What are the properties of divisibility for an arbitrary commutative ring?

Question

Given a commutative ring $R$, we can define a divisibility relation by $a|b$ on the elements of $R$ iff there exists $c \in R$ such that $b = ca$. What are the properties of this relation in general? Is it, like the case of the integers a lattice?

I know for certain cases (e.g. integral domains) this is indeed the case, but what about more general rings? In particular, I am interested in cases where the nilradical of $R$ is non-trivial. In particular, I would like to know in what instances can we say in general that such a preorder is a meet semilattice.

score 4 · Answer 1 · answered Jun 27 '17 at 22:16

4

The divisibility poset of a ring $R$ is a lattice iff every pair of elements has a gcd and a lcm.

This does not happen in every commutative ring.

For instance, in the ring $\mathbb Z[\sqrt{-5}]$ there is no gcd for $6$ and $2(1+\sqrt{-5})$ (see this).

A class of rings that have this property is GCD domains, which generalize UFDs.

answered Jun 27 '17 at 22:16

lhf

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See also https://math.stackexchange.com/questions/1229945/looking-for-an-example-of-a-gcd-domain-which-is-not-a-ufd. – lhf Jun 27 '17 at 22:18
Are you aware of if any non-integeral domain examples exist? – Nathan BeDell Jun 28 '17 at 03:34
@Sintrastes, not right now. – lhf Jun 28 '17 at 10:55

What are the properties of divisibility for an arbitrary commutative ring?

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