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Let $R$ be a commutative ring with 1. The ideals of $R$ form a lattice with inclusion as order relation. Let me call it the ideal lattice $L(R)$ of $R$.

Given an arbitrary lattice $L$, there are some typical operations to obtain new lattices from $L$. I wonder whether there are rings that have these lattices as their ideal lattices, and can be easily constructed from $R$.

  • Is there a ring $R'$ so that $L(R')$ is (isomorphic to) the dual lattice of $L(R)$, i.e. the same lattice but with reversed lattice order.
  • For ideals $I,J\in L(R)$ with $I\subseteq J$, one can form the interval $$[I,J]:=\{K\subseteq R\text{ an ideal}\mid I\subseteq K\subseteq J\}.$$ This is again a lattice. Is there a ring with an ideal lattice isomorphic to $[I,J]$?

As an example, I know that for some ideal $I\in L(R)$, the interval $[I,R]$ is isomorphic to the ideal lattice of the quotient ring $R/I$. There is an inclusion-preserving one-to-one correspondence between the ideals of $R/I$ and the ideals of $R$ that contain $I$.

M. Winter
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  • Same reason, $[I,J]$ is the ideal lattice of the non-unital ring $J/I$. – user10354138 Jun 23 '19 at 14:23
  • @user10354138 Thank you! Now of course I wonder, whether each ideal lattice of a commutative ring is also the ideal lattice of a commutative unital ring. Or more general, what we can learn about the ring from just knowing its ideal lattice. – M. Winter Jun 23 '19 at 14:26
  • @M.Winter: It is not necessarily true that $[I,J]$ is isomorphic to the ideal lattice of the nonunital ring $J/I$. If $k$ is a field, $R=k[x]$, $J=(x)$, $I=(x^2)$, then $[I,J]$ is a two-element chain while the lattice of ideals of the nonunital ring $J/I$ is isomorphic to the lattice of additive subgroups of $k$. – Keith Kearnes Jan 19 '23 at 16:00

1 Answers1

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Is there a ring $R'$ so that $L(R')$ is (isomorphic to) the dual lattice of $L(R)$, i.e. the same lattice but with reversed lattice order.

No, not sticking to rings with identity least. The easy way to see this is just to note that in a ring with identity, there are always maximal ideals, but not always minimal ideals. So if you took a ring with no minimal ideals (like $\mathbb Z$) then there is no ring with identity having a lattice of ideals isomorphic to the dual lattice.

I don't know for sure that your conjecture isn't true, but I do know the problem of representing lattices as ideal lattices of rings is not an easy problem. I think the most famous result along these lines is when von Neumann represented certain types of lattices as lattices of ideals in what are now called von Neumann regular rings for the purposes of studying 'continuous geometry.'

There is an important class of rings for which the ideal lattice is self-dual: they are called (commutative) quasi-Frobenius rings.

Or more general, what we can learn about the ring from just knowing its ideal lattice?

Well, many ring properties are reflected or even defined by their ideal lattice. The definition of "Noetherian" and "Artinian" in terms of chain conditions are exactly statements about how their lattices of (left/right/twosided) ideals behaves.

But a lot of information is lost, of course. As far as their lattices of left/right/twosided ideals go, all division rings look identical.

See also this previous question: What do ideals of a ring say about its inner structure

rschwieb
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  • Thats was very interesting and I should actually have thought about the argument with the maximal ideal. Can you quickly provide an example of a non-unital ring without maximal ideal? I just cannot see right now how the usual Zorn's lemma proof of the existence of maximal ideals uses the unit of the ring. – M. Winter Jun 23 '19 at 16:30
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    @M.Winter The existence of identity is explicitly needed in the zorn’s lemma argument, because otherwise a union of proper ideals may not be a proper ideal. The idea is that your poset of ideals consists of those which do not contain 1. – rschwieb Jun 23 '19 at 21:09
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    @M.Winter If you want examples of rings without maximal ideals, you should search the site first. There is already a good post on them. – rschwieb Jun 23 '19 at 21:33