Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 3 / 6 / a

A nonzero finite-dimensional quiver representation is a brick module when ${\mathrm{End}}_Q(X)=k{1}_X$. 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 every brick is indecomposable.

For the requested converse counterexample, take the dual-number algebra $A=k[\epsilon ]/({\epsilon }^2)$ and its left regular module $A$. The algebra is local with maximal ideal $(\epsilon )$, 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 the regular module is not a brick. Thus indecomposable need not mean brick without the positive-definiteness hypothesis used below.

Use the allowed Ringel lemma on bricks in the following precise form: a finite-dimensional indecomposable quiver representation that is not a brick module contains a nonzero brick $B$ with ${\mathrm{Ext}}_Q^1(B,B)\ne 0$.

Suppose such a nonbrick indecomposable existed, and let $\mathbf{b}=\dim B$. The Tits form of a quiver is $q_Q(\mathbf{b})=⟨\mathbf{b},\mathbf{b}{⟩}_Q$, so the Ringel form identity would givePositive definiteness gives $q_Q(\mathbf{b})>0$ for nonzero $\mathbf{b}$, a contradiction. Therefore every indecomposable is a brick.

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 $q_Q(\mathbf{n})=1$.

An indecomposable of dimension vector $\mathbf{n}$ is a brick module and has $q_Q(\mathbf{n})=1$ by the previous parts. Its automorphism group of a quiver representation has dimension one. ThereforeIt follows that its orbit is open and dense in the irreducible quiver representation space. Two different indecomposables of the same dimension vector would give two disjoint nonempty open orbits. Nonempty open subsets of an irreducible space must intersect, so this is impossible. Hence an indecomposable is uniquely determined by its dimension vector, up to isomorphism. This is the positive definite Tits form indecomposable classification.

If $Q_0$ is empty, there are no nonzero indecomposables and the conclusion is immediate. Otherwise positive definiteness of the Tits form of a quiver gives a constant $c>0$ with $q_Q(v)\ge c\Vert v{\Vert }^2$ for every real vector $v$: take the minimum of $q_Q$ on the compact unit sphere. Every positive root of a quiver therefore satisfies $\Vert \mathbf{n}{\Vert }^2\le 1/c$. Only finitely many nonnegative integer vectors lie in this bounded set. Parts (ii) and (iii) give exactly one indecomposable isomorphism class for each such root, and no others. Consequently $Q$ has finite representation type.