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 ?.
Best Answer
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 that $M=\mathfrak m R[[X_1,\dots,X_n]]+(X_1,\dots,X_n)$ 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 I did before.