# Does this kind of non-noetherian bimodule exist?

Question: Do there exist simple rings $R$ and $S$ (i.e., rings with no proper nonzero ideals) and an $(R,S)$-bimodule $M$ such that $M$ is finitely generated both as a left $R$-module and a right $S$-module, but $M$ has a sub-bimodule that is not finitely generated (i.e., $M$ does not satisfy the ACC on sub-bimodules)?

Of course, for this to work $R$ cannot be left noetherian and $S$ cannot be right noetherian, for then $M$ would satisfy ACC on one-sided submodules. Kudos if we can also take $R = S$.

Motivation: A well-known result of I.S. Cohen in commutative algebra states that if every prime ideal of a commutative ring is finitely generated, then all ideals of that ring are finitely generated.

I am interested in constructing counterexamples to this statement when commutativity of the ring in question is dropped. Suppose that $R$ and $S$ are simple rings and $M$ is an $(R,S)$-bimodule. Then the "triangular matrix" ring $$T = \begin{pmatrix} R & M \\ 0 & S\end{pmatrix}$$ has exactly two prime ideals: $$P_1 = \begin{pmatrix} 0 & M \\ 0 & S\end{pmatrix} \quad \mbox{and} \quad P_2 = \begin{pmatrix} R & M \\ 0 & 0\end{pmatrix}.$$

If $M$ is finitely generated as a bimodule but has a submodule that is not finitely generated, then its prime ideals $P_i$ are finitely generated as ideals of $T$ but there is an ideal in $T$ that is not finitely generated. (Such an example would be $R = S = k$ a field and $M = V$ an infinite dimensional vector space.)

Even "better" (or worse), if $M$ is finitely generated as a left $R$-module but contains a subbimodule that is not finitely generated, then its prime ideals $P_i$ are finitely generated as left ideals but $T$ has an ideal that is not finitely generated. (Such an example would be $R$ any simple ring that is not left noetherian, $S = Z(R)$ the field that is its center, and $M = R$.)

By now you can see what I am after: if one has a bimodule $M$ as in the question, then the ring $T$ has its prime ideals $P_i$ all finitely generated both as left ideals and as right ideals, but it contains an ideal that is not finitely generated.

Update January 13, 2015: George Bergman suggested to me the following reduction of the problem, which he kindly agreed to let me share here.

First note that we may reduce to the case where $M$ is cyclic as a left $R$-module; for if ${}_R M$ is generated by $n$ elements, then the left $\mathbb{M}_n(R)$-module of column vectors $M^n$ is cyclic, it is still finitely generated as a right $S$-module, and the ring $\mathbb{M}_n(R)$ remains simple.

Now assuming ${}_R M = Rm$ is cyclic, if we let $L$ be the left annihilator of the generating element $m \in M$ then $M \cong R/L$ as left $R$-modules. Now $S$ may be viewed as a subring of the endomorphism ring $\mathrm{End}_R(R/L) \cong I(L)/L$ where $$I(L) = \{x \in R \mid Lx \subseteq L\}$$ denotes the idealizer of $L$ (the largest subring of $R$ in which $L$ is an ideal). Thus there is a subring $S_0 \subseteq I(L) \subseteq R$ such that $S$ acts on the right of $R/L$ via $S \cong S_0/L$.

Now since $M \cong R/L$ is finitely generated as a right $S$-module, there exist finitely many elements $\{r_1, \dots, r_n\} \subseteq R$ such that every element of $R$ is congruent modulo $L$ to some $\sum r_i s_i$ with $s_i \in S_0$. But because $L \subseteq S_0$, if we extend the generating set to include $1 \in R$, then we find that $R$ itself is finitely generated as a right $S_0$-module.

Thus we have:

($*$) A simple ring $R$ containing a left ideal $L$ and a subring $S_0$ containing $L$ as a two-sided ideal, such that $R$ is finitely generated as a right $S_0$-module and $S = S_0/L$ is a simple ring.

Conversely, if we have the data above then $M = R/L$ is an $(R,S)$-bimodule that is left cyclic and right finitely generated. Thus the question reduces to:

Reduced Question: In the situation ($*$) above, must the $(R,S)$-bimodule $M = R/L$ satisfy ACC on subbimodules?

I had originally approached my question working under the assumption ($*$) while taking $L = 0$, so that $M = R$ and $S = S_0$ is a subring of $R$. My instinct was to take $R$ to be some sort of "infinite matrix ring" (like the endomorphisms of a countably infinite vector space modulo the finite rank operators, or the direct limit $\varinjlim \mathbb{M}_{2^n}(\mathbb{C})$ that gives the purely algebraic version of the "hyperfinite $\mathrm{II}_1$ factor from von Neumann algebra theory). There is perhaps some justification to this intuition as the reduction above makes $R$ into a matrix ring. Alas, I have yet to find what I seek.

• Manny, have you found objects satisfying (*)? – Pace Nielsen Jan 27 '15 at 16:07
• Pace, yes, these are pretty easy to produce even when $L = 0$. The most trivial example would be to take $R = S$ a simple ring, so that the module $M = R$ is cyclic on both the left and right. Of course, this is useless as a counterexample, since it has no proper nontrivial sub-bimodules (i.e., ideals). Less trivially, if $R = M_2(S)$ for some simple ring $S$, then $M = R$ is left $R$-cyclic and right finitely $S$-generated. (Here I have considered "very infinite" rings where $R \cong M_2(R)$, but with no success yet.) – Manny Reyes Jan 27 '15 at 17:39
• I should have said: these are easy to produce especially when $L = 0$. First note that for $R_{S_0}$ to be finitely generated, it is necessary that $R$ be finitely generated as a right module over the idealizer ring $I(L) \supseteq S_0$. I can't say that it seems terribly easy to begin with $R$ and then find a nonzero $L \subseteq R$ such that $R_{I(L)}$ is finitely generated! – Manny Reyes Jan 28 '15 at 17:16