How to show $P^1\times P^1$ (as projective variety by Segre embedding) is not isomorphic to $P^2$? I am a beginner.
This is an exercise from Hartshorne Chapter 1, 4.5. By his hint, it seems this can be argued that there are two curves in image of Segre embedding that do not intersect with each other while in $P^2$ any two curves intersect.
I feel this solution is very special. I would like to know more.
Is there any invariant to detect whether two birational equivalent varieties are iso or not?
Thanks!

The canonical bundles of these varieties both only have the zero section, so these sections cannot be used to prove the non-isomorphism of the varieties.
The anticanonical bundles $K^\ast=\Lambda^2T$ of these varieties however  have spaces of global sections of different dimensions $h^0$ over the base field and prove that non-isomorphism:
$$h^0(\mathbb P^2, K^\ast_{\mathbb P^2})=h^0(\mathbb P^2,\mathcal O_{\mathbb P^2}(3))=10$$ whereas 
$$h^0(\mathbb P^1\times \mathbb P^1, K^\ast_{{\mathbb P^1\times \mathbb P^1}})=h^0(\mathbb P^1\times \mathbb P^1,\mathcal O_{\mathbb P^1 }(2)\boxtimes \mathcal O_{\mathbb P^1 }(2))=9$$ (Whew, that was close: 10 and 9  !)  
Edit: notation
Given the projections $p,q:\mathbb P^1\times \mathbb P^1\to \mathbb P^1$ of $\mathbb P^1\times \mathbb P^1$ onto its two factors, the tensor product $p^\ast\mathcal O_{\mathbb P^1}(a)\otimes_{\mathcal O_{\mathbb P^2}} q^\ast\mathcal O_{\mathbb P^1}(b)$  of the  pull-backs $p^\ast\mathcal O_{\mathbb P^1}(a)$  and  $q^\ast\mathcal O_{\mathbb P^1}(b)$ is denoted  by $\mathcal O_{\mathbb P^1}(a)\boxtimes \mathcal O_{\mathbb P^1}(b)$ or simply by  $\mathcal O_{\mathbb P^2}(a,b)$.
This convention is used in other answers too.
