The Stacks project

15.13 Lifting and henselian pairs

In this section we mostly combine results from Sections 15.9 and 15.11.

Lemma 15.13.1. Let $(R, I)$ be a henselian pair. Let $\overline{P}$ be a finite projective $R/I$-module. Then there exists a finite projective $R$-module $P$ such that $P/IP \cong \overline{P}$.

Proof. This follows from the fact that we can lift the finite projective $R/I$-module $\overline{P}$ to a finite projective module $P'$ over some $R'$ étale over $R$ with $R/I = R'/IR'$, see Lemma 15.9.11. Then, since $(R, I)$ is a henselian pair, the étale ring map $R \to R'$ has a section $\tau : R' \to R$ (Lemma 15.11.6). Setting $P = P' \otimes _{R', \tau } R$ finishes the proof. $\square$

Lemma 15.13.2. Let $(A, I)$ be a henselian pair. The functor $B \to B/IB$ determines an equivalence between finite étale $A$-algebras and finite étale $A/I$-algebras.

Proof. Let $B, B'$ be two $A$-algebras finite étale over $A$. Then $B' \to B'' = B \otimes _ A B'$ is finite étale as well (Algebra, Lemmas 10.141.3 and 10.35.13). Now we have $1$-to-$1$ correspondences between

  1. $A$-algebra maps $B \to B'$,

  2. sections of $B' \to B''$, and

  3. idempotents $e$ of $B''$ such that $B' \to B'' \to eB''$ is an isomorphism.

The bijection between (2) and (3) sends $\sigma : B'' \to B'$ to $e$ such that $(1 - e)$ is the idempotent that generates the kernel of $\sigma $ which exists by Algebra, Lemmas 10.141.8 and 10.141.9. There is a similar correspondence between $A/I$-algebra maps $B/IB \to B'/IB'$ and idempotents $\overline{e}$ of $B''/IB''$ such that $B'/IB' \to B''/IB'' \to \overline{e}(B''/IB'')$ is an isomorphism. However every idempotent $\overline{e}$ of $B''/IB''$ lifts uniquely to an idempotent $e$ of $B''$ (Lemma 15.11.6). Moreover, if $B'/IB' \to \overline{e}(B''/IB'')$ is an isomorphism, then $B' \to eB''$ is an isomorphism too by Nakayama's lemma (Algebra, Lemma 10.19.1). In this way we see that the functor is fully faithful.

Essential surjectivity. Let $A/I \to C$ be a finite étale map. By Algebra, Lemma 10.141.10 there exists an étale map $A \to B$ such that $B/IB \cong C$. Let $B'$ be the integral closure of $A$ in $B$. By Lemma 15.11.5 we have $B'/IB' = C \times C'$ for some ring $C'$ and $B'_ g \cong B_ g$ for some $g \in B'$ mapping to $(1, 0) \in C \times C'$. Since idempotents lift (Lemma 15.11.6) we get $B' = B'_1 \times B'_2$ with $C = B'_1/IB'_1$ and $C' = B'_2/IB'_2$. The image of $g$ in $B'_1$ is invertible. Then $B_ g = B'_ g = B'_1 \times (B_2)_ g$ and this implies that $A \to B'_1$ is étale. We conclude that $B'_1$ is finite étale over $A$ (integral étale implies finite étale by Algebra, Lemma 10.35.5 for example) and the proof is done. $\square$

Lemma 15.13.3. Let $A = \mathop{\mathrm{lim}}\nolimits A_ n$ be a limit of an inverse system $(A_ n)$ of rings. Suppose given $A_ n$-modules $M_ n$ and $A_{n + 1}$-module maps $M_{n + 1} \to M_ n$. Assume

  1. the transition maps $A_{n + 1} \to A_ n$ are surjective with locally nilpotent kernels,

  2. $M_1$ is a finite projective $A_1$-module,

  3. $M_ n$ is a finite flat $A_ n$-module, and

  4. the maps induce isomorphisms $M_{n + 1} \otimes _{A_{n + 1}} A_ n \to M_ n$.

Then $M = \mathop{\mathrm{lim}}\nolimits M_ n$ is a finite projective $A$-module and $M \otimes _ A A_ n \to M_ n$ is an isomorphism for all $n$.

Proof. By Lemma 15.11.3 the pair $(A, \mathop{\mathrm{Ker}}(A \to A_1))$ is henselian. By Lemma 15.13.1 we can choose a finite projective $A$-module $P$ and an isomorphism $P \otimes _ A A_1 \to M_1$. Since $P$ is projective, we can successively lift the $A$-module map $P \to M_1$ to $A$-module maps $P \to M_2$, $P \to M_3$, and so on. Thus we obtain a map

\[ P \longrightarrow M \]

Since $P$ is finite projective, we can write $A^{\oplus m} = P \oplus Q$ for some $m \geq 0$ and $A$-module $Q$. Since $A = \mathop{\mathrm{lim}}\nolimits A_ n$ we conclude that $P = \mathop{\mathrm{lim}}\nolimits P \otimes _ A A_ n$. Hence, in order to show that the displayed $A$-module map is an isomorphism, it suffices to show that the maps $P \otimes _ A A_ n \to M_ n$ are isomorphisms. From Lemma 15.3.4 we see that $M_ n$ is a finite projective module. By Lemma 15.3.5 the maps $P \otimes _ A A_ n \to M_ n$ are isomorphisms. $\square$


Comments (2)

Comment #4015 by on

In the proof of the lemma 09ZL, "Moreover, if B′′/IB′′→… is an isomorphism, then B′→eB′′ is an isomorphism too", I think it should be "Moreover, if B'/IB' →… is an isomorphism …".


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 0D49. Beware of the difference between the letter 'O' and the digit '0'.