# Stalks of the tensor product presheaf of two sheaves

Let $(X, \mathscr{O})$ be a ringed space and $\mathscr{F}, \mathscr{G}$ be sheaves of $\mathscr{O}$-modules on $X$.

Define $\mathscr{H}(U) = \mathscr{F}(U) \otimes_{\mathscr{O}(U)} \mathscr{G}(U)$. I am stuck trying to prove that $\mathscr{H}_p \cong \mathscr{F}_p \otimes_{\mathscr{O}_p} \mathscr{G}_p$ as $\mathscr{O}_p$-modules.

I know that if $X$ is a topological space, $\mathscr{F}, \mathscr{G}$ are presheaves of abelian groups on $X$ and $\mathscr{H}(U) = \mathscr{F}(U) \otimes_{\mathbb{Z}} \mathscr{G}(U)$ then $\mathscr{H}_p \cong \mathscr{F}_p \otimes_{\mathbb{Z}} \mathscr{G}_p$. This is just a consequence of the fact that tensor products commute with direct limits. But I don't know how to deal with the case when the base ring is changing

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Doesn't the same proof work? The underlying fact that makes both things true is that colimits commute with colimits. – Mariano Suárez-Alvarez Jan 17 '11 at 2:02

## 2 Answers

I figured out how to prove it.

Let $(X, \mathscr{O})$ be a ringed space. Let $\mathscr{F}, \mathscr{G}$ be sheaves of $\mathscr{O}$-modules. Define $\mathscr{H}(U) = \mathscr{F}(U) \otimes_{\mathscr{O}(U)} \mathscr{G}(U)$. Fix $p \in X$. Assume $U$ is an open n.h of $p$.

The $\mathscr{O}_p$-module structure on $\mathscr{F}_p \otimes_{\mathscr{O}_p} \mathscr{G}_p$ induces a $\mathscr{O}(U)$-module structure on $\mathscr{F}_p \otimes_{\mathscr{O}_p} \mathscr{G}_p$ .

Define $\alpha_U : \mathscr{F}(U) \times \mathscr{G}(U) \to \mathscr{F}_p \otimes_{\mathscr{O}_p} \mathscr{G}_p, \quad (s,t) \mapsto s_p \otimes t_p$. This map is $2$-linear over $\mathscr{O}(U)$. Therefore $\alpha_U$ induces an $\mathscr{O}(U)$-module homomorphism from $\mathscr{F}(U) \otimes_{\mathscr{O}(U)} \mathscr{G}(U)$ to $\mathscr{F}_p \otimes_{\mathscr{O}_p} \mathscr{G}_p$. We shall abuse notation and also call this map $\alpha_U$.

Now forget about the $\mathscr{O}(U)$ module structure on the sections of $\mathscr{H}$. The $\alpha_U$s form a co-cone over the $\mathscr{H}(U)$s with $p \in U$ (that is they make the appropriate diagrams commute), therefore they induce a homomorphism of abelian groups $h : \mathscr{H}_p \to \mathscr{F}_p \otimes_{\mathscr{O}_p} \mathscr{G}_p$

Define $\psi : \mathscr{F}_p \times \mathscr{G}_p \to \mathscr{H}_p$ by $(s_p, t_p) \mapsto (s|_{U \cap V} \otimes t|_{U \cap V})_p$ where $s$ is a section of $\mathscr{F}$ over $U$ and $t$ is a section of $\mathscr{G}$ over $V$. It is easily verified that $\psi$ is 2-linear over $\mathscr{O}_p$. Therefore $\psi$ induces a $\mathscr{O}_p$ homomorphism from $\mathscr{F}_p \otimes_{p} \mathscr{G}_p$ to $\mathscr{H}_p$. It is easily verified that this map and $h$ are inverses.

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Alternatively you can prove that the tensor product commutes with colimits not only in both factors, but also over the base ring. This follows easily from the adjunction.

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