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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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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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    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.2012-06-25