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 $k$.

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 $R\subseteq I$. 
