On Gaussian sums over finite fields (Q2708422)

Let \(p > 2\) be a prime number, \(F_{q}\) a finite field with \(q=p^{s}\) elements, \( \chi(x)\) the non-trivial quadratic multiplicative character, \( \psi (x)=\exp (2 \pi i tr (x)/p)\) the canonical additive character of \(F_{q}\), and NEWLINE\[NEWLINE \tau_{s} = \sum_{x \in F_{q}^{*}} \chi (x) \psi (x) NEWLINE\]NEWLINE the normed quadratic Gaussian sum over \(F_{q}\). Since \( \tau_{s}^{2}= (-1)^{(q-1)/2}q\) it follows that either \( \tau_{s}= \pm \sqrt{q}\) or \( \tau_{s}= \pm i \sqrt{q}\). A classical result of Gauss implies that NEWLINE\[NEWLINE \tau_{1}= \begin{cases} \sqrt{p} & \text{if } p \equiv 1 \pmod 4; \\ i \sqrt{p} & \text{if } p \equiv 3 \pmod 4. \end{cases} NEWLINE\]NEWLINE Firstly the author presents well-known Kronecker's proof of this result and then considers the problem on determination of the sign of \( \tau_{s}\) in the case when \(s > 1\). For any positive integer \(s\), the sign of \( \tau_{s}\) is easily determined from the Davenport-Hasse relations: \( \tau_{s}= -(- \tau_{1})^{s}\). The author's result concerning the sign of \( \tau_{s}\) does not agree with the result which follows from these relations.