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(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?

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2 Answers 2

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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 $\mathbb Z[\sqrt{−5}]$.

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  • $\begingroup$ How do we know its true in every Dedekind domain? $\endgroup$ Sep 24, 2015 at 1:28
  • $\begingroup$ @goblin, see math.stackexchange.com/questions/103692/…. $\endgroup$
    – lhf
    Sep 24, 2015 at 1:37
  • $\begingroup$ Okay. Do you know if "Dedekind domain" characterizes such rings? $\endgroup$ Sep 24, 2015 at 6:17
  • $\begingroup$ @goblin, I don't. Perhaps you could ask a separate question. $\endgroup$
    – lhf
    Sep 24, 2015 at 9:58
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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.

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  • $\begingroup$ By a "domain", what precisely do you mean? In particular, are your domains commutative? $\endgroup$ Sep 25, 2015 at 13:17
  • $\begingroup$ @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.) $\endgroup$ Sep 25, 2015 at 13:32
  • $\begingroup$ It was 80% clear. $\endgroup$ Sep 25, 2015 at 13:33
  • $\begingroup$ Your notes assume that $R$ is a Noetherian domain: I assume that is what you mean here? $\endgroup$ Sep 25, 2015 at 13:41
  • $\begingroup$ @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. $\endgroup$ Sep 25, 2015 at 14:14

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