Durham UniversityDepartment of Mathematics

MMath Thesis

The Polynomial MethodA Combinatorial method for bounding solution sets

Samuel JP McStaySupervised byDan EvansApril 29, 2016

Abstract

During this study we will survey the Polynomial method, going from its origins withinCombinatorics, to its most recent applications in Discrete geometry. Through use of

worked examples we will explore the various tools which comprise the method,discussing their applications in a number of significant papers, and considering the

potential extensions for future research.

Declaration

This piece of work is a result of my own work except where it forms an assessmentbased on group project work. In the case of a group project, the work has been

prepared in collaboration with other members of the group. Material from the work ofothers not involved in the project has been acknowledged and quotations and

paraphrases suitably indicated.

Contents

1 Introduction 31.1 Outlining the Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Nullstellensatz . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2.1 Proof of the Combinatorial Nullstellensatz . . . . . . . . . . . . . . 51.3 Using the Combinatorial Nullstellensatz . . . . . . . . . . . . . . . . . . . 71.4 Algebraic Geometry: Connecting the Nullstellensatze . . . . . . . . . . . . 81.5 Bezout’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

I Sumsets and Finite Fields 14

2 Chevalley-Warning Theorem 152.1 An Interesting History . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.2 The Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162.3 Proving the Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2.3.1 The Reduced Polynomial . . . . . . . . . . . . . . . . . . . . . . . 192.3.2 Uses of the Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 21

3 Sumsets 223.1 Introducing Sumsets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

3.1.1 Spin to win - Forming simple sumsets . . . . . . . . . . . . . . . . 223.2 Cauchy-Davenport Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 24

3.2.1 Proving the Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 243.3 The Erdos-Heilbronn Problem . . . . . . . . . . . . . . . . . . . . . . . . . 25

3.3.1 Proving the EHP . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

4 The Erdos-Ginzburg-Ziv Theorem 284.1 Understanding Zero-sum Sequences . . . . . . . . . . . . . . . . . . . . . . 28

4.1.1 Just Sum Dice Game . . . . . . . . . . . . . . . . . . . . . . . . . . 294.1.2 Implications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

4.2 The Erdos-Ginzburg-Ziv Theorem . . . . . . . . . . . . . . . . . . . . . . 314.2.1 Proving EGZ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

4.3 Further Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

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4.3.1 Zn ` Zn and beyond . . . . . . . . . . . . . . . . . . . . . . . . . . 34

5 Polynomial Testing 355.1 The DeMillo-Lipton-Schwartz-Zippel Lemma . . . . . . . . . . . . . . . . 35

5.1.1 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

6 Kakeya 386.1 The Kakeya needle problem . . . . . . . . . . . . . . . . . . . . . . . . . . 386.2 The Finite Field Analogue . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

6.2.1 The Proof of the Finite field Kakeya conjecture . . . . . . . . . . . 41

II Euclidean Space 44

7 Distances 457.1 The s-Distances Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

7.1.1 The Two Distance Problem . . . . . . . . . . . . . . . . . . . . . . 457.2 Bounding Pspnq . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

7.2.1 Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

8 Lines and Joints 518.1 The Joints Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 518.2 Solving the Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

8.2.1 The General Solution . . . . . . . . . . . . . . . . . . . . . . . . . 538.3 Extensions of the Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

8.3.1 The Erdos Distinct Distances Problem . . . . . . . . . . . . . . . . 56

III Graph Theory 57

9 Graph Theory 589.1 Polynomials of Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 589.2 The Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 589.3 Finding a p-regular Subgraph . . . . . . . . . . . . . . . . . . . . . . . . . 599.4 Graph Colouring and the use of Ideals . . . . . . . . . . . . . . . . . . . . 61

9.4.1 k-Colourability - An Ideal ending . . . . . . . . . . . . . . . . . . . 64

10 Concluding Remarks 66

Bibliography 68

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Chapter 1

Introduction

The Polynomial method is a collection of combinatorial tools used in mathematicalproofs. The ideas behind it can be found in writings from over a century ago but it roseto the forefront of mathematical study in 2008. This recognition was largely due to thework of Dvir, who solved the Finite field Kakeya conjecture using the method [1]. Sincethen, the method has been applied to many more problems within Combinatorics togreat success, and its full potential is still unknown. To date, its main applications havebeen in Incidence geometry but it has also been used widely in Coding theory, Graphtheory and Additive number theory. Despite its remarkable simplicity, it remains anessential tool in modern mathematical innovation.

1.1 Outlining the Method

As the name suggests the method is built around the structures of polynomials. Remark-ably simple to work with, we can view polynomials in two ways; either as evaluationmaps, or as formal objects in a ring. The well defined nature of polynomials means theyhave a number of practical properties and useful structure. The Polynomial method at-tempts to take unknown information and carefully embed it into a polynomial. By thenstudying this polynomial, we can achieve a better understanding of our original structure.While this method has a basic form, we can construct more powerful tools and theo-rems from it. By using these more complex constructions we can apply the Polynomialmethod to a problem in a more efficient and effective way. Our basic method, as seenin [2], is derived from the following statements concerning single-variable polynomials.

Theorem 1.1.1 (The Fundamental Theorem of Algebra). A non-zero, single-variable,polynomial of degree d has no more than d distinct roots, and has exactly d roots whencounted with multiplicity over its algebraic closure.

Theorem 1.1.2. For any set S of cardinality d there exists a non-zero polynomial f ofdegree at most d which vanishes on S, that is, fpsq � 0 for all s P S.

Given a polynomial of fixed degree, the first theorem upper bounds the cardinalityof its vanishing set. The second says conversely that every set is the vanishing set of

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some polynomial, with degree bounded by the cardinality of the set. In conjunction,these statements allow us to apply the method in both directions. We can bound thenumber of roots of any polynomial based on its vanishing points, and we can boundthe cardinality of a set, by the degree of polynomial vanishing on it. The multi-variableanalogues of these statements form the basis of our Polynomial method.

More generally we look at properties such as degree and dimension from which, byusing counting arguments, we can obtain bounds on the complexity of a set. We boundand from below by showing that no polynomials of low degree vanish on our set, andfrom above by finding a polynomial which does. In this way we can bound the solutionsto a variety of problems. We can further generalise these ideas to systems of equationsthrough tools such as the Combinatorial Nullstellensatz which, as the name suggests, islinked to the Hilbert Nullstellensatz. These extensions require careful analysis of vanish-ing sets and applications of our basic Polynomial method. This focus on vanishing setsmeans the topic is closely related to, and reliant upon, results from Algebraic geometry.

1.2 Nullstellensatz

Nullstellensatz is a German word meaning “zero-locus theorem” immediately denoting itsassociation to vanishing sets (see [3]). First published in 1893 [4], Hilbert’s Nullstellensatzis derived by focussing on ideals. The statement is a powerful tool concerning thesolutions of any homogeneous system of polynomial equations over an algebraically closedfield (most usually C in Algebraic geometry). The two statements (as seen in [5] and [6]respectively) are equivalent however for convenience we denote them separately. We firstdefine some notation with which we can state the theorem.

Definition 1.2.1. For any set of polynomials P � tp1, . . . , pku define

ZpP q :� tps1, . . . , snq P Fn | pps1, . . . , snq � 0 @ p P P u,

the vanishing set of P .

Theorem 1.2.2 (Hilbert’s Strong Nullstellensatz [5], [6]). Take F to be an algebraicallyclosed field, P � tp1, . . . , pku to be a set of polynomials, and f, p1, . . . , pk to be polyno-mials over Frx1, . . . , xns. Then

I. Take an ideal J in Frx1, . . . , xns. If f vanishes on ZpJq then there exists an integerr ¡ 0 such that

f r P J.

II. If f vanishes on ZpP q then there exists an integer r and polynomials g1, . . . , gk PFrx1, . . . , xns such that

f r �k

i�1gipi.

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This theorem is powerful by itself, proving a fundamental link between Geometryand Algebra. It is therefore the basis of much of Algebraic geometry.

It was from this theorem that Noga Alon made a notable breakthrough to create theCombinatorial Nullstellensatz, which he first published in 1999, [6] (though evidence ofit can be seen in his works as early as 1995, [7]). Similar to the general Hilbert casebut with further restrictions, his Nullstellensatz makes stronger statements about thenature of this set-polynomial construction. Having a more combinatoric approach to thestructures, this Nullstellensatz is derived using our Polynomial method and is a powerfultool for applying the method to problems.

Following Alon’s original line of thought, [6], we take the case as above with the re-striction that k � n and that each pi is a single-variable polynomial factoring completely.Set Si :� Zpfiq, then we have the following:

Theorem 1.2.3 (The Combinatorial Nullstellensatz [6]). Take F to be an algebraicallyclosed field and f in Frx1, . . . , xns. Then

I. Take a set of polynomials P � tp1, . . . pnu with

pipx1, . . . , xnq :�¹sPSi

pxi � sq

and all Si non-empty. If f vanishes on ZpP q � S1�. . .�Sn, there exist polynomialsg1, . . . , gn P Frx1, . . . , xns with deg pi � deg gi ¤ deg f such that

f �n

i�1gipi.

Furthermore if f, p1, . . . , pn lie in a subring of F then g1, . . . , gn also lie in thesubring.

II. Suppose deg f � °ni�1 αi for some non-negative integers αi and the coefficient of±n

i�1 xαii in f non-zero. Then if S1, . . . , Sn are subsets of F such that |Si| ¡ αi

then there exists an element s � ps1, . . . , snq P S1 � . . .� Sn such that fpsq � 0.

This theorem, posed as two separate statements, allows us to apply the Polynomialmethod to a whole system of equations simultaneously. This makes it very useful insome of the most profound works of the Polynomial method. We shorten many proofsby applying the method to a whole system in a single step. To prove this theoremrequires careful use of our basic Polynomial method.

1.2.1 Proof of the Combinatorial Nullstellensatz

We will prove the Combinatorial Nullstellensatz, as in Alon’s original proof (seen in [6],and [5]) using three steps. We first prove the following lemma, then prove each statementseparately.

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Lemma 1.2.4. Suppose f P Frx1, . . . , xns is such that the degree of f in xi is at mostαi, for all i P t1, . . . , nu, and let Si � F have cardinality at least αi � 1. Then if fvanishes on S1 � . . .� Sn, it is the zero polynomial.

Proof. We prove this by induction. For n � 1 assume we have a non-zero single-variablepolynomial of degree at most αi but with at least αi�1 roots. By Theorem 1.1.1 we musthave the zero-polynomial. Since it is true for n � 1 we can now perform our inductivestep, assuming the lemma holds for n� 1, n ¥ 2, and from this showing it holds for n.

Take a polynomial fpx1, . . . , xnq with sets Si as in the statement of the lemma. Wecan write f as a polynomial in xn taking fipx1, . . . , xn�1q to be the coefficient of xin suchthat

f �αn

i�0fipx1, . . . , xn�1qxin.

Fixing the values of x1, . . . , xn as some value s1 � . . .� sn�1 P S1 � . . .� Sn�1 gives usa polynomial in one variable that vanishes on Sn. Since we know the lemma holds forn � 1 this polynomial is identically 0. This means the coefficients are identically zeroimplying that fipx1, . . . xn�1q vanishes S1� . . .�Sn�1. By our inductive assumption thelemma holds for n� 1 and therefore fi � 0 for all i implying f � 0.

We can now prove the first statement of the Combinatorial Nullstellensatz.

Proof of Combinatorial Nullstellensatz I. We again take f P Frx1, . . . , xns vanishing onthe non-empty set S1 � . . .�Sn � Fn and set αi � |Si| � 1. Then for each i P t1, . . . , nudefine

pipxiq :�¹sPSi

pxi � sq � xαi�1i �

αi

j�0pcijxji q

where cij is simply the coefficient of xji . By construction this vanishes on Si and therefore

sαi�1 �αi

j�0pcijsjq

for all s P Si. By applying this relation we can create a reduced polynomial of maximumdegree αi in xi by writing higher powers as linear combinations of lower order terms.

We apply this process to f to get a reduced form f . Note that where this relationholds f will equal f and therefore

fps1, . . . snq � fps1, . . . snq for all ps1, . . . snq P S1 � . . .� Sn.

This gives us that fpx1, . . . xnq also vanishes on S1 � . . .� Sn meaning by Lemma 1.2.4f � 0.

We now look closely at the process taking f to f . We note that the xi terms arereduced by subtracting multiples of the pi of the form gipi where gi P Frx1 . . . xns. Since

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the aim is to cancel out terms of f with multiples of the pi, the coefficients of the gi arewithin the smallest ring containing all the coefficients of f, p1, . . . , pn. We also see that

deg gipi ¤ deg f

which implies thatdeg gi � deg pi ¤ deg f.

as in the statement of the theorem. Finally we see that

f �n

i�1gipi � f � 0

and thereforef �

n

i�1gipi.

We now prove the second statement of the Combinatorial Nullstellensatz.

Proof of Combinatorial Nullstellensatz II. We again set pipxiq �±sPSipxi � sq and set

|Si| � αi�1 with αi is the degree of f in xi. Assume for contradiction that f vanishes onS1� . . .�Sn. We can therefore apply Combinatorial Nullstellensatz I to get polynomialsgi P Frx1 . . . xns with deg gi ¤

°ni�1 αi � deg pi such that

f �n

i�1gipi.

On the left hand side this equation has a maximal degree term±ni�1 x

αii with a non-zero

coefficient hence so does the right hand side. However the RHS is composed of terms ofthe form gipi hence we can write it as

gi¹sPSi

pxi � sq � gi

�xαi�1i �

αi

j�0pcijxji q

�.

The maximal degree terms here must all be divisible by xαi�1i and there can be no term

of form±ni�1 x

αii . This gives a contradiction, so f does not vanish on S1 � . . .� Sn.

This completes the proof of the Combinatorial Nullstellensatz.

1.3 Using the Combinatorial Nullstellensatz

To give an example of using the Combinatorial Nullstellensatz we look at an elegantexample studied in the Gazeta Mathematica [8]. The problem comes from the 2007International Mathematical Olympiad and has a beautiful combinatoric solution, foundby only 5 of the participants.

