2
$\begingroup$

Let $A$ be a Noetherian ring, $I=(a_1,\ldots,a_n)$ an ideal of $A$. I want to show that completion of $A$ by $I$ (denote that by $\hat{A}$) is isomorphic to $A[[x_1,\ldots,x_n]]/(x_1-a_1,\ldots,x_n-a_n)$, hence Noetherian (since the formal power series is Noetherian, and so is the quotient).

I've seen some different proofs for this (both in Matsumura's Commutative Ring Theory and Atiyah, MacDonalds Introduction to Commutative Algebra), but I want to explicitly show we have a isomorphism $\hat{A}\cong A[[x_1,\ldots,x_n]]/(x_1-a_1,\ldots,x_n-a_n)$. I've been trying to construct a map doing this, but so far without any luck.

$\endgroup$
1
  • 1
    $\begingroup$ Hint: if $f(x_1,...,x_n)\in A[[x_1,...,x_n]]$, then $f(a_1,...,a_n)$, viewed as a series in $\hat A$, converges in the $I$-adic topology. $\endgroup$ Commented Apr 18, 2018 at 22:57

1 Answer 1

2
$\begingroup$

This is more or less the proof in Matsumura, Theorem 8.12. Let $\mathfrak{m} := (x_{1},\dotsc,x_{n}) \subset A[x_{1},\dotsc,x_{n}]$ be the ideal. Note that $$ \textstyle \hat{A} = \varprojlim_{\ell \in \mathbb{N}} A/I^{\ell} $$ and $$ \textstyle A[[x_{1},\dotsc,x_{n}]]/(x_{1}-a_{1},\dotsc,x_{n}-a_{n}) \simeq \varprojlim_{\ell \in \mathbb{N}} A[x_{1},\dotsc,x_{n}]/(\mathfrak{m}^{\ell},x_{1}-a_{1},\dotsc,x_{n}-a_{n}) $$ where in the above isomorphism we use that $A$ is Noetherian, show that the $A$-algebra map $$ A[x_{1},\dotsc,x_{n}]/(\mathfrak{m}^{\ell},x_{1}-a_{1},\dotsc,x_{n}-a_{n}) \to A/I^{\ell} $$ sending $x_{i} \mapsto a_{i}$ is an isomorphism, then take projective limit.

The map $\varphi : A[x_{1},\dotsc,x_{n}] \to A/I^{\ell}$ sending $x_{i} \mapsto a_{i}$ is surjective, and $(\mathfrak{m}^{\ell},x_{1}-a_{1},\dotsc,x_{n}-a_{n}) \subseteq \ker \varphi$. Suppose $f(x_{1},\dotsc,x_{n}) \in \ker \varphi$. Reducing modulo $x_{1}-a_{1},\dotsc,x_{n}-a_{n}$, we may assume that $f = a \in A$ is a constant; then $a \in \ker\varphi$ means $a \in I^{\ell}$, namely we can write $a = \sum_{\mathbf{e}} c_{\mathbf{e}}a_{1}^{e_{1}} \dotsb a_{n}^{e_{n}}$ where $\mathbf{e} = (e_{1},\dotsc,e_{n})$ ranges over elements of $(\mathbb{Z}_{\ge 0})^{\oplus n}$ such that $e_{1} + \dotsb + e_{n} = \ell$. Then $a$ is equivalent modulo $x_{1}-a_{1},\dotsc,x_{n}-a_{n}$ to $\sum_{\mathbf{e}} c_{\mathbf{e}}x_{1}^{e_{1}} \dotsb x_{n}^{e_{n}}$, which lies in $\mathfrak{m}^{\ell}$.

$\endgroup$
5
  • $\begingroup$ Is it a general fact, that taking projective limit of an isomorphism, preserves the isomorphism? $\endgroup$
    – njlieta
    Commented Apr 20, 2018 at 9:27
  • 1
    $\begingroup$ Yes, use either the universal property or the explicit construction as a subset of the infinite product. $\endgroup$ Commented Apr 20, 2018 at 21:04
  • 1
    $\begingroup$ I'm not sure if it's obvious but I don't think it's tricky. $\endgroup$ Commented Apr 22, 2018 at 19:17
  • $\begingroup$ I think I might have been a little too quick without thinking it through. I'm not sure if exactly understand the map. How does the elements of $A[x_1,\ldots,x_n]/(m^l,x_1-a_1,\dots,x_n-a_n)$ resp. $A/I^l$ look like (why does it makes sense to take $x_i$ to $a_i$? I can't really picture it) $\endgroup$
    – njlieta
    Commented Apr 24, 2018 at 11:25
  • 1
    $\begingroup$ I added the argument to my answer. $\endgroup$ Commented Apr 24, 2018 at 20:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .