Every indecomposable representation is a brick if $Q$ has no cycles I'm currently following a course on quiver representations, and stumbled across the definition of a brick, which is a quiver representation $M$ with the property that $\mathrm{End}(M)\cong k$, where $k$ is the field over which our representations are defined. 
There is a theorem which states that a representation $M$ is indecomposable if and only if its endomorphism space is a local ring. Because every field is a local ring, every brick is indecomposable. 
However, there is no mention that the converse holds, which by my idea should hold because the endomorphism space of a representation of a quiver always is a vector space over $k$ hence it should be isomorphic to $k^n$ for some $n$, and because $k^n$ is not local if $n>1$  we have that it is isomorphic to $k$ if it is indecomposable, because a nontrivial quiver representation should have a nontrivial endomorphism space. 
I was wondering if my arguing is right or if I miss some weird counterexample, or if I might have made another mistake. 
 A: The correct converse is that $M$ is a brick if it's indecomposable and it admits no self-extensions. That is, we also need to have $\mathrm{Ext}_k(M,M) = 0$ for an indecomposable quiver representation $M$ to be a brick. The flaw in your reasoning is that you've got the wrong multiplication on $\mathrm{End}_k(M)$. Sure, $\mathrm{End}_k(M)$ will just be $k^n$ as a vector space, but the ring structure on $\mathrm{End}_k(M)$ will not be the same as the one on $k^n$.   
For an example of an indecomposable quiver representation that is not a brick, consider the Jordan quiver $J$ — a quiver having one node and one arrow. Representations of this quiver model endomorphisms of a single vector space. Consider the representation $M$ of $J$ consisting of a two-dimensional vector space over $\mathbf{C}$, and a nilpotent endomorphism  
$$\begin{pmatrix}0&1\\0&0\end{pmatrix}\,.$$
This representation is indecomposable, the endomorphism being a single Jordan block. But this representation is not a brick. Elements of $\mathrm{End}_\mathbf{C}(M)$ look like
$$\begin{pmatrix}a&b\\0&a\end{pmatrix}$$
for some $a,b \in \mathbf{C}$, and so, having two parameters, $\mathrm{End}_\mathbf{C}(M)$ is isomorphic to $\mathbf{C}^2$ as a vector space, but the multiplication in $\mathrm{End}_\mathbf{C}(M)$ isn't point-wise like it would be in $\mathbf{C}^2$. 
This example agrees with the correct converse I first stated, since this example does admit a nontrivial self-extension: a four dimensional vector space with an endomorphism consisting of a single Jordan block of all zeros. More details, including an example on a quiver without an oriented cycle (which is morally the same as the example I've given here) can be found in these notes on Vera Serganova's website. 
