Jacobson radical is independent of handedness ($\bigcap L_{\max }=\bigcap I_{\max }$)

The intersection of maximal left ideals is the intersection of annihilators of simple modules on the left, hence is a two-sided ideal. If $j$ belongs to it, $1-rj$ has a left inverse for every $r$. Its left inverse also differs from $1$ by a radical element and has a left inverse; this makes the original inverse two-sided. The identity $(1-ba{)}^{-1}=1+b(1-ab{)}^{-1}a$ then shows that $1-jr$ is invertible too. If $j$ were outside a maximal right ideal $I$, one would have $1-jr\in I$ for some $r$, a contradiction. Applying the same argument to the opposite ring proves equality.