# Gauss sum in character theory

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.

• What is $Tr(x)$ Commented Feb 28 at 15:59
• I wrote in the question what I suppose the trace to be, but I'm not sure Commented Feb 28 at 16:12

$$Tr(x)$$ 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.