi(T )).

For any t ∈ GF(q) the values f1(t), ..., fq(t) are all distinct, this is obvious from(Xq−ε − σ1(t)Xq−ε−1 + σ2(t)Xq−ε−2 − ... ± σq−ε(t)

)((X − fq−ε+1(t))...(X −

fq(t)))

= Xq −X.

For j = q − ε+ 1, ..., q let the plane Ej be defined by X4 = ajX1 + bjX2 + cjX3.We are going to prove that {Ej : j = 1, ..., q} is a flock.

First we check the case “t = ∞”: we have to check whether the intersection pointsEi∩C on the plane at infinity X1 = 0, i.e. the values c1, ..., cq−ε︸︷︷︸

Γ

; cq−ε+1, ..., cq︸︷︷︸Γ∗

are

all distinct (for Γ we know it). (Note that even if q planes partition the affine part

27. Flocks 213

of C∗ then this might be false for the infinite part of C∗.) From (1∗), consideringthe leading coefficients in each defining equality, we have

σ1(Γ∗)=−σ1(Γ); σ2(Γ∗)= σ1(Γ)2−σ2(Γ); σ3(Γ∗)=−σ1(Γ)3+2σ1(Γ)σ2(Γ)−σ3(Γ);

etc., so

Xq−X =(Xq−ε−σ1(Γ)Xq−ε−1+σ2(Γ)Xq−ε−2−...±σq−ε(Γ)

)(Xε−σ1(Γ∗)Xε−1+

σ2(Γ∗)Xε−2− ...± σq−ε(Γ∗)),

which we wanted to prove.Now we want to get rid of the δj ’s, i.e. we are going to prove that δq−ε+1, ..., δq =0. Let s be the maximal T -exponent appearing in any of δq−ε+1, ..., δq, so eachδj(T ) = djT

s + ... (for j = q − ε + 1, ..., q; also 2 ≤ s ≤ d − 1 and there exists adj 6= 0). In the equationG(X,T ) = Xε − σ∗1(T )Xε−1 + σ∗2(T )Xε−2 − ...± σ∗ε (T ) =

ε∏i=1

(X − aq−ε+i − bq−ε+iT − cq−ε+iTd − δq−ε+i(T )),

the coefficient of Xε−jT d(j−1)+s, j = 1, ..., ε is zero on the left hand side (i.e. thecoefficient of T d(j−1)+s in σ∗j , it can be seen by induction from (1∗) for instance),and it is

σj−1(Γ∗\{cq−ε+1})dq−ε+1 + σj−1(Γ∗\{cq−ε+2})dq−ε+2 +...+ σj−1(Γ∗\{cq})dq

on the right hand side. Hence we have a system of homogeneous linear equationsfor dq−ε+1, ..., dq with the elementary symmetric determinant∣∣∣∣∣∣∣∣∣∣∣

1 1 ... 1σ1(Γ∗ \ {cq−ε+1}) σ1(Γ∗ \ {cq−ε+2}) ... σ1(Γ∗ \ {cq})σ2(Γ∗ \ {cq−ε+1}) σ2(Γ∗ \ {cq−ε+2}) ... σ2(Γ∗ \ {cq})

...σε−1(Γ∗ \ {cq−ε+1}) σε−1(Γ∗ \ {cq−ε+2}) ... σε−1(Γ∗ \ {cq})

∣∣∣∣∣∣∣∣∣∣∣=

∏1≤i<j≤ε(cq−ε+i − cq−ε+j), which is non-zero as the ci’s are pairwise distinct.

Hence the unique solution is dq−ε+1, ..., dq = 0 and fj(T ) = aj + bjT + cjTd for

each j = 1, ..., q.Our final and the last missing argument we need is that for j = 1, ..., q the planeEj intersects C in {(1, t, td, fj(t)) : t ∈ GF(q)}∪{(0, 0, 1, cj)}, so these intersectionsare pairwise disjoint, E1, ..., Eq is a flock of C.


28 Spreads and ovoids

In this section we show, very briefly, how spreads and ovoids can be handled bypolynomials. The first, very basic part is from Ball [7], while the second one isbased on [10].

28.1 Spreads

A 1-spread of PG(3, q) is a collection of q2 + 1 lines that partitions the space.Consider the representation of PG(3, q) with (q3 +q2 +q+1)-th roots of unity (i.e.(q − 1)-th powers) in GF(q4)∗. (Hyper)planes are given by the zeros of equationsof the form

aq2+q+1Xq2+q+1 + aq+1Xq+1 + aX + 1 = 0,

where a is also a (q3 + q2 + q+ 1)-th root of unity. Lines are given by the zeros ofpolynomials of the form

Lαβ(X) = Xq+1 − αX + β,

where α and β satisfy certain conditions which we shall calculate. All the pointsx of PG(3, q) satisfy xq

3+q2+q+1 − 1 = 0 so

Lq2

αβXq+1 − (Xq3+q2+q+1 − 1) + αq

2XLqαβ − (βq

2− αq

2+q)Lαβ

= (αq2βq + αβq

2− αq

2+q+1)X − (βq2+1 − αq

2+qβ − 1)

is identically zero since it is a polynomial of degree at most 1 and has q+1 distinctzeros corresponding to the points on the line Lαβ . By manipulating the coefficientswe have that Lαβ is a line precisely when

βq3+q2+q+1 = 1 and αq+1 = βq − βq

2+q+1.

If βq2+1 = 1 then α = 0, if βq

2+1 6= 1 then there are q + 1 possibilities for α.This gives (q3 + q)(q + 1) + q2 + 1 = (q2 + 1)(q2 + q + 1) lines as needed, so theserestrictions are sufficient as well as necessary.

Symplectic spreads

For every point λ of PG(3, q), viewed as a (q3 + q2 + q + 1)-st root of unity inGF(q4), let

ωλ(X) = (−λq2)q

2+q+1Xq2+q+1 + (−λq2)q+1Xq+1(−λq

2)X + 1

be a polynomial over GF(q4) whose zeros correspond to the points of a hyperplane.It is easy to verify that

ωλ(ε)λq+1 = ωε(λ)εq+1 and ωλ(λ) = 0

28. Spreads and ovoids 215

and it follows from this that ω defines a symplectic polarity on PG(3, q). Theplanes ωλ(X) and ωε(X) intersect in the line given by the zeros of the equation

Xq+1 + ελ(λq+1 − εq+1)q

λ− εX − ελq+1 − λεq+1

λ− ε= 0.

The line joining the points λ and ε is given by the zeros of the equation Xq+1 −αX + β = 0 where

α =λq+1 − εq+1

λ− εand β =

ελq+1 − λεq+1

λ− ε.

The two lines coincide whenever αqβ = −α and the lines for which this conditionholds are the totally isotropic lines.

A symplectic 1-spread is a spread whose elements are totally isotropic lines.

28.2 Ovoids

Generalised quadrangles

A generalised quadrangle is a polar space of rank 2 and consists of points and lineswhich have the following properties. (Q1) Two points lie on at most one line (Q2)If L is a line, and p a point not on L, then there is a unique point of L collinearwith p. (Q3) No point is collinear with all others. The axioms (Q1)-(Q3) are self-dual; the dual of a generalised quadrangle is a generalised quadrangle. Let Q bea finite generalised quadrangle. Each line is incident with 1 + s points and eachpoint is incident with 1 + t lines, for some s and t, and we say Q is a generalisedquadrangle of order (s; t). If s = t then Q is said to have order s. An ovoid of ageneralised quadrangle is a set of points O such that each line contains exactlyone point of O. A spread of a generalised quadrangle is a set S of lines such thateach point is incident with exactly one line of S. An ovoid O and a spread S of Qsatisfy

|O| = |S| = st+ 1.

The set of lines dual to the ovoid O form a spread in the generalised quadrangledual to Q. The set of points dual to the spread S form an ovoid in the generalisedquadrangle dual to Q.

Let q = ph for some prime p and integer h. Let Sp(4, q) denote the symplecticgeneralised quadrangle of order q. The points of Sp(4, q) are the points of PG(3, q)and the lines are the totally isotropic lines of a symplectic polarity. Recall that asymplectic polarity is induced by an alternating bilinear form b. An alternatingbilinear form on a vector space V satisfies b(v, v) = 0 for all v ∈ V . This impliesb(v;w) = −b(w; v) for all v, w ∈ V . (Expand b(v + w, v + w) = 0.) Hence if thecharacteristic is 2 then any alternating form is symmetric.


LetO(5, q) denote the generalised quadrangle of order q whose points are the pointsof a non-singular quadric in PG(4, q) and whose lines are the lines contained inthat quadric. Later we will examine ovoids in O(5, q). To define O(5, q) one maychoose the quadratic form Q(x) = x0x4 + x1x3 + x2

2 on V(5, q); in this case notethat any ovoid containing (0; 0; 0; 0; 1) may be written in the form

O(f) = {(0; 0; 0; 0; 1)} ∪ {(1, x, y, f(x, y),−y2 − xf(x, y)) : x, y ∈ GF(q)}.

