The Stacks project

Lemma 17.25.4. Let $(X, \mathcal{O}_ X)$ be a ringed space. Any locally free $\mathcal{O}_ X$-module of rank $1$ is invertible. If all stalks $\mathcal{O}_{X, x}$ are local rings, then the converse holds as well (but in general this is not the case).

Proof. The parenthetical statement follows by considering a one point space $X$ with sheaf of rings $\mathcal{O}_ X$ given by a ring $R$. Then invertible $\mathcal{O}_ X$-modules correspond to invertible $R$-modules, hence as soon as $\mathop{\mathrm{Pic}}\nolimits (R)$ is not the trivial group, then we get an example.

Assume $\mathcal{L}$ is locally free of rank $1$ and consider the evaluation map

\[ \mathcal{L} \otimes _{\mathcal{O}_ X} \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{L}, \mathcal{O}_ X) \longrightarrow \mathcal{O}_ X \]

Looking over an open covering trivialization $\mathcal{L}$, we see that this map is an isomorphism. Hence $\mathcal{L}$ is invertible by Lemma 17.25.2.

Assume all stalks $\mathcal{O}_{X, x}$ are local rings and $\mathcal{L}$ invertible. In the proof of Lemma 17.25.2 we have seen that $\mathcal{L}_ x$ is an invertible $\mathcal{O}_{X, x}$-module for all $x \in X$. Since $\mathcal{O}_{X, x}$ is local, we see that $\mathcal{L}_ x \cong \mathcal{O}_{X, x}$ (More on Algebra, Section 15.117). Since $\mathcal{L}$ is of finite presentation by Lemma 17.25.2 we conclude that $\mathcal{L}$ is locally free of rank $1$ by Lemma 17.11.6. $\square$

Comments (0)

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Your email address will not be published.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 0B8M. Beware of the difference between the letter 'O' and the digit '0'.