# Two definitions of Jacobson Radical

I have in my notes that the Jacobson radical of a ring $R$ is:
$J(R) = \cap${$I$ | $I$ primitive ideal of $R$}
$= \cap$ {$Ann_R M$ | $M$ simple $R$-module}.

I have now seen elsewhere that
$J(R) = \{x ∈ R: xy-1 ∈ R^\times \text{ for all } y ∈R\}$

I was just wondering would anyone be able to provide an explanation as to how to get from one definition to the other? Thanks.

#### Solutions Collecting From Web of "Two definitions of Jacobson Radical"

Let R be a commutative ring. then:
$$J(R) = \cap\{I | \text{I is primitive ideal of R}\} = \cap\{Ann_R M | \text{M is simple R-module}\}=\cap\{m| \text{m is maximal ideal of R}\}=\{x\in R|\forall y\in R, \text{1-xy is unit in R}\}.$$
Proof of the last equality:

assume $x\in J(R)$. If $1-xy$ is not unit of $R$. Then there exists a maximal ideal $m$ of $R$ such that $1-xy \in m$. But by definition of $J(R)$, $x\in m$. So $1\in m$, a contradiction.
conversly:
Suppose that, for each $y\in R$, it is the case that $1-xy$ is a unit of$R$. Let $m$ be a maximal ideal of $R$ such that $x\notin m$. Since $$m\subset m+(x)\subset R$$ by the maximality of $m$, we have $m+(x)=R$. So that there exist $u\in m$, $y\in R$ such that $u+xy=1$. Hence $1-xy\in m$, and so cannot be a unit, contradiction.

If $S$ is a simple (right) $R$-module and $x\in S$, $x\ne0$, then $xR=S$ and so $\def\Ann{\operatorname{Ann}}\Ann_R(x)$ is a maximal right ideal.

Since $\Ann_R(S)=\bigcap_{x\in S}\Ann_R(x)$ we see that $J(R)$ is an intersection of maximal right ideals.

Conversely, since $\Ann_R(R/I)\subseteq I$, for every right ideal $I$, we see that every maximal right ideal contains a primitive ideal.

Therefore the Jacobson radical $J(R)$ is the intersection of all maximal right ideals.

An element $x\in R$ is called right (resp. left) quasiregular if $1-x$ is (left) right invertible. A right ideal is called quasiregular if it consists of right quasiregular elements. In this case, every element is also left quasiregular. Indeed, suppose all elements of $I$ are right quasiregular. Let $x\in I$; then $(1-x)(1-y)=1$ for some $y\in R$, that is, $x+y-xy=0$. In particular $y=xy-x\in I$ and therefore $y$ is right quasiregular; let $(1-y)(1-z)=0$. Now $1-y$ is invertible, so $1-x=1-z=(1-y)^{-1}$ and hence $x=z$, showing that $1-x$ is left quasiregular.

We want now to show that every $x\in J(R)$ is right quasiregular. If not, $(1-x)R\ne R$, so there exists a maximal right ideal $I$ such that $(1-x)R\subseteq I$. In particular $1-x\in I$, so $x\notin I$: this contradicts $x$ belonging to all maximal right ideals.

So, if $x\in J(R)$, we know that $xy\in J(R)$ for every $y\in R$ and hence $1-xy$ is left and right quasiregular; thus $1-xy$ is invertible.

Conversely, suppose $1-xy$ is invertible for every $y\in R$. We want to show that $x$ belongs to all maximal right ideals.

Let $I$ be a maximal right ideal. If $x\notin I$, we know that $xR+I=R$, so $xy+z=1$ for some $y\in R$ and $z\in I$. This is impossible, because $z=1-xy$ is invertible by assumption on $x$.