Ecole polytechnique

Promotion X-2012

SHINDE Sudarshan

RAPPORT DE STAGE DE RECHERCHE

Autour de la conjecture de Snevily

……………………………………………………………………………………

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NON CONFIDENTIEL

PUBLICATION

Département : Mathématiques

Champ: MAT594 : Théorie des nombres

Directeur de stage : M. Gaëtan Chenevier

Maître de stage : M. Stéphane Vinatier

Dates du stage : le 16 mars 2015 – le 3 juillet 2015.

Nom et adresse de l'organisme : Université de Limoges, 33, rue F. Mitterrand, 87000,

Limoges FRANCE.

Declaration d’integrite relative au plagiat

Je soussigne Sudarshan SHINDE certifie sur l’honneur :

1. Que les resultats decrits dans ce rapport sont l’aboutissement de mon travail.

2. Que je suis l’auteur de ce rapport.

3. Que je n’ai pas utilise des sources ou resultats tiers sans clairement les citeret les referencer selon les regles bibliographiques preconisees.

Je declare que ce travail ne peut etre suspecte de plagiat.

Date : 3 Juillet 2015

Signature :

1

Abstract

Snevily’s conjecture says that for G, an abelian group of odd order and for any two k-element subsets{a1, ..., ak} and {b1, ..., bk} of G, there exists a permutation π ∈ Sk such that ai + bπ(i) are pairewisedistinct. This conjecture was proved by Arsovski ([2]) for abelain groups of odd order and Alon([8]) for cyclic groups of prime order. However Arsovaski’s proof does not extend to multisets wherewe allow repetition of set elements and Alon’s result is not valid in other cyclic groups. This factraises a question on the nature of the multisets {a1, ..., ap} in Fp for which there is no ordering ofais such that

∑i iai = 0. This question was answered by Gacs et al ([4]). This result also has

consequences on the polynomials of prescribed range. Gacs et al also came up with a conjectureabout the polynomials of prescribed range in the same paper. In order to better understand theconjecture and its hypothesis, we studied a paper by Biro ([6]).

Resume

La conjecture de Snevily dit que etant donne un groupe abelien d’ordre impair G et deux sousensemble des k elements {a1, ..., ak} et {b1, ..., bk} de G, il existe une permutation π ∈ Sk telleque ai + bπ(i) sont deux-a-deux distincts. Cette conjecture etait demontree par Arsovski ([2]) pourles groupes abelien d’ordre impair et par Alon ([8]) pour les groupe cycliques d’ordre premier.Cependant le resultat d’Arsovski ne s’applique pas au ”multisets” ou on autorise repetition deselements dans le sousensemble que l’on considere et celui d’Alon ne s’applique pas aux groupescycliques d’ordre non-premier. Ce fait a motive une question sur la nature des ’multisets’ {a1, ..., ap}en Fp pour lesquels il n’existe pas une ordre des ais telle que

∑i iai = 0. Cette question ete resolue

par Gacs et al ([4]). Cette resultat a aussi une consequence sur les polynomes dont image estpredifini. De plus, Gacs et al aussi propose une conjecture sur ce sujet. Afin de mieux comprendrecette conjecture, nous avons etudie une article de Biro ([6]).

Contents

1 Snevily’s conjecture 21.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 A proof by Arsovski . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3 A proof by Alon for cyclic groups of prime order . . . . . . . . . . . . . . . . . . . . 4

1.3.1 Proof of Theorem 1.3.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.4 Dyson’s conjecture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2 Permutations on finite fields 92.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.2 Proof of Theorem 2.1.1 when q = p, an odd prime . . . . . . . . . . . . . . . . . . . 10

2.2.1 Some elementary results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102.2.2 The essential part of the proof in the case q = p, an odd prime . . . . . . . . 122.2.3 Proof of Theorem 2.1.1 when p = q, an odd prime . . . . . . . . . . . . . . . 152.2.4 General case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.2.5 Proof for q even . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2.3 Consequences of Theorem 2.1.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.3.1 Polynomials of prescribed range . . . . . . . . . . . . . . . . . . . . . . . . . . 172.3.2 Hyperplanes of a vector space over Fq . . . . . . . . . . . . . . . . . . . . . . 18

3 Biro’s result 213.1 Notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213.2 Introduction and the main theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213.3 Some examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233.4 An example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

Appendices 29

A Combinatorial Nullstellensatz 30

B Python code used in Chapter 3 32

C Lagrange interpolation 35

Acknowledgement 36

1

Chapter 1

Snevily’s conjecture

1.1 Introduction

In this chapter of the report, we will state and prove a result conjectured by Hunter Snevily in 1999in [1]. We will present two proofs. The first one, which is more general and uses linear algebra anda result from representation theory, was given by Bodan Arsovski in 2011 in [2]. And the secondone, which is limited only to cyclic groups of prime order and uses Combinatorial Nullstellensatz,was given by Noga Alon in 2000 in [8]. Alon’s proof was later modified by in 2001 Dasgupta et alfor all finite cyclic groups in [3].

We will start with the statement of Snevily’s conjecture.

Conjecture 1. (Snevily) Let G be a finite abelian group of odd order. Then for any positive integerk ≤ |G| and for any two k-element subsets {a1, ..., ak} and {b1, ..., bk} of G, there exists a permutationπ ∈ Sk such that ai + bπ(i) are pairewise distinct.

Remark 1.1.1. Snevily’s conjecture is true even for the abelian groups of even order provided thesubsets we consider are not subgroups of even order (or their translations).

1.2 A proof by Arsovski

We will need following result from representation theory.

Proposition 1. Let G be a finite group of exponent n and F be a finite field such that n divides|F∗|. Then the system of characters from G to F∗ forms a basis of the space of functions from G toF.

Proof. (of Conjecture 1) We will prove it by contradiction. Let G be a finite abelian group of oddorder m and exponent n. Suppose that there exist two k-element subsets {a1, ..., ak} and {b1, ..., bk}of of G such that ∀π ∈ Sk, one can find a pair of indices i 6= j such that ai + bπ(i) = aj + bπ(j).

Let Fq be a finite field of q elements and of characteristic 2. We will suppose that n divides q− 1and q > m.

2

Lemma 1.2.1. For any system of characters φ1, ..., φk : G −→ F∗q , we have∑π∈Sk

det∣∣φπ(i)(ai + bj)

∣∣ = 0,

where∣∣φπ(i)(ai + bj)

∣∣ is a k × k matrix.

Proof. As we are considering a representation on a field of characteristic 2, we will not distinguishbetween plus and minus signs while expanding the determinant.

∑π∈Sk

det∣∣φπ(i)(ai + bj)

∣∣ =∑π∈Sk

(∑τ∈Sk

k∏i=1

φπ(i)(ai + bτ(i))

)

=∑π∈Sk

(∑τ∈Sk

k∏i=1

φi(aπ−1(i) + bτ(π−1(i)))

)

=∑π∈Sk

(∑τ∈Sk

k∏i=1

φi(aπ(i) + bτ(π(i)))

)

=∑τ∈Sk

(∑π∈Sk

k∏i=1

φi(aπ(i) + bτ(π(i)))

)=∑τ∈Sk

det∣∣φi(aj + bτ(j))

∣∣As we have supposed that Snevily’s conjecture is not true for {a1, ..., ak} and {b1, ..., bk}, every

determinant on the right side has a pair of equal columns forcing each of them to be 0. This concludesthe proof of Lemma 1.2.1.

Lemma 1.2.2. Let φ be any map from G to Fq then

det |φ(ai + bj)| = 0.

Proof. We use the fact that characters φ1, ..., φm : G −→ F∗q form a basis for the vestor space of allsuch maps. Thus we can write φ as a linear combination of all the characters. Hence, there existλ1, ..., λm ∈ Fq such that

φ = λ1φ1 + ...+ λmφm.

As determinant function is multilinear, we obtain,

det |φ(ai + bj)| = det

∣∣∣∣∣m∑s=1

λsφs(ai + bj)

∣∣∣∣∣=

∑(s1,...,sk)∈{1,2,...,m}k

(

k∏i=1

λsi) det |φsi(ai + bj)| .

If any two sis are equal then by using multiplicativity of characters φsi , we obtain that two corre-sponding rows of |φsi(ai + bj)| are proportional. In order to see this, consider si = si′. Then in twocorresponding rows will look like;

3

φsi(ai)φsi(b1) φsi(ai)φsi(b2) ... φsi(ai)φsi(bk)

andφsi′(ai′)φsi′(b1) φsi′(ai′)φsi′(b2) ... φsi′(ai′)φsi′(bk).

Clearly as si = si′, second row = first row× φsi (ai′φsi (ai)

and this forces the determinant of |φsi(ai + bj)|to be zero. We can thus write by putting an order on sis and then summing over all the permutations,

m∑s1,...,sk=1

(

k∏i=1

λsi) det |φsi(ai + bj)| =∑

1≤s1<...<sk≤m

(

k∏i=1

λsi)(∑π∈Sk

det∣∣φsπ(i)

(ai + bj)∣∣).

Finally by Lemma 1.2.1, each determinant on the right side becomes 0. Thus we conclude theproof of Lemma 1.2.2.

Lemma 1.2.3. Let A be a k×k matrix, each of whose entries is one of the formal variables z1, ..., zm.We also suppose that any two entries in the same row or column are distinct. Then these formalvariables can be assigned values from Fq in such a way that det(A) 6= 0.

Proof. We will prove it by induction on k. For k = 1, any non-zero value will suffice. Let k > 1.One can always suppose that z1 occurs as one of the entries in A. Then det(A) will be a polynomialin z1 of degree at most k < |Fq| where the leading coefficient of this polynomial comes from thedeterminant of a submatrix of A. By induction, remaining formal variables can be assigned valuesin such a way that this coefficient is non-zero. Thus the polynomial that one obtains in z1 is not azero polynomial and thus one can assign a value to z1 in such a way that det(A) 6= 0. This ends theproof of Lemma 1.2.3.

Now we continue with the proof of Conjecture 1. we associate with every g ∈ G a formal variablez(g). Then all entries in the same row or column of |z(ai + bj | are distinct. By Lemma 1.2.3, eachof these entries can be assigned a value in such a way that the determinant of the resulting matrixφ(ai + bj) is non-zero. This contradicts Lemma 1.2.2 which says that the determinant of φ(ai + bj)must be zero.This concludes the proof given by Arsovski of Conjecture 1.

