## Tag `0BSK`

## 10.150. 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 10.149.6 and the functorial behaviour pointed out in Lemma 10.149.5 while reading this material.

Lemma 10.150.1. 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 10.141.3. 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 10.141.8. 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 we can find a $g \in S$, $g \not \in \mathfrak q$ and $g \in \mathfrak q_i$ for $i = 1, \ldots, r$, see Lemma 10.14.2. 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 10.141.5. 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 $f$ module $\mathfrak q$.

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. $\square$

Lemma 10.150.2. 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 10.150.1. 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 10.150.3. Let $(R, \mathfrak m, \kappa)$ be a local ring.

- The local ring map $R \to R^h$ constructed in Lemma 10.150.1 is called the
henselizationof $R$.- Given a separable algebraic closure $\kappa \subset \kappa^{sep}$ the local ring map $R \to R^{sh}$ constructed in Lemma 10.150.2 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 henselizationof $R$ if it is isomorphic to one of the local ring maps constructed in Lemma 10.150.2

The maps $R \to R^h \to R^{sh}$ are flat local ring homomorphisms. By Lemma 10.149.6 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 15.42.

Remark 10.150.4. We can also construct $R^{sh}$ from $R^h$. Namely, for any finite separable subextension $\kappa \subset \kappa' \subset \kappa^{sep}$ there exists a unique (up to unique isomorphism) finite étale local ring extension $R^h \subset R^h(\kappa')$ whose residue field extension extension reproduces the given extension, see Lemma 10.148.7. Hence we can set $$ R^{sh} = \bigcup\nolimits_{\kappa \subset \kappa' \subset \kappa^{sep}} R^h(\kappa') $$ The arrows in this system, compatible with the arrows on the level of residue fields, exist by Lemma 10.148.7. This will produce a henselian local ring by Lemma 10.149.7 since each of the rings $R^h(\kappa')$ is henselian by Lemma 10.148.4. By construction the residue field extension induced by $R^h \to R^{sh}$ is the field extension $\kappa \subset \kappa^{sep}$. Hence $R^{sh}$ so constructed is strictly henselian. By Lemma 10.149.2 the $R$-algebra $R^{sh}$ is a colimit of étale $R$-algebras. Hence the uniqueness of Lemma 10.149.6 shows that $R^{sh}$ is the strict henselization.

Lemma 10.150.5. 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 10.148.11. $\square$Lemma 10.150.6. 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 10.149.5. $\square$Here is a slightly different construction of the henselization.

Lemma 10.150.7. 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 10.141.3. 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 10.141.8. 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 10.150.1 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 10.141.3. $\square$

Lemma 10.150.8. 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 10.150.6 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 10.150.7 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 10.141.3). By Lemma 10.149.4 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 10.149.6. $\square$Lemma 10.150.9. Let $R \to S$ be a ring map. Let $\mathfrak q$ be a prime of $S$ lying over $\mathfrak p$ in $R$. Assume $R \to S$ is quasi-finite at $\mathfrak q$. The commutative diagram $$ \xymatrix{ R_{\mathfrak p}^h \ar[r] & S_{\mathfrak q}^h \\ R_{\mathfrak p} \ar[u] \ar[r] & S_{\mathfrak q} \ar[u] } $$ of Lemma 10.150.6 identifies $S_{\mathfrak q}^h$ with the localization of $R_{\mathfrak p}^h \otimes_{R_{\mathfrak p}} S_{\mathfrak q}$ at the prime generated by $\mathfrak q$.

Proof.Note that $R_{\mathfrak p}^h \otimes_R S$ is quasi-finite over $R_{\mathfrak p}^h$ at the prime ideal corresponding to $\mathfrak q$, see Lemma 10.121.6. Hence the localization $S'$ of $R_{\mathfrak p}^h \otimes_{R_{\mathfrak p}} S_{\mathfrak q}$ is henselian, see Lemma 10.148.4. As a localization $S'$ is a filtered colimit of étale $R_{\mathfrak p}^h \otimes_{R_{\mathfrak p}} S_{\mathfrak q}$-algebras. By Lemma 10.150.8 we see that $S_\mathfrak q^h$ is the henselization of $R_{\mathfrak p}^h \otimes_{R_{\mathfrak p}} S_{\mathfrak q}$. Thus $S' = S_\mathfrak q^h$ by the uniqueness result of Lemma 10.149.6. $\square$Lemma 10.150.10. Let $R$ be a local ring with henselization $R^h$. Let $I \subset \mathfrak m_R$. Then $R^h/IR^h$ is the henselization of $R/I$.

