The title pretty much says it all: suppose $R$ is a commutative integral domain such that every countably generated ideal is principal. Must $R$ be a principal ideal domain?

More generally: for which pairs of cardinals $\alpha < \beta$ is it the case that: for any commutative domain, if every ideal with a generating set of cardinality at most $\alpha$ is principal, then any ideal with a generating set of cardinality at most $\beta$ is principal?

Examples: Yes if $2 \leq \alpha < \beta < \aleph_0$; no if $\beta = \aleph_0$ and $\alpha < \beta$: take any non-Noetherian Bezout domain (e.g. a non-discrete valuation domain).

My guess is that valuation domains in general might be useful to answer the question, although I promise I have not yet worked out an answer on my own.

  • $\begingroup$ Do you know if infinite well-ordered groups exist? $\endgroup$ Dec 6, 2009 at 22:57
  • 4
    $\begingroup$ Well-ordered groups do not exist: if x>0 is an element, then either it has finite order, in which case we get a "cycle" of comparisons 0<x<x^2<..<x^n=0, or it has infinite order, in which case the set of its negative powers has no minimal element. $\endgroup$ Dec 7, 2009 at 0:20

4 Answers 4


No such ring exists.

Suppose otherwise. Let $I$ be a non-principal ideal, generated by a collection of elements $f_\alpha$ indexed by the set of ordinals $\alpha<\gamma$ for some $\gamma$. Consider the set $S$ of ordinals $\beta$ with the property that the ideal generated by $f_\alpha$ with $\alpha<\beta$ is not equal to the ideal generated by $f_\alpha$ with $\alpha\leq \beta$.

$I$ is generated by the $f_\beta$ with $\beta \in S$, so if $S$ is finite, then $I$ is finitely generated and thus is principal.

On the other hand, if $S$ is infinite, then take a countable subset $T= \{\beta_1<\beta_2<\dots\}$ of $S$. If the ideal generated by the corresponding set of $f_\beta$'s were principal, its generator would have to be in some $\langle f_{\beta_k} \mid k\leq i \rangle$ for some $i$ (since any element of $\langle f_{\beta}\mid \beta \in T\rangle$ is a finite combination of $f_\beta$'s and therefore lies in some such ideal). Now no $\beta_j$ with $j>i$ could be in $T$.


The same argument shows that all rings for which any countably generated ideal is finitely generated, have all their ideals finitely generated.


Corrected thanks to David's questions.

  • $\begingroup$ Hugh -- I am having trouble following what you wrote, but it seems to me that you are showing the following stronger result: Suppose f_{\alpha} is a set such that the ideal generated by any countable subset of f_{alpha} is principal. Then the ideal generated by f_{alpha} is principal. $\endgroup$ Dec 7, 2009 at 3:05
  • $\begingroup$ Do I understand this correctly? $\endgroup$ Dec 7, 2009 at 3:09
  • $\begingroup$ Yes, I think that's right. Sorry if my explanation is unclear; I'd be happy to improve it if you can suggest where it's failing. $\endgroup$ Dec 7, 2009 at 3:23
  • $\begingroup$ You write "If the ideal generated by the corresponding set of f_{\beta}'s were principal, its generator would have to be in some f_{beta_i}..." I assume you mean "would be some f_{beta_i}", since the f's are ring elements, not sets. But I don't follow why this is true. Why couldn't the generator by something else entirely? $\endgroup$ Dec 7, 2009 at 3:39
  • 4
    $\begingroup$ Thanks to Hugh for a fantastic answer. A bit of personal trivia: he was the TA for the first abstract algebra class I took, 14 years ago. $\endgroup$ Dec 7, 2009 at 15:53

The question is fully settled by Hugh Thomas' anwer, but let me mention this related interesting fact.

Theorem. There is a ring R and ideal I on R, such that every countable subset of I is contained in a principal subideal of I, but I is not principal.

Proof. Let I be the ideal of nonstationary subsets of ω1, in the power set P(ω1), which is a Boolean algebra and hence a Boolean ring. That is, I consists of those subsets of ω1 that are disjoint from a closed unbounded subset of ω1. It is an elementary set-theoretic fact that the intersection of any countably many closed unbounded subsets of ω1 is still closed and unbounded, and thus the union of countably many non-stationary sets remains non-stationary. Thus, every countable subset of I is contained in a principal subideal of I. But I is not principal, since the complement of any singleton is stationary. QED

In the previous example, the ideal I is not maximal. If one assumes the existence of a measurable cardinal (a large cardinal notion), however, then the example can be made with I maximal.

Theorem. If there is a measurable cardinal, then there is a ring R with a maximal ideal I, such that every countable subset of I is contained in a principal sub-ideal of I, but I is not principal.

Proof. Let κ be a measurable cardinal, which means that there is a nonprincipal κ-complete ultrafilter U on the power set P(κ), which is a Boolean algebra and thus a Boolean ring. The ideal I dual to U is also κ-complete, which means that I is closed under unions of size less than κ. In particular, since kappa is uncountable, this means that the union of any countably many elements of I remains in I, and this union set generates a principal subideal of I containing the given countable set. The ideal I is maximal since U was an ultrafilter. QED

I'm not sure at the moment whether the situation of this last theorem requires a measurable cardinal or not, but I'll think about it.


Sorry to dig up an old question, but in case anybody else randomly lands here, here's a quick side note about a way that this can be generalized.

Theorem: If every countably generated ideal of a ring $R$ is finitely generated, then $R$ is Noetherian. Hence, if $n < \infty$ and every countably generated ideal is $n$-generated, then every ideal is $n$-generated.

Proof: By contrapositive. If $R$ is not Noetherian, then we can make an infinite properly ascending chain $I_1 \subsetneq I_2 \subsetneq \cdots$ of finitely generated ideals. The union of this chain is a countably generated ideal, and it cannot be finitely generated, because that would cause the chain to terminate at some point.


Such rings exist. Here is a lemma which might be useful in proving such rings exist:

Lemma: Given any ordered abelian group $A$, there is a valuation ring $k[[A]]$ with valuation group isomorphic to $A$.

Construction: The ring $k[[A]]$ is the ring of formal sums $$\sum_{i_1,i_2, \ldots, i_r=0}^{\infty} k_{i_1 i_2 \cdots i_r} t^{ i_1 a_1 + i_2 a_2+ \cdots i_r a_r}$$ where $a_1$, $a_2$, ..., $a_r$ are allowed to be any elements in $A_{>0}$. The key point to notice is that, if such a sum has nonzero leading term, then its multiplicative inverse is also such a sum.

This lemma let's you transfer the problem from rings to groups. But my attempted construction of a group with the needed property is wrong. I'm not sure if this is salvageable.

Take $P$ to be a totally ordered set in which every countable subset has a minimal element, but which does not itself have a minimal element. (See here.) Let $A$ be the free abelian group generated by $P$, with lexicographic ordering. That is to say, $\sum_{p \in S \subset P} c(p) p$ is positive if $c(p_0)>0$ for $p_0 = \max S$.

I think that $A$ also has the property that every countable subset has a minimal element. No, it doesn't. Consider the set $e$, $e-f$, $e-2f$, $e-3f$, etcetera, where $0 < f < e$.

Oh well, maybe this will give somebody else a better idea.

  • $\begingroup$ David -- in fact, this is the sort of construction I had in mind, hence my allusion to valuation rings above. But at the same time I had some inarticulate feeling that it would not work. So you see why I asked the question... $\endgroup$ Dec 7, 2009 at 18:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.