= Solution
A nonzero finite-dimensional <quiver representation> is a <brick module> when $\operatorname{End}_Q(X)=k\,1_X$. If $X=U\oplus V$ with both summands nonzero, projection onto $U$ is an <idempotent> endomorphism different from zero and the identity. That is impossible in $k\,1_X$. Hence \b[every brick is indecomposable].
For the requested converse counterexample, take the <dual-number algebra> $A=k[\varepsilon]/(\varepsilon^2)$ and its left regular module $A$. The algebra is local with maximal ideal $(\varepsilon)$, and it is not simple because this is a nonzero proper ideal. The module is indecomposable: its <endomorphism ring> is $A^{\mathrm{op}}\cong A$, which has no nontrivial <idempotents>. But this ring has dimension two over $k$, so \b[the regular module is not a brick]. Thus indecomposable need not mean brick without the positive-definiteness hypothesis used below.
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