## 10.154 Filtered colimits of étale ring maps

This section is a precursor to the section on ind-étale ring maps (Pro-étale Cohomology, Section 60.7). The material will also be useful to prove uniqueness properties of the henselization and strict henselization of a local ring.

Lemma 10.154.1. Let $R \to A$ and $R \to R'$ be ring maps. If $A$ is a filtered colimit of étale ring maps, then so is $R' \to R' \otimes _ R A$.

Proof. This is true because colimits commute with tensor products and étale ring maps are preserved under base change (Lemma 10.143.3). $\square$

Lemma 10.154.2. Let $A \to B \to C$ be ring maps. If $A \to B$ is a filtered colimit of étale ring maps and $B \to C$ is a filtered colimit of étale ring maps, then $A \to C$ is a filtered colimit of étale ring maps.

Proof. We will use the criterion of Lemma 10.127.4. Let $A \to P \to C$ be a factorization of $A \to C$ with $P$ of finite presentation over $A$. Write $B = \mathop{\mathrm{colim}}\nolimits _{i \in I} B_ i$ where $I$ is a directed set and where $B_ i$ is an étale $A$-algebra. Write $C = \mathop{\mathrm{colim}}\nolimits _{j \in J} C_ j$ where $J$ is a directed set and where $C_ j$ is an étale $B$-algebra. We can factor $P \to C$ as $P \to C_ j \to C$ for some $j$ by Lemma 10.127.3. By Lemma 10.143.3 we can find an $i \in I$ and an étale ring map $B_ i \to C'_ j$ such that $C_ j = B \otimes _{B_ i} C'_ j$. Then $C_ j = \mathop{\mathrm{colim}}\nolimits _{i' \geq i} B_{i'} \otimes _{B_ i} C'_ j$ and again we see that $P \to C_ j$ factors as $P \to B_{i'} \otimes _{B_ i} C'_ j \to C$. As $A \to C' = B_{i'} \otimes _{B_ i} C'_ j$ is étale as compositions and tensor products of étale ring maps are étale. Hence we have factored $P \to C$ as $P \to C' \to C$ with $C'$ étale over $A$ and the criterion of Lemma 10.127.4 applies. $\square$

Lemma 10.154.3. Let $R$ be a ring. Let $A = \mathop{\mathrm{colim}}\nolimits A_ i$ be a filtered colimit of $R$-algebras such that each $A_ i$ is a filtered colimit of étale $R$-algebras. Then $A$ is a filtered colimit of étale $R$-algebras.

Proof. Write $A_ i = \mathop{\mathrm{colim}}\nolimits _{j \in J_ i} A_ j$ where $J_ i$ is a directed set and $A_ j$ is an étale $R$-algebra. For each $i \leq i'$ and $j \in J_ i$ there exists an $j' \in J_{i'}$ and an $R$-algebra map $\varphi _{jj'} : A_ j \to A_{j'}$ making the diagram

$\xymatrix{ A_ i \ar[r] & A_{i'} \\ A_ j \ar[u] \ar[r]^{\varphi _{jj'}} & A_{j'} \ar[u] }$

commute. This is true because $R \to A_ j$ is of finite presentation so that Lemma 10.127.3 applies. Let $\mathcal{J}$ be the category with objects $\coprod _{i \in I} J_ i$ and morphisms triples $(j, j', \varphi _{jj'})$ as above (and obvious composition law). Then $\mathcal{J}$ is a filtered category and $A = \mathop{\mathrm{colim}}\nolimits _\mathcal {J} A_ j$. Details omitted. $\square$

Lemma 10.154.4. Let $R$ be a ring. Let $A \to B$ be an $R$-algebra homomorphism. If $A$ and $B$ are filtered colimits of étale $R$-algebras, then $B$ is a filtered colimit of étale $A$-algebras.

