1
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.