Rings such that there are finitely many modules of a given length

Question

I'm interested in rings $R$ (commutative or non-commutative) such that for any $n\in \Bbb N$, there exist only finitely many isomorphism classes of $R$-modules of length $n$. By Krull-Schmidt it's enough to check this for indecomposable modules.

Easy examples I found are PIDs with finitely many irreducible elements and division algebras. The class of rings with this property is closed under finite products and Morita equivalence. Therefore it contains all semisimple rings by Artin-Wedderburn (thanks @rschwieb), which includes group algebras $k[G]$ for finite groups $G$ in the case that $\operatorname{char}(k) \not \mid |G|$. $k[G]$ also has this property if $\operatorname{char}(k) = p$ and the $p$-Sylow subgroup is cyclic. I don't know if $k[G]$ has this property in the other case.

I would be glad to see anything in that direction, whether it's concrete examples, generalizations of the examples I gave or whole classes of rings or necessary conditions for rings to have that property.

Answer 1

This is not an answer, but a long comment which addresses the question Do you think it extends to Artinian rings? (Answer: No.)

If $k$ is an infinite field, then $R=k[x,y]/(x,y)^2$ is a $3$-dimensional, commutative, local $k$-algebra with infinitely many isotypes of length two modules. This makes it a commutative Artinian ring with infinitely many isotypes of modules of length two. (It exactly one isotype of simple module, namely a $1$-dimensional $k$-space with both $x$ and $y$ acting like zero.)

To show how to produce lots of modules of length two, let $V=k^2$ and let $x, y$ act on $V$ via multiplication by the matrices $$ X = \left[\begin{matrix}0&1\\0&0\end{matrix}\right],\quad Y= \left[\begin{matrix}0&\lambda\\0&0\end{matrix}\right], \quad(\lambda\in k, \;\textrm{$\lambda$ fixed}). $$ This makes $V$ an $R$-module annihilated by $y-\lambda x$, but $V$ is not annihilated by $y-\mu x$ for $\mu\neq \lambda$. Thus different $\lambda$'s yield different modules of length two.

Answer 2

The answer can depend fairly delicately on how big $R$ is. As seen in Keith Kearnes' answer, even if $R$ is a finite-dimensional algebra over a field $k$ it can happen that $R$ has finite length modules which vary in parameters living in $k$, and so the number of such modules depends delicately on how big $k$ is.

Here is a basic discussion, inducting on length. First, a necessary condition is that there are finitely many length $1$ modules, or equivalently simple modules. This is true if $R$ is artinian, or more generally if $R/J(R)$ is artinian; actually I think it's true iff $R/J(R)$ is artinian.

Next, let's consider the length $2$ modules. These are extensions $S_1 \to M \to S_2$ where the $S_i$ are simple modules, and so are roughly classified by $\text{Ext}^1(S_2, S_1)$ (this isn't quite the classification up to isomorphism but I think we get the classification up to isomorphism by dividing out by the actions of $\text{Aut}(S_1)$ and $\text{Aut}(S_2)$). If $R$ is an algebra over a field $k$ then these are $k$-vector spaces, which is the basic phenomenon responsible for isomorphism classes of modules varying in parameters living in $k$. If all of these Ext groups vanish then by induction every finite length module is semisimple; this happens if $R$ is semisimple and I want to say it happens iff $R$ is semisimple but I'm not sure how to show that.

In any case this typically doesn't occur and Exts are typically nonzero; you can rig up examples with as many simple modules and as many Exts between them as you want by taking $R$ to be a quiver algebra, in which case the simple modules can be identified with the vertices of the quiver and Exts between them have dimension the number of edges.

From here I'm going to assume for simplicity that $R$ is a finite-dimensional algebra over a field $k$. Then we can calculate $\text{Ext}^n(M, N)$ using a free resolution of $M$ in which every term is a finite free $R$-module, hence a finite-dimensional $k$-vector space; it follows that if $M, N$ are finite-dimensional (equivalently, finite-length) modules then $\text{Ext}^n(M, N)$ is also finite-dimensional. From this we roughly deduce that we expect finite-length modules to vary in finitely many parameters living in $k$, and hence we don't generally expect there to be finitely many such modules unless $k$ is finite.

In this case it's clear without any homological algebra that there are finitely many modules of a given finite length, because there are finitely many homomorphisms $R \to M_n(k)$. Somewhat more generally, $R$ can be any finite ring.

Answer 3

The Second Brauer-Thrall Conjecture is that a finite dimensional algebra of infinite representation type over an infinite field has for infinitely many $d\in\mathbb N$ infinitely many indencomposables of length $d$.

This has been proved by Raimundo Bautista for algebras over an algebraically closed field — a very difficult result. See

• R. Bautista, On algebras of strongly unbounded representation type, Comment. Math. Helv. 60 (1985), no. 3, 392–399.

Nazarova and Roijter later proved this for perfect fields, and Ringel announced the general case but that seems not to have been published. There's a nice survey by Ringel here

---

This means that over algebraically closed fields the only algebras that satisfy your condition are in fact those that have finitely many indecomposable modules (up to iso, of course)