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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?

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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
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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
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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).

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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
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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.

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