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.

cusp local ring not regular

Example

For every field $k$ , the cusp local ring $R = (k [x, y] / (y^{2} - x^{3}))_{(x, y)}$ has dimension one and embedding dimension two, hence is not 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.

[F1]

quotient and lifting regularity across a regular element: Let $(R, m)$ be nonzero Noetherian local. If $x \in m$ is a nonzerodivisor and $R / (x)$ is regular, then $R$ is regular and $x \notin m^{2}$ . For every nonzerodivisor $x \in m$ , $\dim (R / (x)) = \dim R - 1$ . If $R$ is regular and $0 \neq x \in m$ , then $R / (x)$ is regular if and only if $x \notin m^{2}$ .

[F2]

associated graded polynomial surjection: Let $(R, m, k)$ be nonzero Noetherian local and let $x_{1}, \dots, x_{e}$ lift a basis of $m / m^{2}$ . There is a surjective graded $k$ -algebra map $ϕ : k [X_{1}, \dots, X_{e}] \to {gr}_{m} R$ , determined by $X_{i} \mapsto x_{i} + m^{2}$ , with every variable of degree one.

[F3]

dimension at most embedding dimension: Every nonzero commutative Noetherian local ring $R$ satisfies $\dim R \leq edim R < \infty$ .

Verification

1.1

The quotient $A = k [x, y] / (y^{2} - x^{3})$ has unique representatives $a (x) + y b (x)$ by division by the monic polynomial in $y$ . Under $x \mapsto t^{2}$ , $y \mapsto t^{3}$ , the two summands have even and odd powers of $t$ , respectively; their vanishing forces both to be zero. Hence $A$ embeds in $k [t]$ and is a domain in every characteristic. The origin ideal remains a proper nonzero maximal ideal after localization.

givenalgebra

2.1

The ambient local ring $S = k [x, y]_{(x, y)}$ has dimension two and cotangent basis $x, y$ : the coordinate chain gives dimension at least two, and its two maximal-ideal generators give the reverse bound. The nonzero $f = y^{2} - x^{3}$ is a nonzerodivisor in this polynomial domain; the dimension-drop argument of the regular-element quotient theorem gives $\dim S / (f) = 1$ . Since $f \in (x, y)^{2}$ , quotienting adds no linear cotangent relation, so the embedding dimension stays two. This proves the claim over any field, including characteristics two and three.

F1F2F3step 1.1algebra∎

Depends on

Used by

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Dependency tree · two levels

12 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)