1.3 A proof by Alon for cyclic groups of prime order

Theorem 1.3.1. (Alon) Let p be an odd prime and Zp be the additive group of integers modulo p.Then for any positive integer k ≤ p and for any two k-element subsets {a1, ..., ak} and {b1, ..., bk} ofZp, there exists a permutation π ∈ Sp such that ai + bπ(i) are pairewise distinct in Zp.

Snevily’s conjecture says that Theorem 1.3.1 is true even when we replace Zp by an abelian groupof odd order. Theorem 1.3.1 is non-trivial only when k < p.1 Furthermore, one can deduce Theorem1.3.1 from following result where we allow {a1, ..., ak} to be a multiset.

Theorem 1.3.2. Let p be a prime and let k be a non-negative integer such that k < p. Let{a1, ..., ak} be a multiset in Zp and let B = {b1, ..., bk} be a k-element subset of Zp. Then thereexists a permutation π ∈ Sk such that the sums ai + bπ(i) are pairewise distinct in Zp.

1If k = p, it suffices to take ai = bi i.e. π = id.

4

Remark 1.3.1. Let us remark that Theorem 1.3.2 is not true if we replace Zp by Zn where n is acomposite number. In order to see this, suppose n = k× s then it suffices to take {a1 = 0, ..., ak−1 =0, ak = s} as a multiset and B = {0, s, 2s, ..., (k−1)s} is a k-element subset of Zp. One can see thatthere does not exist any permutation in Sk for which ai + bπ(i) are pairewise distinct in Zp. AlsoTheorem 1.3.2 is not true when k = p, it suffices to take {a1 = 0, ..., ap−1 = 0, ap = 1} as a multisetand B = {0, 1, 2, ..., p− 1} as a k-element subset Zp.

In order to prove Theorem 1.3.2 which will imply Theorem 1.3.1, we will need a result known asCombinatorial Nullstellensatz. It could be found in [7]. We will also talk about it in Appendix A.

Theorem 1.3.3. (Alon) Let F be any field and let P be a polynomial in F[Z1, ..., Zk]. Suppose

deg(P ) =∑ki=1 ti, where tis are non-negative integers. Furthermore we will suppose that the coeffi-

cient of∏ki=1 Z

tii in P is non-zero. Then, if S1, ..., Sk are subsets of F such that #(Si) > ti, there

are s1 ∈ S1, ..., sk ∈ Sk such that P (s1, ..., sk) 6= 0.

Note that in this report the notation Sk is also used for the symmetric group of k symbols.However the context should clarify its usage.

1.3.1 Proof of Theorem 1.3.2

Consider the following polynomial in k variables over Zp;

P (X1, ..., Xk) =∏

1≤i<j≤k

(Xi −Xj)∏

1≤i<j≤k

((ai +Xi)− (aj +Xj)).

We want to calculate the coefficient of the monomial∏ki=1 x

k−1i in P . As the total degree of P is

equal to that of given monomial, the coefficient of∏ki=1 x

k−1i in P must be equal to the coefficient

of∏ki=1 x

k−1i in ∏

1≤i<j≤k

(Xi −Xj)∏

1≤i<j≤k

(Xi −Xj) =∏

1≤i<j≤k

(Xi −Xj)2.

On the other hand, the following Vandermonde identity implies that upto plus or minus sign thiscoefficient must be equal to k!.∏

1≤i<j≤k

(Xi −Xj) = ±det |xi−1j | =∑π∈Sk

(−1)σ(π)k∏i=1

xk−iπ(i).

In order to see this, we take squares;∏1≤i<j≤k

(Xi −Xj)2 =

∑π∈Sk

k∏i=1

x2k−2iπ(i) +∑

π 6=τ∈Sk

(−1)σ(π)+σ(τ)k∏i=1

xk−iπ(i)xk−iτ(i).

Now one can see that the monomial∏ki=1 x

k−1i does not appear in the first term in the right-

hand side because when i becomes k, one of the Xis disappear as its degree vanishes, thus themonomial

∏ki=1 x

k−1i has to appear in the second term of the right-hand side. We will now show

that for any permutation π ∈ Sk there exists an unique permutation τ 6= π in Sk such that∏ki=1 x

k−1i =

∏ki=1 x

k−iπ(i)x

k−iτ(i). Let π be a given permutation. We need to construct τ . Let us

first expand∏ki=1 x

k−iπ(i)x

k−iτ(i) to obtain;

k∏i=1

xk−iπ(i)xk−iτ(i) = xk−1π(1)x

k−1τ(1)x

k−2π(2)x

k−2τ(2)...x

1π(k−1)x

1τ(k−1)x

0π(k)x

0τ(k).

5

This expansion suggests that τ(k) = π(1) and τ(1) = π(k). Similarly we also notice that τ(k− 1) =π(2) and τ(2) = π(k−1). We can continue in the same manner to conclude that τ is the permutationsatisfying π(i) = τ(k + 1 − i). Thus the coefficient we are looking for is k! = #Sk upto a plus orminus sign.2

To conclude the proof of Theorem 1.3.2, we remark that the coefficient we obtained is never 0modulo p as k < p thus by Theorem 1.3.3 applied when t1 = ... = tk = k− 1 and S1 = ... = Sk = B,we obtain that there exist bi ∈ Si = B such that

P (b1, ..., bk) =∏

1≤i<j≤k

(bi − bj)∏

1≤i<j≤k

((ai + bi)− (aj + bj)) 6= 0.

As all the elements in B are distinct, the sums ai + bi must be pairwise distinct too. This ends theproof of Theorem 1.3.2 which implies Theorem 1.3.1.

We end this section by noting that Theorem 1.3.1 can be generalized using Dyson’s conjecturein its generality. These generalised extensions of Theorem 1.3.1 can be found in [8].

1.4 Dyson’s conjecture

This proof of Dyson’s conjecture was given by I. J. Good in [10]. We will prove an equivalentreformulation.

Theorem 1.4.1. Let x = (x1, ..., xn) and a = (a1, ..., an) where ai ≥ 0 ∀1 ≤ i ≤ n. Let G(a) bethe constant term in the expansion of

F (x; a) =∏i 6=j

(1− xjxi

)aj , 1 ≤ i, j ≤ n.

Then

G(a) =(a1 + ...+ an)!

a1!...an!.

Proof. Let g be a function which is equal to 1 for all xis. We will apply Lagrange interpolation tog. Thus we have n data points, (x1, 1), ..., (xn, 1), and Lagrange polynomial associated to these ndata points;

L(X) =

n∑j=1

n∏i=1,i6=j

X − xixj − xi

.

As a consequence, it is the unique polynomial of degree at most n which describes g as a function.But we notice that the constant function g = 1 suffices. Thus we must have L(X) = 1. In particular,L(0) = 1. Thus,

n∑j=1

n∏i=1,i6=j

−xixj − xi

=

n∑j=1

n∏i=1,i6=j

(1− xjxi

)−1 = 1.

Now we claim thatF (x; a) =

∑j

F (x; a1, ..., aj−1, aj − 1, aj+1, ..., an).

2Actually this coefficient is −1

(k2

)k! but as we are only going to use the fact that it is not 0 modulo p, we don’t

need to calculate the sign of the coefficient.

6

In order to prove it, suppose aj 6= 0,∀1 ≤ j ≤ n. Also define a(j)j′ = aj when j′ 6= j anda(j)j′ = aj − 1 when j′ = j. We then see,

∑j

F (x; a1, ..., aj−1, aj − 1, aj+1, ..., an) =∑j

∏j′

∏i 6=j′

(1− xj′

xi

)a(j)j′

=∑j

∏j′

∏i 6=j′

(1− xj′

xi

)aj′∏i 6=j′

(1− xj

xi

)−1= F (x; a)×

∑j

∏i 6=j

(1− xj

xi

)−1= F (x; a)

Then we can compute the constant term,

G(a) =

n∑j=1

G(a1, ..., aj−1, aj − 1, aj+1, ..., an). (1.1)

Now suppose aj = 0 for some 1 ≤ j ≤ n then the terms of the form(

1− xjxi

)ajwill not appear

in the product for all is. Thus, xj will come only from the denominator and hence will have onlynegative powers in the expansion of F (x; a) so G(a) must be equal to the constant term in

F (x1, ..., xj−1, xj , xj+1, ..., xn; a1, ..., aj−1, aj , aj+1, ..., an),

where xj or aj means that xj and aj are absent as variables of F . We then obtain if aj = 0,

G(a) = G(a1, ..., aj−1, aj , aj+1, ..., an). (1.2)

And we also have for n-tuple; 0 = (0, ..., 0),

G(0) = 1 (1.3)

Thus we notice that G(a) can be defined recursively. It remains to see that the constant termclaimed in Theorem 1.4.1 also satisfies above three equations. Let us denote the constant term byM(a).Clearly as 0! = 1, M(0) = 1. Now if aj = 0 for some 1 ≤ j ≤ n then

M(a) =(a1 + ...aj−1 + aj+1 + ...+ an)!

a1!...aj−1!aj+1!...an!.

We thus see that if aj = 0 then M(a) does not depend on aj . In order to verify Equation 1.1, wecalculate M(a) for a = (a1, ..., aj−1, aj − 1, aj+1, ..., an). We will need to show that after summingover js, we obtain M(a) for a = (a1, a2, ..., an).

M(a1, ..., aj−1, aj − 1, aj+1, ..., an) =(a1 + ...+ aj−1 + (aj − 1) + aj+1 + ...+ an)!

a1!...aj−1!(aj − 1)!aj+1!...an!

=((∑ni=1 ai)− 1)!× aj∏n

i=1 ai!

=(∑ni=1 ai)!× aj

(∑ni=1 ai)×

∏ni=1 ai!

7

Now we sum it over all the js to obtain,

n∑j=1

M(a1, ..., aj−1, aj − 1, aj+1, ..., an) =(∑ni=1 ai)!×

∑nj=1 aj

(∑ni=1 ai)×

∏ni=1 ai!

=(∑ni=1 ai)!∏ni=1 ai!

= M(a1, ..., an).

Thus we conclude G(a) = M(a).

