Subsets of $\mathbb{F}^*_p$ with only small products or ratios

TL;DR

This paper investigates the structure of subsets of the finite field multiplicative group with restrictions on their products or ratios, providing estimates for the number of elements satisfying certain modular product conditions.

Contribution

It introduces new bounds and estimates for the number of elements in subsets of finite fields with small product or ratio properties, extending previous results.

Findings

Derived bounds for subset sizes with small product conditions

Extended analysis to ratio-based subset properties

Generalized results to broader classes of finite field subsets

Abstract

Let $p$ be a fixed prime. We estimate the number of elements of a set $A \subseteq \mathbb{F}^*_p$ for which $$ s_1s_2 \equiv a \pmod{p} \quad \mbox{for some}\quad a \in [-X,X] \quad \mbox{for all}\quad s_1,s_2 \in A. $$ We also consider variations and generalizations.

Equations

$$s_1{s}_2\equiv a(\mathrm{mod}p)\text{mbox}forsomea\in [-X,X]\text{mbox}forall{s}_1,s_2\in A.$$

$$|\alpha {|}_p:=\underset{k\in \mathbb{Z}}{\min }|\alpha +kp|$$

$$\Bigl{|}\frac{s_{1}}{s_{2}}\Bigr{|}_{p}\leq X\ \mbox{and/or}\ \Bigl{|}\frac{s_{2}}{s_{1}}\Bigr{|}_{p}\leq X\ \mbox{for each}\ (s_{1},s_{2})\in A^{2}.$$

$$S(X):=\underset{\begin{array}{c}A\in {\mathbb{F}}_p^{\ast }\\ A\in \mathcal{S}(X)\end{array}}{\max }|A|.$$

$$|s_1\cdots s_n{|}_p\le X\text{mbox}forallpairwisedistinct{s}_1,\dots ,s_n\in A.$$

$$R_n(X):=\underset{\begin{array}{c}A\in {\mathbb{F}}_p^{\ast }\\ A\in {\mathcal{R}}_n(X)\end{array}}{\max }|A|.$$

$${\tau }_n(m):=|\{(d_1,\dots ,d_n)\in {\mathbb{N}}^n:d_1\cdots d_n=m\}|.$$

$$S(X){\ll }_{ϵ}\min \left(X^{ϵ}+\frac{X^{2+ϵ}}{p},p^{1/2}\right)$$

$$R_n(X){\ll }_{ϵ,n}\min \left(X^{1/n+ϵ}+\frac{X^{n/(n-1)+ϵ}}{p^{1/(n-1)}},\underset{k\in \{2,\dots ,n-2\}}{\min }{X}^{1/k+ϵ}+\frac{X^{1+1/k+ϵ}}{p^{1/k}},p^{1/n+ϵ}\right)$$

$$\left|\sum _{a_1\in A_1,\dots ,a_n\in A_n}{e}_p(a_1\cdots a_n)\right|\le p^{1/2}(|A_1|\cdots |A_n|{)}^{\frac{n-1}{n}}.$$

$$\underset{m\in {\mathbb{F}}_p^{\ast }}{\max }\left|\sum _{a_1\in A_1,a_2\in A_2}{e}_p(m{a}_1{a}_2)\right|\le (p|A_1||A_2|{)}^{1/2},$$

$$\biggl{|}\sum_{a_{1}\in A_{1},\dots,a_{n}\in A_{n}}e_{p}(a_{1}\cdots a_{n})\biggr{|}<p^{-\delta^{\prime}}|A_{1}|\cdots|A_{n}|.$$

$$\frac{s}{s_1}\equiv a(\mathrm{mod}p)\text{mbox}and\frac{s_2}{s}\equiv b(\mathrm{mod}p)\text{mbox}with0<|a|,|b|\le X.$$

$$s_{1}a\equiv \frac{s_2}{b}(\mathrm{mod}p)\Rightarrow ab\equiv \frac{s_2}{s_1}\equiv :\alpha (\mathrm{mod}p)\text{mbox}with0<|a|,|b|,|\alpha |\le X.$$

$$|A|\le 4|A_2|\ll X^{ϵ}\left(1+\frac{X^2}{p}\right).$$

$$|T_1|+|T_2|:=\left|\sum _{s_1,s_2\in A}{e}_p\left(\frac{s_1}{s_2}\right)\right|+\left|\sum _{s_1,s_2\in A}{e}_p\left(\frac{2s_1}{s_2}\right)\right|\le 2p^{1/2}|A|.$$

