Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26
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.

For a finite module, support is the set of primes containing the annihilator

Statement

If M is a finitely generated left R-module, then

SuppR(M)={p:AnnR(M)p}.

Facts & Assumptions

Given: A commutative ring R and a finitely generated left R-module M.

[L1]

A prime ideal lies in SuppR(M) exactly when some element of M has annihilator inside it (A prime lies in the support exactly when some element has annihilator inside it).

[L2]

If m1,,mr generate M, then SuppR(M)=iSuppR(R/AnnR(mi)) (A finite module has the union of its generator-cyclic supports).

[L3]

The annihilator of M is AnnR(M)={rR:rm=0 for every mM} (Annihilators, torsion elements and the torsion subset of a module).

Proof

technique · direct
1.1

If pSuppR(M), [L1] gives mM with AnnR(m)p. Since every element of AnnR(M) kills every element of M, one has AnnR(M)AnnR(m)p.

L1L3
1.2

Choose generators m1,,mr of M. If AnnR(M)p and no AnnR(mi) is contained in p, choose tiAnnR(mi)p for every i. Then t=t1trp, but t annihilates every generator and hence all of M, so tAnnR(M)p, a contradiction. Thus AnnR(mi)p for some i, and [L2] gives pSuppR(M).

L2L3choose
2.1

Steps 1.1 and 1.2 prove the support-annihilator formula.

step 1.1step 1.2

Depends on

Used by

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