Lemma 10.148.7. Let $(R, \mathfrak m, \kappa )$ be a henselian local ring. The category of finite étale ring extensions $R \to S$ is equivalent to the category of finite étale algebras $\kappa \to \overline{S}$ via the functor $S \mapsto S/\mathfrak mS$.

**Proof.**
Denote $\mathcal{C} \to \mathcal{D}$ the functor of categories of the statement. Suppose that $R \to S$ is finite étale. Then we may write

with $A_ i$ local and finite étale over $S$, use either Lemma 10.148.5 or Lemma 10.148.3 part (10). In particular $A_ i/\mathfrak mA_ i$ is a finite separable field extension of $\kappa $, see Lemma 10.141.5. Thus we see that every object of $\mathcal{C}$ and $\mathcal{D}$ decomposes canonically into irreducible pieces which correspond via the given functor. Next, suppose that $S_1$, $S_2$ are finite étale over $R$ such that $\kappa _1 = S_1/\mathfrak mS_1$ and $\kappa _2 = S_2/\mathfrak mS_2$ are fields (finite separable over $\kappa $). Then $S_1 \otimes _ R S_2$ is finite étale over $R$ and we may write

as before. Then we see that $\mathop{\mathrm{Hom}}\nolimits _ R(S_1, S_2)$ is identified with the set of indices $i \in \{ 1, \ldots , n\} $ such that $S_2 \to A_ i$ is an isomorphism. To see this use that given any $R$-algebra map $\varphi : S_1 \to S_2$ the map $\varphi \times 1 : S_1 \otimes _ R S_2 \to S_2$ is surjective, and hence is equal to projection onto one of the factors $A_ i$. But in exactly the same way we see that $\mathop{\mathrm{Hom}}\nolimits _\kappa (\kappa _1, \kappa _2)$ is identified with the set of indices $i \in \{ 1, \ldots , n\} $ such that $\kappa _2 \to A_ i/\mathfrak mA_ i$ is an isomorphism. By the discussion above these sets of indices match, and we conclude that our functor is fully faithful. Finally, let $\kappa \subset \kappa '$ be a finite separable field extension. By Lemma 10.141.15 there exists an étale ring map $R \to S$ and a prime $\mathfrak q$ of $S$ lying over $\mathfrak m$ such that $\kappa \subset \kappa (\mathfrak q)$ is isomorphic to the given extension. By part (1) we may write $S = A_1 \times \ldots \times A_ n \times B$. Since $R \to S$ is quasi-finite we see that there exists no prime of $B$ over $\mathfrak m$. Hence $S_{\mathfrak q}$ is equal to $A_ i$ for some $i$. Hence $R \to A_ i$ is finite étale and produces the given residue field extension. Thus the functor is essentially surjective and we win. $\square$

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