# Line bundles on $\mathbb{P}^2 \times \mathbb{P}^2$

I'm curious as to what are all the line bundles on $\mathbb{P}^2 \times \mathbb{P}^2$? I might be mistaken but I believe they are classified as $$\mathcal{O}_{\mathbb{P}^2 \times \mathbb{P}^2}(p,q) := \pi_1^* \mathcal{O}_{\mathbb{P}^2}(p)\otimes \pi_2^* \mathcal{O}_{\mathbb{P}^2}(q)$$ where $\pi_i: \mathbb{P}^2 \times \mathbb{P}^2 \rightarrow \mathbb{P}^2$ is the map onto the i-th factor. Is this correct? If so, then why is this true? Is it true for just line bundles or any rank vector bundle?

Also, if I have a hypersurface of this space, say, $H$, then are the line bundles on $H$ given by $$\mathcal{O}_{H}(p,q):= \mathcal{O}_{\mathbb{P}^2 \times \mathbb{P}^2}(p,q)\vert_{H} \quad ?$$

I believe this is true as well and has something to do with the Lefshetz Hyperplane Theorem but I'm just not seeing how to put the pieces together.

Oh, and everything is complex.

Thanks for the help.

• All you say is true for line bundles. Larger rank bundles of course are another story. – Mohan Aug 28 '15 at 18:18
• There are general statements to the effect that taking the Picard group of a projective bundle over $X$ is $\operatorname{Pic} X \times \mathbf Z$. In general I don't know how to this without high powered theorems from cohomology and base change but when $X$ is regular it shouldn't be bad. I think the proof should be a lot like the proof for $\mathbf P^1 \times \mathbf P^1$. – Hoot Aug 28 '15 at 18:56
• @Hoot : The case of the trivial bundle is perhaps a little tedious but not that difficult and is exercise II.6.1 in Hartshorne. It should be fairly accessible for user46348. – KReiser Aug 28 '15 at 22:36
• @KReiser Thanks for digging that up! I was looking in II.7 for a reference. Whoops. – Hoot Aug 28 '15 at 22:37