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Invertible fractional ideals are exactly the rank-one projective modules
Statement
Assume the Axiom of Choice. Let be a domain with fraction field . A nonzero fractional ideal of is invertible if and only if, as an -module, it is finite projective and becomes after tensoring with . Equivalently, the finite projective -modules of constant rank one are precisely the invertible fractional ideals inside .
Facts & Assumptions
Given: A domain with fraction field .
Invertibility of a fractional ideal means (Invertible fractional ideals).
A fractional ideal is invertible exactly when it is finite projective and locally principal at every maximal ideal (Equivalent characterizations of invertible fractional ideals).
Every finite projective module with is isomorphic to a fractional ideal (A finite rank-one projective module embeds as a fractional ideal).
Proof
Let be an invertible fractional ideal. Then [L1] makes finite projective and locally free of rank one at every maximal ideal. After tensoring with the field , the module is a one-dimensional -vector space, hence isomorphic to . Thus invertible fractional ideals are finite rank-one projectives.
Conversely, let be a finite projective module with . By [L2], it is isomorphic to a fractional ideal . For every maximal ideal , the localisation is a free rank-one module over the local ring , so is principal. Therefore [L1] makes invertible.
Steps 1.1 and 1.2 prove the equivalence between invertible fractional ideals and finite projective modules of constant rank one.
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Used by
Dependency tree · two levels
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Sources
- J. P. May, Notes on Dedekind Rings (standard reference, not scraped)
- The Stacks Project, Section 10.78: Finite projective modules (standard reference, not scraped)