Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedverified 2026-09-26 (gpt-6-sol)
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.

associated graded ring of a regular local ring

Statement

Assume the Axiom of Choice (The Axiom of Choice). If (R,m,k) is regular local of dimension d, any cotangent basis induces a graded isomorphism k[X1,…,Xd]≅gr⁡mR. Conversely, if the associated graded ring of a nonzero Noetherian local ring is isomorphic as a graded k-algebra to k[X1,…,Xd] with standard grading, then R is regular of dimension d.

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]

associated graded polynomial surjection: Let (R,m,k) be nonzero Noetherian local and let x1,…,xe lift a basis of m/m2. There is a surjective graded k-algebra map ϕ:k[X1,…,Xe]→gr⁡mR, determined by Xi↦xi+m2, with every variable of degree one.

[F2]

regular local graded surjection has zero kernel: Assume the Axiom of Choice. For the graded map ϕ:k[X1,…,Xe]↠gr⁡mR defined by a cotangent basis in a nonzero Noetherian local ring, if e=dim⁡R, then ker⁡ϕ=0.

[F3]

The degree of the Hilbert-Samuel polynomial equals the dimension of the support: Assume the Axiom of Choice. Let (R,m) be a Noetherian local ring, let M≠0 be a finite R-module, and let I be an ideal of definition for M. Then the Hilbert-Samuel polynomial PI,M has degree deg⁡PI,M=dim⁡Supp⁡(M).

Proof

1.1F1F2

In a regular local ring the cotangent dimension equals d. The polynomial map is surjective and has zero kernel, hence is the claimed isomorphism.

2.1F3givenalgebra∎

Conversely, the degree-one component of a supplied graded isomorphism has dimension d, so edim⁡R=d. Its cumulative graded dimensions are (n+dd), also when d=0, where the count is one. These are the lengths of R/mn+1. The Hilbert–Samuel dimension theorem gives dim⁡R=d, proving regularity.

Depends on

Used by

Dependency tree · two levels

14 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