The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

10.149 Filtered colimits of étale ring maps

This section is a precursor to the section on ind-étale ring maps (Pro-étale Cohomology, Section 56.7). The material will also be useful to prove uniqueness properties of the henselization and strict henselization of a local ring.

Lemma 10.149.1. 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 10.141.3). $\square$

Lemma 10.149.2. 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 10.126.4. 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 10.126.3. By Lemma 10.141.3 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 10.126.4 applies. $\square$

Lemma 10.149.3. 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 10.126.3 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 10.149.4. 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 10.126.3. The factorization $A_ i \to B_ j$ is étale by Lemma 10.141.8. Since $A \to A \otimes _{A_ i} B_ j$ is étale (Lemma 10.141.3) 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 10.149.5. 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 10.148.11. Set $f = \mathop{\mathrm{colim}}\nolimits f_ i$. $\square$

Lemma 10.149.6. 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 10.149.5. $\square$

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

Lemma 10.149.7. 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. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0BSG. Beware of the difference between the letter 'O' and the digit '0'.