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.
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 #8241 by Ryo Suzuki on
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