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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.

Marcos
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  • One way to do this is the fact that $\Bbb P^1\times\Bbb P^1$ admits a nonconstant map to $\Bbb P^1$ while $\Bbb P^2$ does not: see here for instance. Is this within the realm of what you're looking for? – KReiser Nov 04 '21 at 21:57
  • @KReiser yeah, this could work, but the proof given there is more sophisticated than my knowledge. Maybe for the case of $n=2$ and $m=1$ the solution can be simplified? – Marcos Nov 04 '21 at 22:11
  • The proofs in the link given by @KReiser -- that every morphism $\mathbb{P}^2 \to \mathbb{P}^1$ is constant -- also rely on Bezout's theorem. – Daniel Schepler Nov 04 '21 at 22:49
  • @DanielSchepler Really? They didn't mention it in the solution... – Marcos Nov 04 '21 at 22:52
  • The solution there by Tony, which is pretty much equivalent to the other one but more explicit, uses it in the discussion of "common vanishing locus of polynomials". – Daniel Schepler Nov 04 '21 at 22:53
  • @DanielSchepler ohh, I see, thanks, maybe there is no other solution other than to use Bezout. – Marcos Nov 04 '21 at 22:56
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    @DanielSchepler I'd say that it's more dimension theory than Bezout, but perhaps this is splitting hairs. – KReiser Nov 04 '21 at 23:04
  • Even if you can’t prove it in full generality, you can try to prove it for some specific fields you know, it should still be an interesting exercise. If I am not mistaken, the proof for finite fields is easy, and for $\mathbb{R}$ and $\mathbb{C}$, a little less easy. – Plop Nov 04 '21 at 23:17
  • @Plop I know how to do it in the case of finite fields, however I don't see why in general there are no other posible proof. – Marcos Nov 04 '21 at 23:35
  • See https://math.stackexchange.com/questions/983568/how-to-show-p1-times-p1-as-projective-variety-by-segre-embedding-is-not-is, Georges posted some cohomology arguments. – Qi Zhu Nov 05 '21 at 08:10
  • Sure, I wasn’t saying that a proof for specific fields should generalize easily. All I am saying is that if you are looking for an elementary solution to a problem that may not admit any, it is still interesting to replace the problem by some easier one! – Plop Nov 05 '21 at 09:10

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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.

orangeskid
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  • There is a way in which we can generalice this for any field $k$? Since counting argument only works in finite fields – Marcos Nov 05 '21 at 07:30
  • The scheme is like this: iso over $k$, implies iso over $\bar k$. If char $k=p$, then iso over $\overline{\mathbb{F}_p}$, so done. If char $ k =0$, still true over all alg closed of char $p$, $p\not \in P_0$. Perhaps the last line is the most interesting. But it's nothing more than elimination of quantifiers, noting that you are dealing with statements about integers being equal to $0$ or not in a field. – orangeskid Nov 05 '21 at 12:58
  • Thanks, I like your solution, and porbably there are no more general solutions without Bezout, Cohomology or somethnig deeper. – Marcos Nov 05 '21 at 15:47