A Mind for Madness

Musings on art, philosophy, mathematics, and physics

Regular Local Rings I

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We have defined and used the associated graded ring G_a(R)=\bigoplus \frak{a}^n/\frak{a}^{n+1}. Now we want to see how it behaves under completions.

By the last post, we have \frak{a}^n/\frak{a}^{n+1}\cong \hat{\frak{a}}^n/\hat{\frak{a}}^{n+1}, so we immediately get that G_a(R)\cong G_{\hat{a}}(\hat{R}).

A great theorem that I’ll skip proving is that if R is Noetherian, and \frak{a} is any ideal, then the completion with respect to the \frak{a}-adic topology is Noetherian. As a corollary we get that for any Noetherian ring R, R[[x_1, \ldots, x_n]] is Noetherian by noting that the completion of the Noetherian ring R[x_1, \ldots, x_n] with respect to the (x_1, \ldots, x_n)-adic topology is R[[x_1, \ldots, x_n]].

After this brief excursion, we’ll come back to the dimension theory we left off from. The next natural place to go is to regular local rings. A local ring (R, \frak{m}, k) is regular if \dim_k(\frak{m}/\frak{m}^2)=\dim R. (Recall that it is always true that \dim_k(\frak{m}/\frak{m}^2)\geq \dim R).

Suppose we have a Noetherian local ring such that \dim R=d. Then the following are equivalent definitions of regular: G_m(A)\cong k[t_1, \ldots, t_d] or \frak{m} can be generated by d elements.

The first condition implies that \dim_k(\frak{m}/\frak{m}^2)=d, so it implies regular. Regular implies that \frak{m} can be generated by d elements, by this post. Lastly, if \frak{m} can be generated by x_1, \ldots, x_d (if you’ve seen the term, this is a system of parameters), then we have a surjective map of graded rings \phi: k[x_1, \ldots, x_d]\to G_m(A) with kernel \cap \frak{m}^n=0. So it is an iso. This finishes up the equivalences.

Last time we saw without proof that (for Noetherian local rings) R is regular if and only if \hat{R} is regular. Now we can prove it.

By the equivalent definition of regular, R is regular iff G_m(R)\cong k[t_1, \ldots, t_n], but we proved that G_m(R)\cong G_{\hat{m}}(\hat{R}), so G_{\hat{m}}(\hat{R})\cong k[t_1, \ldots, t_d] but this happens iff \hat{R} is regular.

We’ll wrap up today with trying to keeping the geometric picture in mind. Regular means non-singular geometrically. So we see that passing to the completion doesn’t introduce any singularities. But since the dimension of the local ring at a point equals the dimension of the variety we actually get that completion of \mathcal{O}_P where P is non-singular is k[[x_1, \ldots, x_n]] where n is the dimension of the variety.

So if we interpret completion of \mathcal{O}_P as the “analytically local” picture, then we see that locally all non-singular points on a variety are analytically isomorphic.

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Author: hilbertthm90

I am a mathematics graduate student fascinated in how all my interests fit together.

2 thoughts on “Regular Local Rings I

  1. “But since the dimension of the local ring at a point equals the dimension of the variety we actually get that completion of \mathcal{O}_P where P is non-singular is k[[x_1, \ldots, x_n]] where n is the dimension of the variety.”

    Can you elaborate on this? In my understanding this is Cohen structure theorem which I guess is nontrivial, but the way you mention it makes it seemingly an apparent fact.

  2. Oops. Sorry. I thought I had mentioned in the previous post the Cohen Structure Theorem as something that is there, but I wasn’t going to prove. Sorry about that. It definitely is not obvious without that.

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