Proof.This is a special case of Lemma 10.150.9. $\square$Lemma 10.150.11. 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 10.148.11. $\square$Lemma 10.150.12. 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 10.149.5. $\square$Lemma 10.150.13. 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 10.141.3. 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 10.141.8. 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 10.150.7. 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 10.148.4 and 10.149.7. Finally, by Lemma 10.141.15 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 10.150.14. 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 10.150.12 identifies $S^{sh}$ with the strict henselization of $R^{sh} \otimes_R S$ at a prime lying over $\mathfrak m^{sh}$ and $\mathfrak q$.

Proof.The proof is identical to the proof of Lemma 10.150.8 except that it uses Lemma 10.150.13 instead of Lemma 10.150.7. $\square$Lemma 10.150.15. Let $R \to S$ be a ring map. Let $\mathfrak q$ be a prime of $S$ lying over $\mathfrak p$ in $R$. Let $\kappa(\mathfrak q) \subset \kappa^{sep}$ be a separable algebraic closure. Assume $R \to S$ is quasi-finite at $\mathfrak q$. The commutative diagram $$ \xymatrix{ R_{\mathfrak p}^{sh} \ar[r] & S_{\mathfrak q}^{sh} \\ R_{\mathfrak p} \ar[u] \ar[r] & S_{\mathfrak q} \ar[u] } $$ of Lemma 10.150.12 identifies $S_{\mathfrak q}^{sh}$ with a localization of $R_{\mathfrak p}^{sh} \otimes_{R_{\mathfrak p}} S_{\mathfrak q}$.

Proof.The residue field of $R_{\mathfrak p}^{sh}$ is the separable algebraic closure of $\kappa(\mathfrak p)$ in $\kappa^{sep}$. Note that $R_{\mathfrak p}^{sh} \otimes_R S$ is quasi-finite over $R_{\mathfrak p}^{sh}$ at the prime ideal corresponding to $\mathfrak q$, see Lemma 10.121.6. Hence the localization $S'$ of $R_{\mathfrak p}^{sh} \otimes_{R_{\mathfrak p}} S_{\mathfrak q}$ is henselian, see Lemma 10.148.4. Note that the residue field of $S'$ is $\kappa^{sep}$ since it contains both the separable algebraic closure of $\kappa(\mathfrak p)$ and $\kappa(\mathfrak q)$. Furthermore, as a localization $S'$ is a filtered colimit of étale $R_{\mathfrak p}^{sh} \otimes_{R_{\mathfrak p}} S_{\mathfrak q}$-algebras. By Lemma 10.150.14 we see that $S_{\mathfrak q}^{sh}$ is a strict henselization of $R_{\mathfrak p}^{sh} \otimes_{R_{\mathfrak p}} S_{\mathfrak q}$. Thus $S' = S_\mathfrak q^h$ by the uniqueness result of Lemma 10.149.6. $\square$Lemma 10.150.16. Let $R$ be a local ring with strict henselization $R^{sh}$. Let $I \subset \mathfrak m_R$. Then $R^{sh}/IR^{sh}$ is a strict henselization of $R/I$.

Proof.This is a special case of Lemma 10.150.15. $\square$Lemma 10.150.17. 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 10.150.4 and the fact that the residue fields are equal. Some details omitted. $\square$