Proof. Write $A = \mathop{\mathrm{colim}}\nolimits A_ i$ and $B = \mathop{\mathrm{colim}}\nolimits B_ j$ as filtered colimits with $A_ i$ and $B_ j$ étale over $R$. For each $i$ we can find a $j$ such that $A_ i \to B$ factors through $B_ j$, see Lemma 10.127.3. The factorization $A_ i \to B_ j$ is étale by Lemma 10.143.8. Since $A \to A \otimes _{A_ i} B_ j$ is étale (Lemma 10.143.3) it suffices to prove that $B = \mathop{\mathrm{colim}}\nolimits A \otimes _{A_ i} B_ j$ where the colimit is over pairs $(i, j)$ and factorizations $A_ i \to B_ j \to B$ of $A_ i \to B$ (this is a directed system; details omitted). This is clear because colimits commute with tensor products and hence $\mathop{\mathrm{colim}}\nolimits A \otimes _{A_ i} B_ j = A \otimes _ A B = B$. $\square$

Lemma 10.154.5. Let $R \to S$ be a ring map with $S$ henselian local. Given

1. an $R$-algebra $A$ which is a filtered colimit of étale $R$-algebras,

2. a prime $\mathfrak q$ of $A$ lying over $\mathfrak p = R \cap \mathfrak m_ S$,

3. a $\kappa (\mathfrak p)$-algebra map $\tau : \kappa (\mathfrak q) \to S/\mathfrak m_ S$,

then there exists a unique homomorphism of $R$-algebras $f : A \to S$ such that $\mathfrak q = f^{-1}(\mathfrak m_ S)$ and $f \bmod \mathfrak q = \tau$.

Proof. Write $A = \mathop{\mathrm{colim}}\nolimits A_ i$ as a filtered colimit of étale $R$-algebras. Set $\mathfrak q_ i = A_ i \cap \mathfrak q$. We obtain $f_ i : A_ i \to S$ by applying Lemma 10.153.11. Set $f = \mathop{\mathrm{colim}}\nolimits f_ i$. $\square$

Lemma 10.154.6. Let $R$ be a ring. Given a commutative diagram of ring maps

$\xymatrix{ S \ar[r] & K \\ R \ar[u] \ar[r] & S' \ar[u] }$

where $S$, $S'$ are henselian local, $S$, $S'$ are filtered colimits of étale $R$-algebras, $K$ is a field and the arrows $S \to K$ and $S' \to K$ identify $K$ with the residue field of both $S$ and $S'$. Then there exists an unique $R$-algebra isomorphism $S \to S'$ compatible with the maps to $K$.

Proof. Follows immediately from Lemma 10.154.5. $\square$

The following lemma is not strictly speaking about colimits of étale ring maps.

Lemma 10.154.7. A filtered colimit of henselian local rings along local homomorphisms is henselian.

Proof. Categories, Lemma 4.21.5 says that this is really just a question about a colimit of henselian local rings over a directed set. Let $(R_ i, \varphi _{ii'})$ be such a system with each $\varphi _{ii'}$ local. Then $R = \mathop{\mathrm{colim}}\nolimits _ i R_ i$ is local, and its residue field $\kappa$ is $\mathop{\mathrm{colim}}\nolimits \kappa _ i$ (argument omitted). Suppose that $f \in R[T]$ is monic and that $a_0 \in \kappa$ is a simple root of $\overline{f}$. Then for some large enough $i$ there exists an $f_ i \in R_ i[T]$ mapping to $f$ and an $a_{0, i} \in \kappa _ i$ mapping to $a_0$. Since $\overline{f_ i}(a_{0, i}) \in \kappa _ i$, resp. $\overline{f_ i'}(a_{0, i}) \in \kappa _ i$ maps to $0 = \overline{f}(a_0) \in \kappa$, resp. $0 \not= \overline{f'}(a_0) \in \kappa$ we conclude that $a_{0, i}$ is a simple root of $\overline{f_ i}$. As $R_ i$ is henselian we can find $a_ i \in R_ i$ such that $f_ i(a_ i) = 0$ and $a_{0, i} = \overline{a_ i}$. Then the image $a \in R$ of $a_ i$ is the desired solution. Thus $R$ is henselian. $\square$

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