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.

embedding dimension versus dimension node

Example

For every field $k$ , the split node $R = (k [x, y] / (x y))_{(x, y)}$ is reduced and has dimension one and embedding dimension two. It is neither a domain nor regular.

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.

[F2]

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.

Verification

1.1

In $k [x, y]$ , $(x y) = (x) \cap (y)$ , as divisibility of monomials shows. It is therefore radical. Every prime of the quotient contains $x$ or $y$ ; in the origin localization either branch is the local line $k [t]_{(t)}$ , with prime chain of length one. Thus $R$ is reduced of dimension one. Both $x$ and $y$ survive and their product is zero, so it is not a domain.

givenalgebra

2.1

The relation $x y$ is quadratic, so $m / m^{2}$ has independent basis $\overset{ˉ}{x}, \overset{ˉ}{y}$ . Its dimension two strictly exceeds the dimension one just computed, and the definition makes $R$ nonregular. Localization does not change these cotangent classes since denominators have nonzero constant term. This works also in characteristic two.

F2step 1.1algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

7 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

Lecture 25, Propositions 25.6–25.8, pp.67–68 (standard reference, not scraped)