# Followup to “Examples of rings with ideal lattice isomorphic to $M_3$, $N_5$”

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

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