Lemma 10.153.7. 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 10.153.5 or Lemma 10.153.3 part (10). In particular $A_ i/\mathfrak mA_ i$ is a finite separable field extension of $\kappa $, see Lemma 10.143.5. 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 10.144.3 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. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)

There are also: