# Absolute value of Gauss sums is $\sqrt m$

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.