= Solution
Take rings and modules to be unital. Let $J_\ell$ denote the intersection of the <maximal left ideals>. For each nonzero vector $v$ of a <simple module> $M$ on the left, the map $R\to M$, $r\mapsto rv$, is onto and its kernel is a <maximal left ideal>. Consequently $J_\ell$ annihilates every <simple module> on the left. Conversely, annihilating $R/L$ forces an element to belong to the <maximal left ideal> $L$. Thus
$$
J_\ell=\bigcap_{M\text{ simple left}}\operatorname{Ann}_R(M),
$$
which shows that $J_\ell$ is a two-sided <ideal>.
For $j\in J_\ell$ and $r\in R$, the <left ideal> $R(1-rj)$ cannot be proper: a containing <maximal left ideal> would also contain $rj$ and hence $1$. Therefore $u(1-rj)=1$ for some $u\in R$. Here $u=1+urj$ also differs from $1$ by an element of $J_\ell$, so it too has a left inverse $v$. Multiplying the first equality by $v$ gives $1-rj=v$; hence $(1-rj)u=1$. We have proved that $1-rj$ is a <unit>, not just left invertible.
The identity
$$
(1-ba)^{-1}=1+b(1-ab)^{-1}a
$$
is verified by multiplying on either side, using $(1-ab)^{-1}(1-ab)=(1-ab)(1-ab)^{-1}=1$. It implies that $1-jr$ is a <unit> as well. If $j$ were outside a <maximal right ideal> $I$, then $I+jR=R$, so $1-jr\in I$ for some $r$. A proper <right ideal> cannot contain a <unit>, giving a contradiction. Thus $J_\ell$ belongs to every <maximal right ideal>. Applying the same argument to the <opposite ring> proves the reverse inclusion. This proves the <Jacobson radical is independent of handedness>:
$$
\boxed{J(R)=\bigcap_{L\text{ maximal left}}L=\bigcap_{I\text{ maximal right}}I.}
$$
For the zero ring the intersections are interpreted as the whole ring, so the equality remains valid.
If a two-sided <ideal> $N$ satisfies $N^m=0$, then for $x\in N$ and $r\in R$, $(rx)^m=0$. The finite <geometric series> $\sum_{i=0}^{m-1}(rx)^i$ inverts $1-rx$. The <unit criterion for the Jacobson radical>, also proved above by the maximal-ideal argument, gives \b[$N\subseteq J(R)$]. This assertion concerns a <nilpotent ideal>; arbitrary <nilpotent elements> of a noncommutative ring need not lie in its <Jacobson radical>.
For the <formal power series ring> over the <p-adic integers>, a series $f=\sum_{i\ge0}a_it^i$ is a <unit> exactly when $a_0$ is a <unit> of $\mathbb Z_p$. Necessity follows by comparing constant coefficients. If $a_0$ is invertible, construct the inverse recursively by $b_0=a_0^{-1}$ and $b_m=-a_0^{-1}\sum_{i=1}^m a_ib_{m-i}$. Therefore the nonunits are exactly the series whose constant coefficient lies in $p\mathbb Z_p$. They form the unique maximal <ideal> $(p,t)$, and the quotient is $\mathbb F_p$. Hence
$$
\boxed{J(\mathbb Z_p[[t]])=(p,t).}
$$
Back to article page
