# 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 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 $3$.

• I don't understand why you assume the characteristic $\ne 2$. In this answer I've shown that $R\subset S$ (in your notation) isn't flat for any field. Am I mistaken? Dec 28, 2014 at 15:29
• @user26857 The original question specified that the extension of function fields should be separable, which it isn't if $\mathrm{char}(k)=2$. Dec 28, 2014 at 16:25

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.

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.

• 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 $S \cong R \oplus (a,b)R$. Aug 1, 2011 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. Aug 1, 2011 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.

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.