In fact, due to the following lemma, ovoids in O(5, q) are dual to spreads in Sp(4, q).

Lemma 28.1. O(5, q) is the dual of Sp(4, q).

Proof: Let O+(6, q) be the Klein quadric of lines of PG(3, q). The image of thelines of Sp(4, q) is the intersection of O+(6, q) with a hyperplane PG(4, q), whichis O(5, q). The lines of Sp(4, q) incident with a given point form a pencil of linesin a plane and therefore their images on O(5, q) lie on a line.

The geometry Sp(4, q) in GF(q4)

We recall the geometry of PG(3, q) represented in GF(q4), see Section 6.1. Condi-tion (∗∗), that we will use, can be found there.

Let Γ be an element of GF(q4) satisfying Γq2−1 = −1. Let b be the alternating

bilinear form defined by

b(X,Y ) := Tr(ΓY q2X) = ΓY q

2X + ΓqY q

3Xq − ΓY Xq2 − ΓqY qXq3

and note that b(X,Y ) = −b(Y,X). The map y 7→ b(X, y) = Tr(Γyq2X) = 0 maps

y to its symplectic hyperplane and defines a symplectic polarity. Let x and y betwo orthogonal elements of GF(q4), b(x, y) = 0, and let L(X) = Xq2 +cXq+eX bethe line that joins them. By elimination from the equations L(x) = 0 and L(y) = 0we can deduce that

(xqy − yqx)e = xq2yq − yq

2xq and (xqy − yqx)c = −(xq

2y − yq

2x)

and(Γc+ Γqecq)(xqy − yqx) = b(x, y) = 0.

The totally isotropic lines of the polarity defined by b(X,Y ) have −γc = ecq whereγ = Γ1−q, as well as the necessary restrictions. In the case when c = 0 there areq2 + 1 lines where each line is given by the set of zeros of an equation of the formXq2 + eX = 0 where eq

2+1 = 1. In the case c is non-zero let d = c−1 and we findthat e = −γdq−1 and

γdq3+q − γ−1dq

2+1 + 1 = 0. (+)

For each d satisfying this equation there is a totally isotropic line which is givenby the set of zeros of an equation of the form dXq2 +Xq − γdqX = 0. The points

28. Spreads and ovoids 217

of Sp(4, q) are the points of PG(3, q) and for this reason we take as before thepoints to be the (q3 + q2 + q + 1)-st roots of unity (alternatively the non-zero(q − 1)-st powers) in GF(q4). Therefore we replace the indeterminate X by Uwhere U = Xq−1. It now follows that the lines of Sp(4, q) are given by the zeros(all (q3+q2+q+1)-st roots of unity) of equations Uq+1+e = 0 for each e satisfyingeq

2+1 = 1 anddUq+1 + U − γdq = 0 (++)

for each d satisfying (+).Remark. The geometry PG(1, q2) has as points the subspaces of rank 1 in V(2, q2).In GF(q4) they are the given by sets of zeros of equations of the form

Xq2 + eX = 0

where eq2+1 = 1. Hence the lines defined by the sets of zeros of equations of the

form Uq+1 + e = 0 are skew and together they form a Desarguesian spread Rof Sp(4, q). A Desarguesian spread is equivalent to a regular spread. The set ofpoints in the generalised quadrangle O(5, q) dual to a regular spread of Sp(4, q)is an elliptic quadric. Hence we need to prove that a spread of Sp(4, q) meets thespread R in 1 modulo p lines.Remark. The equations (**) and −γc = ecq imply that

c2 = γ−1eq+1(eq2+1 − 1).

When q is even we can take square roots and parameterize the lines using (q3+q2+q + 1)-st roots of unity. Moreover when q is even we can assume that γ = 1 sincethe alternating form is also symmetric. Thus we have that the totally isotropiclines of Sp(4, q) are given by the zeros of equations of the form

Uq+1 + (e(q2+q+2)/2 + e(q+1)/2)U + e = 0

and one can check that if the point x lies on the line parameterized by e then elies on the line parameterized by x2q. Hence we see that Sp(4, q) is self-dual whenq is even, a fact first noted by Tits in 1962.

Ovoids of O(5, q)

Theorem 28.2. An ovoid in O(5, q) meets an elliptic quadric in 1 modulo p points.

This result was proved for q even by Bagchi and Sastry in 1987, who proved thatan ovoid meets not only an elliptic quadric but also a Tits ovoid in an odd numberof points.Recall that q = ph for some prime p and integer h. In this section we prove thata spread of Sp(4, q) has 1 modulo p lines in common with the regular spread R.This regular spread is entirely arbitrary.


Proof of the theorem. Let S be a spread of Sp(4,q) and let the sets D and E besuch that for d ∈ D the line

dUq+1 + U − γdq = 0

is in S and for e ∈ E the line Uq+1 + e = 0 is in S. Clearly |D|+ |E| = q2 + 1. Theaim will be to show that |D| = 0 modulo p and then the result will follow.The bilinear form b(X,Y ) can be rewritten for the points of PG(3, q) by replacingXq−1 by U and Y q−1 by V . Hence for a fixed point u in PG(3, q) the zeros of thepolynomial

(u, V ) := γuq+1 − γV q+1 + uq2+q+1V − uV q

2+q+1

are the points that are orthogonal to u, i.e. lie on the symplectic hyperplanethrough u. Let v be the point of Sp(4, q) (i.e. PG(3, q)) that is the intersection ofthe line of Sp(4, q) of the form

dV q+1 + V − γdq = 0

with the plane (u, V ) = 0, assuming that the line is not contained in the plane.We can calculate directly or check by substitution that

vq = −u(duq+1 + u− γdq)q−1.

Similarly, if v is the point of intersection of the line of Sp(4, q) of the form V q+1 +e = 0 with the plane (u, V ) = 0 then

vq = γ−1ue(uq+1 + e)q−1.

The coefficient of V q2+q in (u, V ) is minus the sum of all the points in the plane

(u, V ) = 0 and is zero. Likewise the sum of all the points on any line is minus thecoefficient of V q in the equation of this line which is also zero. Hence the sum ofall points lying in an affine plane is also zero. Thus the sum of all the points ofintersection of the spread S with the plane (u, V ) = 0 is zero and we have

0 =∑

v =∑

vq = −∑d∈D

u(duq+1 + u− γdq)q−1 +∑e∈E

γ−1ue(uq+1 + e)q−1.

Note that of course one of the lines of the spread contains the point u and theterm in the sum corresponding to this line will be zero. The polynomial∑

d∈D

U(dUq+1 + U − γdq)q−1 −∑e∈E

γ−1Ue(Uq+1 + e)q−1

is zero for all points of PG(3, q) and since it’s degree is only q2 it is identicallyzero. However the coefficient of Uq is |D| and hence |D| = 0 modulo p.

29. Non-geometrical applications 219

29 Non-geometrical applications

There is an extensive number of applications of polynomials throughout math-ematics, here I show some which are close to my combinatorial-algebraic taste.E.g. the (unfortunately still unpublished) book of Babai and Frankl [4] containsa chapter devoted to spaces of polynomials. I learned some of the following appli-cations from Tamas Szonyi’s paper [127], some others from Alon’s paper [1]. I amalso grateful for the advice of Gy. Karolyi.

29.1 Normal factorizations of Abelian groups

Let G = Zp × Zp (we are going to use the additive notation) and suppose thatG = A+B with A,B ⊂ G, (0, 0) ∈ A,B and for each g ∈ G there is a unique wayto write it as g = a + b, a ∈ A, b ∈ B. Then we say that G = A + B is a normalfactorization of G.

Theorem 29.1. In every normal factorization of G = Zp × Zp, either A or B is asubgroup.

Proof: [91] One can identify G and AG(2, p) in the obvious way; then the sub-groups of order p correspond to lines through the origin. The key idea is to showthat no direction is determined by both A and B.

Suppose to the contrary that some direction (suppose without loss of generalitythat the horizontal direction, i.e. (0)) is determined by both A and B. Let α(g)be the first coordinate of g ∈ G, and ω a primitive p-th root of unity. Then

(∑a∈A

ωα(a))(∑

b∈B

ωα(b))

=∑

a∈A,b∈B

ωα(a+b) = p

p−1∑i=0

ωi = 0,

as each element of G occurs precisely once in the form a + b. So one of the twofactors of the product above must be zero, suppose this is the first one (corre-sponding to A). It means that the set of numbers {α(a) : a ∈ A} contain all theresidues modulo p, hence A does not determine the horizontal direction. In generalit means that for any direction (m), at least one of A and B does not determine(m). So either A or B determines at most p+1

2 directions, hence, by Theorem 18.1,it is a line. As they contain the origin, that line is a subgroup.

