Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 3 / 1

A representation of a quiver assigns a $k$-vector space $X_i$ to each vertex $i\in Q_0$ and a linear map $f_{\rho }:X_{s(\rho )}\to X_{t(\rho )}$ to each arrow $\rho \in Q_1$. Here $s(\rho )$ and $t(\rho )$ are its source and target. A quiver representation morphism $\theta :X\to Y$ is a family of linear maps ${\theta }_i:X_i\to Y_i$ such that every arrow square commutes:Composition and identities are defined vertex by vertex. A quiver representation morphism is an isomorphism precisely when every ${\theta }_i$ is invertible.

Use the following Jacobson radical facts for a unital algebra: a nilpotent ideal is contained in the Jacobson radical; under a surjective algebra homomorphism the image of the Jacobson radical is contained in the radical of the quotient; and a finite product of fields has zero Jacobson radical. For completeness, if $I^N=0$ and $x\in I$, then $ax\in I$ for every $a\in A$, and $1-ax$ is invertible with inverse ${\sum }_{j=0}^{N-1}(ax{)}^j$. The usual unit criterion for the Jacobson radical therefore gives $I\subseteq J(A)$. The quotient assumption gives $J(A)\subseteq I$. HenceThis is the nilpotent ideal with semisimple quotient radical criterion.

For the finite quiver under consideration, let $R$ be the arrow ideal of a path algebra, spanned by paths of positive length. If $Q$ has $r$ vertices and no oriented cycle, a path cannot repeat a vertex, so $R^r=0$. Meanwhile $kQ/R\cong {\prod }_{i\in Q_0}k$, with the constant paths giving the coordinate idempotents. The criterion just proved yields $J(kQ)=R$.

The condition on cycles is necessary. A quiver with one loop has path algebra $k[t]$, whose arrow ideal is $(t)$. But $J(k[t])=0\ne (t)$: the maximal ideals $(t-a)$, $a\in k$, have intersection zero, since the algebraically closed field $k$ is infinite and a nonzero polynomial has only finitely many roots. Thus the arrow ideal need not be the Jacobson radical when oriented cycles are present.

Rows in the following matrix index the source, columns the target, in the order $X,Y,Z$. Solving the arrow-square equation for a quiver representation morphism givesIn particular, the nonzero morphisms between distinct representations are $X\to Z$ and $Z\to Y$, each a one-dimensional family of scalar multiples of the vertexwise inclusion or projection. A map $Y\to Z$ is forced to vanish at the source by the identity arrow of $Z$; similarly a map $Z\to X$ is forced to vanish at the target. Maps between $X$ and $Y$ are zero. Each endomorphism ring is $k$.

The extension complex of quiver representations for the one-arrow quiver isIts cokernel is ${\mathrm{Ext}}_Q^1(U,V)$. Substitution of the three representations givesHere the first argument is the quotient endpoint of a short exact sequence. Thus the only possible nonsplit endpoint pair is subobject $X$, quotient $Y$. The sequenceis nonsplit, since $Z$ is indecomposable. More explicitly, all extensions with these endpoints have a middle arrow $k\stackrel{\lambda }{\to }k$: $\lambda =0$ gives the split short exact sequence, while each $\lambda \ne 0$ gives a middle representation isomorphic to $Z$. With endpoint identifications fixed the extension classes form $k$; up to endpoint automorphisms all nonzero classes give this same nonsplit sequence. There are no other nonsplit sequences with the listed endpoints, including equal endpoints.