8

Chapter 2

Permutations on finite fields

2.1 Introduction

In this section of the report we will prove a result proved by Gacs et al in [4]. We will follow theirscheme of the proof.

Theorem 2.1.1. Let {a1, a2, ..., aq} be a multiset in Fq. Then there are no distinct elements b1, ..., bq(that is an ordering of the field elements) in Fq such that

∑i aibi = 0 if and only if q − 2 of the ais

are equal (to say a) and remaining two are of the form a + b and a − b for some b 6= 0 i.e. after asuitable permutation a1 = a2 = ... = aq−2 = a, aq−1 = a+ b and aq = a− b for some b 6= 0 in Fq.

One can explain where this theorem comes from. In general Arsovski’s result does not extendto multisets. Also Alon’s result for cyclic groups of prime order can not be extended to other cyclicgroups. In order to see that, let p be a prime and let b1, ..., bp−1 be the elements of F∗p. Supposethat for a multiset {a1, ..., ap−1} of F∗p which is a multiplicative cyclic group, we have a permutationπ ∈ Sp−1 such that aibπ(i) are pairwise distinct. Then by taking sum of these products, we see that∑i aibπ(i) = 0 as we are adding all non-zero elements of a finite field Fp. In other words, for any

multiset of size p, {a1, ..., ap = 0} in Fp, we could find distinct field elements b1, ..., bp = 0 such that∑i aibi = 0 and this would not produce an exceptional case in Theorem 2.1.1 when considered for

Fp, p prime.

We will prove the theorem in full detail in case of q = p, an odd prime. When q = ph, h > 1, theproof follows similar algorithm but is more complicated and involves comparatively more calculationsand manipulations thus we will only provide some indications. We will deal with the fields of evencharacteristics in the end.

In order to prove this theorem, we will require two results, namely, Lucas’ theorem and a con-sequence of Alon’s Combinatorial Nullstellensatz. For the sake of completeness we will state thesetheorems now.

Theorem 2.1.2 (Lucas). Let p be a prime. For two integers n and k having p-adic expansionsrespectively n1 + n2p+ ...+ nrp

r−1 and k1 + k2p+ ...+ krpr−1, we have;(

n

k

)≡(n1k1

)...

(nrqr

)mod p.

9

Theorem 2.1.3 (Alon). Let G(Y1, ..., Yk) be a polynomial over Fq which vanishes everywhere. Thenone can write G in the following form:

G(Y1, ..., Yk) = (Y q1 − Y1)f1 + ...+ (Y qk − Yk)fk,

where fis are polynomials of degree at most deg(G)− q.

2.2 Proof of Theorem 2.1.1 when q = p, an odd prime

From now on, we will suppose that q ≥ 11. Other cases can be treated by explicit calculations. Wewill start with some elementary results.

2.2.1 Some elementary results

Lemma 2.2.1. If for a multiset {a1, a2, ..., aq}, there is no permutation of the elements b1, ..., bq ofFq such that

∑i aibi = 0 then the same is true for {a1 + d, a2 + d, ..., aq + d} where d ∈ Fq and for

{ca1, ca2, ..., caq} for any non-zero c ∈ Fq.

Proof. Easy to see.

Lemma 2.2.2. If ais admit at most 3 different values then Theorem 2.1.1 is true.

Proof. If all the ais are equal then we can by the above lemma, translate it to a multiset havingq 0s then any ordering of the elements of Fq will give us

∑i aibi = 0. So we can suppose that there

are at least two different values of ais.

After translation, we can suppose that 0 is the value with the largest multiplicity and other twovalues (possibly equal) are 1 and a 6= 0.

If a = 1 then we can suppose that there are m 1s and q −m 0s. We will construct an orderingb1, ..., bq of Fq recursively. Let b1 take any non-zero value. Then we can let b2 = −b1. For b3, weassign a value that has not already been taken and let b4 = −b3. We continue in this manner. If mis even then we are done as we can take random values for remaining bis. If m is odd then we letbm = 0 and other values are again to be taken randomly.

If a 6= 1 then suppose that first m ais are 1 then next l ais are a and remaining are zero. If atmost one of m and l is odd then we follow the same procedure as above to conclude.

Now we shall consider the case when both of them are odd. Then by letting b1 = −a andbm+1 = 1, and then we can follow the same method. As we want to construct distinct bis, thismethod won’t work if −a = 1 i.e. a = −1. If m = l = 1 then we obtain that there are q − 2 0s,one 1 and one −1 in our multiset {a1, a2, ..., aq} which precisely is the exceptional case in Theorem2.1.1, after reduction by using Lemma 2.2.1. If one of them is at least 3 (say m) then we letb1 = A, b2 = B, b3 = C and b4 = A+B + C with well chosen A,B and C and for remaining bis, welet b5 = d where d has not already been taken and b6 = −d. As l − 1 is even, we can continue inthis way.

Now we state a technical lemma.

10

Lemma 2.2.3. Let the multiset {a1, a2, ..., ak} contain at least 3 different values and let l be themaximal multiplicity. Let m1,m2,m3 be natural numbers such that m1 + 2m2 + 3m3 = k. Then wecan partition the ais into m3 classes of size 3, m2 classes of size 2 and m1 classes of size 1 in sucha way that elements in the same class are pairwise different, provided we have one of the followingcases.

1. m2 = 0,m1 = 1 and l ≤ m3;

2. m2 = 1,m1 = 0 and l ≤ m3 + 1;

3. m3 = 0 and l ≤ m1 +m2;

4. m3 = 1,m2 = 0 and l ≤ m1;

5. m3 = 1,m2 = 1 and l ≤ m1 + 1.

Proof. After some permutation we can suppose that the ais are arranged in such a way that equalelements have consecutive indices hence if |i− j| ≥ l then ai 6= aj as l is the largest multiplicity.

1. We have k = 3m3 + 1 and l ≤ m3. It suffices to take the i-th class consisting of ai, ai+m3and

ai+2m3 for i = 1, 2, ...,m3 and let ak be the class of the size 1.

2. We have k = 3m3 + 2 and l ≤ m3 + 1. It suffices to take the i-th class consisting of ai, ai+m3+1

and ai+2m3+2 for i = 1, 2, ...,m3 and let am3+1 and a2m3+2 be the class of the size 2.

3. We have k = 2m2 +m1 and l ≤ m1 +m2. It suffices to take the i-th class consisting of ai andai+m1+m2

for i = 1, 2, ...,m2 and let the rest of the classes of size 1 be arbitrary.

4. As the multiset has at least 3 different values, this case becomes evident.

5. If there are at least 4 different values, then one can easily see that such an arrangement ispossible. If there are exactly three different values then at least two of them occur at leasttwice as we have at least 5 elements and then it is easy to find a required arrangement.

Now we will prove a result which will serve us as a tool to prove Theorem 2.1.1.

Theorem 2.2.1. Let a1, a2, ..., ak be non-zero field elements such that there are no distinct fieldelements b1, b2, ..., bk such that

∑i aibi = 0. Furthermore let G be the polynomial in k variables

defined by:G(Y1, ..., Yk) = ((Y1 + ...+ Yk)q−1 − 1)D(Y1, ..., Yk)

where D is the discriminant; ∣∣∣∣∣∣∣∣ak−11 ak−21 Y1 ak−31 Y 2

1 ... Y k−11

: : : : :: : : : :

ak−1k ak−2k Yk ak−3k Y 2k ... Y k−1k

∣∣∣∣∣∣∣∣ .Then

G(Y1, ..., Yk) = (Y q1 − Y1)f1 + (Y q2 − Y2)f2...+ (Y qk − Yk)fk.

Here fis are polynomials in Y1, ..., Yk of degree at most deg(G)− q.

11

Proof. Let us consider a polynomial

F (X1, ..., Xk) = ((a1X1 + ...+ akXk)q−1 − 1)∏

1≤i<j≤k

(Xi −Xj).

We claim that this polynomial is zero everywhere. Indeed as the factor∏

1≤i<j≤k(Xi −Xj) is non-zero if and only if X1, ..., Xk are pairwise distinct. And we know by hypothesis that for such pairwisedistinct Xis, we must have a1X1+...akXk 6= 0. Thus by a consequence of Fermat’s little theorem, wemust have (a1X1+ ...+akXk)q−1−1) = 0. And one the other hand if (a1X1+ ...+akXk)q−1−1) 6= 0then we must have a1X1 + ...+ akXk = 0 which means there are repetitions in Xis which will force∏

1≤i<j≤k(Xi −Xj) to be 0.

Now we will do a change of variables in F . Let Yi := aiXi for all is. As∏

1≤i<j≤k(Xi −Xj) isa Vandermonde determinant (upto a plus or minus sign), we claim that F is zero everywhere if andonly if

((Y1 + ...+ Yk)q−1 − 1)D1(Y1, ..., Yk)

is zero everywhere, where D1 is the determinant:∣∣∣∣∣∣∣∣1 (Y1/a1) (Y1/a1)2 ... (Y1/a1)k−1

: : : : :: : : : :1 (Yk/ak) (Yk/ak)2 ... (Yk/ak)k−1

∣∣∣∣∣∣∣∣ .The polynomial G can be obtained by multiplying i-th row by ak−1i for all is. Thus G becomes 0 forall the substitutions. Finally we use Theorem 2.1.3 to conclude that G has the required form.

Remark 2.2.1. By the above theorem, we see that if there is a term in G of maximal degree, i.e.

degree of G itself which is (q − 1) + k(k−1)2 , that term must have at least one Yi with degree at least

q. This remark will be useful in order to prove Theorem 2.1.1.

2.2.2 The essential part of the proof in the case q = p, an odd prime

We will begin with a lemma. Let A := {a1, a2, ..., aq} be a multiset in Fq.

Lemma 2.2.4. Suppose that there does not exist any ordering of the elements b1, ..., bp of Fp suchthat

∑i aibi = 0. Then at least p+2

3 of the ais are equal.

Proof. Suppose that 0 does not occur in A.1 Let us consider the polynomial G defined in Theorem2.2.1 with k = p. We will follow an algorithm in order to prove this lemma.Step 1 Consider a monomial Y in Y1, ..., Yp of degree deg(G) such that deg(Yi) < p for all 1 ≤ i ≤ p.By Remark 2.2.1, such term won’t appear in G i.e. its coefficient will be 0.Step 2 Furthermore one can compute the coefficient of Y in terms of ais by using the fact that Gis a product of a Vandermonde determinant and a polynomial of degree q − 1.Step 3 We then equate this expression of the coefficient with 0 and obtain certain relations on theais. As this will work for any permutation (of the elements of A) that we begin with, we will useLemma 2.2.3 to conclude.

