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 $(X_1,…,X_n)$.
Best Answer
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 $xy = 1 + (x_1y + x_0y - 1) + \sum_{i = 2}^{\infty} x_iy,$ 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.