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.
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 $x$.