TITLE: Flatness of normalization
QUESTION [18 upvotes]: Let $X$ be a noetherian integral scheme and let $f \colon X' \to X$ be the normalization morphism. It is known that, if non trivial, $f$ is never flat (see Liu, example 4.3.5).
What happens if we suppose $X$ normal, and we take the normalization in a finite (separable) extension of the function field of $X$? Note that in the easiest case, namely $X=\rm{Spec}(R)$, with $R$ a Dedekind domain, we have that $f$ is flat.

REPLY [4 votes]: If $f: X' \to X$ is flat, then $X$ tends to inherit nice properties of $X'$ (for example, being regular). So if you arrange for $X'$ to be "nicer" than $X$, as in David and one of Karl's examples, then $f$ can't be flat.
Here is a quick and dirty proof when "nice" = "regular". The claim is that if $R\to S $ is a finite flat local homomorphism of Noetherian local rings and $S$ is regular, then $R$ is regular as well. 
Let $m$ be the maximal ideals of $R$. Then as $S$ is regular, $S/mS$ has finite flat dimension (in fact, projective dim) over $S$. But $S$ is flat over $R$, so $S/mS$ has finite flat dimension over $R$. But as $R$-modules, $S/mS$ is direct sum of copies of  $k=R/m$, so $k$  has finite flat dimension over $R$, which characterizes regularity.