Existence of an isomorphism $\mathbb{P}^n\times\mathbb{P}^m \rightarrow \mathbb{P}^{n+m}$

Question

There exist an isomorphism of varieties? $$\mathbb{P}^n\times\mathbb{P}^m \rightarrow \mathbb{P}^{n+m}$$

I am considering $\mathbb{P}^n\times\mathbb{P}^m$ as the product in the category of varieties.

Answer 1

I) The varieties $\mathbb P^1\times \mathbb P^1$ and $\mathbb P^2$ have subvarieties behaving differently and can thus not be isomorphic:

a) In $\mathbb P^1\times \mathbb P^1$ there exist two irreducible subvarieties which are are not points but are nevertheless disjoint, namely $\{a\}\times \mathbb P^1$ and $\{b\}\times \mathbb P^1$ for $a\neq b\in \mathbb P^1$.
b) If you take two different irreducible subvarieties $X,Y\subset \mathbb P^2_k$ which are not reduced to points they have non-empty intersection: $X\cap Y\neq \emptyset$.

Assertion b) results from elementary elimination theory and does not require Bézout's theorem with its subtle notion of intersection multiplicity. It requires however that $k$ be algebraically closed.

II) If $k$ is not algebraically closed you can go over to an algebraic closure of $k$ but this "change of basis" is generally not studied in elementary algebraic geometry.
For the non algebraically closed field $k=\mathbb R$ however, the topological manifolds underlying $\mathbb P^{n+m}_\mathbb R$ and $\mathbb P^n_\mathbb R\times \mathbb P^m_\mathbb R$ are not even homeomorphic, since their fundamental groups are respectively $\mathbb Z/2$ and $\mathbb Z/2 \times \mathbb Z/2 $.
So these algebraic varieties are a fortiori not isomorphic.

III) For general values of $n,m$ more sophisticated methods are necessary over general fields.
For example one can use Picard groups: $\text {Pic} (\mathbb P^n\times \mathbb P^m)=\mathbb Z\times \mathbb Z$, whereas $\text {Pic} (\mathbb P^{m+n})=\mathbb Z$

Edit
The result $\text {Pic} (\mathbb P^{m+n})=\mathbb Z$ is proved in detail in Hartshorne's Algebraic Geometry, Chapter II, Proposition 6.4.
The result $\text {Pic} (\mathbb P^n\times \mathbb P^m)=\mathbb Z\times \mathbb Z$ follows from the general result of Chapter III, Exercise 12.6, page 292. See also here.
There are probably more elementary references and I would be grateful to any user who would indicate one.

Answer 2

These spaces are not isomorphic, so you won't find an isomorphism.

I'm not sure what the best way to see this is if you don't know much algebraic geometry. Heuristically, you can note that, topologically, $\mathbb{RP}^1 \times \mathbb{RP}^1$ is a torus, whereas $\mathbb{RP}^2$ is a non-orientable surface, but I guess you could object that that doesn't actually prove anything in algebraic geometry...

You can embed $\mathbb{P}^1 \times \mathbb{P}^1$ in $\mathbb{P}^3$ via the Segre embedding and compare it to $\mathbb{P}^2$, but that comparison may take more machinery than you're familiar with.

Answer 3

A partial answer ; I'll get to the rest.

If by $\mathbb P $ you refer to the case of Real projective space, then , if both m,n are odd, their sum is even, which cannot happen, since Real projective spaces $\mathbb RP^n$ are orientable iff n is odd; the product of orientable manifolds is orientable, e.g., by Kunneth's theorem, the top homology over the integers is $\mathbb Z$.

You can rule out many other cases where both $m,n$ are even, since then $m+n$ will be even, and, by, e.g., Wikipedia, $H_p (\mathbb P^n(\mathbb R))$ = (over the integers)

1) $\mathbb Z/2$ , for $p<n $ , when p is odd.

2) $\mathbb Z$ , for n , when n is odd

3) $0$ for n, when n is even.

4)$\mathbb Z$ , for $p=0$;

Then, for $m+n$>$k$> $max$ {$n,m$} and k odd,

We will have $H_k(\mathbb (P^n \times \mathbb P^m, \mathbb Z))= \mathbb Z/2(+)Z/2$ , $\neq$ $H^k(\mathbb P^{n+m}, \mathbb Z )=\mathbb Z/2$ .

So the two are not even homotopically-equivalent , at least most of the time.