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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.