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In this post: Examples of rings with ideal lattice isomorphic to $M_3$, $N_5$ a nice example was given of a non-distributive ring. The lattice of ideals turned out to be the Diamond lattice $M_3$ with the biggest ideal $R$ appended above the top of the Diamond.

Question: Is there a similar example whose lattice ideal looks like the same lattice upside-down? (That is, $R$ would now be in the Diamond, and the $\{0\}$ ideal would be the only ideal outside of the Diamond.)

Edit: I had not intended for anyone to assume commutativity, but I had forgotten I put that in the tags aways back. Jack Schmidt's answer below reminds us why there is no example in the commutative case.

Edit 2: As I was reading Jack Schmidt's second answer below, I realized that $R=M_2(\mathbb{F_2})$ is already a very good example, since both the lattices of right and of left ideals are already precisely the Diamond!

In order to keep going with the original question though, I wanted to bring up the following strategy. By taking an $R-R$ bimodule $B$ and forming the triangular ring $\begin{pmatrix}R&B\\0&R\end{pmatrix}$, and taking the subring $T=\{ \begin{pmatrix}x&y\\0&x\end{pmatrix}\mid x\in R, y\in B\}$, then one has obtained a ring $T$ with $rad(T)=\begin{pmatrix}0&B\\0&0\end{pmatrix}$ such that $T /rad(T)\cong R$. We would understand the structure above $rad(T)$, and the rest of the ideal structure would be determined within $B$. If $B$ could be simple as a right module, then we would be done, but my gut says this is impossible.

If anyone can explain in the comments, I would be grateful. Usually when I ask anything about a bimodule, the answer is "No, because (simple reason)." Followed by: "This is, of course, the 0-th Hochschild cohomology."

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I don't think B exists. I show that R/n=M2(2) in my answer, so B is a bi-module over M2(2), so has dimension a multiple of 4, but a simple (one-sided) module has dimension 2. – Jack Schmidt Jun 6 '12 at 21:52
@JackSchmidt Apparently my sense for bimodules is pretty low: Why must a bimodule over $M_2(F_2)$ have dimension a multiple of 4? – rschwieb Jun 6 '12 at 22:01
@JackSchmidt Ok, I think I see. Let $R=M_2(F_2)$. Then $B$ is a semisimple $S=R\otimes R^{op}$ module, and since the simple modules of $S$ are 4-dimensional, $B$'s dimension is a multiple of 4. – rschwieb Jun 6 '12 at 22:42
up vote 1 down vote accepted

Here are a few thoughts on the non-commutative, one-sided ideal case: Drop the assumption that R is commutative, and ask for the poset of nonzero left ideals to be M3.

The left ideals are the maximal left ideals m1, m2, m3 and their intersection, n. n is the nilradical and R/n is semisimple with three maximal left ideals. R/n is a direct product of matrix rings over division rings. The maximal left ideals of a matrix ring over a division ring are in 1-1 correspondence with the maximal submodules of the natural module. In particular, if the division ring is infinite, there are infinitely many maximal submodules. If the division ring has q elements and the matrix ring is degree n, then there are $(q^n-1)/(q-1)$ maximal submodules. Solving $(q^n-1)/(q-1)=3$ gives $q=n=2$.

The direct product of rings has left ideals precisely the direct products of left ideals of the factors, and so R/n is in fact a simple ring (R/n, m1/n, m2/n, m3/n, n/n is again not a "product" of lattices).

In particular, R is an algebra over a field of characteristic 2 with a 2-dimensional simple module (its only simple module up to iso).

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Thank you, those observations on the "top" are really helpful. In fact, isn't $M_2(\mathbb{F}_2)$ the optimal noncommutative answer to this question? I mean that the entire lattice of right (and of left) ideals is the Diamond. (Last thing, I don't think standard usage of "local" applies to this ring.) – rschwieb Jun 6 '12 at 20:07
Yes, M2(F2) has the diamond as its lattice of right and left ideals, I just assumed you wanted to append something at the bottom. I haven't found such a ring, but I don't have a systematic way to do this (I suspect one could find such a 5 dimensional algebra). "Local": you are correct. I've forgotten the term for R/n is simple, which is all I meant. – Jack Schmidt Jun 6 '12 at 20:41
I was distracted by the fact that there was no example for the commutative case, and I didn't bother to check the most obvious candidate for the noncommutative version... thanks for jarring that loose in my head. I hope you find time to check the addendum to the question I wrote a little while ago. It contains an idea about adding onto the bottom. – rschwieb Jun 6 '12 at 20:49

Suppose R is a commutative, associative, unital ring whose poset of nonzero ideals is M3. Then R has finitely many ideals, and so is commutative artinian. Such a ring is a direct product of commutative local artinian rings. The maximal ideals of such a ring are products of a maximal ideal from one factor with the whole ring from all other factors. In particular, R has 3 maximal ideals and so must have 3 direct factors: R = R1 × R2 × R3. If Ri has Ki ideals, then R has K1×K2×K3 ideals namely the products of the K1 ideals of R1 with the K2 ideals of R2 with the K3 ideals of R3). Each Ki is greater than or equal to 2, since Ri must have a maximal ideal, and so R has at least 2×2×2 = 8 ideals, a contradiction to it only having 6.

This is basically the same as my previous answer.

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I had thought of this, and it is really unfortunate that I put the "commutative algebra" tag on this way back when I asked it. I did not intend to require the ring to be commutative. Thanks for bringing this up here, though. – rschwieb Jun 6 '12 at 1:21

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