Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $(X, O_X)$ be an arbitrary scheme and $\mathcal{F}$ a presheaf of $O_X$-modules. The presheaf $\mathcal{F}$ is said to be separated if the natural map to it's sheafification $\mathcal{F}^+$ is an injection. That is, for each open set $U$, $\mathcal{F}(U) \hookrightarrow \mathcal{F}^+(U)$.

Let $\mathcal{F}, \mathcal{G}$ be two sheaves of $O_X$ modules. Is the presheaf tensor product $\mathcal{F} \otimes_{O_X} \mathcal{G}$ a separated presheaf? Is this true more generally if $(X,O_X)$ is just a ringed space?

This tells you for example, that if $\mathcal{F}, \mathcal{G}$ have some global sections then so does their sheaf tensor product (assuming the presheaf tensor product has global sections).

I wasn't able to find this in the stacks project. Is this proved in books on sheaf theory?

share|improve this question
    
Is it true when X is affine? –  Martin Brandenburg Sep 8 '12 at 14:24

1 Answer 1

up vote 9 down vote accepted

If I understand correctly your question, then the answer is "no, the presheaf tensor product need not be separated." Let $(X,\mathcal{O}_X)$ be the projective line $(\mathbb{P}^1,\mathcal{O}_{\mathbb{P}^1})$ over a base ring $R$. Let $\mathcal{F}$ and $\mathcal{G}$ both be the invertible sheaf $\mathcal{O}_{\mathbb{P}^1}(1)$. Since these are sheaves, they are separated. Consider the presheaf tensor product $\mathcal{E}$. In particular, the space of global sections of this presheaf is $$ H^0(\mathbb{P}^1,\mathcal{E}) = H^0(\mathbb{P}^1,\mathcal{O}_{\mathbb{P}^1}(1))\otimes_R H^0(\mathbb{P}^1,\mathcal{O}_{\mathbb{P}^1}(1)).$$ This is a free $R$-module of rank $4$. On the other hand, the sheafification of the presheaf tensor product is $\mathcal{O}_{\mathbb{P}^1}(2)$, whose space of global sections is a free $R$-module of rank $3$.

share|improve this answer

Your Answer

 
discard

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.