Does this characterize rings in which every ideal is principal?

Question

(All my rings are commutative.)

Let $R$ denote a ring. Then given ideals $I,J$ and $K$ of $R$, from $J=IK$, we may deduce that $J \subseteq I$. In some rings, the converse doesn't necessarily hold. For example, let $R = \mathbb{R}[x,y]$. Then:

$$xR \subseteq xR+yR$$

but there is no ideal $K$ of $R$ satisfying $xR = (xR+yR)K$

However, suppose that every ideal of $R$ is principal. Then from $J \subseteq I,$ we can conclude the existence of $K$ satisfying $J=IK$. For example, in the integers, from $12\mathbb{Z} \subseteq 2\mathbb{Z}$, we may deduce that $12\mathbb{Z} = (2\mathbb{Z})K$ for some ideal $K$, namely $K = 6\mathbb{Z}$.

Question. Does this characterize rings in which every ideal is principal? Explicitly: suppose we're given a ring $R$ such that for all ideals $I,J$ of $R$, from $J \subseteq I$ we may deduce the existence of an ideal $K$ of $R$ such that $J=IK$. Does it follow that every ideal of $R$ is principal?

lhf · Accepted Answer · 2015-09-24T01:23:31.077

4

No, it does not follow that $R$ is a principal ideal ring, not even if $R$ is a domain.

"To contain is to divide" is true in every Dedekind domain, but not all Dedekind domains are principal ideal domains. One example is $\mathbb Z[\sqrt{−5}]$.

edited Sep 24 '15 at 01:23

answered Sep 24 '15 at 01:18

lhf

221,500

How do we know its true in every Dedekind domain? – goblin GONE Sep 24 '15 at 01:28
@goblin, see http://math.stackexchange.com/questions/103692/divides-means-contains-for-dedekind-domains-by-treating-pids-and-localising. – lhf Sep 24 '15 at 01:37
Okay. Do you know if "Dedekind domain" characterizes such rings? – goblin GONE Sep 24 '15 at 06:17
@goblin, I don't. Perhaps you could ask a separate question. – lhf Sep 24 '15 at 09:58

score 3 · Answer 2 · edited Jun 23 '22 at 18:52

3

The Noetherian domains which have this property are precisely the Dedekind domains: see e.g. Proposition 20.9 of my commutative algebra notes. One way to say it is that this condition buys precisely that all ideals are locally principal, not that they are principal.

What about an arbitrary commutative ring? Certainly rings in which each ideal is principal have this property. Off the top of my head I don't know whether requiring that every ideal be locally principal is either necessary or sufficient.

edited Jun 23 '22 at 18:52

Community

1

answered Sep 25 '15 at 10:18

Pete L. Clark

100,402

By a "domain", what precisely do you mean? In particular, are your domains commutative? – goblin GONE Sep 25 '15 at 13:17
@goblin: I mean precisely a commutative ring without nonzero zero-divisors. (Was it not clear that we are in the context of commutative algebra? You mentioned it and so did I.) – Pete L. Clark Sep 25 '15 at 13:32
It was 80% clear. – goblin GONE Sep 25 '15 at 13:33
Your notes assume that $R$ is a Noetherian domain: I assume that is what you mean here? – goblin GONE Sep 25 '15 at 13:41
@goblin: That's a good catch, thanks. When I wrote my answer I was thinking this was true for all domains, but upon reflection I believe there are non-Noetherian domains with this property. – Pete L. Clark Sep 25 '15 at 14:14

Does this characterize rings in which every ideal is principal?

2 Answers2