On the field of fractions of certain kind of integral domain

Question

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.

Answer 1

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.

Answer 2

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).