Sums and products in finite fields: an integral geometric viewpoint
| dc.creator | Hart, Derrick | |
| dc.creator | Iosevich, Alex | |
| dc.date | 2007-05-29 | |
| dc.date | 2007-06-27 | |
| dc.date.accessioned | 2026-07-07T08:12:22Z | |
| dc.date.available | 2026-07-07T08:12:22Z | |
| dc.description | We prove that if $A \subset {\Bbb F}_q$ is such that $$|A|>q^{{1/2}+\frac{1}{2d}},$$ then $${\Bbb F}_q^{*} \subset dA^2=A^2+...+A^2 d \text{times},$$ where $$A^2=\{a \cdot a': a,a' \in A\},$$ and where ${\Bbb F}_q^{*}$ denotes the multiplicative group of the finite field ${\Bbb F}_q$. In particular, we cover ${\Bbb F}_q^{*}$ 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 $$|dA^2| \ge q \cdot \frac{C^2_{size}}{C^2_{size}+1}.$$ Thus $dA^2$ contains a positive proportion of the elements of ${\Bbb F}_q$ under a considerably weaker size assumption.We use the geometry of ${\Bbb F}_q^d$, 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. | |
| dc.identifier | https://arxiv.org/abs/0705.4256 | |
| dc.identifier | http://arxiv.org/abs/0705.4256 | |
| dc.identifier.uri | http://salesiana.dossiersoluciones.com/handle/123456789/132431 | |
| dc.subject | Number Theory | |
| dc.subject | Classical Analysis and ODEs | |
| dc.title | Sums and products in finite fields: an integral geometric viewpoint | |
| dc.type | text |