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.
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$\begingroup$ $r^2 x = r$ 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.) $\endgroup$– Amateur_AlgebraistCommented May 20 at 15:06
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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.
