## 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. The map

\[ P \longrightarrow P/IP \]

induces a bijection between the sets of isomorphism classes of finite projective $R$-modules and finite projective $R/I$-modules. In particular, any finite projective $R/I$-module is isomorphic to $P/IP$ for some finite projective $R$-module $P$.

**Proof.**
We first prove the final statement. Let $\overline{P}$ be a finite projective $R/I$-module. We can find a finite projective module $P'$ over some $R'$ étale over $R$ with $R/I = R'/IR'$ such that $P'/IP'$ is isomorphic to $\overline{P}$, 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 $P = P' \otimes _{R', \tau } R$ we conclude that $P/IP$ is isomorphic to $\overline{P}$. Of course, this tells us that the map in the statement of the lemma is surjective.

Injectivity. Suppose that $P_1$ and $P_2$ are finite projective $R$-modules such that $P_1/IP_1 \cong P_2/IP_2$ as $R/I$-modules. Since $P_1$ is projective, we can find an $R$-module map $u : P_1 \to P_2$ lifting the given isomorphism. Then $u$ is surjective by Nakayama's lemma (Algebra, Lemma 10.20.1). We similarly find a surjection $v : P_2 \to P_1$. By Algebra, Lemma 10.16.4 the map $v \circ u$ is an isomorphism and we conclude $u$ is an isomorphism.
$\square$

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 10.143.3 and 10.36.13). Now we have $1$-to-$1$ correspondences between

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

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

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 10.143.8 and 10.143.9. 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 10.20.1). 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 10.143.10 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 10.36.5 for example) and the proof is done.
$\square$

Lemma 15.13.3. Let $R$ be a ring and $S$ a smooth $R$-algebra. Assume that $A$ is an $R$-algebra and $(A,I)$ is a henselian pair. Then any $R$-algebra map $S \to A/I$ can be lifted to an $R$-algebra map $S \to A$.

**Proof.**
Let $\tau : S \to A/I$ be an $R$-algebra map. Observe that $S \otimes _ R A$ is a smooth $A$-algebra by Algebra, Lemma 10.137.4. Thus by Lemma 15.9.14 we can lift the induced map $S \otimes _ R A \to A/I$ to an $A$-algebra homorphism $S \otimes _ R A \to A'$ where $A \to A'$ is étale and induces an isomorphism $A/I \to A'/IA'$. Since $(A, I)$ is henselian there is an $A$-algebra map $A' \to A$, see Lemma 15.11.6. The composition $S \to S \otimes _ R A \to A' \to A$ is the desired lift.
$\square$

Lemma 15.13.4. 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

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

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

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

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)

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