Let $R$ be an integral domain, let $\mathrm{Frac}(R)$ denote the field of fractions of $R$. Then as $\mathrm{Frac}(R)$ contains $R$ it is an $R$-module in an usual way. Now suppose every proper submodule of the $R$-module $\mathrm{Frac}(R)$ is cyclic. Then what can we say about $R$ ? Can such integral domains be characterized in some way ?

If I wanted the field of fractions $\mathrm{Frac}(R)$ itself to be cyclic (or even finitely generated) as an $R$-module then I know that $R$ would become a field. But my condition is only on proper submodules.


Any proper ideal of $R$ is a proper submodule of $\mathrm{Frac}(R)$, so $R$ must be a PID. If $p$ is any prime element of $R$, then $M=\{r/p^n:r\in R, n\in\mathbb{N}\}$ is a submodule of $\mathrm{Frac}(R)$ which is not cyclic. Thus $M$ must be all of $\mathrm{Frac}(R)$, which means every element of $R$ is divisible by $p^n$ for some $n$. This means $p$ is the only prime of $R$, so $R$ has at most one prime and hence is either a DVR or a field.

Conversely, if $R$ is a field, your statement clearly holds. If $R$ is a DVR with uniformizer $p$, let $M$ be a proper submodule of $\mathrm{Frac}(R)$. If $M$ has elements of arbitrarily small valuation, then $M$ is all of $\mathrm{Frac}(R)$. Otherwise, let $d\in\mathbb{Z}$ be the minimal valuation of an element of $M$; then $M$ is generated by $p^d$ and hence is cyclic.

So every proper submodule of $\mathrm{Frac}(R)$ is cyclic iff $R$ is either a field or a DVR.


This means the ideal class monoid is trivial. In particular it is a (trivial) group, so $R$ is a P.I.D.

One of the ways to prove a ring of integers is a P.I.D. consists in proving its ideal class group is trivial (usually with the help of Minkowski's lemma).

  • $\begingroup$ how to see that the ideal class monoid of $R$ is trivial ? And perhaps if PID can be said then something more can be said because for example, $\mathbb Z$ does not satisfy my condition $\endgroup$ – user Jul 9 '17 at 20:50
  • $\begingroup$ But the ideal class monoid of $\mathbf Z$ is trivial! Which condition do you mean? $\endgroup$ – Bernard Jul 9 '17 at 20:57
  • $\begingroup$ yes definitely the ideal class monoid of the ring of integers is trivial as it is a PID , what I meant is that it does not satisfy the criteria in my question ... from your reasoning I get that if a domain satisfies the condition in my question, then it is a PID , and I was hoping to get something more too . $\endgroup$ – user Jul 9 '17 at 21:02

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