= Solution
{c}
Let $L$ be any <maximal left ideal>. If $I\not\subseteq L$, maximality gives $L+I=R$, because the two-sided <ideal> $I$ is in particular a <left ideal>. Thus $1=\ell+i$ for some $\ell\in L$ and $i\in I$. But
$$
\ell=1-i\in1+I
$$
is a <unit> by hypothesis. A proper <left ideal> cannot contain a <unit>, since left multiplication by its inverse would put $1$ in the <left ideal>. This contradiction proves $I\subseteq L$ for every <maximal left ideal> $L$, and therefore
$$
\boxed{I\subseteq J(R).}
$$
Together with <1/a/ii>, this is the ideal form of the <unit criterion for the Jacobson radical>.
Back to article page
