# Rings all of whose torsion modules are cyclic

Let us call a (possibly non-commutative) ring $R$ "very good" if every finitely generated torsion left $R$-module is cyclic. Here is an example of such a ring:

Let $k=\mathbb{C}((t))$ and let $R=k[\partial]$ denote the ring of differential polynomials. The addition in this ring is defined as usual, but the multiplication is adjusted by the rule $$\partial a = a\partial + \frac{d}{dt} a,\quad \quad \forall a\in k.$$

Then it is known that every finitely generated $R$-module is a direct sum of a free module together with a cyclic module. In other words, every finitely generated torsion left $R$-module is cyclic. This fact is known as the cyclic vector theorem and has important consequences for the study of linear differential equations.

Question 1: What other examples of "very good" rings are known?

Question 2: Is there a general abstract property we can put on a ring which guarantees that it is "very good"?

In regards to Question 2, it helps to have the following fact in mind: if we require $R$ to be a PID, then every finitely generated module over $R$ will be a direct sum of cyclic modules. But $\mathbb{Z}$ is an example of a PID which is not "very good"! ($\mathbb{Z}/{2}\oplus \mathbb{Z}/4$ is not cyclic). Thus in some sense, $k[\partial]$ is better than $\mathbb{Z}$. I would like to know if there is a way to formalise this observation using (non-commutative) ring theory.

• Such a ring $R$ is necessarily noncommutative. Certainly $R$ cannot be a field. If $R$ is commutative it has a nontrivial maximal ideal $m$, and the finitely generated torsion module $R/m \oplus R/m$ is not cyclic because it has dimension $2$ as an $R/m$-vector space. This argument shows more generally that such a ring $R$ cannot admit a nonzero map into a commutative ring. – Qiaochu Yuan Nov 5 '15 at 1:50

Here is a condition on a ring $R$ that is equivalent to every finitely generated module (right or left, they are equivalent) being cyclic: there exists a finitely generated (fg) free module on more than one generator that is a direct summand of $R$ (equivalently, there exists an idempotent $e$ in the ring such that $eR$ is free on more than one generator).

To see this, if $F$ is free on $k > 1$ generators which is a direct summand of $R$, then so is $F_i:= F^{k^i}$ for all $i$. Every fg free module is a direct summand of one of these, hence is a factor of one of them; but every direct summand of $R$ is cyclic; thus all $F_i$ are cyclic, and therefore so are all fg free modules. It follows immediately that all fg modules are cyclic. The converse is straightforward.

There are lots of rings (necessarily noncommutative) that satisfy this, e.g., Cuntz algebras, the endomorphism ring of infinite dimensional vector spaces (this example was due to Baer), von Neumann algebras that are properly infinite (in the sense of vN algebras), maximal rings of quotients of almost any prime ring (except the nice ones), .... I worked on things like this in my PhD thesis (forty years ago).

This condition is slightly stronger than the negation of directly finite.

For noncommutative rings, we have to be careful about what is meant by a torsion module. One candidate refers to singular torsion (the annihilator on the appropriate side of every nonzero element is an essential right or left ideal (depending on the appropriate side). I really don't know off-hand how to deal with every fg torsion module being cyclic.

With this definition of torsion, every field or division ring satisfies this property, because the only finitely generated torsion module is zero.

• Thanks David. Is this condition (existence of a nontrivial free submodule) satisfied by the ring $k[\partial]$ mentioned above? – Dr. Evil Nov 7 '15 at 8:43
• No, because $k[\partial]$ is a domain (no zero divisors, let alone idempotents). It is also noetherian (I think), which also by itself precludes that condition. [By the way, existence of a nontrivial free submodule is not enough: take the free ring on more than one generator. To get every fg to be cyclic, need the free on more than one generator to be a direct summand of $R$.] – David Handelman Nov 7 '15 at 15:19
• Thanks! It would be nice to have some general reasoning that would explain, in particular, $k[\partial]$. – Dr. Evil Nov 10 '15 at 6:58

Suppose that $R$ satisfies the following properties:

1. $R$ is a simple ring : it has no non-trivial two-sided ideals,
2. $R$ has left Krull dimension equal to $1$. This means that $R$ has at least one infinite strictly descending chain of left ideals, and for every chain $I_0 \supseteq I_1 \supseteq I_2 \supseteq \cdots$ of left ideals, there is an integer $N \geq 0$ such that the $R$-module $I_n / I_{n+1}$ has finite length for all $n \geq N$,
3. $R$ is left Noetherian.

Then every finitely generated torsion left $R$-module is cyclic.

Already if $R$ is a simple ring which is not left Artinian (implied by 1) and 2) above), then every left $R$-module of finite length is cyclic. This result is due to J.C.Robson - see Corollary 5.7.3(i) of his book Noncommutative Noetherian Rings with J.C.McConnell.

If $R$ satisfies 1) and 3), then there is a good notion of "torsion module": a finitely generated left $R$-module $M$ is said to be torsion if for every $m \in M$ there is a non-zero-divisor $s \in R$ such that $s \cdot m = 0$. This is a far-reaching and well-behaved generalisation of the well-known notion of torsion module over a commutative integral domain. At the heart of this theory is (a special case of) Goldie's Theorem: the set of non-zero-divisors in a prime left Noetherian ring is a left Ore set. Note that every simple ring is automatically prime.

Finally, if we suppose further that 2) holds, then every finitely generated torsion left $R$-module actually has finite length. This follows from Proposition 6.3.11(iii) of the book mentioned above.

Although it may appear that these conditions are rather restrictive, it turns out that there are many interesting examples of Noetherian rings "occurring in nature" that satisfy them.

• Let $X$ be a smooth connected affine curve over a field $k$ of characteristic zero, and let $R = \mathcal{D}(X)$ be the $k$-algebra of global differential operators on $X$. Then $R$ satisfies all of these conditions.
• This includes as a special case the first Weyl algebra $A_1(k) = \mathcal{D}(\mathbb{A}^1_k) = k[x][\partial; d/dx]$.
• Each one of 1)-3) is a Morita-invariant property, so in particular, if $R$ satisfies them, then so does every matrix algebra $M_n(R)$.
• If $R$ satisfies 1)-3), and $S$ is a left Ore set in $R$ such that the Ore localisation $S^{-1} R$ is not left Artinian, then $S^{-1}R$ again satisfies 1)-3).
• $\mathbb{C}((t))[\partial; d/dt]$ also satisfies 1)-3).