发表机构
Queens College, CUNY; Xiamen University Malaysia（纽约市立大学皇后学院; 厦门大学马来西亚分校）
机构由 AI 辅助整理，请以论文原文为准。AI 中文总结
本文研究有限域上高斯和相等何时蕴含特征为 Frobenius 共轭，利用 Stickelberger 分解给出可检验判据，并证明阶为 q-1 或 (q-1)/2 时成立。
AI 中文摘要
设 $\F_q$ 为素数 $p$ 的 $q = p^f$ 阶有限域。若 $G(χ)$ 是 $\F_q^\times$ 上乘法特征 $χ$ 对应的高斯和，则 $G(χ) = G(χ^p)$。我们研究其逆问题：高斯和相等 $G(χ_2) = G(χ_1)$ 何时蕴含存在整数 $j$ 使得 $χ_2 = χ_1^{p^j}$。若 $χ_2 = χ_1^{p^j}$，则称 $χ_1$ 与 $χ_2$ 为 Frobenius 共轭。我们利用分圆域中理想的 Stickelberger 分解，基于 $p$-进数字展开，给出一个易于检验的高斯和相等判据。作为应用，我们发展若干条件，在这些条件下 $\F_q^\times$ 上特征之间的高斯和相等可由 Frobenius 共轭解释。例如，若 $χ_1$ 的阶为 $q-1$ 或 $\frac{1}{2}(q-1)$ 且 $G(χ_2) = G(χ_1)$，则 $χ_1$ 与 $χ_2$ 为 Frobenius 共轭。我们还包含若干例子，基于 R.J. Evans 对某些“纯”高斯和的显式求值。
英文摘要
Let $\F_q$ be the field of order $q = p^f$ for prime $p$. If $G(χ)$ is the Gauss sum attached to a multiplicative character $χ$ on $\F_q^\times$, then $G(χ) = G(χ^p)$. We investigate the converse question: when does the equality of Gauss sums $G(χ_2) = G(χ_1)$ imply that $χ_2 = χ_1^{p^j}$ for some integer $j$. If $χ_2 = χ_1^{p^j}$, we say that $χ_1$ and $χ_2$ are Frobenius-conjugate. We use the Stickelberger factorization of ideals in cyclotomic fields to give an easily testable criterion for equality of Gauss sums, based on $p$-adic digit expansion. As an application, we develop several conditions under which Gauss sum equalities between characters on $\F_q^\times$ are explained by Frobenius-conjugacy. For example, if $χ_1$ has order $q-1$ or $\frac{1}{2}(q-1)$ and $G(χ_2) = G(χ_1)$, then $χ_1$ and $χ_2$ are Frobenius-conjugate. We also include several examples, based on the explicit evaluation of certain {\em pure} Gauss sums by R.J. Evans.
CommentsAccepted by Springer Proceedings in Mathematics and Statistics: Representations of p-adic groups and noncommutative geometry
