Is a morphism with finite fibers birational?

Question

Let $f: X \rightarrow Y$ be a morphism of projective varieties such that its fibers have finitely many points. Is $f$ birational on its image?

Thanks.

A birational map is generically one-to-one, by definition. – Jun 01 '14 at 18:33 — , Jun 01 '14 at 18:33
Ok, but which are the isomorphic open sets? – rla Jun 01 '14 at 19:31 — rla, Jun 01 '14 at 19:31
Dear rla, I don't understand your question. – Jun 02 '14 at 07:24 — , Jun 02 '14 at 07:24

score 3 · Accepted Answer · answered Jun 02 '14 at 03:02

3

The answer is no in general. $X$ and $Y$ can be two smooth projective curves of different genera which are by no means birational.

answered Jun 02 '14 at 03:02

rfauffar

Dear @Robert you are right. I made a tremendous confusion. I am trying to show that a certain divisor is big, what is sufficient to show that its associated morphism has finite fibers. This guarantees that the morphism associated to a tensor power of the divisor is birational and not necessarily the morphism associated to the divisor. Thanks. – rla Jun 02 '14 at 20:21

1 Answers1