在距离集的商集上

Q4 Mathematics
A. Iosevich, D. Koh, Hans Parshall
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引用次数: 11

摘要

让 ${\Bbb F}_q$ 是一个有限的有序域 $q.$ 我们证明如果 $d\ge 2$ 是偶数且 $E \subset {\Bbb F}_q^d$ 有 $|E| \ge 9q^{\frac{d}{2}}$ 然后 $$ {\Bbb F}_q=\frac{\Delta(E)}{\Delta(E)}=\left\{ \frac{a}{b}: a \in \Delta(E), b \in \Delta(E) \backslash \{0\} \right\},$$ 在哪里 $$ \Delta(E)=\{||x-y||: x,y \in E\}, \ ||x||=x_1^2+x_2^2+\cdots+x_d^2.$$ 如果维度 $d$ 是奇数和 $E\subset \mathbb F_q^d$ 有 $|E|\ge 6q^{\frac{d}{2}},$ 然后 $$ \{0\}\cup\mathbb F_q^+ \subset \frac{\Delta(E)}{\Delta(E)},$$ 在哪里 $\mathbb F_q^+$ 中的非零二次残的集合 $\mathbb F_q.$ 一般来说,这两个结果都是最好的,包括关于奇维非零二次残的结论。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
On the quotient set of the distance set
Let ${\Bbb F}_q$ be a finite field of order $q.$ We prove that if $d\ge 2$ is even and $E \subset {\Bbb F}_q^d$ with $|E| \ge 9q^{\frac{d}{2}}$ then $$ {\Bbb F}_q=\frac{\Delta(E)}{\Delta(E)}=\left\{ \frac{a}{b}: a \in \Delta(E), b \in \Delta(E) \backslash \{0\} \right\},$$ where $$ \Delta(E)=\{||x-y||: x,y \in E\}, \ ||x||=x_1^2+x_2^2+\cdots+x_d^2.$$ If the dimension $d$ is odd and $E\subset \mathbb F_q^d$ with $|E|\ge 6q^{\frac{d}{2}},$ then $$ \{0\}\cup\mathbb F_q^+ \subset \frac{\Delta(E)}{\Delta(E)},$$ where $\mathbb F_q^+$ denotes the set of nonzero quadratic residues in $\mathbb F_q.$ Both results are, in general, best possible, including the conclusion about the nonzero quadratic residues in odd dimensions.
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来源期刊
Moscow Journal of Combinatorics and Number Theory
Moscow Journal of Combinatorics and Number Theory Mathematics-Algebra and Number Theory
CiteScore
0.80
自引率
0.00%
发文量
21
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