Lemma 17.25.8. Let $(X, \mathcal{O}_ X)$ be a ringed space. There exists a set of invertible modules $\{ \mathcal{L}_ i\} _{i \in I}$ such that each invertible module on $X$ is isomorphic to exactly one of the $\mathcal{L}_ i$.

Proof. Recall that any invertible $\mathcal{O}_ X$-module is locally a direct summand of a finite free $\mathcal{O}_ X$-module, see Lemma 17.25.2. For each open covering $\mathcal{U} : X = \bigcup _{j \in J} U_ j$ and map $r : J \to \mathbf{N}$ consider the sheaves of $\mathcal{O}_ X$-modules $\mathcal{F}$ such that $\mathcal{F}_ j = \mathcal{F}|_{U_ j}$ is a direct summand of $\mathcal{O}_{U_ j}^{\oplus r(j)}$. The collection of isomorphism classes of $\mathcal{F}_ j$ is a set, because $\mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ U}(\mathcal{O}_ U^{\oplus r}, \mathcal{O}_ U^{\oplus r})$ is a set. The sheaf $\mathcal{F}$ is gotten by glueing $\mathcal{F}_ j$, see Sheaves, Section 6.33. Note that the collection of all glueing data forms a set. The collection of all coverings $\mathcal{U} : X = \bigcup _{j \in J} U_ i$ where $J \to \mathcal{P}(X)$, $j \mapsto U_ j$ is injective forms a set as well. For each covering there is a set of maps $r : J \to \mathbf{N}$. Hence the collection of all $\mathcal{F}$ forms a set. $\square$

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