Let $A = kQ/I$ be a bound quiver algebra for some algebraically closed field $k$, $Q$ a finite connected quiver without oriented cycles, and $I$ an admissible ideal. Say that $I'$ is also an admissible ideal such that $I \subseteq I'$.


My question is this: if $A$ is representation-finite, is $A' = kQ/I'$ also representation-finite? 


It seems like this should be true, and it seems like there should be an "easy" way to prove this (maybe if I knew more algebra or representation theory!), but I can't seem to find this in the literature as a proposition/remark/exercise. Specifically, I've been looking at "Elements of the Representation Theory of Associative Algebras, Vol I" by Assem, Simson, and Skowronski, and at "Quiver Representations" by Schiffler.


Does anyone know if this is a known result? Or is there an easy way to see if this is true/false?



Is a quotient of a bound quiver algebra of finite representation type also representation-finite?
There is indeed an easy way to see this.


Let A be a ring and let I be an ideal of A.
If A has only finitely many isomorphism classes of indecomposable modules then the ring A/I also has only finitely many isomorphism classes of indecomposable modules. The point is that an A/I-module is indecomposable iff it is indecomposable as an A-module, and two A/I-modules are isomorphic as A/I-modules iff they are isomorphic as A-modules.