The “stability version” of this result is the following: let G = Zp×Zp and A,B ⊂G, (0, 0) ∈ A,B, |A| > p − ε, |B| > p − ε, where ε ≤ cp for a small constant c.Suppose that if for some g ∈ G g = a1 + b1 = a2 + b2, ai ∈ A, bi ∈ B then a1 = a2

and b1 = b2. (We may say that A+B ⊆ G is a partial normal factorization of G.)


Exercise 29.2. In every partial normal factorization of G = Zp × Zp, either A orB can be extended to a subgroup of order p, if ε ≤ ... (determine!).

Exercise 29.3. (Redei) Let G = Zp × Zp and let A ⊂ G, |A| = p. Assume that Ais not a subgroup. Then there are at most p−1

2 subsets B for which |B| = p, 0 ∈ Band G = A+B holds.

Note that the factorizations of Abelian groups has a rather complicated theory;neither this statement above can be extended to groups of order pn, n > 2.A motivating problem for Lovasz and Schrijver [91] was the following application.It was conjectured by van Lint (and proved in the prime case by van Lint andWilliams) that if X ⊂ GF(q2), 0, 1 ∈ X, |X| = q and the difference of any twoelements of X is a square element of GF(q2), then X = GF(q).

Exercise 29.4. Prove this conjecture in the case q = p prime, using the represen-tation AG(2, p) ∼ GF(p2) and Theorem 18.1.The conjecture for general q was proved by Blokhuis [29] with polynomials. Hisproof and his method was generalized for the analogous statement concerning d-thpowers instead of squares of GF(q2) in [113]:

Theorem 29.5. If 1 < d|(q + 1), X ⊂ GF(q2), 0, 1 ∈ X, |X| = q and the differenceof any two elements of X is always a d-th power of GF(q2), then X = GF(q).

Exercise 29.6. Prove either of these results using Theorem 18.14.Again, the q-subsets of GF(q2) above (or the q-cliques of the Paley graph Pq2,d,see below) are quite “stable”:

Exercise 29.7. [113] If 1 < d|(q + 1), X ⊂ GF(q2), 0, 1 ∈ X, |X| = q − ε, whereε < (1− 1

d )√q. Suppose that the difference of any two elements of X is always a

d-th power of GF(q2), then X ⊆ GF(q) ⊂ GF(q2).

29.2 The Paley graph

The Paley graph Pq has GF(q) as vertex set, and (a, b) is an edge if and only if b−ais a square element of GF(q). It is undirected iff −1 is a square, i.e. q = 4k + 1.Let’s examine the automorphisms of Pq. Note that the maps x 7→ a2x + b, a, b ∈GF(q), a 6= 0 are always automorphisms.Let f : GF(q) → GF(q) be an automorphism and consider the difference quotientsf(y)−f(x)

y−x . If (x, y) was an edge then both the denominator and the numeratorare squares, while if (x, y) was not an edge then both the denominator and the


numerator are nonsquares; in both cases the difference quotient, i.e. the directiondetermined by (x, f(x)) and (y, f(y)) is a square. In fact we have proved that f isan automorphism if and only if f is bijective and the directions it determines areall squares in GF(q).

Hence f determines at most q−12 directions and Theorems 18.1 and 18.14 can be (0) cannot be

determined as fis a permutationapplied.

So for example if q = p we immediately have that f is linear, so of form x 7→ a2x+b.The non-prime case is more difficult.

Exercise 29.8. Prove that the automorphism group of the Paley graph Pq (q = ph)is {f(X) = a2Xpj + b : a ∈ GF(q)∗, b ∈ GF(q), j = 0, ..., h− 1}

Exercise 29.9. Prove that the Paley graph Pq2 has clique number ω(Pq2) = q.

Exercise 29.10. Generalize the notion of Paley graph, using d-th powers instead ofsquares. Pq,d

Exercise 29.11. Formulate Theorem 29.5 in the language of Paley graphs.

29.3 Representing systems

Another application in Redei’s book [103] (Section 37) is about common repre-senting systems of subgroups. Let G be a group, H a subgroup of it, then R ⊂ G isa representing system of H if R intersects each coset of H in precisely one element. RH = G,

|R| · |H| = |G|

Let G = Zp×Zp again, and let H1, ...,Hk be its subgroups of order p. We want tofind a common representing system R, so for which (0, 0) ∈ R (it can be achievedw.l.o.g.), |R| = p (this is obvious) and R+Hi = G (by definition). Note that anyfurther subgroup (different from H1, ...,Hk) will do, so we require that R is not asubgroup. In the usual representation (as above), every Hi is a line through theorigin; and R is a representing system for Hi if ∀r1 6= r2 ∈ R (r1 − r2) 6∈ Hi, inother words when the directions determined by R are different from the slopes ofthe given lines Hi. Hence, by 18.1, there is no such common representing systemif k > p−1

2 , and if k = p−12 then R can be described:

Theorem 29.12. Suppose that the subgroups H1, ...,Hk ≤ G = Zp × Zp have acommon representing system R, (0, 0) ∈ R, which is not a subgroup. Then k ≤ p−1

2 ,and if k = p−1

2 then, after a suitable change of the two factors Zp, R is the set so a change ofcoordinatesdescribed in Theorem 19.1, and the subgroups Hi are lines of form {(x,mx) : x ∈

GF(p)}, where each (m) is a direction not determined by R.

Note that it does not mean that for any set of (p−12 or less) subgroups there is a

common representing system.


29.4 Wielandt’s visibility theorem

The following theorem was proved first by Wielandt using very complicated meth-ods; later Blokhuis and Seidel [38] has pointed out that it was an easy consequenceof Redei’s theorem.

Theorem 29.13. (Wielandt) Let G be a permutation group acting on the points ofAG(2, p), p prime, containing all the translations. Let G0 denote the stabilizer ofthe origin. Let S be the set of k lines through the origin, 1 ≤ k ≤ p−1

2 . If G0 mapsthe union of the lines in S, as a pointset, onto itself, then any g ∈ G0 maps thelines of S to lines (of S).

Proof: [38] Let g ∈ G0 and let’s denote by τ(v) the translation by the vectorv ∈ GF(p)× GF(p); (g is also a map GF(p)2 → GF(p)2). Let x and y be points ona line ` of S. The vector (x− y) shows the slope of `. Translate the point (x− y)by y, then let g act on it, finally translate it by −g(y): we get(τ(−g(y)) g τ(y)

)(x−y) =

(τ(−g(y)) g

)(x) =

(τ(−g(y))

)(g(x)) = g(x)−g(y).

Note that the two translations do not change the directions, and g turned thewe want g(`) tobe a line ∀` ∈ S direction of x− y to the direction of a(nother) line of S.

It means that for any line ` in S, the direction determined by any two points ong(`), coincides with the slope of one of the lines in S. Hence g(`) determines atmost p−1

2 directions, so by Theorem 18.1 it is a line.

Note that the permutations of G do not necessarily preserve the lines. Naturally,as G = G0T (where T is the group of translations), all the elements of G mapthe lines in S into lines. To make the theorem above (which sounds somewhatpeculiar) more “motivated”, we remark that the translations form a regular, but(in general) not normal subgroup T of order p2, and the lines through the originare subgroups of T , each of order p.

29.5 Burnside’s theorem

Dress, Klin and Muzichuk [60] and, independently, Ott [96] has realized that afamous theorem of Burnside can be proved in an elementary way, using Redei’sresult (in fact the prime case was re-proved independently by Muzichuk and alsoby others a few times). Here, for historical reason, we just quote the theorem,without proof, for the details see [60].

Theorem 29.14. (Burnside) Let G be a transitive permutation group of degree p(p prime). Then either G is doubly transitive or it is isomorphic to a subgroup ofthe affine transformation group {x 7→ ax+ b : a, b ∈ GF(p), a 6= 0}.


We also mention an application of Blokhuis’ Theorem 21.4, given by A. Bereczky.His result is about the existence of fixed-point free p-elements of a permutationgroup (a p-element is a permutation whose order is a power of p).

Theorem 29.15. (A. Bereczky [25]) Let p be an odd prime, a ≥ 1. If p + 1 ≤b < 3

2 (p + 1) then every transitive permutation group of degree pab contains afixed-point free p-element.

29.6 Jaeger’s conjecture

In [2] Alon and Tarsi use Redei polynomials to prove Jaeger’s conjecture for q nonprime. Jaeger’s conjecture states that when q ≥ 5, for all matrices M ∈ GL(n, q)there exists a vector y ∈ GF(q)n with the property that neither y nor My haveany zero coordinate. (Jaeger conjectured this for q = 5 only.)This problem can be formulated in many equivalent ways:

• the points of the vectorspace V(n, q) cannot be covered by two sets of nindependent linear hyperplanes; independent:

they intersectin {0} only

• given two bases of V(n, q), {vi : i = 1, ..., n} and {wi : i = 1, ..., n} then thereexists a vector u ∈ V(n, q), u =

∑αivi =

∑βiwi, with each αi, βj being

nonzero coordinates;

• PG(n− 1, q) cannot be covered by (the union of the faces of) two simplices.

