110.31 A nonfinite module with finite free rank 1 stalks
Let R = \mathbf{Q}[x]. Set M = \sum _{n \in \mathbf{N}} \frac{1}{x - n}R as a submodule of the fraction field of R. Then M is not finitely generated, but for every prime \mathfrak p of R we have M_{\mathfrak p} \cong R_{\mathfrak p} as an R_{\mathfrak p}-module.
An example of a similar flavor is R = \mathbf{Z} and M = \sum _{p \text{ prime}} \frac{1}{p} \mathbf{Z} \subset \mathbf{Q}, which equals the set of fractions \frac{a}{b} with b nonzero and squarefree.
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Comment #3777 by Remy on
Comment #3907 by Johan on