# A finite-type quiver has no self-loops

I am reading through Etingof et al's notes on representation theory, and they assert in Exercise 5.4(c) on page 80 that a finite-type quiver has no self-loops. I think the way to show this is to consider the simplest case: a quiver with one vertex and one edge - the self-loop. Representations of this quiver correspond to a choice of vector space $V$ and any linear endomorphism $T: V \to V$. I should be able to simply write down an infinite family of nonisomorphic indecomposable representations, but I'm not very good at seeing which representations are indecomposable/nonisomorphic. Can someone give me a hint?

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Here is one infinite family of pairwise non-isomorphic indecomposables. Take $V$ to be $F^n$ (your base field $F$) a nilpotent matrix with a single Jordan block for $T$: any submodule must contain the (unique up to scalars) vector in the kernel, so these are indecomposable.
Here is another, if the base field is infinite: take for $T$ the scalar matrix $T=a$ for $a \in F$, acting one a one-dimensional $V=F$. These are simple and non-isomorphic. The real issue here is the infinite dimensionality of the path algebra...
Thank you for your help. I have a question about your first example. Why must every submodule $W \leq V$ contain the kernel? Does it have to do with the filtration $W \supseteq T(W) \supseteq T^2(W) \supseteq \cdots \supseteq 0$? –  JHF Jun 18 '13 at 1:51
A submodule is necessarily stable by $T$, and since $T$ is nilpotent, a submodule must therefore contain a vector $v$ with $Tv=0$ (just start with any $v$ and apply $T$ til you get $0$). –  Stephen Jun 18 '13 at 2:04