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.



Flatness of normalization
A characteristic zero example: Let $k$ be a field of characteristic zero (or anything not $2$.) Let $L$ be the field $k(x,y)$ and let $K$ be the subfield $k(x^2, xy, y^2)$, so $[L:K]=2$. Let $S \subset L$ be the ring $k[x,y]$, and let $R = S \cap K = k[x^2, xy, y^2]$. 


Then $S$ is the normalization of $R$ in $L$. I claim that $\mathrm{Spec} \ k[x,y] \to \mathrm{Spec} \ k[x^2, xy, y^2]$ is not flat. Proof: the map is generically $2 \to 1$. However, the fiber above the origin is $k[x,y] /(x^2, xy, y^2)$, which has length $3$.


