Alphabeta Math
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.

associated graded polynomial map singular kernel

Example

For the cusp local ring R=(k[x,y]/(y2x3))(x,y) with maximal ideal m, the associated graded ring is k[X,Y]/(Y2). Thus the polynomial map defined by the cotangent classes has kernel exactly (Y2).

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]

associated graded polynomial surjection: Let (R,m,k) be nonzero Noetherian local and let x1,,xe lift a basis of m/m2. There is a surjective graded k-algebra map ϕ:k[X1,,Xe]grmR, determined by Xixi+m2, with every variable of degree one.

Verification

1.1

In the ambient coordinate local ring S=k[x,y](x,y), the associated graded ring is k[X,Y]: a rational function with denominator of nonzero constant term has initial form equal to its numerator initial form divided by that constant. This identifies each graded piece and respects products. Therefore orders add on products of nonzero elements of S. In particular for f=y2x3, in(f)=Y2, and in(hf)=in(h)Y2 for every nonzero hS.

givenalgebra
2.1

The degree-n kernel of grSgr(S/(f)) consists of classes of ann with a(f)+nn+1. Write a=hf+b with bnn+1. If its degree-n class is nonzero, it is exactly the initial form of hf, hence a multiple of Y2. Conversely every homogeneous multiple of Y2 is the initial form of a polynomial multiple of f. Thus the graded kernel is exactly (Y2) and the surjective polynomial map of the cotangent-basis lemma has the stated quotient.

F1step 1.1algebra

Depends on

Used by

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Dependency tree · two levels

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Sources