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Completion of a Noetherian ring is Noetherian

Statement

Assume the Axiom of Choice.

Let $R$ be a Noetherian commutative ring and let $I \subseteq R$ be an ideal. Then the $I$ -adic completion $\hat{R}$ is a Noetherian ring.

Facts & Assumptions

Given: The Axiom of Choice, a Noetherian commutative ring $R$ , and an ideal $I \subseteq R$ .

[L1]

Quotients of Noetherian rings are Noetherian (Every quotient and every localisation of a Noetherian ring is Noetherian).

[L2]

Every ideal of a Noetherian ring is finitely generated (A commutative ring is Noetherian exactly when every ideal is finitely generated, exactly when its ideals satisfy the ascending chain condition, and exactly when every nonempty set of ideals has a maximal member).

[L3]

A finite-variable polynomial ring over a Noetherian ring is Noetherian (If $R$ is Noetherian then $R [x_{1}, \dots, x_{n}]$ is Noetherian for every $n \in N$ ).

[L4]

For a finite module, completion commutes with quotients and ideal powers (Completion commutes with finite quotients and induced submodules).

Proof

technique · direct

1.1

By [L2], choose generators $f_{1}, \dots, f_{r}$ of $I$ . The quotient $R / I$ is Noetherian by [L1], hence the polynomial ring $(R / I) [T_{1}, \dots, T_{r}]$ is Noetherian by [L3].

L1L2L3choose

2.1

The graded ring ${gr}_{I} (R) : = ⨁_{n \geq 0} I^{n} / I^{n + 1}$ is a quotient of $(R / I) [T_{1}, \dots, T_{r}]$ via $T_{j} \mapsto \overline{f_{j}}$ , so ${gr}_{I} (R)$ is Noetherian.

step 1.1algebra

3.1

By part 3 of Completion commutes with finite quotients and induced submodules, $\hat{R} / I^{n} \hat{R} ≅ R / I^{n}$ for every $n$ , and part 2 identifies $I^{n} \hat{R}$ with the completion of $I^{n}$ . Therefore $I^{n} \hat{R} / I^{n + 1} \hat{R} ≅ I^{n} / I^{n + 1}$ for all $n$ , so the associated graded ring ${gr}_{I \hat{R}} (\hat{R})$ is canonically isomorphic to ${gr}_{I} (R)$ . Hence it is Noetherian.

L4step 2.1

4.1

Let $J \subseteq \hat{R}$ be an ideal. Since ${gr}_{I \hat{R}} (\hat{R})$ is Noetherian, [L2] makes the graded ideal $gr (J) : = ⨁_{n \geq 0} \frac{J \cap I^{n} \hat{R}}{J \cap I^{n + 1} \hat{R}}$ finitely generated. Taking homogeneous components of a finite generating set, choose homogeneous generators ${\overline{g}}_{1}, \dots, {\overline{g}}_{m}$ , where $g_{j} \in J \cap I^{d_{j}} \hat{R}$ .

L2step 3.1choosealgebra

5.1

We claim that $g_{1}, \dots, g_{m}$ generate $J$ . Put $r_{0} = x \in J$ . Inductively, suppose $r_{n} \in J \cap I^{n} \hat{R}$ . Express its degree- $n$ class in $gr (J)$ as ${\overline{r}}_{n} = \sum_{d_{j} \leq n} {\overline{a}}_{j, n} {\overline{g}}_{j},$ with ${\overline{a}}_{j, n}$ homogeneous of degree $n - d_{j}$ . Lift it to $a_{j, n} \in I^{n - d_{j}} \hat{R}$ , and put $a_{j, n} = 0$ when $d_{j} > n$ . Then $r_{n + 1} : = r_{n} - \sum_{j} a_{j, n} g_{j} \in J \cap I^{n + 1} \hat{R} .$ The assumed Choice principle supports this countable recursion. Consequently, for every $N \geq 0$ , $x - \sum_{n = 0}^{N} \sum_{j} a_{j, n} g_{j} = r_{N + 1} \in J \cap I^{N + 1} \hat{R} .$

step 4.1inductionchoosealgebra

6.1

The completion ring $\hat{R}$ is complete for the $I \hat{R}$ -adic topology because it is already the inverse limit of the quotients $R / I^{n}$ and step 3.1 identifies these with $\hat{R} / I^{n} \hat{R}$ . For fixed $j$ , step 5.1 has $a_{j, n} = 0$ for $n < d_{j}$ and $a_{j, n} \in I^{n - d_{j}} \hat{R}$ thereafter, so its partial sums are Cauchy and converge to some $A_{j} \in \hat{R}$ . For $N \geq d_{j}$ , the tail satisfies $A_{j} - \sum_{n = 0}^{N} a_{j, n} \in I^{N + 1 - d_{j}} \hat{R} .$ Multiplying by $g_{j} \in I^{d_{j}} \hat{R}$ and using step 5.1 gives $x - \sum_{j} A_{j} g_{j} \in I^{N + 1} \hat{R}$ for every sufficiently large $N$ . Completeness includes separatedness, so the intersection of the powers is $0$ and therefore $x = \sum_{j} A_{j} g_{j}$ . Thus $J = (g_{1}, \dots, g_{m})$ is finitely generated.

step 3.1step 5.1algebra

7.1

Every ideal $J \subseteq \hat{R}$ is finitely generated, so $\hat{R}$ is Noetherian by the ideal characterization.

L2step 6.1∎

Depends on

Used by

Dependency tree · two levels

23 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

A. Altman and S. Kleiman, A Term of Commutative Algebra, 13th ed., Theorem 22.30 (standard reference, not scraped)
The Stacks Project, Lemmas 10.97.5-10.97.6 (standard reference, not scraped)