The Radical of a Ring

Definition 5.1   For a left-ideal $R$ define $(I:R)=\{ b\in R:bR\subset I\}$.

It is easy to see that $I\subset(I:R)$ and $(I:R)$ is a two-sided ideal. Also $(I:R)$ contains every two-sided ideal contained in $I$. Also, if $M=R/I$, then $b\in\textrm{Ann}_{R}M$ iff $bR=0+I$ iff $bR\subset I$ iff $b\in(I:R)$ hence $\textrm{Ann}_{R}M=(I:R)$.

If $M$ is an irreducible module, then for every $x\neq0$ in $M$, the mapping $r\to rx$ is onto and is a $R$-modules homomorphism with kernel $\textrm{Ann}_{R}x$. It follows that $M\sim R/\textrm{Ann}_{R}x$. Since $M$ is irreducible we must have that $\textrm{Ann}_{R}x$ is a maximal left ideal. Conversely, if $I$ is a maximal left-ideal then $R/I$ is an irreducible $R$ module with annihilator $(I:R)$.

Definition 5.2   The Jacobson Radical $\textrm{Rad}(R)$ of the ring $R$ is the intersection of the kernels the irreducible representations of $R$.

Every irreducible module is isomorphic to $R/I$ for some maximal left-ideal $I$, and the kernel of the corresponding representation is $(I:R)$, hence $\textrm{Rad}(R)=\cap(I:R)$, where the intersection is over all maximal left ideals. If $M$ is an irreducible module and $x\in M$, then $\textrm{Ann}_{R}x=\{ x\in R:rx=0\}$ is a maximal left-ideal and $\textrm{Ann}_{R}M=\cap\textrm{Ann}_{R}x=\ker\rho_{M}$. Hence $\textrm{Rad}(R)$ is an intersection of some maximal left ideals. On the other hand, if $I$ is a maximal left ideals then $R/I$ is an irreducible module and $\textrm{Ann}_{R}R/I=(I:R)\subset I$ and therefore $\textrm{Rad}(R)\subset\cap(I:R)\subset\cap I$ where the intersections are taken over all maximal left-ideals. This shows that $\textrm{Rad}(R)$ is the intersection of all maximal left ideals of $R$.

Definition 5.3   An element $x\in R$ is called nilpotent if there exists a positive integer $n$ such that $x^{n}=0$.

 

Definition 5.4   An element $x\in R$ is called left (right) quasi-regular if $1-x$ has a left (right) inverse.

An ideal is called quasi-regular if all its elements are quasi-regular. An ideal is nil if all its elements are nilpotent. Note that nilpotent elements are both left and right quasi-regular, since $(1-x)(1+x+\cdots+x^{n-1})=1-x^{n}=1$, hence nil ideals are quasi-regular ideals.

We have the following:

Theorem 5.5   $\textrm{Rad}(R)$ is a left quasi-regular two-sided ideal that contains all left-quasi regular ideals.

Proof. $\textrm{Rad}(R)$ is intersections of kernels of ring homomorphisms hence a two-sided ideal. Let $x\in\textrm{Rad}(R)$. If $x$ is not left quasi-regular then $R(1-x)\neq R$, therefore there is a maximal left ideal $I$ that contains $R(1-x)$. But, since all maximal left ideals contain $x$, then $1=x+(1-x)\in I$ and $I=R$. The contradiction shows that $\textrm{Rad}(R)$ is left quasi-regular.

Let $B\subset R$ be a left quasi-regular ideal and assume that $B\not\subset\textrm{Rad}(R)$, then there is a maximal left ideal such that $B\not\subset I$. Let $b\in B$ such that $b\not\in I$. We have $Rb+I\supsetneq I$ and therefore $Rb+I=R$. It follows that there is $r\in R$ and $z\in I$ such that $rb+z=1$, but then $rb\in B$ and therefore $z=1-rb$ is invertible, therefore $I=R$. Contradiction. $\qedsymbol$