TITLE: Tensor product of invertible sheaves
QUESTION [22 upvotes]: Given two invertible sheaves $\mathcal{F}$ and $\mathcal{G}$, one can define their tensor product, but in this definition $\mathcal{F} \otimes \mathcal{G} (U)$ is (apparently) not simply equal to $\mathcal{F} (U) \otimes \mathcal{G}(U)$ for an open set $U$. This latter structure is a presheaf; can someone give me an example for when it is not a sheaf? Is there a characterization for when this actually does give a sheaf?

REPLY [2 votes]: Sorry for resurrecting such an old thread yet again...
This following paragraph is basically subsumed in Professor Emerton's and Smith's terrific answers, albeit in my own words. Let $p$ be a point of $\mathbb{P}^1$, and let $\mathcal{O}_{\mathbb{P}^1}(-p)$ denote the sheaf of functions which have a zero at $p$ (in other words, this is the ideal sheaf of $p$ in $\mathbb{P}^1$). Then $\mathcal{O}_{\mathbb{P}^1}(-p)$ has no global sections, since the only global sections are constants, so that the naïve definition of the tensor product has no global sections. However, the tensor product of these two sheaves is nothing more than $\mathcal{O}_{\mathbb{P}^1}$, the sheaf of holomorphic functions (the idea being that the two conditions, pole at $p$ and zero at $p$ cancel each other), which has plenty of global sections (the constants).
This following paragraph is not subsumed, though. For a very explicit example of where the naïve definition of the tensor product of two sheaves is not necessarily a sheaf, let $X = \{a, b\}$ be a two point space, with open sets $\{\emptyset, \{a\}, \{b\}, \{a, b\}\}$. Define $\mathcal{F}$ by$$\mathcal{F}(\{a\}) = \mathbb{Z}/2\mathbb{Z},\text{ }\mathcal{F}(\{b\}) = \mathbb{Z}/2\mathbb{Z},\text{ }\mathcal{F}(\{a, b\}) = \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z},$$and $\mathcal{G}$ by$$\mathcal{G}(\{a\}) = \mathbb{Z}/2\mathbb{Z},\text{ }\mathcal{G}(\{b\}) = \mathbb{Z}/3\mathbb{Z},\text{ }\mathcal{G}(\{a, b\}) = \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/3\mathbb{Z}.$$Then the naïve definition of the tensor product would give$$(\mathcal{F} \otimes \mathcal{G})(\{a\}) = \mathbb{Z}/2\mathbb{Z},\text{ }(\mathcal{F} \otimes \mathcal{G})(\{b\}) = 0,\text{ }(\mathcal{F} \otimes \mathcal{G})(\{a, b\}) = \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z},$$which is clearly not a sheaf.