The Stacks project

15.3 Stably free modules

Here is what seems to be the generally accepted definition.

Definition 15.3.1. Let $R$ be a ring.

  1. 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.

  2. 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

  1. 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,

  2. 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

In Definition 15.3.1. (1), "if there exist n,m≥0" should be "if there exist m≥0" as there is no 'n'.

Comment #2485 by on

OK, the mistake was that one of the direct sums should be and the other should be . Thanks for pointing this out. The fix is here.

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0BC2. Beware of the difference between the letter 'O' and the digit '0'.