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

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.
