# Prove that $\mathbb{P}^2$ is not isomorphic to $\mathbb{P}^1\times\mathbb{P}^1$ without Bezout

algebraic-geometryprojective-geometry

Studying algebraic geometry I found the following exercise:

Prove that $$\mathbb{P}^2$$ is not isomorphic to $$\mathbb{P}^1\times\mathbb{P}^1$$.

Here $$\mathbb{P}^1\times\mathbb{P}^1$$ is defined as the image of the Segre embedding.

I have seen many proofs of this fact on internet, however all of them uses Bezout theorem (the usual proof is to use Bezout's theorem to show that every two curves on $$\mathbb{P}^2$$ interesect on at least one point). I am wondering if there exists a proof of this fact without the use of Bezout's theorem (a proof that every two curves on $$\mathbb{P}^2$$ interesect on at least one point without Bezout) since it is a theorem I haven't studied yet. If not feel free to close this question.

A map from $$\mathbb{P}^2 \to \mathbb{P}^3$$ is given by a set of homogenous polynomials of some degree $$d$$
$$[x_0, x_1, x_2] \mapsto [P_i(x_0, x_1, x_2)]_{0\le i \le 3}$$
We also want the image to be in the quadric $$y_0 y_3 - y_1 y_2=0$$, so that's an extra condition on the $$P_i$$. Moreover, we want it to be a isomorphism of projective varieties.
Now, say we are working over an algebraically closed field $$k$$. The existence of such a map is (by the elimination of quantifiers) equivalent to an elementary condition for the field $$k$$. If it holds for the field $$\mathbb{C}$$, then it will hold for algebraically closed fields of characteristic $$p$$, for $$p \not \in P_0$$, where $$P_0$$ is finite. Choose such a prime $$p$$. Then we will have an isomorphism between $$\mathbb{P}^2$$ and $$\mathbb{P}^1\times \mathbb{P}^1$$ defined over a field $$\mathbb{F}_{p^m}$$, and so a bijection between the $$\mathbb{F}_{p^m}$$ points. But a simple counting argument that you mentioned gets a contradiction.