1This assumption is not restrictive as we can always translate a multiset by adding the opposite of an elementwhich is not there in the multiset. This won’t work if A = Fp as a set but in that case, we do have an ordering of theelements of Fp such that

∑i aibi = 0. It suffices to take ai = bi.

12

Let us first suppose that p ≡ 1 mod 3. We consider following monomial;

Y = (Y1Y2Y3)p−1(Y4Y5Y6)p−4...(Yp−3Yp−2Yp−1)3.

Total degree of this term is (p−1)(p+2)2 which is equal to the degree of G. By Remark 2.2.1, the

coefficient of this term in G is 0. Now we will express this coefficient in a different way.

Claim 2.2.1. The coefficient of Y upto multiplication by a non-zero scalar is

(a1 − a2)(a2 − a3)(a3 − a1).(a4 − a5)(a5 − a6)(a6 − a4)...

(ap−3 − ap−2)(ap−2 − ap−1)(ap−1 − ap−3).

Proof. The monomial Y can be written as a product of two monomials. One coming from theVandermonde determinant, D and another coming from the polynomial of degree p− 1. Note thatthe terms of D are of the form Y p−1π(1) Y

p−2π(2) ...Y

0π(p), where π is a permutation of the indices. In order

to obtain Y , one such monomial needs to be multiplied by a term of the form Y n11 Y n2

2 ...Ynpp which

come from the polynomial of degree p − 1. By looking at the degrees on the both sides, we obtainnp = 0 and for all 0 ≤ i ≤ p−4

3 , we have {n1+3i, n2+3i, n3+3i} = {0, 1, 2}. So the factor coming fromD is of the form

r−1∏i=o

Y3(r−i)−n1+3i

1+3i Y3(r−i)−n2+3i

2+3i Y3(r−i)−n3+3i

3+3i ,

where r = p−13 . Thus the terms from D that contribute to Y can be seen in the determinant;∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

a21Yp−31 a1Y

p−21 Y p−11

a22Yp−32 a2Y

p−22 Y p−12

a23Yp−33 a3Y

p−23 Y p−13

a54Yp−64 a44Y

p−54 a34Y

p−44

a55Yp−65 a45Y

p−55 a35Y

p−45

a56Yp−66 a46Y

p−56 a36Y

p−46

a87Yp−97 a77Y

p−87 a67Y

p−77

a88Yp−98 a78Y

p−88 a68Y

p−78

a89Yp−99 a79Y

p−89 a69Y

p−79

:ap−1p

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

.

After putting Yi = 1,∀ i, we obtain that the part of the coefficient coming from the VandermondeD upto multiplication by a non-zero scaler is

(a1 − a2)(a2 − a3)(a3 − a1).(a4 − a5)(a5 − a6)(a6 − a4)...

(ap−3 − ap−2)(ap−2 − ap−1)(ap−1 − ap−3).

Thus we conclude the proof of Claim 2.2.1.

After considering Remark 2.2.1, we note that this part coming from the Vandermonde D mustbe equal to 0. All that we have done so far remains valid even if we start with some permutation ofthe elements of A, thus for any permutation π of indices, we should have,

13

(aπ(1) − aπ(2))(aπ(2) − aπ(3))(aπ(3) − aπ(1)).(aπ(4) − aπ(5))(aπ(5) − aπ(6))(aπ(6) − aπ(4))..

(aπ(p−3) − aπ(p−2))(aπ(p−2) − aπ(p−1))(aπ(p−1) − aπ(p−3)) = 0.

Now suppose the maximal multiplicity in A be l ≤ p−13 . We apply Lemma 2.2.3(1) then we

obtain that first three elements are different, so are the second three etcetera. This contradicts theequation above. Thus l ≥ p+2

3 i.e. at least p+23 of the ais are same. This concludes the proof for the

case p ≡ 1 mod 3.

Let us now turn to the case where p ≡ 2 mod 3. As we did earlier, here also we will considerthe following monomial of maximal degree;

Y = (Y1Y2Y3)p−1(Y4Y5Y6)p−4...(Yp−4Yp−3Yp−2)4Yp−1Yp.

We notice that all the Yis have degree strictly less than p. Thus the coefficient of this term mustbe 0. Furthermore we claim that the coefficient of this monomial is a non-zero term (coming from(Y1 + ...+Yp)

p−1− 1) times a term coming from the Vandermonde D. This term coming from D is;

(a1 − a2)(a2 − a3)(a3 − a1).(a4 − a5)(a5 − a6)(a6 − a4)...

(ap−4 − ap−3)(ap−3 − ap−2)(ap−2 − ap−4)(ap−1 − ap).Here again we follow the same argument and by using the second part of Lemma 2.2.3, we obtain

the result. This concludes the proof of Lemma 2.2.4.

Finally we will need one last lemma to prove the main theorem.

Remark 2.2.2. In order to prove Theorem 2.1.1, we will only need the part where p = q in Lemma2.2.5. The part where q = ph where h > 1 is needed when we provide indications for the generalcase.

Lemma 2.2.5. Suppose a1, ..., ak are non-zero elements of Fq with k < 2q3 if q = p prime and

k ≤ q−32 if q = ph, h > 1, we suppose such that there are at least three different values of ais and

no ai has multiplicity more than q − k. Then either there are different elements b1, ..., bk such that∑i aibi = 0 or k = 3.

Proof. Let G be the usual polynomial.

• Case 1If 4 ≤ k ≤ q+3

2 then we consider following monomial of maximal degree:

Y = (Yq−52 +k

1 + Yq−52 +k

2 ) + (Y k−33 + Y k−34 )Y k−55 Y k−66 ...Y 0k .

Terms contributing to Y come from the Vandermonde and the polynomial of degree q − 1.

Those who come from the polynomial are of the form YiYq−12

j Yq−32

k where i = 3 or 4 and

{j, k} = {1, 2} and have coefficient (q − 1)( q−2

(q−1)2

)which is non-zero. As the coefficient of Y is

zero for it is a term of maximal degree, the part of the coefficient coming from the Vandermondemust be 0, i.e. (a1 − a2)(a3 − a4) = 0. Furthermore this is true for all the permutations ofindices. Thus by lemma 2.2.3, there is an ai with multiplicity at least k− 1 which contradictsthe hypothesis that there are at least three different values of ais.

14

• Case 2Now suppose k > q+3

2 .

As we suppose that k ≤ q−32 if q = ph, h > 1, the case k > q+3

2 will make sense only whenq = p. Here we will again make two cases.

– p ≡ 1 mod 3Consider the following monomial.

(Yk+ p−7

31 Y

k+ p−73

2 Yk+ p−7

33 )Y k−44 Y k−55 ...Y 0

k .

This a monomial of maximal degree thus of the coefficient 0. Its coefficient is a productof a non-zero term and (a1 − a2)(a2 − a3)(a3 − a1). Thus we must have,

(a1 − a2)(a2 − a3)(a3 − a1) = 0.

Once again, we must have a value with multiplicity at least k − 2 which produces acontradiction to our assumption that no ai has multiplicity more than p− k.

– p ≡ 2 mod 3Consider the following monomial.

(Yk+ p−8

31 Y

k+ p−83

2 Yk+ p−8

33 )(Y k−44 Y k−45 )Y k−66 Y k−77 ...Y 0

k .

Similar argument tells us that (a1−a2)(a2−a3)(a3−a1)(a4−a5) = 0, which again yieldsa contradiction.

2.2.3 Proof of Theorem 2.1.1 when p = q, an odd prime

Proof. Let a1, ..., ap be a given multiset in Fp. By Lemma 2.2.2, we know that if there are at mostthree different values of ais and as p here is odd, the main theorem is true. So we suppose thatthere are at least 4 different values and there is no ordering b1, ..., bp of the elements of Fp such that∑i aibi = 0. We will produce a contradiction.

Firstly, we will transform (Lemma 2.2.1) the ais and suppose that 0 is not among them. Thenwe will use Lemma 2.2.4 to obtain that p+2

3 of the ais are the same. Then again by translation, wemake this value equal to 0. Then apply lemma 2.2.5 to the remaining non-zero values. There areat most 2p−2

3 of them. One can verify that we can apply this lemma as 2p−23 < 2p

3 and as p > 10,no ai has multiplicity more than p − k. Thus by Lemma 2.2.5, either there are different elementsbis such that

∑i aibi = 0 or k = 3. But the sum cannot be 0 as we can assign arbitrary values

from Fp to extend the bis to an ordering of Fp. Thus we must have k = 3 i.e. there are p− 3 zeros,and remaining three non-zero values are distinct, say a, b and c. If a + b = 0, we can produce anordering and if a− b = 0, we have ba+ (−a)b+ 0c = 0 and thus again we can produce an ordering,which gives us the contradiction. Thus there are at most three different values of the ais and thenwe apply Lemma 2.2.2 to conclude.

2.2.4 General case

The proof presented above works even in case of Fq where q = ph, h > 1. We will only have toreplace Lemma 2.2.4 by Lemma 2.2.6. As the case of Fq where q = ph, h > 1 demands too manycalculations, we will just give some indications. A rigorous proof can be found in [4].

15

Lemma 2.2.6. Let q = ph > 9, h > 1 where p is an odd prime. Suppose that there does not existany ordering of the elements b1, ..., bq of Fq such that

∑i aibi = 0. Then at least q+3

2 of the ais areequal.

In order to prove this theorem we follow the same algorithm. Let us consider the followingmonomial in G.

Y = (

q∏i=1

Y i−1i ).(Y2p−1Y2p...Y2p−3)p(Y1Y3...Y2p−3)(Yp2+1Yp2+2...Yp2+p−1)p2

...

.(Yph−1+1Yph−1+2...Yph−1+p−1)ph−1

.

