Quadratic Gauss sum in a prime cyclotomic field

Let $p$ be an odd prime, let ${\zeta }_p$ be a fixed primitive $p$-th root of unity (The cyclotomic extension $K({\mu }_n)$ as a splitting field of $t^n-1$), and let $(\cdot \mathrm{/}p)$ be the Legendre symbol (The Legendre symbol, including its zero value). The quadratic Gauss sum attached to ${\zeta }_p$ is

${\tau }_p:={\sum }_{a<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>p}\left(\frac{a}{p}\right){\zeta }_p^a={\sum }_{a=1}^{p-1}\left(\frac{a}{p}\right){\zeta }_p^a\in \mathbb{Z}[{\zeta }_p]\mathrm{.}$

The sum lies in $\mathbb{Z}[{\zeta }_p]$ and is an algebraic integer. The term $a=0$ vanishes because $(0\mathrm{/}p)=0$, so the sum is finite over the classes $a=1,\dots ,p-1$; each ${\zeta }_p^a$ is a root of $t^p-1$ and hence integral over $\mathbb{Z}$ (Integral elements over a commutative ring and algebraic integers), and the integral elements of $\mathbb{C}$ form a subring, so ${\tau }_p$ is an algebraic integer lying in $\mathbb{Z}[{\zeta }_p]\subseteq {\mathcal{O}}_{\mathbb{Q}({\zeta }_p)}$ (Ring of integers).

The chosen root is part of the data. The notation ${\tau }_p$ always refers to the sum built from the explicitly chosen ${\zeta }_p$. Replacing ${\zeta }_p$ by another primitive $p$-th root changes ${\tau }_p$ up to a sign: indeed ${\sigma }_b({\zeta }_p)={\zeta }_p^b$ gives ${\sum }_a(a\mathrm{/}p){\zeta }_p^{ab}=(b\mathrm{/}p){\tau }_p$ for every $b\equiv ̸0(\mathrm{m}\mathrm{o}\mathrm{d}p)$. The square ${\tau }_p^2=p^{\ast }=(-1{)}^{(p-1)\mathrm{/}2}p$ and the field $\mathbb{Q}({\tau }_p)$ are unchanged by that replacement (Square of the quadratic Gauss sum, Quadratic subfield generated by the Gauss sum), but the sign of ${\tau }_p$ is not fixed until the primitive root (equivalently, a complex embedding) is fixed.