## 10.155 Henselization and strict henselization

In this section we construct the henselization. We encourage the reader to keep in mind the uniqueness already proved in Lemma 10.154.7 and the functorial behaviour pointed out in Lemma 10.154.6 while reading this material.

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

$R^ h$ is henselian,

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

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

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

**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 $Q \in S[T]$. Set $S' = S[T]/(Q)$ and let $\mathfrak q' \subset S'$ be the maximal ideal $\mathfrak q' = \mathfrak qS' + (T - a')S'$ where $a' \in S$ is any element lifting $a_0$. By construction $S \to S'$ is étale at $\mathfrak q'$ and $\kappa = \kappa (\mathfrak q')$. 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.
$\square$

Lemma 10.155.2. Let $(R, \mathfrak m, \kappa )$ be a local ring. Let $\kappa \subset \kappa ^{sep}$ be a separable algebraic closure. There exists a commutative diagram

\[ \xymatrix{ \kappa \ar[r] & \kappa \ar[r] & \kappa ^{sep} \\ R \ar[r] \ar[u] & R^ h \ar[r] \ar[u] & R^{sh} \ar[u] } \]

with the following properties

the map $R^ h \to R^{sh}$ is local

$R^{sh}$ is strictly henselian,

$R^{sh}$ is a filtered colimit of étale $R$-algebras,

$\mathfrak m R^{sh}$ is the maximal ideal of $R^{sh}$, and

$\kappa ^{sep} = R^{sh}/\mathfrak m R^{sh}$.

**Proof.**
This is proved by exactly the same proof as used for Lemma 10.155.1. The only difference is that, instead of pairs, one uses triples $(S, \mathfrak q, \alpha )$ where $R \to S$ étale, $\mathfrak q$ is a prime of $S$ lying over $\mathfrak m$, and $\alpha : \kappa (\mathfrak q) \to \kappa ^{sep}$ is an embedding of extensions of $\kappa $.
$\square$

Definition 10.155.3. Let $(R, \mathfrak m, \kappa )$ be a local ring.

The local ring map $R \to R^ h$ constructed in Lemma 10.155.1 is called the *henselization* of $R$.

Given a separable algebraic closure $\kappa \subset \kappa ^{sep}$ the local ring map $R \to R^{sh}$ constructed in Lemma 10.155.2 is called the *strict henselization of $R$ with respect to $\kappa \subset \kappa ^{sep}$*.

A local ring map $R \to R^{sh}$ is called a *strict henselization* of $R$ if it is isomorphic to one of the local ring maps constructed in Lemma 10.155.2

The maps $R \to R^ h \to R^{sh}$ are flat local ring homomorphisms. By Lemma 10.154.7 the $R$-algebras $R^ h$ and $R^{sh}$ are well defined up to unique isomorphism by the conditions that they are henselian local, filtered colimits of étale $R$-algebras with residue field $\kappa $ and $\kappa ^{sep}$. In the rest of this section we mostly just discuss functoriality of the (strict) henselizations. We will discuss more intricate results concerning the relationship between $R$ and its henselization in More on Algebra, Section 15.45.

Lemma 10.155.5. Let $R \to S$ be a local map of local rings. Let $S \to S^ h$ be the henselization. Let $R \to A$ be an étale ring map and let $\mathfrak q$ be a prime of $A$ lying over $\mathfrak m_ R$ such that $R/\mathfrak m_ R \cong \kappa (\mathfrak q)$. Then there exists a unique morphism of rings $f : A \to S^ h$ fitting into the commutative diagram

\[ \xymatrix{ A \ar[r]_ f & S^ h \\ R \ar[u] \ar[r] & S \ar[u] } \]

such that $f^{-1}(\mathfrak m_{S^ h}) = \mathfrak q$.

