The Stacks project

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

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 18.32.2.

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 18.32.2 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

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 10.78.2 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 18.40.1 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. $\square$