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Let $R=k[X_1,...,X_n]$ be a polynomial ring over the field $k$ and let $\mathfrak m$ be a maximal ideal of $R$.

Question: Is the $\mathfrak m$-adic completion of $R$ a local ring ?

If $\mathfrak m=(X_1-a_1,\ldots,X_n-a_n)$ with $a_i \in k$ then the $\mathfrak m$-adic completion of $R$ is the local ring $k[[X_1-a_1,\ldots,X_n-a_n]]=k[[X_1,\ldots,X_n]]$. In particular, the question has an affirmative answer if $k$ is algebraically closed.

A related question is When is the completion of a ring a local ring ?.

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In general, if $R$ is a noetherian ring and $\mathfrak m$ a maximal ideal of $R$, then $\hat R$ (the $\mathfrak m$-adic completion of $R$) is a noetherian local ring.

Write $\mathfrak m=(a_1,\dots,a_n)$. Then $$\hat R\simeq R[[X_1,\dots,X_n]]/(X_1-a_1,\dots,X_n-a_n).$$ A maximal ideal of $R[[X_1,\dots,X_n]]$ has the form $$M=\mathfrak n R[[X_1,\dots,X_n]]+(X_1,\dots,X_n)$$ with $\mathfrak n\subset R$ a maximal ideal. Since $(X_1-a_1,\dots,X_n-a_n)\subseteq M$ we must have $a_i\in\mathfrak n$ for all $i$, that is, $\mathfrak m=\mathfrak n$. As a consequence we get $$M=\mathfrak m R[[X_1,\dots,X_n]]+(X_1,\dots,X_n)$$ and this is the only maximal ideal of $R[[X_1,\dots,X_n]]$ containing $(X_1-a_1,\dots,X_n-a_n)$.

Added in proof. I've found here, on page $6$, a more general result: the $I$-adic completion $\hat R$ is quasi-local iff $R/I$ is quasi-local. (Quasi-local means local, but not necessarily noetherian.) In the noetherian case the proof goes exactly as before.

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  • $\begingroup$ Great. Thanks. The paper says there is a bijection between the max. ideals of the $I$-adic completion $\hat{R}$ of $R$ and $R/I$. So, don't we even have an iff-condition: $\hat{R}$ is local iff $R/I$ is local ? $\endgroup$
    – Ralph
    Dec 22, 2012 at 1:07

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