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Example 1.3.1 (Covering the unit grid except p0, 0, 0q). Let n be a positive integer.Consider

S � tpx, y, zq | px, y, zq P t0, 1, . . . , nu3, x� y � z ¡ 0ua set of pn� 1q3 � 1 points in R3. Determine the smallest number of planes, the unionof which contains S but does not include p0, 0, 0q.Solution. We begin by upper bounding our answer by 3n. This is attained by the solution

3n¤i�1

αi where αi :� x� y � z � i

By contradiction we now show that this is minimal. Assume we have a solution using only3n�1 planes, π1, . . . π3n�1. Each plane can be defined by an equation aix�biy�ciz�di �0 with di � 0 since we are avoiding the origin. We define

gpx, y, zq �3n�1¹i�1

paix� biy � ciz � diq

and note that if we have a solution then g, the product of our planes, will vanish on Sbut not at the origin since all our di are non-zero.

We now form a new polynomial fpx, y, zq from gpx, y, zq which vanishes on S Y 0,

fpx, y, zq � gpx, y, zq � kn¹i�1px� iqpy � iqpz � iq

with k � gp0,0,0qp�13nqpn!q3 so that fp0, 0, 0q � 0. Since g has degree 3n� 1, and the coefficient

of xnynzn is k � 0 we see that f has degree 3n. We are now in a case where we canapply our Combinatorial Nullstellensatz II to achieve our contradiction.

We are working over the field R with f P Rrx, y, zs, deg f � 3n � n� n� n, and thecoefficient of xnynzn in f , is k � 0. We then take S1 � S2 � S3 � t0, 1, . . . , nu � R eachof cardinality n�1 such that |S| ¡ n. Hence by our Combinatorial Nullstellensatz thereexists an ps1, s2, s3q P S1 � S2 � S3 such that fpsq � 0 contradicting that f vanishes onS Y 0.

This sample problem, studying a polynomial formed by the products of planes, fur-ther suggests the link between the Nullstellensatze and Algebraic geometry.

1.4 Algebraic Geometry: Connecting the Nullstellensatze

In appearance the two Nullstellensatze are quite similar. The important difference beingthat by looking at a more specialised case, we can replace f r by f . Removing thispower allows the Combinatorial Nullstellensatz to be applied effectively in a number ofmore combinatorial situations where the Hilbert Nullstellensatz isn’t applicable. This isbecause we can make much more rigorous statements a polynomial than we can about a

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polynomial raised to an unknown power. On the other hand the significance of Hilbert’sNullstellensatz is its foundational importance to Algebraic geometry. In fact we can nowclarify definition 1.2.1 to reflect this.

Definition 1.4.1 (Algebraic Set). For any P � Frx1, . . . , xns and ZpP q as in Definition1.2.1, we may call ZpP q an algebraic set.

Remark. For further insight into this section the reader is directed to a set of lecturenotes by Gathmann [9].

These algebraic sets are the fundamental objects of study in Algebraic geometry. Aswe have already seen our Polynomial method is very closely linked to these sets andtherefore Algebraic geometry.

Theorem 1.4.2. For any algebraically closed field F, there is a one-to-one correspon-dence between the radical ideals of Frx1, . . . , xns and the algebraic sets of Fn.

This theorem is the basis of Algebraic geometry, proving the link between our alge-braic objects, the radical ideals, and our geometric structures, the algebraic sets. Theproof relies on Hilbert’s Nullstellensatz and is perhaps the most important use of it.While the proof is not strictly an application of the Polynomial method it bears someresemblance and demonstrates the use of Hilbert’s Nullstellensatz effectively. This proofis adapted from the work of Jones in [5].

Proof.

I. We define two maps and note some of their properties.

(a) Our first map takes a set from Fn to the collection of polynomials vanishingon it.

IpSq :� tf P Frx1, . . . xns | fps1, . . . , snq � 0 @ ps1, . . . , snq P Su

(b) Note that IpSq is closed under addition and under multiplication by any poly-nomial, hence it is the ideal of S. Furthermore we note that if gkps1, . . . , snq �0 then gps1, . . . , snq � 0. This means if a power of an element, g, is in J thenso is g. Hence IpSq is in fact a radical ideal.

(c) Our second map is Z as before,

ZpJq :� tps1, . . . , snq P Fn | fps1, . . . , snq � 0 @ f P Jubut with J a radical ideal. By Definition 1.4.1 this maps to an algebraic set.

We now look at the composition of these two maps.

II. ZpIpSqq:(a) S � ZpIpSqq: Take an algebraic set S. All f P IpSq vanish on S, and therefore

s P ZpIpSqq for any s P S.

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(b) ZpIpSqq � S: Now take ps1, . . . , snq P ZpIpSqq. Then for all f P IpSq we havethat fps1, . . . , snq � 0. Therefore ps1, . . . , snq P S.

(c) These two statements imply that ZpIq is the identity on algebraic sets.

III. IpZpJqq:

(a) pJ � IpZpJqq: For all f P J , f vanishes on ZpJq. Therefore f P IpZpJqq.(b) pIpZpJq � Jq: Now take f P IpZpJqq. This means f vanishes on ZpJq. We

now apply Hilbert’s Strong Nullstellensatz I implying that f r P J for someinteger r ¡ 0. Therefore f P J because J is radical.

(c) These two statements imply that IpZq is the identity on radical ideals.

We therefore have a one-to-one correspondence as defined by these maps, completingthe proof.

We will now work through an example to help explain the theorem using the I andZ functions as seen in the proof.

Example 1.4.3. Take the algebraically closed field C and take n � 1 so that we workover Crzs. Define our algebraic set to be

S � t0, 1, 2u � C.

For a polynomial to vanish on this set it clearly must be of form gpzqfpzq where gpzq isa polynomial in Crzs and

fpzq � zpz � 1qpz � 2q.Therefore the ideal of S, the set of polynomials vanishing on it, is generated by fpzq.That is

IpSq � th P Crzs|hpzq � 0 @ z P Su� tgpzqfpzq|gpzq P Crzsu� xfy.

If we now look at where this ideal vanishes, we see that ZpIpSqq � S. This is becauseall of IpSq shares f as a common factor but no other root. Therefore we see ZpIpSqq isthe identity on S.

Conversely, with f as above, start with the ideal

J � xfy.We know from before that this vanishes only on S � ZpJq. Therefore the ideal of this is

IpZpJqq � IpSq� xfy� J

Which shows that IpZpJqq is the identity on J .

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The close resemblance between our method and the fundamentals of Algebraic ge-ometry, means that often the two areas cross over. While the Polynomial method hasapplications in many other areas, this one is of particular note. We therefore consideranother important result from Algebraic geometry.

1.5 Bezout’s Theorem

One of the most important theorems addressing the size of vanishing sets is Bezout’sTheorem. Given a homogeneous system of polynomials, Bezout’s Theorem bounds thenumber of intersecions. The ability to do this, with only minimal knowledge about ourfunctions, will be important in later problems, particularly within Euclidean Space. Webegin by reflecting on this example (seen in [3]).

Example 1.5.1. Take the following two polynomials acting on the plane R2:

fpx, yq � px� 1q . . . px� nqgpx, yq � py � 1q . . . py �mq

and consider the solutions of f � g � 0. Note that deg f � n, deg g � m and that

Zpfq � tpa, yq P R2 | a P t1, 2, . . . , nuuZpgq � tpx, bq P R2 | b P t1, 2, . . . ,muu.

Representing these sets graphically we can easily count the number of intersections,|Zpfq X Zpgq|. Note that this is well defined (finite) since f and g do not share anon-constant common component.

Zpfq

Zpgq

Figure 1.1: A graphical representation of the intersections of our two curves.

Clearly there are precisely nm intersections and therefore

|Zpfq X Zpgq| � deg f � deg g.

In fact by Bezout’s Theorem this is maximal.

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Theorem 1.5.2 (Bezout’s Theorem). Let f, g P Frx, ys be polynomials with no non-constant common components, and of degree n and m respectively with m,n ¥ 1. Then

|Zpfq X Zpgq| ¤ mn � deg f � deg g.

Furthermore if F is algebraically closed, and we count intersections properly, that isincluding intersections at infinity and counting with multiplicity, then

|Zpfq X Zpgq| � mn.

This theorem acts a multivariate extension of our Fundamental Theorem of Algebra1.1.1 and can be used to prove the Joints problem in Rn. I therefore include a proof ofthis form of the theorem. The format of this proof is adapted from a proof by Hilton [10].

Remark. While this proof is essentially complete I have chosen to limit the detail onthe construction of the resultant and its properties as it is not further relevant to thisstudy. Hilton and Guth provide more insight into this concept, [10, 11].

Proof. (Sketch).

I. Take f , g P Frx, ys polynomials over an algebraically closed field with no commoncomponents. Then as previously

Zpfq � tpx, yq P F2 | fpx, yq � 0uZpgq � tpx, yq P F2 | gpx, yq � 0u

II. Define fi and gi as the homogeneous components of f and g of degree i such that

f �m

i�0fipx, yq

g �n

i�0gipx, yq

III. Homogenise our equations by adding a dummy variable z to get equations for Fand G homogeneous of degree m and n respectively.

F �m

i�0fiz

m�i

g �n

i�1giz

n�i

IV. We now calculate RespF,Gq, the resultant of these two polynomials. That is thedeterminant of the Sylvester matrix:

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RespF,Gq �

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

f0 f1 � � � fm 0 � � � 00 f0 f1 � � � fm 0 � � � 00 0 f0 f1 � � � fm 0 � � � 0... . . . . . . . . . . . . . . . ...0 � � � 0 f0 f1 � � � fm 00 � � � 0 0 f0 f1 � � � fmg0 g1 � � � gn 0 � � � 00 g0 g1 � � � gn 0 � � � 0... . . . . . . . . . . . . . . . ...0 � � � 0 g0 g1 � � � gn 00 � � � 0 g0 g1 � � � gn

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

.

The resultant has the remarkable property of being zero only where F and G sharea solution in z. Note also that all terms of the resultant have degree mn meaningour resultant is a homogeneous polynomial in x and y of degree mn, [10].

V. Since our field is algebraically closed we see, by Theorem 1.1.1, that the resultantfactors into mn linear factors. Each factor corresponds to a line from the origin to apoint of intersection of the curves, hence we have exactly mn points of intersection(counting properly).

Note that the first case follows from the algebraically closed case. If there are exactlymn solutions within the algebraic closure F then there are at most mn solutions withina sub-field of F.

As is key when using the Polynomial method, this theorem bounds the complexity ofvanishing sets. By cleverly applying this we can get new results, particularly when usedin conjunction with other aspects of the method. In essence this theorem allows us toextend the principle of looking for roots, (where a function intersects the line y � 0), tolooking for intersections with more complicated structures. As such Bezout’s Theoremis an extension of Theorem 1.1.1 and so it is considered part of the toolbox that is ourPolynomial method.

Remark. For further reading on Bezout’s Theorem, and the topic in general, the readeris directed to the lecture notes of Larry Guth [11]. A higher level, but comprehensiveresource, on this and the Polynomial method in general, is a survey by Terence Tao,[12]. This is an excellent resource for much of this topic but notably discusses Bezout’sTheorem in further detail.

13

Part I

Sumsets and Finite Fields

14

Chapter 2

Chevalley-Warning Theorem

Remark. Throughout this chapter the reader is directed to the works of Pete Clarkin [13], [14] for further details and information.

2.1 An Interesting History

This theorem has controversial origins as the results arose from the work of two com-peting mathematicians. This led to a collection of very similar statements proposed andproved by the pair. 1934 was a tumultuous time for Emil Artin and his half-Jewish wife,as residents of what was Nazi Germany. Despite his strong anti-Nazi views he continuedto work there before his emigration to the U.S.A in 1937. During this period he wasworking on finite fields, supervising a PhD student Ewald Warning. Much of their workconcerned the following concept formulated by Artin [13], [15].

Definition 2.1.1. A C1-field is a field F such that given a homogeneous polynomial

fpx1, . . . , xnq P Frx1, . . . , xns

of degree d, with d ¡ n, there exists a non-zero point, s, of Fn such that fpsq � 0.

It is now known that all fields are C1-fields. However at the time, for finite fields,this was an open problem and Artin decided to task Warning to solve this. It was at thispoint that Claude Chevalley visited Artin and, after discussing the ongoing work, wentaway and released a proof of the conjecture himself, [16]. Not to be outdone, Warningfinished his thesis and furthered Chevalley’s result. The title of his work even translatesas “Remarks preceding work of Mr. Chevalley” and his introduction comments that hisresults further Chevalley’s work, [17]. Compiling their works gives us the Chevalley-Warning Theorem.

15

2.2 The Theorems

Theorem 2.2.1 (Chevalley-Warning Theorem [13], [14]). Take a set S of m polynomialsover a finite field,

S � tf1, . . . , fmu � Fqrx1, . . . , xnswith deg pfiq � αi ¥ 1 and q � pk for some prime p. Then if

m

i�1αi � d   n

we have that

I. (Chevalley’s Theorem). If ZpSq � ∅ then |ZpSq| ¥ 2.

II. (Warning’s first Theorem). |ZpSq| � 0 pmod pq.III. (Warning’s second Theorem). If ZpSq � ∅ then |ZpSq| ¥ qn�d.

In line with our Polynomial method these statements look at the vanishing sets ofsystems of polynomials. They allow us to bound the cardinality based on the degrees offreedom coming from the number of variables and degrees of the polynomials. Withinfinite fields these theorems give us a very good way to apply the Polynomial method.To understand this theorem we will use it to prove Artin’s original conjecture.

Proof of Artin’s conjecture. Take a single homogeneous polynomial f P Frx1, . . . , xnswith deg pfq   n and F finite. We then set S � tfu. Since it is homogeneous

fp0, . . . , 0q � 0

and thereforeZpfq � ∅.

By Chevalley’s Theorem|Zpfq| ¥ 2

implying there exists another point where f vanishes which must be non-zero, provingthe conjecture.

Remark. While here we used Chevalley’s Theorem, we can easily also use Warning’sTheorem to state that that there are at least p solutions, where p ¥ 2 is the characteristicprime of the field.

While this is only a simple proof, the theorem can be used as part of our Polynomialmethod to prove various results including, as we see in Chapter 4, the Erdos-Ginzburg-ZivTheorem.

16

2.3 Proving the Theorem

The first two statements can be proved relatively quickly by using our Polynomial methodas their proofs are equivalent. However Warning’s Second Theorem is a stronger state-ment than we require, stated because of its historical relevance rather than it’s applica-tion, hence I leave it without proof. (For more on this theorem the reader is deirected toan article by Heath, [18]).

