# Why is the absolute value of this Gauss sum obvious?

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?

• It does not say that it's obvious, but that it's easy, meaning that the computation does not require any specific insight, it's just a computation. – Captain Lama Apr 29 '16 at 16:52
• Please ask a selfcontained question. – quid Apr 29 '16 at 16:53
• Mathematicians use "easy" or "obvious" as code for "I don't feel like proving this". – carmichael561 Apr 29 '16 at 16:56
• I don't see how the computation proceeds, though. – Vik78 Apr 29 '16 at 16:56

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 $p$.