# Von Neumann regular rings and its ideals

Let R be a strongly regular Von Neumann ring, this is, that for every $$r \in R$$ there exists $$x \in R$$ such that $$r^2x=r$$. From here, how to prove that $$R$$ is strongly Von Neumann regular if and only if every principal ideal $$I$$ of $$R$$ is generated by an idempotent element? If the ring were to be commutative, then the element $$rx$$ would be idempotent because $$(rx)^2=r(xr)x=r(rx)x=(rr)xx=r^2xx=rx$$, and proving the statement from here is rather simple. But I am unable to find a way that does not require commutativity. Any help is apreciated, thaks in advance.

• $r^2 x = r$ implies $r(r-rxr) = (r-r^2x)r = 0$, so $(r-rxr)^2=0$, and $r = rxr$, since the ring is reduced. (See this in M. P. Drazin, "Rings with Central Idempotent or Nilpotent Elements" (1956), just before Theorem 7.) Commented May 20 at 15:06

I will use abbreviations VNR and SVNR for brevity.

how to prove that $$R$$ is strongly Von Neumann regular if and only if every principal ideal $$I$$ of $$R$$ is generated by an idempotent element?

As far as I can interpret this, it does not sound correct. A ring is VNR iff every principal right ideal is a summand, i.e. generated by an idempotent. So the "if" statement here is false, but the "only if" part is true.

So I suspect you meant the "only if" part here, or in other words

How to prove that SVNR implies VNR?

Then one can approach this by observing first that SVNR $$\implies$$ reduced (no nonzero nilpotents, equivalent, nothing nonzero squares to zero.) If anything satisfies $$r^2=0$$, then $$r=0$$ using that equation.

Secondly then, if you have $$x$$ such that $$r^2x=r$$, then fortuitously it also occurs that $$rxr=r$$. For, $$(rxr-r)^2=rxrrxr-rxrr-rrxr+r^2=rxrr-rxrr-r^2+r^2=0$$ by using the reduction $$r^2x=r$$. That means $$rxr-r$$ is nilpotent, and as we know then it must be $$0$$. Hence, $$rxr=r$$.

So it is useful to remember that VNR+reduced = SVNR. Another useful characterization is that VNR+Abelian=SVNR, where "Abelian" means that all idempotents are central.