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

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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? –  Artie Prendergast-Smith Nov 23 '13 at 22:21

1 Answer 1

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.

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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
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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
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@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
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@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

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