Jacobson radical as greatest quasinilpotent ideal ($J(A)=\max \{I◃A:I\subseteq \{x:r(x)=0\}\}$)

The unit criterion for the Jacobson radical implies that every radical element is a quasinilpotent element, because $1-x/\lambda$ is invertible for every nonzero $\lambda$. Conversely, if a two-sided ideal consists of quasinilpotent elements, then $yx$ is quasinilpotent for every member $x$ and every $y\in A$. Hence $1+yx$ is invertible and $x$ lies in the Jacobson radical. The maximality statement concerns ideals contained in the quasinilpotent set, not the whole set itself.