In Chevalley’s original proof he used his own form of the Combinatorial Nullstel-lensatz I and an early form of the Polynomial method, long before the method wascategorised. However before we can prove this theorem we discuss the concept of thereduced polynomial and an important corollary of Lagrange’s Theorem.

Theorem 2.3.1 (Lagrange’s Theorem). Given a finite group, G, the order of any sub-group will divide the order of G.

We prove this using the concept of a left coset as seen in [19].

Proof. Take our group G with H a subgroup of it. If H � G or H � ∅ then the theoremholds trivially. So assume H is non-trivial and choose an element g1 P G�H and fromit create a left coset by left multiply every element of H with g1,

g1H � tg1h | h P Hu.Because g1 is invertible we have a bijection,

hÑ g1hÑ g�11 g1h � h,

meaning the cardinality of this set is |H|. Now assume there is an element

k � g1h1 P H Y g1H,

and thereforeg1 � kh�1

1 P H.However since we assumed g1 R H we have a contradiction meaning H and g1H aredisjoint. If

H Y g1H � G

then our theorem holds with|G| � 2 |H| .

If not then we repeat our method using g2 P G � pH Y g1Hq. We use this to generateg2H which will again have cardinality |H| and is disjoint from H as before, but alsofrom g1H. If there was an element

k � g1h1 � g2h2 P g1H Y g2H

theng2 � g1h1h

�12 P g1H.

17

This is a contradiction since we assumed g2 R g1H. If

H Y g1H Y g2H � G

then our theorem holds with|G| � 3 |H| .

Otherwise we continue to repeat this method taking

gi P G� pH Y g1H Y g2H Y . . .Y gi�1Hq.We form giH of cardinality |H| which will be disjoint from H and all gjH. If we assumethere is an element

k � gihi � gjhj P giH Y gjH

thengi � gjhjh

�1i P gjH

which gives us a contradiction since by this process gi R gjH. Because G is finite theprocess will eventually end and we get that

H Y g1H Y . . .Y gnH � G

which are all disjoint groups of cardinality |H| meaning that

|G| � pn� 1q |H| .This completes our proof that

|H| �� |G| .With this theorem we can prove the following important Corollary, which is used

throughout this paper.

Corollary 2.3.1.1 (An Important Relation). Over a finite field order q � pk with pprime and a P Fqzt0u,

aq�1 � 1.

The proof of this corollary is simple when we consider the multiplicative view of a finitefield.

Remark. The reader seeking more explanation on this multiplicative structure is directedto a short set of lecture notes by Ryan Vinroot [20].

Proof. Take an element β of our multiplicative group F�q � Fqzt0u and use it to generatea subgroup

xβy � tβ, β2, . . . u.By Lagrange this set will be finite, with some cardinality cβ. If cβ � q � 1 the theoremholds. If not then we have a subgroup and cβ divides q � 1, meaning q � 1 � dcβ forsome integer d. Since β generates a subgroup of cardinality cβ we have that

βcβ � 1

18

and thereforeβq�1 � βdcβ � 1d � 1

finishing the proof since β was any non-zero element of our field.

Remark. More generally this result is that the order of any element of a finite groupdivides the cardinality of the group. This is because any element will generate a cyclicgroup with order dividing the group cardinality. Therefore by essentially the same ar-gument as above, raising the element to the power of the group order minus one gives1G. By extending this logic one can prove both Euler’s Theorem and Fermat’s LittleTheorem.

This corollary is a powerful tool within finite fields as it allows us to reduce thedegree of a polynomial without changing what it evaluates to. We use this relation toremove any powers larger than q � 1 since we know this power gives the identity. Withthis process we can restrict the number, and degrees of, polynomials we need to considerin our work over finite fields.

2.3.1 The Reduced Polynomial

Given any polynomial, by repeatedly using our corollary, we can form a reduced poly-nomial. This is a polynomial that evaluates in exactly the same way but has a degreein each variable of less than q. This means if the polynomial is in n variables then itstotal degree is bounded by npq � 1q.Example 2.3.2 (Reducing a polynomial). To reduce a polynomial F , we find everypower higher than pq�1q and rewrite it as a� bpq�1q where a   pq�1q. We then applyour relation to these terms,

xa�bpq�1qi � xai � xbpq�1q

i � xai � 1b � xai .

After this process the degree of the reduced polynomial in each variable will be less thanpq � 1q.

degxipF q ¤ pq � 1qWhile this changes our polynomial’s structure, it maintains its evaluation at every pointmeaning that for F the reduced form of F ,

F pxq � F pxq � 0 @ x P Fnq .

Clearly by construction our reduced polynomial has degree not more than the originalpolynomial.

We now seek to show that the map from all polynomials in Fqrx1, . . . xns to thereduced polynomials, via this method, is many-to-one.

Lemma 2.3.3. Every polynomial maps to a single reduced polynomial.

19

Remark. We prove this based on the work of Clark [14].

Proof. Begin by noting that the reduced polynomials are spanned by the monomials ofthe form

xα11 . . . xαnn , αi P Fq

of which there are qn. Therefore we can count the total number of such functions.To define a reduced polynomial means choosing a coefficient of Fq for each of the qnmonomials. Therefore there are exactly qqn reduced polynomials.

We now seek to show they all evaluate differently. We count the total number ofdifferently evaluating functions f : Fnq Ñ Fq. Clearly we can send any of the qn points ofour codomain to any of q values of our domain. Hence there are qqn differently evaluatingfunctions of this form. Now for any function f (not necessarily a polynomial) we definea polynomial,

Rf px1, . . . , xnq :�¸yPFnq

fpyqn¹i�1p1� pxi � yiqq�1q

This is a reduced polynomial in n variables which evaluates exactly to f . This means thereduced polynomials from Fnq to Fq evaluate in qq

n different ways. Therefore, there areqqn different reduced polynomials, which collectively evaluate in all of the qqn different

ways, hence a one-to-one relation. This means that for any function (and thereforepolynomial) there is a unique reduced polynomial which evaluates identically to it.

This idea of the reduced polynomial is important within finite field work and essentialin the following proof of Warning’s first Theorem, based on the works of Pete Clark [14].Note that by proving Warning’s first Theorem we also prove Chevalley’s Theorem.

Proof of Warning’s first Theorem. Take the case as is our theorem. A set of polynomials

S � tf1, . . . , fmu � Fqrx1, . . . , xns

with°ri�1 deg pfiq � d   n and q � pk with p prime. We look to create an indicator

function IZ which indicates whether a point of Fnq is in ZpSq:

IZpxq �#

1, if x P ZpSq0, if x R ZpSq.

We can write this function explicitly by using our relation from Corollary 2.3.1.1,

F pxq :�m¹i�1p1� fipxqq�1q.

This is an indicator since the product has a zero term for all x R ZpSq and if x P ZpSqall fipxq are zero and we get 1.

20

We now look to create a reduced polynomial from F pxq, which we know to be uniqueand of degree at most degpF pxqq. One way to write this reduced form is

F pxq :�¸yPFnq

F1pyqn¹i�1p1� pxi � yiqq�1q

which does indeed evaluate identically to F . Since F vanishes outside of ZpSq we ignorethose terms and look at where it evaluates to 1, writing it as

F pxq :�¸sPZ

1n¹i�1p1� pxi � siqq�1q.

The trick here now is to look at the degrees of our functions. We see that

degpF q � dpq � 1q   npq � 1q.

However within our reduced polynomial the coefficient of xq�11 . . . xq�1

n is p�1qn |ZpSq|.Assume that this coefficient is not zero modulo p and so non zero. This gives us that

degpF q � dpq � 1q   npq � 1q ¤ degpF q.

Therefore we have a contradiction and

p�1qn |ZpSq| � 0 pmod pq

which gives our result that|ZpSq| � 0 pmod pq.

Remark. As mentioned this also proves Chevalley’s Theorem since p ¥ 2.

2.3.2 Uses of the Theorem

The Chevalley-Warning Theorem is a powerful combinatoric tool which can be usedto prove various results particularly within Additive combinatorics and the study ofsumsets. It is a useful tool which we use to apply the Polynomial method. One ofthe most notable uses is the proof of the Erdos-Ginzburg-Ziv Theorem which we willintroduce later.

21

Chapter 3

Sumsets

3.1 Introducing Sumsets

One of the major areas in which the Polynomial method has brought breakthroughsis within the study of Additive number theory. A relatively new area of study, lyingwithin Combinatorial number theory, it is best described as the study of sums of sets ofintegers. A variety of interlinked problems in this area, some of which stood for over 30years, have been solved by applying our method. Through both the Chevalley-WarningTheorem and the Combinatorial Nullstellensatz some results have been given new proofsand others proved for the first time. One of the most notable of these, is the Erdos-Heilbronn problem, which stood for over 30 years. However we will work our way up tothis from basics and see some of the other works in this area. Throughout this chapterwe will define a sumset as follows:

Definition 3.1.1. Given subsets A and B of an additive group G, a sumset is

A�B :� ta� b | a P A, b P Bu.

This definition easily extends to other group operations but in this chapter we focuson additive groups.

Remark. A concise resource giving some insight into this chapter is a set of notes onthe Polynomial method by Matt Devos [21], particularly chapter 9.

3.1.1 Spin to win - Forming simple sumsets

Imagine you have a spinner, divided into 25 sections. Each section contains an outernumber and an inner number, each between 1 and 5. After each spin you record theresult to form a growing set of outer numbers, and also a set of inner numbers. Howmany spins does it take such that, by adding one value from each set, we can form anynumber pmod 5q?

22

Figure 3.1: A spinner with 25 sections, each containing two numbers between 1 and 5.

Example 3.1.2 (Trial run). Imagine our first result is p1, 2q, where the first number isthe outer number. Recording this we see our sumset consists of only 3 as we have onlyone choice from each set. We therefore continue to spin, generating a table of results.

Spin to Win - Sample GameSpin # Result Outer set Inner set Sumset pmod 5q

1 p1, 2q t1u t2u t3u2 p2, 4q t1, 2u t2, 4u t0, 1, 3, 4u3 p1, 3q t1, 2u t2, 3, 4u t0, 1, 3, 4u4 p5, 2q t1, 2, 5u t2, 3, 4u t0, 1, 2, 3, 4u

We see that after 4 spins we have achieved a complete sumset of 5 elements and completeour trial run.

This example highlights some interesting points about set addition. Firstly we seethat adding an element to our subset does not necessarily increase the cardinality ofour sumset, we see this in our example where we move from our second to our thirdspin. We also see that to complete our sumset required us to have 6 elements in oursubsets and that 5 can be inadequate. It is these sort of points that our theorems address.

One such theorem is the Cauchy-Davenport Theorem, this states that if the sumsetdoes not have cardinality p � 5, a complete set, then it has cardinality at least |O|�|I|�1.For our example this means if we have 6 elements between our two subsets we areguaranteed a complete sumset. This requires between 3 and 5 spins assuming we getdistinct results (else the game could go indefinitely).Remark. This result generalises from 6 elements to p� 1 elements for any prime p.

23

3.2 Cauchy-Davenport Theorem

The theorem was first proved in 1813 by Cauchy, however it was independently proven byDavenport over 100 years later in 1935. Since Davenport was unaware of Cauchy’s proofthe name credits both for their contributions and it is arguably the origin of Additivenumber theory.

Theorem 3.2.1. Given a cyclic group of prime order Zp, if A,B � Zp are non-emptythen

|A�B| ¥ mintp, |A| � |B| � 1u.As we saw in our example this result allows us to comment on the size of sumsets.

The theorem is a powerful tool and we will use it to prove a number of results, it wasalso used in the original proof of the Erdos-Ginzburg-Ziv Theorem, [22], a theorem westudy in the next chapter.

3.2.1 Proving the Theorem

The proofs of Cauchy and Davenport were different but both relied on induction on thesize of the sets. Cauchy used a form of transform to arrive at his proof in 1813, [23], givinga new way to prove Lagrange’s four squares Theorem, on the other hand Davenport’sproof came from his work on residue classes, [24], and was seen as a discrete analogue towork on Schnirelmann density by Khintchine. Both of these approaches led to furtherresults but it is perhaps Alon’s proof, [6], using the Combinatorial Nullstellensatz whichis most easily generalised. With a few small modifications the following proof will giveus a number of results concerning sumsets, and much more concisely than proofs byother methods.

Remark. A good source for the reader seeking more information on the earlier proofsis an article by Adhikari, Chintamani, Geeta and Moriya, which summarises a numberof the proofs as well as discussing some further works, [25].

Proof. Given a prime p and sets A,B � Zp, we first take the case where |A| � |B| ¡ p.Take an element k P Zp, then note that

k P A�B ðñ AX pk �Bq � ∅.

Since k � B has equal cardinality to B we have that |A| � |k �B| ¡ p implying theirintersection is non-empty. As this is true for all k we have that A�B � Zp.

This leaves the more difficult case where |A|�|B| ¤ p. We assume, for contradiction,that

|A�B| ¤ |A| � |B| � 2.Take a subset C which contains A � B and has cardinality precisely |A| � |B| � 2 andcleverly create our polynomial

fpx1, x2q �¹cPC

px1 � x2 � cq

24

of degree |C| such that fpa, bq � 0, @ a P A, b P B. Set α1 � |A| � 1 and α2 � |B| � 1such that deg pfq � α1 � α2, and set A � S1, B � S2. The coefficient of our element,xα1

1 xα22 is

� |C||A|�1

�which is non-zero since 0   |C|   p. With these conditions we apply

the Combinatorial Nullstellensatz II to find an element such that fpa, bq � 0 giving ourcontradiction and completing the proof.

Using this style of proof, and this theorem, we will be able to show a number ofsimilar results from this area of mathematics.

Remark. The Cauchy-Davenport Theorem is a very important tool, particularly in Ad-ditive number theory. It is also even more important because it generalises to be true forall (not just abelian) finite groups. For more about the extensions into Group theory ashort but useful resource is an article by Jeffrey Wheeler, [26].

3.3 The Erdos-Heilbronn Problem

Remark. The reader is again directed to Chapter 9 of the notes of Matt Devos for moreon this area, [21].

Perhaps the most significant result in this area, relating to the Polynomial method,is the proof of the Erdos-Heilbronn problem (EHP). It is similar in many ways to theCauchy-Davenport Theorem but with the restriction to the summation of distinct ele-ments. We therefore define the following notation:

A`B :� ta� b pmod pq | a P A, b P B and a � bu`kA :� tpa1 � a2 � . . .� akq pmod pq | ai P A, ai � aj for all i � ju

with which we can state the problem.

