Let $Q$ be a finite quiver. Then the following hold: (a) If $KQ$ is semisimple, then $|Q_1| = 0$.

If, moreover, $Q $ is connected, show that:

(b)$KQ$ is local only if $|Q_0| = 1$ and $|Q_1| = 0$,

  • $\begingroup$ Where exactly are you having problems? What have you tried? $\endgroup$ Nov 3, 2012 at 0:21
  • $\begingroup$ What definition of semisimple do you take? Is your quiver allowed to have oriented cycles? $\endgroup$ Nov 3, 2012 at 1:02
  • $\begingroup$ $KQ$ is semisimple as an algebra, the quiver $Q$ is allowed to have oriented cycles. $\endgroup$
    – Aimin Xu
    Nov 3, 2012 at 6:28

1 Answer 1


For (a) I guess it depends on the definition of semisimple. If you take Jacobson radical vanishes as the definition. Your claim is not true, since if you take the quiver with one vertex and a loop, you get the polynomial ring in one variable $K[X]$, whose Jacobson radical vanishes, but it has an arrow.

In Assem, Simson, Skowronski however, which seems to be the book you use (it is helpful if you include this data in your question), if I recall correctly, semisimple is only defined for finite-dimensional algebras. Then $KQ$ finite-dimensional implies that $Q$ has not oriented cycles. And in this case the ideal spanned by the arrows coincides with the Jacobson radical. Hence for the algebra to be semisimple there must not be arrows, hence $|Q_1|=0$.

For (b) if there are more than two vertices, then pick one of them, say $i$. Call the other $j$. Then $e_i$ is not invertible (since it is a zero divisor since $e_j\cdot e_i=0$). Similarly $1-e_i$ is not invertible. A contradiction to the property that for each local ring $x$ or $1-x$ is invertible. Thus $|Q_0|=1$. Now you are left with a polynomial ring (although with non-commuting variables). And you can use the excellent answers to your former question to conclude that there also must not be arrows.

  • $\begingroup$ If $KQ$ is finite-dimensional, the proof is easy as you give. If it is infinitely dimensional, I can not prove (a) in this case. $KQ$ is semisimple in case $KQ$ is semisimple as a left $KQ$-module. $\endgroup$
    – Aimin Xu
    Nov 3, 2012 at 13:01
  • $\begingroup$ I think it is not true in the infinite-dimensional case, see e.g. math.stackexchange.com/questions/133737/…, math.uni-bielefeld.de/~sek/select/rw1.pdf $\endgroup$ Nov 3, 2012 at 13:24
  • $\begingroup$ Dear Julian, can you explain why "Jacobson radical of $K[X]$ vanishes"? Isn't it the ideal generated by $X$? $\endgroup$
    – Aaron
    Nov 5, 2012 at 12:41
  • $\begingroup$ The Jacobson radical is the intersection of all maximal left ideals (since we are in the commutative setting, the left can be omitted). There are infinitely many of them, for infinite fields the $(X-\lambda)$ for $\lambda\in K$ are examples. For finite fields, see e.g. math.stackexchange.com/questions/70127/… . Hence their intersection is $0$. $\endgroup$ Nov 5, 2012 at 12:52
  • 1
    $\begingroup$ If $D$ is a division ring, then the polynomial ring $R=D[x_1,x_2,\ldots x_n]$ is $J$-semisimple, that is, the radical $J(R)$ of $R$ is zero. For if $f\in J(R)$, then $1_{D}-f$ is a unit in $R$ . Since the only units in $R$ are the nonzero elements of $D$ , it follows that $f\in D$. Thus $J(R)$ is an ideal of $D$, whence $J(R)=0$ or $J(R)=D$ by the simplicity of $D$. Therefore $J(R)=0$ since $-1_{D}\notin J(R)$ ($1_{D}-1_{D}=0 $ is not a unit) $\endgroup$
    – Aimin Xu
    Nov 8, 2012 at 1:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .