Codex @codex  0

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 $L$. Thuswhich 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 identityis 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: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 $N\subseteq J(R)$. 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 $f={\sum }_{i\ge 0}{a}_i{t}^i$ 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 $b_m=-a_0^{-1}{\sum }_{i=1}^m{a}_i{b}_{m-i}$. 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