 Solution

The Jacobson radical isHere 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 $u=1-uj\in 1+J(R)$. Applying the same argument to $u$ supplies $v\in R$ with $vu=1$. Thenand hence $au=vu=1$. ThusFor the zero ring the conclusion holds directly; the maximal left ideal argument concerns the nonzero case.