Note that for q = 2 the matrix(

1 11 0

), while for q = 3 the matrix

(1 11 −1

)are

counterexamples. Also if the conjecture is false for some (q, n0) then it is false for add a zerocolumn and anindependentrow to thecounterexamplematrix M

any (q, n), n ≥ n0.Let us see how Redei polynomials can be used to tackle this problem. If theconjecture is not true then there is a matrix M ∈ GL(n, q) with the propertythat My has a zero component, for all y ∈ GF(q)n with no zero component.The set of n points A that is made up of the rows of M is a set of n points inPG(n− 1, q) that spans the whole space and has the property that the hyperplaney1X1 + y2X2 + ...+ ynXn = 0 is incident with a point of A for all y ∈ GF(q)n withno zero component. In other words, if we define a Redei polynomial

R(Y1, Y2, ..., Yn) :=∏

(x1,x2,...,xn)∈A

(x1Y1 + x2Y2 + ...+ xnYn)

then for all vectors y which have no zero component R(y1, y2, ..., yn) = 0. Thisalgebraic property of R implies that we can write

R(Y1, Y2, ..., Yn) = (Y q−11 − 1)f1 + (Y q−1

2 − 1)f2 + ...+ (Y q−1n − 1)fn, (∗)


where f1, f2, ..., fn ∈ GF(q)[Y1, Y2, ..., Yn] are polynomials of degree at most n −(q − 1). The matrix M is non-singular and so has an inverse M−1 = (aij). If we Hence n ≥ q − 1

or contradictionmake the substitution Yi =∑nj=1 aijZj we have

Z1Z2...Zn =n∑i=1

(( n∑j=1

aijZj

)q−1

− 1)fi.

Now on the right-hand side when we try to find a term Zj1Zj2 ...Zjq−1 from one

of the(∑

aijZj

)q−1

terms, as we must since the left-hand side is Z1Z2...Zn, thecoefficient is a multiple of (q − 1)!. If q is not prime then this is always zero andwe have a contradiction.

We can explore this situation a little bit. First note that if q is a prime we have

Z1Z2...Zn = −n∑i=1

( ∑j1<j2<...<jq−1

aij1aij2 ...aijq−1Zj1Zj2 ...Zjq−1

)fi =

= −∑

j1<j2<...<jq−1

Zj1Zj2 ...Zjq−1

n∑i=1

(aij1aij2 ...aijq−1

)fi.

A typical term of fi is

Yi1Yi2 · · ·Yin−q+1 =( n∑j=1

ai1j Zj

)( n∑j=1

ai2j Zj

)· · ·( n∑j=1

ain−q+1j Zj

),

where i1, ..., in−q+1 are not necessarily pairwise distinct. What is the coefficient ofZ1Z2 · · ·Zn−q+1 for instance on the right hand side? It is∑

π∈Sym(1,2,...,n−q+1)

ai1π(1)ai2π(2) · · · ain−q+1π(n−q+1).

Let’s examine the Redei polynomial of the counterexamples in the q = 2 and q = 3cases above! They are

R(Y1, Y2) = (Y1 + Y2)Y1 = (Y 11 − 1)Y1 + (Y 1

2 − 1)Y1 = Y 11 · Y1 + Y 1

2 · Y1

and

R(Y1, Y2) = (Y1 + Y2)(Y1 − Y2) = (Y 21 − 1)1 + (Y 2

2 − 1)(−1) = Y 21 · 1 + Y 2

2 · (−1).

In (∗) consider the terms of degree ≤ (n− q+1). They are f1 +f2 + ...+fn, hence

f1+f2+...+fn ≡ 0 and R(Y1, Y2, ..., Yn) = Y q−11 f1+Y

q−12 f2+...+Y q−1

n fn, (∗∗)


where f1, ..., fn are homogeneous polynomials of degree n − q + 1. We claim thatif (y1, ..., yn) is a zero-free point which is orthogonal to at least two rows of thematrix M then fj(y1, ..., yn) = 0 for each j = 1, ..., n. Let e.g. j = 1 and considerR(Y1) = R(Y1, y2, ..., yn) = (Y q−1

1 − 1)f1(Y1, y2, ..., yn). Since R splits into linearfactors, so does f1(Y1, y2, ..., yn) as well; and as Y1 = y1 is a multiple root of R, itis a root of f1(Y1, y2, ..., yn) (with multplicity decreased by one).Consequently each fi vanishes on the same set of points S ⊆ (GF(q)∗)n.

29.7 Graphs containing p-regular subgraphs

Theorem 29.16. Alon-Friedland-Kalai [1] For any prime p, any loopless graphG(V,E) with average degree bigger than 2p − 2 and maximum degree at most2p− 1 contains a p-regular subgraph.

The most interesting case is p = 3, see [1] for the details.Proof: Let (av,e)v∈V,e∈E be the incidence matrix of G, i.e. av,e = 1 or 0 as v ∈ eor not. Associate a variable Xe to each edge e, and define the polynomial

F =∏v∈V

(1− (

∑e∈E

av,eXe)p−1

)−∏e∈E

(1−Xe)

over GF(p). degF = |E| as (p − 1)|V | < |E| by assumption. The coefficient of∏e∈E Xe is (−1)|E|+1 6= 0. Hence, by Theorem 5.17, there is a vector x with

coordinates xe ∈ {0, 1} such that F (xe : e ∈ E) 6= 0. This x is not the zero vectoras F (0) = 0. Also for each v ∈ V ,

∑e∈E av,exe = 0 mod p since otherwise F would

vanish at x. So in the subgraph consisting of the edges with xe = 1 every degreeis precisely p (as the maximum degree is ≤ 2p− 1).

29.8 Blokhuis’ proof of Bollobas’ theorem

Theorem 29.17. Let A1, ..., Ah and B1, ..., Bh be collections of sets with ∀i : |Ai| =r, |Bi| = s and Ai ∩Bj = ∅ if and only if i = j. Then h ≤

(r+ss

).

Proof: [30] W.l.o.g. we can assume that every set is a subset of R. Define thepolynomials

ai(X) =∏α∈Ai

(X − α), i = 1, ..., h; bj(X) =∏β∈Bj

(X − β), j = 1, ..., h.

By the properties of resultants, see Section 9.5, and the conditions on the sets AiandBj , we get that R(ai, bj) 6= 0 if and only if i = j. Let c(X) = csX

s+...+c1X+c0


be an arbitrary polynomial of degree s, then for a fixed i we have that R(ai, c) isa homogeneous polynomial of degree s in the coefficients (considered as variables)c0, ..., cs of c. We claim that the polynomials R(ai, c), i = 1, ..., h are independentin the

(r+ss

)-dimensional vectorspace of homogeneous polynomials of degree r in

the variables c0, ..., cs (and hence h ≤(r+ss

)). To prove this independence, suppose

that∑i λiR(ai, c) = 0. Then substituting bj for the polynomial c we get λj = 0.

29.9 Alon-Furedi

Theorem 29.18. Let F be a field. Suppose that in Fd the union of the hyperplanesHi, i = 1, ...,m cover the vertices of the hypercube {0, 1}d except the origin. Thenm ≥ d.

Proof: Let the equation of Hi be∑dj=1 aijXj − 1 = 0. Then

F (X1, ..., Xd) =m∏i=1

( d∑j=1

aijXj − 1)

vanishes in every x = (x1, ..., xd), xj ∈ {0, 1}, x 6= (0, 0, ..., 0), but F (0, 0, ..., 0) 6= 0.Now

G(X1, ..., Xd) =m∏i=1

( d∑j=1

aijXj − 1)− (−1)d+m

d∏j=1

(Xj − 1)

vanishes on {0, 1}d. Theorem 5.17 states that if X1X2 · · ·Xd is a leading term thenG cannot vanish on a set of form S1 × S2 × ... × Sd, with each |Si| > 1. HenceX1X2 · · ·Xd is not a leading term and m ≥ d.

Note that in case of the equality m = d the coefficient of X1X2 · · ·Xd must bezero in

G(X1, ..., Xd) =d∏i=1

( d∑j=1

aijXj − 1)−

d∏j=1

(Xj − 1)

hence Per(aij) = 1.

29.10 Chevalley-Warning

This theorem of Chevalley and Warning holds for any finite field; for simplicitywe prove it over GF(p) only.

Theorem 29.19. Let f1, ..., fm be polynomials from GF(p)[X1, ..., Xn]. If n >∑mi=1 deg(fi) and the polynomials have a common zero c = (c1, ..., cn) then they

have another common zero.


