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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