A Polynomial Method Approach to Zero-Sum Subsets in $\mathbb{F}_{p}^{2}$

TL;DR

This paper proves that subsets of _p^2 meeting all lines through the origin contain zero-sum subsets, using a polynomial method inspired by the Combinatorial Nullstellensatz, advancing understanding of Olson constants.

Contribution

The paper introduces a polynomial method approach to establish the existence of zero-sum subsets in _p^2 under specific geometric conditions, extending Olson constant results.

Findings

Every subset meeting all lines through the origin has a zero-sum subset.

The proof employs a polynomial method inspired by the Combinatorial Nullstellensatz.

Extends Olson constant results for _p^2.

Abstract

In this paper we prove that every subset of $\mathbb{F}_p^2$ meeting all $p+1$ lines passing through the origin has a zero-sum subset. This is motivated by a result of Gao, Ruzsa and Thangadurai which states that $OL(\mathbb{F}_{p}^{2})=p+OL(\mathbb{F}_{p})-1$, for sufficiently large primes $p$. Here $OL(G)$ denotes the so-called Olson constant of the additive group $G$ and represents the smallest integer such that no subset of cardinality $OL(G)$ is zero-sum-free. Our proof is in the spirit of the Combinatorial Nullstellensatz.

Equations

$$\Sigma (A)=\left\{\sum _{x\in B}x|B\subset A\right\},{\Sigma }^{\ast }(A)=\left\{\sum _{x\in B}x|B\subset A,B\ne \varnothing \right\}.$$

$$OL(G)\le \sqrt{2|G|}+O(|G{|}^{1/3}\log |G|),$$

$$|\Sigma (A)|\ge \min \left\{p,d(d+1)/2+1\right\}\text{and}|{\Sigma }^{\ast }(A)|\ge \min \left\{p,d(d+1)/2\right\}.$$

$$OL({\mathbb{F}}_p^2)=p+OL({\mathbb{F}}_p)-1$$

$$OL({\mathbb{F}}_p^d)=p+OL({\mathbb{F}}_p^{d-1})-1$$

$$\sum _{i\in I}{x}_i=0.$$

$$f(s_1,\dots ,s_n)\ne 0.$$

$${\mathbb{I}}_f:=\sum _{(x_1,\dots ,x_n)\in {\left\{0,1\right\}}^n}(-1{)}^{x_1+\cdots +x_n}f(x_1,\dots ,x_n).$$

$$g(x_1,\dots ,x_n):=f(x_1,\dots ,x_n)-c\prod _{i=1}^n{x}_i$$

$$\sum _{(x_1,\dots ,x_n)\in \{0,1{\}}^n}(-1{)}^{x_1+\cdots +x_n}{x}_1^{i_1}\dots x_n^{i_n}$$

$$P_1=P_1(x_1,\dots ,x_n),\dots ,P_m=P_m(x_1,\dots ,x_n)$$

$$n>\sum _{j=1}^m{d}_j,$$

$$n>1+\sum _{i=1}^k(p^{{\alpha }_i}-1),$$

$$f_j(x_1,\dots ,x_n):=\sum _{i=1}^n{a}_{ij}{x}_i^{p-1}$$

$$\sum _{i\in I}{v}_i=0,$$

$$\sum _{i\in I}{x}_i=0.$$

$$A=\left\{(a_1,a_1),\dots ,(a_{p-1},(p-1)a_{p-1}),(a_p,0),(0,a_{p+1})\right\}$$

$$f(x_1,\dots ,x_{p+1})=(x_1+\dots +x_p{)}^{p-1}(x_1+2x_2+\dots +p{x}_p+x_{p+1}{)}^{p-1}$$

$$[x_1^{c_1}\dots x_{p+1}^{c_{p+1}}](f)={\left(c_1\cdot \dots \cdot c_{p+1}\right)}^{-1}[x_1^{c_1-1}\dots x_{p+1}^{c_{p+1}-1}](g),$$