Proof: Suppose that the statement is false and define

f(X1, ..., Xn) =m∏i=1

((1− fi(X1, ..., Xn))p−1

)− δ

n∏j=1

∏α∈GF(p)\{cj}

(Xj − α),

where δ 6= 0 is chosen such that f(c) = 0.

Observe that f(x) = 0 for any x ∈ GF(p)n. It holds for x = c; by assumption thereexists some fj for which fj(x) 6= 0 hence 1− fj(x)p−1 = 0. Similarly, as for somei xi 6= ci, so we have

∏α∈GF(p)\{cj}(Xi − α) = 0 and then f(x) = 0.

Let ti = p − 1 for i = 1, ..., n; then the coefficient of∏ni=1X

tii in f is −δ 6= 0.

Hence by Theorem 5.17 with Si = GF(p) for all i, we have that there exists ans ∈ GF(p)n for which f(s) 6= 0, a contradiction.

A Chevalley-Warning type application of the punctured Nullstellensatz

This nice result of Ball-Serra [22] can be now posed as an exercise.

Exercise 29.20. Let f1, f2, ..., fm be polynomials of GF(q)[X1, X2, ..., Xn] and letd = |D1|+ ...+ |Dn|, where Di is the set of elements c of GF(q) where there existsa common zero of f1, f2, ..., fm with i-th coordinate c. If d 6= 0, in other words ifthe polynomials f1, f2, ..., fm have a common zero, then

m∑i=1

deg(fi) ≥nq − d

q − 1.

29.11 Cauchy-Davenport

Theorem 29.21. Let A,B ⊆ (Zp,+) be two subsets of size |A| = k, |B| = l. Then|A+B| ≥ min{k + l − 1, p}.

Proof: Let Zp = (GF(p),+). If k+ l > p then everything is obvious: for any g ∈ Gthe sets A and g−B will intersect, so g ∈ A+B. Suppose that |A+B| ≤ k+ l−2and choose C ⊆ G such that A + B ⊆ C and |C| = k + l − 2. Define F (X,Y ) =∏c∈C(X + Y − c), it vanishes on A × B. Here degF = (k − 1) + (l − 1) and the

coefficient of Xk−1Y l−1 is(k+l−2k−1

)6= 0 (as k+ l−2 < k+ l−1 ≤ p). It contradicts

Theorem 5.17.

The very similar proof of the Eliahou-Kervaire-Bollobas-Leader theorem is


Exercise 29.22. Alon [1] Let βp(r, s) denote the smallest n for which the triple(r, s, n) satisfies the Hopf-Stiefel condition, i.e.

(nk

)≡ 0 (mod p) for every n− r <

k < s.If A and B are nonempty subsets of the vectorspace V(d, p), |A| = r, |B| = s, then|A+B| ≥ βp(r, s).Dias da Silva and Hamidoune proved that for A ⊂ Zp, |A| = k, p a prime, |A+A| =|{a1 + a2 : a1 6= a2 ∈ A}| ≥ min{2|A| − 3, p}, settling a problem of Erdos andHeilbronn. The proof below contains a slight simplification by Karolyi.Suppose to the contrary that a set GF(p) ⊃ C ⊇ A+A, with |C| = 2k−4. Consider

f(X,Y ) = (X−Y )2∏c∈C

(X+Y−c) = (X−Y )2(X+Y )2k−4+ terms of lower degree.

Then f(a1, a2) = 0 for all a1, a2 ∈ A. deg f = 2k− 2, the coefficient of Xk−1Y k−1

is(2k−4k−3

)− 2(2k−4k−2

)+(2k−4k−1

)= −2(2k−4)!(k−1)

(k−1)!(k−1)! , which is nonzero if 0 < 2k − 4 < p

(otherwise we are done). Thus, by Theorem 5.17 we can find a1, a2 ∈ A such thatf(a1, a2) = 0, contradiction.

29.12 Another application

Let H be a fixed graph. The classical problem from which extremal graph theoryhas originated is to determine the maximum number of edges a graph on n verticescan have without containing a copy of H. This maximum value is the Turannumber of H and is customarily denoted by ex(n,H).It is particularly interesting to determine the Turan numbers when H is bipartite,as in most cases even the order of magnitude is open. The Zarankiewicz problemconcerns the complete bipartite graph, so ex(n,Kt,s), where t ≤ s.Kovari, T. Sos and Turan gave the upper bound for arbitrary fixed t:

ex(n,Kt,s) ≤ ct,sn2− 1

t ,

where ct,s > 0 is a constant depending on t and s. The right hand side is con-jectured to give the correct order of magnitude. However the best general lowerbound, obtained by the probabilistic method, is

c′n2− s+t−2st−1 ≤ ex(n,Kt,s),

where c′ > 0 is an absolute constant, but for all 2 ≤ t ≤ s s+t−2st−1 > 1

t holds; sothe lower bound is always of lower order of magnitude.The optimality of the upper bound (up to a constant factor) is proved for t = 2, 3and all s ≥ t. The incidence graphs of projective planes demonstrate this order of


magnitude for t = 2. In this case, however, even the asymptotic order of magnitudeis known: ex(n,K2,s) =

√s−12 n3/2 +O(n4/3) (Furedi [65]).

The optimality for t = 3 was established by Brown [50]. Later Furedi proved thatBrown’s construction is asymptotically optimal: ex(n,K3,3) = Θ(n5/3).

The construction of Brown is the following “unit distance graph” in a 3-dimensional affine space: the vertices of our graph will be the points of AG(3, q).Let k1, k2 be such that E : X2 + k1Y

2 + k2Z2 = 1 is an elliptic quadric in

AG(3, q). Then the vertices (x, y, z) and (u, v, w) are connected by an edge iff(x− u)2 + k1(y − v)2 + k2(z − w)2 = 1.

Proposition 29.23. This graph has ≈ n5/3 edges and does not contain a K3,3.

Proof: E intersects the ideal plane in the conic X2 + k1Y2 + k2Z

2 = 0, so it hasq2 + 1 − (q + 1) = q2 − q affine points and the same is true for all its translates.The neighbours of the vertex are on a translate of E , so the number of edges is12q

3(q2 − q) ≈ (q3)5/3.

Let U(u1, u2, u3), V (v1, v2, v3) and W (w1, w2, w3) be three points and EU , EV , EWthe corresponding quadrics, i.e. the (equations of the) neighbours of U, V and Wresp., so (X − u1)2 + ... = 1, (X − v1)2 + ... = 1 and (X −w1)2 + ... = 1. Subtractthe first one from the second and the third, we get two linear equations which donot coincide, so they define either a line or an empty set. As EU does not containa line, this line (if exists), intersects it in at most two points, so U, V and W hasat most 2 common neighbours.

The idea of this construction is that we define a “surface” “around” each point ofthe space, such that (i) they are translates of each other (so it is easy to handlethem); (ii) they have “many” points (so the graph will contain many edges); and(iii) any three translates intersect in a bounded number of points (so any triple ofvertices has a bounded number of common neighbours). The next result, due toKollar, Ronyai and Szabo [87] generalizes the same idea.

Let the set of vertices of our graph be the elements of GF(qt), and recall thatNormqt→q(a) = a · aq · ... · aqt−1

= a(qt−1)/(q−1). Now two vertices a and b areadjacent iff Normqn→q(a + b) = 1. The number of solutions of Normqn→q(x) = 1is qt−1

q−1 in GF(qt) so the number of edges is at least 12qt( q

t−1q−1 − 1) ≥ 1

2 (qt)2−1t .

Theorem 29.24. This graph Gq,t contains no subgraph isomorphic to Kt,t!+1.

Corollary 29.25. For t ≥ 2 and s ≥ t! + 1 we have ex(n,Kt,s) ≥ ctn2− 1

t , wherect > 0 depends on t only, we may choose ct = 2−t. For every t and s ≥ t, theinequality holds with c = 1/2 for infinitely many values of n.


The Corollary follows from the Theorem and from the fact that there is a primepower between 1

2n1/t and n1/t. The union of [ nqt ] disjoint copies of Gq,t will have

the appropriate number of edges.The Theorem is a direct consequence of the following: if d1, d2, ..., dt are distinctelements of GF(qt) then the system of equations

Normqt→q(X + d1) = (X + d1)(Xq + dq1)...(Xqt−1

+ dqt−1

1 ) = 1

Normqt→q(X + d2) = (X + d2)(Xq + dq2)...(Xqt−1

+ dqt−1

2 ) = 1

...

Normqt→q(X + dt) = (X + dt)(Xq + dqt )...(Xqt−1

+ dqt−1

t ) = 1

has at most t! solutions x ∈ GF(qt). They prove it by considering a more generalsystem of equations:

Lemma 29.26. Let K be a field and aij , bi ∈ K for 1 ≤ i, j ≤ t such that aij1 6= aij2if j1 6= j2. Then the system of equations

(X1 − a11)(X2 − a21)...(Xt − at1) = b1

(X1 − a12)(X2 − a22)...(Xt − at2) = b2

...

