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I am looking for a proof of the following statement:

A Noetherian ring $R$ is regular if and only if $R[x]$ is regular if and only if $R[[x]]$ is regular.

I am trying to understand the properties of polynomial rings over DVRs, and I came across this result. Since, DVRs are regular local rings, this would mean the polynomial and power series rings over it are regular too. So I was looking to understand the proof of the above statement.

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The only place I've seen it so far is as an exercise in Eisenbud. –  John M Jul 26 '11 at 2:31

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up vote 5 down vote accepted

A noetherian local ring $(A,\mathfrak m,k)$ is regular if $dim(A)=dim_k\mathfrak m / \mathfrak m^2$. The left-hand side refers of course to Krull dimension and the number on the right-hand side is also the smallest numbr of generators of $\mathfrak m$ (by Nakayama's lemma).
In dimension zero a local noetherian ring is regular iff it is a field. So for example $k[T]/(T^2)$ is a noetherian non regular local ring of dimension zero.
In dimension one a noetherian local ring is regular iff it is a PID, just as you said.
In higher dimension regularity is to be thought as being manifold-like and is related to (but not identical with) smoothness.
Finally a noetherian ring $R$ is said to be regular if its localizations $R_\mathfrak m$ at all maximal ideals $\mathfrak m$ are regular in the sense above.

Let's change the subject.

The projective dimension $pd(M)$ of an $R$-module $M$ is the smallest integer $r$ ( or $\infty) $ such that there exists a resolution $0 \to P_r \to \ldots \to P_1 \to P_0 \to M\to 0 $ of length $r$ with all $P_i$'s projective $R$-modules.
The global (or homological) dimension $gldim(R)$ of $R$ is the supremum of the projective dimensions $pd(M)$ over all $R$-modules $M$ (it is an integer or $\infty$) .

And now for the surprise: I haven't really changed the subject! Serre, Auslander and Buchsbaum proved that a local noetherian ring is regular if and only if it has finite global dimension.
The equivalences you require then follow quickly :
$$A \; regular \iff A[X] \; regular $$ $$A \; regular \iff A[[X]] \; regular $$

For a proof see Bruns-Herzog, Cohen-Macaulay rings, 2.2.13 page 69.

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Thanks a lot for the insightful answer. –  B M Jul 30 '11 at 16:48

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