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\section{Henselization and strict henselization}
\label{section-henselization}
\noindent
In this section we construct the henselization. We encourage the reader
to keep in mind the uniqueness already proved in
Lemma \ref{lemma-uniqueness-henselian}
and the functorial behaviour pointed out in
Lemma \ref{lemma-map-into-henselian-colimit}
while reading this material.
\begin{lemma}
\label{lemma-henselization}
Let $(R, \mathfrak m, \kappa)$ be a local ring. There exists a
local ring map $R \to R^h$ with the following properties
\begin{enumerate}
\item $R^h$ is henselian,
\item $R^h$ is a filtered colimit of \'etale $R$-algebras,
\item $\mathfrak m R^h$ is the
maximal ideal of $R^h$, and
\item $\kappa = R^h/\mathfrak m R^h$.
\end{enumerate}
\end{lemma}
\begin{proof}
Consider the category of pairs $(S, \mathfrak q)$ where $R \to S$ is an
\'etale 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
$$
R^h = \colim_{(S, \mathfrak q)} S.
$$
Let us show that the category of pairs is filtered, see
Categories, Definition \ref{categories-definition-directed}.
The category contains the pair $(R, \mathfrak m)$ and hence is not empty,
which proves part (1) of
Categories, Definition \ref{categories-definition-directed}.
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 \'etale by
Lemma \ref{lemma-etale}.
This proves part (2) of
Categories, Definition \ref{categories-definition-directed}.
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 \'etale ring maps by
Lemma \ref{lemma-map-between-etale}.
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 \'etale maps is \'etale, composition of
\'etale maps is \'etale) we see that $S''$ is \'etale 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 \ref{categories-definition-directed}.
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 $\varphi(f) = \varphi'(f')$.
\medskip\noindent
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 we can find a $g \in S$, $g \not \in \mathfrak q$ and
$g \in \mathfrak q_i$ for $i = 1, \ldots, r$, see
Lemma \ref{lemma-silly}. 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 \'etale, we have
$\mathfrak mS_{\mathfrak q} = \mathfrak qS_{\mathfrak q}$, see
Lemma \ref{lemma-etale-at-prime}.
Hence we actually get that $\mathfrak mS = \mathfrak q$.
\medskip\noindent
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 $f$ module $\mathfrak q$.
\medskip\noindent
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 \'etale. Moreover, $\kappa = \kappa(\mathfrak q')$.
Pick $g \in S'$, $g \not \in \mathfrak q'$ such that
$S'' = S'_g$ is \'etale 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.
\end{proof}
\begin{lemma}
\label{lemma-strict-henselization}
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
\begin{enumerate}
\item the map $R^h \to R^{sh}$ is local
\item $R^{sh}$ is strictly henselian,
\item $R^{sh}$ is a filtered colimit of \'etale $R$-algebras,
\item $\mathfrak m R^{sh}$ is the
maximal ideal of $R^{sh}$, and
\item $\kappa^{sep} = R^{sh}/\mathfrak m R^{sh}$.
\end{enumerate}
\end{lemma}
\begin{proof}
This is proved by exactly the same proof as used for
Lemma \ref{lemma-henselization}.
The only difference is that, instead of pairs, one uses triples
$(S, \mathfrak q, \alpha)$ where $R \to S$ \'etale,
$\mathfrak q$ is a prime of $S$ lying over $\mathfrak m$, and
$\alpha : \kappa(\mathfrak q) \to \kappa^{sep}$ is an embedding
of extensions of $\kappa$.
\end{proof}
\begin{definition}
\label{definition-henselization}
Let $(R, \mathfrak m, \kappa)$ be a local ring.
\begin{enumerate}
\item The local ring map $R \to R^h$ constructed in
Lemma \ref{lemma-henselization}
is called the {\it henselization} of $R$.
\item Given a separable algebraic closure $\kappa \subset \kappa^{sep}$
the local ring map $R \to R^{sh}$ constructed in
Lemma \ref{lemma-strict-henselization}
is called the