(X1 − a1t)(X2 − a2t)...(Xt − att) = bt

has at most t! solutions (x1, x2, ..., xt) ∈ Kt.

In the proof of the Lemma first polynomials fj = (X1 − a1j)(X2 − a2j)...(Xt −atj) are defined, then the regular map F : Kt → Kt by F (X1, X2, ..., Xt) :=(f1, f2, ..., ft). Then the Lemma states that |F−1(b)| ≤ t! holds for every b ∈ Kt.It is obvious for b = 0. The rest of the proof involves some commutative algebraand we omit it.

Remark 29.27. One gets Theorem 29.24 by putting

K = GF(qt), aij = −dqi−1

j , bj = 1.

Chapter 3

Hints and short solutions to theexercises

1 Of Chapter I

Exercise 5.1

In the polynomial ring GF(q)[X] the equality (1 + X)n =∏∞i=0(1 + X)nip

i

=∏∞i=0(1 + Xpi)ni holds. In other words

∑∞k=0

(nk

)Xk =

∏∞i=0

∑p−1ki=0

(niki

)Xkip

i

=∑p−1k0,k1,...=0

(n0k0

)(n1k1

)...Xk0+k1p+k2p

2+... and the result follows from comparison ofcoefficients.

Exercise 5.2

If ω is a generator element of GF(qn) then ωqn−1q−1 generates GF(q). Now let N :

GF(qn) → GF(q) be a multiplicative map with N(ω) = α ∈ GF(q), then N(ωk) =

αk. If α = ωqn−1q−1 ·a then N(ωk) = Norm((ωk)a).

Exercise 5.5

−∑x∈GF(q) f(x)k is on the one hand the coefficient of Xq−1, on the other hand it

is∑a∈GF(q) a

k.

Exercise 5.6

If b = 0 then the equation is solvable as x 7→ x2 is an automorphism. Multiplyingthrough by a/b2 and introducing Y = aX/b we get Y 2 + Y + ac/b2 = 0. As Tr isadditive we have Tr(Y 2)+Tr(Y )+Tr(ac/b2) = 0. As, by definition, Tr(Y 2) = Tr(Y ),if the equation is solvable then Tr(ac/b2) = 0.

231

232 Hints and solutions

Conversely, if Tr(ac/b2) = 0 then the equation is solvable: the sets {y2 + y : y ∈GF(2h)} and {z : Tr2h→2(z) = 0} are identical; ⊆ is obvious and either of them isthe image space of a linear mapping with one-dimensional kernel.

Exercise 5.7

Let {deg(f) : f ∈ V } = {d1, d2, ..., dn}, where d1 > d2 > ... > dn. Choose a monicfi with deg(fi) = di, i = 1, ..., n. We prove that {f1, ..., fn} is a base of V . Letg ∈ V , then define g0 = g, and for k = 1, ..., n let ak be the coefficient of Xdk ingk−1, finally let gk = gk−1 − akfk. Observe that deg(gk) < dk and gk ∈ V . Nowgn = 0, so g =

∑aifi ∈ 〈f1, ..., fn〉.

Exercise 5.8

(X − a)λ =∑λr=0

(λr

)(−a)λ−rXr; use Lucas’ theorem (Exercise 5.1).

For the converse prove that∑a∈GF(q) a

q−1+r−λ(X−a)λ = (−1)λ−r−1(λr

)Xr unless

λ = q − 1 and r = 0 or q − 1.

Exercise 5.9

If f(aY + b, Y ) = 0 then f(X,Y ) = 0 modulo X − aY − b, and hence X − aY −b | f(X,Y ). It follows that

∏(a,b)∈S(X − aY − b) | f(X,Y ). Since the degree of

the left hand side is |S|, the result follows.

Exercise 5.13

Let A = {Pi : i = 1, ..., n+t−1} be an arc in PG(n−1, q). Then fi(X) = Pi ·X = 0is the equation of a hyperplane Hi through the origin in AG(n, q). Any point ofAG(n, q) \ 0 is covered at most (n− 1) times by these hyperplanes, because of thearc property.

Now fi(X)q−1 − 1 vanishes precisely in the points of AG(n, q) \ Hi; hence for∏n+t−1i=1 (fi(X)q−1−1), at any point of AG(n, q)\0 at least (n+ t−1)− (n−1) = t

factors will be zero. This is a polynomial of degree (n+ t− 1)(q − 1).

Exercise 5.14

Let Ci be defined by the polynomial fi(X,Y ), then consider∏i fi and use 5.12.

Exercise 5.21

Let gi(Xi) =∏si∈Si(Xi − si), and li(Xi) =

∏di∈Di(Xi − di). By Theorem 5.20

we can write f =∑ni=1 higi+w, and w = u

∏ni=1

gili

for some non-zero polynomialu, and the degree in Xi of w is less than |Si|.

Exercise 5.23

Suppose that f has a zero of degree at least t at all elements of S1×S2× ...×Sn.By Theorem 5.22 there are polynomials hτ ∈ F[X1, X2, ..., Xn] with the propertythat f =

∑τ∈T gτ(1)...gτ(t)hτ , and deg hτ ≤ deg(f) −

∑i∈τ deg(gi). On the right

1. Of Chapter I 233

hand side of this equality the terms of highest degree are divisible by∏i∈τ X

|Si|i

for some τ . So there is a τ for which ri ≥∑i∈τ |Si| for all i ∈ τ , a contradiction.

Exercise 5.26f is singular iff f(X) = 0 has more than one solution. For any solution x ∈ GF(q)one can consider the pk-th power 0 = f(x)p

k

. Using xph

= x and re-ordering, onegets the row of the matrix corresponding to this exponent. Hence for any solution x,the vector (x, xp, xp

2, ..., xp

h−1)> is a solution of the homogeneous matrix equation.

The other direction is similar.

Exercise 5.27As in the proof of Theorem 5.25, the “if” part is obvious. Consider GF(q) as an h

e -dimensional vectorspace over GF(pe), then f is a GF(pe)-linear map. The numberof such maps is (pe)

he . The polynomials of form Trq→pe(aX) are among these maps,

the number of these polynomials is q = (pe)he .

Exercise 5.32Multiply the sum by χ(4a2) = 1, we get χ(a)

∑x χ((2ax + b)2 − (b2 − 4ac)).

Replacing 2ax+ b by y and b2−4ac by d yields χ(a)∑y χ(y2−d). The case d = 0

is clear, for d 6= 0 we have to count the solutions of Y 2 − d = Z2. It defines ahyperbola with q − 1 affine points. If d is a nonsquare then for q−1

2 values y wehave χ(y2−d) = 1 and for the other q+1

2 it is −1. If d is a square then for y = ±√d

we have χ(y2 − d) = 0; for q−32 values y we have χ(y2 − d) = 1 and for the other

q−12 it is −1.

Exercise 6.1The number of k-dimensional (linear) subspaces in V(n, q) is

[nk

]q.

The number of k-dimensional (affine) subspaces in AG(n, q) is qn−d[nk

]q.

The number of k-dimensional projective subspaces in PG(n, q) is[n+1k+1

]q.

Exercise 7.6Xq−1 means that (1, 0, 0) is the nucleus in the even case, every line except thosethrough the nucleus, intersects the parabola in 0 mod 2 points. Adding (1, 0, 0) toS the *-polynomial becomes zero.In the odd case the tangents of the conic are the solutions [x, y, z] of X2 − 4Y Z.

Exercise 9.5Each factor of the right hand side should divide the determinant. The degree ofthe two sides is equal.

Exercise 9.7


Similar to Exercise 9.5. The ‘variables’ are x1, ..., xn only.

Exercise 9.13Let F (T ) =

∏x(T − f(x)). The Newton-Waring-Girard formulae (Ex. 9.3, N1-3)

imply that all the coefficients, i.e. the elementary symmetric polynomials σi arezero if i ≤ q−2 and p 6 |i. Hence F (T ) = G(T )p+ cT , where c =

∑x f(x)q−1. Here

c = 0 means case (ii), while c 6= 0 implies F ′(T ) = −c 6= 0, so F has no multipleroot, f is a permutation polynomial, F (T ) = T q − T , c = −1.

Exercise 9.14Let V (f) = {a1, ..., aK} and mi = |f−1(ai)| the “multiplicity” of ai. If wf < ∞then

m1 +m2 + ...+mK = 0a1m1 + a2m2 + ...+ aKmK = 0a21m1 + a2

2m2 + ...+ a2KmK = 0

...

awf1 m1 + a

wf2 m2 + ...+ a

wfK mK = β,

where β 6= 0. If K < wf + 1 then the first K equations has the trivial solutionm1 = ... = mK = 0 only, contradicting the last one.

Exercise 9.22Use Exercise 9.8!

Exercise 9.25Consider

1 − t 1 ... 1y1 y2 ... yty21 y22 ... y2t...

yt−11 y

t−12 ... y

t−1t

111

.

