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. We know that every left primitive ideal is thelargest ideal contained in a maximal left ideal. So the intersectionof all left primitive ideals will be contained in the intersection of allmaximal left ideals.

2. 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 $1-rt$ is not left invertible; therefore there exists a maximalleft ideal $M_0$ of $R$ such that $R(1-rt)\subseteq M_0$.

   Note then that $1-rt\in M_0$. Also, by definition of $t$, we have$rt\in M_0$. Therefore $1\in M_0$; this contradiction implies $1-rt$ isleft invertible.

3. We claim that 3) satisfies the condition of 4).

   Let $K=\{t\in R:1-rt\text{ is left invertible for all }r\in R\}$.

   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

   $$1-r(t_1+t_2)=(1-r{t}_1)-r{t}_2$$

   Now there exists $u_1$ such that $u_1(1-r{t}_1)=1$, hence

   $$u_1((1-r{t}_1)-r{t}_2)=1-u_{1}r{t}_2$$

   Similarly, there exists $u_2$ such that $u_2(1-u_{1}r{t}_2)=1$, therefore

   $$u_2{u}_1(1-r(t_1+t_2))=1$$

   Hence $t_1+t_2\in K$.

   Now if $t\in K,r\in R$, to show that $tr\in K$ it suffices toshow that $1-tr$ is left invertible. Suppose $u(1-rt)=1$, hence$u-urt=1$, then $tur-turtr=tr$.

   So $(1+tur)(1-tr)=1+tur-tr-turtr=1$.

   Therefore $K$ is an ideal.

   Now let $v\in K$. Then there exists $u$ such that $u(1-v)=1$, hence$1-u=-uv\in K$, so $u=1-(1-u)$ is left invertible.

   So there exists $w$ such that $wu=1$, hence $wu(1-v)=w$, then $1-v=w$.Thus $(1-v)u=1$ and therefore $1-v$ is a unit.

   Let $J$ be the largest ideal such that, for all $v\in J$, $1-v$ is aunit. 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 $1-(rx+sy)=(1-rx)-sy$, and since $rx\in K$, then $1-rx=u$ for someunit $u\in R$.

   So $1-(rx+sy)=u-sy=u(1-u^{-1}sy)$, and clearly $u^{-1}sy\in J$ since$y\in J$. Hence $1-u^{-1}sy$ is also a unit, and thus $1-(rx+sy)$ isa unit.

   Thus $1-v$ is a unit for all $v\in K+J$. But this contradicts theassumption that $J$ is the largest such ideal. So we must have$K\subseteq J$.

4. We must show that if $I$ is an ideal such thatfor all $u\in I$, $1-u$ 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 maximalleft 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 thereexist $u\in I$ and $v\in J$ such that $u+v=1$. Then $v=1-u$, so$v$ is a unit and $J=R$. But since $J$ is a proper left ideal, thisis a contradiction.