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 $(X_1,...,X_n)$.

  • $\begingroup$ @BenjaLim: My original title was "When is the completion of a ring local ?" Was there linguistically something wrong with it ? $\endgroup$
    – Ralph
    Dec 21, 2012 at 23:36
  • $\begingroup$ I believe it is called a local ring. en.wikipedia.org/wiki/Local_ring $\endgroup$
    – user38268
    Dec 21, 2012 at 23:37
  • 1
    $\begingroup$ 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, $\endgroup$
    – Matt E
    Dec 22, 2012 at 0:46

3 Answers 3


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 $x_i \in m^i$. 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 $xy = 1 + (x_1y + x_0y - 1) + \sum_{i = 2}^{\infty} x_iy,$ 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. This completes the proof.

  • $\begingroup$ 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. $\endgroup$
    – Ralph
    Dec 22, 2012 at 1:40
  • $\begingroup$ $\def\m{\widehat{m^n}} \def\R{\hat{R}}$Denote $y=x-1\in\R$. The series $\sum_{i=0}^{+\infty}(-y)^i$ converges in $\R$ by using that $\R$ is complete (as a topological ring) and the fact that the sequence of partials sums of this series is Cauchy in $\R$: indeed, it suffices to see that $y^n\in\m$. Since the product is continuous in $\R$, $$y^n=\prod_{j=1}^n\left(\lim_N\sum_{i=1}^Nx_i\right)=\lim_N\underbrace{\prod_{j=1}^n\sum_{i=1}^Nx_i}_{\in m^n\R\subset\m}.$$ Hence $y^n$ is a limit point of $\m$. But $\m$ is closed in $\R$ (Atiyah, MacDonald, p. 102, last paragraph), so $y^n\in\m$. $\endgroup$ Sep 18, 2023 at 15:51

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.

  • $\begingroup$ In general, the $m\hat{R}$-adic topology is not the topology on $\hat{R}$ (see equation (1) from here and discussion below). A sufficient condition when this property holds is when $R$ is Noetherian; see e.g. Atiyah, MacDonald, Introduction to Commutative Algebra, (10.15) iv). Even in the Noetherian case, proving the fact that $m\hat{R}\subset\operatorname{rad}(\hat{R})$ is not immediate and requires some work (cf. ibid.). $\endgroup$ Sep 18, 2023 at 12:41

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