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If we have two rational varieties (i.e varieties which are birational to some projective space) is their product also a rational variety? would this rely on the fact that the Zariski topology is finer than the product topology and that $\mathbb{P}^{r} \times \mathbb{P}^{s}$ is birational to $\mathbb{P}^{r+s}$?

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up vote 7 down vote accepted

What you write is correct but it's even simpler than that: projective space needn't be invoked, nor the product topology.

To say that $X$ (resp. $X'$) is a rational variety means that some non-empty open subset $U\subset X$ (resp. $U'\subset X'$) is isomorphic to some open subset $V\subset \mathbb A^n$ (resp. $V'\subset \mathbb A^{n'}$).
But then the open subset $U\times U'\subset X\times X'$ is isomorphic to the open subset $V\times V'\subset \mathbb A^{n}\times \mathbb A^{n'}=\mathbb A^{n+n'}$, proving that $X\times X'$ is rational.

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thanks, I've noticed you know a lot about algebraic geometry. I'm a beginner and I would like to learn about schemes, sheaves, etc. What book(s) do you recommend for self study? (I find Hartshorne kinda hard sometimes) –  user10 Jun 25 '12 at 15:14
    
Dear user 10, Hartshorne is very hard. It contains an impressive lot of material but it is best used as a reference book.Shafarevich's Basic Algebraic Geometry (in two volumes) is the only book that develops classical varieties, schemes and holomorphic complex manifolds. The pace is leisurely and there is an astonishing wealth of classical examples in the text.The style is a little old-fashioned, though: proofs are sometimes a bit long-winded because commutative algebra is very little used. The historical remarks are very interesting and the covers are aesthetically very pleasing, too! –  Georges Elencwajg Jun 25 '12 at 21:56
    
The book is at the other extreme from Bourbaki : there are no long chains of logical dependence and the second volume can be seen as a set of complementary chapters that you can dip into according to your taste. –  Georges Elencwajg Jun 25 '12 at 22:02

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