$$g(x_1,\dots ,x_{p+1}):=\frac{{\partial }^{n}f(x_1,\dots ,x_{p+1})}{\partial x_1\dots \partial x_{p+1}}.$$

$$P(x_1,\dots ,x_{p+1})=\left[1-{\left(\sum _{i=1}^p{a}_i{x}_i\right)}^{p-1}\right]\left[1-{\left(\sum _{j=1}^{p+1}j{a}_j{x}_j\right)}^{p-1}\right].$$

$$P(0,\dots ,0)=1\text{and}P(x_1,\dots ,x_{p+1})=0,$$

$$\sum _{(x_1,\dots ,x_{p+1})\in \{0,1{\}}^{p+1}}(-1{)}^{x_1+\cdots +x_{p+1}}P(x_1,\dots ,x_{p+1})=1.$$

$$Q(x_1,\dots ,x_{p+1})={\left(\sum _{i=1}^p{a}_i{x}_i\right)}^{p-1}{\left(\sum _{j=1}^{p+1}j{a}_j{x}_j\right)}^{p-1},$$

$$\sum _{(x_1,\dots ,x_{p+1})\in \{0,1{\}}^{p+1}}(-1{)}^{x_1+\cdots +x_{p+1}}{x}_1^{i_1}\dots x_{p+1}^{i_{p+1}}$$

$$\sum _{(x_1,\dots ,x_{p+1})\in \{0,1{\}}^{p+1}}(-1{)}^{x_1+\cdots +x_{p+1}}\left(P(x_1,\dots ,x_{p+1})-Q(x_1,\dots ,x_{p+1})\right)=0.$$

$$\sum _{(x_1,\dots ,x_{p+1})\in \{0,1{\}}^{p+1}}(-1{)}^{x_1+\cdots +x_{p+1}}Q(x_1,\dots ,x_{p+1})=1.$$

$$\sum _{(x_1,\dots ,x_{p+1})\in \{0,1{\}}^{p+1}}(-1{)}^{x_1+\cdots +x_{p+1}}{x}_1^{j_1}\dots x_{p+1}^{j_{p+1}}=0,$$

$$\sum _{(x_1,\dots ,x_{p+1})\in \{0,1{\}}^{p+1}}(-1{)}^{x_1+\cdots +x_{p+1}}Q(x_1,\dots ,x_{p+1})=0,$$

Full text

1. Introduction

For a set $A$ in an additive group $G$, denote

[TABLE]

Define the Olson constant $OL(G)$ of a finite additive group $G$ to be the minimal $d$ such that every subset $A\subset G$ of cardinality $d$ satisfies $0\in\Sigma^{*}(A)$.

It turns out this is a difficult quantity to compute for general groups. First, Erdős and Heilbronn initially proved in [5] that there exists an absolute constant $c$ such that $OL(\mathbb{F}_{p})\leq c\sqrt{p}$, where $p$ is a prime. They also conjectured that this bound extends to arbitrary additive groups $G$, and years later Szemerédi showed in [12] that $OL(G)\leq k\sqrt{|G|}$ for some $k>0$. However, both results were far from being optimal. Olson [11] also came into the picture later and proved that that $OL(G)\leq 2\sqrt{|G|}$, which has been subsequently improved by Hamidoune and Zemor [7], who proved that

[TABLE]

conjecturing that $c=\sqrt{2}$ is perhaps optimal.

