# Why tensor product of two sheaves of modules is a sheaf of modules?

Let $(X, \mathcal O_X)$ be a ringed topological space. Consider two $\mathcal O_X$ modules, $\mathcal F$ and $\mathcal G$. First we define the tensor product presheaf $\mathcal F \otimes_{p,\mathcal O_X} \mathcal G$, which assigns every open set $U$ in $X$ the $\mathcal O_X(U)$-module $\mathcal F(U) \otimes_{\mathcal O_X(U)} \mathcal G(U)$. Now, $\mathcal F \otimes_{p,\mathcal O_X} \mathcal G$ is a presheaf of abelian groups. We take the the sheafication of the $\mathcal F \otimes_{p,\mathcal O_X} \mathcal G$ to obtain a sheaf of abelian groups, $\mathcal F \otimes_{\mathcal O_X} \mathcal G$. My question is for any open set $U$ in $X$ what is the $\mathcal O_X(U)$-module on $(\mathcal F \otimes_{\mathcal O_X} \mathcal G)(U)$?

If $(X,\mathcal O_X)$ is a scheme and if $U\subset X$ is is an open affine subscheme, then we have for all quasi-coherent sheaves $\mathcal F,\mathcal G$ of $\mathcal O_X$-modules the extremely pleasant equality of $\mathcal O_X(U)$-modules:

$$(\mathcal F \otimes_{ \mathcal O_X} \mathcal G)(U)= \mathcal F(U) \otimes_{\mathcal O_X(U)} \mathcal G(U)$$

However if $U$ is not affine all hell can break loose:
For example if $X=\mathbb P^1_\mathbb C$ is the complex projective line and if $\mathcal F=\mathcal O_X(1), \mathcal G=\mathcal O_X(-1)$, then for these quasi-coherent sheaves $\mathcal F \otimes_{ \mathcal O_X} \mathcal G=\mathcal O_X$, so that for $U=X$: $$(\mathcal F \otimes_{ \mathcal O_X} \mathcal G)(X)= \mathcal O_X(X) =\mathbb C \neq \mathcal F(X)\otimes _{\mathcal O_X(X)}\mathcal G(X)=\mathbb C^2\otimes_\mathbb C 0=0$$

Reference
The first displayed equality is proved in Qing-Liu, chapter 5, Proposition 1.12 (b), page 162.

• Great answer, thanks. Just to double check, $\mathcal F(X)$ is the space of quotients of homogenous polynomials in two variables, with the top one degree more than the bottom, and the bottom shouldn't vanish on $\mathbb P^1$. So the bottom ends up being a nonzero constant (by projective Nullstellensatz), and the top is just a degree 1 homogenous polynomial, which can be identified with $\mathbb C^2$ as you said. Oct 18 '16 at 21:08
• Dear hwong, your comment is perfectly correct. More genrally the set of global sections $\mathcal O_{\mathbb P^n}(d)(\mathbb P^n)$ of $\mathcal O_{\mathbb P^n}(d)$ consists of the polynomials in $n+1$ indeterminates which are homogeneous of degree $d$. (In particular if $d\lt 0$ the only global section is zero!) The dimension of that vector space of sections $\mathcal O_{\mathbb P^n}(d)(\mathbb P^n)$ is $\binom {n+d}{d}$, which is indeed equal to $2$ for $n=d=1$ Oct 18 '16 at 21:38

Are you asking for the $\mathcal O_X(U)$ action on $(\mathcal F \otimes_{\mathcal O_X} \mathcal G)(U)$? If so, it works on the level of stalks. Let $f \in \mathcal O_X(U)$ and let $s \in (\mathcal F \otimes_{\mathcal O_X} \mathcal G)(U)$. $s$ can be thought of as a collection of compatible germs $s_p \in \mathcal F_p \otimes_{\mathcal O_{x, p}} \mathcal G_p$. Then $f \cdot s$ is the collection of germs $f_p \cdot s_p$, where the dot is the $\mathcal O_{x, p}$-module action on $\mathcal F_p \otimes_{\mathcal O_{x, p}} \mathcal G_p$. One just needs to verify that the collection of $f_ps_p$ is still a collection of compatible germs, which they are.

• Yes. I actually just want to prove $(\mathcal F \otimes_{\mathcal O_X} \mathcal G)(U)$ is a $\mathcal O_X(U)$. Is there any way to see that without going to the level of stalks?
– grok
Oct 18 '16 at 21:32
• Dear grok: given any presheaf of $\mathcal O_X$-modules, its sheafification is automatically an $\mathcal O_X$-module too. Apply this to the presheaf of $\mathcal O_X$-modules $\mathcal F \otimes_{p,\mathcal O_X} \mathcal G$ whose sheafification is $\mathcal F \otimes_{\mathcal O_X} \mathcal G$ Oct 18 '16 at 21:41
• @grok. I agree with Georges' comment above, but as far as I know, the only way to see that $(\mathcal F \otimes_{\mathcal O_X} \mathcal G)$ is an $\mathcal O_X(U)$-module is to go through stalks. The sheafification is in terms of stalks so I don't know how to get around it. Oct 18 '16 at 21:51