**Proof.**
This is a special case of Lemma 10.153.11.
$\square$

Lemma 10.155.6. Let $R \to S$ be a local map of local rings. Let $R \to R^ h$ and $S \to S^ h$ be the henselizations. There exists a unique local ring map $R^ h \to S^ h$ fitting into the commutative diagram

\[ \xymatrix{ R^ h \ar[r]_ f & S^ h \\ R \ar[u] \ar[r] & S \ar[u] } \]

**Proof.**
Follows immediately from Lemma 10.154.6.
$\square$

Here is a slightly different construction of the henselization.

Lemma 10.155.7. Let $R$ be a ring. Let $\mathfrak p \subset R$ be a prime ideal. Consider the category of pairs $(S, \mathfrak q)$ where $R \to S$ is étale and $\mathfrak q$ is a prime lying over $\mathfrak p$ such that $\kappa (\mathfrak p) = \kappa (\mathfrak q)$. This category is filtered and

\[ (R_{\mathfrak p})^ h = \mathop{\mathrm{colim}}\nolimits _{(S, \mathfrak q)} S = \mathop{\mathrm{colim}}\nolimits _{(S, \mathfrak q)} S_{\mathfrak q} \]

canonically.

**Proof.**
A morphism of pairs $(S, \mathfrak q) \to (S', \mathfrak q')$ is given by an $R$-algebra map $\varphi : S \to S'$ such that $\varphi ^{-1}(\mathfrak q') = \mathfrak q$. Let us show that the category of pairs is filtered, see Categories, Definition 4.19.1. The category contains the pair $(R, \mathfrak p)$ and hence is not empty, which proves part (1) of Categories, Definition 4.19.1. Suppose that $(S, \mathfrak q)$ and $(S', \mathfrak q')$ are two pairs. Note that $\mathfrak q$, resp. $\mathfrak q'$ correspond to primes of the fibre rings $S \otimes \kappa (\mathfrak p)$, resp. $S' \otimes \kappa (\mathfrak p)$ with residue fields $\kappa (\mathfrak p)$, hence they correspond to maximal ideals of $S \otimes \kappa (\mathfrak p)$, resp. $S' \otimes \kappa (\mathfrak p)$. Set $S'' = S \otimes _ R S'$. By the above there exists a unique prime $\mathfrak q'' \subset S''$ lying over $\mathfrak q$ and over $\mathfrak q'$ whose residue field is $\kappa (\mathfrak p)$. The ring map $R \to S''$ 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

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

Arguing as above (base change of étale maps is étale, composition of étale maps is étale) we see that $S''$ is étale over $R$. The fibre ring of $S''$ over $\mathfrak p$ is

\[ F'' = (F' \otimes _{\varphi , F, \psi } F') \otimes _{F' \otimes _{\kappa (\mathfrak p)} F'} F' \]

where $F', F$ are the fibre rings of $S'$ and $S$. Since $\varphi $ and $\psi $ are morphisms of pairs the map $F' \to \kappa (\mathfrak p)$ corresponding to $\mathfrak p'$ extends to a map $F'' \to \kappa (\mathfrak p)$ and in turn corresponds to a prime ideal $\mathfrak q'' \subset S''$ whose residue field is $\kappa (\mathfrak p)$. The canonical map $S' \to S''$ (using the right most factor for example) is a morphism of pairs $(S', \mathfrak q') \to (S'', \mathfrak q'')$ which equalizes $\varphi $ and $\psi $. This proves part (3) of Categories, Definition 4.19.1. Hence we conclude that the category is filtered.

