5

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

user 1
  • 7,447
  • 7
    You don't lose anything from having your entire question in the title. Not having a clear indication of what the question is about just makes it less likely that the people who might be able to help you will take a look. – Qiaochu Yuan Jan 06 '11 at 13:58
  • 5
    (It never hurts to make explicit the fact that your rings are commutative!) – Mariano Suárez-Álvarez Jan 06 '11 at 14:16

2 Answers2

4

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!)

Mariano Suárez-Álvarez
  • 135,076
  • 1
    I can't see the direction you say Mariano. Can you give me some other hint? –  Jan 07 '11 at 12:31
3

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.

user26857
  • 52,094
Matt E
  • 123,735
  • Is the 'Noetherian' assumption necessary? If $R$ is commutative and $\mathfrak{a}=(a_1,\ldots,a_n)$, can $\widehat{R}_\mathfrak{a}\ncong R[[x_1,\ldots,x_n]]/(x_1-a_1,\ldots,x_n-a_n)$ happen? – Leo Oct 03 '14 at 00:07