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. $\square$

Comment #4660 by Tim Holzschuh on

Typo: "The residue field is κ since we can define $R^h/mR^h$$\kappa$ by mapping a triple $(S,\mathfrak q,f)$ to the residue class of $f$ module $\mathfrak q$." I guess that should be "$f$ modul$\textbf{o}$ \mathfrak q" not "module".

Also, although the current argument is working of course: It is argued that $\mathfrak q \not\subset \mathfrak q_i$ by quasi-finiteness of etale ring maps. However, it's already stated in the beginning that $\mathfrak q$ has to be maximal since $\kappa(\mathfrak q) = \kappa$ which is an easier argument I guess.

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