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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...

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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.

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  • $\begingroup$ Thank you, the first thing was really easy (I thought to "complex"...;-)). $\endgroup$ – sBs Oct 21 '14 at 19:39
  • $\begingroup$ 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. $\endgroup$ – sBs Oct 21 '14 at 19:43
  • $\begingroup$ Yes, sorry, I fixed it. $\endgroup$ – Dietrich Burde Oct 21 '14 at 20:07
  • $\begingroup$ Okay, yes, your argument works, thank you! $\endgroup$ – sBs Oct 21 '14 at 20:29
  • $\begingroup$ $G(1,\chi)=\sum_t \chi(t)e^{\frac{2\pi i}{p}}$ should be $G(1,\chi)=\sum_t \chi(t)e^{\frac{2\pi i {\color{red}{t}}}{p}}$? $\endgroup$ – mike Aug 5 at 1:52

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