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Let $f: Y \to X$ be a morphism of varieties (proper if necessary).

I read from a paper that if all the fibres of $f$ are of the same dimension then $f$ is flat. This seems skeptical for me, and I was wondering more conditions are need to have the flatness

user26857
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Li Yutong
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2 Answers2

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This is true provided that the source is Cohen--Macaulay (e.g. regular) and the base is regular. This is sometimes called miracle flatness. (It can be found as Thm. 23.1 of Matsumura's CRT.)

It generalizes the fact that a non-constant map from a reduced curve to a smooth curve is necessarily flat.

(In the case when the fibre dimension is zero, it is related to the Auslander--Buchsbaum theorem, which says that a f.g. faithful Cohen--Macaulay module over a regular local ring is necessarily free.)

Added: As Georges Elencwajg points out in a comment below, in the context of miracle flatness I should require that we are working with locally finite type schemes over a field.

Matt E
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    Dear Matt, aren't the schemes in question required to be locally algebraic over a field ? ( The references I checked, including Matsumura, make that hypothesis). – Georges Elencwajg Feb 04 '14 at 09:56
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    @GeorgesElencwajg: Dear Georges, Quite possibly, if that's what the references suggest. (I've never carefully thought about this result or its proof; even my remark about the relative dimension zero case was a bit off-the-cuff, although I do think there's a connection.) Best wishes, Matt – Matt E Feb 04 '14 at 23:56
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    @GeorgesElencwajg: P.S. I should have added: thanks for pointing this out! I've made a correction to this effect. – Matt E Feb 04 '14 at 23:58
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You are right to be skeptical because the result is false.
For example if $X\subset \mathbb A^2$ is the cusp defined by $y^2=x^3$, its normalization $f:\mathbb A^1=Y\to X: t\mapsto (t^2,t^3)$ is not flat although $f$ is proper and all of its fibers consist of exactly one point and thus have the same dimension, namely $0$ .

Georges Elencwajg
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