The degree of this monomial is(q2

)+ q− 1 which is maximal. One can also prove that all the Yis

in this monomial have degree at most q− 1. We note that if we exchange Y1 with Y3, the monomialremains unchanged. We first identify all such pairs. This gives us some information about thepermutation of the indices of the term coming from Vandermonde. For example, we will have onerelation {π(1), π(2)} = {1, 2}. Similarly the term coming from the polynomial part starts withYπ(1)Yπ(3)...Yπ(2p−3). Associated coefficient starts with (q − 1)(q − 2)...(q − p+ 1). Remaining partof the coefficient depends on the degrees of remaining Yis. If any of those degrees is not divisible byp, Lucas theorem tells us that the coefficient must be 0. In order to produce a non-zero coefficient2,the term Yπ(1)Yπ(3)...Yπ(2p−3) must go on with all the Yi having degree divisible by p. After havingdealt with the terms from the Vandermonde upto degree 2p− 3, we focus on the terms coming fromthe Vandermonde of degrees between 2p− 2 and 4p− 4. We already know that corresponding termscoming from the polynomial part must have degrees divisible by p, we get the relations of the form{π(2p− 1), π(3p− 1)} = {2p− 1, 3p− 1} etcetera.

Only terms that remain are the ones with unique degrees, (Y 4p−34p−2 Y

4p−24p−1 ...Y

p2−1p2 ),. Thus the

Vandermonde must have them.

We see that the terms coming from the polynomial start with p − 1 terms of degree 1 thenthere are p − 1 terms of degree p. Again in order to get a non-zero coefficient, remaining termsmust be divisible by p2. We go on by induction to obtain the form of the terms coming from theVandermonde:

q∏i=1

Y i−1π(i) ,

where π is the permutation which fixes almost all the is.

Finally we note that the the coefficient of Y is something non-zero times the coefficient from theVandermonde. This is a product of the 2× 2 determinants of the form:∣∣∣∣∣ aq−1−ki Y ki aq−1−k−p

m

i Y k+pm

i

aq−1−kj Y kj aq−1−k−pm

j Y k+pm

j

∣∣∣∣∣ .We then divide by (aiaj)

q−1−k−pm 6= 0, we get a similar relation between the coefficients as wegot in the first case. Also one can see that these relations between the ais hold for any permutationof ais that we begin with. Then we conclude by the usual argument.

2Because a non-zero coefficient will give us necessary relations between the ais.

16

2.2.5 Proof for q even

In this case as well, the proof is similar thus we won’t prove these results here. They follow the samealgorithm and can be found in the original article [4]. However we shall need to modufy certainlemmas when we consider the case q even.

Lemma 2.2.1 works even in this case. We will replace Lemma 2.2.2 by the following lemma.

Lemma 2.2.7. If the multiset in question has only 1 or 2 different values and q is even then themain theorem is true.

Furthermore Lemma 2.2.3 and Theorem 2.2.1 remains true even when q is even. We will justreplace Lemma 2.2.6 which assumes that q is odd by the following lemma.

Lemma 2.2.8. Let q = 2h > 8. Suppose that there is no ordering of the field elements b1, ..., bqsuch that

∑i aibi = 0. Then at least q+3

2 of the ais are equal.

Finally instead of Lemma 2.2.5, we use the following.

Lemma 2.2.9. Suppose that a1, ..., ak are non-zero elements of Fq, q even and 1 < k < q2 . Then

either there are different field elements b1, ..., bq such that∑i aibi = 0 or all the ais are the same.

In order to prove the main theorem when q is even, these lemmas are sufficient.

2.3 Consequences of Theorem 2.1.1

2.3.1 Polynomials of prescribed range

Lemma 2.3.1. Let f(x) = cq−1xq−1 + ...+ c0 be a polynomial over Fq. Then

1.∑x∈Fq f(x) = −cq−1

2.∑x∈Fq xf(x) = −cq−2

Proof. It suffices to apply the fact that for any finite field Fq,∑x∈Fq x

k = 0 when 1 ≤ k ≤ q − 2

and∑x∈Fq x

q−1 = −1.

Let M be a multiset of size q. We say that M is the range of a polynomial if M = {f(x)|x ∈ Fq}as a multiset. Furthermore let us recall that, due to Lagrange’s interpolation formula, one can writeany function on a finite field as a polynomial of the degree at most q−1. We will note it as ’reduced’degree of the associated function.

Suppose given a multiset M , we want to find out a smaller degree polynomial having M as itsrange. By Lemma 2.3.1, if the sum of elements of M is not zero then any reduced polynomial havingM as its range will be of reduced degree q − 1. And if the sum is 0 then the degree will be at mostq − 2.

Theorem 2.3.1. Let M = {a1, ..., aq} be a multiset in Fq with∑i ai = 0. There is no polynomial,

having M as its range, of reduced degree at most q − 3 if and only if q − 2 elements of M are equal(to say, a) and remaining two are of the form a+ b and a− b where a, b ∈ Fq and b 6= 0.

17

Proof. By Lemma 2.3.1, a polynomial with M as its range will have reduced degree q−1 if and onlyif∑i ai 6= 0. As here we have

∑i ai = 0, Lemma 3.1(2), shows that a polynomial with range M has

reduced degree q − 3 if and only if∑x∈Fq xf(x) = −cq−2 = 0.

We also see that there is a bijective correspondence between polynomials with range M andthe permutations of the elements b1, ..., bq of Fq. A permutation (b1, . . . , bq) gets mapped to thefunction f(bi) = ai. And under this correspondence, having

∑x∈Fq xf(x) = 0 is equivalent to

having∑aibi = 0. Thus Theorem 2.1.1 gives the result.

It is also natural to look for the polynomials having degree less than q−3 with prescribed range.It is thought that the only reason why a multiset M with sum of elements = 0 is not a range of apolynomial of reduced degree less than q − k is that there is a value in M of multiplicity at leastq − k. This results into the following conjecture.

Conjecture 2. Let M = {a1, ..., aq} be a multiset in Fq such that a1 + ...+ aq = 0, where q = ph,p prime. Let k <

√p. If ∀ai ∈ M , the multiplicity of ai < q − k then there is a polynomial with

range M of degree less than q − k.

We can explain why one needs an upper bound on k by using the main result of Biro in [6]. Wewill come back to this in the next chapter.

2.3.2 Hyperplanes of a vector space over FqTheorem 2.1.1 deals with the multisets of size q. We can extend it to smaller multisets. In order toit, let us take a multiset {a1, ..., an) where n ≤ q − 1. Let us also assume that there are no distinctfield elements {b1, ..., bn} such that

∑i aibi = 0. Now we add n − q 0s to have a new multiset

{a1, ..., an, an+1 = 0, ..., aq = 0}. One can notice that our assumption remains valid even for thisnew multiset. By Theorem 2.1.1, we know that after a suitable permutation, a1 = ... = aq−2 = afor some field element a, aq−1 = a+ b and aq = a− b where b is another non-zero field elements.If a = 0, then by removing extra zeros, we immediately obtain a1 = a2 = ... = an−2 = 0, an−1 = band an = −b where b 6= 0.If a 6= 0 then we could not have possibly added more than 2 extra zeros. Thus we must haven ≥ q − 2. Then either n = q − 2 or n = q − 1.If n = q − 2 then we have a1 = ... = aq−2 = a, a+ b = a− b = 0. Thus 2a = 2b = 0. As both a andb are non-zero, we must have a field of characteristic two i.e. q even.If n = q − 1 then we have a1 = ... = aq−2 = a, aq−1 = a+ b and aq = a− b = 0 which implies a = bi.e. aq−1 = 2a.

We can also see that if we are in one of these conditions, we cannot produce distinct bis suchthat

∑i aibi = 0. For example, if we have n = q − 2, where q is even and if for a multiset (of size

q − 2) {a, ..., a} where a 6= 0 there are distinct field elements b1, ..., bq−2 such that∑i abi = 0 then

as a 6= 0 we immediately have∑q−2i=1 bi = 0. As the sum of field elements in Fq where q > 2 is 0, we

must have bq−1 = −bq = bq, a contradiction. Other cases can be treated similarly.

We summarize this discussion in the following corollary.

Corollary 2.3.1.1. Suppose {a1, ..., an} is a multiset in Fq with n ≤ q − 1. Then there are nodistinct field elements b1, ..., bn such that

∑i aibi = 0 if and only if we have one of the following

conditions.

18

1. n ≤ q − 1 and after a suitable permutation of the indices a1 = a2 = ... = an−2 = 0, an−1 = band an = −b for some field element b 6= 0.

2. n = q − 1 and after suitable a permutation of the indices a1 = a2 = ... = aq−2 = a, aq−1 = 2afor some field element a 6= 0.

3. n = q − 2, q even, and after a suitable permutation of the indices a1 = a2 = ... = aq−2 = a,for some field element a 6= 0.

Proof. It follows from the discussion above.

Remark 2.3.1. In [4], this corollary is incomplete. In the first case, authors have supposed that nis strictly less than q − 1 whereas the result works even when n = q − 1.

Let V be a vector space of dimension n over Fq. We will denote by Hij a vector subspace ofV defined by the equation xi = xj where i 6= j. We are interested in hyperplanes fully containedin ∪i 6=jHij . If n > q, the whole space will be contained in this union. So the question becomesnon-trivial when n ≤ q.

Theorem 2.3.2. Suppose n ≤ q and H ⊂ ∪i6=jHij is a hyperplane in V , H 6= Hij for any i 6= j.Then one of the following conditions.

1. n = q, H = {(x1, ..., xn)|c(xj − xk) +∑i xi = 0} for some non-zero c in Fq and some indices

j 6= k.

2. n = q − 1, H = {(x1, ..., xn)|xj +∑i xi = 0} for an index j.

3. n = q − 2, q even. H = {(x1, ..., xn)|∑i xi = 0}.

Remark 2.3.2. In the original article ([4]), Theorem 2.3.2 is incomplete. Authors don’t mentionthe third case at all.

Proof. Let H =< (a1, ..., an) >⊥. The demand that H be contained in ∪i 6=jHij is equivalent toexpect that whenever (x1, ..., xn) is in H i.e. whenever a1x1 + ...+ anxn = 0, two of the xis shouldbe same i.e. there should exist i 6= j such that xi = xj . This is equivalent to say that there are nodistinct x1, ..., xn such that a1x1 + ...+ anxn = 0. Now we can apply Theorem 2.1.1 and Corollary2.3.1.1. Then we must have the following possibilities.

