TITLE: Completion of a Noetherian ring R at the ideal $ (a_1,\ldots,a_n)$
QUESTION [5 upvotes]: How can we prove that if $R$ is a commutative Noetherian ring, $\mathfrak{m} = (a_1,\ldots,a_n)$ is an ideal, then the completion of $R$ at $\mathfrak{m}$ is isomorphic to $R[[x_1,\ldots,x_n]]/(x_1-a_1,\ldots,x_n-a_n)$?

REPLY [3 votes]: Here is an argument (maybe the same as the one that Mariano sketched, but with a little more detail):
Let $S = R[[x_1,\ldots,x_n]]$, and let $T = R[[x_1,\ldots,x_n]]/(x_1-a_1,\ldots,x_n-a_n).$  The ring $S$ is Noetherian (since $R$ is), and is complete with respect to the ideal $(x_1,\ldots,x_n)$.  By Artin--Rees (for example) the finitely generated $S$-module $T$ is also complete with respect to this ideal.
Thus $T=\varprojlim T/(x_1,\ldots,x_n)^i = \varprojlim R/(a_1,\ldots,a_n)^i.$
Thus $T$ is the completion of $R$ with respect to $(a_1,\ldots,a_n)$, as claimed.