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. $\square$

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