Proposition 5

— There exists a unique isomorphism of topological groups $\pi_K : \mathbf{Z} \to \mathrm{Gal}(\bar{K}/K)$ such that $\pi_K(1) = \sigma_q$.

Let $\Gamma$ be the subgroup of $\mathrm{Gal}(\bar{K}/K)$ generated by $\sigma_q$. For every integer $m > 0$ we have $\sigma_q^m(x) = x^{q^m}$ for all $x \in \bar{K}$, hence the set of fixed points of $\sigma_q^m$ is equal to $K$. Since $K \neq \bar{K}$, we have $\sigma_q^m \neq 1$. Hence there is an isomorphism $\pi_0$ of $\mathbf{Z}$ onto $\Gamma$ which maps 1 to $\sigma_q$.

The field of invariants of $\Gamma$ consists of the $x \in K$ such that $x^q = x$, hence is equal to $K$. Therefore (V, p. 67, Lemma 2) the group $\Gamma$ is dense in $\mathrm{Gal}(\bar{K}/K)$. Since every subextension of $\bar{K}$ of finite degree over $K$ is one of the fields $K_m$, the subgroups $\mathrm{Gal}(\bar{K}/K_m)$ form a fundamental system of neighbourhoods of 1 in $\mathrm{Gal}(\bar{K}/K)$. It is clear that $\Gamma \cap \mathrm{Gal}(\bar{K}/K_m)$ is the cyclic group generated by $\sigma_q^m$, hence equal to $\pi_0(m\mathbf{Z})$.

From the above remarks about the topology of $\mathbf{Z}$ the isomorphism $\pi_0 : \mathbf{Z} \to \Gamma$ extends in a unique fashion to an isomorphism of topological groups $\pi_K : \mathbf{Z} \to \mathrm{Gal}(\bar{K}/K)$.

Let $m \geq 1$ be an integer; it is clear that the Frobenius automorphism of $\bar{K}$ relative to $K$ is $\sigma_q^m$. Hence we obtain the relation