Alphabeta Math
ExampleConstruction: AI-generatedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

completion regularity invariance

Example

The ring R=k[x,y](x,y) and its completion k[ ⁣[x,y] ⁣] both have dimension and embedding dimension two. For every q1, their quotients by the qth powers of the maximal ideals agree and have basis the monomials of total degree less than q.

Facts & Assumptions

Given: The objects and hypotheses in the example. 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/m2m^/m^2. In particular their embedding dimensions agree.

[F2]

completion preserves regular local rings: A nonzero Noetherian local ring R is regular if and only if its maximal-adic completion R^ is regular.

[F3]

Completion of a Noetherian local ring is local with the same residue field: Assume the Axiom of Choice. Let (R,m) be a Noetherian local ring, and let R^ be its m-adic completion. 1. R^ is a Noetherian local ring with maximal ideal mR^. 2. The residue field is unchanged: R^/mR^R/m. 3. The completion map RR^ is faithfully flat.

[F4]

Completion preserves dimension and Hilbert-Samuel data: Assume the Axiom of Choice. Let (R,m) be a Noetherian local ring, let M0 be a finitely generated R-module, and let R^, M^ denote the m-adic completions. 1. For every n0, 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.

[F5]

dimension at most embedding dimension: Every nonzero commutative Noetherian local ring R satisfies dimRedimR<.

Verification

1.1

Degree truncation identifies series modulo (x,y)q with polynomials modulo that ideal. In the truncated polynomial ring, every denominator allowed in R is a unit, by a finite geometric-series expansion of its nonconstant part. Thus both quotients have the stated monomial basis, and their inverse limit is k[ ⁣[x,y] ⁣], identifying it as the maximal-adic completion.

givenalgebra
2.1

The coordinate chain and two maximal-ideal generators give dimR=edimR=2. The cotangent-completion theorem preserves embedding dimension and the completion dimension theorem preserves dimension independently. The completion is Noetherian local and regular. At q=1 the quotient is k; at q=2 the basis 1,x,y exhibits the two cotangent classes.

F1F4F3F2F5step 1.1

Depends on

Used by

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Dependency tree · two levels

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Sources