= Jacobson radical is independent of handedness
{c}
{title2=$\bigcap L_{\max}=\bigcap I_{\max}$}
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.
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