# Completion of Noetherian ring $A$ by $I=(a_1,\ldots,a_n)$ is isomorphic to $A[[x_1,\ldots,x_n]]/(x_1-a_1,\ldots,x_n-a_n)$

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.

• 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. – Pierre-Yves Gaillard Apr 18 '18 at 22:57

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 $\mathfrak{m}^{\ell}$.
• 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) – jta Apr 24 '18 at 11:25