$$T_{1,1}+T_{1,2}+T_{1,3}:=\sum _{\begin{array}{c}{s}_1,s_2\in A\\ (s_1,s_2)\in W\end{array}}{e}_p\left(\frac{s_1}{s_2}\right)+\sum _{\begin{array}{c}{s}_1,s_2\in A\\ |\frac{s_1}{s_2}{|}_p>\frac{3p}{8}\\ (s_1,s_2)\notin W\end{array}}{e}_p\left(\frac{s_1}{s_2}\right)+\sum _{\begin{array}{c}{s}_1,s_2\in A\\ |\frac{s_1}{s_2}{|}_p\le \frac{3p}{8}\\ (s_1,s_2)\notin W\end{array}}{e}_p\left(\frac{s_1}{s_2}\right).$$

$$\Re (T_{1,1}+T_{1,2}+T_{1,3})\le \frac{|A{|}^2}{400}$$

$$u_2\ge u_1\cos \left(\frac{\pi }{6}\right)-u_3\cos \left(\frac{\pi }{4}\right)-\frac{1}{400}.$$

$$u_1\cos \left(\frac{\pi }{3}\right)+u_2\cos \left(\frac{\pi }{2}\right)\ge u_3+\frac{1}{400}.$$

$$u_2\ge \frac{\frac{1}{2}\left(\cos \left(\frac{\pi }{6}\right)-\cos \left(\frac{\pi }{4}\right)\right)+\frac{1}{2|A|}\left(\cos \left(\frac{\pi }{6}\right)+\cos \left(\frac{\pi }{4}\right)\right)-\frac{1}{400}}{1-\cos \left(\frac{\pi }{4}\right)}.$$

$$\left(1+\frac{1}{|A|}\right)\cos \left(\frac{\pi }{3}\right)+\frac{\cos \left(\frac{\pi }{6}\right)-\cos \left(\frac{\pi }{4}\right)+\frac{1}{|A|}\left(\cos \left(\frac{\pi }{6}\right)+\cos \left(\frac{\pi }{4}\right)\right)-\frac{1}{200}}{1-\cos \left(\frac{\pi }{4}\right)}\ge 1-\frac{1}{|A|}+\frac{1}{200}$$

$$\cos \left(\frac{\pi }{3}\right)+\frac{\cos \left(\frac{\pi }{6}\right)-\cos \left(\frac{\pi }{4}\right)-\frac{1}{200}}{1-\cos \left(\frac{\pi }{4}\right)}\ge 1+\frac{1}{200}.$$

$$|A|^{n}\ll\Biggl{|}\sum_{\begin{subarray}{c}s_{1},\dots,s_{n}\in A\\ (s_{1},\dots,s_{n})\ admissible\end{subarray}}e_{p}(s_{1}\cdots s_{n})\Biggr{|}-O(|A|^{n-1})\ll\Biggl{|}\sum_{s_{1},\dots,s_{n}\in A}e_{p}(s_{1}\cdots s_{n})\Biggr{|}\leq p^{-\delta^{\prime}}|A|^{n}$$

$$\pm \alpha :=\underset{\begin{array}{c}{r}_1,\dots ,r_n\in A\\ (r_1,\dots ,r_n)admissible\end{array}}{\max }|r_1\cdots r_n{|}_p,$$

$$r_1\cdots r_n\equiv c(\mathrm{mod}p)$$

$$\Rightarrow$$

$$|A^{\prime }{|}^n$$

$$r{r}_2\cdots r_n\equiv c(\mathrm{mod}p)$$

$$a{a}_2\cdots a_n=c{\alpha }^{n-1}+Kp\Rightarrow a^{\prime }{a}_2\cdots a_n\equiv K^{\prime }p(\mathrm{mod}\beta )$$

$$|A^{\prime }{|}^{n-1}\ll \sum _{\begin{array}{c}(c,K)\\ c{\alpha }^{n-1}+Kp\ne 0\end{array}}{\tau }_{n-1}\left(\frac{c{\alpha }^{n-1}+Kp}{a}\right)\ll \frac{X^{n+ϵ}}{p}$$

Full text

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We also consider variations and generalizations.

AMS Subject Classification numbers: 11A07, 11B75

Key words: congruences, special sets $\pmod{p}$

1 Introduction and notation

Let $p$ be a fixed prime number. For any member $\alpha$ of an equivalence class of $\mathbb{Z}/p\mathbb{Z}$, we write

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and for any finite set $A$ we write $|A|:=\#A$ which should not be confused with the norm of a complex number. Inspired by the paper [2], we are interested by the cardinality of a set $A\subseteq\mathbb{F}^{*}_{p}$ that satisfies some property. Precisely, for each $X\geq 1$ we let $\mathcal{S}(X)$ be the set of all subsets $A\subseteq\mathbb{F}^{*}_{p}$ that satisfy

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We thus define

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Similarly, for each integer $n\geq 2$ and $X\geq 1$ we let $\mathcal{R}_{n}(X)$ be the set of all subsets $A\subseteq\mathbb{F}^{*}_{p}$ that satisfy

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Then, we consider the quantity

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For any $m,n\in\mathbb{N}$, we write

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We will often use the well known fact that $\tau_{n}(m)\ll_{n,\epsilon}m^{\epsilon}$ for each $n\geq 2$ and $\epsilon>0$. We also write e_{p}(z):=\exp\bigl{(}\frac{2\pi iz}{p}\bigr{)} for any $z\in\mathbb{C}$.

