 Solution

If $X$ is indecomposable, part (i) makes it a brick module. For its nonzero dimension vector of a quiver representation $\mathbf{n}$,The Tits form of a quiver takes integer values on integer vectors, so $q_Q(\mathbf{n})=1$ and $X$ has no self-extensions.

Conversely, suppose $\mathbf{n}$ has nonnegative integer entries and $q_Q(\mathbf{n})=1$. Choose a representation $M$ of that dimension vector whose orbit has maximal dimension. Such an orbit exists because dimensions are integers bounded by $\dim {\mathrm{Rep}}_Q(\mathbf{n})$. If $M=U\oplus V$ with nonzero $U,V$, a nonzero extension in either direction would, by part 5(c), produce a middle representation of the same dimension vector with larger orbit. Hence ${\mathrm{Ext}}_Q^1(U,V)={\mathrm{Ext}}_Q^1(V,U)=0$.

Writing $\mathbf{u}=\dim U$ and $\mathbf{v}=\dim V$, the Ringel form identity then givesThe last inequality uses positive integral values of $q_Q$ at both nonzero vectors. This contradiction makes $M$ indecomposable. Thus the indecomposable dimension vectors are exactly the positive roots $q_Q(\mathbf{n})=1$.