Recall that in the proof of Lemma 10.155.1 we constructed $(R_{\mathfrak p})^ h$ as the corresponding colimit but starting with $R_{\mathfrak p}$ and its maximal ideal $\mathfrak pR_{\mathfrak p}$. Now, given any pair $(S, \mathfrak q)$ for $(R, \mathfrak p)$ we obtain a pair $(S_{\mathfrak p}, \mathfrak qS_{\mathfrak p})$ for $(R_{\mathfrak p}, \mathfrak pR_{\mathfrak p})$. Moreover, in this situation

\[ S_{\mathfrak p} = \mathop{\mathrm{colim}}\nolimits _{f \in R, f \not\in \mathfrak p} S_ f. \]

Hence in order to show the equalities of the lemma, it suffices to show that any pair $(S_{loc}, \mathfrak q_{loc})$ for $(R_{\mathfrak p}, \mathfrak pR_{\mathfrak p})$ is of the form $(S_{\mathfrak p}, \mathfrak qS_{\mathfrak p})$ for some pair $(S, \mathfrak q)$ over $(R, \mathfrak p)$ (some details omitted). This follows from Lemma 10.143.3.
$\square$

Lemma 10.155.8. Let $R \to S$ be a ring map. Let $\mathfrak q \subset S$ be a prime lying over $\mathfrak p \subset R$. Let $R \to R^ h$ and $S \to S^ h$ be the henselizations of $R_\mathfrak p$ and $S_\mathfrak q$. The local ring map $R^ h \to S^ h$ of Lemma 10.155.6 identifies $S^ h$ with the henselization of $R^ h \otimes _ R S$ at the unique prime lying over $\mathfrak m^ h$ and $\mathfrak q$.

**Proof.**
By Lemma 10.155.7 we see that $R^ h$, resp. $S^ h$ are filtered colimits of étale $R$, resp. $S$-algebras. Hence we see that $R^ h \otimes _ R S$ is a filtered colimit of étale $S$-algebras $A_ i$ (Lemma 10.143.3). By Lemma 10.154.5 we see that $S^ h$ is a filtered colimit of étale $R^ h \otimes _ R S$-algebras. Since moreover $S^ h$ is a henselian local ring with residue field equal to $\kappa (\mathfrak q)$, the statement follows from the uniqueness result of Lemma 10.154.7.
$\square$

Lemma 10.155.9. Let $\varphi : R \to S$ be a local map of local rings. Let $S/\mathfrak m_ S \subset \kappa ^{sep}$ be a separable algebraic closure. Let $S \to S^{sh}$ be the strict henselization of $S$ with respect to $S/\mathfrak m_ S \subset \kappa ^{sep}$. Let $R \to A$ be an étale ring map and let $\mathfrak q$ be a prime of $A$ lying over $\mathfrak m_ R$. Given any commutative diagram

\[ \xymatrix{ \kappa (\mathfrak q) \ar[r]_{\phi } & \kappa ^{sep} \\ R/\mathfrak m_ R \ar[r]^{\varphi } \ar[u] & S/\mathfrak m_ S \ar[u] } \]

there exists a unique morphism of rings $f : A \to S^{sh}$ fitting into the commutative diagram

\[ \xymatrix{ A \ar[r]_ f & S^{sh} \\ R \ar[u] \ar[r]^{\varphi } & S \ar[u] } \]

such that $f^{-1}(\mathfrak m_{S^ h}) = \mathfrak q$ and the induced map $\kappa (\mathfrak q) \to \kappa ^{sep}$ is the given one.

**Proof.**
This is a special case of Lemma 10.153.11.
$\square$

Lemma 10.155.10. Let $R \to S$ be a local map of local rings. Choose separable algebraic closures $R/\mathfrak m_ R \subset \kappa _1^{sep}$ and $S/\mathfrak m_ S \subset \kappa _2^{sep}$. Let $R \to R^{sh}$ and $S \to S^{sh}$ be the corresponding strict henselizations. Given any commutative diagram

\[ \xymatrix{ \kappa _1^{sep} \ar[r]_{\phi } & \kappa _2^{sep} \\ R/\mathfrak m_ R \ar[r]^{\varphi } \ar[u] & S/\mathfrak m_ S \ar[u] } \]

There exists a unique local ring map $R^{sh} \to S^{sh}$ fitting into the commutative diagram

\[ \xymatrix{ R^{sh} \ar[r]_ f & S^{sh} \\ R \ar[u] \ar[r] & S \ar[u] } \]

and inducing $\phi $ on the residue fields of $R^{sh}$ and $S^{sh}$.