2 Statement of theorems

Theorem 1**.**

For each $1\leq X\leq\frac{p}{12}$, we have

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for each fixed $\epsilon>0$.

Theorem 2**.**

For each integer $n\geq 2$ and $1\leq X\leq 0.24p$, we have

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for each fixed $\epsilon>0$.

3 Preliminary lemmas

There are a number of interesting results in the literature concerning multilinear exponential sums; see [1], [3], [4] and [5] for example and other references. We will need the following two.

Lemma 1**.**

Let $A_{1},\dots,A_{n}\subseteq\mathbb{F}^{*}_{p}$ $(n\geq 2)$ be subsets. Then

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Proof.

We assume that $|A_{1}|\geq|A_{2}|\geq\cdots\geq|A_{n}|$. The inequality follows from the well known result

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see [4, (2)]. ∎

Lemma 2**.**

*Let $0<\delta<1/4$ and $n\in\mathbb{Z}_{+}$. There is an effectively computable $\delta^{\prime}=\delta^{\prime}(\delta)>0$ such that if $p$ is a sufficiently large prime and $A_{1},\dots,A_{n}\subset\mathbb{F}_{p}$ satisfy

\begin{array}[]{ll}(i)&|A_{i}|>p^{\delta}\ \mbox{for}\ 1\leq i\leq n;\\ (ii)&\prod_{i=1}^{n}|A_{i}|>p^{1+\delta};\end{array}

then there is the exponential sum bound*

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Proof.

It follows from Theorem A of the paper [1]. ∎

4 Proof of Theorem 1

We assume throughout the proof that $A\in\mathcal{S}(X)$ and satisfies $S(X)=|A|$. We begin with the first inequality. We choose $s_{1}\in A$ that realizes (1.1) with every element of $A$ by being at least $\frac{|A|}{2}$ times at the denominator. We denote by $A_{1}$ the set of values that are thereby at the numerator. Restricting our attention to $A_{1}$, we choose $s_{2}\in A_{1}$ that realizes (1.1) with every element of $A_{1}$ by being at least $\frac{|A_{1}|}{2}$ times at the numerator and we denote by $A_{2}$ the set of values that are thereby at the denominator.

Now, for each value $s\in A_{2}$ we have two representations. Indeed,

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We deduce that

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We thus have $ab=\alpha+Kp$ with 0\leq|K|\leq\bigl{\lfloor}\frac{2X^{2}}{p}\bigr{\rfloor}. For each fixed value of $K$, the number of solutions $(a,b)$ is at most $2\tau_{2}(\alpha+Kp)\ll X^{\epsilon}$ and we deduce that

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We now turn to the second inequality. From Lemma 1 with $n=2$, we know that

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The result will follow if we can show that $|T_{1}|+|T_{2}|\geq\frac{|A|^{2}}{400}$. We will assume that $|T_{1}|\leq\frac{|A|^{2}}{400}$. We denote by $W$ the set of pairs $(s_{1},s_{2})\in A^{2}$ for which we know from the hypothesis $A\in\mathcal{S}(X)$ that 0<\bigl{|}\frac{s_{1}}{s_{2}}\bigr{|}_{p}\leq X. In particular, $|W|=\frac{|A|^{2}+|A|}{2}$. We divide $T_{1}$ into

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From now, we denote by $U_{i}$ the number of pairs $(s_{1},s_{2})$ in the sum $T_{1,i}$. We have $U_{1}=|W|=\frac{|A|^{2}+|A|}{2}$ and $U_{2}+U_{3}=\frac{|A|^{2}-|A|}{2}$ and we write $U_{i}:=u_{i}|A|^{2}$. From

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and the hypothesis $1\leq X\leq\frac{p}{12}$, we get to the inequality

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But we will have $|T_{2}|\geq\frac{|A|^{2}}{400}$ if

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We deduce from (4.1), $u_{1}=\frac{1}{2}+\frac{1}{2|A|}$ and $u_{2}+u_{3}=\frac{1}{2}-\frac{1}{2|A|}$ that

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Using inequality (4.3) and the fact that $\cos\left(\frac{\pi}{2}\right)=0$, we see that (4.2) is satisfied if

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which holds since

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The proof is complete.

