Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-31
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.

Every nonzero fractional ideal of a Dedekind domain is invertible

Statement

Assume the Axiom of Choice. Every nonzero fractional ideal of a Dedekind domain is invertible.

Facts & Assumptions

Given: A Dedekind domain R and a nonzero fractional ideal I of R.

[F1]

A Dedekind domain is a Noetherian integrally closed domain of dimension 1 (Dedekind domains).

[L1]

A nonzero-prime localisation of a Dedekind domain is a DVR (Localizing a Dedekind domain at a nonzero prime gives a DVR).

[L2]

Every nonzero ideal of a DVR is principal (Ideals in a DVR are powers of the maximal ideal).

[L3]

A nonzero finitely generated fractional ideal is invertible exactly when all maximal localisations are principal (Equivalent characterizations of invertible fractional ideals).

Proof

technique · direct
1.1

Choose 0dR with J:=dIR. Then J is a nonzero integral ideal of R, so [F1] makes it finitely generated; hence the fractional ideal I=d1J is finitely generated as well. Let m be a maximal ideal. If Jm, then Jm=Rm, so Im=d1Rm is principal. If Jm, then m is a nonzero prime, [L1] makes Rm a DVR, and [L2] makes the integral ideal Jm principal. Hence Im=d1Jm is principal in either case. Thus I is finitely generated and every maximal localisation of I is principal.

F1L1L2givenchoosealgebra
2.1

Applying [L3] to step 1.1 shows that I is invertible.

L3step 1.1

Depends on

Used by

Dependency tree · two levels

21 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