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

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@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

2 Answers 2

up vote 5 down vote accepted

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

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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

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.

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