Jacobson radical

Definition

The Jacobson radical of a commutative ring $R$ is the intersection of all maximal ideals of $R$.

Qualification of elements

An element $r\in R$ is in the Jacobson radical $J$ if and only if $1+rs$ is a unit for all $s\in R$.

Proof

Let $r\in J$. It is important to note that a maximal ideal cannot contain a unit, or it would not be a proper subset of $R$. Additionally, every element that is not a unit is in some maximal ideal, as one can always take the ideal generated by the element (which is a proper subset of $R$ since it is not a unit). If this ideal is not maximal, take the maximal ideal that it is contained in.

Assume for a contradiction that $1+rs$ is not a unit. Then $1+rs\in I$ for some maximal ideal $I\subset R$. We know that $r\in I$ by definition since $r\in J$. Therefore, $-rs\in I$ for every $s\in R$ and

$$(1+rs)-rs=1\in I$$

this is a contradiction as $I$ is a maximal ideal. Hence, $1+rs$ is a unit.

Next, let $1+rs$ be a unit for every $s\in R$. Just for sanity’s sake, I will show that $r$ cannot be a unit. If $r$ were a unit, then we could set $s=r^{-1}(-1)$. Since $1+rs$ is a unit then multiplying by its inverse $x$ would give

$$\begin{array}{r}(1+r(-r^{-1}))x=1\\ x+-r(r^{-1})x=1\\ x-x=1\\ 0=1.\end{array}$$

So $r$ is contained in some maximal ideal, but it remains to show that it is contained in all of them. Assume otherwise for contradiction, and let $m$ be a maximal ideal such that $r\notin m$. Then we can take the ideal adding $r$ as a generator, $m+(r)$. This means that $m\subset m+(r)$, but $m$ is maximal therefore,

$$m+(r)=R$$

Thus, we can pick some element $\tilde{m}\in m$ and $rn\in (r)$ such that

$$1=\tilde{m}+rn⟹\tilde{m}=1+r(-n).$$

However, $\tilde{m}\in m$ and $1+r(-n)$ is a unit by the hypothesis which is a contradiction. Hence, $r\in m$ for every maximal ideal $m$ and thus $r\in J$.

Non-commutative case… eventually…