 Solution

A nonzero finite-dimensional representation is a brick module when its endomorphism ring is a division algebra. Over the algebraically closed field $k$, this means ${\mathrm{End}}_Q(X)=k$: for any endomorphism $f$, an eigenvalue $\lambda$ makes $f-\lambda I$ noninvertible, hence zero in a division algebra.

For a counterexample to the converse of “brick implies indecomposable”, take the one-loop representation $k^2$ with nilpotent Jordan block $N=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$. Its endomorphism ring is $k[N]\cong k[t]/(t^2)$, a local endomorphism ring of dimension two. It has no nontrivial idempotents, so the module is indecomposable, but it is not a brick.

For the one-arrow quiver, splitting the kernel, image and target complement decomposes any representation into copies of $k\to 0$, $0\to k$ and $k\stackrel{1}{\to }k$. Each has endomorphism ring $k$. Therefore every indecomposable of the one-arrow quiver is a brick.