ExampleConstruction: AI-generatedVerification: AI-adaptedPipeline-generatedverified 2026-09-26 (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.

hypersurface regularity at a rational point

Example

Assume the Axiom of Choice. Let $k$ be any field, $a \in k^{n}$ , and $0 \neq f \in k [x_{1}, \dots, x_{n}]$ with $f (a) = 0$ . The local hypersurface ring at $a$ is regular if and only if at least one formal partial derivative $\partial f / \partial x_{i}$ is nonzero at $a$ .

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]

localisation and polynomial extension of regular rings: Localizations and finite polynomial extensions of a commutative regular Noetherian ring are regular. Regularity can equivalently be tested at maximal ideals. For every nonzero such ring, $gldim R = \dim R$ , allowing infinity. More generally, for a finite module over any commutative Noetherian ring, projective dimension is the supremum of its prime-local projective dimensions. Dedekind domains and their finite polynomial extensions are regular.

[F2]

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 $x \notin m^{2}$ .

Verification

1.1F1F2

Translate coordinates $u_{i} = x_{i} - a_{i}$ . The ambient local ring $S = k [u_{1}, \dots, u_{n}]_{(u_{1}, \dots, u_{n})}$ is regular, with cotangent basis the $u_{i}$ . It is a domain, so the nonzero polynomial $f$ is a nonzerodivisor. The quotient regularity criterion says $S / (f)$ is regular exactly when $f \notin m_{S}^{2}$ . The hypotheses cannot hold for n=0, because then f is a nonzero constant.

2.1step 1.1algebra∎

Monomial expansion after translation gives $f \equiv \sum_{i} (\partial f / \partial x_{i}) (a) u_{i} (m o d m_{S}^{2})$ ; the constant term vanishes. Independence of the cotangent basis makes this class nonzero precisely when at least one coefficient is nonzero. This proves both implications over every characteristic. It is a rational-point hypersurface statement and makes no assertion about smoothness over arbitrary residue-field extensions.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

23 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

Exercise 12.17, p.117; rational-point calculation removes the unnecessary algebraic-closure hypothesis (standard reference, not scraped)