Let $H$ denote a quadric hypersurface in $\mathbb P^2\times \mathbb P^2$. In the Chow ring of $\mathbb P^2\times \mathbb P^2$, we have $H\equiv 2H_1+2H_2$, where $H_1,H_2$ are the classes of linear forms in each copy of $\mathbb P^2$. I want to prove that $H$ is very ample as a divisor.
This should mean that there are ''enough'' global sections $s$ in $\Gamma(\mathbb P^2\times \mathbb P^2,\mathcal O(H))$. By this I mean that there should exist global sections $s_1,\ldots,s_n$ that do not vanish simultaneously, and the map $[s_1:\cdots:s_n]$ from $\mathbb P^2\times \mathbb P^2$ to $\mathbb P^{n-1}$ is isomorphic onto its image. But I can't see how to proceed from there.
The line bundle on $\mathbb{P}^2 \times \mathbb{P}^2$ you're considering is $$ \mathcal{O}(2) \boxtimes \mathcal{O}(2) = p_1^\ast\mathcal{O}_{\mathbb{P}^2}(2) \otimes p_2^\ast \mathcal{O}_{\mathbb{P}^2}(2). $$ This is evidently the square of the line bundle $\mathcal{O}(1) \boxtimes \mathcal{O}(1)$, defining the Segre embedding $\mathbb{P}^2 \times \mathbb{P}^2 \hookrightarrow \mathbb{P}^8$ which is a closed immersion (i.e. $\mathcal{O}(1) \boxtimes \mathcal{O}(1)$ is ample). Since a positive power of an ample line bundle is ample we win :)
Hope this helps!