5 Proof of Theorem 2

We assume throughout the proof that $A\in\mathcal{R}_{n}(X)$ and satisfies $R_{n}(X)=|A|$. Also, for any $k\geq 1$, we say that $(s_{1},\dots,s_{k})$ is an admissible $k$-tuple if the $s_{j}$ are pairwise distinct $(j=1,\dots,k)$. There are exactly $|A|\cdot(|A|-1)\cdots(|A|-k+1)$ admissible $k$-tuples in $A^{k}$. We can assume that $|A|$ is large enough since otherwise there is nothing to prove.

We begin with the third inequality. Assuming that $X\leq 0.24p$ and that $|A|>p^{1/n+\delta}$ for some fixed $0<\delta<1/2n$, we get

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for some $\delta^{\prime}>0$, from Lemma 2. This is a contradiction for $p$ large enough and we deduce that $|A|\ll p^{1/n+\epsilon}$ for each $\epsilon>0$.

For the first inequality, we define $\alpha$ by

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and we assume that $\alpha\equiv s_{1}\cdots s_{n}\pmod{p}$ (with $(s_{1},\dots,s_{n})$ admissible). We now define a change of variable according to this choice. In the set $A^{\prime}:=A\setminus\{s_{1},\dots,s_{n}\}$, we can write an element $r$ as $r\equiv a_{j}\frac{s_{j}}{\alpha}\pmod{p}$ for some $0<|a_{j}|\leq X$ $(j=1,\dots,n)$.

Any of the $|A^{\prime}|\cdot(|A^{\prime}|-1)\cdots(|A^{\prime}|-n+1)$ admissible $n$-tuples $(r_{1},\dots,r_{n})$ gives rise to

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where $0<|c|,|a_{1}|,\dots,|a_{n}|\leq X$ and $0\leq|K|\leq\lfloor\frac{2X^{n}}{p}\rfloor$. From there, we distinguish two cases.

Case 1: $K=0$ for more than half of the admissible $n$-tuples. In this case, we have

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for each fixed $\epsilon\geq 0$.

Case 2: $K\neq 0$ for at least half of the admissible $n$-tuples. In this case, we fix a value of $r=r_{1}\equiv a\frac{s_{1}}{\alpha}(\not\equiv 0)\pmod{p}$ that is in $\gg|A^{\prime}|^{n-1}$ admissible $n$-tuples $(r,r_{2},\dots,r_{n})$ in (5) that lead to (5.2) with $K\neq 0$. Then, we consider the equation

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with $0<|c|,|a_{2}|,\dots,|a_{n}|\leq X$ and $0<|K|\leq\lfloor\frac{2X^{n}}{p}\rfloor$. Now, we write $d:=gcd(a,\alpha^{n-1})$ and $a^{\prime}:=\frac{a}{d}$, $\beta:=\frac{\alpha^{n-1}}{d}$ and $K^{\prime}:=\frac{K}{d}$. We find that

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so that a fixed value of $K^{\prime}$ gives at most $d$ values of $a_{2}\cdots a_{n}\pmod{\alpha^{n-1}}$. There are $\ll\frac{X^{n}}{dp}$ possible values for $K^{\prime}$ and since $0<|a_{2}\cdots a_{n}|\leq\alpha^{n-1}$ we get that we have in fact at most $2d$ values of $a_{2}\cdots a_{n}$ for each. That is, we have at most $\ll\frac{X^{n}}{p}$ possible values of $(c,K)$. We get

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for each fixed $\epsilon>0$. For $n=2$ we have in fact $\epsilon=0$ in this last inequality. The result follows.

For the second inequality, let’s write

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for each $k=1,\dots,n-1$. For a fixed value of $k$ we can split each admissible $n$-tuple $(s_{1},\dots,s_{n})\in A^{\prime n}$ into $s_{1}\cdots s_{n}\equiv bc\equiv a\pmod{p}$, i.e. respectively $s_{1}\cdots s_{n-k}\equiv b\pmod{p}$ and $s_{n-k+1}\cdots s_{n}\equiv c\pmod{p}$. This leads to

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Now, for any fixed $m\in\mathbb{F}^{*}_{p}$ we use the change of variable stated above to write

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As previously, we deduce that

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Overall, we get to

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for any $k=1,\dots,n-1$.

6 Concluding remarks

The set

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shows that $S(X)\gg\log(2X)$. Also, the set

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shows that $R_{n}(X)\gg X^{1/n}$. We conjecture that both $S(X)\ll_{\epsilon}X^{\epsilon}$ and $R_{n}(X)\ll_{\epsilon,n}X^{1/n+\epsilon}$ hold for each $\epsilon>0$ when $X\leq\left(\frac{1}{2}-t\right)p$ for a fixed $t>0$ as $p\rightarrow\infty$.