# Do there exist non-PIDs in which every countably generated ideal is principal?

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 <= \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.

-
Do you know if infinite well-ordered groups exist? – Qiaochu Yuan Dec 6 '09 at 22:57
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. – Andrew Critch Dec 7 '09 at 0:20

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.

-
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. – David Speyer Dec 7 '09 at 3:05
Do I understand this correctly? – David Speyer Dec 7 '09 at 3:09
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. – Hugh Thomas Dec 7 '09 at 3:23
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? – David Speyer Dec 7 '09 at 3:39
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. – Pete L. Clark Dec 7 '09 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.

-

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.

-
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... – Pete L. Clark Dec 7 '09 at 18:11