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

In dimension 2, a rational map becomes a morphism after a sequence of blow-ups. Does this still hold in higher dimensions?

share|cite|improve this question
In light of the comments under abx's answer, it might be a good idea to clarify the question a bit; in particular, do you require the blow-ups to be along smooth centers? – user5117 Nov 23 '13 at 22:21

Yes, but this is a very difficult result, due to Hironaka. To be precise : given a rational map $f:X --> Y$, there exists a birational morphism $b:\hat{X}\rightarrow X$, obtained as the composition of successive blown-up with smooth centers, such that $f\circ b$ is a morphism.

share|cite|improve this answer
Thank you. Do you have a reference? Or is this just in Hironaka's article on resolution of singularities? – Timo Keller Nov 23 '13 at 12:59
Actually, you do not need Hironaka at all. Let $X$ and $Y$ be projective schemes. Let $f$ be a morphism from a dense open subset $U$ of $X$ to $Y$. Let $\Gamma_f$ be the graph of $f$ inside $X\times Y$. Let $\widehat{X}$ be the Zariski closure of $\Gamma_f$. Then the projection, $b:\widehat{X}\to X$, is a projective, birational morphism. Thus, by a theorem in Chapter II, Section 7 of Hartshorne's "Algebraic Geometry", the morphism $b$ is (isomorphic to) the blowing up of $X$ along a coherent sheaf of ideals. – Jason Starr Nov 23 '13 at 13:03
@Jason: that's right, of course, but I believe the OP meant blow-ups with smooth centers (and rational map of smooth varieties). – Serge Lvovski Nov 23 '13 at 14:56
@Timo Keller: Yes, this is in Hironaka's paper (Resolution of singularities of an algebraic variety over a field of characteristic zero", Ann. Of Math. (2) 79, 109–326. The statement is explained in CH. 0, §5. – abx Nov 23 '13 at 17:40
@Serge and abx: Okay, that makes sense to me. – Jason Starr Nov 23 '13 at 18:14

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.