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Let $f:X\rightarrow Y$ a $k$-morphism of finite type from a Cohen-Macaulay scheme to a normal scheme such that all the fibers have the same dimension, do we have that $f$ is flat?

We know, that it's true for $Y$ smooth. Can we weaken this condition?

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See also… – J.C. Ottem Dec 22 '12 at 20:14

No, the smoothness is really needed !

To construct a counter-example, let us choose $X=\mathbb{A}^2$. Consider the action of $\mathbb{Z}/2\mathbb{Z}$ by $(x,y)\mapsto (-x,-y)$. The quotient of $X$ by this action is the quadric cone $Y\subset \mathbb{A}^3$ defined by $tv=u^2$, where the quotient map $f:X\to Y$ is given by $(x,y)\mapsto (t,u,v)=(x^2,xy,y^2)$.

Now $X$ is Cohen-Macaulay, because it is regular, $Y$ is normal by Serre's criterion, and the fibers of $f$ all have dimension $0$ ($f$ is finite). However, $f$ is not flat. Indeed, if it were, since it is proper, all of its fibers would have the same length. But the fibers of $f$ have length $2$, except the fiber over $(0,0,0)$ that has length $3$.

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"$Y$ is normal by Serre's criterion": true, but more simply, it is immediate that if a group acts on an integrally closed domain, the subring of invariants is integrally closed. – Laurent Moret-Bailly Dec 22 '12 at 21:17

Let $X=Y$ be of characteristic $p$ and let $f:X\to Y$ be the (absolute) Frobenius morphism. Kunz's Regularity Criterion states that $f$ is flat if and only if $X$ is regular. In particular, taking $X$ to be Cohen-Macaulay and normal but not regular/smooth gives another counterexample.

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