The story is not much better for groups of the form $\mathbb{F}_{p}^{d}$, where $p$ is an odd prime. For $d=1$, Hamidoune and Zemor were indeed correct; in [8], Nguyen, Szemeredi and Vu improved upon the Erdős-Heilbronn theorem and showed that for sufficiently large primes $p$, $OL(\mathbb{F}_{p})\leq\sqrt{2p}$. This was subsequently also shown by Balandraud in [2] for all primes $p$, who in fact actually proved that for $A\subset\mathbb{F}_{p}$, with $|A|=d$ and such that $A\cap(-A)=\emptyset$, we have that

[TABLE]

Balandraud’s original proof made use of some impressive recursions for binomial determinants, but it has been recently recasted by the same author in [3] as a natural consequence of the Combinatorial Nullstellensatz due to Alon [2]. For $d=2$, Gao, Ruzsa and Thangadurai showed in [7] that

[TABLE]

holds for $p>4.67\times 10^{34}$. Their argument was subsequently updated by Bhowmik and Schlage-Puchta to show that the equality holds for all primes $>6000$; thus, given the result for $\mathbb{F}_{p}$, the Olson constant constant of $\mathbb{F}_{p}^{2}$ is also determined for large primes. For higher dimensions, Gao, Ruzsa and Thangadurai conjectured that

[TABLE]

should hold in general, but little is known in this case. We refer to [9] for an account of the state of the art.

In this paper we shall prove the following

Theorem 1. Let $p$ be an odd prime and let $\left\{x_{1},\ldots,x_{n}\right\}\subset\mathbb{F}_{p}^{2}\ \backslash\left\{0,0\right\}$ be a set of vectors spanning all $p+1$ nonzero directions from [math] in $\mathbb{F}_{p}^{2}$. We prove that there exists a subset $I\subset\left\{1,\ldots,n\right\}$ such that

[TABLE]

This is meant to complement the Gao-Rusza-Thangadura-Bhowmik-Schlage-Puchta theorem. Here, we shall say that two vectors $(a,b),(c,d)\in\mathbb{F}_{p}^{2}$ have the same direction from zero if there is some $t\in\mathbb{F}_{p}\ \backslash\left\{0\right\}$ such that $(a,b)=(tc,td)$. The above result thus says that a set $A\subset\mathbb{F}_{p}^{2}$, with $(0,0)\notin A$ and with the property that it meets all $p+1$ lines of the form $ax+by=0$, must contain a non-empty subset $B\subset A$ with zero sum.

2. Preliminaries

Although not implicitly used in our proof, we first recall the statement of the Combinatorial Nullstellensatz.

Theorem 2. Let $\mathbb{F}$ be an arbitrary field, and let $f=f(x_{1},\ldots,x_{n})$ be a polynomial in $F[x_{1},\ldots,x_{n}]$. Suppose the degree $\deg(f)$ of $f$ is $\sum_{i=1}^{n}{t_{i}}$, where each $t_{i}$ is a nonnegative integer, and suppose the coefficient of $\prod_{i=1}^{n}{x_{i\ }^{t_{i}}}$ in $f$ is nonzero. Then, if $S_{1},\ldots,S_{n}$ are the subsets of $F$ with $|S_{i}|>t_{i}$, there are $s_{1}\ldots S_{1}$,$\ldots$, $s_{n}\in S_{n}$ so that

[TABLE]

This is usually proven in general by induction on the number of variables (see for instance [1]), but whenever $S_{i}=\left\{0,1\right\}$, $t_{i}=1$ for each $i\in\left\{1,\ldots,n\right\}$, there’s a very neat way of thinking about this, which will be relevant to our proof of Theorem 1. One can simply consider the following “integral” of $f$ on $\left\{0,1\right\}^{n}$:

[TABLE]

If $f$ vanishes everywhere on $S_{1}\times\ldots\times S_{n}$, then clearly $\mathbb{I}_{f}=0$.

On the other hand, if $c$ is the nonzero coefficient of $\prod_{i=1}^{n}{x_{i}}$ in $f$, then

[TABLE]

is a polynomial of degree less than $n$ in $\mathbb{F}[x_{1},\ldots,x_{n}]$. In particular, every monomial appearing in $g\in\mathbb{F}[x_{1},\ldots,x_{n}]$ with nonzero coefficient has to be of the form $x_{1}^{i_{1}}\ldots x_{n}^{i_{n}}$, with $i_{j}\in\left\{0,1\right\}$ not all $1$. For such a monomial, where say $i_{K}=0$ for some $K\in\left\{1,\ldots,n\right\}$, we then note that

[TABLE]

Therefore, $\mathbb{I}_{g}=0$, which yields $\mathbb{I}_{f}=\mathbb{I}_{c\prod_{i=1}^{n}{x_{i}}}=\pm c\neq 0$, a contradiction.

