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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-31
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Invertible fractional ideals are exactly the rank-one projective modules

Statement

Assume the Axiom of Choice. Let R be a domain with fraction field K. A nonzero fractional ideal of R is invertible if and only if, as an R-module, it is finite projective and becomes K after tensoring with K. Equivalently, the finite projective R-modules of constant rank one are precisely the invertible fractional ideals inside K.

Facts & Assumptions

Given: A domain R with fraction field K.

[F1]

Invertibility of a fractional ideal means I(R:I)=R (Invertible fractional ideals).

[L1]

A fractional ideal is invertible exactly when it is finite projective and locally principal at every maximal ideal (Equivalent characterizations of invertible fractional ideals).

[L2]

Every finite projective module with PRKK is isomorphic to a fractional ideal (A finite rank-one projective module embeds as a fractional ideal).

Proof

technique · direct
1.1

Let IK be an invertible fractional ideal. Then [L1] makes I finite projective and locally free of rank one at every maximal ideal. After tensoring with the field K, the module IRK is a one-dimensional K-vector space, hence isomorphic to K. Thus invertible fractional ideals are finite rank-one projectives.

F1L1givenalgebra
1.2

Conversely, let P be a finite projective module with PRKK. By [L2], it is isomorphic to a fractional ideal IK. For every maximal ideal m, the localisation Pm is a free rank-one module over the local ring Rm, so Im is principal. Therefore [L1] makes I invertible.

L1L2givenalgebra
2.1

Steps 1.1 and 1.2 prove the equivalence between invertible fractional ideals and finite projective modules of constant rank one.

step 1.1step 1.2

Depends on

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