= Solution
{c}
The <Jacobson radical> is
$$
J(R)=\bigcap_{L\text{ maximal left ideal}}L
=\bigcap_{V\text{ simple left module}}\operatorname{Ann}_R(V).
$$
Here the second intersection ranges over isomorphism classes of nonzero <simple modules>. For the equality, every nonzero $v$ in a <simple module> gives a <surjective> <module homomorphism> $R\to V$, $r\mapsto rv$, whose <kernel> is a <maximal left ideal>. Conversely, every <maximal left ideal> $L$ gives the <simple module> $R/L$. This description by <annihilators> shows that $J(R)$ is a two-sided <ideal>.
Let $j\in J(R)$ and put $a=1+j$. If the <left ideal> $Ra$ were proper, it would lie in a <maximal left ideal> $L$. Both $a$ and $j$ would belong to $L$, giving $1\in L$, a contradiction. Therefore there is $u\in R$ with $ua=1$.
A left inverse alone need not be a two-sided inverse in a general <ring>, so one more step is necessary. The equality $u(1+j)=1$ gives $u=1-uj\in1+J(R)$. Applying the same argument to $u$ supplies $v\in R$ with $vu=1$. Then
$$
v=v(ua)=(vu)a=a,
$$
and hence $au=vu=1$. Thus
$$
\boxed{1+J(R)\subseteq R^\times.}
$$
For the <zero ring> the conclusion holds directly; the <maximal left ideal> argument concerns the nonzero case.
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