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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 $g_1(\chi)^m \in \mathbb{Q}(\zeta_m)$?

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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 $\mathbb{Q}(\zeta_m)$.

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