= Jacobson radical as greatest quasinilpotent ideal
{c}
{title2=$J(A)=\max\{I\triangleleft A:I\subseteq\{x:r(x)=0\}\}$}
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.
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