proof of characterizations of the Jacobson radical
First, note that by definition a left primitive ideal is the annihilator^{} of an irreducible left $R$module, so clearly characterization^{} 1) is equivalent^{} to the definition of the Jacobson radical^{}.
Next, we will prove cyclical containment. Observe that 5) follows after the equivalence of 1)  4) is established, since 4) is independent of the choice of left or right ideals^{}.

1) $\subset $ 2)
We know that every left primitive ideal is the largest ideal contained in a maximal left ideal. So the intersection^{} of all left primitive ideals will be contained in the intersection of all maximal left ideals.

2) $\subset $ 3)
Let $S=\{M:M\text{a maximal left ideal of}R\}$ and take $r\in R$. Let $t\in {\cap}_{M\in S}M$. Then $rt\in {\cap}_{M\in S}M$.
Assume $1rt$ is not left invertible; therefore there exists a maximal left ideal ${M}_{0}$ of $R$ such that $R(1rt)\subseteq {M}_{0}$.
Note then that $1rt\in {M}_{0}$. Also, by definition of $t$, we have $rt\in {M}_{0}$. Therefore $1\in {M}_{0}$; this contradiction^{} implies $1rt$ is left invertible.

3) $\subset $ 4)
We claim that 3) satisfies the condition of 4).
Let $K=\{t\in R:1rt\text{is left invertible for all}r\in R\}$.
We shall first show that $K$ is an ideal.
Clearly if $t\in K$, then $rt\in K$. If ${t}_{1},{t}_{2}\in K$, then
$$1r({t}_{1}+{t}_{2})=(1r{t}_{1})r{t}_{2}$$ Now there exists ${u}_{1}$ such that ${u}_{1}(1r{t}_{1})=1$, hence
$${u}_{1}((1r{t}_{1})r{t}_{2})=1{u}_{1}r{t}_{2}$$ Similarly, there exists ${u}_{2}$ such that ${u}_{2}(1{u}_{1}r{t}_{2})=1$, therefore
$${u}_{2}{u}_{1}(1r({t}_{1}+{t}_{2}))=1$$ Hence ${t}_{1}+{t}_{2}\in K$.
Now if $t\in K,r\in R$, to show that $tr\in K$ it suffices to show that $1tr$ is left invertible. Suppose $u(1rt)=1$, hence $uurt=1$, then $turturtr=tr$.
So $(1+tur)(1tr)=1+turtrturtr=1$.
Therefore $K$ is an ideal.
Now let $v\in K$. Then there exists $u$ such that $u(1v)=1$, hence $1u=uv\in K$, so $u=1(1u)$ is left invertible.
So there exists $w$ such that $wu=1$, hence $wu(1v)=w$, then $1v=w$. Thus $(1v)u=1$ and therefore $1v$ is a unit.
Let $J$ be the largest ideal such that, for all $v\in J$, $1v$ is a unit. We claim that $K\subseteq J$.
Suppose this were not true; in this case $K+J$ strictly contains $J$. Consider $rx+sy\in K+J$ with $x\in K,y\in J$ and $r,s\in R$. Now $1(rx+sy)=(1rx)sy$, and since $rx\in K$, then $1rx=u$ for some unit $u\in R$.
So $1(rx+sy)=usy=u(1{u}^{1}sy)$, and clearly ${u}^{1}sy\in J$ since $y\in J$. Hence $1{u}^{1}sy$ is also a unit, and thus $1(rx+sy)$ is a unit.
Thus $1v$ is a unit for all $v\in K+J$. But this contradicts the assumption^{} that $J$ is the largest such ideal. So we must have $K\subseteq J$.

4) $\subset $ 1)
We must show that if $I$ is an ideal such that for all $u\in I$, $1u$ is a unit, then $I\subset \mathrm{ann}({}_{R}M)$ for every irreducible left $R$module ${}_{R}M$.
Suppose this is not the case, so there exists ${}_{R}M$ such that $I\not\subset \mathrm{ann}({}_{R}M)$. Now we know that $\mathrm{ann}({}_{R}M)$ is the largest ideal inside some maximal left ideal $J$ of $R$. Thus we must also have $I\not\subset J$, or else this would contradict the maximality of $\mathrm{ann}({}_{R}M)$ inside $J$.
But since $I\not\subset J$, then by maximality $I+J=R$, hence there exist $u\in I$ and $v\in J$ such that $u+v=1$. Then $v=1u$, so $v$ is a unit and $J=R$. But since $J$ is a proper left ideal, this is a contradiction.
Title  proof of characterizations of the Jacobson radical 

Canonical name  ProofOfCharacterizationsOfTheJacobsonRadical 
Date of creation  20130322 12:48:56 
Last modified on  20130322 12:48:56 
Owner  rspuzio (6075) 
Last modified by  rspuzio (6075) 
Numerical id  31 
Author  rspuzio (6075) 
Entry type  Proof 
Classification  msc 16N20 