{\it strict henselization of $R$ with respect to
$\kappa \subset \kappa^{sep}$}.
\item A local ring map $R \to R^{sh}$ is called a {\it strict henselization}
of $R$ if it is isomorphic to one of the local ring maps constructed in
Lemma \ref{lemma-strict-henselization}
\end{enumerate}
\end{definition}
\noindent
The maps $R \to R^h \to R^{sh}$ are flat local ring homomorphisms.
By Lemma \ref{lemma-uniqueness-henselian} 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 \'etale $R$-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 \ref{more-algebra-section-permanence-henselization}.
\begin{remark}
\label{remark-construct-sh-from-h}
We can also construct $R^{sh}$ from $R^h$. Namely, for any finite separable
subextension $\kappa \subset \kappa' \subset \kappa^{sep}$
there exists a unique (up to unique isomorphism) finite \'etale local
ring extension $R^h \subset R^h(\kappa')$
whose residue field extension extension reproduces the given extension, see
Lemma \ref{lemma-henselian-cat-finite-etale}.
Hence we can set
$$
R^{sh} =
\bigcup\nolimits_{\kappa \subset \kappa' \subset \kappa^{sep}}
R^h(\kappa')
$$
The arrows in this system, compatible with the arrows on the level
of residue fields, exist by
Lemma \ref{lemma-henselian-cat-finite-etale}.
This will produce a henselian local ring by
Lemma \ref{lemma-colimit-henselian}
since each of the rings
$R^h(\kappa')$ is henselian by
Lemma \ref{lemma-finite-over-henselian}.
By construction the residue field extension induced by
$R^h \to R^{sh}$ is the field extension $\kappa \subset \kappa^{sep}$.
Hence $R^{sh}$ so constructed is strictly henselian.
By Lemma \ref{lemma-composition-colimit-etale} the $R$-algebra
$R^{sh}$ is a colimit of \'etale $R$-algebras. Hence the uniqueness
of Lemma \ref{lemma-uniqueness-henselian} shows that $R^{sh}$
is the strict henselization.
\end{remark}
\begin{lemma}
\label{lemma-henselian-functorial-prepare}
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 \'etale 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 $f^{-1}(\mathfrak m_{S^h}) = \mathfrak q$.
\end{lemma}
\begin{proof}
This is a special case of Lemma \ref{lemma-map-into-henselian}.
\end{proof}
\begin{lemma}
\label{lemma-henselian-functorial}
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
$$
\xymatrix{
R^h \ar[r]_f & S^h \\
R \ar[u] \ar[r] & S \ar[u]
}
$$
\end{lemma}
\begin{proof}
Follows immediately from Lemma \ref{lemma-map-into-henselian-colimit}.
\end{proof}
\noindent
Here is a slightly different construction of the henselization.
\begin{lemma}
\label{lemma-henselization-different}
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 \'etale and $\mathfrak q$ is a prime lying over $\mathfrak p$
such that $\kappa(\mathfrak p) = \kappa(\mathfrak q)$.
This category is filtered and
$$
(R_{\mathfrak p})^h = \colim_{(S, \mathfrak q)} S
= \colim_{(S, \mathfrak q)} S_{\mathfrak q}
$$
canonically.
\end{lemma}
\begin{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 \ref{categories-definition-directed}.
The category contains the pair $(R, \mathfrak p)$ and hence is not empty,
which proves part (1) of
Categories, Definition \ref{categories-definition-directed}.
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 \'etale by
Lemma \ref{lemma-etale}.
This proves part (2) of
Categories, Definition \ref{categories-definition-directed}.
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 \'etale ring maps by
Lemma \ref{lemma-map-between-etale}. Consider
$$
S'' = (S' \otimes_{\varphi, S, \psi} S')
\otimes_{S' \otimes_R S'} S'
$$
Arguing as above (base change of \'etale maps is \'etale, composition of
\'etale maps is \'etale) we see that $S''$ is \'etale 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 \ref{categories-definition-directed}.
Hence we conclude that the category is filtered.
\medskip\noindent
Recall that in the proof of
Lemma \ref{lemma-henselization}
