Lemma 58.6.3. Let $K$ be a field and set $X = \mathop{\mathrm{Spec}}(K)$. Let $\overline{K}$ be an algebraic closure and denote $\overline{x} : \mathop{\mathrm{Spec}}(\overline{K}) \to X$ the corresponding geometric point. Let $K^{sep} \subset \overline{K}$ be the separable algebraic closure.

1. The functor of Lemma 58.2.2 induces an equivalence

$\textit{FÉt}_ X \longrightarrow \textit{Finite-}\text{Gal}(K^{sep}/K)\textit{-Sets}.$

compatible with $F_{\overline{x}}$ and the functor $\textit{Finite-}\text{Gal}(K^{sep}/K)\textit{-Sets} \to \textit{Sets}$.

2. This induces a canonical isomorphism

$\text{Gal}(K^{sep}/K) \longrightarrow \pi _1(X, \overline{x})$

of profinite topological groups.

Proof. The functor of Lemma 58.2.2 is the same as the functor $F_{\overline{x}}$ because for any $Y$ étale over $X$ we have

$\mathop{\mathrm{Mor}}\nolimits _ X(\mathop{\mathrm{Spec}}(\overline{K}), Y) = \mathop{\mathrm{Mor}}\nolimits _ X(\mathop{\mathrm{Spec}}(K^{sep}), Y)$

Namely, as seen in the proof of Lemma 58.2.2 we have $Y = \coprod _{i \in I} \mathop{\mathrm{Spec}}(L_ i)$ with $L_ i/K$ finite separable over $K$. Hence any $K$-algebra homomorphism $L_ i \to \overline{K}$ factors through $K^{sep}$. Also, note that $F_{\overline{x}}(Y)$ is finite if and only if $I$ is finite if and only if $Y \to X$ is finite étale. This proves (1).

Part (2) is a formal consequence of (1), Lemma 58.3.11, and Lemma 58.3.3. (Please also see the remark below.) $\square$

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