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Let $A$ be an integral domain, and that nonzero ideals of $A$ under ideal multiplication is a free commutative monoid, then $A$ is a Dedekind domain.

By Dedekind domain we mean a Noetherian normal domain of dimension $1$, or equivalently a domain whose fractional ideals form a group, or equivalently every nonzero ideal is a product of prime ideals.

This appears in Larsen, Max D., and Paul J. McCarthy. Multiplicative theory of ideals. Academic press, 1971., Exercise VI.14, and also somewhere on Zariski-Samuel as exercise.

There are several attempting ways to do this. Firstly it is likely that being a free commutative monoid will imply $I \supset J \Rightarrow I \mid J$, but recall even we know a priori the basis consisting of prime ideals, this is not so direct.

Secondly I know (from various properties of Prüfer domains) that a Noetherian domain satisfying the condition is a Dedekind domain, so one may proceed by showing every (prime) ideal is invertible, or merely finitely generated. Still I can't make it through.

rschwieb
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fyx1123581347
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    Which definition of dedekind domain are you using? There are close to $100$ known (if not more), e.g. see see this answer. – Bill Dubuque Oct 17 '22 at 13:19
  • Can you show that every ideal $I$ of the monoid is inversible (there exists $J$ such that $IJ$ is principal) – reuns Oct 17 '22 at 13:57
  • @BillDubuque OK, so you may feel better if I wrote: show that the basis must consists of prime ideals. – fyx1123581347 Oct 17 '22 at 14:07
  • @reuns If I could, then I would have not post the question. Perhaps there is something I missed, but I feel hard to deduce that prime ideals are invertible and thus finitely generated, which is sufficient for the purpose. I guess one perhaps need to go through stuffs on Prufer rings and reproduce some of the proof, but I am not enthusiastic enough to do this and wish someone would give me a detailed reference (i.e., not as an exercise) or an answer. – fyx1123581347 Oct 17 '22 at 14:30
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    Given $I$ take $a\in I-0$ then use the free commutative monoid property to construct $J$ such that $IJ=(a)$. It implies that $A$ is a Dedekind domain (by one of the non-trivial equivalent definitions of Dedekind domains) – reuns Oct 17 '22 at 14:34
  • @reuns The only candidate is $((a):I)$ but why it works? – fyx1123581347 Oct 17 '22 at 14:51
  • Let us continue this discussion in chat. – fyx1123581347 Oct 17 '22 at 15:21
  • I overlooked that $a\in I$ doesn't mean $(a)\ge I$ on the free commutative monoid side – reuns Oct 18 '22 at 08:15
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    While it is excellent you sourced the problem, please don't link to pdf copies of books. It is almost always a bootleg pdf and therefore unethical to distribute. We would not like such distribution to happen on the site. (I've replaced the link with an appropriate citation.) – rschwieb Oct 18 '22 at 14:45