I have a followup to the following question: [Flatness of normalization](https://mathoverflow.net/questions/64776/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!



Flatness of Normalization of regular schemes
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!


