Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-6-sol)audited 2026-09-27
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.

A quadratic-cubic plane complete intersection has eventual Hilbert value six

Example

Let k be a field and let S=k[x0,x1,x2]/(x02,x13) carry the standard grading. Then HS⁡S(t)=(1+t)(1+t+t2)1−t, and the Hilbert function of S is 1,3,5,6,6,6,… in degrees 0,1,2,3,4,5,…: it is constantly 6=2⋅3 from degree 3 on. The quotient S is not a finite-dimensional k-algebra; only its graded pieces are computed here.

Facts & Assumptions

Given: A field k and the standard graded quotient S=k[x0,x1,x2]/(x02,x13).

[L1]

The Hilbert function of a graded module records dim⁡kMn, and its Hilbert series is the formal power series ∑ndim⁡kMntn, whose coefficients are read off by coefficient extraction (The Hilbert function and formal Hilbert series of a graded module with finite-length pieces, Formal power series over a commutative ring and the coefficient-extraction functional [xn], Nonnegatively graded rings and modules, homogeneous elements, and twists).

[L2]

For two plane forms F,G with no common nonconstant factor, of positive degrees d and e, the pair (F,G) is a regular sequence and HS⁡k[x0,x1,x2]/(F,G)(t)=(1+t+⋯+td−1)(1+t+⋯+te−1)/(1−t), with Hilbert function constantly de in every degree n≥d+e−2 (Coprime positive-degree plane forms form a regular sequence, Hilbert series and eventual Hilbert value of a two-form plane complete intersection, homogeneous polynomial and homogeneous ideal).

[L4]

Assuming the Axiom of Choice, for coprime plane forms of degrees d,e, the total length of Proj⁡(k[x0,x1,x2]/(F,G)) equals de (Two coprime projective plane forms meet in total length equal to their degree product, Total length of a zero-dimensional projective scheme).

Verification

technique · direct
1.1

The two forms x02 and x13 are coprime in k[x0,x1,x2] because they involve distinct variables, and have degrees 2 and 3, so [L2] gives HS⁡S(t)=(1−t2)(1−t3)/(1−t)3=(1+t)(1+t+t2)/(1−t), using (1−t2)=(1−t)(1+t) and (1−t3)=(1−t)(1+t+t2).

L2algebra
1.2

Equivalently, the monomials x0ax1bx2c with 0≤a≤1, 0≤b≤2, c≥0 form a k-basis of S by [L3], and their generating function by total degree is (1+t)(1+t+t2)(1+t+t2+⋯ )=(1+t)(1+t+t2)/(1−t).

L3algebra
2.1

By step 1.2, dim⁡kSn=#{(a,b):0≤a≤1, 0≤b≤2, a+b≤n}, which is 1,3,5,6,6,6,… for n=0,1,2,3,4,5,…; equivalently these are the partial sums of the coefficients 1,2,2,1 of (1+t)(1+t+t2), in agreement with [L1].

L1step 1.2algebra
3.1

Since Sn≠0 for every n by step 2.1, the k-vector space S is infinite-dimensional, so S is not a finite-dimensional k-algebra; no Artinian claim is made.

step 2.1
4.1

If the Axiom of Choice is assumed, then as a consistency check [L4] gives len⁡k(Proj⁡S)=2⋅3=6, which is exactly the eventual value of the Hilbert function computed in step 2.1. The Hilbert-series and Hilbert-function claims above hold over every field without this additional assumption.

L4step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

48 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