Theorem 3.3.1 (The Erdos-Heilbronn Problem). Take non-empty subsets A,B of Zpwith p prime. Then

|A`B| ¥ mintp, |A| � |B| � 3u.First conjectured by Erdos in 1963 and discussed in his lecture courses and various

later works of his, (for further details see [27], page 106). The conjecture suggests thatif we exclude the summation of duplicate elements the bound of the Cauchy-DavenportTheorem changes only slightly. The problem stood for over 30 years until 1995. Thefirst breakthrough came in 1994 through the work of Dias da Silva and Hamidoune.

Theorem 3.3.2 (Dias da Silva, Hamidoune). Take a non-empty subset A of Zp with pprime. Then

|`kA| ¥ mintp, k |A| � k2 � 1u.By setting k � 2 in this theorem the conjecture is proved for A � B. Their proof

of this used Linear algebra and Representation theory relating to the symmetric group.However only a year later, Alon, Nathanson and Ruzsa, using an early CombinatorialNullstellensatz, produced an elegant proof of not only this theorem, but also the case ofA � B, [7].

25

Theorem 3.3.3 (Alon, Nathanson, Ruzsa). Take non-empty subsets A,B of Zp with pprime and |A| � |B|. Then

|A`B| ¥ mintp, |A| � |B| � 2u.

In 1996 they also generalised their work to other values of k proving the full resultof Dias da Silva and Hamidoune. We will focus on their work to demonstrate the use ofthe Polynomial method in this area.

Remark. We restrict ourselves to the setting of the EHP, with k � 2, however their fullworks can be found in [7, 28].

3.3.1 Proving the EHP

To prove the conjecture we will use our Polynomial method, as done by Alon, Nathansonand Ruzsa [7]. We first prove their result, and from this we derive our other results.

Remark. This proof drawn from the original work, along with the notes of Matt Devos[21] to take full advantage of the modern Combinatorial Nullstellensatz.

Proof of Theorem 3.3.3. We prove this by contradiction. Take non-empty subsets A,Bof Zp of differing cardinalities, with

|A`B|   mintp, |A| � |B| � 2u

such that |A| � |B| is minimal, and set |A| � α and |B| � β. Without loss of generality,we will also assume that 1 ¤ β   α ¤ p.

We now note that if α � β � 2 ¡ p then we can take A, a subset of A that hascardinality α � p� β � 2 such that β � α� 2 � p. We see that if the theorem holds forA, B, it also holds for A and B as

|A`B| ¡ ��A`B�� � mintp, α� β � 2u � p.

Hence to satisfy our minimality condition we must have α� β � 2 ¤ p.We therefore take a set C � A`B such that, |C| � α�β�3. From this we construct

the following polynomial, which vanishes at all pa, bq where a P A, b P B

fpx, yq � px� yq¹cPC

px� y � cq.

Note that the coefficient of the term xα�1yβ�1 is�α� β � 3α� 2

��α� β � 3α� 1

� pα� 1qpα� β � 3q!

pα� 1q!pβ � 1q! � 0 pmod pq.

The first binomial comes from choosing the x outside our product, and the second fromchoosing the p�yq. We get that this is non-zero pmod pq due to the fact that p is prime

26

and that all the factors are non-zero and less than p (and since we are in a field thereare no zero divisors other than zero).

We can now apply our Combinatorial Nullstellensatz II. We have that degpfq �α�β�2 � pα�1q�pβ�1q and the coefficient of xα�1yβ�1 is non-zero. Then by settingS1 � A and S2 � B and we find that there exists an element pa, bq P A � B such thatfpa, bq � 0. This contradicts that C � A`B giving the result that

|A`B| ¥ mintp, |A| � |B| � 2u.

From this theorem we can now easily prove the Dias da Silva, Hamidoune Theoremfor the case k � 2.

Proof of Theorem 3.3.2. If A has only one element the result is trivial. If A has morethan one element then define B � Aztau for a some element of A. Then we have that

|A`A| ¥ |A`B| ¥ mintp, |A| � |B| � 2u � mintp, 2 |A| � 3u.

Remark. We can also begin to see that we could generalise the polynomial in our firstresult to prove the Dias da Silva, Hamidoune Theorem for all k. This and other exten-sions of these results, were explored in much more detail by Alon in 1999 when he hadfully formed the Combinatorial Nullstellensatz, [6].

We now have all we need to prove the EHP.

Proof of the Erdos-Heilbronn Problem. If A � B the result holds by the Dias da Silva,Hamidoune Theorem so we assume A � B. Then if |A| � |B| the result immediatelyfollows from the result of Alon, Nathanson and Ruzsa. We therefore assume that A � Band |A| � |B|. In this case we set A � Aztau for some a P A. Then we have that

|A`B| ¥ ��A`B�� ¥ mintp, ��A��� |B| � 2u � mintp, |A| � |B| � 3u

completing the proof.

This result is perhaps one of the most significant breakthroughs from the Polynomialmethod as it proved that a long standing problem, when approached with this method,could be proved within only a few pages.

Remark. The tendency of the method to give powerful results with minimal work issignificant and a recurring theme throughout this survey.

27

Chapter 4

The Erdos-Ginzburg-Ziv Theorem

Remark. Throughout this chapter the reader is again directed to the works of PeteClark, specifically in [14]. Another excellent source is the work compiled in 1993 by Alonand Dubiner, this was to celebrate the 80th birthday of Erdos [29]. Five different proofsof the the Erdos-Ginzburg-Ziv Theorem are discussed including the proof we use, whichrelies on the Chevalley-Warning Theorem.

The Erdos-Ginzburg-Ziv Theorem, which I will now refer to as EGZ, is an interestingand useful tool used in Combinatorics. EGZ looks more closely at sums of sequencesand in particular zero-sum subsequences. While here we focus on sums within Zn, thetheorem does in fact extend in various ways such as to sums within other finite groups offixed order. Many proofs for this theorem exist, however notably they tend to work byeither direct or indirect application of the Polynomial method. In this section we applyit by using our previous result, the Chevalley-Warning Theorem.

4.1 Understanding Zero-sum Sequences

Throughout this chapter we use the following definitions.

Definition 4.1.1. A zero-sum sequence modulo n is a sequence S � ts1, . . . , sku,such that ¸

iPIsi � 0 pmod nq

Definition 4.1.2. Given a sequence S � ts1, . . . , sku we define a zero-sum subse-quence modulo n to be a subsequence S � S such that¸

sPSs � 0 pmod nq.

To give some background to these definitions and show how we arrive at theoremssuch as EGZ, we consider the following dice game.

28

4.1.1 Just Sum Dice Game

Imagine you are playing a dice game. The game is played using 3 standard dice eachnumbered from 1 to 6. Dice rolls are scored by summing the values and taking off 3 togive a number from 0 to 15. The first roll determines the target number, which I referto as T . After this the dice are repeatedly rolled and the result written down each time.The aim of the game is to find a subset of the numbers rolled that sums to a multipleof your target number. The first player to do so wins.

Example 4.1.3 (Playing the game). The first dice roll is a 9, therefore the target is setat T � 6.

Ñ T � 6Target roll:

Rolls then continue until a subset summing to a multiple of 6 is found:

Ñ 4Roll 1:

Ñ 9Roll 2:

Ñ 8Roll 3:

After three rolls it is spotted that 4� 8 � 12 � 2� 6 ending the game.

While this game seems quite simple it does have deeper mathematical meaning andraises some interesting questions. For example can the game always be won? And howmany rolls must be performed to guarantee a winning solution exists? Studying thesequestions leads onto some interesting theorems about subsequences and particularly ourzero-sum subsequences.

To study this game mathematically we will generalise it by assuming we play withnot just 3 but D, some large integer number of dice. We set the scores accordinglybetween 0 and M � 5D, using this we pose the following lemma.

Lemma 4.1.4. For target value T a solution always exists after T (or fewer) rolls.Furthermore pT � 1q rolls may be insufficient.

Proof. We start by proving the second statement. For small T this is clear as if T � 1we clearly require one roll, for T � 2 we require two rolls if the first is odd. Furthermoreif we simply roll 1 on pT � 1q occasions they can’t sum to a multiple of T .

29

We now move to prove that T rolls are always sufficient. We here define Sk as thesum of the values of our first k rolls modulo T , that is

Sk �k

i�1ri k P t1 . . . T u pmod T q

where ri is the value of our ith roll. After T rolls we have T such Sk with T possiblevalues, implying by the pigeonhole principle that either we have some l such that

Sl � 0 pmod T q

giving us our solution, or some values m and n such that

Sm � Sn pmod T q, m � n.

In this case we have that (assuming m   n upto relabelling)

0 � Sn � Sm � rm�1 � . . .� rn pmod T q

which is a solution.

4.1.2 Implications

Through this proof we actually not only show that a solution exists, but also a solutionmade from consecutive rolls. However this isn’t necessarily the best solution in termsof the game. In Example 4.1.3 from the scores 4, 9, 8 we could construct a zero-sumsubsequence pmod 6q but a consecutively rolled solution hadn’t appeared yet. Formal-izing this we see we have proved the following theorem.

Theorem 4.1.5. Given a set of integers S � ta1, . . . , anu , there exists a zero-sumsubsequence. That is that there exists I � t1, . . . , nu non-empty, such that¸

iPIai � 0 pmod nq.

Remark. We can also consider the game where we do not allow repeated rolls. In thiscase we cannot take a sequence of 1s to show T � 1 rolls can be inadequate. With thisrestriction the condition is weakened such that only when

M ¥ pT � 2qT � 1 � pT � 1q2

is satisfied can T � 1 rolls be insufficient. When this condition is met we can roll thedistinct values

1, T � 1, 2T � 1, . . . , pT � 2qT � 1

which yields no solution as all these values are 1 when considered pmod T q and hencewe require T rolls.

30

We now consider a final variation of our game by playing our original game with theadded restriction that we must sum exactly T values in our solution. Again we can askif the game is winnable? And at most how many rolls are required? From our originalgame we know at least T rolls are required. In fact, we can show we require at least2T � 1 rolls.

Proof. Assume we have 2T � 2 rolls in which we have T � 1 instances of 0 pmod T q andT � 1 instances of 1 pmod T q. In this case selecting any T values gives us a numberbetween 1 and pT � 1q � �1 pmod T q. Therefore at least 2T � 1 rolls are required for aguaranteed result.

In fact by the Erdos-Ginzburg-Ziv Theorem, this is a strict lower limit on the numberof rolls required.

4.2 The Erdos-Ginzburg-Ziv Theorem

The theorem was first proved in 1961 by the trio of mathematicians by using the Cauchy-Davenport Theorem, our previous result concerning sumsets [22]. A later proof wasshown using the Chevalley-Warning Theorem [29]. These early proofs would not neces-sarily be considered an application of the Polynomial method, however through Alon’swork in the 90’s both these theorems have been given new, Polynomial method basedproofs [6]. As such they are now considered tools for applying our Polynomial method.We will focus on the Chevalley-Warning Theorem as this more readily allows for ex-tensions of this area further into Group theory, and is a more direct application of ourmethod.

Theorem 4.2.1 (The Erdos-Ginzburg-Ziv Theorem). Let J be a sequence of elementsfrom Zn of length 2n� 1. Then there exists I � J such that |I| � n and¸

aPIa � 0 pmod nq

In other words there exists a zero-sum subsequence pmod nq of length n.

Applying EGZ to our restricted game, with n � T , gives that there must exist azero-sum subsequence pmod T q of length T , after 2T � 1 rolls. To show this is the casehowever we must prove the theorem.

4.2.1 Proving EGZ

We will prove this in two stages, first proving that EGZ holds for primes, and secondlythat if the theorem holds for two numbers then it also holds for their product (as seenin [2], [14]).

31

Remark. Note that the reason this work features in our study of finite fields is thatfor p a prime Zp � Fp. This allows us to apply the finite field forms of our Polynomialmethod such as the Chevalley-Warning Theorem. For more explanation of the congruencesee [20].

Lemma 4.2.2. EGZ holds for n � p where p P Z is prime.

Proof. Firstly we take a prime number p P Z and a sequence J � tr1, . . . , r2p�1u. Wethen consider two different polynomials over the field Fprx1, . . . , x2p�1s,

P1px1, . . . , x2p�1q �2p�1

i�1rix

p�1i ,

P2px1, . . . , x2p�1q �2p�1

i�1xp�1i .

Note thatP1p0, . . . , 0q � P2p0, . . . , 0q � 0

and thatdeg pP1q � deg pP2q � 2p� 2   2p� 1

Therefore by Chevalley-Warning Theorem there exists a non-zero s � ps1, . . . , s2p�1q PF2p�1p such that

2p�1

i�1ris

p�1i � 0,

2p�1

i�1sp�1i � 0.

By Corollary 2.3.1.1 we know that sp�1i � 1 for all si � 0 and 0 if si � 0. Therefore we

need only consider the values in the set I where

I � tri P J with i such that si � 0u.

This gives us the following equations¸riPI

ri � 0 pmod pq,¸riPI

1 � 0 pmod pq.

The second equation is simply the cardinality of I pmod pq and since 0 ¤ |I| ¤ 2p�1we have that |I| � p. Therefore I is a zero-sum subsequence of J of length p, completingthe proof of our first lemma.

We now generalise this statement to all integers by showing the property is main-tained under multiplication.

Lemma 4.2.3. If EGZ holds for integers p and q, then it holds for n � pq.

32

Proof. We do this proof by induction. We assume the theorem holds for p and q. Thentake cp � 1 integers for some c ¥ 2, tr1, . . . , rcp�1u. Now for c � 2 we have the caseabove and there exists a set which we will label I1 of size p satisfying the following (withj � 1), ¸

iPIjri � 0 pmod pq. (4.1)

Now for c � 3 we will have 3p � 1 integers. We can treat this as the case where c � 2with an additional p integers added. As such we choose any 2p � 1 of them and find aset as before I1. Doing so, since I1 has cardinality p, we are left with 2p�1 integers. Wecan therefore form a second set I2 which also satisfies the equation above, with j � 2.Additionally I1 and I2 will be disjoint simply by construction, as all elements of I1 wereremoved before we formed the second subset. By the same principle, using induction onc, we see that we can in fact always construct c� 1 pairwise disjoint subsets I1, . . . Ic�1satisfying Equation 4.1, each of cardinality p.

