Let $R$ be a commutative noetherian ring. I know that an $R$-module is invertible iff it is finitely generated and locally free of rank one. I presume then that there are examples of non-finitely generated, rank one projective modules which are not invertible. Can someone help me out by providing a specific example? Additionally, is there any extension of the notion of the Picard group that includes non-finitely generated rank one projectives?



An example of a rank one projective R-Module that is not invertible
A rank one projective module $M$ over a commutative noetherian ring is necessarily finitely generated. Indeed assume otherwise. Let $a\_i$ be the minimal idempotents of $R$ (there are finitely many since $R$ is commutative noetherian), so that $R=\bigoplus\_i a\_iR$. Then $a\_iM$ is projective over the connected (=indecomposable) noetherian ring $a\_iR$. Bass (Illinois Math J, 1963) showed that a infinitely generated projective module over a connected noetherian commutative ring is necessarily free. So here $a\_iM$ is in addition free of rank one over $a\_iR$. If follows that $M=\bigoplus\_i a\_iM$ is finitely generated (actually: free of rank one), contradiction. 


Remarks:
1) I assume any reasonable definition of "[locally free of] rank one", for instance the (weak) assumption that $M\otimes R\_P$ is free of rank one for every prime $P$ in $R$.


2) Tom Goodville gave a nice example of a module, locally free of rank one, that is not projective. [Recall that a f.g. locally free module has to be projective.] This is the $\mathbf{Z}$-submodule of $\mathbf{Q}$ consisting of fractions $a/b$ with squarefree $b$. It is locally free in the above weak sense (localization at primes), but not in the stronger sense, where the strong sense of locally free would be: free over every open subset in some open covering of the spectrum (is this equivalent to being projective? I'm not sure in the infinitely generated case).


