Alphabeta Math
ExampleConstruction: AI-generatedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-10
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.

koszul euler characteristic empty sequence

Example

Assume AC. For any finite-length module M over a commutative Noetherian local ring (R,m), the empty sequence satisfies χ(K(;M))=e0(0,M)=R(M). For the concrete instance M=R/m, both numbers are 1. For M=0, both are 0.

Facts & Assumptions

Given: AC, a commutative Noetherian local ring (R,m), a finite-length R-module M, and the empty sequence. The displayed special instances are M=R/m and M=0.

[A1]

We assume The Axiom of Choice for the bridge theorem cited below.

[F1]

The empty sequence has complex M[0], ideal zero, and constant polynomial: koszul euler characteristic and degree indexed multiplicity.

[F2]

The coefficient indexed by the sequence length equals the Koszul Euler characteristic: degree-r Hilbert–Samuel coefficient as Koszul Euler characteristic.

Verification

technique · direct
1.1

The empty Koszul complex has K0=M, all other terms zero and all differentials zero. Thus H0=M and Hi=0 for i0, giving χ(K)=R(M).

F1given
2.1

For every n0, 0n+1M=0, so M/0n+1M=M and P(T)=R(M). Hence e0=0![T0]P=R(M). A composition series also makes M finitely generated: take a lift of one nonzero generator of each simple factor; induction through the finite series shows these finitely many lifts generate M. The module-relative hypothesis is exactly finite length of M, so the bridge theorem with r=0 applies under AC and agrees with this direct calculation.

A1F1F2step 1.1
3.1

In particular take M=k=R/m. Its only submodules are zero and k, since any nonzero vector spans this one-dimensional k-space; its length is 1. We obtain H0=k, P=1 and χ=e0=1. With M=0 the empty series has length zero and the same calculation gives P=χ=e0=0.

F1step 1.1step 2.1algebra

Remarks

This is a locally calculated design example, not a named example attributed to a source. The empty-sequence convention is consistent with Hochster printed p.166 (the zero-generator case) and with Stacks 43.15.4 at index zero. The general bridge is only a consistency check here; the homology and polynomial were calculated directly.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

11 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