What is an example of a birational morphism between $\mathbb{P}^{n} \times \mathbb{P}^{m} \rightarrow \mathbb{P}^{n+m}$?
Birational map between product of projective varieties
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algebraic-geometry
1 Answers
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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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0why $\mathbb{A}^{n} \times \mathbb{A}^{m}$ and $\mathbb{A}^{n+m} $ are open in $\mathbb{P}^{n} \times \mathbb{P}^{m}$? Can you please explain more this part? – 2012-05-19
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2Dear 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. – 2012-05-19
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0ah, missed the finer thing. I've been stuck with this problem, can you have a look please? http://math.stackexchange.com/questions/146377/birational-map-from-a-variety-to-projective-line – 2012-05-20