TITLE: Rings such that there are finitely many modules of a given length
QUESTION [7 upvotes]: 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.

REPLY [4 votes]: 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)