$\mathbb P^n\times \mathbb P^m\not\cong \mathbb P^{m+n}$

Question

I know that $\mathbb P^{1}\times \mathbb P^1$, as a projective variety, is not isomorphic to $\mathbb P^2$.

Now, I think that also $\mathbb P^n\times \mathbb P^m$ is not isomorphic to $\mathbb P^{m+n}$ for any $m,n$, but why? For the above statement I use the fact that any two lines in $\mathbb P^2$ intersect, but in general?

Thank you in advance.

Answer 1

The universal cover of $\mathbb{P}^n\times \mathbb{P}^m$ is $S^n\times S^m$ and the universal cover of $\mathbb{P}^{n+m}$ is $S^{n+m}$.

$S^n\times S^m$ and $S^{n+m}$ are not diffeomorphic if $n,m>0$ to see this compute the cohomology, $H^n(S^n\times S^m)\neq 0$ (Kunneth) $H^n(S^{n+m})=0$, (use cellular cohomology).

Answer 2

I'm no algebraic geometer, but I think the following argument works over any field (maybe it has to be algebraically closed).

First, $\operatorname{Pic}(\mathbb{P}^n) = \mathbb{Z}$ generated by $\mathcal{O}_{\mathbb{P}^n}(1)$. Second, $\operatorname{Pic}(X\times\mathbb{P}^n) = \operatorname{Pic}(X)\times\mathbb{Z}$. So we see that $\operatorname{Pic}(\mathbb{P}^m\times\mathbb{P}^n) = \operatorname{Pic}(\mathbb{P}^m)\times\mathbb{Z} = \mathbb{Z}\times\mathbb{Z}$ while $\operatorname{Pic}(\mathbb{P}^{m+n}) = \mathbb{Z}$. Therefore $\mathbb{P}^m\times\mathbb{P}^n$ and $\mathbb{P}^{m+n}$ are not isomorphic.

Answer 3

Another proof: if $X,Y$ are two subvarieties of $\Bbb P^r$ such that $\dim X + \dim Y \geq r$, then they must intersect nontrivially in $\Bbb P^r$. At least one of $2m$ and $2n$ must be greater than or equal to $m+n$ - without loss of generality, suppose $2m\geq m+n$. Then any two copies of $\Bbb P^m$ inside $\Bbb P^{n+m}$ must intersect, while two copies of $\Bbb P^m$ inside $\Bbb P^m\times \Bbb P^n$ embedded as $\Bbb P^m\times \{x\}$ and $\Bbb P^m\times \{y\}$ for distinct points $x,y$ do not intersect.