# stalk of a sheaf

I think this question is pretty easy, but I really don't see it at the moment.

If I have a scheme $X$ and a point x in $X$, then I have a natural map $i: Spec(O_{X,x})\rightarrow X$. Given a sheaf of modules $F$ on X, then why is $\Gamma(Spec(O_{X,x}), i^*F)$ equal to $F_x$ ?

Thank you!

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Proposition. Let $X$ be a scheme, with structure sheaf $\mathscr{O}_X$. Let $P \in X$, and let $B = \mathscr{O}_{X,P}$. Then, there is a natural map $i : \textrm{Spec}(B) \to X$ such that for any $\mathscr{O}_X$-module $\mathscr{F}$, $\Gamma(\textrm{Spec}(B), i^* \mathscr{F}) \cong \mathscr{F}_P$.

Proof. Let $U$ be an affine open neighbourhood of $P$, with $U \cong \textrm{Spec}(A)$. Then $P$ corresponds to some prime ideal $\mathfrak{p}$ of $A$, and $B \cong A_{\mathfrak{p}}$. It is then clear that there is a natural map $i : \textrm{Spec}(B) \to X$, which factors through $U \hookrightarrow X$ and is moreover an isomorphism onto its image. Let $Y = \textrm{Spec}(B)$, and consider the open neighbourhoods of $i(Y)$. These are necessarily open neighbourhoods of $P$. Conversely, any open neighbourhood of $P$ must be an open neighbourhood of $i(Y)$, since if $\mathfrak{p} \nsupseteq \mathfrak{a}$ then certainly $\mathfrak{p}' \nsupseteq \mathfrak{a}$ for any $\mathfrak{p}' \subseteq \mathfrak{p}$. Hence the global sections of $i^{-1} \mathscr{F}$ are precisely the elements of $\mathscr{F}_P$. Note that $i^{-1} \mathscr{O}_X (Y) \cong \mathscr{O}_{X,P}$, so $i^{-1} \mathscr{F} (Y) \cong i^* \mathscr{F} (Y)$. The result follows.

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very nice comment, zhen! –  Descartes Aug 9 '11 at 10:04

Assume $X$ is a scheme, $\mathscr{F}$ is a sheaf of $\mathscr{O}_X$-modules and $x \in X$.

Let $U$ be an open affine neighbourhood of $X$. Let $\Gamma(U) \to \mathscr{O}_{X,x} \to \kappa(X)$ be the natural map. It induces a morphism $i: {\rm spec} \kappa(x) \to U \rightarrow X$. Then $i^{-1} \mathscr{F} = \mathscr{F}_x$. Tensoring with $\mathscr{O}_{X,x}$ gives $i^{*} \mathscr{F} = \mathscr{F}_x$.

This doesn't solve your question, but it seems related somehow.

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