**Proof.**
Follows immediately from Lemma 10.154.6.
$\square$

Lemma 10.155.11. Let $R$ be a ring. Let $\mathfrak p \subset R$ be a prime ideal. Let $\kappa (\mathfrak p) \subset \kappa ^{sep}$ be a separable algebraic closure. Consider the category of triples $(S, \mathfrak q, \phi )$ where $R \to S$ is étale, $\mathfrak q$ is a prime lying over $\mathfrak p$, and $\phi : \kappa (\mathfrak q) \to \kappa ^{sep}$ is a $\kappa (\mathfrak p)$-algebra map. This category is filtered and

\[ (R_{\mathfrak p})^{sh} = \mathop{\mathrm{colim}}\nolimits _{(S, \mathfrak q, \phi )} S = \mathop{\mathrm{colim}}\nolimits _{(S, \mathfrak q, \phi )} S_{\mathfrak q} \]

canonically.

**Proof.**
A morphism of triples $(S, \mathfrak q, \phi ) \to (S', \mathfrak q', \phi ')$ is given by an $R$-algebra map $\varphi : S \to S'$ such that $\varphi ^{-1}(\mathfrak q') = \mathfrak q$ and such that $\phi ' \circ \varphi = \phi $. Let us show that the category of pairs is filtered, see Categories, Definition 4.19.1. The category contains the triple $(R, \mathfrak p, \kappa (\mathfrak p) \subset \kappa ^{sep})$ and hence is not empty, which proves part (1) of Categories, Definition 4.19.1. Suppose that $(S, \mathfrak q, \phi )$ and $(S', \mathfrak q', \phi ')$ are two triples. Note that $\mathfrak q$, resp. $\mathfrak q'$ correspond to primes of the fibre rings $S \otimes \kappa (\mathfrak p)$, resp. $S' \otimes \kappa (\mathfrak p)$ with residue fields finite separable over $\kappa (\mathfrak p)$ and $\phi $, resp. $\phi '$ correspond to maps into $\kappa ^{sep}$. Hence this data corresponds to $\kappa (\mathfrak p)$-algebra maps

\[ \phi : S \otimes _ R \kappa (\mathfrak p) \longrightarrow \kappa ^{sep}, \quad \phi ' : S' \otimes _ R \kappa (\mathfrak p) \longrightarrow \kappa ^{sep}. \]

Set $S'' = S \otimes _ R S'$. Combining the maps the above we get a unique $\kappa (\mathfrak p)$-algebra map

\[ \phi '' = \phi \otimes \phi ' : S'' \otimes _ R \kappa (\mathfrak p) \longrightarrow \kappa ^{sep} \]

whose kernel corresponds to a prime $\mathfrak q'' \subset S''$ lying over $\mathfrak q$ and over $\mathfrak q'$, and whose residue field maps via $\phi ''$ to the compositum of $\phi (\kappa (\mathfrak q))$ and $\phi '(\kappa (\mathfrak q'))$ in $\kappa ^{sep}$. The ring map $R \to S''$ is étale by Lemma 10.143.3. Hence $(S'', \mathfrak q'', \phi '')$ is a triple dominating both $(S, \mathfrak q, \phi )$ and $(S', \mathfrak q', \phi ')$. This proves part (2) of Categories, Definition 4.19.1. Next, suppose that $\varphi , \psi : (S, \mathfrak q, \phi ) \to (S', \mathfrak q', \phi ')$ 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