• n = q and after a permutation of the indices a1 = a2 = ... = aq−2 = a where a is a fieldelement, aq−1 = a + b and aq = a − b for some field element b 6= 0. We need to show thatthere exists non-zero a c ∈ Fq, such that for any arbitrary vector (x1, ..., xq) in H, we havec(xj − xk) +

∑i xi = 0 for some indices j 6= k. As

∑i axi + bxq−1 − bxq = 0, we have∑

i xi + ba (xq−1 − xq) = 0. We let c = b

a which is clearly non-zero. Note that a priori a canbe 0 but here if a = 0, H = Hij where i = q− 1 and j = q which is not allowed as we supposeH 6= Hij for any i 6= j. Thus we can divide by a.

• n ≤ q − 1 and after a permutation of the indices a1 = a2 = ... = an−2 = 0, an−1 = b andan = −b for some field element b 6= 0. Let (x1, ..., xn) be a vector in H. We then obtainbxn−1− bxn = 0. As b 6= 0, we can divide by it to obtain xn−1 = xn thus H is of the form Hij

where i 6= j, which is not allowed and thus this case does not occur.

• n = q− 1 and after a permutation of the indices a1 = a2 = ... = aq−2 = a, aq−1 = 2a for somefield element a 6= 0. Let (x1, ..., xq) be a vector in H. We have a

∑i xi + axq−1 = 0. As a 6= 0,

we divide this equation by a to obtain the condition that∑i xi + xq−1 = 0.

19

• n = q − 2, q even, and after a permutation of the indices a1 = a2 = ... = aq−2 = a, for somefield element a 6= 0. Let (x1, ..., xn) be a vector in H. We have a

∑i xi = 0. As a 6= 0, we can

divide by a to obtain the condition that∑i xi = 0.

Another example of hyperplanes contained in ∪i6=jHij comes from affine hyperplanes.

Theorem 2.3.3. All affine hyperplanes contained in ∪i 6=jHij are linear. Only exception is whenn = q and in this case the hyperplane is a translate of (1, 1, ..., 1)⊥.

Proof. Let an affine hyperplane {(x1, ..., xn)|a1x1 + ... + anxn = c} be contained in ∪i 6=jHij . Firstwe will choose random distinct field elements x1, ..., xn. As they are distinct, we have d := a1x1 +... + anxn 6= c. If d 6= 0, we obtain that a scaler multiple of (x1, ..., xn); ( cdx1, ...,

cdxn) is in our

hyperplane which is possible only when c = 0. Thus c = 0.

Now if d = 0, we have a1x1 + ...+ anxn = 0. We interchange xi and xj suitably in order to geta1x1 + ... + anxn = (ai − aj)(xj − xi) 6= 0. This will always be possible unless all the ais are thesame. For the time being we suppose that ais are not same. Then after the interchange, we geta1x1 + ...+ aixj + ...+ ajxi + ...+ anxn 6= 0. We then use the previous argument.

Now suppose all the ais are equal to say 1. When n < q, we can find distinct x1, ..., xn such thatx1 + ... + xn 6= 0 and then we use the previous argument. If n = q, we obtain the exceptional casestated in the theorem.

20

Chapter 3

Biro’s result

3.1 Notations

In this chapter p will denote an odd prime. q = ph where h is a natural number > 1. Fp willdenote the field of p elements and F∗p = Fp \ {0}. Similarly Fq will denote the field of q elements andF∗q = Fq \ {0}. For a set X, |X| will denote its cardinality i.e. the number of distinct elements in X.Also we will use Lagrange interpolation technique at many occasions. This can be found in AppendixC.

3.2 Introduction and the main theorem

In the previous chapter we stated Conjecture 2 concerning the multisets in Fq. In this chapter, wewill mostly deal with the case q = p for Conjecture 2, so we will restate it here for q = p.

Conjecture 3. Let M = {a1, ..., ap} be a multiset in Fp such that a1 + ... + ap = 0. Let k <√p.

If there is no polynomial with range M of degree less than p − k, then M contains an element ofmultiplicity at least p− k.

Equivalently, the above conjecture says that if ∀ ai ∈ M , the multiplicity of ai < p − k thenthere exists a polynomial of degree strictly less than p− k having M as its range.

If a value a in a multiset M occurs m times then any polynomial of range M must have degree atleast m because the associated reduced polynomial f is such that f − a has m roots. So Conjecture3 says that the only reason why a multiset cannot be the range of a polynomial of degree strictlyless than p− k is that there is at least one value in the multiset with multiplicity ≥ p− k.

We notice the condition k <√p in Conjecture 3. In order to explain it, we studied a paper by

Biro ([6]). This chapter is based on that paper.

We will perform some calculations in Section 3.3 which are not present in the original paper ([6]).

The main result of this chapter is the following theorem which says that if a polynomial f ∈ Fp[X]takes exactly two values on F∗p then normally deg(f) ≥ 3

4 (p−1). This problem was raised by AndrasGacs in [5].

21

Theorem 3.2.1. Let f ∈ Fp[X], deg(f) < p− 1. Let us also assume that |f(F∗p)| = 2. Then one ofthe following three assertions is true.

1. f(X) = a+ bXp−12 , for some a and non-zero b in Fp.

2. p ≡ 1 mod 3 and f is a polynomial of the form aX2 p−13 + bX

p−13 + c where a and b are both

non-zero and a3 = b3.

3. deg(f) ≥ 34 (p− 1).

Remark 3.2.1. If a polynomial anXn + an−1X

n−1 + ... + a0, an 6= 0 takes exactly two values onF∗p then Xn + an−1X

n−1 + ... + a1X also takes exactly two values on F∗p. Furthermore if Xn +an−1X

n−1 + ...+a1X takes exactly two values on F∗p then so does anXn+an−1X

n−1 + ...+a1X+a0for all an ∈ F∗p and a0 ∈ Fp.

We will need the above remark in Section 3.3 where we compute certain two-valued polynomialson F∗5 and F∗7.

Lemma 3.2.1. If f is a two-valued polynomial of degree at most p− 2 on F∗p then f is three-valuedon Fp and at least f(0) occurs exactly once.

Proof. Let f be a two-valued polynomial on F∗p. Let f(F∗p) = {a, b} as a set for some a 6= b in Fp.Suppose in f(F∗p), the multiplicity of a is α and the multiplicity of b is β where α > 0 and β > 0such that α+ β = p− 1 in Z so in Fp, α+ β = −1.We need to prove that f(0) 6= a and f(0) 6= b.

We will prove it by contradiction. Suppose f(0) = a. By using Lemma 2.3.1, we obtain,

(α+ 1)a+ βb = 0 in Fp.

By replacing α+ 1 by −β, we obtain a = b which produces a contradiction to our assumption thatf is two-valued on F∗p.

Similarly we get a contradiction if f(0) = b. Thus f(0) 6= a and f(0) 6= b. This concludes theproof of Lemma 3.2.1.

Theorem 3.2.1 helps us to understand why we need an upper bound on k in Conjecture 3. Forexample, if we consider a multiset M of the form {a, . . . , a, b, . . . , b, c} where a occurs α times, boccurs β times and c occurs once and furthermore we have aα + bβ + c = 0. So by Lemma 2.3.1,every polynomial f of range M is such that deg(f) ≤ p− 2 and by Lemma 3.2.1, f(0) = c. Thus byTheorem 3.2.1, we have, except some particular cases, deg(f) ≥ 3

4 (p−1). This means that normallythere does not exist a polynomial of range M having degree strictly less than p − k once we havek ≥ p

4 . So the conjecture is not valid when k ≥ p4 . So we must have k < p

4 . We note however thatGacs et al don’t provide exact details explaining why we need to suppose that k <

√p in their paper

[4].

A simple example of a polynomial taking exactly two values on F∗p is the Legendre symbol,

f(X) = Xp−12 which is either 1 or −1 on F∗p. Furthermore if d is a divisor p−1, then the polynomial

g(X) =

d−1∑j=1

Xj p−1d

22

also takes exactly two values on F∗p. Indeed as g is a finite geometric series, we can write,

g(x) =

{d− 1 if x

p−1d = 1

−1 otherwise.

For f , a polynomial which takes exactly two values on F∗p,we will be interested in the quotient

δ = deg(f)p−1 which enables us to consider different coefficient fields Fp. For example, one can see

that the Legendre symbol defined above gives the smallest possible value of δ which is 1/2. Indeedbecause if a polynomial f takes exactly two values, say {a, b} on F∗p then the polynomial (f−a)(f−b)is identically zero on F∗p. Thus Xp−1 − 1 must divide (f − a)(f − b). By comparing their degrees,

we see that the smallest possible degree of f is p−12 and thus the smallest possible value of δ is 1

2and it is given by the Legendre symbol.

For the polynomial g(X) =∑d−1j=1 X

j p−1d , clearly the value of δ is d−1

d . So 12 , 23 , 34 , 45 ,... are some

possible values of δ. The main theorem of this chapter says that the smallest three values of δ are12 , 23 , 34 .

Remark 3.2.2. One might be tempted to imagine that the next possible value of δ is 45 but it is not

the case as we will see in an example, that we have taken from [6], at the end of this chapter.

3.3 Some examples

Using Python (code in Appendix B), we could determine explicitly the number of monic polynomialsof degree 3 and 5 with no constant term taking exactly two values on F∗5 and on F∗7. For thesepolynomials, δ = 3

4 on F∗5 and δ = 56 on F∗7.

In order to simplify the calculations, we will use Remark 3.2.1 and denote Xn + an−1Xn−1 +

...+ a1X by (1, an−1, ..., a1) in the following tables.

The following table lists down all the polynomials of the form X3 + aX2 + bX in F5[X] takingexactly two values on F∗5. And δ associated with these polynomials is 3

4 .

f f(1) f(2) f(3) f(4)

(1, 0, 2) 3 2 3 2(1, 0, 3) 4 4 1 1(1, 1, 1) 3 4 4 4(1, 2, 4) 2 4 2 2(1, 3, 4) 3 3 1 3(1, 4, 1) 1 1 1 2

Remark 3.3.1. 1. There are 6 monic polynomials of degree 3 with no constant term in F5[X]which take exactly two values on F∗5.

2. If both a and b occur with the multiplicity 5−12 = 2 in f(F∗5) where f ∈ F5[X] is a monic

polynomial with no constant term and which is two-valued on F∗5, then a+ b ≡ 0 mod 5.

