# Group of divisibility

We know that the necessary condition for any partially ordered group to be a group of divisibility is that the group must be a directed group. What is the sufficient condtion for partially ordered group to be a group of divisibility?

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For reference: Google tells me the divisibility group of a commutative domain is the group of non-zero principal fractional ideals under multiplication, ordered by reverse inclusion. – Mariano Suárez-Alvarez Mar 7 '12 at 7:08
Thanks Mariano. This happens naturally. I am looking the situation is that what kind of partially ordered group is a group of divisibility? If I have a partially ordered group $G$ then when can I say that there exist a domain whose group of divisibility is ordered isomorphic to $G$. If we take lattice-ordered group and totally ordered group they are always group of divisibility. – Rajnish Mar 7 '12 at 7:47

## 1 Answer

Let me begin by saying I am far from an expert on this area; however, I attended a seminar in which M. Zafrullah gave a few talks on this subject. It appears this question you posted has not been answered in quite some time, so I don't feel too terribly about at least posting this in hopes that it might help. At worst, hopefully it is better than nothing.

I read a couple of articles back when he was talking just to get some background in this subject area for his talks. One I found particularly insightful and seemed to answer your question at least partly was one by Joe L. Mott entitled "Convex directed subgroups of a group of divisibility."

In it he mentions $D$ is a UFD if and only if $G(D)$ is a cardinal sum of copies of $\mathbb{Z}$.

$D$ is a valuation ring if and only if $G(D)$ is totally ordered.

$D$ is a GCD domain if and only if $G(D)$ is lattice ordered.

But I think this is not exactly what you are asking. You want to know, given any partially ordered group, can I tell if it arises as the group of divisibility of some ring $R$.

It sounds like this is not an easy question to answer, and I think from your comment it appears you know of Krull's theorem that answers this question in the affirmative for totally ordered groups. And in more generality, it is true for abelian lattice ordered groups by the theorem of Jaffard, Kaplansky and Ohm theorem, that they occur as groups of divisibility of Bezout domains that can be constructed.

There is a nice paper, that I came across by Yi Chuan Yang in 2008 called "Some remarks on almost l-groups" where they define an almost l-group, and show that almost GCD domains have group of divisibility which is an almost l-group, and seek to answer the converse. They answer in the negative and construct a counter-example which shows that that converse is not true. There is an almost GCD domain which is not an almost l-group. He also provides references to several other possibly ways of generalizing this converse question.

I hope this may at least point you in a positive direction!

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Thank you CPM. This makes sense. – Rajnish Oct 11 '12 at 18:30