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The completion of a Noetherian ring is flat

Statement

Assume the Axiom of Choice.

Let $R$ be a Noetherian commutative ring and let $I \subseteq R$ be an ideal. Then the completion map

R \to \hat{R}

makes $\hat{R}$ into a flat $R$ -module.

Facts & Assumptions

Given: A Noetherian commutative ring $R$ and an ideal $I \subseteq R$ .

[L1]

For a Noetherian ring, completion identifies finite modules with extension of scalars, and it carries ideal multiples to the corresponding multiples after tensoring (Completion of a finite module is extension of scalars, Completion commutes with finite quotients and induced submodules).

[L2]

An $R$ -module is flat exactly when $J \otimes_{R} M \to M$ is injective for every ideal $J \subseteq R$ (Flatness is equivalent to preserving injections and to the ideal and finitely generated ideal tests).

[L3]

Completion is exact on short exact sequences of finite modules over a Noetherian commutative ring (Adic completion is exact on finite modules over a Noetherian ring).

[L4]

Every ideal of a Noetherian commutative 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).

Proof

technique · direct

1.1

Let $J \subseteq R$ be an ideal. It is a finite $R$ -module by [L4]. Hence [L1] identifies $J \otimes_{R} \hat{R}$ with the completion $\hat{J}$ , and under this identification the multiplication map $J \otimes_{R} \hat{R} \to \hat{R}$ becomes the natural map $\hat{J} \to \hat{R} .$

L1L4

2.1

The map $\hat{J} \to \hat{R}$ is injective because [L3] applies to the short exact sequence $0 \to J \to R \to R / J \to 0.$ Therefore $J \otimes_{R} \hat{R} \to \hat{R}$ is injective for every ideal $J$ .

L3step 1.1

3.1

Applying [L2], we conclude that $\hat{R}$ is flat as an $R$ -module.

L2step 2.1∎

Depends on

Used by

Jacobson-adic completion is faithfully flat Theorem

Dependency tree · two levels

27 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., Corollary 22.23 (standard reference, not scraped)
J. S. Milne, A Primer of Commutative Algebra, Proposition 24.6 (standard reference, not scraped)
The Stacks Project, Lemma 10.97.2 (standard reference, not scraped)