Alphabeta Math
LemmaStatement: Literature-sourcedProof: 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.

embedding dimension is minimal maximal ideal generator number

Statement

For a nonzero Noetherian local ring (R,m,k), edimR is the least number of generators of m.

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]

embedding dimension and regular local ring: For a nonzero commutative Noetherian local ring (R,m,k), define edimR=dimk(m/m2). The ring is regular local when edimR=dimR. The cotangent space is intrinsic, and is finite-dimensional because m is finitely generated.

[F2]

Assuming the Axiom of Choice, minimal generators over a local ring are exactly residue-field bases: Assume the Axiom of Choice. Let (R,m) be a local ring with residue field k=R/m, and let M be a finitely generated left R-module. A finite generating set x1,,xr of M is minimal if and only if the images of x1,,xr in M/mM form a k-basis. In particular every minimal generating set of M has the same cardinality.

[F3]

Assuming the Axiom of Choice, generators modulo an ideal in the Jacobson radical lift to generators: Assume the Axiom of Choice. Let R be a commutative ring, let IR satisfy IJ(R), and let M be a finitely generated left R-module. If elements x1,,xrM generate M/IM, then x1,,xr generate M.

Proof

1.1

Write e=dimkm/m2. Lift a basis to x1,,xem. Since m is finite and m=J(R), Nakayama gives m=(x1,,xe). If e=0, the same assertion gives m=0.

F1F3
2.1

Any generating tuple of m spans its quotient by m2, so its length is at least e. The lifted basis is a minimal generating tuple by the local generator criterion. Thus the least length is e.

F2step 1.1algebra

Depends on

Used by

Dependency tree · two levels

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Sources