Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedverified 2026-09-24 (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.

regular local domain induction

Statement

Assume the Axiom of Choice (The Axiom of Choice). Every regular local ring is an integral domain.

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 ring of a regular local ring: 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.

[F2]

The Krull intersection is the (1−a)-torsion submodule, and it vanishes in the Jacobson-radical case: The first clause below is choice-free; the second uses the published Jacobson-radical unit criterion and therefore inherits its Axiom-of-Choice boundary. Let R be a Noetherian commutative ring, let I⊂R be an ideal, and let M be a finite R-module. Put K:=⋂n≥0InM. Then: 1. K is exactly the set of elements m∈M for which (1−a)m=0 for some a∈I; 2. if I⊆J(R), then K=0.

Proof

1.1F2given

Under the assumed AC, the Jacobson-radical clause of Krull intersection [F2] makes the maximal-adic filtration separated. For each nonzero a∈R there is therefore a largest integer r≥0 with a∈mr; its class in⁡(a) in degree r is nonzero.

2.1F1step 1.1algebra∎

For nonzero a,b of orders r,s, their initial classes have nonzero product in the graded polynomial ring, which is a domain: multiplying leading monomials proves this over the field k. This product is the class of ab in mr+s/mr+s+1, so ab≠0. The same argument includes r=0, s=0, and dimension zero.

Depends on

Used by

Dependency tree · two levels

11 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

  • 10.106.2 (standard reference, not scraped)