In the case of this Lemma we are looking at a set of 2pq�1 integers, so we set c � 2qas in the above. Therefore there exist 2q � 1 pairwise disjoint subsets, Ij , each of size pand satisfying Equation 4.1. For each of these sets we define

sj �¸iPIj

ri, rsj � sjp, @ j P t1, . . . , 2q � 1u.

There are 2q � 1 such rsj . Since the theorem holds for q and we can find a zero-sumsubsequence K of size q such that,¸

rskPKrsk � 0 pmod qq

Finally we set I � �kPK Ik, such that |I| � pq � n. We then see that

¸iPIri �

¸kPK

¸iPIk

ri �¸kPK

sk

�¸kPK

prsk � p¸kPK

rsk� 0 pmod pqq.

We have therefore found a set I of cardinality n � pq such that it is a zero-sumsubsequence pmod nq. This proves that EGZ holds for n � pq if it holds for p and q.

We have now shown that EGZ holds for all primes and for all products of numbersthat it holds for. By a simple argument we can now show that EGZ holds for all of Zby using induction on the number of primes in the decomposition of n P Z.

33

Proof. We begin our induction by noting that EGZ holds for compound numbers madefrom two primes n � p1p2 by Lemma 4.2.3 and can therefore start our induction. Assumethat n � p1...pm with all the pi prime numbers (they need not be distinct) and thatEGZ holds for compound numbers formed from m� 1 primes. Therefore EGZ holds forq � p1...pm�1 by our assumption, and for p � pm by Lemma 4.2.2 since pm is prime. Assuch it holds for their product completing the induction.

4.3 Further Applications

EGZ Theorem is an interesting statement about modular mathematics, which logicallyseems very reasonable, however we can look to extend EGZ further into Group theory.Rather than only using the cyclic groups, Zn, we can extend it to, for example, otherfinite abelian groups. One extension is to find the minimum sequence length to guaranteea zero-sum subsequence pmod nq of length n, in Zn ` Zn.

4.3.1 Zn ` Zn and beyond

The case of Zn`Zn is actually a very recently resolved one. The following conjecture ofKemnitz [30], which is the strict extension of EGZ to these groups was recently provedby Christian Reiher in 2004, [31].Conjecture 4.3.1 (Kemnitz’ conjecture). Let J be a sequence of elements from Zn`Zn,

J � tpa1, b1q, . . . , pa4n�3, b4n�3qu.Then there exists I � J such that |I| � n and¸

aiPIai � 0 pmod nq¸

biPIbi � 0 pmod nq.

In other words there exists a zero-sum subsequence in Zn ` Zn of length n.It is quite clear this is a sensible extension of EGZ. To bound this from below we

take pn � 1q of each of the following, p0, 0q, p1, 0q, p0, 1q, and p1, 1q, giving us a set of4n�4 elements which has no zero-sum subsequence of length n. However bounding fromabove is a much more complicated work, though it once again uses our combinatorialtools including the Chevalley-Warning Theorem, [31].

While this theorem was a significant step forwards little is known about the exactbounds of the higher dimensional forms such as Zn`Zn`Zn and beyond. As we go intothese higher dimensions, we end up more geometrically analysing these sets as latticestructures. This underlying lattice structure crops up in a number of areas of study inwhich the Polynomial method is applied. With more study being done in the area, andthe intricacies of these combinatorial tools still yet to be fully understood, it is possiblewe will see more breakthroughs in this area coming from our Polynomial method.

34

Chapter 5

Polynomial Testing

An interesting direction for this work is to consider adding a probabilistic element toour considerations. We can in fact use some of the results and theorems in this areato carry out Polynomial identity testing. Given two polynomials f1 and f2 over a fieldFrx1, . . . , xns, we ask whether

f1ptq � f2ptq?We reduce this question to asking whether

fptq � f1ptq � f2ptq � 0.

We can test this by simply trying values of t. Should we find a non-zero value of fptqclearly it is non-zero, if it is zero we have found a root. If can bound the probability offinding a root then we can say with some certainty whether it is identically zero. Onepowerful tool we can derive for doing this is the DeMillo-Lipton-Schwartz-Zippel Lemma.This Polynomial testing has applications in various areas but most notably in Codingtheory.

5.1 The DeMillo-Lipton-Schwartz-Zippel Lemma

This lemma has an interesting history coming out of the work of four mathematicians.In fact the strong result, which is what we will be using, can reasonably be called theSchwartz Lemma as of the four mathematicians he was the only one to find this form ofthe result. Therefore to distinguish it I will refer to this as the Schwartz Lemma, afterthe late Jack Schwartz, though the work of the other 3 mathematicians also played arole in bringing us this result (see Ray Lipton’s writings on these events, [32]). We beginwith the following form of the statement.

Lemma 5.1.1 (The DeMillo-Lipton-Schwartz-Zippel Lemma). Take a non-zero polyno-mial fpx1, . . . , xnq of degree d over a field Frx1, . . . , xns. Then for any S � F such that|S| ¥ d,

|Zpfq X Sn| ¤ d|S|n�1.

There can be at most d|S|n�1 roots of f in Sn.

35

This lemma is a useful extension of Theorem 1.1.2 hence we apply it in Polynomialmethod proofs. In fact we use this to prove Dvir’s result on the Finite field Kakeyaconjecture in the next chapter. We can also reformulate this lemma to add a probabilisticconsideration. We do this by noting that the statement implies that at most d|S|n�1

elements in Sn are roots. Therefore if we randomly select an element from Sn we saythe following.

Lemma 5.1.2 (Schwartz Lemma). Take a non-zero polynomial fpx1, . . . , xnq of degreed over a field Frx1, . . . , xns and a subset S � F such that |S| ¥ d. Then

Prrfps1, . . . , snq � 0s ¤ d

|S|where the si are randomly selected elements from S.

These lemmas concern the same sort of ideas as our Combinatorial Nullstellensatzbut with S � S1 � . . . � Sn all of size greater than d. As such, the CombinatorialNullstellensatz is in many ways a refinement, however for our Polynomial testing thisform is more useful. We prove these lemmas using only our basic Polynomial method,note that we need only prove the first statement as the second is a simple reformulation.Like a number of proofs in this subject we will use induction on the number of variables(as seen in [2]).

Proof. Set f to be a non-zero polynomial of degree d. With n � 1 we can apply ourFundamental Theorem of Algebra 1.1.1 to say there are at most d distinct roots andtherefore at most d elements of S are roots. The theorem therefore holds and we can useinduction with the assumption that the theorem holds for pn� 1q variables, with n ¥ 2.We now write f based on the powers of xn such that

f � f0 � xnf1 � . . .� xknfk

where the fi are polynomials in the first pn� 1q variables and fk is non-zero. We boundthe number of zeroes in Sn by splitting the cases based on whether fkps1, . . . sn�1q � 0.

I. fkps1, . . . sn�1q � 0: Since f is not identically zero and has degree d we see thatfk has degree at most d � k (as it is is also non-zero). Therefore, since it is apolynomial in n� 1 variables, by our induction hypothesis it vanishes on at mostpd� kq|S|n�2 points. We then have |S| choices for xn and can therefore say thereare at most pd� kq|S|n�1 roots of f in Sn where fkps1, . . . , sn�1q also vanishes.

II. fkps1, . . . sn�1q � 0: In this case we fix our first n� 1 variables and consider f as afunction in only xn of degree k. This is not identically zero since the coefficient ofxkn is non-zero. This is our one dimensional case so there can be at most k roots.Since we can choose each of our first pn � 1q variables from S there are at mostk|S|n�1 roots of f in Sn where fkps1, . . . sn�1q does not vanish.

36

Combining these two cases we have that there are at most

pd� kq|S|n�1 � k|S|n�1 � d|S|n�1

roots of f in Sn.

Remark. To additionally prove our Schwartz Lemma we simply note that this impliesthat, as a proportion, at most d|S|n�1

|S|n � d|S| elements of Sn are roots.

5.1.1 Applications

As mentioned before we can use this Schwartz Lemma to perform Polynomial testing,checking if two polynomials evaluate identically. Again we use f1 and f2 over a fieldFrx1, . . . , xns and check whether

f1ptq � f2ptqby analysing

fptq � f1ptq � f2ptq.By our lemma, if r1, . . . , rm are randomly generated elements of Sn with |S| ¥ d then

Prrfpriq � 0 | i P t1, . . . , nus ¤�d

|S|m

.

The advantage of this method is that we never need to explicitly write the polynomial.As such this method can often be done rapidly by computers to efficiently determine, toa high probability, whether the polynomial is identically zero. We can see the methodin action by considering the following.

Example 5.1.3. Assume we are working in Fp with p a large prime. Then take poly-nomials f1 and f2 each of degree n where n ¤ ?

p in Fprx1, . . . , xns, and set S � Fp.We then test to see if f � f1 � f2, which will also have degree at most ?p, is identicallyzero. Assume our random element is r, then by our lemma if f1 � f2

Prrf1prq � f2prqs ¤ n

p¤ 1?

p.

Therefore if there is a difference the likelihood of it going undetected is less than 1?p . If

we repeat the test 2m times then the probability of an incorrect identification is at mostp�m, which tends to 0 very quickly as m increases.

We can use this for error detection by using the data string as the coefficients ofa polynomial in n variables, and evaluating at a predetermined point. Comparing theresults determines with some certainty whether an error has occurred. Repeating theprocess with a new random value will make the result even more certain. This exampleshows the benefit of a large alphabet of codewords, large p, and the downside to longerdata strings, large n. Achieving a balance between these values is a problem withinCoding theory, a problem which the Polynomial method helps resolve.

37

Chapter 6

Kakeya

When discussing the Polynomial method there are a few key figures associated with it’sprominence. One of these is most definitely Zeev Dvir for his breakthrough work on theFinite field Kakeya conjecture, [1]. His solution to this problem highlighted the power ofthis method to solve complex, and long standing problems, in short concise ways. Hissolution to the problem was announced in 2008, but to approach the problem we needto go back over 90 years.

6.1 The Kakeya needle problem

In 1917 Soichi Kakeya first posed this problem, which asked what the minimum area of aregionR on the Euclidean plane can be, if a unit length needle may be rotated continuoslywithin it such that it returns to its original position with opposite orientation, [33]. Themost obvious solution is a circle of unit diameter and area π{4. A slightly better solutionis an equilateral traingle of height 1 and area 1{?3, this triangle is in fact the optimalsolution for convex sets as was shown by Pal in 1921 (discussed in [34], though originallyproved (in German) in [35]). However if we do not restrict ourselves to convex shapesthen this is still not minimal, as for example the deltoid of area π{8 is also a Kakeyaneedle set.

Figure 6.1: The circle radius 12 , the equilateral triangle of height 1, and the deltoid of

area π8 are all Kakeya sets in R2.

38

The answer was actually a corollary to the work already done by Besicovitch in 1919along with that of Pal. However due to the isolation of Russia from the western worldthis would not be discovered until 1928. Besicovitch was actually working on Riemannintegration in the plane, which led him to seek to develop compact shapes with a linein every direction but of Lebesgue measure (area) 0. From his work we can derive thefollowing construction known as a Perron tree after the mathematician Oskar Perron.The idea is to divide the equilateral triangle into 2n equal parts. Then by overlayingthese we can achieve a set, still with a line in every direction, but with much lower area.In fact by taking a large enough n we may form a tree of arbitrarily small area.

Figure 6.2: The construction of a Perron Tree using four parts

By carefully combining and overlaying these Besicovitch sets we can arrive at Kakeyaneedle sets of arbitrarily small area. While this solves the problem in the plane, we caneasily extend these ideas to higher dimensions. This brings us to the following openconjecture, [2].

Conjecture 6.1.1 (The Kakeya Conjecture). A Kakeya set in Rn must have Hausdorffdimension at least n, where a Kakeya set in Rn is defined to be a set with a line in everydirection.

This is still an open problem with nothing more than partial result for any n ¡ 2.Because of the difficulties with solving this problem in 1999 Wolff proposed a finite fieldanalogue to the problem, a problem which would come to the attention of Zeev Dvir, [36].

Remark. For more info on some of the partial results and other open problems aroundthis conjecture the reader is directed to the writings of Markus Furtner in [34] and theoriginal article by Wolff, [36].

6.2 The Finite Field Analogue

In 1999 in an article looking at recent developments concerning the Kakeya Conjecture,Wolff posed the problem over a finite field, an analogue which was almost completelyunstudied at the time. The analogue, as originally stated, [36], is as follows.

39

Definition 6.2.1. A Kakeya set K, is a subset of Fnq containing a line in every direc-tion. Alternatively

@m P Fnq zt0u D c P K such that c�mx � K @ x P Fq.

This is much like our real case however there only exist a finite number of lines. Wealso note that clearly such a set does exist since Fnq must contain all possible lines, henceit is a reasonable analogue. With this definition Wolff posed the following conjecture.Conjecture 6.2.2 (The Finite field Kakeya conjecture). Every Kakeya set has cardi-nality at least Cnqn, where Cn is only dependent on n. That is

|K| ¥ Cnqn.

This conjecture prompted a lot of work in the area which pushed forward our un-derstanding of the links betweens sums and products in finite fields, [37]. Ultimatelythrough the work of Dvir, prompted by Alon and Tao, a very straight forward Polyno-mial method proof of the conjecture was constructed, cementing Dvir’s name alongsidethe method.Theorem 6.2.3 (Dvir, [1]). Every Kakeya set has cardinality at least Cnqn, where Cnis only dependent on n. More precisely

|K| ¥�q � n� 1

n

¥ qn

n! .

Before we prove this conjecture we will consider F25 as a proof of concept.

Example 6.2.4 (Kakeya sets in F25). The space of Fnq can be thought of the integer

points of an n-dimensional hypercube with side length q. This can be a helpful way ofconsidering it when in lower dimensions. For example F2

5 my be viewed as a 5�5 integergrid on the plane. We can use this view to construct our Kakeya set.

Figure 6.3: Calculating a Kakeya set in F25

40

To be a Kakeya set means it contains a line in every direction, in this case there are6 lines to consider. Because of this any union of 6 such lines is a Kakeya set. Achievingthe minimum set size is equivalent to maximising the number of intersections. By simpletrial and improvement we can achieve this minimisation of our set cardinality. Aftersome testing we see that 17 is minimal in F2

5, as seen in Figure 6.3. However this bruteforce method is very inefficient so we require a more general statement such as Dvir’s.

Dvir’s bound for F25 gives a minimum cardinality of 15, however our example shows

that the Kakeya set actually contains at least 17 points. This suggests the constantis good but not optimal. The important point however, is that the constant is not qdependent.

