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I have a followup to the following question: Flatness of normalization.

Suppose that $X$ is a regular scheme (of finite type over a $\mathbb{C}$ if one wants) and $X'$ is the normalization of $X$ in a finite separable extension of $K(X)$. Then is $X'\rightarrow X$ flat?

The motivation is that I am trying how to understand normalizations of etale covers on complements of divisors.

Thanks!

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1 Answer 1

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I guess not. Take an affine integral normal $X'$ which is not Cohen-Macaulay. Find a Noether normalization map $f:X'\to X=\mathbb{A}^n$. This is a finite map, and $X'$ is the normalization of $X$ inside its field of rational functions. If $f$ were flat, by duality for finite flat morphisms $X'$ would be Cohen-Macaulay.

On the other hand, if $X'$ is Cohen-Macaulay, then $f$ is automatically flat!

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    $\begingroup$ Sorry to resurrect this question, but do you have any thoughts on whether the answer to the OP would be positive if we assumed that the normalization map $X'\rightarrow X$ is etale above the complement of an integral effective Cartier divisor of $X$? $\endgroup$
    – Will Chen
    Jul 20, 2016 at 0:53

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