3. If we consider any polynomial of degree at most 3 in F5[X] taking exactly two values, say{a, b} ⊂ F5 such that a 6= b and a and b both occur with the same multiplicity (which has to be2) on F∗5 then f(0) = a+b

2 .

23

We note that the third point in Remark 3.3.1 is true for every p. So the following lemma,

Lemma 3.3.1. If both a and b occur with the multiplicity p−12 in f(F∗p) where f is any two-valued

polynomial on F∗p of degree ≤ p− 2 then f(0) = a+b2 .

Proof. Let the multiplicity of a and that of b be equal to p−12 in f(F∗p), where f is any two-valued

polynomial on F∗p of degree ≤ p− 2. As deg(f) ≤ p− 2, we can apply Lemma 2.3.1 to obtain

f(0) +p− 1

2(a+ b) = 0.

So in Fp, f(0) = a+b2 .

The table below lists down all the polynomials of the form X5 +aX4 + bX3 + cX2 +dX in F7[X]taking exactly two values on F∗7. δ associated with these polynomials is 5

6 .

f f(1) f(2) f(3) f(4) f(5) f(6)

(1, 0, 3, 0, 1) 5 2 5 2 5 2(1, 0, 5, 0, 2) 1 6 6 1 1 6(1, 0, 6, 0, 4) 4 4 4 3 3 3(1, 1, 0, 4, 5) 4 4 4 6 4 6(1, 1, 1, 1, 1) 5 6 6 6 6 6(1, 1, 2, 5, 4) 6 1 1 6 1 6(1, 1, 5, 1, 1) 2 3 2 3 2 2(1, 2, 0, 4, 3) 3 2 2 2 3 2(1, 2, 1, 5, 1) 3 3 4 4 3 4(1, 2, 4, 1, 2) 3 6 3 3 3 3(1, 2, 6, 1, 2) 5 1 1 5 1 1(1, 3, 0, 3, 6) 6 6 6 2 2 6(1, 3, 2, 6, 4) 2 2 4 2 2 2(1, 3, 3, 6, 4) 3 3 3 3 1 1(1, 3, 4, 2, 2) 5 5 2 2 2 5(1, 4, 0, 4, 6) 1 5 5 1 1 1(1, 4, 2, 1, 4) 5 5 5 3 5 5(1, 4, 3, 1, 4) 6 6 4 4 4 4(1, 4, 4, 5, 2) 2 5 5 5 2 2(1, 5, 0, 3, 3) 5 4 5 5 5 4(1, 5, 1, 2, 1) 3 4 3 3 4 4(1, 5, 4, 6, 2) 4 4 4 4 1 4(1, 5, 6, 6, 2) 6 6 2 6 6 2(1, 6, 0, 3, 5) 1 3 1 3 3 3(1, 6, 1, 6, 1) 1 1 1 1 1 2(1, 6, 2, 2, 4) 1 6 1 6 6 1(1, 6, 5, 6, 1) 5 5 4 5 4 5

As we can see, there are 27 monic polynomials of degree 5 with no constant term in F7[X] whichtake exactly two values on F∗7. Furthermore if f ∈ F7[X] is a polynomial of degree 5 such thatf(0) = 0 and f(F∗7) = {a, a, a, a, a, b} as a multiset then b is equal to 2a.

24

This is true for any p and any polynomial f of degree p− 2 taking exactly two values on F∗p oneof which is of multiplicity one, such that f(0) = 0. To see that, let f(F∗p) = {a, ..., a, b} as a multisetwhere a occurs p − 2 times and b occurs once and f(0) = 0. Then as deg(f) = p − 2 we can applyLemma 2.3.1 to obtain (p− 2)a + b = 0 which implies b = 2a. In this case f(0) = 0, so by Lemma3.2.1, neither a nor b is 0. Furthermore by Theorem 2.3.1, a polynomial taking exactly two valueson F∗p one of which is of multiplicity one is essentially of degree p−2. It is a polynomial of Lagrangeinterpolation on p points of Fp. Its degree is p− 2 by Lemma 2.3.1.

So for k = 2, Conjecture 3 is true by Theorem 2.3.1 of Gacs et al. As k is required to be lessthan

√p and also less than p

4 , we need p ≥ 11 when k = 2.

We end this section by the following result which gives the exact number of monic, two-valued(on F∗p) polynomials of degree ≤ p− 2 with no constant term in Fp[X]. Let us define,

Sp = {f ∈ Fp[X] |deg(f) ≤ p− 2, f(0) = 0 and |f(F∗p)| = 2}

andS(m)p = Sp ∩ {f ∈ Fp[X] |f is monic}.

Furthermore we havef ∈ S(m)

p ⇒ ∀a ∈ F∗p, af ∈ Spand

f ∈ Sp ⇒ ∃!a ∈ F∗p, af ∈ S(m)p .

Note that a in the second equation above is the multiplicative inverse of the leading coefficient off ∈ Sp. So we have a function,

ϕ : Sp → S(m)p

f 7→ c(f)−1f

and as for all f ∈ S(m)p , there are p − 1 preimages of f by ϕ, we have the following result on

relating the cardinalities of Sp and S(m)p .

|Sp| = (p− 1)|S(m)p |, (3.1)

Claim 3.3.1.|S(m)p | = 2p−2 − 1.

Proof. For f ∈ Sp, we have M := f(F∗p) = {a, ..., a, b, ..., b} as a multiset where a occurs α times andb occurs β times in M and α > 0 and β > 0 such that;

α+ β = p− 1 in Z. (3.2)

Also as deg(f) ≤ p− 2 and f(0) = 0, by Lemme 2.3.1, we have,

aα+ bβ = 0 in Fp (3.3)

Furthermore, as α < p − 1, α + 1 is never 0 and both a and b are non-zero by Lemma 3.2.1 andf(0) = 0.

b =aα

1 + αin Fp and β = p− 1− α in Z. (3.4)

25

Note that a = b does not occur in 3.4 as it would imply 1 = 0 which is not possible.

Furthermore by 3.4, given (a, α) ∈ F∗p × {1, ..., p − 2}, ∃!(b, β) ∈ F∗p × {1, ..., p − 2} where b 6= asuch that the multiset {a, ..., a, b, ..., b} (where a occurs α times and b occurs β times) is equal tof(F∗p) for some polynomial f in Sp. To obtain this polynomial, we fix an order on {a, ..., a, b, ..., b}.We can then use Lagrange interpolation on p points of Fp to obtain a unique polynomial l such thatl(F∗p) = {a, ..., a, b, ..., b} and l(0) = 0. As aα + bβ = 0, the degree of l is at most p − 2 by Lemma2.3.1. And we have l(F∗p) = {a, ..., a, b, ..., b}. So l is in Sp.

So in order to count the polynomials in Sp, it suffices to count the number of possible (p−1)-tuplesof the above form.

Now suppose that the minimal multiplicity in a given (p − 1)-tuple is 1. Thus the (p − 1)-tuple is of the form (b, ..., b, a) upto a permutation. As we know fixing a and its multiplicity α willautomatically fix b and its multiplicity β, we just need to calculate in how many ways we can choosea from F∗p and in how many ways we can place a in our (p− 1)-tuple. Clearly there are (p− 1)

(p−11

)choices to do so.

We can continue in this way upto when the minimal multiplicity is p−32 because till this point

α 6= β which ensures that no two (p− 1)-tuples are repeated. And thus so far we have

(p− 1)

p−32∑

α=1

(p− 1

α

),

(p− 1)-tuples.

When α = β then necessarily α = β = p−12 in Z and by Lemma 3.3.1, in this case b = −a. As

a and −a give the same (p − 1)-tuple twice, the number possible (p − 1)-tuples in this case getsreduced by the factor of 1

2 , which is

(p− 1)

{1

2

(p− 1p−12

)}.

26

So to obtain |Sp|, we sum over minimal multiplicities.

|Sp| = (p− 1)

p−32∑

α=1

(p− 1

α

)+

1

2

(p− 1p−12

)=p− 1

2

2

p−32∑

α=1

(p− 1

α

)+

(p− 1p−12

)=p− 1

2

p−32∑

α=1

(p− 1

α

)+

p−32∑

α=1

(p− 1

p− 1− α

)+

(p− 1p−12

)=p− 1

2

p−32∑

α=1

(p− 1

α

)+

p−2∑α= p+1

2

(p− 1

α

)+

(p− 1p−12

)=p− 1

2

{p−2∑α=1

(p− 1

α

)}

=p− 1

2(2p−1 − 2) as

p−1∑α=0

(p− 1

α

)= 2p−1

= (p− 1)(2p−2 − 1)

So by Equation 3.1,|S(m)p | = 2p−2 − 1.

Remark 3.3.2. In S(m)p , one polynomial is always there namely the Legendre symbol, X

p−12 . If

p ≡ 1 mod 3, we get 3 more polynomials of the form X2 p−13 + bX

p−13 where b ∈ {1, ω, ω2}, ω is a

primitive cube root of unity in Fp.

Using Claim 3.3.1 and the remark above, we obtain |S(m)5 | = 7 and |S(m)

7 | = 31 which is expected.

3.4 An example

Now we will give a numerical example of a polynomial taking exactly two values on F∗29 such thatthe associated δ is strictly contained between 3/4 and 4/5.

Let p = 29. Assume that B ⊂ F∗29 such that −B = B, 1 ≤ |B| ≤ 14 and∑x∈B2

x =∑x∈B2

x2 = 0, (3.5)

where B2 is the set (without repetition of elements) of the squares of elements of B.1 So if B ={±a1, ...,±an} where 0 ≤ n ≤ 7. As −B = B,

∑x∈B x

2n+1 = 0 for every n ∈ N. Also the secondand the fourth power sum also vanish because of the condition 3.5.

1In the end, we will give an example of such B. This example is taken from [6]

27

Let f ∈ F29[X] be the unique polynomial with deg(f) ≤ 27 such that

f(x) =

{1 if x ∈ B0 if x ∈ F∗29 \B.

Such f can be obtained by Lagrange interpolation on F∗29. As the first fifth power sums vanish,we obtain deg(f) ≤ 22.

If deg(f) < 22 then it must be ≤ 20 as deg(f) is even because −B = B.

