# When is the completion of a ring a local ring ?

Let $R$ be a commutative ring with unit and let $m$ be a maximal ideal of $R$. Are there known conditions on $R$ or $m$ such that the $m$-adic completion $\hat{R}$ of $R$ is a local ring.

Since the completion of a Noetherian local ring is again local, I'm primarily interested in cases where $R$ itself is not local.

An example is the polynomial ring $R=k[X_1,...,X_n]$ ($k$ a field) with $m=(X_1,...,X_n)$ where the power series ring $\hat{R}=k[[X_1,...,X_n]]$ is local with maximal ideal $(X_1,...,X_n)$.

• @BenjaLim: My original title was "When is the completion of a ring local ?" Was there linguistically something wrong with it ? – Ralph Dec 21 '12 at 23:36
• I believe it is called a local ring. en.wikipedia.org/wiki/Local_ring – user38268 Dec 21 '12 at 23:37
• Dear Ralph, As you likely know, there was nothing linguisitcally wrong with your original title, although a sentence ending with an adjective like this which applies not to the noun immediately preceding it ("a ring") but to a noun introduced earlier ("the completion") can be occassionally confusing on a quick reading, and perhaps BenjaLim misinterpreted its meaning. Regards, – Matt E Dec 22 '12 at 0:46

## 3 Answers

The completion of a ring with respect to a maximal ideal is always local.

Proof: If $x \in \hat{R}$, then we may write $x = \sum_{i=0}^{\infty} x_i,$ where $x_i \in \mathfrak m^i$. If $x_0 \not\in m,$ then I claim that $x$ is a unit. Indeed, in this case we may find $y \in R$ such that $x_0 y \equiv 1 \mod m,$ and so $xy = 1 + (x_1 + xy - 1) + \sum_{i = 2}^{\infty} x_i,$ and so it suffices to show that $\sum_{i=0}^{\infty} x_i$ is a unit under the additional assumption that $x_0 = 1$. But then we can construct an explicit inverse for $x$ using the formula for a geometric series: $x^{-1} = 1 + (x_1 + x_2 + \cdots) + (x_1 + x_2 + \cdots )^2 + \cdots.$

Thus the kernel of the map $x \mapsto x_0 \bmod m$ (i.e. the kernel of the natural projection $\hat{R} \to R/m$) has the property that its complement consists of units, and so it must be the unique maximal ideal of $\hat{R}$, and so $\hat{R}$ is local. QED

• Thanks for your answer. I learned this result from YACP's answer here: math.stackexchange.com/questions/263468/…. But if you have an alternative reference, I would also appreciate it. – Ralph Dec 22 '12 at 1:40

With Matt E's simple answer, the following answer may be of little help. But I'll post it nevertheless.

Recall that the completion is the inverse limit $\hat{R}=\varprojlim (R/m^i)$ which can be described as a subring of the product $\prod (R/m^i)$. $R/m^i$ is a local ring($m/m^i$ is a nilpotent maximal ideal) in which every element outside $m/m^i$ is a unit (let $w$ be in the complement, together with $m/m^i$ it generates the unit ideal, so $vw+n=1$ for some $n\in m/m^i$ and some $v$; thus, $vw$ being a sum of a unit and nilpotent is itself a unit).

The ideal $M\subseteq \hat{R}$ given by $M=\{(x_1,x_2,...)\in \hat{R}|x_1= 0\}$ is maximal since it is the kernel of the surjective homomorphism $\hat{R}\rightarrow R/m$ onto a field. Further we can see that any element of $\hat{R}$ outside $M$ is a unit as follows. Given $(x_1,x_2,...)$ with $x_1\neq 0$ we have an inverse $y_1\in R/m$. Now $x_2$ is non nilpotent since its homomorphic image $x_1$ is. Hence, $x_2$ is a unit by the observation in the previous answer. Moreover the image of $y_2$ in $R/m$ is $y_1$ (if $\varphi:R/m^2\rightarrow R/m$ is the canonical surjection, $x_1\varphi(y_2)=\varphi(x_2y_2)=1$ so $\varphi(y_2)=y_1x_1\varphi(y_2)=y_1$). Proceeding inductively with the same method, we get $y_i$ as the inverse of $x_i$ with the $y_i$'s satisfying the required compatibility and thus $y=(y_1,y_2,...)\in \hat{R}$ as the inverse of $(x_1,x_2,...)\notin M$.

Thus $M\subseteq\hat{R}$ is a maximal ideal with every element in the complement as a unit, hence $\hat{R}$ is local.

It is easy to show $m\hat{R}$ is maximal ideal of $\hat{R}$. Now we know that topology on $\hat{R}$ is same as $m\hat{R}$-adic topology on $\hat{R}$ as an $\hat{R}$- module. so it is complete with respect to the later one and so $m\hat{R}$ is contained in rad$(\hat{R})$. therefore it is unique maximal ideal.

## protected by user26857Sep 20 '15 at 18:50

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?