Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-28
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 primes of a submodule lie in those of the ambient module

Statement

If

0⟶M′⟶M⟶M′′⟶0

is a short exact sequence of left R-modules, then

Ass⁡R(M′)⊆Ass⁡R(M).

Facts & Assumptions

Given: A commutative ring R and a short exact sequence 0→M′→M→M′′→0 of left R-modules.

[L1]

A prime p is associated to a module exactly when R/p embeds in that module (Associated primes are exactly primes of embedded cyclic residue modules).

[L2]

In a short exact sequence, the left map is injective (Exact sequences and short exact sequences of modules).

Proof

technique · direct
1.1

Let p∈Ass⁡R(M′). By [L1], there is an embedding R/p↪M′. Composing with the injective map M′↪M from [L2] gives an embedding R/p↪M.

L1L2
2.1

Applying [L1] again, the embedding from step 1.1 shows that p∈Ass⁡R(M). Therefore Ass⁡R(M′)⊆Ass⁡R(M).

L1step 1.1∎

Depends on

Used by

Dependency tree · two levels

6 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