Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
2  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.