TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01

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Finite modules over complete Noetherian rings are complete

Statement

Assume the Axiom of Choice.

Let $R$ be a Noetherian commutative ring, let $I \subseteq R$ be an ideal, and assume that $R$ is $I$ -adically complete. Then every finitely generated $R$ -module $M$ is $I$ -adically complete.

Facts & Assumptions

Given: A Noetherian commutative ring $R$ , an ideal $I \subseteq R$ , and a finitely generated $R$ -module $M$ , with $R$ $I$ -adically complete.

[L1]

Completion of a finite module is extension of scalars: $M \otimes_{R} \hat{R} \tilde{\to} \hat{M}$ (Completion of a finite module is extension of scalars).

[L2]

An $I$ -adically complete ring is identified with its own completion (Separated and complete filtered modules, The $I$ -adic completion of a module).

Proof

technique · direct

1.1L2algebra

Since $R$ is $I$ -adically complete, [L2] gives an isomorphism $κ_{R} ⁣ : R \tilde{\to} \hat{R} .$ Tensoring with the finite module $M$ yields an isomorphism $M \otimes_{R} R \tilde{\to} M \otimes_{R} \hat{R} .$ Using the unit isomorphism $M \otimes_{R} R ≅ M$ , we get an isomorphism $M \tilde{\to} M \otimes_{R} \hat{R} .$

2.1L1step 1.1∎

Composing the isomorphism of step 1.1 with [L1] gives an isomorphism $M \tilde{\to} \hat{M} .$ By construction this composite sends $m$ to the compatible residue class system $(m m o d I^{n} M)_{n}$ , so it is exactly the completion map $κ_{M}$ . Therefore $κ_{M}$ is an isomorphism and $M$ is $I$ -adically complete.

Depends on

Used by

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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., Lemma 22.27 (standard reference, not scraped)
The Stacks Project, Lemma 10.96.11 (standard reference, not scraped)