MathOverflow is a question and answer site for professional mathematicians. 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

Let $V, \tilde{V}$ be smooth algebraic varieties over $\mathbb{C}$ and $f \colon \tilde{V} \rightarrow V$ a projective (or proper) birational morphism. Assume that the exceptional locus $E \subset \tilde{V}$ has codimension $\ge 2$.

Question Is $f$ an isomorphism?

share|cite|improve this question
Yes, such an $f$ is an isomorphism. Consider the pullback map on relative differentials, $f^*:f^*\Omega^1_V \to \Omega^1_{\tilde{V}}$. This is a map of locally free sheaves of the same rank. It is an isomorphism if and only if the associated determinant is an isomorphism, i.e., it is everywhere nonzero considered as a section of the associated Hom sheaf. This Hom sheaf is invertible, so this section is zero on a Cartier divisor. Your hypotheses imply this Cartier diviser is empty. Hence $f^*$ is everywhere an isomorphism. – Jason Starr Jun 14 '12 at 10:47
Also see Sándor's answer to this question.… – Karl Schwede Jun 14 '12 at 12:02
@ Jason Starr, Thank you very much for the answer. I think that this answers my question. – tarosano Jun 14 '12 at 12:54
@ Karl Schwede, thank you very much for teaching me the related question. – tarosano Jun 14 '12 at 12:55
I think this is Zariski's "main theorem", factorial or smooth case, as in Shafarevich BAG vol. 1, p.120, or Mumford's red book, SLN 1358, 2nd ed. p.210. – roy smith Jun 15 '12 at 16:25

Yes. Suppose $f$ contracts a curve $C$. Then for any ample divisor $D$, we have $D\cdot C>0$. But $D=f^*f_*D$ by your hypotheses on the exceptional locus, and so $D\cdot C=f_*D\cdot f_*C=0$, a contradiction.

share|cite|improve this answer
@ John, thank you for the comment. Actually, I'm not assuming that $f$ is extremal. I thought contractions like flopping contractions might cause a problem, but it seems there does not exist such ones between smooth ones. – tarosano Jun 14 '12 at 12:58
Hi, I edited the answer. I think you can make it work in $\mathbb{Q}$-factorial case as well. – J.C. Ottem Jun 14 '12 at 15:58
@Ottem -- What if the varieties / the morphism are not projective? – Jason Starr Jun 14 '12 at 16:35

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.