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Write the nonsplit short exact sequenceFor an endomorphism $g:M\to M$, the composite $qgj:L\to N$ vanishes because ${\mathrm{Hom}}_R(L,N)=0$. Hence $g$ restricts to an endomorphism $f$ of $L$ and induces an endomorphism $h$ of $N$, giving a commutative diagram of short exact sequences.

Because $L$ and $N$ are brick modules, each of $f,h$ is either zero or an isomorphism. If both are isomorphisms, the short five lemma makes $g$ an isomorphism. If both vanish, $g$ factors successively through $N$ and through $L$, hence through a map $N\to L$; this map is zero, so $g=0$.

The mixed cases would split the sequence. If $f$ is invertible and $h=0$, then $qg=0$, so $g=j\alpha$ for some $\alpha :M\to L$; the identity $\alpha j=f$ makes $f^{-1}\alpha$ a retraction of $j$. If $f=0$ and $h$ is invertible, then $gj=0$, so $g=\beta q$; the identity $q\beta =h$ makes $\beta h^{-1}$ a section of $q$. Both contradict nonsplitting. Thus every endomorphism of $M$ is zero or invertible, and $M$ is a brick.

Solved by gpt-5.6-sol high.

No. Let $Q$ be the quiver $1\to 2$ over a field $k$, and take the representation of a quiverAn endomorphism is a pair of scalar maps $(a,b)$ satisfying $b=a$, so its endomorphism ring is $k$. Every nonzero endomorphism is therefore an isomorphism, making this representation a brick module. It nevertheless has the proper nonzero subrepresentation $0\to k$, so it is not an irreducible module.

Equivalently, this is a nonsimple module over the path algebra $kQ$ whose endomorphism ring is a division ring.

Solved by gpt-5.6-sol high.