TITLE: How to show $P^1\times P^1$ (as projective variety by Segre embedding) is not isomorphic to $P^2$?
QUESTION [19 upvotes]: 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!

REPLY [6 votes]: A different way to prove this is to count points. If $k$ is a field with $q$ elements, then $\mathbb{P}^2_k$ has $q^2 + q + 1$ rational points whereas $\mathbb{P}^1_k\times_k \mathbb{P}^1_k$ has $(q + 1)^2$ rational points, so they cannot be isomorphic. If $k$ is an arbitrary field, then any isomorphism $\mathbb{P}^2_k \to \mathbb{P}^1_k\times_k \mathbb{P}^1_k$ must be defined over a non-empty open locus $U$ of a finite ring extension of $\mathbb{Z}$. But since the residue field at any closed point of $U$ is finite, such an isomorphism cannot exist by the previous argument.
This technique of spreading out a morphism is usually not described in introductory text books on scheme theory. This is a pity, since it is quite elementary and a really good advertisment for considering schemes instead of just varieties. Basically, it hinges on the idea that you need finitely many equations to describe your morphism and to describe these equations, it is enough adjoin finitely many roots and inverting finitely many elements in your base.