Lemma 15.3.5. Let $R$ be a ring. Let $I \subset R$ be an ideal. Assume that every element of $1 + I$ is a unit (in other words $I$ is contained in the Jacobson radical of $R$). If $P$ and $P'$ are finite projective $R$-modules, then

if $\varphi : P \to P'$ is an $R$-module map inducing an isomorphism $\overline{\varphi } : P/IP \to P'/IP'$, then $\varphi $ is an isomorphism,

if $P/IP \cong P'/IP'$, then $P \cong P'$.

**Proof.**
Proof of (1). As $P'$ is projective as an $R$-module we may choose a lift $\psi : P' \to P$ of the map $P' \to P'/IP' \xrightarrow {\overline{\varphi }^{-1}} P/IP$. By Nakayama's lemma (Algebra, Lemma 10.19.1) $\psi \circ \varphi $ and $\varphi \circ \psi $ are surjective. Hence these maps are isomorphisms (Algebra, Lemma 10.15.4). Thus $\varphi $ is an isomorphism.

Proof of (2). Choose an isomorphism $P/IP \cong P'/IP'$. Since $P$ is projective we can choose a lift $\varphi : P \to P'$ of the map $P \to P/IP \to P'/IP'$. Then $\varphi $ is an isomorphism by (1).
$\square$

## Comments (2)

Comment #3652 by Brian Conrad on

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