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
$$
S_{\mathfrak p} = \colim_{f \in R, f \not \in \mathfrak p} S_f.
$$
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 \ref{lemma-etale}.
\end{proof}
\begin{lemma}
\label{lemma-henselian-functorial-improve}
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 \ref{lemma-henselian-functorial} identifies $S^h$
with the henselization of $R^h \otimes_R S$ at the unique prime
lying over $\mathfrak m^h$ and $\mathfrak q$.
\end{lemma}
\begin{proof}
By Lemma \ref{lemma-henselization-different} we see that $R^h$, resp.\ $S^h$
are filtered colimits of \'etale $R$, resp.\ $S$-algebras.
Hence we see that $R^h \otimes_R S$ is a filtered colimit of
\'etale $S$-algebras $A_i$ (Lemma \ref{lemma-etale}). By
Lemma \ref{lemma-colimits-of-etale} we see that $S^h$ is a
filtered colimit of \'etale $R^h \otimes_R S$-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 \ref{lemma-uniqueness-henselian}.
\end{proof}
\begin{lemma}
\label{lemma-quasi-finite-henselization}
Let $R \to S$ be a ring map.
Let $\mathfrak q$ be a prime of $S$ lying over $\mathfrak p$ in $R$.
Assume $R \to S$ is quasi-finite at $\mathfrak q$.
The commutative diagram
$$
\xymatrix{
R_{\mathfrak p}^h \ar[r] & S_{\mathfrak q}^h \\
R_{\mathfrak p} \ar[u] \ar[r] & S_{\mathfrak q} \ar[u]
}
$$
of
Lemma \ref{lemma-henselian-functorial}
identifies $S_{\mathfrak q}^h$ with the localization of
$R_{\mathfrak p}^h \otimes_{R_{\mathfrak p}} S_{\mathfrak q}$
at the prime generated by $\mathfrak q$.
\end{lemma}
\begin{proof}
Note that $R_{\mathfrak p}^h \otimes_R S$ is quasi-finite over
$R_{\mathfrak p}^h$ at the prime ideal corresponding to $\mathfrak q$, see
Lemma \ref{lemma-four-rings}. Hence the localization $S'$ of
$R_{\mathfrak p}^h \otimes_{R_{\mathfrak p}} S_{\mathfrak q}$ is henselian, see
Lemma \ref{lemma-finite-over-henselian}. As a localization $S'$ is a filtered
colimit of \'etale
$R_{\mathfrak p}^h \otimes_{R_{\mathfrak p}} S_{\mathfrak q}$-algebras.
By Lemma \ref{lemma-henselian-functorial-improve} we see that
$S_\mathfrak q^h$ is the henselization of
$R_{\mathfrak p}^h \otimes_{R_{\mathfrak p}} S_{\mathfrak q}$.
Thus $S' = S_\mathfrak q^h$ by the uniqueness
result of Lemma \ref{lemma-uniqueness-henselian}.
\end{proof}
\begin{lemma}
\label{lemma-quotient-henselization}
\begin{slogan}
Henselization is compatible with quotients.
\end{slogan}
Let $R$ be a local ring with henselization $R^h$.
Let $I \subset \mathfrak m_R$.
Then $R^h/IR^h$ is the henselization of $R/I$.
\end{lemma}
\begin{proof}
This is a special case of
Lemma \ref{lemma-quasi-finite-henselization}.
\end{proof}
\begin{lemma}
\label{lemma-strictly-henselian-functorial-prepare}
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 \'etale 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.
\end{lemma}
\begin{proof}
This is a special case of Lemma \ref{lemma-map-into-henselian}.
\end{proof}
\begin{lemma}
\label{lemma-strictly-henselian-functorial}
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 $S^{sh}$.
\end{lemma}
\begin{proof}
Follows immediately from Lemma \ref{lemma-map-into-henselian-colimit}.
\end{proof}
\begin{lemma}
\label{lemma-strict-henselization-different}
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 \'etale, $\mathfrak q$ 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
$$
(R_{\mathfrak p})^{sh} =
\colim_{(S, \mathfrak q, \phi)} S =
\colim_{(S, \mathfrak q, \phi)} S_{\mathfrak q}
$$
canonically.
\end{lemma}
\begin{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 \ref{categories-definition-directed}.
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 \ref{categories-definition-directed}.
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 \'etale by
Lemma \ref{lemma-etale}.
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 \ref{categories-definition-directed}.
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 \'etale ring maps by
