Lemma 20.6.1. Let (X, \mathcal{O}_ X) be a ringed space. If all stalks \mathcal{O}_{X, x} are local rings, then there is a canonical isomorphism
of abelian groups.
The Picard group of a ringed space is defined in Modules, Section 17.25.
Lemma 20.6.1. Let (X, \mathcal{O}_ X) be a ringed space. If all stalks \mathcal{O}_{X, x} are local rings, then there is a canonical isomorphism
of abelian groups.
Proof. Let \mathcal{L} be an invertible \mathcal{O}_ X-module. Consider the presheaf \mathcal{L}^* defined by the rule
This presheaf satisfies the sheaf condition. Moreover, if f \in \mathcal{O}_ X^*(U) and s \in \mathcal{L}^*(U), then clearly fs \in \mathcal{L}^*(U). By the same token, if s, s' \in \mathcal{L}^*(U) then there exists a unique f \in \mathcal{O}_ X^*(U) such that fs = s'. Moreover, the sheaf \mathcal{L}^* has sections locally by Modules, Lemma 17.25.4. In other words we see that \mathcal{L}^* is a \mathcal{O}_ X^*-torsor. Thus we get a map
We omit the verification that this is a homomorphism of abelian groups. By Lemma 20.4.3 the right hand side is canonically bijective to H^1(X, \mathcal{O}_ X^*). Thus we have to show this map is injective and surjective.
Injective. If the torsor \mathcal{L}^* is trivial, this means by Lemma 20.4.2 that \mathcal{L}^* has a global section. Hence this means exactly that \mathcal{L} \cong \mathcal{O}_ X is the neutral element in \mathop{\mathrm{Pic}}\nolimits (X).
Surjective. Let \mathcal{F} be an \mathcal{O}_ X^*-torsor. Consider the presheaf of sets
where the action of f \in \mathcal{O}_ X^*(U) on (s, g) is (fs, f^{-1}g). Then \mathcal{L}_1 is a presheaf of \mathcal{O}_ X-modules by setting (s, g) + (s', g') = (s, g + (s'/s)g') where s'/s is the local section f of \mathcal{O}_ X^* such that fs = s', and h(s, g) = (s, hg) for h a local section of \mathcal{O}_ X. We omit the verification that the sheafification \mathcal{L} = \mathcal{L}_1^\# is an invertible \mathcal{O}_ X-module whose associated \mathcal{O}_ X^*-torsor \mathcal{L}^* is isomorphic to \mathcal{F}. \square
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Comment #5992 by Gabriel Ribeiro on
Comment #5993 by Johan on