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In this question, I gave an example of a ring whose lattice of two-sided ideals is order-isomorphic to $\omega+1$. I've been playing a bit with trying to find rings with a given lattice of ideals since, and one case I found interesting and difficult at the same time. (Difficult for me of course. I've learned here that many questions I can't answer turn out to be trivial.)

Is there a unital ring whose lattice of two-sided ideals is order-isomorphic to $(\omega+1)^{\operatorname{op}}?$

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Hi: can you elaborate in the question on how the ideals in that example correspond to $\omega +1$, and how this new example looks? I'm just a little dense tonight, and I could use a little help :) –  rschwieb Jun 12 '12 at 0:44
    
@rschwieb The ideals in that example are the sets of endomorphisms with rank $\leq\aleph_i$ for $i=1,2,\ldots,\omega.$ If we denote the set of these ideals by $\mathscr I$, then the function $f:\mathscr I\to(\omega+1),$ $f(\{\varphi\in R\,|\,\operatorname{rank}(\varphi)\leq\aleph_i\})=i$ is an order isomorphism. –  user23211 Jun 12 '12 at 1:06
    
OK, so here the idea is to find an example whose ideals are like this, except indexed so that they form a descending chain? –  rschwieb Jun 12 '12 at 11:03
    
@rschwieb Yes, that's it. It's difficult for me because I don't know any general methods of constructing rings with a given lattice of ideals. It's pretty much random guessing for me most of the time. –  user23211 Jun 12 '12 at 11:20
    
That's how it normally is, with a few technical tricks accumulated through experience. The problem in general of realizing a given latice (not every lattice, of course) as an ideal lattice is a very hard one. –  rschwieb Jun 15 '12 at 13:26
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Maybe you can tell me if this is what you're looking for.

What about the power series $R=\mathbb{F}[[x]]$? Doesn't $R\supset(x)\supset(x^2)\supset\dots$ make the chain you are looking for?

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I think it is an example! :-) Thank you very much. That every ideal is of this form has to do with the fact that $\mathbb F[[x]]$ is a discrete valuation ring, right? –  user23211 Jun 15 '12 at 1:51
    
@ymar Yes, it's pretty easy to see this way: 1) An element of the ring is a unit iff it has nonzero constant term. 2) Every(nonzero)thing is a unit times a power of $x$! If you like valuation theory better then I'm sure you can come up with a snazzier description that way too. –  rschwieb Jun 15 '12 at 1:52
    
I know nothing about valuation theory except some random facts without proofs. I understand your argument though. Thanks again! –  user23211 Jun 15 '12 at 1:54
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