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

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.

• What is true, however, is that $P$ is a prime ideal if and only if $R/P$ is a prime ring. So, it suffices to show that any prime von Neumann regular ring is a simple ring. Commented Jan 4 at 4:02
• If I’m not mistaken, there can be no Noetherian counterexamples, since Noetherian von Neumann regular rings are semisimple and prime semisimple rings are simple. Commented Jan 4 at 5:41

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