# Dirichlet character modulo p

How can I prove that if $\chi$ is a non-principal character modulo $p$ prime, then $\chi (-1) = \overline{\chi} (-1)= \pm 1$ and $\sum_{x=1}^p \chi (x) e^{2\pi i x}=0$?

For the first question, I just know that $\chi(-1)\overline{\chi} (-1)=1$, but they could be complex?

And for the second question: no idea...

We have $$1=\chi(1)=\chi((-1)^2))=\chi(-1)^2$$. Since the square roots of $$1$$ are $$1$$ and $$-1$$, we obtain $$\chi(-1)=\pm 1$$. Also, $$\overline{\chi (-1)}=\chi(-1)^{-1}=\pm 1$$. For the second part, I show that another sum is zero, i.e., $$S=\sum_t\chi(t)$$. Note that there is an $$a\neq 0$$ such that $$\chi(a)\neq 1$$, because $$\chi$$ is non-principal. Let $$S=\sum_t\chi(t)$$. Then $$\chi(a)S=\sum_t\chi(a)\chi(t)=\sum_t\chi(at)=S.$$ So we have $$S(\chi(a)-1)=0$$. Since $$\chi(a)\neq 1$$, it follows $$S=0$$. This is also used for proving that the Gauss sum $$G(1,\chi)=\sum_t \chi(t)e^{\frac{2\pi it}{p}}$$ has absolute value $$\sqrt{p}$$. Your sum is similar.
• But the second: It is not the Gauss sum itself, the Gauss sum is $\sum_{x=1}^{p-1}\chi (x) e(x/p)$, but I have no denominator $p$ in my sum - and on my way proving that the Gauss sum has absolute value $\sqrt{p}$, I have to prove that my sum is zero. – sBs Oct 21 '14 at 19:43
• $G(1,\chi)=\sum_t \chi(t)e^{\frac{2\pi i}{p}}$ should be $G(1,\chi)=\sum_t \chi(t)e^{\frac{2\pi i {\color{red}{t}}}{p}}$? – mike Aug 5 at 1:52