Why is the absolute value of this Gauss sum obvious?

Question

I came across the Gauss sum discussed in the following post in a problem from my Galois theory course: https://mathoverflow.net/a/71282. Why exactly is the square of its norm obvious?

Answer 1

Let $G=\sum_{a}\Big(\frac{a}{p}\Big)\zeta_p^a$. Here's a proof that $G^2=\Big(\frac{-1}{p}\Big)p$, which in particular shows that $|G|^2=p$.

As $a$ runs over $(\mathbb{Z}/p\mathbb{Z})^{\times}$, so does $ab$ for fixed $b\neq0$, so we have: $$ G^2=\sum_{a,b}\left(\frac{a}{p}\right)\left(\frac{b}{p}\right)\zeta_p^{a+b}=\sum_{a,b}\left(\frac{ab}{p}\right)\zeta_p^{a+b}=\sum_{a,b}\left(\frac{ab^2}{p}\right)\zeta_p^{b(a+1)}=\sum_{a,b}\left(\frac{a}{p}\right)\zeta_p^{b(a+1)} $$ $$=\sum_{b}\left(\frac{-1}{p}\right)+\sum_{a\neq-1}\left(\frac{a}{p}\right)\sum_{b}\zeta_p^{b(a+1)} $$

Moreover, $1+\zeta_p+\dots+\zeta_p^{p-1}=0$, so $\displaystyle\sum_{b}\zeta_p^{b(a+1)}=-1$, and

$$ G^2=(p-1)\left(\frac{-1}{p}\right)-\sum_{a\neq-1}\left(\frac{a}{p}\right)=p\left(\frac{-1}{p}\right)-\sum_{a}\left(\frac{a}{p}\right)=p\left(\frac{-1}{p}\right) $$ since there are as many quadratic residues as non-residues mod $p$.