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.
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 61.7). The material will also be useful to prove uniqueness properties of the henselization and strict henselization of a local ring.
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 I be a directed set. Let i \mapsto (R_ i \to A_ i) be a system of arrows of rings over I. Set R = \mathop{\mathrm{colim}}\nolimits R_ i and A = \mathop{\mathrm{colim}}\nolimits A_ i. If each A_ i is a filtered colimit of étale R_ i-algebras, then A is a filtered colimit of étale R-algebras.
Proof. This is true because A = A \otimes _ R R = \mathop{\mathrm{colim}}\nolimits A_ i \otimes _{R_ i} R and hence we can apply Lemma 10.154.3 because R \to A_ i \otimes _{R_ i} R is a filtered colimit of étale ring maps by Lemma 10.154.1. \square
Lemma 10.154.5. 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 10.154.6. Let R \to S be a ring map with S henselian local. Given
an R-algebra A which is a filtered colimit of étale R-algebras,
a prime \mathfrak q of A lying over \mathfrak p = R \cap \mathfrak m_ S,
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 10.154.7. 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 10.154.6. \square
The following lemma is not strictly speaking about colimits of étale ring maps.
Lemma 10.154.8. A filtered colimit of (strictly) henselian local rings along local homomorphisms is (strictly) henselian.
Proof. Categories, Lemma 4.21.5 says that this is really just a question about a colimit of (strictly) 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). It is easy to see that \mathop{\mathrm{colim}}\nolimits \kappa _ i is separably algebraically closed if each \kappa _ i is so; thus it suffices to prove R is henselian if each R_ i is henselian. 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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