Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedverified 2026-09-24 (gpt-6-sol)
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completion preserves regular local rings

Statement

Assume the Axiom of Choice (The Axiom of Choice). A nonzero Noetherian local ring R is regular if and only if its maximal-adic completion R^ is regular.

Facts & Assumptions

Given: The objects and hypotheses in the statement. We work with the Axiom of Choice; cited dependent-choice and resolution-existence hypotheses are retained.

[F1]

completion preserves embedding dimension: For a nonzero Noetherian local ring (R,m,k), its maximal-adic completion R^ has maximal ideal m^=mR^, residue field k, and a canonical isomorphism m/m2≅m^/m^2. In particular their embedding dimensions agree.

[F2]

Completion preserves dimension and Hilbert-Samuel data: Assume the Axiom of Choice. Let (R,m) be a Noetherian local ring, let M≠0 be a finitely generated R-module, and let R^, M^ denote the m-adic completions. 1. For every n≥0, M^/mn+1M^≅M/mn+1M. In particular the Hilbert-Samuel functions of M and M^ agree. 2. The Hilbert-Samuel multiplicity of M equals that of M^. 3. The support dimensions of M and M^ are equal.

Proof

1.1F1F2given

Completion preserves the embedding dimension. Under the assumed AC, apply [F2] to the nonzero finite module R: the completion dimension theorem gives dim⁡R^=dim⁡R, because the support of a ring over itself is its entire spectrum.

2.1step 1.1algebra∎

Thus edim⁡R=dim⁡R holds exactly when edim⁡R^=dim⁡R^. These are the two regularity conditions. The argument also applies when the common dimension or embedding dimension is zero.

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Dependency tree · two levels

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Sources