Lemma 15.3.5 (0BC6)—The Stacks project

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 10.20.1) $\psi \circ \varphi $ and $\varphi \circ \psi $ are surjective. Hence these maps are isomorphisms (Algebra, Lemma 10.16.4). 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 October 20, 2018 at 14:08

In the statement, replace "radical" with "Jacobson radical" (this is a recurring typo all over the place, keeping in mind that someone may look at a result in isolation without reading lots of surrounding text -- admittedly this isn't truly confusing if one looks at the content of the proof, but nonetheless...)

Comment #3747 by Johan on October 24, 2018 at 09:07

OK, I replaced "radical" by "Jacobson radical" in all places where appropriate and whenever we used the notation $\text{rad}(R)$ I have added text saying this denotes the Jacobson radical. See the corresponding changes here.

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2 comment(s) on Section 15.3: Stably free modules

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