 Nilpotent ideal with semisimple quotient radical criterion

If $I$ is a nilpotent ideal and $J(A/I)=0$, then $I=J(A)$. Every $1-ax$, $x\in I$, has a finite geometric-series inverse, giving $I\subseteq J(A)$. The radical's image lies in $J(A/I)=0$, giving the reverse inclusion. A semisimple algebra as quotient supplies the required zero radical.