I am studying representation theory at the moment and I was wondering the following (Since I found no source discussing it):
Let $Q$ be a quiver and let $KQ$ be its path algebra. Let Rep($Q$) be the category of representations of $Q$ and Mod($KQ$) be the category of modules over $KQ$.
Now in representation theory one finds that there is an equivalence between Rep($Q$) and Mod($KQ$).
Now my question is: Given a indecomposable representation of $Q$, is the corresponding module over $KQ$ also indecomposable? It would make sense to me but the following example shows otherwise.
Here is a contradictory example: Let $K \stackrel{id_K}{\rightarrow} K \rightarrow 0$ be a indecomposable representation of the quiver $1 \rightarrow 2 \rightarrow3$. By the equivalence theorem the corresponding module is $K^2$ which is decomposable.
Actually, $K^2$ is an indecomposable module over $KQ$. What is your proposed direct sum decomposition? If you are proposing $K \oplus K$ (with the factors chosen the "natural" way) this doesn't work, because the first factor is taken to the second factor by multiplication by the element $1 \to 2$ of the path algebra. I think you were just thinking that $K^2$ is decomposable as a $K$-module, which of course is true.
The difference between a module over $KQ$ and an element of your $\text{Rep}(Q)$ is really just expanding some definitions; you'll find that any proof that $K \xrightarrow{\text{id}_K} K \to 0$ is indecomposable directly translates to a proof that the module $K^2$ you've defined is indecompsable, and vice versa. This correspondence will extend to any more complicated (sensibly-defined) property that quiver representations and $KQ$-modules can have.