I'm following the proof by Marcus regarding the absolute value of the Gauss Sum.

He defines the Gauss sum $\tau_k(\chi)=\sum_{a\in \mathbb Z^*_m}\chi(a)\omega^{ak}$

Consider now $\tau_1(\chi)=\tau(\chi)$ and $\chi$ primitive nontrivial character mod m

$|\tau(\chi)|^2=\sum_{a,b\in\mathbb Z^*_m}\chi(a)\overline\chi(b)\omega^{a-b}=\sum_{b,c\in\mathbb Z^*_m}\chi(c)\omega^{(c-1)b}$ [and this is fine. From now on I do not understand.]

Moreover for $b\in\mathbb Z_m^*$ we have $\sum_{c\in\mathbb Z^*_m}\chi(c)\omega^{(c-1)b}=\omega^{-b}\tau_b(\chi)=0$ [Why is zero and why is important??]

Hence $|\tau(\chi)|^2=\sum_{c\in\mathbb Z_m^*}\chi(c)\sum_{b=0}^{m-1}\omega^{(c-1)b}$ [why the inner sum is over $\{0,...,m-1\}$ instead of $\mathbb Z_m^*$??]

Finally, $\omega^{c-1}$ is a nontrivial $mth$ root of $1$ for $c\not=1$, hence the inner sum vanishes for $c\not=1$ and we obtain the thesis. [this is fine]


I think that in the second line you should have for $b\notin \mathbb{Z}_m^*$ instead of $\in$. In general you have that if $(c,m)=1$ and $c\bar{c}=1$, then $$\chi(\bar c)\tau_b(\chi)=\sum_{a\in \mathbb Z^*_m}\chi(a\bar c)\omega^{ab}=\sum_{a\in \mathbb Z^*_m}\chi(a)\omega^{acb}=\tau_{cb}(\chi)$$ Choose such $c$ which satisfies $cb\equiv b$ mod m, so that $\chi(\bar c)\tau_b(\chi)=\tau_{b}(\chi)$. From the primitivity of $\chi$ you can choose such $c$ so that $\chi(c)\neq 1$, so that $\tau_b(\chi)=0$. Now you can add these terms to the equation in the first line in order to get the equation in the third line.

  • $\begingroup$ Yes, after some time, I thought it too. In practice you have to add zero terms to the sum. It's weird to find such a mistake in Marcus. Thank you. $\endgroup$ – Richard Jan 23 '17 at 14:08
  • $\begingroup$ Please, take a look also to this question:math.stackexchange.com/questions/2110578/… $\endgroup$ – Richard Jan 23 '17 at 17:41

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