We will consider such an integral in our proof of Theorem 1 as well. Before doing so however, let us mention first that one of the first certificates of the usefulness of the Combinatorial Nullstellensatz in [1] was a proof of the following theorem due to Chevalley and Warning.

Theorem 3. Let $p$ be a prime and let $\mathbb{F}$ be a finite field. Let

[TABLE]

be $m$ polynomials in $\mathbb{F}[x_{1},\ldots,x_{n}]$ such that

[TABLE]

where $d_{j}$ is the total degree of $f_{j}$. Then, if the polynomials $P_{i}$ have a common zero $(c_{1},\ldots,c_{n})$, they must have another common zero.

We mention this because the one can use the Theorem 3 as a “black-box” to show something in the spirit of Theorem 1.

Theorem 4. Let $p$ be a prime and let $n$ be integers such that $n>2(p-1)$. Then, among any $n$ elements $v_{1},\ldots,v_{n}$ of $\mathbb{F}_{p}^{2}$ there exists a a nonempty subsequence with a zero-sum.

In fact, it is a result of Olson [10] that if $G$ is an abelian $p$-group of the form $G=\mathbb{Z}_{p^{\alpha_{1}}}\oplus\ldots\oplus\mathbb{Z}_{p^{\alpha_{k}}}$, and $n$ is any integer such that

[TABLE]

then among every sequence of elements of $G$ with length $n$ there is a nonempty subsequence with a zero sum. We sketch the proof for the scenario described in Theorem 4.

Proof of Theorem 4. Assume without loss of generality that $n=2p-1$, and furthermore let $v_{i}=(a_{i1},a_{i2})$, for each $1\leq i\leq n$. Define

[TABLE]

for each $j\in\left\{1,2\right\}$. Note that these polynomials fulfll the hypothesis of the Chevalley-Warning theorem: the sum of their degrees is $2(p-1)$, which is less than $n=2p-1$, the number of variables. Also, note that the system $f_{1}=f_{2}=0$ has the trivial solution $x_{1}=\ldots=x_{n}=0$. By the Chevalley-Warning theorem, it thus has another one, say $(\alpha_{1},\ldots,\alpha_{n})$. Setting $I:=\left\{i\ |\ \alpha_{i}\neq 0\right\},$ this is non-empty and it is easy to check that

[TABLE]

as desired.

$\hfill\square$

3. Proof of Theorem 1

Let $A$ be a set of vectors $\left\{x_{1},\ldots,x_{n}\right\}\subset\mathbb{F}_{p}^{2}\ \backslash\left\{0,0\right\}$ spanning all $p+1$ nonzero directions from [math] in $\mathbb{F}_{p}^{2}$. Recall that we want to prove that there exists a subset $I\subset\left\{1,\ldots,n\right\}$ such that

[TABLE]

Without loss of generality, assume that $n=p+1$ and that

[TABLE]

where $a_{1},\ldots,a_{p+1}$ are non-zero elements in $\mathbb{F}_{p}$. We claim that there exists a non-empty subset of $A$ with sum of elements equal to zero.

We will first prove a crucial Lemma.

Lemma 5. The polynomial $f\in\mathbb{F}_{p}[x_{1},\ldots,x_{p+1}]$ defined by

[TABLE]

contains no monomial where variables $x_{1},\ldots,x_{p+1}$ all have positive exponents.

