= 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 $\operatorname{End}_Q(X)=k$: 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 $N=\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)$. 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 $k\to0$, $0\to k$ and $k\xrightarrow{1}k$. Each has endomorphism ring $k$. Therefore \b[every indecomposable of the one-arrow quiver is a brick].
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