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Let $R$ be a principal ideal domain with $|R|=\infty$. Suppose $M$ is an $R$-module such that $2 \le |M| < \infty$. What properties does this imply about $R$?

Background: I was hoping that $R\cong \mathbb{Z}$ is the only possibility, since $M$ is (also) a finitely generated $\mathbb{Z}$-module.

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    $\begingroup$ If $R=\mathbb C$, then the answer is no. $\endgroup$
    – Mariano Suárez-Álvarez
    Jul 11, 2015 at 3:24
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    $\begingroup$ If you don't like that example because it is a field, then let $R=\mathbb C[t]$. $\endgroup$
    – Mariano Suárez-Álvarez
    Jul 11, 2015 at 3:26
  • $\begingroup$ I have significantly altered the question. (I was trying to avoid a vague question like the present one.) $\endgroup$
    – nknight
    Jul 11, 2015 at 3:34
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    $\begingroup$ The ring of integers in any number field of finite degree over $\mathbb Q$ also works, so $\mathbb Z$ is not the only example; likewise, the integers in the finite extensions of the $p$-adic numbers also work, as do power series in one variable over a finite field. &c. $\endgroup$
    – Mariano Suárez-Álvarez
    Jul 11, 2015 at 3:37
  • $\begingroup$ OK, the theory is clearly deeper than I realized. I also doubt this question is answerable in its present form; is the protocol to delete it? $\endgroup$
    – nknight
    Jul 11, 2015 at 3:46

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If I understand well you have an infinite PID which has a finite non-zero module, say $M$. In particular, $M$ is finitely generated and torsion, so it is a (finite) direct sum of indecomposable torsion modules, that is, of modules of the form $R/(p^m)$ with $p\in R$ prime, $p\ne0$. This shows that you have to impose on $R$ the following condition:

$R/(p)$ is finite for some prime $p\in R$, $p\ne0$

which is equivalent to $R/(p^m)$ is finite for some prime $p\in R$, $p\ne0$ and $m\ge1$.

For instance, $R=\mathbb Z[i]$ is such an example (other than $\mathbb Z$).

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