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.

fields as regular local rings

Example

Every field $k$ is regular local with dimension and embedding dimension zero. Conversely every zero-dimensional regular local ring is a field. Its regular system is empty and its residue field has a free resolution concentrated in degree zero.

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]

embedding dimension and regular local ring: For a nonzero commutative Noetherian local ring $(R, m, k)$ , define $edim R = \dim_{k} (m / m^{2})$ . The ring is regular local when $edim R = \dim R$ . The cotangent space is intrinsic, and is finite-dimensional because $m$ is finitely generated.

[F2]

Assuming the Axiom of Choice, Nakayama's lemma: Assume the Axiom of Choice. Let $R$ be a commutative ring, let $I ⊴ R$ satisfy $I \subseteq J (R)$ , and let $M$ be a finitely generated left $R$ -module. If $I M = M$ , then $M = 0$ .

Verification

1.1

A field has the single prime and maximal ideal $(0)$ , so its dimension and cotangent dimension are both zero. It is regular by definition. The empty tuple generates its maximal ideal and $k \overset{1}{\to} k$ is its length-zero augmented resolution.

F1algebra

2.1

For a zero-dimensional regular local ring, $m / m^{2} = 0$ . Nakayama applied to the finite ideal $m$ gives $m = 0$ . Every element outside the maximal ideal is a unit, so the ring is a field.

F2step 1.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

8 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

Example 12.6, p.115 (standard reference, not scraped)