8
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Manny, have you found objects satisfying (*)? $\endgroup$ – Pace Nielsen Jan 27 '15 at 16:07
  • 1
    $\begingroup$ 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.) $\endgroup$ – Manny Reyes Jan 27 '15 at 17:39
  • $\begingroup$ 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! $\endgroup$ – Manny Reyes Jan 28 '15 at 17:16

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.