Endomorphisms of a quiver I know that when $k$ is algebraically closed any indecomposable representation of a quiver over $k$ has endomorphism ring isomorphic to $k$ (this is because by Schur's lemma it is a division algebra).
If $k$ is not assumed to be algebraically closed, could you please give me examples (as many as you can!) of indecomposable quivers that possess representations with endomorphisms algebra that are not commutative? 
I do not have many examples of interesting indecomposable representations: I tried those of the $A$ and $D$ type quivers but of course these don't work, and I also tried those of the Jordan quiver (1 vertex, 1 loop).
 A: It's not true that the endomorphism ring of an indecomposable representation is a division ring, even over an algebraically closed field. Schur's lemma only applies to irreducible representations.
But for an arbitrary field $k$ there are indecomposable representations with any local finite dimensional $k$-algebra as endomorphism ring.
You won't find (finite dimensional, at least) representations of the Jordan quiver with noncommutative endomorphism rings, since its path algebra is $k[t]$, and by the classification of finitely generated modules for a PID, every indecomposable module is of the form $k[t]/I$ for some ideal, whose endomorphism algebra is $k[t]/I$ and hence commutative.
Here's one way of getting any finite dimensional algebra $A$ over $k$ as the endomorphism ring of some representation.
Let $\{a_1,\dots,a_n\}$ be a set of $k$-algebra generators of $A$. Take the quiver with one vertex and $n$ loops, and the representation with $A$, considered as a vector space over $k$, at the unique vertex, with the loops acting by right multiplication by $a_1,\dots,a_n$. 
Or for a quiver with no directed cycles, take the quiver with two vertices and $n+1$ arrows from the first vertex to the second vertex, and the representation with $A$ at both vertices, with the arrows acting by right multiplication by $1,a_1,\dots,a_n$.
If $A$ is a local algebra, then both these examples will be indecomposable. If $A$ is a division ring then the first one will even be irreducible.
