# When can a finite map be blown up to a flat one?

Let $f:X\to Y$ be a generically finite proper morphism of varieties. There is some locus in $Y$ over which the fiber of $f$ is positive dimensional, so we blow it up, along with the preimage of it in $X$ to get a map $\tilde{f}:\tilde{X}\to\tilde{Y}$ which has finite fibers.

Are there any nice conditions that will guarantee that the map $\tilde{f}$ is flat?

• Charles, do you mean $f$ is generically finite? I think any finite morphism is already quasi-finite, i.e., has finite fibers. Aug 3, 2011 at 18:52
• As for general criteria, one thing I have found useful is this [Hartshorne Exercise III.10.9]: if $f$ has equidimensional fibers, $Y$ smooth, and $X$ Cohen-Macaulay, then $f$ is flat. (Perhaps in your situation you can achieve this by blowing up.) Aug 3, 2011 at 19:11

Since the blowups are proper and $f$ is proper, the "property P argument" shows that $\tilde f$ is proper. A proper quasi-finite morphism is finite (EGA IV 18.12.4), so $\tilde f$ is finite.
This (more or less) reduces to the case when $f$ is finite to begin with, so no blowups are needed. Then the only condition I know to ensure flatness is the one Dave Anderson cited, [Matsumura's Commutative Ring Theory, Theorem 23.1]: if $Y$ is regular and $X$ is Cohen-Macaulay, then $f$ is flat.
Basically, there exists a blow-up $Y' \to Y$ (you can assume $Y'$ is normal, ie normalize the blow-up) such that the appropriate component(s) of the fiber product $Y'' \to Y' \times_Y X$ give us a map $Y'' \to Y'$ which is flat. This is proven in the modern setting by Hilbert-Scheme arguments usually (also see for example various papers talking about universal flattening).
• If the morphism is projective, you do not need the full strength of Raynaud-Gruson. Just embed $X$ (locally over $Y$) in $Y\times \mathbb{P}^n$. Then think of the morphism as a rational map to the Hilbert scheme of $\mathbb{P}^n$. The closure of the graph gives a blowing up of $Y$ which does the trick. Aug 3, 2011 at 21:08