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A unipotent algebraic group admits a faithful linear representation in which every group element is a unipotent matrix. For $E={\mathrm{End}}_Q(X)$, invertibility is the nonvanishing condition ${\prod }_i\det h_i\ne 0$. Therefore ${\mathrm{Aut}}_Q(X)=E^{\times }$ is a nonempty Zariski-open subset of the vector space $E$.

Take a Krull-Schmidt decomposition $X\cong {⨁}_{a=1}^r{M}_a^{\oplus m_a}$, with pairwise nonisomorphic indecomposable modules. The Fitting lemma makes each ${\mathrm{End}}_Q(M_a)$ a local endomorphism ring. Its residue division algebra is $k$: over an algebraically closed field, every element of a finite-dimensional division algebra has an eigenvalue and hence must be scalar. The semisimple quotient of a module endomorphism algebra consequently gives, for the Jacobson radical $J=J(E)$,is surjective with kernel $1+J$. The nilpotence of $J$ makes $(1+j{)}^{-1}=1-j+j^2-\cdots$ finite and each $1+j$ unipotent. The kernel is closed and normal. Acting on the multiplicity spaces embeds the product of general linear groups back into $E^{\times }$ and splits this quotient. This proves the Levi decomposition of a quiver automorphism groupSince $1+J$ is the unipotent radical, a nonzero $X$ is indecomposable exactly when ${\mathrm{Aut}}_Q(X)/R_u({\mathrm{Aut}}_Q(X))\cong {\mathbb{G}}_m$: the product has a single factor of size one.

For the base change action on quiver representations, the orbit map is $g\mapsto (g_j{x}_{\rho }{g}_i^{-1}{)}_{\rho :i\to j}$. Substituting $g_i=I+ϵu_i$, with ${ϵ}^2=0$, shows its differential isIts kernel is ${\mathrm{End}}_Q(X)$. The stabilizer is smooth because it is open in that vector space. Hence the differential has rank $\dim \mathrm{GL}(\mathbf{n})-\dim {\mathrm{Aut}}_Q(X)=\dim {\mathcal{O}}_X$, and its image is the Zariski tangent space $T_x{\mathcal{O}}_X$. The normal space to a quiver orbit is thereforeThe ambient quiver representation space is an irreducible affine space, and orbits are locally closed. An orbit is open exactly when its dimension equals that ambient dimension, equivalently when ${\mathrm{Ext}}_Q^1(X,X)=0$. This proves that rigid quiver representations have open orbits. Such an orbit is dense and unique, since two nonempty open subsets of an irreducible space intersect.

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.

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 a finite-dimensional module over an algebraically closed field, write a Krull-Schmidt decomposition $X={⨁}_a{M}_a^{\oplus m_a}$ with distinct indecomposable types. Each endomorphism ring of $M_a$ is a local endomorphism ring with residue division algebra $k$, by the finite-dimensional division algebra over an algebraically closed field result. Modulo the Jacobson radical of $\mathrm{End}(X)$, the blocks of one type become $M_{m_a}(k)$ and all maps between different types vanish. A composite through a different indecomposable type cannot be invertible, since that would make one type a direct summand of the other. This yields the displayed product and the Levi decomposition of a quiver automorphism group.