Let $p$ be a prime number, and consider a Dirichlet character $\chi : (\mathbf Z/p\mathbf Z)^\times \to \mathbf C^\times$. Its image lands in the group $\mu_{p-1}$ of $(p-1)$-st roots of unity.

The Gauss sum is defined as $\eta(\chi) := \sum_{a=1}^p \chi(a) \zeta_p^a$, where $\zeta_n = \exp(2\pi i/n)$. By the preceding remark, $\eta(\chi) \in \mathbf Q(\zeta_{p(p-1)})$. Now, since $p$ is relatively prime with $p-1$, $$\text{Gal} (\mathbf Q(\zeta_{p(p-1)})/\mathbf Q) \simeq \text{Gal}(\mathbf Q(\zeta_p)/\mathbf Q) \times \text{Gal}(\mathbf Q(\zeta_{p-1})/\mathbf Q),$$ (which identifies with $(\mathbf Z/p\mathbf Z)^\times \times (\mathbf Z/(p-1)\mathbf Z))^\times$ in the usual way).

Thus, we can let $\text{Gal}(\mathbf Q(\zeta_p)/\mathbf Q) = (\mathbf Z/p\mathbf Z)^\times$ act on the Gauss sum $\eta(\chi)$. An easy calculation shows that, for $a \in (\mathbf Z/p\mathbf Z)^\times$, the corresponding automorphism $\sigma_a$ acts on the Gauss sum by

$$\eta(\chi)^{\sigma_a} = \overline{\chi}(a) \eta(\chi).$$

The map $\sigma_a \mapsto \eta(\chi)/\eta(\chi)^{\sigma_a}$ is (by definition) a $1$-coboundary of the $(\mathbf Z/p\mathbf Z)^\times$-module $\mathbf Q(\zeta_{p(p-1)})^\times$, and the fact that $\eta(\chi)/\eta(\chi)^{\sigma_a} = \chi(a)$ shows that this coboundary actually lands in the trivial $(\mathbf Z/p\mathbf Z)^\times$-module $\mu_{p-1}$. Therefore, the Gauss sum determines a cohomology class in

$$H^1((\mathbf Z/p\mathbf Z)^\times, \mu_{p-1}) = \hom((\mathbf Z/p\mathbf Z)^\times, \mu_{p-1}) = \widehat{(\mathbf Z/p\mathbf Z)^\times},$$

and the formula $\eta(\chi)/\eta(\chi)^{\sigma_a} = \chi(a)$ shows that this cohomology class is precisely the Dirichlet character we started with.

Now, I am puzzled. Have I gone full circle, or is there actually something going on? Why does the Gauss sum "realize" the Dirichlet character as a coboundary coming from the larger module $\mathbf Q(\zeta_{p(p-1)})/\mathbf Q)$? I'm not so sure what is the right way to think about this. Any insight appreciated!


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