.

.1

=

000

.

.

.0

,

hence the determinant should be zero:V dM(y1, y2, ..., yt)− ty2y3 · · · ytV dM(y2, y3, ..., yt) = 0. Sot = V dM(y1,y2,...,yt)

y2y3···ytV dM(y2,y3,...,yt)= (y1−y2)(y1−y3)···(y1−yt)

y2y3···yt .

Exercise 10.1The number of coefficients is

(n+2

2

). So the “coefficient vectors” form the points

of an(n+2

2

)− 1 dimensional projective space over GF(q).

Exercise 10.2Each point (u, v) ∈ S gives a linear equation for the coefficients of the desiredpolynomial f . Hence we have a homogeneous system of |S| such equations. Whendeg f ≥

√2|S| − 1 then the number of coefficients is larger than |S|.

2. Of Chapter II 235

Exercise 10.8Consider f ∩ f (q), where f (q) denotes

∑ijk

(αijk)qXiY jZk when f(X,Y, Z) =∑ijk

αijkXiY jZk; then use Bezout’s theorem.

Exercise 10.10Consider the curve F (X,Y ) = f(X) − Y 2, it has roughly 2q points (a bit less infact). By Weil it cannot be irreducible. Write it as the product of two factors, it isnot too difficult to see that it must be of form f(X)−Y 2 = (g(X)−Y )(h(X)+Y ),and then g = h and f = g2.

Exercise 10.11Stand in a point P of the curve and look around! If P has multiplicity m ≥ 1 andthere is no linear component then on each line we can see at most n −m points,counted with intersection multiplicities, plus the one we are standing in. It gives≤ m+ (q + 1)(n−m) = qn+ n− qm ≤ q(n− 1) + n.

Exercise 10.12Note that 1 ≤ k ≤ n−1. (i) If we have a k-secant line ` and P ∈ `∩C then lookingaround from P we get the bound Nq ≤ (n−1)q+k. (ii) We say that a point P cansee the tangency at Q if Q is on the curve and (one of) the tangent line(s) at Q goesthrough P . (P = Q is allowed.) Now the points of the curve can see at least Nqktangencies, so there is at least one point P of the curve seeing at least k tangencies.Counting the points of the curve looking around from P we “lose” at least one(from the total number) at each tangency that P can see, which gives (n−1)q+n−kas an upper bound. Finally min{(n− 1)q + k, (n− 1)q + n− k} ≤ (n− 1)q + n

2 .

Exercise 10.16Use Theorem 10.9 by Stohr and Voloch!

Exercise 10.18Suppose that f is irreducible. By a coordinate transformation, which only per-mutes the elements of I, one can achieve f(X ′, Y ′, Z ′) = X ′2 − Y ′Z ′. But nowfp+12 6∈ I, contradiction.

2 Of Chapter II

Exercise 12.3Let {(ai, bi) : i = 1, ..., k} be an affine blocking set and suppose that (a1, b1) =(0, 0). Consider F (X,Y ) =

∏ki=2(aiX+biY+1). As {(ai, bi) : i = 2, ..., k} blocks all

the lines not through the origin, F (x, y) = 0 for all (x, y) 6= (0, 0). Now G(X,Y ) =


∏ki=2(aiX + biY + 1) − (Xq−1 − 1)(Y q−1 − 1) vanishes on GF(q) × GF(q), hence

Xq−1Y q−1 cannot be (one of) its leading term(s) by Theorem 5.17. Hence k−1 ≥2q − 2.

Exercise 12.5If P ∈ B is essential with t tangents through it then choose the coordinate systemso that P ∈ `∞ and `∞ is a tangent to B. Putting one point on each tangentexcept `∞ results in an affine blocking set of size |B| − 1 + t − 1, which is, byTheorem 12.2, at least 2q − 1, hence t ≥ 2q + 1− |B|.

Exercise 13.2If x ∈ S is of multiplicity m then from x in each direction one can see 1−m mod pfurther points. Hence (1) the number of points (with multiplicity) ism+(1−m)(q−1) ≡ 1 (mod p); (2)

∑s∈S(x− s)q−1 = (1−m)

∑εq+1=1 ε = 0, so again x is a root

of∑s∈S(X − s)q−1 of degree q − 1.

Exercise 13.3n = 1 is obvious, n = 2 is Proposition 13.1. Suppose that the statement is provedfor n − 1 ≥ 2 and consider a (multi)set S in AG(n, q). Let P ∈ H∞ be a pointat infinity (in other words, a direction) not determined by S. Then the quotientgeometry at P is ' PG(n − 1, q), the points of S go into points of a multisetS′ ⊂ PG(n − 1, q) in a natural way without any further coincidence; there is ahyperplane (i.e. the quotient of H∞) disjoint from S′, so S′ ⊂ AG(n− 1, q); finallyevery hyperplane of AG(n− 1, q) intersects S′ in either 0 or 1 mod p points, so bythe (induction) hypothesis we are done.

Exercise 13.4Counting the points of S on the lines through some fixed point s ∈ S we have|S| ≡ r (mod p). After the AG(2, q) ↔ GF(q2) identification define

f(X) =∑s∈S

(X − s)q−1,

it is not identically zero as the coefficient of Xq−1 is r. If x ∈ S then every directionwill occur with the same multiplicity r − 1 mod p, hence for each point s ∈ S,f(s) = (r− 1)

∑αq+1=1 α = 0, so they are roots of f , which is of degree q− 1. The

biggest value ≡ r mod p below q is q− p+ r. However, a set of size ≤ q+1 alwayshas a 1-secant, so the corollary is that r = 1.

Exercise 14.1See the proof of Exercise 10.11.

Exercise 14.2Let `i be the axis Xi = 0 and mj the line X2 = ajX1 + bj . So for the intersectionswith the axes we have GF(q)∗ = {aj : j = 1, ..., q − 1} = {bj : j = 1, ..., q − 1} =


{ajbj : j = 1, ..., q − 1}. By Wilson’s theorem (Section 9), we get −1 =∏j

aj =∏j

bj =∏j

ajbj

=Qj ajQj bj

= 1.

If q is even then the lines {X2 = aX1 + a2 : a ∈ GF(q)∗} suffice. In fact, if{`i}∪{mj} is a dual hyperoval then it works. Another solution is the following: letP be a point not on the triangle. Then the q−2 lines through P , not containing thevertices of the triangle, plus the line joining the three missing non-vertex pointson `1, `2 and `3 form such a set.

Exercise 14.7Let Q = L1∩L2 and let F and G two curves through the given points. As neither Fnor G contains Q, a suitable linear combination H of them does, hence H containsthe given points and Q as well. Then L1 is a component of H, so H intersectseach of the other lines Li also in at least n + 1 points, so

∏ki=1 Li divides H. As

degH = n < k, it follows that H = 0 and F = λG.

Exercise 15.1The condition is equivalent to saying that there is no collinear triple of points onthe graph of f .

Exercise 15.2As f(X) is a permutation polynomial, deg f ≤ q − 2. Calculate Fs(0) = f ′(s), itvanishes for every s ∈ GF(q) by Exercise 15.1.

Exercise 15.3Xr is a permutation polynomial iff (r, q − 1) = 1. F0(X) = Xr−1 gives (r− 1, q −1) = 1 similarly. For s 6= 0 Fs(X) = (X+s)r−sr

X = 1sr−1

(X/s+1)r−1X/s , which is a

permutation polynomial if and only if (Y+1)r−1Y is.

Exercise 15.4Use Exercise 15.3.

Exercise 15.11 Its size is 1 + (n− 1)(q+ 1), 1 for F0 and q+ 1 for the others. Thelines through (0, 0, 1) intersect each Fλ in one point. For the other affine lines ofequation aX + Y + bZ, b 6= 0, calculate the intersections and use that H is anadditive subgroup.

Exercise 17.13Let us count now the weighted index of points, that is let i(P ) =

∑ktn+k(P ),

where tn+i(P ) denotes the number of (n + i)-secants through P . If t0(P ) = q/n,then the index of P is just ε. In general, 0 ≤ i(P ) ≡ ε (mod n). In particular,for ε < n, the indices are always at least ε. On the other hand, adding up indicesalong a 0-secant gives the total (weighted number) of lines longer than n. For a


point of S, the index is ε, hence the total (weighted) number of long lines is atmost ε|S|/(n+ 1) < ε(q + 1). This is a contradiction.

Exercise 18.3Stand into a point P ∈ `\B and look around, you have to see at least one (affine)point of B on each line 6= `.In case of equality, if a direction P ∈ ` is determined by two affine points of B \ `then, as there are q affine lines through P and at least one of them contains atleast two of the q affine points of B, there exists an affine line through P which isnot blocked by the affine points of B hence P should be added. A non-determinedpoint means precisely one point per affine line through it, hence it is unnecessaryto put it to B.

