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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 IR 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 IR, and a finitely generated R-module M, with R I-adically complete.

[L1]

Completion of a finite module is extension of scalars: MRR^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
1.1

Since R is I-adically complete, [L2] gives an isomorphism κR ⁣:RR^. Tensoring with the finite module M yields an isomorphism MRRMRR^. Using the unit isomorphism MRRM, we get an isomorphism MMRR^.

L2algebra
2.1

Composing the isomorphism of step 1.1 with [L1] gives an isomorphism MM^. By construction this composite sends m to the compatible residue class system (mmodInM)n, so it is exactly the completion map κM. Therefore κM is an isomorphism and M is I-adically complete.

L1step 1.1

Depends on

Used by

Nothing in the library uses this result yet.

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