MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $f:X\to Y$ 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?

share|cite|improve this question
Charles, do you mean $f$ is generically finite? I think any finite morphism is already quasi-finite, i.e., has finite fibers. – Dave Anderson Aug 3 '11 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.) – Dave Anderson Aug 3 '11 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.

share|cite|improve this answer

Have you seen "Critères de platitude et de projectivité. Techniques de "platification'' d'un module" by Raynaud-Gruson (1971)? In particular 5.2.2. I think it is very close to what you want (this was explained to me by Bhargav Bhatt not too long ago). It doesn't say that any blow-up works, but there is one that's ok.

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).

share|cite|improve this answer
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. – Jason Starr Aug 3 '11 at 21:08

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.