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!

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    $\begingroup$ you can ask the question more clearly as: «if $J(R)=0$ and all right ideals of $R$ are idempotent, is $R$ von Neumann regular?» by replacing $R$ by $R/J(R)$, which you can since $R$ plays no useful role in your question. $\endgroup$ – Mariano Suárez-Álvarez Aug 2 '16 at 3:53

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:

  1. every ideal of $R$ satisfies (SI);
  2. $R$ is a finite direct sum of simple rings with identity;
  3. every right ideal of $R$ satisfies (SI)
  4. 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.

  • $\begingroup$ Thanks a lot! Since the ring is simple, $J(R)=0$ or $J(R)=R$, and since it is a domain then $0$ is a maximal two-sided ideal. Why $J(R)=0$? $\endgroup$ – karparvar Aug 2 '16 at 14:25
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    $\begingroup$ @karparvar $J(R)\neq R$ in any ring with unity... $\endgroup$ – rschwieb Aug 2 '16 at 15:04
  • $\begingroup$ After logging in DRT, how and where could I search for a certain ring? Thanks! $\endgroup$ – karparvar Aug 2 '16 at 17:20
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    $\begingroup$ @karparvar You mean search via properties or just browse? There are links for both options on the front page, as well as in the nav menu. Also, you don't have to be logged in :) Logging in is useful if you want to suggest additions to the site, though. $\endgroup$ – rschwieb Aug 2 '16 at 18:06

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