The Stacks project

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. $\square$

Comments (0)

There are also:

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

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Your email address will not be published.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 0BC4. Beware of the difference between the letter 'O' and the digit '0'.



View Lemma 15.3.2 as pdf