Let $\chi\_0$ be the unique Dirichlet character $\text{mod }1$ (i.e. $\chi\_0(n) = 1$ for all $n$), $\zeta\_p$ be a primitive $p$th root of $1$, and for any $a \in \mathbb{F}\_p^\times$ and any Dirichlet character $\chi$ modulo $p$ (or $\chi\_0 \text{ mod }1$) define the Gauss sum$$g\_a(\chi) = \sum\_{x \in \mathbb{F}\_p} \chi(x)\zeta\_p^{ax}.$$


Let $m$ be an integer prime to $p$ such that $\chi^m = \chi\_0$ on elements of $\mathbb{F}\_p^\times$. We let $\zeta\_m$ be a primitive $m$th root of unity. For $b$ any integer prime to $m$ define $\sigma\_b \in \text{Gal}(\mathbb{Q}(\zeta\_m, \zeta\_p)/\mathbb{Q})$ by $\sigma\_b(\zeta\_m) = \zeta\_m^b$, $\sigma\_b(\zeta\_p) = \zeta\_p$. How do I see that $$g\_1(\chi)^{b - \sigma\_b} = {{g\_1(\chi)^b}\over{\sigma\_b(g\_1(\chi))}}$$belongs to $\mathbb{Q}(\zeta\_m)$ and that $g\_1(\chi)^m \in \mathbb{Q}(\zeta\_m)$?



Property of Dirichlet character
By Galois Theory, it is enough to show that 
$$
{{g\_1(\chi)^b}\over{\sigma\_b(g\_1(\chi))}}, \ \ g\_1(\chi)^m $$
are fixed by all $\tau\_a$, which is defined for $(a,p)=1$, $\tau\_a(\zeta\_p)=\zeta\_p^a$, $\tau\_a(\zeta\_m)=\zeta\_m$. 


In fact, 
$$
\tau\_a({{g\_1(\chi)^b}\over{\sigma\_b(g\_1(\chi))}})={{g\_a(\chi)^b}\over{g\_a(\chi^b)}}={{(\overline{\chi}(a)g\_1(\chi))^b}\over \overline{\chi^b}(a)g\_1(\chi^b)}={{g\_1(\chi)^b}\over{g\_1(\chi^b)}}={{g\_1(\chi)^b}\over{\sigma\_b(g\_1(\chi))}},$$
and
$$
\tau\_a(g\_1(\chi)^m)=g\_a(\chi)^m=(\overline{\chi}(a)g\_1(\chi))^m=g\_1(\chi)^m.$$
Therefore, the elements in question are fixed by any $\tau\_a$, thus belong to $\mathbb{Q}(\zeta\_m)$. 


