The Stacks project

Lemma 10.155.1. Let $(R, \mathfrak m, \kappa )$ be a local ring. There exists a local ring map $R \to R^ h$ with the following properties

  1. $R^ h$ is henselian,

  2. $R^ h$ is a filtered colimit of étale $R$-algebras,

  3. $\mathfrak m R^ h$ is the maximal ideal of $R^ h$, and

  4. $\kappa = R^ h/\mathfrak m R^ h$.

Proof. Consider the category of pairs $(S, \mathfrak q)$ where $R \to S$ is an étale ring map, and $\mathfrak q$ is a prime of $S$ lying over $\mathfrak m$ with $\kappa = \kappa (\mathfrak q)$. A morphism of pairs $(S, \mathfrak q) \to (S', \mathfrak q')$ is given by an $R$-algebra map $\varphi : S \to S'$ such that $\varphi ^{-1}(\mathfrak q') = \mathfrak q$. We set

\[ R^ h = \mathop{\mathrm{colim}}\nolimits _{(S, \mathfrak q)} S. \]

Let us show that the category of pairs is filtered, see Categories, Definition 4.19.1. The category contains the pair $(R, \mathfrak m)$ and hence is not empty, which proves part (1) of Categories, Definition 4.19.1. For any pair $(S, \mathfrak q)$ the prime ideal $\mathfrak q$ is maximal with residue field $\kappa $ since the composition $\kappa \to S/\mathfrak q \to \kappa (\mathfrak q)$ is an isomorphism. Suppose that $(S, \mathfrak q)$ and $(S', \mathfrak q')$ are two objects. Set $S'' = S \otimes _ R S'$ and $\mathfrak q'' = \mathfrak qS'' + \mathfrak q'S''$. Then $S''/\mathfrak q'' = S/\mathfrak q \otimes _ R S'/\mathfrak q' = \kappa $ by what we said above. Moreover, $R \to S''$ is étale by Lemma 10.143.3. This proves part (2) of Categories, Definition 4.19.1. Next, suppose that $\varphi , \psi : (S, \mathfrak q) \to (S', \mathfrak q')$ are two morphisms of pairs. Then $\varphi $, $\psi $, and $S' \otimes _ R S' \to S'$ are étale ring maps by Lemma 10.143.8. Consider

\[ S'' = (S' \otimes _{\varphi , S, \psi } S') \otimes _{S' \otimes _ R S'} S' \]

with prime ideal

\[ \mathfrak q'' = (\mathfrak q' \otimes S' + S' \otimes \mathfrak q') \otimes S' + (S' \otimes _{\varphi , S, \psi } S') \otimes \mathfrak q' \]

Arguing as above (base change of étale maps is étale, composition of étale maps is étale) we see that $S''$ is étale over $R$. Moreover, the canonical map $S' \to S''$ (using the right most factor for example) equalizes $\varphi $ and $\psi $. This proves part (3) of Categories, Definition 4.19.1. Hence we conclude that $R^ h$ consists of triples $(S, \mathfrak q, f)$ with $f \in S$, and two such triples $(S, \mathfrak q, f)$, $(S', \mathfrak q', f')$ define the same element of $R^ h$ if and only if there exists a pair $(S'', \mathfrak q'')$ and morphisms of pairs $\varphi : (S, \mathfrak q) \to (S'', \mathfrak q'')$ and $\varphi ' : (S', \mathfrak q') \to (S'', \mathfrak q'')$ such that $\varphi (f) = \varphi '(f')$.

Suppose that $x \in R^ h$. Represent $x$ by a triple $(S, \mathfrak q, f)$. Let $\mathfrak q_1, \ldots , \mathfrak q_ r$ be the other primes of $S$ lying over $\mathfrak m$. Then $\mathfrak q \not\subset \mathfrak q_ i$ as we have seen above that $\mathfrak q$ is maximal. Thus, since $\mathfrak q$ is a prime ideal, we can find a $g \in S$, $g \not\in \mathfrak q$ and $g \in \mathfrak q_ i$ for $i = 1, \ldots , r$. Consider the morphism of pairs $(S, \mathfrak q) \to (S_ g, \mathfrak qS_ g)$. In this way we see that we may always assume that $x$ is given by a triple $(S, \mathfrak q, f)$ where $\mathfrak q$ is the only prime of $S$ lying over $\mathfrak m$, i.e., $\sqrt{\mathfrak mS} = \mathfrak q$. But since $R \to S$ is étale, we have $\mathfrak mS_{\mathfrak q} = \mathfrak qS_{\mathfrak q}$, see Lemma 10.143.5. Hence we actually get that $\mathfrak mS = \mathfrak q$.

Suppose that $x \not\in \mathfrak mR^ h$. Represent $x$ by a triple $(S, \mathfrak q, f)$ with $\mathfrak mS = \mathfrak q$. Then $f \not\in \mathfrak mS$, i.e., $f \not\in \mathfrak q$. Hence $(S, \mathfrak q) \to (S_ f, \mathfrak qS_ f)$ is a morphism of pairs such that the image of $f$ becomes invertible. Hence $x$ is invertible with inverse represented by the triple $(S_ f, \mathfrak qS_ f, 1/f)$. We conclude that $R^ h$ is a local ring with maximal ideal $\mathfrak mR^ h$. The residue field is $\kappa $ since we can define $R^ h/\mathfrak mR^ h \to \kappa $ by mapping a triple $(S, \mathfrak q, f)$ to the residue class of $f$ modulo $\mathfrak q$.

We still have to show that $R^ h$ is henselian. Namely, suppose that $P \in R^ h[T]$ is a monic polynomial and $a_0 \in \kappa $ is a simple root of the reduction $\overline{P} \in \kappa [T]$. Then we can find a pair $(S, \mathfrak q)$ such that $P$ is the image of a monic polynomial $Q \in S[T]$. Since $S \to R^ h$ induces an isomorphism of residue fields we see that $S' = S[T]/(Q)$ has a prime ideal $\mathfrak q' = (\mathfrak q, T - a_0)$ at which $S \to S'$ is standard étale. Moreover, $\kappa = \kappa (\mathfrak q')$. Pick $g \in S'$, $g \not\in \mathfrak q'$ such that $S'' = S'_ g$ is étale over $S$. Then $(S, \mathfrak q) \to (S'', \mathfrak q'S'')$ is a morphism of pairs. Now that triple $(S'', \mathfrak q'S'', \text{class of }T)$ determines an element $a \in R^ h$ with the properties $P(a) = 0$, and $\overline{a} = a_0$ as desired. $\square$


Comments (3)

Comment #4660 by Tim Holzschuh on

Typo: "The residue field is κ since we can define by mapping a triple to the residue class of module ." I guess that should be " modul \mathfrak q" not "module".

Also, although the current argument is working of course: It is argued that by quasi-finiteness of etale ring maps. However, it's already stated in the beginning that has to be maximal since which is an easier argument I guess.

Comment #8576 by on

Writing the ideal as is not precise, because is an element of , while is an ideal of .

There are also:

  • 8 comment(s) on Section 10.155: Henselization and strict henselization

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.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 04GN. Beware of the difference between the letter 'O' and the digit '0'.