Jacobson radical ($J(R)$)

The Jacobson radical is the intersection of all maximal left ideals, equivalently the intersection of the annihilators of all simple left modules.

The unit criterion for the Jacobson radical implies that every radical element is a quasinilpotent element, because $1-x/\lambda$ is invertible for every nonzero $\lambda$. Conversely, if a two-sided ideal consists of quasinilpotent elements, then $yx$ is quasinilpotent for every member $x$ and every $y\in A$. Hence $1+yx$ is invertible and $x$ lies in the Jacobson radical. The maximality statement concerns ideals contained in the quasinilpotent set, not the whole set itself.

The intersection of maximal left ideals is the intersection of annihilators of simple modules on the left, hence is a two-sided ideal. If $j$ belongs to it, $1-rj$ has a left inverse for every $r$. Its left inverse also differs from $1$ by a radical element and has a left inverse; this makes the original inverse two-sided. The identity $(1-ba{)}^{-1}=1+b(1-ab{)}^{-1}a$ then shows that $1-jr$ is invertible too. If $j$ were outside a maximal right ideal $I$, one would have $1-jr\in I$ for some $r$, a contradiction. Applying the same argument to the opposite ring proves equality.

The Jacobson radical is the intersection of the ideals ${\mathrm{Ann}}_R(S)$ over all simple modules, where $\mathrm{Ann}$ is the annihilator of a module, so it is a two-sided ideal. If $y\in J(R)$, the left ideal $R(1-y)$ cannot be proper, since a containing maximal left ideal would also contain $y$ and hence $1$. Thus $c(1-y)=1$ for some $c$. Now $c=1+cy$ also differs from $1$ by a radical element and has a left inverse $d$. Multiplying $c(1-y)=1$ by $d$ gives $d=1-y$, and hence $(1-y)c=1$. This proves the unit assertion. Conversely, if $x$ is outside a maximal left ideal $L$, then $ax+l=1$ for some $a\in R$, $l\in L$. Consequently $1-ax\in L$ cannot be a unit, proving the reverse direction.

If $I$ is a nilpotent ideal and $J(A/I)=0$, then $I=J(A)$. Every $1-ax$, $x\in I$, has a finite geometric-series inverse, giving $I\subseteq J(A)$. The radical's image lies in $J(A/I)=0$, giving the reverse inclusion. A semisimple algebra as quotient supplies the required zero radical.

For an integral extension $A\subseteq B$ of commutative rings,Contraction sends maximal ideals of $B$ to maximal ideals of $A$, while the Lying-over theorem puts a maximal ideal of $B$ over every maximal ideal of $A$.

A commutative ring $R$ is Jacobson when every prime ideal is the intersection of the maximal ideals containing it. Equivalently, for every ideal $I$, its radical is the intersection of the maximal ideals containing $I$.

For a finite-type integer algebra, a nonnilpotent element $r$ gives a nonzero localization of a ring $R[1/r]$. A maximal ideal there has finite residue field by the finite-field theorem for finitely generated integer algebras. The image of $R$ in that field is a finite integral domain, hence a field. Its kernel is therefore maximal in $R$ and avoids $r$. Nilpotents lie in every maximal ideal, proving the radical equality. Applying the same argument to every prime quotient shows that these algebras are Jacobson rings. Merely contracting a localized maximal ideal without the finite-image argument would not prove maximality.

Every ring integral over a Jacobson ring is Jacobson. After quotienting by a prime, an integral equation for a nonzero element has nonzero constant term; choose a maximal ideal of the base avoiding that term and apply the Lying-over theorem.