6.2.1 The Proof of the Finite field Kakeya conjecture

Remark. This proof is adapted from the work of Dvir and Tao who were in close com-munication working to solve the problem and each presented Dvir’s work separately,see [1,38]. However another good resource which discusses this proof, following the samemethod, is Extremal Combinatorics by Jukna, see [2].

This relatively short proof once again highlights how powerful and concise our Poly-nomial method can be. To prove the conjecture we first establish a few small lemmasconcerning fields which we use both now and in later chapters.

Lemma 6.2.5. Given a field F, a polynomial f P Frx1, . . . , xns of degree d contains atmost

�n�dn

�distinct monomials.

Remark. This proof relies on using using ‘stars and bars’ notation. For more info onthis, particularly in the case of proving Lemma 6.2.5 see this online article by BrendanMurphy, [39].

Proof. A monomial m of f has the form

m �¸

0¤i1�...�in¤dxi11 x

i22 . . . x

inn ,

where i1, . . . , in are all non-negative integers, and

degpmq � i1 � i2 � . . .� in.

We can treat this problem as placing balls in boxes (an extension of the pigeon holeprinciple) by using ‘stars and bars’ notation. We use � to represent a ball, and | torepresent the division between boxes. With this notation a monomial of degree k requiresus to place n� 1 lines and k balls within them. For example, with this notation

x21x2 P F3 Ø | � �| � ||.

There are two balls in the x1 box, one in the x2 box and none in the x3 box. Since thefirst and last symbol must be lines we are in fact choosing k symbols from k�n�1 to be

41

stars giving�n�k�1

k

�options. To calculate the number of monomials of degree at most d

we add an extra box for the ‘unused’ stars, and count the monomials of degree exactlyd (this is equivalent to homogenising the polynomial by adding an extra variable). Wetherefore have n� 1 variables giving our result, that there are precisely�

n� d

d

��n� d

n

monomials of degree less than or equal to d.

With this simple combinatorial fact we can prove the following lemma.

Lemma 6.2.6. Take a subset S of Fn where F is some field. If |S|   �n�dn �, for somepositive integer d, then there exists a non-zero polynomial f P Frx1, . . . , xns, vanishingon S with degree at most d.

Remark. While we include this in the finite field section of our work there is no needfor F to be finite. As such we use this result with F � R in Chapter 8.

Note that by setting n � 1 we see this lemma is precisely Theorem 1.1.2 as�1�d

1� �

d � 1 so |S| ¤ d. We therefore consider this as a higher dimensional form of our basicPolynomial method, extending Theorem 1.1.2 to Fn.

Proof. By Lemma 6.2.5 such a polynomial f , can have at most r � �n�dn

�distinct

monomials to which we will give coefficients c1, . . . , cr. Therefore to have f vanish onthe whole of S we have to resolve |S| equations in r variables. Since |S|   r a non-trivialsolution exists corresponding to the coefficients of f such that it vanishes on S.

Lemma 6.2.7. If a polynomial f P Fqrx1, . . . , xns of degree at most q� 1 vanishes on aKakeya set K, then it is identically zero.

We will prove this by breaking down our non-zero polynomial f , into homogeneouscomponents. Then by applying our methods, including Lemma 5.1.1 from the lastchapter, we derive a contradiction giving us that f � 0.

Proof. Assume that f P Fqrx1, . . . , xns is both non-zero and vanishes on a Kakeya setK. We write f as a sum of homogeneous components,

f �i�d

i�0fi � f0 � f1 � . . .� fd

where degpfq � d ¤ q � 1 with fd � 0. We then take some fixed non-zero point b P Fnq .Since K is a Kakeya set it contains a line in all directions so there exists a point a P Fnqsuch that

ta � b � t | t P Fqu � K

and thereforefpa � b � tq � 0 @ t P Fq.

42

Since a and b are fixed the left hand side of this equation is a polynomial of degreeat most q � 1 in a single variable, t P Fq. However it has q ¡ q � 1 roots meaning byTheorem 1.1.1 it is identically zero, and hence all of the coefficients are zero. Howeverthe coefficient of td is simply fdpbq, which therefore must be zero. However our choiceof b was arbitrary meaning that fd vanishes on all points of Fnq (note that though wedo not include b � 0 in K the statement clearly still holds in this case since K is non-empty). We can now apply our DeMillo-Lipton-Schwartz-Zippel Lemma 5.1.1 from thelast chapter, to bound the number of roots that fd may have. Set S � Fq, then since fdhas degree d   pq � 1q we get that

dqn�1 ¤ pq � 1qqn�1   qn � ��Fnq �� .Therefore fd � 0 which is a contradiction giving us that f � 0.

We now have all we need to finish Dvir’s proof of the Finite field Kakeya conjecture,Theorem 6.2.3.

Proof of Theorem 6.2.3. We assume for contradiction that

|K|  �q � n� 1

n

.

Then by Lemma 6.2.6 there exists a non-zero polynomial f of degree at most q�1 whichvanishes on our Kakeya set K. However this contradicts Lemma 6.2.7 completing theproof. Hence since �

q � n� 1n

¥ qn

n!

the Finite field Kakeya conjecture is proven with Cn � 1n! .

This proof of this theorem inspired people to look further at the applications of thePolynomial method and gave it the credibility and reputation that it has today. LarryGuth and Nets Hawk Katz are two mathematicians who were particularly inspired byDvir. They would go on to do significant work on the Joints problem which led onto amuch improved bound for the Erdos distinct distances problem. As stated in their ownwork on the topic

Both our proofs are adaptations of Dvir’s argument for the Finite fieldKakeya problem. (L. Guth, N. Katz, [40])

Interestingly their work, while related, is not within finite fields but Euclidean space.We therefore shift our focus to reflect this, coming to the Joints problem in Chapter 8.

43

Part II

Euclidean Space

44

Chapter 7

Distances

We now look to consider some more of the applications of this method outside of finitefields. Notably in this section we consider some problems within Euclidean space. It ishere that we begin to see more of our links to Algebraic geometry and later, can makeuse of tools such as Bezout’s Theorem 1.5.2.

7.1 The s-Distances Problem

The s-Distances Problem is a problem within Discrete geometry. To approach the prob-lem we begin by defining the following.

Definition 7.1.1. Given a discrete set of points S � Rn, we define the Set of distancesof S as

DpSq � t‖x� y‖ | x,y P S, x � yuwhere ‖�‖ is the standard Euclidean norm on Rn. That is

‖v‖ �bv2

1 � . . . v2n for any v � pv1, . . . , vnq P Rn

With this notation we can ask the following, what is the maximum cardinality of thediscrete set S � Rn such that |DpSq| � s? We denote this value Pspnq as it is a functionof both the number of distances, s, and the number of dimensions, n.

7.1.1 The Two Distance Problem

Remark. For an accessible introduction to this problem the reader is directed to a shortchapter on the subject by the late Jirı Matousek, [41].

First we note that clearly Pspnq ¤ Ps�1pnq since s   s � 1. We can therefore get alower bound for P2pnq using P1pnq. This is simply n� 1 points arranged as the verticesof an n-simplex. Before attempting to calculate P2pnq for general n we will considersome simple cases. For n � 1 we see that P2p1q � 3, and for n � 2, by consideringgeometric constructions with circles, we can derive that P2p2q � 5. However for n ¡ 2

45

Figure 7.1: The optimal arrangements for some small values of s and n.

using such constructions would involve spheres and is much more difficult. We thereforeseek to find a general way of perhaps bounding, rather than directly solving, our result.We can use our Polynomial method to do this.

7.2 Bounding Pspnq

We can in fact bound the solution size quite effectively both from above and below. Webegin by stating the following result first proved by Aart Blokhuis, [42].

Theorem 7.2.1 (Blokhuis).

Pspnq ¤�n� s

s

.

This theorem, gives us that the bound is of order ns (or nn if s ¡ n). Howeverto determine the sharpness of this bound requires either a complete solution or a closelower bound. We can in fact easily get a very close lower bound.

Theorem 7.2.2. If s ¤ pn� 1q{2 then

Pspnq ¥�n� 1s

.

Remark. This simple proof expands upon the work of Matousek in [41]. Since the proofwe use only holds for s ¤ pn�1q{2, then where it is true the bound has order strictly ns.

Proof. Take ej � p0, . . . , 0, 1, 0, . . . , 0q P Rn, the zero vector with a single 1 in the jthcolumn. Then set

S � ts

i�1epjiq | 1 ¤ j1   j2 . . .   js ¤ nu

46

the set of sums of any s of these vectors. This is an s-distance set of cardinality�ns

�.

However we can embed S into Rn�1 since S lies within the hyperplane°nk�1 xk � s.

Since the cardinality is unchanged we achieve our result that Pspnq ¥�n�1s

�.

We therefore have a close lower bound when s ¤ pn� 1q{2. We will prove the upperbound for s � 2 however the proof for general s uses much the same method but withmore complex linear algebra. In fact Blokhuis extended this result not just to s-distancesets in Euclidean space but also in Hyperbolic space, as well as various other results.The interested reader can find all of these in his book, [43]. To prove the theorem we willuse a slightly modernised version of the original proof attaining various weaker resultsenroute, (see both [41] and [42]).

Theorem 7.2.3.P2pnq ¤

�n� 2

2

.

Remark. The original proof of this was released in 1984, long before the Polynomialmethod was fully categorised, however this slightly modernised proof applies our tools ina different way and as such it is fitting to feature it in a study of the Polynomial method.

Proof. Suppose we have a maximal two distance set

S � tp1, . . . ,pku � Rn

such that |S| � k and‖pi � pj‖ P td1, d2u for all i � j.

Now for each i we cleverly define the following polynomial over Rrx1, . . . , xns,

fipxq :� 1d2

1d22

�‖x� pi‖2 � d2

1

�‖x� pi‖2 � d2

2

.

We consider these functions as vectors within the space of real-valued functions fromRn Ñ R. Notice that fippjq � 0 for all i � j and 1 for i � j. Due to this we see that thefi are all linearly independent as follows. We assume we can find non-zero coefficientssuch that

0 � λ1f1pxq � . . .� λkfkpxq.If this is the case then it is true for all of Rn which includes pi for any i,

λ1f1ppiq � . . .� λkfkppiq � 0ùñ 0� . . .� λi � . . .� 0 � 0ùñ λi � 0.

Therefore λi � 0 for all i, giving a contradiction implying the fi are linearly independentover our vector space. This means we have k linearly independent vectors which thereforespan a linear subspace, V , of dimension k. Therefore if we can bound the dimension ofV we can bound the cardinality of S. To do this we need to find a set of functions, abasis, which generates V .

47

Remark. This is in fact an alternative form of our Polynomial method. Rather thanusing degree we can consider the dimension property. This form of our method is usedmore in Euclidean space where we tend to discuss objects more in the language of vectorspaces (though this does apply over finite fields as well). We use the fact that a vectorspace of dimension k is spanned by exactly k linearly independent functions. This sort ofrestriction on the complexity of a set of functions is what leads our method into Algebraicgeometry and links it to results such as Bezout’s Theorem 1.5.2.

The simplest set is to use all monomials of degree at most 4 of which there are�n�4

4�.

However this has quartic order so we seek a much more effective count, we thereforeexpand the fi. To do this we note that

p‖x� q‖2 � d2�q � px1 � q1q2 � . . . pxn � qnq2 � d2

�� x2

1 � 2q1x1 � q21 � . . . x2

n � 2qnxn � q2n � d2

�

� ‖x‖2 � 2n

i�1qixi �

n

i�1q2i � d2

�.

We see that this is spanned by the terms

‖x‖2, xi, 1

and therefore the fi are all linear combinations of the following terms (i � j),

‖x‖4, xi‖x‖2, x2i , xixj , xi, 1.

Remark. Note that the x4i terms can only appear within ‖x‖4 however all of the x2

i

terms appear independently. As such, since ‖x‖2 falls within their span, we do not countthis separately.

Counting the number of terms we see there are 1 � n � n � npn � 1q{2 � n � 1 �pn�1qpn�4q{2 linearly independent terms. This gives us a much better bound as provedby Larman Rogers and Seidel, [44], that

P2pnq � k � dimV ¤ 12pn� 1qpn� 4q

however we can still improve this slightly (proceeding as in [42]). We do this by showingthat the fi, xi and 1 are all linearly independent. We assume we can find non-zerocoefficients such that

k

i�1λifipxq �

n

j�1γjxj � α � 0. (7.1)

We first evaluate Equation 7.1 at Cej � p0, . . . , C, . . . 0q, by using our previous expansion(with pi � ppi1 , . . . , pinq),

1d2

1d22

k

i�1λipC2 � 2Cpij � ‖pi‖2 � d2

1qpC2 � 2Cpij � ‖pi‖2 � d22q � Cγj � α � 0.

48

Since this is identically zero for all C we can view it as a function of C in which allthe coefficients must be zero. Notably the coefficients of C4 and C3 are equal to zerowhich gives us that

k

i�1λi � 0,

k

i�1λipij � 0. (7.2)

We secondly evaluate Equation 7.1 at pi to give us

λi �n

j�1γjpij � α � 0.

We now multiply this by λi,

λ2i � λi

n

j�1γjpij � αλi � 0

and then sum over all pi to get

k

i�1λ2i �

k

i�1λi

n

j�1γjpij � α

k

i�1λi � 0.

We now apply the relations in Equation 7.2 to see that

k

i�1λ2i � 0 ùñ λi � 0 @ i P t1, . . . , ku

We can therefore rewrite Equation 7.1 accounting for this asn

j�1γjxj � α � 0. (7.3)

Evaluating Equation 7.3 at the zero-vector yields that α � 0 and thereforen

j�1γjxj � 0. (7.4)

Finally we evaluate Equation 7.4 at ej to get that γj � 0 for any j P t1, . . . , nu givingus our contradiction, and that the set of fi, xi and 1 are all linearly independent. Wecan therefore remove n� 1 functions from our previous bound as they are not requiredto span V . We therefore have the following inequality, and our result,

k ¤ 12pn� 1qpn� 4q � pn� 1q,

ùñ P2pnq � k ¤ 12pn� 1qpn� 2q �

�n� 2

2

.

49

While our upper bound holds for all n we have that for n ¥ 3,�n� 1

2

¤ P2pnq ¤

�n� 2

2

.

