## 15.3 Stably free modules

Here is what seems to be the generally accepted definition.

Definition 15.3.1. Let $R$ be a ring.

Two modules $M$, $N$ over $R$ are said to be *stably isomorphic* if there exist $n, m \geq 0$ such that $M \oplus R^{\oplus m} \cong N \oplus R^{\oplus n}$ as $R$-modules.

A module $M$ is *stably free* if it is stably isomorphic to a free module.

Observe that a stably free module is projective.

Lemma 15.3.2. Let $R$ be a ring. Let $0 \to P' \to P \to P'' \to 0$ be a short exact sequence of finite projective $R$-modules. If $2$ out of $3$ of these modules are stably free, then so is the third.

**Proof.**
Since the modules are projective, the sequence is split. Thus we can choose an isomorphism $P = P' \oplus P''$. If $P' \oplus R^{\oplus n}$ and $P'' \oplus R^{\oplus m}$ are free, then we see that $P \oplus R^{\oplus n + m}$ is free. Suppose that $P'$ and $P$ are stably free, say $P \oplus R^{\oplus n}$ is free and $P' \oplus R^{\oplus m}$ is free. Then

\[ P'' \oplus (P' \oplus R^{\oplus m}) \oplus R^{\oplus n} = (P'' \oplus P') \oplus R^{\oplus m} \oplus R^{\oplus n} = (P \oplus R^{\oplus n}) \oplus R^{\oplus m} \]

is free. Thus $P''$ is stably free. By symmetry we get the last of the three cases.
$\square$

Lemma 15.3.3. 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 $R$). For every finite stably free $R/I$-module $E$ there exists a finite stably free $R$-module $M$ such that $M/IM \cong E$.

**Proof.**
Choose a $n$ and $m$ and an isomorphism $E \oplus (R/I)^{\oplus n} \cong (R/I)^{\oplus m}$. Choose $R$-linear maps $\varphi : R^{\oplus m} \to R^{\oplus n}$ and $\psi : R^{\oplus n} \to R^{\oplus m}$ lifting the projection $(R/I)^{\oplus m} \to (R/I)^{\oplus n}$ and injection $(R/I)^{\oplus n} \to (R/I)^{\oplus m}$. Then $\varphi \circ \psi : R^{\oplus n} \to R^{\oplus n}$ reduces to the identity modulo $I$. Thus the determinant of this map is invertible by our assumption on $I$. Hence $P = \mathop{\mathrm{Ker}}(\varphi )$ is stably free and lifts $E$.
$\square$

Lemma 15.3.4. 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 $R$). Let $M$ be a finite flat $R$-module such that $M/IM$ is a projective $R/I$-module. Then $M$ is a finite projective $R$-module.

**Proof.**
By Algebra, Lemma 10.78.5 we see that $M_\mathfrak p$ is finite free for all prime ideals $\mathfrak p \subset R$. By Algebra, Lemma 10.78.2 it suffices to show that the function $\rho _ M : \mathfrak p \mapsto \dim _{\kappa (\mathfrak p)} M \otimes _ R \kappa (\mathfrak p)$ is locally constant on $\mathop{\mathrm{Spec}}(R)$. Because $M/IM$ is finite projective, this is true on $V(I) \subset \mathop{\mathrm{Spec}}(R)$. Since every closed point of $\mathop{\mathrm{Spec}}(R)$ is in $V(I)$ and since $\rho _ M(\mathfrak p) = \rho _ M(\mathfrak q)$ whenever $\mathfrak p \subset \mathfrak q \subset R$ are prime ideals, we conclude by an elementary argument on topological spaces which we omit.
$\square$

The lift of Lemma 15.3.3 is unique up to isomorphism by the following lemma.

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 #2442 by Peng Du on

Comment #2485 by Johan on