Lemma 18.39.7. Let $(\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O})$ be a ringed topos. Any locally free $\mathcal{O}$-module of rank $1$ is invertible. If $(\mathcal{C}, \mathcal{O})$ is locally ringed, then the converse holds as well (but in general this is not the case).

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

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

Given any object $U$ of $\mathcal{C}$ and restricting to the members of a covering trivializing $\mathcal{L}$, we see that this map is an isomorphism (details omitted). Hence $\mathcal{L}$ is invertible by Lemma 18.31.2.

Assume $(\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O})$ is locally ringed. Let $U$ be an object of $\mathcal{C}$. In the proof of Lemma 18.31.2 we have seen that there exists a covering $\{ U_ i \to U\}$ such that $\mathcal{L}|_{\mathcal{C}/U_ i}$ is a direct summand of a finite free $\mathcal{O}_{U_ i}$-module. After replacing $U$ by $U_ i$, let $p : \mathcal{O}_ U^{\oplus r} \to \mathcal{O}_ U^{\oplus r}$ be a projector whose image is isomorphic to $\mathcal{L}|_{\mathcal{C}/U}$. Then $p$ corresponds to a matrix

$P = (p_{ij}) \in \text{Mat}(r \times r, \mathcal{O}(U))$

which is a projector: $P^2 = P$. Set $A = \mathcal{O}(U)$ so that $P \in \text{Mat}(r \times r, A)$. By Algebra, Lemma 10.77.2 the image of $P$ is a finite locally free module $M$ over $A$. Hence there are $f_1, \ldots , f_ t \in A$ generating the unit ideal, such that $M_{f_ i}$ is finite free. By Lemma 18.39.1 after replacing $U$ by the members of an open covering, we may assume that $M$ is free. This means that $\mathcal{L}|_ U$ is free (details omitted). Of course, since $\mathcal{L}$ is invertible, this is only possible if the rank of $\mathcal{L}|_ U$ is $1$ and the proof is complete. $\square$

## 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Your email address will not be published.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 0B8Q. Beware of the difference between the letter 'O' and the digit '0'.