Prove that if $P$ is a prime ideal of a non-commutative von Neumann regular ring $R$, then $P$ is a maximal ideal

Question

Let $R$ be a ring with identity elements (not necessarily commutative). A ring $R$ is called a von Neumann regular ring if for every $a\in R$ there is $b\in R$ such that $a=aba$. How do I prove that if $R$ is a von Neumann regular ring (not necessarily commutative rings) and $P$ is a prime ideal of $R$, then $P$ is a maximal ideal?

For the commutative ring case, this question has been answered on the page Prove that if $R$ is von Neumann regular and $P$ a prime ideal, then $P$ is maximal

In the case of a non-commutative ring, I think if $P$ is a prime ideal, then $R/P$ is not necessarily a domain. Moreover, ideal $P$ is a maximal ideal if and only if $R/P$ is a simple ring.

Answer 1

This cannot be proven. There are examples of VNR rings which make this fail spectacularly.

One famous example would be $End(V_k)$ where $V$ is an infinite dimensional vector space over a field $k$. Its zero ideal is known to be prime, and its maximal ideal is prime.

In fact more is true: the two-sided ideals of such a ring are always linearly ordered, and the proper ones are all prime. When the dimension of $V$ is just $\aleph_0$ you only get two proper ideals, but if you choose it to be larger cardinalities you get more between.

The question amounts to asking what can ensure a prime VNR ring is a simple ring.

Here are a few alternative variations that would work:

1. A completely prime ideal in a VNR ring is always maximal (although, there is no guarantee such an ideal exists in a given VNR ring.)

2. A prime ideal of a strongly regular ring (meaning a VNR ring with no nonzero nilpotent elements) is maximal