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
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