Exercise 18.6A 3/2-transitive group is either primitive or a Frobenius group. As D ⊆ GF(pe),Frobenius: tran-

sitive, not regu-lar, ∀g 6=id has≤ 1 fixed points

we have that GF(pe) is a block of FD, hence FD is not primitive. As F linD ⊆ FD, it

follows that F linD = FD.

Exercise 18.8Use Exercise 18.5.

Exercise 18.12(i) If bi = bj then the parabolas meet at X = 0. (ii) The equation ai−aj

bi−bj = −1/x2

must not have a solution x ∈ GF(q). (iii) f defines ≤ p+12 directions, hence its

graph is a line. If it is non-horizontal then it intersect the horizontal axis.

Exercise 23.3Use that the eigenvalues of the incidence matrix are q + 1 and ±√q.

Exercise 24.6Assume that 0 is in the blocking set, and π is the dual translation plane withrespect to 0. The dual π is a translation plane of order q2 with kernel of order q.Each line of π can be regarded as a plane of AG(4, q) and by Jamison or Theorem24.5 with n = 4, k = 2, t = 1 we are done.

Exercise 24.8Let l1, l2, ..., ln be n lines of PG(n, q) incident with the same point x of A andspanning PG(n, q). Let H be the hyperplane which is incident with no point of Aand set Si = li \ {x} and Di = li ∩H. Theorem 24.7 implies |A| − 1 ≥ (t− 1)(q −1) + n(q − 1).

Exercise 24.13The binomial coefficient in Corollary 24.12 with n = 2 and j = k = t − (t, q) is(−1−(t,q)t−(t,q)

), which is non-zero by Lucas Theorem.


Exercise 27.5Use the Stohr-Voloch bound (Exercise 10.16).

Exercise 27.6Use σk(T ) def= σk(T ) instead of σ∗k(T ) for k = 1, ..., ε in the proof; everythingremains the same, like in Section 11.

Exercise 27.7The secant joining (c, c2) and (d, d2) is Y = (c + d)X − cd. If the missing pointsare (γ, γ2) and (δ, δ2) then we need α = γ + δ and β = −γδ. As

∑z∈GF(q) z = 0

and∏z∈GF(q)∗ z = −1, we have α = −

∑ai and β = ±

(∏ q−32

i=1 bi

)−1

.

Exercise 29.3By the preceding, every such set B must be a subgroup, hence a line through theorigin. G = A + B implies that every line parallel to B must intersect A, hencethe direction of B is not determined by A. By Theorem 18.1, there are at mostp−12 such directions.

Exercise 29.4For x 6= y ∈ GF(p2), observe that x − y is a square in GF(p2)∗ iff (x − y)q−1 is asquare in the multiplicative subgroup of the (p+1)-th roots of unity (⊂ GF(p2)∗).So after the GF(p2) ↔ AG(2, p) identification, X ⊂ AG(2, p) determines at mostp+12 directions, hence, by Theorem 18.1, it is a line containing 0 and 1.

Exercise 29.8Use Theorem 18.7.

Exercise 29.20Define f =

∏mi=1(1 − fi(X1, X2..., Xn)q−1) and note that f is non-zero only pre-

cisely at the common zeros of f1, f2, ..., fm. If there is a common zero then Theorem5.20 implies that deg f = (q − 1)

∑mi=1 deg(fi) ≥ nq −

∑ni=1 |Di|.

Exercise 29.22Let’s identify V(d, p) with GF(pd). Let C = A + B, suppose |C| = n < βp(r, s).Define f(X,Y ) =

∏c∈C(X + Y − c), then f vanishes on A×B. By the definition

of βp(r, s) there exists some n − r < k < s such that(nk

)6≡ 0 (mod p). So the

coefficient of Xn−kY k in f is nonzero, and (as |A| = r > n − k, |B| = s > k), byTheorem 5.17 it is a contradiction.

Chapter 4

Glossary of concepts

Here one can find the most important definitions, in alphabetical order.

A (k, n)-arc of PG(2, q) is a pointset of size k, meeting every line in at most npoints. An arc is a (k, 2)-arc. A (k, n)-arc is complete if it is not contained in a(k + 1, n)-arc.

A blocking set (w.r.t. lines) is a pointset meeting every line. In general, a blockingset in PG(n, q) w.r.t. k-dimensional subspaces (sometimes it is called an (n − k)-blocking set) is a pointset meeting every k-subspace. A t-fold blocking set meetsevery k-subspace in at least t points. A point P of B is essential if B \ {P} is nolonger a blocking set, i.e. there is a 1-secant k-space through P . B is minimal ifevery point of it is essential. A blocking set B of PG(2, q) is small if |B| < 3

2 (q+1),in general, a blocking set B in PG(n, q) w.r.t. k-dimensional subspaces is small if|B| < 3

2qn−k + 1.

Given B ⊂ PG(2, q), the point P 6∈ B is a t-fold nucleus of B if all the linesthrough P intersect B in at least t points. P 6∈ B is a t-fold lower nucleus of B ifall the lines through P intersect B in at most t points. So from a nucleus B seemsto be a t-fold blocking set while from a lower nucleus B seems to be a (|B|, t)-arc.If P is a point of B then the similar notions are called internal nuclei.

A blocking set B ⊂ PG(n, q), with respect to k-dimensional subspaces, is of Redeitype, if it has precisely qn−k points in the affine part AG(n, q) = PG(n, q) \H∞.

A semioval is a pointset of PG(2, q) with the property that there is a unique tangentline at each point of it. It is regular, if it is a pointset of type (0, 1, a), i.e. all thesecants are of the same length a.

A k-spread of PG(n, q) is a partition of the space into k-dimensional subspaces.

241

242

A subgeometry of Π = PG(n, q) is a copy of some Π′ = PG(n′, q′) embedded in it,so the points of Π′ are points of Π and the k-dimensional subspaces of Π′ are justthe intersections of some k-subspaces of Π with the pointset of Π′. It follows thatGF(q′) must be a subfield of GF(q).

The type of a pointset of PG(2, q) is the set of its possible intersection numberswith lines. In particular, an arc is a set of type (0, 1, 2), a set of even type is apointset with each intersection numbers being even, etc.

A unital is a pointset of PG(2, q2) of size q3 + 1 intersecting each line in either 1or q+1 points. It has a unique 1-secant through each point of it so it is a minimalblocking set.

An untouchable set is a set without tangents.

Chapter 5

Notation

V(n,F) denotes the n-dimensional vector space with coordinates from the field F.If F = GF(q) then we write V(n, q) instead.

AG(n,F) denotes the n-dimensional affine space with coordinates from the field F.If F = GF(q) then we write AG(n, q) instead.

PG(n,F) denotes the n-dimensional projective space with coordinates from thefield F. If F = GF(q) then we write PG(n, q) instead.[ab

]q

= (qa−1)(qa−1−1)...(qa−b+1−1)(qb−1)(qb−1−1)...(q−1)

(the q-binomials or Gaussian binomials, thenumber of b-dimensional linear subspaces of V(a, q)).

If the order q of a plane or space is fixed we writeθi =

[i+11

]q

= qi+1−1q−1 = qi + qi−1 + ...+ q + 1.

Trqn→q(X) = X + Xq + Xq2 + ... + Xqn−1is the trace function from GF(qn) to

GF(q).

Normqn→q(X) = XXqXq2Xqn−1is the norm function from GF(qn) to GF(q).

In some cases ω will denote a primitive element of GF(qn) (and sometimes it isnice to assume that (ω, ωq, ωq

2, ..., ωq

n−1) form a trace-orthogonal base over GF(q)

if possible, see 5.2).

Jt is the ideal 〈(Xq1−X1)i1(X

q2−X2)i2 ...(Xq

n−Xn)in : 0 ≤ i1+i2+ ...+in = t〉 inGF(q)[X1, ..., Xn] of polynomials vanishing everywhere with multiplicity at least t.

H∞ is the hyperplane at infinity in the projective space PG(n, q) when an “affinepart” is fixed, i.e. H∞ = PG(n, q) \ AG(n, q). When n = 2, it is called the “line atinfinity” `∞.

243

244

Mf : given the polynomial f ∈ GF(q)[X], the number of elements a ∈ GF(q) forwhich f(X) + aX is a permutation polynomial.

Df : given the polynomial f ∈ GF(q)[X], Df = { f(x)−f(y)x−y : x 6= y ∈ GF(q)}, the

set of directions determined by the graph of f .

Nf = |Df |.

wf : given the polynomial f ∈ GF(q)[X], wf = min{k :∑x∈GF(q) f(x)k 6= 0}.

Wf : given the polynomial f ∈ GF(q)[X], Wf = min{k+ l :∑x∈GF(q) x

kf(x)l 6= 0}.

N(B) denotes the set of (external) nuclei of the pointset B.

IN(B) denotes the set of internal nuclei of the pointset B.

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