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

Arguing as above (base change of étale maps is étale, composition of étale maps is étale) we see that $S''$ is étale over $R$. The fibre ring of $S''$ over $\mathfrak p$ is

\[ F'' = (F' \otimes _{\varphi , F, \psi } F') \otimes _{F' \otimes _{\kappa (\mathfrak p)} F'} F' \]

where $F', F$ are the fibre rings of $S'$ and $S$. Since $\varphi $ and $\psi $ are morphisms of triples the map $\phi ' : F' \to \kappa ^{sep}$ extends to a map $\phi '' : F'' \to \kappa ^{sep}$ which in turn corresponds to a prime ideal $\mathfrak q'' \subset S''$. The canonical map $S' \to S''$ (using the right most factor for example) is a morphism of triples $(S', \mathfrak q', \phi ') \to (S'', \mathfrak q'', \phi '')$ which equalizes $\varphi $ and $\psi $. This proves part (3) of Categories, Definition 4.19.1. Hence we conclude that the category is filtered.

We still have to show that the colimit $R_{colim}$ of the system is equal to the strict henselization of $R_{\mathfrak p}$ with respect to $\kappa ^{sep}$. To see this note that the system of triples $(S, \mathfrak q, \phi )$ contains as a subsystem the pairs $(S, \mathfrak q)$ of Lemma 10.155.7. Hence $R_{colim}$ contains $R_{\mathfrak p}^ h$ by the result of that lemma. Moreover, it is clear that $R_{\mathfrak p}^ h \subset R_{colim}$ is a directed colimit of étale ring extensions. It follows that $R_{colim}$ is henselian by Lemmas 10.153.4 and 10.154.8. Finally, by Lemma 10.144.3 we see that the residue field of $R_{colim}$ is equal to $\kappa ^{sep}$. Hence we conclude that $R_{colim}$ is strictly henselian and hence equals the strict henselization of $R_{\mathfrak p}$ as desired. Some details omitted.
$\square$

Lemma 10.155.12. Let $R \to S$ be a ring map. Let $\mathfrak q \subset S$ be a prime lying over $\mathfrak p \subset R$. Choose separable algebraic closures $\kappa (\mathfrak p) \subset \kappa _1^{sep}$ and $\kappa (\mathfrak q) \subset \kappa _2^{sep}$. Let $R^{sh}$ and $S^{sh}$ be the corresponding strict henselizations of $R_\mathfrak p$ and $S_\mathfrak q$. Given any commutative diagram

\[ \xymatrix{ \kappa _1^{sep} \ar[r]_{\phi } & \kappa _2^{sep} \\ \kappa (\mathfrak p) \ar[r]^{\varphi } \ar[u] & \kappa (\mathfrak q) \ar[u] } \]

The local ring map $R^{sh} \to S^{sh}$ of Lemma 10.155.10 identifies $S^{sh}$ with the strict henselization of $R^{sh} \otimes _ R S$ at a prime lying over $\mathfrak q$ and the maximal ideal $\mathfrak m^{sh} \subset R^{sh}$.

**Proof.**
The proof is identical to the proof of Lemma 10.155.8 except that it uses Lemma 10.155.11 instead of Lemma 10.155.7.
$\square$

Lemma 10.155.13. Let $R \to S$ be a ring map. Let $\mathfrak q \subset S$ be a prime lying over $\mathfrak p \subset R$ such that $\kappa (\mathfrak p) \to \kappa (\mathfrak q)$ is an isomorphism. Choose a separable algebraic closure $\kappa ^{sep}$ of $\kappa (\mathfrak p) = \kappa (\mathfrak q)$. Then

\[ (S_\mathfrak q)^{sh} = (S_\mathfrak q)^ h \otimes _{(R_\mathfrak p)^ h} (R_\mathfrak p)^{sh} \]

**Proof.**
This follows from the alternative construction of the strict henselization of a local ring in Remark 10.155.4 and the fact that the residue fields are equal. Some details omitted.
$\square$

## Comments (8)

Comment #1726 by Matthieu Romagny on

Comment #1765 by Johan on

Comment #2991 by Peng Du on

Comment #3114 by Johan on

Comment #4167 by fherzig on

Comment #4369 by Johan on

Comment #6223 by eunwoo Heo on

Comment #6224 by eunwoo Heo on