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It follows from the Hopkins–Levitzki theorem that if a ring satisfies the DCC on left ideals, then it also satisfies the ACC on left ideals. I've been trying to find a counterexample to the following statement.

If a ring satisfies the DCC on two-sided ideals, then it also satisfies the ACC on two-sided ideals.

The best I could come up with is not really good. My example is of large cardinality and since I don't know much about set theory, I can't be sure if it's correct. I use this. I take the ring $R$ of endomorphisms of an $\aleph_\omega$-dimensional vector space over a field $\mathbb F,$

$$R=\operatorname{End}\left(\bigoplus_{i\in\aleph{\omega}}\mathbb F\right).$$

From the linked answer, I know that the two-sided ideals in $R$ are the sets of endomorphisms of rank $\kappa$, for each infinite cardinal $\kappa \leq \aleph_\omega.$ In ZFC, these ideals form a lattice isomorphic to $\mathbb N\cup \{\infty\}$ with the natural order. This lattice satisfies the DCC, but not the ACC.

This is a silly example, and (a) I'm not sure it's correct, (b) I have never seen the proofs of the facts relevant to it.

I have three questions.

(1) Is the above correct?

(2) (Changed) Is there an example of a smaller cardinality (at most $\mathfrak c$), and preferably uncomplicated? I'm quite sure there must be, but I can't think of one.

(3) Is it possible to construct a simple example of a ring whose lattice of two-sided ideals is isomorphic to $\mathbb N\cup\{\infty\}?$

EDIT After the the discussion in comments in which my ignorance in set theory became obvious, I would like to add a restriction in (2) and (3) that the examples can proven to be examples without the use of the axiom of choice. I'm not sure it's a good restriction, but I don't see one that would fit my needs better.

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    The infinite cardinals $\le\aleph_{\omega}$ are ordered in type $\omega+1$ in ZFC; you don’t need GCH for that.2012-05-19
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    @BrianM.Scott Oh, I'm not really sure now why I wanted to use the GCH... Thanks, I'll edit the question.2012-05-19
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    @BrianM.Scott So this example works in ZFC? Or is there a mistake?2012-05-19
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    You can drop question (2) now, I think, or at least the first part of it. As far as (3) goes, I think that this example *is* pretty simple. As for (1), I’ve not really thought about the answer to the other question, but modulo that it appears to be right.2012-05-19
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    @BrianM.Scott OK, thank you! I've changed (2) to ask for examples of smaller cardinality. This one seems very unnatural to me, because I have never really encountered any rings of cardinality greater than $\mathfrak c$ in ring theory.2012-05-19
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    You can always force $\frak c=\aleph_{\omega+2}$... :-)2012-05-19
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    Be aware that "at most $\mathfrak{c}$" need not be a smaller cardinality than $\aleph_\omega$.2012-05-19
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    @AsafKaragila I don't understand... I thought it could be proven in ZFC that $\mathfrak c<\aleph_\omega$... (Not that I know the proof.)2012-05-19
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    Not even close. The reals can have almost any cardinality you'd like them to have. That is every uncountable cardinal whose cofinality is uncountable can be the real numbers.2012-05-19
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    @AsafKaragila Oh! I took my statement from [Wikipedia](http://en.wikipedia.org/wiki/Aleph_number#Aleph-.CF.89). (Maybe I misunderstood what they wrote there?) There seems to be pretty little left of my question then... Asking for an example "without any set theory" seems a pretty bad idea...2012-05-19
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    @ymar: Yes, you misunderstood. It is provable that $\frak c\neq\aleph_\omega$ for reasons of cofinality (as mentioned in my previous comment). Cohen and Easton did a wonderful work to ensure that $\kappa\to2^\kappa$ is completely independent of ZFC as long as it obeys two rules: (1) Increase in size; (2) increase in cofinality. [Regardless to this, I think we are digressing to a whole other question in these comments... we better stop now.]2012-05-19
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    @AsafKaragila OK, thanks. I'll think if I can make this question precise, and if not, I'll probably delete it then. (Unless someone answers it, in which case I can't do that.)2012-05-19
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    @AsafKaragila I'm sorry to bother you again, but do you think it would make sense to ask for an example for which the AC is not necessary? I'm trying to express my intuition of a "usual ring-theoretic example" and I'm failing badly. The AC is a standard ring-theoretic tool, but I think it's not so often used in counter-examples.2012-05-19
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    Hmmm... I will have to think about it. I'm not too good with non-commutative algebra.2012-05-19

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