Lemma 10.153.4 (04GH)—The Stacks project

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Table of contents
Part 1: Preliminaries
Chapter 10: Commutative Algebra
Section 10.153: Henselian local rings
Lemma 10.153.4 (cite)

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Lemma 10.153.4. Let $(R, \mathfrak m, \kappa )$ be a henselian local ring.

If $R \to S$ is a finite ring map then $S$ is a finite product of henselian local rings each finite over $R$ .
If $R \to S$ is a finite ring map and $S$ is local, then $S$ is a henselian local ring and $R \to S$ is a (finite) local ring map.
If $R \to S$ is a finite type ring map, and $\mathfrak q$ is a prime of $S$ lying over $\mathfrak m$ at which $R \to S$ is quasi-finite, then $S_{\mathfrak q}$ is henselian and finite over $R$ .
If $R \to S$ is quasi-finite then $S_{\mathfrak q}$ is henselian and finite over $R$ for every prime $\mathfrak q$ lying over $\mathfrak m$ .

Proof. Part (2) implies part (1) since $S$ as in part (1) is a finite product of its localizations at the primes lying over $\mathfrak m$ by Lemma 10.153.3 part (10). Part (2) also follows from Lemma 10.153.3 part (10) since any finite $S$ -algebra is also a finite $R$ -algebra (of course any finite ring map between local rings is local).

Let $R \to S$ and $\mathfrak q$ be as in (3). Write $S = A \times B$ with $A$ finite over $R$ and $B$ not quasi-finite over $R$ at any prime lying over $\mathfrak m$ , see Lemma 10.153.3 part (11). Hence $S_\mathfrak q$ is a localization of $A$ at a maximal ideal and we deduce (3) from (1). Part (4) follows from part (3). $\square$

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