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PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24
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.

Torsion elements and p-primary elements form submodules over a domain

Statement

Let R be an integral domain and M an R-module. The torsion subset Tor⁡(M) is a submodule. For every irreducible p∈R, the p-primary component M[p∞] is a submodule of Tor⁡(M).

Facts & Assumptions

Given: An integral domain R, an R-module M, the p-primary definition of The p-primary component of a module over a domain, and the submodule test of Submodule of a module.

[F1]

If R is an integral domain, an element m∈M is a torsion element when rm=0M for some nonzero r∈R (Annihilators, torsion elements and the torsion subset of a module).

Proof

technique · direct
1.1F1algebra

The zero element is torsion. If am=0 and bn=0 with a,b≠0, then ab≠0 and ab(m+n)=b(am)+a(bn)=0; if c∈R, then a(cm)=c(am)=0. Closure under negatives is the scalar case c=−1, so Tor⁡(M) is a submodule.

2.1givenalgebra∎

The zero element lies in M[p∞]. If pkm=0 and pℓn=0 with k,ℓ≥1, then pmax⁡(k,ℓ)(m+n)=0, and pk(cm)=0 for every c∈R. Thus M[p∞] is a submodule contained in Tor⁡(M). The proof includes k=1, the zero module, and replacing p by an associate.

Depends on

Used by

Cited to discharge well-definedness by The p-primary component of a module over a domain.

Dependency tree · two levels

8 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