Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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
up vote 6 down vote accepted

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.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.