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.

regular flat local map with singular closed fibre

Example

For every field $k$ , the local map $k [s]_{(s)} \to k [t]_{(t)}$ , $s \mapsto t^{2}$ , is finite free of rank two between regular DVRs. Its closed fibre is $k [t] / (t^{2})$ and 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]

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]

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

Before localization every polynomial in $t$ has a unique expression $a (t^{2}) + t b (t^{2})$ , so $k [t]$ is free on $1, t$ over $k [s]$ . After localizing the base at $(s)$ , call the resulting rank-two free algebra $C$ . If $h (t)$ has nonzero constant term, write $h = a (s) + t b (s)$ . Then $(a + t b) (a - t b) = a (s)^{2} - s b (s)^{2}$ is a unit of the base, since its constant term is $a (0)^{2} \neq 0$ . Thus every such $h$ is a unit of $C$ , proving $C = k [t]_{(t)}$ , also in characteristic two.

givenalgebra

2.1

The source and target are regular one-dimensional coordinate local rings (their elements are units times powers of their variable, giving DVRs). The maximal ideal contracts correctly, and freeness makes the map flat. Modulo the source maximal ideal the fibre is $k [t] / (t^{2})$ , whose only prime is $(t)$ , with zero square and one-dimensional cotangent space. Its Krull dimension is zero, so it is singular by the regularity definition.

F1F2step 1.1algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

18 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 12.40(iii), pp.124–125 (standard reference, not scraped)