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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.

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4 Answers 4

up vote 4 down vote accepted

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 $3$.

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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?

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 $\leq 3$.

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Hi Karl - why exactly is $S$ indecomposable over $R$? Or are you saying that $S \cong M^n$ for a single indecomposable $M$? Either way, I'm confused. I think that if $f = xz-y^2$, then $R \cong k[[a^2,ab,b^2]] \subset k[[a,b]] = S$, and $S \cong R \oplus (a,b)R$. –  Graham Leuschke Aug 1 '11 at 18:39
Graham, I probably didn't say what I meant to above very clearly. I am not claiming either I think. What I am claiming is that $S \cong R \oplus M$, as an $R$-module, where $M$ has no $R$-summands. –  Karl Schwede Aug 1 '11 at 22:23

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.

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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.

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