Also 20 < (3/4)(28) = 21. So by Theorem 3.2.1, there is just one possibility for f i.e. f(X) =a + bX14 where b 6= 0. And also |B| = 14 as f is zero on F∗29 \ B so it has at least 28 − |B| rootsand we also have 1 ≤ |B| ≤ 14.Now if deg(f) = 22, then clearly we obtain

δ =deg(f)

p− 1=

11

14,

which is strictly contained between 3/4 and 4/5.

The explicit set can be given as

B = {±1,±3,±4,±6,±7,±11}.

So |B| = 12 and B2 = {1, 5, 7, 9, 16, 20}. One see that Equation 3.5 is satisfied. We note thatG = {±1} and in this case we obtain β = 3/7, γ = 1/14.

28

Appendices

29

Appendix A

Combinatorial Nullstellensatz

In Chapters 1 and 2, we used two different versions of a result known as combinatorial nullstellensatzin Theorems 1.3.3 and 2.1.3 respectively. In this appendix, we will give the actual result from whichthose two different versions follow. The proof of this theorem can be found in [7].

Theorem A.0.1. (Alon) Let F be a finite field and let f = f(X1, ..., Xn) be a polynomial inF[X1, ..., Xn]. Let S1, ..., Sn be non-empty subsets of F and define gi(Xi) =

∏s∈Si(Xi − s). If f

vanishes over all the common zeros of g1, ..., gn (that is; if f(s1, ..., sn) = 0 for all si ∈ Si), thenthere are polynomials h1, ..., hn in F[X1, ..., Xn] satisfying deg(hi) ≤ deg(f) − deg(gi) for all is sothat

f =

n∑i=1

higi.

Let us state Theorems 1.3.3 once again.

Theorem A.0.2. Let F be any field and let f be a polynomial in F[X1, ..., Xn]. Suppose deg(f) =∑ni=1 ti, where tis are non-negative integers. Furthermore we will suppose that the coefficient of∏ni=1X

tii in f is non-zero. Then, if S1, ..., Sn are subsets of F such that |Si| > ti, there are s1 ∈

S1, ..., sn ∈ Sn such that f(s1, ..., sn) 6= 0.

Proof. Let us assume that |Si| = ti + 1 for all i. Suppose the above result is false. So f vanishesat over all the common zeros of g1, ..., gn. Let us define gi(Xi) =

∏s∈Si(Xi − s). So by Theorem

A.0.1, there are polynomials h1, ..., hn in F[X1, ..., Xn] satisfying deg(hi) ≤∑ni=1 ti − deg(gi) for all

is so that

f =

n∑i=1

higi.

By our assumption, the coefficient of∏ni=1X

tii in the left hand side is non-zero so it must be non-zero

even on the right hand side. But the degree of higi = hi∏si∈Si(Xi − s) is at most deg(f). And if

there are monomials of degree deg(f) in it, they must be divisible by Xti+1i . It then follows that

the coefficient of∏ni=1X

tii in the right hand side is zero which is a contradiction.

Theorem A.0.3. Let G(X1, ..., Xn) be a polynomial over Fq which vanishes everywhere. Then onecan write G in the following form:

G(X1, ..., Xk) = (Xq1 −X1)f1 + ...+ (Xq

k −Xk)fk,

where fis are polynomials of degree at most deg(G)− q.

30

Proof. It suffices to apply directly Theorem A.0.1 for f = G, S1 = ... = Sn = Fq and gi(Xi) =∏si∈Fq (Xi − s) = Xq

i −Xi because Xqi −Xi vanishes everywhere on Fq and thus can be factorized

in the required form.

31

Appendix B

Python code used in Chapter 3

In this appendix we shall give a simple Python code which counts the total number of monicpolynomials of degree p− 2 with no constant term taking exactly two values on F∗p. It also gives usthe range of these polynomials.

For p = 5:

Python 2 . 7 . 1 0 ( de fau l t , May 23 2015 , 0 9 : 4 0 : 3 2 ) [MSC v .1500 32 b i t ( I n t e l) ] on win32

Type ” copyr ight ” , ” c r e d i t s ” or ” l i c e n s e ( ) ” f o r more in fo rmat ion .>>> from sympy import ∗>>> a , b , c , d , x , y , z = symbols ( ’ a b c d x y z ’ )>>> de f g ( a , b , x ) :

r e s u l t = ( x∗∗3+a ∗( x∗∗2)+b∗x ) % 5return r e s u l t

>>> de f T1( a , b) :r e s u l t = [ g ( a , b , 1 ) , g ( a , b , 2 ) , g ( a , b , 3 ) , g ( a , b , 4 ) ]r e turn r e s u l t

>>> de f crunch ( input1 ) :output1 = [ ]f o r z in input1 :

i f z not in output1 :output1 . append ( z )

re turn output1

>>> de f rangetwo ( input2 ) :output2 = [ ]f o r z in input2 :

i f l en ( z ) == 2 :output2 . append ( z )

re turn output2

>>> f i r s t l i s t = [ ]

32

>>> f o r a in range (0 , 5 ) :f o r b in range (0 , 5 ) :

f i r s t l i s t . append (T1( a , b) )

>>> f i n a l l i s t = [ ]>>> f o r i in range (0 , l en ( f i r s t l i s t ) ) :

f i n a l l i s t . append ( crunch ( f i r s t l i s t [ i ] ) )

>>> rangetwo ( f i n a l l i s t )[ [ 3 , 2 ] , [ 4 , 1 ] , [ 3 , 4 ] , [ 2 , 4 ] , [ 3 , 1 ] , [ 1 , 2 ] ]>>> l en ( f i n a l l i s t )25>>> l en ( rangetwo ( f i n a l l i s t ) )6>>>

For p = 7:

Python 2 . 7 . 1 0 ( de fau l t , May 23 2015 , 0 9 : 4 0 : 3 2 ) [MSC v .1500 32 b i t ( I n t e l) ] on win32

Type ” copyr ight ” , ” c r e d i t s ” or ” l i c e n s e ( ) ” f o r more in fo rmat ion .>>> from sympy import ∗>>> a , b , c , d , x , y , z = symbols ( ’ a b c d x y z ’ )>>> de f g ( a , b , c , d , x ) :

r e s u l t = ( x∗∗5+a ∗( x∗∗4)+b∗( x∗∗3)+c ∗( x∗∗2)+d∗x ) % 7return r e s u l t

>>> de f T1( a , b , c , d ) :r e s u l t = [ g ( a , b , c , d , 1 ) , g ( a , b , c , d , 2 ) , g ( a , b , c , d , 3 ) , g ( a , b , c , d , 4 ) , g (

a , b , c , d , 5 ) , g ( a , b , c , d , 6 ) ]r e turn r e s u l t

>>> de f crunch ( input1 ) :output1 = [ ]f o r z in input1 :

i f z not in output1 :output1 . append ( z )

re turn output1

>>> de f rangetwo ( input2 ) :output2 = [ ]f o r z in input2 :

i f l en ( z ) == 2 :output2 . append ( z )

re turn output2

33

>>> f i r s t l i s t = [ ]>>> f o r a in range (0 , 7 ) :

f o r b in range (0 , 7 ) :f o r c in range (0 , 7 ) :

f o r d in range (0 , 7 ) :f i r s t l i s t . append (T1( a , b , c , d ) )

>>> f i n a l l i s t = [ ]>>> f o r i in range (0 , l en ( f i r s t l i s t ) ) :

f i n a l l i s t . append ( crunch ( f i r s t l i s t [ i ] ) )

>>> rangetwo ( f i n a l l i s t )[ [ 5 , 2 ] , [ 1 , 6 ] , [ 4 , 3 ] , [ 4 , 6 ] , [ 5 , 6 ] , [ 6 , 1 ] , [ 2 , 3 ] , [ 3 , 2 ] , [ 3 , 4 ] ,

[ 3 , 6 ] , [ 5 , 1 ] , [ 6 , 2 ] , [ 2 , 4 ] , [ 3 , 1 ] , [ 5 , 2 ] , [ 1 , 5 ] , [ 5 , 3 ] , [ 6 ,4 ] , [ 2 , 5 ] , [ 5 , 4 ] , [ 3 , 4 ] , [ 4 , 1 ] , [ 6 , 2 ] , [ 1 , 3 ] , [ 1 , 2 ] , [ 1 , 6 ] ,[ 5 , 4 ] ]

>>> l en ( f i n a l l i s t )2401>>> l en ( rangetwo ( f i n a l l i s t ) )27>>>

For p > 7, an ordinary computer could not do the calculations.

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Appendix C

Lagrange interpolation

In Chapter 3, we often used the technique of Lagrange interpolation in order to obtain a polynomialpassing through certain given points. In this appendix, we will state the theorem of Lagrangeinterpolation.

Theorem C.0.4. Let (x0, ..., xn) and (a1, ..., an) be two ordered (n+ 1)-tuples of real numbers suchthat xi 6= xj when i 6= j. Then there exists a unique polynomial P ∈ R[X] of degree at most n suchthat

P (xi) = ai, ∀ 0 ≤ i ≤ n.

This polynomial P can be given by the following formula,

P (X) =

n∑j=0

aj

n∏i=0,i6=j

X − xixj − xi

.

Proof. Let Lj [X] =∏ni=0,i6=j

X−xixj−xi . Clearly

Lj(xi) =

{1 if i = j0 if i 6= j.

We then have P (xi) = ai, ∀ 0 ≤ i ≤ n. Also degree of P defined above is at most n. We willnow prove the uniqueness. Let Q be another polynomial having desired properties. Let

R = P −Q.

Clearly degree of R is at most n as R is a difference between two polynomials of degree at most n.

Furthermore R(xi) = 0, ∀ 0 ≤ i ≤ n. As xis are distinct, R has n+ 1 distinct roots. So we canwrite,

R = α

n∏i=0

(X − xi).

If α is not zero then we obtain a contradiction to the fact that the degree of R is at most n. Soα = 0 which proves the uniqueness.

35

Acknowledgement

I would like to sincerely thank my professor, Mr. Stephane Vinatier for all his help and patienceduring this internship. I would like to note that this internship would not have been possible withouthis constant support and encouragement.

I would also like to thank Mr. Gaetan Chenevier and Mrs. Anna Cadoret for finding thisinternship for me.

36

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