Without Gabriel's theorem, how do you classify indecomposable reps of $\mathbb{A}_n$? Given the Dynkin quiver of type $\mathbb{A}_n$ without orientation,
$$
\begin{aligned}
  1\!-\!2\!-\dotsb-\!n 
\end{aligned}
$$
let $M(i,j)$ be a representation of $\mathbb{A}_n$, such that $V_l=K$ for all $i \leq l \leq j$ and $0$ otherwise.
It is a well know result that $\mathbb{A}_n$ is representation finite. In particular every indecomposable representation of $\mathbb{A}_n$ is isomorphic to some $M(i,j)$. This means that up to isomorphism there are $n(n+1)/2$ indecomposable representations of $\mathbb{A}_n$, which are given by the $M(i,j)$.
My question is, how can one show this without using Gabriels Theorem? Is there an author who has done that already? Gabriels Theorem is very long and involves reflection functors, but at the same time its statement is very general and applies to different kinds of Dynkin quivers.
 A: Since your looking at representations of $A_n$ first pick some orientation of the edges as arrows. And if you'd like, without loss of generality choose the orientation where they're all pointing in the same direction, to the right.
Sketch. Think in terms of strings. In a representation $M$ let $i$ be the left-most vertex such that $\dim M_i >0$. Pick a basis for $M_i$ and let $x$ be an element of that basis. Let $M_{f_i}, M_{f_{i+1}}, \dotsc, M_{f_n}$ be the linear maps to the right of the vertex $i$. Follow $x$ along this chain of maps, forming a string of one-dimensional vector spaces spanned by the successive images of $x$. There must be some left-most $k$ between $i$ and $n$ such that you get $$M_{f_k} \circ\dotsb \circ M_{f_{i+1}} \circ M_{f_{i}} (x) = 0$$ This is where the string stops. The idea is that you can peel off this string of one-dimensional spaces
$$ 0 \rightarrow \dotsb \rightarrow \langle x \rangle \rightarrow \langle M_{f_i}(x)\rangle \rightarrow \langle M_{f_{i+1}}(M_{f_i}(x))\rangle \rightarrow \dotsb \rightarrow \langle M_{f_{k-1}} \circ\dotsb \circ M_{f_{i+1}} \circ M_{f_{i}} (x)\rangle \rightarrow 0 \rightarrow \dotsb \rightarrow 0$$
as a direct summand of $M$. The details of are too annoying for me to want to type up here, but I hope this conveys the idea well enough.
