Define a non trivial additive character $\psi: \mathbb F_q \rightarrow \mathbb C$ s.t $\psi(x)= \xi^{Tr(x)}$ where $\xi=e^{2\pi i /p}$, where $Tr(x):\mathbb F_q \rightarrow \mathbb F_p$ .
First question: is it correct that $Tr(x)= x+ x^p + ... + x^{p^{r-1}}$, with $q=p^r$?
Define now a multiplicative character $\chi:\mathbb F_q^* \rightarrow \mathbb C^*$ and extende it so that $\chi(0)=0$.
Define finally the Gauss sum of $\chi$ as $g(\chi)=\displaystyle \sum_{x \in \mathbb F_q}\chi(x)\psi(x)$.
The goal now is to prove that this sum for the trivial character $\epsilon$ is equal to $-1$.
Clearly $g(\epsilon)=\displaystyle \sum_{x \in \mathbb F_q}\psi(x)$, but I don't know how to calculate this sum.
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$\begingroup$ What is $Tr(x)$ $\endgroup$– Infinity_hunterCommented Feb 28 at 15:59
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$\begingroup$ I wrote in the question what I suppose the trace to be, but I'm not sure $\endgroup$– GiuseppeCommented Feb 28 at 16:12
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1 Answer
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$Tr(x)$ is the field trace mapping $\mathbb{F}_q\to\mathbb{F}_p$, where $\mathbb{F}_q$ is viewed as a field extension of $\mathbb{F}_p$. The automorphisms of a finite field send $x\mapsto x^{p^r}$, and so the formula for the trace you indicated is indeed correct.
Now, since the trace function maps $\mathbb{F}_q\to\mathbb{F}_p$, the sum is really just a sum over the $p$th roots of unity, besides $1$, since $\epsilon(0)=0$. The sum of the $p$th roots of unity is well known to be $0$ including $1$, so without $1$ the sum has to be $-1$, as desired. Some people will define $\epsilon(0)=1$ specifically so that the Gauss some of the trivial character is $0$. This of course has the disadvantage that the trivial character is no longer multiplicative.
