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Homeproof of characterizations of the Jacobson radical

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# 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 $1-rt$ is not left invertible; therefore there exists a maximal left 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 $1-rt$ is left invertible.

- 3) $\subset$ 4)
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

$1-r(t_{1}+t_{2})=(1-rt_{1})-rt_{2}$ Now there exists $u_{1}$ such that $u_{1}(1-rt_{1})=1$, hence

$u_{1}((1-rt_{1})-rt_{2})=1-u_{1}rt_{2}$ Similarly, there exists $u_{2}$ such that $u_{2}(1-u_{1}rt_{2})=1$, 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 to show 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 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 $1-(rx+sy)=(1-rx)-sy$, and since $rx\in K$, then $1-rx=u$ for some unit $u\in R$.

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 $1-(rx+sy)$ is a unit.

Thus $1-v$ 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 $u\in I$, $1-u$ is a unit, then $I\subset\operatorname{ann}({}_{R}M)$ for every irreducible left $R$-module ${}_{R}M$.

Suppose this is not the case, so there exists ${}_{R}M$ such that $I\not\subset\operatorname{ann}({}_{R}M)$. Now we know that $\operatorname{ann}({}_{R}M)$ 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 $\operatorname{ann}({}_{R}M)$ 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 $v=1-u$, so $v$ is a unit and $J=R$. But since $J$ is a proper left ideal, this is a contradiction.

## Mathematics Subject Classification

16N20*no label found*

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