Is Kaplansky’s radical same as Jacobson radical

Question

For a ring $R$, Kaplansky defines $x\circ y= x+y+xy$ and an element $x$ is s.t.b. right quasi-regular (r.q.r) if $x\circ y=0$ for some element $y$. He defines radical to be set theoretic join (sum) of all right quasi regular (r.q.r) ideals where an ideal is r.q.r if all its elements are r.q.r.

Jacobson radical $J(R)$ is the intersection of maximal left (or right) ideals of a ring.

Does either of these radical implies other or are they completely unrelated? I can't see any connection here.

Answer 1

I'm assuming for the time being that you intended for all these rings to have identity, because the maximal left ideal definition needs to be refined when identity is gone. I'm pretty sure everything holds without identity, it just takes more care.

Here's some observations to work through:

1. $x$ is r.q.r as you defined iff $x+1$ is right invertible in $R$.
2. The statement that "For $x\in R$, there exists $y$ such that $x+y-xy=0$" is equivalent to $x-1$ being right invertible.

3. These statements are equivalent:

   1. $xr-1$ is right invertible for every $r\in R$
   2. $xr+1$ is right invertible for every $r\in R$.
   3. $1+xr$ is right invertible for every $r\in R$.
   4. $1-xr$ is right invertible for every $r\in R$.

4. From the wiki article you can see that an element $x\in R$ is in $J(R)$ iff $1-xr$ is right invertible for all $r\in R$.

So from the last point, you can see that $J(R)$ itself has the property that all elements are r.q.r., and furthermore any element $x$ which lies in a r.q.r. ideal must have the property "$1-xr$ is right invertible for every $r\in R$" and therefore such an ideal is contained in $J(R)$.

Then the sum of all r.q.r. ideals would have to be equal to $J(R)$.