Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 3 / 6

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.

For the Kronecker quiver representation with arrows $I$ and $N=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$, an endomorphism $(P,Q)$ satisfies $Q=P$ and $PN=NP$. Solving the second equation givesThus this representation is indecomposable but is not a brick: its endomorphism ring is a local endomorphism ring, while the nonzero endomorphism $N$ is nilpotent.

For a general indecomposable non-brick, the proof of Ringel lemma on bricks finds a proper indecomposable submodule with nonzero self-extensions. Repetition in strictly decreasing dimension reaches a brick module $Y\subset X$ with ${\mathrm{Ext}}_Q^1(Y,Y)\ne 0$. The linked proof supplies the minimal-rank, retraction and hereditary-extension steps.

Now assume the Tits form of a quiver is positive definite. If an indecomposable $X$ were not a brick, this $Y$ would give the contradictionHence $X$ is a brick. For its nonzero dimension vector $\mathbf{n}$, positivity and integrality then implyThus every indecomposable in this case is a rigid brick. This deduction uses the Ringel lemma on bricks and the Ringel form, without requiring the full Gabriel theorem.