Distinguishing Gauss sums

Source abstract

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.