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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?

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How do you define the rank in this case? Dimension of the residue fields? Of course your alluded generalized Picard group will be no group. –  Martin Brandenburg Jul 2 '11 at 19:40
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Dear Karl, I am not able to extract an answer to Andrew's question from your link. Could you please state a precise statement there that answers his question? –  Georges Elencwajg Jul 3 '11 at 8:23
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Andrew, what definition of "invertible" you are using? The ones I know would force $M$ to be finitely generated automatically, so the answer is trivially no. –  Hailong Dao Jul 3 '11 at 15:51
    
@Hailong: the actual question does not ask $M$ to be invertible, in fact it asks that it is not that, so it seems that that could not force anything. But it does ask that $M$ be of rank $1$ and perhaps this is what you have in mind. So my question is the same as Martin's: @Andrew: How do you define rank? –  Sándor Kovács Jul 3 '11 at 17:32
    
@S\andor: you are right, I meant to say, if $M$ is not f.g. (as the OP wanted), then it is trivially not invertible. So I am not sure what the question asked. Thank you for the correction. –  Hailong Dao Jul 3 '11 at 18:41

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up vote 10 down vote accepted

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).

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@Yves, thanks for your clarification re projectivity & FG. I was considering exactly what you are referencing in your first remark, that $M \otimes R_P \equiv R_P$ for all primes $P$ as my 'definition' for invertibility. –  Andrew Parker Jul 5 '11 at 17:23

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