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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.2012-03-10
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    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?2012-03-12
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    function field of a dense open subset is of course the same as the function field of the whole variety2012-03-12
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    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.2012-03-12
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    glad to be of help.2012-03-13

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