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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?

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    $\begingroup$ Every $A'$-module is an $A$-module by pullback, and this preserves non-isomorphism of modules. Thus, there are no more isomorphism types of indecomposable $A'$-modules than of $A$-modules. $\endgroup$
    – Ben Webster
    Dec 16, 2015 at 13:52

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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.

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