Galois action on the quadratic Gauss sum

Let $p$ be an odd prime, let ${\zeta }_p$ be a fixed primitive $p$-th root of unity, let ${\tau }_p={\sum }_{a<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>p}(a\mathrm{/}p){\zeta }_p^a$ be the quadratic Gauss sum attached to ${\zeta }_p$, and let $b$ be an integer not divisible by $p$. For the automorphism ${\sigma }_b$ of $\mathbb{Q}({\zeta }_p)$ with ${\sigma }_b({\zeta }_p)={\zeta }_p^b$ one has ${\sigma }_b({\tau }_p)=\left(\frac{b}{p}\right){\tau }_p\mathrm{.}$

• No analytic sign is used. The argument is a finite rearrangement of the defining sum together with the multiplicativity of the Legendre symbol; it never chooses a complex embedding of $\mathbb{Q}({\zeta }_p)$ or a sign of ${\tau }_p$. Later, together with the theorem that ${\tau }_p^2=p^{\ast }\mathrm{\ne }0$, this identity shows that exactly the square classes $b$ fix ${\tau }_p$, which identifies the quadratic subfield generated by ${\tau }_p$.
• Convention. As in the definition, ${\sigma }_b(\zeta )={\zeta }^b$ is the arithmetic power map. Its inverse $\zeta \mapsto {\zeta }^{b^{-1}}$ has the same action on ${\tau }_p$, since $(b^{-1}\mathrm{/}p)=(b\mathrm{/}p)$ by [F3].