Lemma \ref{lemma-map-between-etale}.
Consider
$$
S'' = (S' \otimes_{\varphi, S, \psi} S')
\otimes_{S' \otimes_R S'} S'
$$
Arguing as above (base change of \'etale maps is \'etale, composition of
\'etale maps is \'etale) we see that $S''$ is \'etale 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 \ref{categories-definition-directed}.
Hence we conclude that the category is filtered.
\medskip\noindent
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 \ref{lemma-henselization-different}.
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 \'etale ring extensions.
It follows that $R_{colim}$ is henselian by
Lemmas \ref{lemma-finite-over-henselian} and
\ref{lemma-colimit-henselian}.
Finally, by
Lemma \ref{lemma-make-etale-map-prescribed-residue-field}
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.
\end{proof}
\begin{lemma}
\label{lemma-strictly-henselian-functorial-improve}
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 \ref{lemma-strictly-henselian-functorial} identifies $S^{sh}$
with the strict henselization of $R^{sh} \otimes_R S$ at a prime
lying over $\mathfrak m^{sh}$ and $\mathfrak q$.
\end{lemma}
\begin{proof}
The proof is identical to the proof of
Lemma \ref{lemma-henselian-functorial-improve}
except that it uses
Lemma \ref{lemma-strict-henselization-different}
instead of
Lemma \ref{lemma-henselization-different}.
\end{proof}
\begin{lemma}
\label{lemma-quasi-finite-strict-henselization}
Let $R \to S$ be a ring map.
Let $\mathfrak q$ be a prime of $S$ lying over $\mathfrak p$ in $R$.
Let $\kappa(\mathfrak q) \subset \kappa^{sep}$ be a separable
algebraic closure. Assume $R \to S$ is quasi-finite at $\mathfrak q$.
The commutative diagram
$$
\xymatrix{
R_{\mathfrak p}^{sh} \ar[r] & S_{\mathfrak q}^{sh} \\
R_{\mathfrak p} \ar[u] \ar[r] & S_{\mathfrak q} \ar[u]
}
$$
of
Lemma \ref{lemma-strictly-henselian-functorial}
identifies $S_{\mathfrak q}^{sh}$ with a localization of
$R_{\mathfrak p}^{sh} \otimes_{R_{\mathfrak p}} S_{\mathfrak q}$.
\end{lemma}
\begin{proof}
The residue field of $R_{\mathfrak p}^{sh}$ is the separable
algebraic closure of $\kappa(\mathfrak p)$ in $\kappa^{sep}$.
Note that $R_{\mathfrak p}^{sh} \otimes_R S$ is quasi-finite over
$R_{\mathfrak p}^{sh}$ at the prime ideal corresponding to $\mathfrak q$, see
Lemma \ref{lemma-four-rings}. Hence the localization $S'$ of
$R_{\mathfrak p}^{sh} \otimes_{R_{\mathfrak p}} S_{\mathfrak q}$
is henselian, see
Lemma \ref{lemma-finite-over-henselian}.
Note that the residue field of $S'$ is $\kappa^{sep}$ since it
contains both the separable algebraic closure of
$\kappa(\mathfrak p)$ and $\kappa(\mathfrak q)$.
Furthermore, as a localization $S'$ is a filtered colimit of \'etale
$R_{\mathfrak p}^{sh} \otimes_{R_{\mathfrak p}} S_{\mathfrak q}$-algebras.
By Lemma \ref{lemma-strictly-henselian-functorial-improve}
we see that $S_{\mathfrak q}^{sh}$ is a strict henselization of
$R_{\mathfrak p}^{sh} \otimes_{R_{\mathfrak p}} S_{\mathfrak q}$.
Thus $S' = S_\mathfrak q^h$ by the uniqueness
result of Lemma \ref{lemma-uniqueness-henselian}.
\end{proof}
\begin{lemma}
\label{lemma-quotient-strict-henselization}
Let $R$ be a local ring with strict henselization $R^{sh}$.
Let $I \subset \mathfrak m_R$.
Then $R^{sh}/IR^{sh}$ is a strict henselization of $R/I$.
\end{lemma}
\begin{proof}
This is a special case of
Lemma \ref{lemma-quasi-finite-strict-henselization}.
\end{proof}
\begin{lemma}
\label{lemma-sh-from-h-map}
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
$$
(S_\mathfrak q)^{sh} =
(S_\mathfrak q)^h \otimes_{(R_\mathfrak p)^h} (R_\mathfrak p)^{sh}
$$
\end{lemma}
\begin{proof}
This follows from the alternative construction of the strict henselization
of a local ring in Remark \ref{remark-construct-sh-from-h} and the
fact that the residue fields are equal. Some details omitted.
\end{proof}
```

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