Finally, we’re going to come to the Kahler criterion for regularity. As far as algebraic geometry is concerned, it states that a variety over an algebraically closed field of characteristic zero is nonsingular precisely when the sheaf of differentials on it (to be defined shortly) is locally free of rank equal to the dimension.

**Theorem 1** *Suppose is a local domain which is a localization of a finitely generated -algebra for a field of characteristic zero, with residue field . Then is a regular local ring if and only if is a free -module of rank . *

First, I claim is finitely generated. This follows because the corresponding claim is true for a polynomial ring, we have a conormal sequence implying it for finitely generated algebras over a field, and taking differentials commutes with localization.

Let be the residue field of . I claim

Then, the theorem will follow from the next lemma:

**Lemma 2**

*Let be a finitely generated module over the local noetherian domain , with residue field and quotient field . Then is free iff*
* *

* *

*
*

The lemma is easy and fun, so let’s prove it here. If is free, the equality is immediate. Suppose the equality holds. In any case, there is an epimorphism for free, say with kernel . Then

is exact (localization is an exact functor) and

is exact. The arrow on the left is injective in view of a dimension count that works because of the hypothesis. This holds for any surjection . In fact, by Nakayama’s lemma, we can construct of rank with an isomorphism, implying , implying by Nakayama again.

Anyway, the equality follows because the two are *isomorphic*; this is a refinement of the conormal sequence in this case.

For the second equality, note that , whose dimension is the transcendence degree of over . This actually equals , because is the localization of a finitely generated -algebra at a maximal ideal (since is also the residue field).

(I should note incidentally that finitely generated domains over a field are very nice. Their dimension is equal to the transcendence degree of their quotient field; this is a conseqeuence of the Noether normalization lemma and the fact that Krull dimension is preserved by integral extensions; cf. e.g. hilbertthm90’s two posts at A Mind for Madness. Also, such rings are catenarian: every maximal chain of prime ideals has length equal to the Krull dimension. In particular, any localization at a maximal ideal has the same dimension.)

You don’t actually need characteristic zero for any of this.

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February 18, 2010 at 12:42 am

The really frustrating thing about this picture is that it only makes sense in the function field case; I don’t think anyone has a good idea of what the right notion of Kahler differential should be in the number field case. (The problem, of course, being that Kahler differentials should be F_1-linear instead of Z-linear, whatever that means.)

February 18, 2010 at 7:02 pm

What do you mean by F_1? (I assume it is “the field with one element” but I really can’t see how it’s relevant in characteristic zero.)

February 20, 2010 at 8:42 pm

Briefly, the problem is this: the localizations of the ring of integers of a number field are all regular local, but one can’t detect this using Kahler differentials because rings of integers aren’t algebras over a field, so the naive definition of Kahler differential gives you nonsense. They “should” be algebras over F_1, so the “right” definition of Kahler differential in this setting should have F_1 as its “field of constants” and should be “F_1-linear.” Of course, until anyone figures out what this actually means it’s nonsense.

A good enough theory of differentiation in the F_1 setting would imply a lot of awesome theorems in the arithmetic setting; most notably it should be possible to prove the abc conjecture in this way by adapting the proof of the Mason-Stothers theorem.

February 20, 2010 at 8:45 pm

Do you have a source for this sort of stuff? It sounds interesting enough.

February 21, 2010 at 12:48 pm

This is mostly hearsay I’ve picked up from random places like Math Overflow, unfortunately.