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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 $\sigma_b \in \text{Gal}(\mathbb{Q}(\zeta_m, \zeta)/\mathbb{Q})$ by $\sigma_b(\zeta_m) = \zeta_m^b$, $\sigma_b(\zeta) = \zeta$. 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 $g_1(\chi)^m \in \mathbb{Q}(\zeta_m)$?

Here, $\chi_0$ is the unique Dirichlet character $\text{mod }1$ (i.e. $\chi_0(n) = 1$ for all $n$), $\zeta$ is 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$$g_a(\chi) = \sum_{x \in \mathbb{F}_p} \chi(x)\zeta^{ax}.$$

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We have$$\sigma_b\left(g_1(x)\right) = \sum_{x \in \mathbb{F}_p} \chi(x)^b \xi^x = g_1(x^b) \text{ for all }a \in \mathbb{Z},\,(a, p) = 1.$$Define $\sigma_a \in \text{Gal}(\mathbb{Q}(\xi_m, \xi)/\mathbb{Q})$ such that$$\sigma_a(\xi_m) = \xi_m,\text{ }\sigma_a(\xi) = \xi^a.$$Then$$g_1(x)^{b - \sigma_a} = {{g_1(x)^b}\over{g_1(x^b)}} \in \mathbb{Q}(\xi_m) \iff \sigma_ag_1(x)^{b - \sigma_a} = g_1(x)^{b - \sigma_b} \text{ for all }a.$$Note that$$\sigma_a g_1(x)^{b - \sigma_a} = {{\sigma_a\left( \sum_{x \in \mathbb{F}_p} \chi(x) \xi^x\right)^b}\over{\sigma_a\left( \sum_{x \in \mathbb{F}_p} \chi^b(x) \xi^x\right)}} = {{\left( \sum_{x \in \mathbb{F}_p} \chi(x) \xi^{ax}\right)^b}\over{\sum_{x \in \mathbb{F}_p} \chi^b(x)\xi^{ax}}}$$$$ = {{g_a(x)^b}\over{g_a(x^b)}} = {{\overline{\chi}(a)^b g_1(x)^b}\over{\overline{\chi}^b(a) g_1(x^b)}} = {{\overline{\chi}(a)^bg_1(x)^b}\over{\overline{\chi}(a)^bg_1(x^b)}} = {{g_1(x)^b}\over{g_1(x^b)}},$$so $g_1(x)^{b - \sigma_b} \in \mathbb{Q}(\xi_m)$, as desired.

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