# Completion of a polynomial ring w.r.t. a maximal ideal

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 ?.

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.
• 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 ? Dec 22, 2012 at 1:07