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Finite modules over complete Noetherian rings are complete
Statement
Assume the Axiom of Choice.
Let be a Noetherian commutative ring, let be an ideal, and assume that is -adically complete. Then every finitely generated -module is -adically complete.
Facts & Assumptions
Given: A Noetherian commutative ring , an ideal , and a finitely generated -module , with -adically complete.
Completion of a finite module is extension of scalars: (Completion of a finite module is extension of scalars).
An -adically complete ring is identified with its own completion (Separated and complete filtered modules, The -adic completion of a module).
Proof
Since is -adically complete, [L2] gives an isomorphism Tensoring with the finite module yields an isomorphism Using the unit isomorphism , we get an isomorphism
Composing the isomorphism of step 1.1 with [L1] gives an isomorphism By construction this composite sends to the compatible residue class system , so it is exactly the completion map . Therefore is an isomorphism and is -adically complete.
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
- 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)