Let $R$ be a ring with identity such that each (right) ideal of $R/J(R)$ is idempotent, where $J(R)$ is the Jacobson radical of $R$. Is $R/J(R)$ necessarily von-Neumann regular?
Certainly, the answer is "yes" in the commutative setting due to the fact that a commutative ring is von-Neumann regular if and only if each ideal of which is idempotent.
Thanks in advance!
I was not previously aware of this paper, but searching for "all right ideals idempotent" drew me to it:
Lanski, Charles. "Idempotent ideals and Noetherian polynomial rings." Can. Math. Bull. 25.1 (1982): 48-53.
Theorem 4 says:
If $R$ is a ring with DCC on right annihilators, then the following are equivalent:
- every ideal of $R$ satisfies (SI);
- $R$ is a finite direct sum of simple rings with identity;
- every right ideal of $R$ satisfies (SI)
- every right ideal of $R$ is idempotent
Furthermore, each of 1-4 implies that every left ideal of $R$ is idempotent
(If you're curious about 1 and 3 and the (SI) condition, refer to the paper.)
At any rate, this gives us a lever to search for a candidate in the Database of Ring Theory.
At present, the hit you get is to a non-Artinian simple domain, which obviously satisfies the annihilator condition. By simplicity, $J(R)=\{0\}$. It can't be von Neumann regular, for a von Neumann regular domain must be a division ring.
