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. $\square$

Comment #7894 by Torsten Wedhorn on

One could more precisely show that $P \mapsto P/IP$ induces a bijection between isomorphism classes of finite projective $R$-modules and isomorphism classes of finite projective $R/I$-modules: The current formulation shows the surjectivity of the map. To show the injectivity, let $P$, $P'$ be finite projective $R$-modules such that $P/IP \cong P'/IP'$. Hence we find an $R$-linear map $P \to P'/IP'$. As $P$ is projective, it lifts to an $R$-linear map $u\colon P \to P'$. As $u$ is an isomorphism modulo $I$, it is surjective by Nakayama's lemma. The ranks of the finite projective $R$-modules $P$ and $P'$ are equal in every open neighborhood of $V(I)$. As $I$ is contained in the radical of $R$, the only open neighborhood of $V(I)$ is $\Spec A$. Hence $u$ is a surjective map of finite projective modules of the same rank. Hence it is an isomorphism.

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