When n is much larger than 2 this is a very tight bound as both bounds are of order n2

but the difference is simply n� 1. The sharpness of this bound is even more noticeablefor larger values of s as the difference is still linear but (so long as s ¤ pn � 1q{2) thebounds both have order ns.

7.2.1 Extensions

We saw earlier that P2p2q � 5 however our upper bound gives that P p2q ¤ 6, as such thebound is not sharp and therefore, this problem remains somewhat open. The same isalso true of course for the lower bound as this does not even apply here. An interestingcase is that where s is of similar magnitude or even much larger than n, for examplethe solutions of Psp2q for all s. There are still a number of open problems relating tothis sort of question and it is highly possible that with continued use of the Polynomialmethod more breakthroughs will be seen in this area. One breakthrough of particularnote was in the Erdos distinct distances problem, which had a new bound establishedvery recently due to the Polynomial method. This problem comes from a different angle,asking for the minimum number of distinct distances if we have a fixed number of points.We discuss this problem more at the end of the next chapter.

50

Chapter 8

Lines and Joints

Remark. For the reader seeking to expand on this chapter the problem is discussed byTao in his survey of the Polynomial method, [12], and by Adam Sheffer in his lecturenotes, [45]. Rene Quilodran also has two papers on the matter, in the first he introducesthe problem, [46], and in the second gives a proof of the problem in Rn as well as disussingthe extensions to algebraic curves, [47].

8.1 The Joints Problem

The Joints problem is based on counting the number of joints formed by a finite set oflines in Rn.

Definition 8.1.1. Given a set of lines L � tl1, . . . , lku � Rn, a joint in Rn is a pointwhere n linearly independent lines l1, . . . , ln P L are concurrent.

The Joints problem is concerned with finding the maximum number of joints thatcan be formed with k lines. The problem was first posed as early as 1990, [48], andstood unsolved for almost 20 years yet its proof is remarkably simple. To solve thisproblem, rather than looking for the exact arrangement to achieve greatest efficiency wesimply look for the order of this maximal arrangement with respect to k. That is, weseek to bound the order of the relationship between the number of lines and the numberof joints.

Example 8.1.2 (A Simple Arrangement). One quite simple, but remarkably efficientarrangement of lines is to form an n-dimensional unit grid - that is to have our lines atunit intervals, with an equal number parallel to each of the axes. In our 3-dimensionalcase we therefore we have k{3 of our lines pointing in the direction of the x-axis arrangedin a square lattice, k{3 in the direction of the y-axis in a square lattice and similarlyfor the z-axis. If we assume k{3 is a square number then we form a perfect cube of sidelength m where m is the square root of k{3. The figure shows an example where m � 6in 3-dimensions.

51

Figure 8.1: A lattice arrangement of 108 lines in R3 with the joints highlighted.

By placing lines unit distances apart we find that we have joints at all of the pointsof

J � tpx, y, zq P N3 | 1 ¤ x, y, z ¤ak{3u.

Therefore if we have k � 3m2 lines, with this lattice arrangement, we have m3 joints.We now consider the order of this solution,

|J | � m3 ��c

k

3

�3

� k3{2?

27� Θpk3{2q.

Remark. Note that we achieve the lower bound for this problem by using the samecubic arrangement but generalised to Rn. By having k{n lines in the direction of eachaxis arranged in a square lattice this grid attains an order of Θpkn{n�1q. This gives thelower bound for our general solution.

We therefore see that arrangements of order at least k3{2 exist in 3-dimensions. Asof December 2008, due to the work of Guth and Katz, this order has been shown to bemaximal.

Theorem 8.1.3 (Guth and Katz, (2008)). The maximum number of joints that can beformed by a set of k lines in R3 has order Θpk3{2q.

8.2 Solving the Problem

While Guth and Katz were the first to have a proof to the problem, a number of paperswere released in quick succession giving both refinements and extensions of their work.Only a few months after, the proof was refined by Elekes, Kaplan, and Sharir, [49].

52

Their work was based on the original proof but introduced some refinements. Shortlyafterwards, in a follow up paper by Kaplan, Sharir, and Shustin, the bound was gener-alised to all n, with the proof refined further still, [50]. Later in 2009 Rene Quilodran,assisted by Fedor Nazarov found a remarkably concise form of the proof, however itremains in essence quite similar. In 2012 in a lecture course of his, Guth also gave aconcise version of his original proof which maintains the combinatoric principles, but isgreatly simplified. This proof appears in the lecture notes of Adam Sheffer, [45]. It isthis refined proof which we will adapt to prove the following for all n.

Theorem 8.2.1 (The Joints Problem in Rn). The maximum number of joints that canbe formed by a set of k lines in Rn has order Θpkn{n�1q.

8.2.1 The General Solution

As was previously mentioned, Guth and Katz were heavily influenced by the work ofDvir on the Finite field Kakeya conjecture. As such the proof of their work, and of thegeneral case, requires the following lemma as seen in Dvir’s proof in Chapter 6, (thoughwe have chosen F � R).

Lemma 6.2.6. Take a subset S of Rn. If |S|   �n�dn �, for some positive integer d, thenthere exists a non-zero polynomial f P Rrx1, . . . , xns, vanishing on S with degree at mostd.

We also make use of Bezout’s Theorem 1.5.2 in this proof to bound the number ofincidences between 2 curves in a plane. It is through this theorem that we apply a higherdimensional variant of our Polynomial method.

Theorem 1.5.2 (Bezout’s Theorem). Let f, g P Rrx, ys be polynomials with no non-constant common components, and of degree n and m respectively with m,n ¥ 1. Then

|Zpfq X Zpgq| ¤ mn � deg f � deg g.

Remark. The reader may wish to familiarise themselves with Example 1.5.1 and [11]to help grasp our use of Bezout’s Theorem.

With these two tools we can prove the following lemma. This is the key step toproving Theorem 8.2.1.

Lemma 8.2.2. Let L be a set of k lines in Rn and J the set of joints formed by L.There exists a line, l P L, such that l contains at most n |J |1{n of the joints.

Proof. For the proof we assume that all lines contain more than N � n |J |1{n joints andtake f to be the minimal degree polynomial vanishing on J .

I. We begin by bounding the degree of f using the following argument,�n�N

n

� 1n!

n¹j�1

pj �Nq ¡ Nn

n! � nn

n! |J | ¡ |J | .

53

This in conjunction with Lemma 6.2.6 proves that, since f has minimal degree,degpfq ¤ N .

II. We now show that f vanishes on all the lines in L.We choose a line l P L and taking some plane π with l � π. With this we defineγ � Zpfq X π which must be non-trivial since π contains some of our joints. Wenow apply a change of basis to Rn such that

tx1, x2, . . . , xnu Ñ ty1, y2, . . . , ynu

with the yi pairwise linearly independent, and y1, y2 spanning the plane. Sincethis is a linear transformation it takes f to some f of equal degree, respectingthe intersections between our functions. In this basis we can express γ and l asfunctions of y1 and y2 since y3, y4, . . . , yn are fixed by π. Fixing a variable eithermaintains or decreases the degree meaning

degpγq ¤ degpfq � degpfq ¤ N.

Figure 8.2: A cross section of a line, function and plane, undergoing a change of basis.This process respects intersections.

Since they are functions in two variables we can apply Bezout’s Theorem 1.5.2 toγ and l. The theorem states that the number of intersections of these curves is atmost the product of the degrees unless they share a common component. That is,

|γ X l| ¤ N � 1

unless they share a component. Since we assumed that l contains more than Njoints they must intersect at least N times and therefore they share a component.Since l is a line and hence only has one component, l � γ. This argument holdsfor every l P L and hence Zpfq contains every line of L.

54

III. We now perform a differentiating step which gives a contradiction of our minimalityassumption on f .Consider p P J incident to l1, l2, . . . , ln which are linearly independent. By theprevious statement l1, l2, . . . ln � Zpfq. If p was a non-singular point the lineswould all be contained within the tangent hyperplane to Zpfq at p. Howeversince they are linearly independent they span the whole of Rn and therefore p isa singular point of f . Furthermore, this means ∇fppq � 0 for all p P J . Since wetook f to be non-trivial it must have a partial derivative that is not identically zerowhich, without loss of generality, we assume to be f1 � Bf

Bx1. Since ∇f vanishes on

J , f1 also vanishes on J . This is a non-zero polynomial of degree strictly less thanf contradicting our minimality condition, proving the lemma.

With this lemma it is simple to prove Theorem 8.2.1, the Joints problem in Rn.

Proof of Theorem 8.2.1. Take L a set of k lines in Rn, J the set of joints of L and fixN � n |J |1{n. We can remove a line from L that is incident to at most N points of Jand update J and L accordingly but keeping N fixed. We can now repeat this processfor our new L, J as Lemma 8.2.2 guarantees such a line exists at every stage. We repeatthis process k times until all lines have been removed. Each iteration removes at mostN joints and all of the joints are removed after k iterations. This gives us the followingbound, leading to our proof,

|J | ¤ kN � kn |J |1{n ,ùñ |J | ¤ pnkqn{n�1 � Θpkn{n�1q.

This bound combined with the n-dimensional generalisation of our work in Subsection8.1.2 proves the maximum order is strictly Θpkn{n�1q.

This proof again shows the remarkable power of the method, a problem which stoodunsolved for nearly 20 years, proved within a few pages.

8.3 Extensions of the Problem

We have now seen the Joints problem generalised to n-dimensions, however there area number of other simple extensions which also use our Polynomial method in theirproofs. We previously defined a joint in n-dimensions as a common intersection pointof n linearly independent lines. However we can for example, consider the number ofintersections of fewer lines, lessen the the constraint of independence, or consider theintersections of curves rather than lines. This allows us to generalise beyond the Jointsproblem and look more into the intersections of general algebraic curves (see [47]). Thesegeneralisations are still very much at the forefront of mathematical research and it ishere that we most closely see our method overlap into Algebraic geometry. In fact italso from this work that Guth and Katz found a much improved result relating to theErdos distinct distances problem.

55

8.3.1 The Erdos Distinct Distances Problem

It would be wrong to discuss the work of Guth and Katz in this area without notingtheir remarkable work on this problem, also recognising the contribution of the lateGyorgy Elekes who’s work help establish the key link between Incidence geometry andthis problem, [51]. This Distances problem expands on our two distance problem posingit conversely as follows.

Given n distinct points in the plane what is gpnq, the minimum number of distinctdistances between them?

This problem was first posed in 1946 by Paul Erdos, perhaps one of the greatestmathematicians of the 20th century. He conjectured that the order of gpnq was greaterthat nc for all c   1. Using a square lattice he proved that

Θ�?n� ¤ gpnq ¤ Θ

�n?

logn

and conjectured the exact value was of a magnitude very close to the upper bound, [52].Of this conjecture he stated, writing for the event of his 80th birthday, that

My most striking contribution to geometry is, no doubt, my problem on thenumber of distinct distances. (P. Erdos, [53])

From a mathematician with such a variety of contributions this is a powerful claim.The problem stood for almost 75 years until 2010 when, by linking the problem to theirwork on the Joints problem, Guth and Katz proved the following, [54].

Theorem 8.3.1 (Guth and Katz, (2010)).

Θ�

n

logn

¤ gpnq ¤ Θ

�n?

logn

This theorem proved the conjecture of Erdos and brings the problem very close to

a full solution as the bounds are now separated by only a factor of?

logn. With themethod still being used to great effect we may perhaps soon see a full solution in thisarea.

Remark. The interested reader can find the original work by Erdos here, [52], andthe proof of the conjecture by Guth and Katz here [54] which uses some interestingextensions of the Polynomial method. Most notable of these is the method of polynomialcell decompostion. A lot of background work was also done By Elekes and Sharir in [51].Another good source of information is an online article by Terence Tao discussing theproof and its consequences, [55].

56

Part III

Graph Theory

57

Chapter 9

Graph Theory

9.1 Polynomials of Graphs

Perhaps one of the most surprising areas of application of the Polynomial method is toGraph theory. While we do not often evaluate these polynomials in a standard sense, wecan still use them to make combinatorial statements. This is particularly notable whenattempting to give graphs labellings.

Example 9.1.1 (A Simple labelling). Assume we have n points, tv1, v2 . . . , vnu, andhave some function f which assigns a label to them. Then how do we check the labellingis distinct? We can actually very easily construct a polynomial which will evaluate tozero unless the labelling is entirely distinct,¹

1¤i j¤npfpviq � fpvjqq .

By extending this sort of idea we can create more complicated polynomials which encodeinformation about graphs. By making the points of interest the roots we can even applyour various tools to arrive at some surprising conclusions.

Remark. This sort of construction is actually remarkably useful in mathematics evenwhen there is no metric or values, for example polynomial constructions are used intopology to help describe knots.

9.2 The Basics

Throughout this chapter we will use the definitions as seen in Wilson’s Introduction toGraph Theory, [56], to which the reader is directed for more information. We assume aminimal understanding of Graph theory but also state some definitions for reference.

• A Graph G � pV,Eq is defined by a set of vertices V � tv1, . . . , vnu, and a set ofedges between them, E � te1, . . . , emu where the ej are pairs of vertices.

58

• H � pW,F q is a subgraph of G � pV,Eq if W � V and F � E.

• A loop is an edge connecting a vertex to itself.

• A simple graph is a graph with no loops and no multiple (repeated) edges.

• An edge between two vertices is incident to those vertices and we say those verticesare incident to the edge.

• The incidence matrix, M, of a graph with n vertices and m edges is an n �mmatrix where ai,j is 1 if vi is incident to ej and 0 otherwise.

• The degree of a vertex is the number of edges incident to it, however loops counttwice as both ends are incident to the vertex.

• A regular graph is a graph in which every vertex has the same degree. Wherethat degree is d we call the graph d-regular.

• A complete graph is a simple graph with an edge between every pair of vertices,the complete graph with n vertices is denoted Kn.

Figure 9.1: The complete graph on 5 vertices, K5. This is also a loopless 4-regular graph.Due to being loopless, the regularity can be seen directly from the incidence matrix M,as precisely four 1s appear in each column meaning each vertex is of degree 4.

With these basic definitions we have all that we need to begin discussing the use ofthe Polynomial method in Graph theory.

9.3 Finding a p-regular Subgraph

In this section we prove a result concerning when a graph has a p-regular subgraph,with p a prime. We essentially state that when our graph is close to p2p� 1q-regular itcontains a p-regular subgraph.

59

Theorem 9.3.1. Given a prime p, if the vertices of a loopless graph G have averagedegree bigger than 2p�2 but degree at most 2p�1, then G contains a p-regular subgraph.

