TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01

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Jacobson-adic completion is faithfully flat

Statement

Assume the Axiom of Choice.

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

R \to \hat{R}

is faithfully flat.

Facts & Assumptions

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

[L1]

The completion $\hat{R}$ is flat over $R$ (The completion of a Noetherian ring is flat).

[L2]

For every maximal ideal $m \supseteq I$ , one has $\hat{R} / m \hat{R} ≅ \hat{R / m} ≅ R / m,$ because completion commutes with finite quotients and the $I$ -adic filtration on $R / m$ is already zero after one step (Completion commutes with finite quotients and induced submodules).

[L3]

A flat ring map is faithfully flat exactly when every maximal ideal of the source has a prime ideal lying over it (A flat ring map is faithfully flat exactly when it detects proper ideals and is surjective on spectra).

Proof

technique · direct

1.1

Let $m \subset R$ be a maximal ideal. Because $I \subseteq J (R)$ , one has $I \subseteq m$ , so [L2] gives $\hat{R} / m \hat{R} ≅ R / m \neq 0.$

L2given

2.1

The ring $R / m$ is a field, so step 1.1 shows directly that $m \hat{R}$ is a maximal, hence prime, ideal of $\hat{R}$ . Its contraction to $R$ contains $m$ ; the contraction is proper because it is the preimage of a prime ideal, so maximality of $m$ forces the contraction to equal $m$ . Thus every maximal ideal of $R$ has a prime ideal of $\hat{R}$ above it.

step 1.1algebra

3.1

The completion map $R \to \hat{R}$ is flat by [L1]. Together with step 2.1, [L3] now shows that $R \to \hat{R}$ is faithfully flat.

L1L3step 2.1∎

Depends on

Used by

Completion of a Noetherian local ring is local with the same residue field Theorem

Dependency tree · two levels

10 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., Exercise 22.24 (standard reference, not scraped)
The Stacks Project, Lemma 10.97.3 (standard reference, not scraped)