# Regular Local Rings I

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.

### Author: hilbertthm90

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### 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.