This idea of finding a subset order p within a set of order 2p � 1 in some waysresembles the Erdos-Ginzburg-Ziv Theorem, suggestive of the fact that our methods canbe applied. In fact a corollary to this theorem is that a p2p� 1q-regular graph containsa p-regular subgraph, which is a very similar statement. We also prove this theorem bycreating a polynomial over Fp, further implying the connection between our Additivenumber theory and Graph theory. We must therefore refer back to an earlier corollaryof Lagrange’s Theorem.

Corollary 2.3.1.1 (An Important Relation). Over a finite field order q � pk with pprime and a P Fqzt0u,

aq�1 � 1.

We make use of this as in earlier chapters to create an indicator function whichvanishes on all non-zero elements. This proof is remarkably short but highlights the useof our method within Graph theory. We will prove this result as done by Alon in [6].

Remark. For this theorem we can assume all vertices of G have degree at least 1. If thisis not the case we simply take a subgraph, H of G for which this is true as the conditionsof the theorem will be satisfied by H if they are satisfied by G. Furthermore if the proofholds for H then it will hold for G, since any subgraph of H is also a subgraph of G.

Proof. Set G � pV,Eq, a graph with n vertices and m edges and take M to be ourincidence matrix as previously defined such that ai,j is 1 if vi is incident to ej and 0otherwise. We now define the following polynomial over Frx1, . . . , xms,

fpx1, . . . , xnq �n¹i�1

��1��

m

j�1ai,jxj

�p�1��� m¹

j�1p1� xjq.

Note that since our average degree is more than 2p�2 and every edge connects precisely2 vertices we have that

|E| ¡ 12p2p� 2q |V |

and equivalentlym ¡ pp� 1qn.

Therefore since the first product is of degree at most pp � 1qn and the second producthas degree m, we have that degpfq � m as the coefficient of

±mj�1 xj � �1. We

can now apply our Combinatorial Nullstellensatz II, with α1 � . . . � αm � 1, andS1 � S2 � . . . � Sm � t0, 1u as this has cardinality greater than 1. By the theoremthere exists a point s � ps1, . . . , smq P t0, 1um such that fpsq � 0. Note that since fdoes vanish on the zero vector our s is non-zero. This means the second product will

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vanish at s, and hence the first product must not. By our Corollary 2.3.1.1, for this tobe the case we must have that

m

j�1ai,jsj � 0 pmod pq

for all i. We therefore take the subgraph formed from the set of edges

E � tek | sk � 0u.

The corresponding vertices must have a multiple of p non-zero elements in their respectivecolumn of the incidence matrix. Since the graph is loopless and the vertices all havedegrees between 1 and 2p we have that all the vertices must be of degree p, giving usthe desired p-regular subgraph.

This idea of applying combinatorial methods to the study of graphs has become verypopular in recent years and has notably led to various results in the topic of graphcolouring and choosability, (though other topics of Graph theory such as spanning treeshave also been addressed).

9.4 Graph Colouring and the use of Ideals

One of the most significant areas of research in Graph theory is that of graph colourabil-ity, and an interesting variant of this problem is the study of graph choosability, whichis a stronger, more general condition.

A proper vertex colouring of a graph is a function from the vertices to a list ofcolours such that the endpoints of every edge are differently coloured. We define a graphto be k-colourable for some integer k if there is a proper vertex colouring using at mostk colours.

In the case of choosability we assign a list of colours to each vertex and ask if we canform a proper vertex colouring by choosing, for each vertex, a colour from its associatedset. For notation we use positive integers to represent colours, such that we have infinitelymany distinct colours t1, 2, 3, . . . u. For a graph G � pV,Eq to be f -choosable means thatthe list of colours assigned to vi has cardinality fpviq. Then, if regardless of the coloursplaced on each list we still have a proper vertex colouring the graph is f -choosable. Ifthis holds for a constant function f � k then we describe the graph as k-choosable.

Remark. Note that k-choosable implies k-colourable as by placing the same set of kcolours on all vertices choosability implies a proper vertex colouring exists. However thecase of the inverse implication is a more interesting one. A line graph LpGq is a graphwhich has a vertex for every edge in G and an edge connecting every pair of verticeswhich came from adjacent edges. With this idea we can state what is perhaps the mostrenowned open problem in this area, the List colouring conjecture, which states that

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for line graphs k-colourable also implies k-choosable, meaning there is an equivalence.A recent result in the area was the proof that this holds for complete graphs of primedegree, [57].

Figure 9.2: K5 is not 4 choosable as if we choose the same set for all 5 vertices thereis no choosable proper vertex colouring. However it is 5 colourable, no matter what setof colours we assign to each vertex there will be a subset of them corresponding to aproper vertex colouring.

Various results concerning choosability can be proved using our Polynomial methods,many of which are discussed in a thesis on the topic by Anne Marie Spencer, [58]. Onetheorem of note is a result which was originally proved by Alon and Tarsi in 1992 [59],and has since had the proof somewhat reformulated in light of Alon’s development ofhis Combinatorial Nullstellensatz. However we first need a few more concepts.

• An orientation, O, of a graph G � pV,Eq gives a direction to every pair of indicesin E such that |O| � |E| and pvi, vjq P O ùñ tvi, vju P E. We say the orientationis even if |tpvi, vjq P O : i ¡ ju| � 0 pmod 2q and that it is odd otherwise.

• The outdegree of v P V in O is |tpv, wq P Ou|.• Given outdegrees d1, d2, . . . , dn for the vertices v1, v2, . . . , vn we define EOGpd1, . . . , dnq

to be the number of even orientations with these outdegrees and defineOOGpd1, . . . , dnqsimilarly for odd orientations.

• The graph polynomial, is a polynomial defined for G � pV,Eq, a graph with nvertices and m edges, as

fGpx1, . . . , xnq �¹tpxi � xjq | i   j, tvi, vju P Eu.

This polynomial has degree precisely m and notably vanishes when px1, . . . , xnq isnot a proper vertex colouring.

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With these ideas now outlined we can state and prove the following theorem con-cerning f -choosability.

Theorem 9.4.1. Take G a graph on n vertices with m edges. If we have a labellingfunction f : V Ñ N such that fpviq � αi�1 and

°ni�1 αi � m, then if EOGpα1, . . . , αnq �

OOGpα1, . . . , αnq then G is f -choosable.

Remark. The proof of this theorem is based on the work of Brad Jones in [5].

Proof. If there are no orientations with outdegrees α1, . . . , αn then the theorem is nullbut holds, so we proceed assuming that orientations with these outdegrees do exist. Wenow consider our graph polynomial

fGpx1, . . . , xnq �¹tpxi � xjq such that i   j, tvi, vju P Eu.

Note that fG has degree m � °ni�1 αi since we multiply over all edges. By induction on

m we now prove that

fG �¸

d1,...,dnPNpEOGpd1, . . . , dnq �OOGpd1, . . . , dnqq

n¹i�1

xdii .

If m � 0 then there is a single even orientation at pd1, . . . , dnq � p0, . . . , 0q and no otherorientations. This gives that us that fG � 1 which is equal to our trivial product withoutany terms. We can therefore begin our induction.

We assume that the result holds for m � 1 edges and prove it for m. Take a singlepoint of E, tv, wu, with v ¡ w and define a graph G � pV,Eztv, wuq for which ourtheorem holds by our assumption. Define O to be the reduction of O where this edgeis removed. To get an even orientation on G, if O is even we must add pw, vq so thatthe parity doesn’t change and we increase the outdegree of w by 1. If O is odd we mustadd pv, wq such that the parity does change and we increase the outdegree of v by 1.Therefore

EOGpd1, . . . , dnq � EOGpd1, . . . , dw � 1, . . . , dnq �OOGpd1, . . . , dv � 1, . . . , dnq,

and by similar argument

OOGpd1, . . . , dnq � EOGpd1, . . . , dv � 1, . . . , dnq �OOGpd1, . . . , dw � 1, . . . , dnq.

Note also that fG � pxw�xvqfG and that we assumed the theorem holds for m�1. Theterms in the following need to have

63

fG �pxw � xvq¸

d1,...,dnPN

�EOGpd1, . . . , dnq �OOGpd1, . . . , dnq

� n¹i�1

xdii

�¸

d1,...,dnPN

�EOGpd1, . . . , dv � 1, . . . , dnq �OOGpd1, . . . , dv � 1, . . . , dnq

�� �EOGpd1, . . . , dw � 1, . . . , dnq �OOGpd1, . . . , dw � 1, . . . , dnq

� n¹i�1

xdii

�¸

d1,...,dnPNpEOGpd1, . . . , dnq �OOGpd1, . . . , dnqq

n¹i�1

xdii

finishing our induction. We can now apply our Combinatorial Nullstellensatz II. Since weassume inequality between the even and odd terms the coefficient of

±ni�1 x

αii is non-zero.

Hence setting |Si| � αi � 1 tells us there is a point s � ps1, . . . , snq P S1 � . . .� Sn suchthat fGpsq � 0. Therefore, by the properties of the graph polynomial, this ps1, . . . , snqgives a proper vertex colouring showing G is f -choosable.

This proof begins to demonstrate both the ways in which we can use our Polynomialmethod in Graph theory and the usefulness of the graph polynomial, a polynomial weuse again in the next section.

9.4.1 k-Colourability - An Ideal ending

We end this study close to where we began, by again considering Ideals and their usewithin our Polynomial method. In doing so we will be able to apply our CombinatorialNullstellensatz I, as opposed II, to prove a result concerning k-colourability. We considerthe Ideal generated by the polynomials xki � 1, and the graph polynomial, defined asbefore.

Theorem 9.4.2. fG lies in the ideal generated by xki � 1 for i P t1, . . . , nu if and onlyif G � ptv1, . . . , vnu, Eq is not k-colourable.

The proof of this, as seen in [6], is a very short, but effective application of ourCombinatorial Nullstellensatz I.

Proof. We first prove the forwards implication. Assume that fG lies in the ideal gen-erated by xki � 1. Therefore all the generators of this ideal vanish on the kth roots ofunity. Therefore if we label the vertices by the kth roots of unity fG will vanish. Thismeans xi � xj for some pair in E and hence G is not k-colourable.

We now prove the reverse implication. Assume G is not k-colourable. Set S � S1 �. . . � Sn � tkth roots of unityu and note that G cannot be properly coloured with these.As in our Combinatorial Nullstellensatz I define pipxiq �

±sPSipxi � sq � xki � 1. Since

fG vanishes over any labelling by k colours it will vanish over Sn and therefore over all

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the common zeros of the pi. Therefore by the Combinatorial Nullstellensatz I there existpolynomials such that fG � °n

i�1 gipi and therefore fG is in the ideal generated by thepi.

This short proof links our work on Graph theory right back to our original statementcoming from Hilbert’s Nullstellensatz and nicely summarises how these combinatorialideas permeate through many areas of mathematics.

Remark. There are many more results in Graph theory due to the Polynomial methodand we have only discussed a very small sample of them here. The reader interestedparticularly in the applications of the method to Graph theory is directed to a chapterof an upcoming book by Doug West which dicusses many of these proofs in considerabledetail, [60]. A number of results are also discussed by Alon in [6]. A problem in a set ofnotes by Matousek, [3], may also be of interest. This links Polynomial identity testing tofinding perfect matchings in graphs, connecting the probabilistic elements of the SchwartzLemma 5.1.2 with our study of Graph theory, a concept not discussed here.

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Chapter 10

Concluding Remarks

We concluded this survey as we began, applying the underlying Algebraic geometrybehind the Polynomial method. Throughout this paper we have seen that these com-binatoric principles can be adapted to produce a broad range of tools, giving us thecapability to address a range of problems each in a specific manner. Whether in theform of the Combinatorial Nullstellensatz, or the Schwartz Lemma, the success of themethod is, at least in part, due to this variety. In a number of cases we saw that a newformulation led to a breakthrough in a previously unsolved area such as the case of theErdos distinct distances problem. Perhaps more remarkably though is the simplicitywith which the method is applied. Even in the seemingly complex problems dating backdecades, such as the Erdos-Heilbronn problem, the proof takes but a matter of para-graphs. This is truly what distinguishes the Polynomial method from other tools andguarantees its central importance to the mathematician. For indeed if the proof is notelegant, then perhaps it is unfinished.

Beauty is the first test: there is no permanent place in this world for uglymathematics. (G.H. Hardy, A Mathematician’s Apology. [61])

While I’ve attempted to include the major works of the method in this paper, it isby no means complete. Areas of interest such as spanning trees and the Szemeredi andTrotter Theorem going overlooked. As more work continues to be done, the Polynomialmethod continues to grow as a topic well beyond the scope of a single paper. A morecomplete picture requires a variety of resources but perhaps the best resource would bethe survey by Terence Tao, [12]. Other good resources for a balanced picture would beAlon’s writings on the method in [6], and perhaps the thesis of Robert Laudone whichcovers a number of areas, such as the Szemeredi and Trotter Theorem, which have notfeatured here, [62].

Turning our attention to the future of the method there are a number of areas wherewe can hope to see advances. Within Graph theory problems such as the List colouringconjecture continue to draw the attention of many great minds. Equipped with an evergrowing understanding of the workings of tools like the Combinatorial Nullstellensatz it

66

is very possible more results will come with time. As recently as March, a paper wasreleased which contained a new result concerning total colourings, a result which wasproved using the Combinatorial Nullstellensatz, [63]. However it is not just Graph the-ory where the method is still making headway as last year a seemingly unrelated resultconcerning partitions was proved in [64] also making use of our Polynomial method. Ofcourse when discussing recent work it is appropriate to include the bound on the Erdosdistinct distances problem. This proof is only a few years old and still very new to theworld of published mathematics. The methods developed by this paved the way for anumber of results, through the correspondence between Discrete and Incidence geometryproblems. It is possible further results will come from this, as people turn their attentionto higher dimensional analogues and similar problems such as the Erdos unit distanceproblem.

Since the relatively recent recognition of the method and its potential, it has con-tinually been applied to new and different problems across broad areas of mathematics.With each new application the potential for a new breakthrough increases. Whether thenext inspiration comes from a young mind like Christian Reiher’s, or from a more es-tablished name such as Alon’s, only time will tell. Despite its remarkable simplicity thePolynomial method remains an essential tool at the forefront of modern mathematicalresearch, and based on my findings there is no reason this won’t continue to be the case.

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