## 109.30 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.

## Comments (2)

Comment #3777 by Remy on

An easier (?) example of a similar flavour is $R = \mathbb Z$ and $M = \sum_{p \text{ prime}} \frac{1}{p}\mathbb Z \subseteq \mathbb Q$, which equals the set of fractions $\frac{a}{b}$ with $b \neq 0$ squarefree.

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