Sums and products in finite fields: an integral geometric viewpoint

Abstract

We prove that if $A\subset {\mathbb{F}}_q$ is such that $$|A|>q^{1/2+\frac{1}{2d}},$$ then $${\mathbb{F}}_q^{\ast }\subset d{A}^2=A^2+...+A^{2}d\text{times},$$ where $$A^2=\{a\cdot a^{\prime }:a,a^{\prime }\in A\},$$ and where ${\mathbb{F}}_q^{\ast }$ denotes the multiplicative group of the finite field ${\mathbb{F}}_q$. In particular, we cover ${\mathbb{F}}_q^{\ast }$ by $A^2+A^2$ if $|A|>q^{3/4}$. Furthermore, we prove that if $$|A|\ge C_{size}^{\frac{1}{d}}{q}^{1/2+\frac{1}{2(2d-1)}},$$ then $$|d{A}^2|\ge q\cdot \frac{C_{size}^2}{C_{size}^2+1}.$$ Thus $d{A}^2$ contains a positive proportion of the elements of ${\mathbb{F}}_q$ under a considerably weaker size assumption.We use the geometry of ${\mathbb{F}}_q^d$, averages over hyper-planes and orthogonality properties of character sums. In particular, we see that using operators that are smoothing on $L^2$ in the Euclidean setting leads to non-trivial arithmetic consequences in the context of finite fields.