Flatness of normalization 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.
 A: 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 $3$.
A: David's answer reminds me also of the following:
For ordinary double points $R = k[[x,y,z]]/f$, $\text{A}_n$, $\text{D}_n$, $\text{E}_6$,  $\text{E}_7$,  $\text{E}_8$.   These all have normalization in some field extension as a regular ring (they are all quotient singularities).  Let $R \subseteq S$ be the usual extension with $S$ regular that $R$ is a quotient of.  Then $S$ has exactly 1 $R$-summand, and so $S$ can't be free (and thus can't be flat either).  
I know one way to to deduce this from a paper of Huneke-Leuschke, but I don't know the proof off the top of my head that there is clearly exactly one summand.  Maybe Graham or Long does?
EDIT:  Here's a quick way to see it.  If $R \subseteq S$ is local and etale in codimension 1, then the trace map generates $\mathrm{Hom}_R(S, R)$ as an $S$-module.  Let's also suppose that $R$ and $S$ have the same residue field.  The trace map is also surjective (at least if $k$ is characteristic zero, or the extension is tamely ramified).  This gives one summand.  If we had another summand, it would be obtained by pre-multiplying the trace map by some non-unit in $S$.  But those non-units live in the maximal ideal of $S$, and the trace map sends the maximal ideal of $S$ into the maximal ideal of $R$, so any such potential map $S \to R$ cannot be surjective.  Thus there are no other splittings.
Finally, I should also add:
Theorem: Let $S$ be a module finite local extension of a regular local ring $R$ (for example, $S$ is the normalization of $R$ in some extension field of $K(R)$), then $S$ is Cohen-Macaulay if and only if $S$ is free = flat as an $R$-module.
Hm, maybe I'll also point out...
Conjecture (Direct summand):  Let $R$ be regular, and $S$ the integral closure of $R$ in some finite extension of $K(R)$.  Then $S$ has at least one $R$-summand (in other words, $R \to S$ splits as a map of $R$-modules).  In particular, for any $R$-module $M$, $M \otimes R \to M \otimes S$ is injective.  
This conjecture is known for rings containing a field and for rings in mixed characteristic of dimension $\leq 3$.
EDIT:  Now due to spectacular work of Yves Andre, this conjecture has been solved.  
A: A silly case:  Suppose that $X$ is characteristic $p$ and normal and suppose that we embed $K(X) \subseteq L$ where $L = (K(X))^{1/p}$ (the stereo-typical inseparable extension).  Then the normalization of $X$ is isomorphic to $X = X'$ again, and the natural map $f : X' \to X$ is Frobenius.  Then
Theorem:  (Kunz)  $f$ is flat if and only if $X$ is regular.
A: 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. 