Proof of Lemma 5. Assume that the coefficient of some monomial $x_{1}^{c_{1}}\dots x_{p+1}^{c_{p+1}}$ appearing in the expansion of $f$ is nonzero modulo $p$, where $c_{i}>0$ for each $i\in\left\{1,\ldots,p+1\right\}$. First, note that each $c_{i}$ is also less than $p$, since otherwise the monomial would have degree at least $2p>2p-2=\deg f$. Moreover, the coefficient $[x_{1}^{c_{1}}\dots x_{p+1}^{c_{p+1}}](f)$ of $x_{1}^{c_{1}}\dots x_{p+1}^{c_{p+1}}$ in $f$ satisfies

[TABLE]

where the latter denotes the coefficient of $x_{1}^{c_{1}-1}\dots x_{p+1}^{c_{p+1}-1}$ in the derivative

[TABLE]

Since $p>c_{i}>0$ for each $i$, this coefficient is also nonzero modulo $p$.

On the other hand, the polynomial $g$ is identically zero. Indeed, in the computation of the multiple derivative, we differentiate $(x_{1}+x_{2}+\dots+x_{p})^{p-1}$ with respect to some $k$ variables ($2\leq k\leq p-1$), including $x_{p}$ but excluding $x_{p+1}$, and we differentiate $(x_{1}+2x_{2}+\dots+px_{p}+x_{p+1})^{p-1}$ with respect to other $p+1-k$ variables (including $x_{p+1}$ but excluding $x_{p}$). It is easy to check that for each summand appearing in $g$, a multiple of the $(p-k)$-th elementary symmetric polynomials in the variables $1,2,\dots,p$ arises for some $k\in\left\{2,\ldots,p-1\right\}$. By Vieta’s formulas, these are all equal to [math] over $\mathbb{F}_{p}$, as $(t-1)\dots(t-p)=t^{p}-t$. We have therefore arrived at a contradiction.

$\hfill\square$

Now, suppose that $A$ contains no non-empty subset with sum of elements equal to zero. Consider the following polynomial in $\mathbb{F}_{p}[x_{1},\ldots,x_{p+1}]$:

[TABLE]

Note that $P(x_{1},\ldots,x_{p+1})=1$ precisely when $\sum_{i=1}^{p}{a_{i}x_{i}}=\sum_{j=1}^{p+1}{ja_{j}x_{j}}=0$ and is [math] otherwise; therefore, when restricted on $\left\{0,1\right\}^{p+1}$, the polynomial $P$ identifies zero-sum subsets of $A$. In particular, given our zero-sum-subset free assumption on $A$, we have that

[TABLE]

for every $(p+1)$-tuple $(x_{1},\ldots,x_{p+1})\in\left\{0,1\right\}^{p+1}\backslash\left\{0,\ldots,0\right\}$. In particular,

[TABLE]

If $Q\in\mathbb{F}_{p}[x_{1},\ldots,x_{p+1}]$ is defined by

[TABLE]

then $P-Q$ is a polynomial of degree $p-1$ in $\mathbb{F}_{p}[x_{1},\ldots,x_{p+1}]$. In particular, there is no monomial in $P-Q$ which contains all variables $x_{1},\ldots,x_{p+1}$, each with positive exponent. For each monomial $x_{1}^{i_{1}}\ldots x_{p+1}^{i_{p+1}}$, for which $i_{K}=0$ for some $K\in\left\{1,\ldots,p+1\right\}$, we then note that

[TABLE]

Since this holds for every monomial appearing in $P-Q$, we get that

[TABLE]

Consequently,

[TABLE]

But this gives a contradiction with the structure of $A$. Lemma 5 says that each monomial appearing in $Q$ must have at least one variable absent; by performing the same trick as above, we can thus argue that for each monomial $x_{1}^{j_{1}}\ldots x_{p+1}^{j_{p+1}}$ appearing in $Q$, we have that

[TABLE]

and therefore

[TABLE]

which is a contradiction.

$\hfill\square$

Acknowledgments

I would like to thank Fedor Petrov for helpful comments on a prior version of this preprint.