Projective representations of 1-loop quiver I study (finite) representations (over an algebraically closed field $k$) of the 1-loop quiver $Q$ which is defined by having a single vertex and a single edge. So, representations of $Q$ are just endomorphisms of finite dimensional $k$-vectorspaces. I need to determine all indecomposeable representations of $Q$, and by using the structure theorem of finitely generated PID's I'm pretty sure in saying that a representation $f : V \to V$ of $Q$ is indecomposeable exactly if the matrix of $f$ is a Jordan block $J_{a,\dim V}$ in some basis for $V$. My question is now: Which of these indecomposeables are projective? Here a projective representation $A : k^n \to k^n$ means that given any commuting diagram of $k$-linear maps as below with $\pi$ surjective, we should be able to construct a $k$-linear map $g$ (denoted by the dotted line below) such that the resulting diagram commutes.

I have tried to deduce some nessecery conditions for the maps in such diagrams if the representation is to be projective, but I have not been getting any wiser as to decide what I should think. My problem is that it seems (to me!) kind of arbitrary when representations of quivers are projective/not-projective so any ideas or hints as to how I could/should approach this problem would be highly appreciated.
Thank you in advance for your time.
 A: Since your algebra is isomorphic to the polynomial algebra $k[x]$, which is a PID, let's answer a slightly more general question:
Question: Let $R$ be any PID.  What are the finitely generated projective $R$-modules?
Answer: The free $R$-modules.
To see why a torsion module cannot be projective, let $M$ be an indecomposable torsion module.  By the structure theorem for finitely generated modules over PIDs, there exists a non-zero, non-invertible element $a$ of $R$ such that $M$ is isomorphic to $R/(a)$.  Then there is a short exact sequence
$$
 0 \to R/(a) \to R/(a^2) \to R/(a) \to 0,
$$
where the first morphism is multiplication by $a$ and the second is the natural projection.  Clearly, this sequence is not split.  Thus the right-most module, $R/(a)$, cannot be projective.

Going back to representations of the 1-loop quiver: over $k[x]$, the finitely generated torsion modules are preciely the modules which are finite-dimensional over $k$.  Thus there are no projective finite-dimensional representations of the 1-loop quiver.
