# Projective varieties: Reduction to the affine case

I ran across a certain type of argument for the second time now. Assume that $f$ is some rational map between projective varieties $X$ and $Y$, then supposedly I may replace $X$ with an open affine subset. It follows that the target space is also affine.

This might be a dumb question, but how exactly do I construct those open sets? I initially just took the standard affine covering of $X$, but I don't see how the image of such an open set under $f$ is affine. Wouldn't the image have to have empty intersection with some hyperplane?

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what you can do is take an affine open subset of $Y$ then take a preimage under the map, then restrict to an open affine subset of the preimage. –  Dima Sustretov Mar 10 '12 at 13:22
Thanks; I know that the function field in the target space will then be isomorphic to the function field of Y, but is the same true for the domain? –  Guest Mar 12 '12 at 12:26
function field of a dense open subset is of course the same as the function field of the whole variety –  Dima Sustretov Mar 12 '12 at 13:11
I see, thank you. That essentially answers my question, so if you want to reformulate your comment as an answer, I'll gladly accept it. –  Guest Mar 12 '12 at 14:31
glad to be of help. –  Dima Sustretov Mar 13 '12 at 23:44