Lemma 17.11.6. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{F}$ be a finitely presented $\mathcal{O}_ X$-module. Let $x \in X$ such that $\mathcal{F}_ x \cong \mathcal{O}_{X, x}^{\oplus r}$. Then there exists an open neighbourhood $U$ of $x$ such that $\mathcal{F}|_ U \cong \mathcal{O}_ U^{\oplus r}$.

Proof. Choose $s_1, \ldots , s_ r \in \mathcal{F}_ x$ mapping to a basis of $\mathcal{O}_{X, x}^{\oplus r}$ by the isomorphism. Choose an open neighbourhood $U$ of $x$ such that $s_ i$ lifts to $s_ i \in \mathcal{F}(U)$. After shrinking $U$ we see that the induced map $\psi : \mathcal{O}_ U^{\oplus r} \to \mathcal{F}|_ U$ is surjective (Lemma 17.9.4). By Lemma 17.11.3 we see that $\mathop{\mathrm{Ker}}(\psi )$ is of finite type. Then $\mathop{\mathrm{Ker}}(\psi )_ x = 0$ implies that $\mathop{\mathrm{Ker}}(\psi )$ becomes zero after shrinking $U$ once more (Lemma 17.9.5). $\square$

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Your email address will not be published.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 0B8J. Beware of the difference between the letter 'O' and the digit '0'.