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What is an example of a birational morphism between $\mathbb{P}^{n} \times \mathbb{P}^{m} \rightarrow \mathbb{P}^{n+m}$?

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up vote 4 down vote accepted

The subset $\mathbb A^n\times \mathbb A^m$ is open dense in $\mathbb P^n\times \mathbb P^m$ and the subset $\mathbb A^{n+m}$ is open dense in $\mathbb P^n\times \mathbb P^m$.
Hence the isomorphism $\mathbb A^n\times \mathbb A^m\stackrel {\cong}{\to} \mathbb A^{n+m}$ is the required birational isomorphism.

The astonishing point is that a rational map need only be defined on a dense open subset , which explains the uneasy feeling one may have toward the preceding argument, which may look like cheating.
The consideration of "maps" which are not defined everywhere is typical of algebraic ( or complex analytic) geometry, as opposed to other geometric theories like topology, differential geometry,...

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why $\mathbb{A}^{n} \times \mathbb{A}^{m}$ and $\mathbb{A}^{n+m} $ are open in $\mathbb{P}^{n} \times \mathbb{P}^{m}$? Can you please explain more this part? – user31509 May 19 '12 at 23:34
Dear user, $\mathbb A^N=\mathbb P^N \setminus H$ where $H$ is the hyperplane $z_0=0$ ( $\mathbb P^N$ has coordinates $[z_0:z_1:...:z_N]$). Since $z_0$ is a homogeneous polynomial (of degree one) , $H$ is closed and its complement $\mathbb A^N$ is open. For your first question use this result plus the fact that the product of two open sets is open : the Zariski topology for $\mathbb P^n\times \mathbb P^m$ is finer than the product topology. – Georges Elencwajg May 19 '12 at 23:49
ah, missed the finer thing. I've been stuck with this problem, can you have a look please?… – user31509 May 20 '12 at 1:12

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