Lemma 17.14.6. Let $(X, \mathcal{O}_ X)$ be a ringed space. If all stalks $\mathcal{O}_{X, x}$ are local rings, then any direct summand of a finite locally free $\mathcal{O}_ X$-module is finite locally free.

Proof. Assume $\mathcal{F}$ is a direct summand of the finite locally free $\mathcal{O}_ X$-module $\mathcal{H}$. Let $x \in X$ be a point. Then $\mathcal{H}_ x$ is a finite free $\mathcal{O}_{X, x}$-module. Because $\mathcal{O}_{X, x}$ is local, we see that $\mathcal{F}_ x \cong \mathcal{O}_{X, x}^{\oplus r}$ for some $r$, see Algebra, Lemma 10.78.2. By Lemma 17.11.6 we see that $\mathcal{F}$ is free of rank $r$ in an open neighbourhood of $x$. (Note that $\mathcal{F}$ is of finite presentation as a summand of $\mathcal{H}$.) $\square$

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Your email address will not be published.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