Why is the arrow ideal $R_Q$ of a finite, connected, acyclic quiver $Q$ equal to the Jacobson radical?

Question

If $Q$ is a finite, connected, acyclic quiver, why does the arrow ideal $R_Q$ equal the Jacboson radical $J$ of the quiver algebra $k(Q)$?

It comes up in showing that the quotient $k(Q)/R_Q$ is a basic algebra, by using the fact that $k(Q)/J\simeq k^n$ for the ground field $k$.

Answer 1

First, since $Q$ is finite and acyclic, it has a maximal path length, so $R_Q^m=0$ for sufficiently large $m$. Thus $R_Q$ is a nilpotent ideal, so is contained in $J$.

To see the reverse containment, note that $k(Q)/R_Q\simeq k^n$ for some $n$. This follows since $R_Q$ contains all nontrivial paths, so after quotienting we're just left with copies of $k$ for each vertex in $Q$.

It turns out, generally that if $I$ is a nilpotent ideal of a $k$-algebra $A$ and $A/I\simeq k^n$, then $I=J(A)$. The maximal ideals of $A/I=k^n$ are the $n$ possible various products of $n-1$ copies of $k$, with one factor of $(0)$ somewhere, so $J(A/I)=0$.

I claim $\pi\colon A\to A/I$ sends $J(A)$ into $J(A/I)$. Let $a\in J(A)$. So $1-ba$ is invertible for all $b\in A$. Then $\pi(1-ba)=1-\pi(b)\pi(a)$ is invertible in $A/I$. Since $\pi$ is surjective, this shows $1-c\pi(a)$ is invertible for all $c\in A/I$, hence $\pi(a)$ is in $J(A/I)=0$. Thus $J(A)\subseteq\ker\pi=I$, so since $I\subset J(A)$ as a nilpotent ideal, $I=J(A)$.

So finally, $$ k(Q)/R_Q=k(Q)/J\simeq k^n $$ which means $k(Q)$ is a basic algebra.

Answer 2

Let $F$ be the arrow ideal. The hypotheses (finite, acyclic) mean that $F$ is nilpotent and $Q/F$ is semisimple. This means that $F={\rm rad}\,Q$, since

Proposition Let $A$ be an Artinian ring, and let $R={\rm rad}\,A$.

$(1)$ $R$ is a nilpotent ideal.

$(2)$ Any nilpotent ideal is contained in $R$.

$(3)$ $A/R$ is semisimple, and if $A/I$ is semisimple, $R\subseteq I$.

Proof

$(1)$ This is a mildly tedious to prove yet standard fact for Artinian rings.

$(2)$ Given $x\in I$ a nilpotent ideal and $y\in A$, $yx\in I$ is nilpotent, so $1-yx$ is invertible and thus is in the radical of $A$.

$(3)$ Artinian J-reduced algebras are semisimple. If $A/I$ is semisimple then ${\rm rad}(A/I)={\rm rad}(A) A/I=R\cdot A/I=(R+I)/I=0$ so $R+I=I$ and $R\subseteq I$.