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 $3$.

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 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.