Completion of a Noetherian ring R at the ideal $ (a_1,\ldots,a_n)$

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 $R[[x_1,\ldots,x_n]]/(x_1-a_1,\ldots,x_n-a_n)$?

Solution 1:

Use the fact that completing is the same thing as tensoring with the completed ring, and that the completed ring is flat over the original ring.

(All this is true in view of your hypothesis, of course!)

Solution 2:

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.