An example of a rank one projective R-Module that is not invertible - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T03:56:21Z http://mathoverflow.net/feeds/question/69353 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/69353/an-example-of-a-rank-one-projective-r-module-that-is-not-invertible An example of a rank one projective R-Module that is not invertible Andrew Parker 2011-07-02T18:47:31Z 2011-07-03T20:43:58Z <p>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?</p> http://mathoverflow.net/questions/69353/an-example-of-a-rank-one-projective-r-module-that-is-not-invertible/69414#69414 Answer by Yves Cornulier for An example of a rank one projective R-Module that is not invertible Yves Cornulier 2011-07-03T20:43:58Z 2011-07-03T20:43:58Z <p>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. </p> <p>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 $R$.</p> <p>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).</p>