In Herstein's "Noncommutative rings" it is proved that any rng $R$ such that for every $x,y\in R$ there exists an integer $n(x,y)>1$ such that $$(xy-yx)^{n(x,y)}=(xy-yx)$$ must be commutative. At the end of the section he says that the previous proof can be generalized replacing $[x,y]=xy-yx$ by higher commutators, e.g. $[[x,y],z]$, $[[[x,y],z],w]]$ and so on. I think I managed to modify his proof in order to prove this fact, but then he says that if $R$ does not contain nil ideals, then $R$ is even commutative. I'm struggling proving this latter fact. Let's take into account the commutator $[[x,y],z]$, then my problem is the following one:
If $R$ is a rng satisfying $[[x,y],z]=0$ for all $x,y,z\in R$ and $R$ does not contain nil ideals, then $R$ is commutative.
I was trying to prove that the ideal generated by all commutators $[x,y]$ with $x,y\in R$ is nil, but that does not seem to be the right path. Do you have any suggestion?
We can prove this when $R$ is free of $2$-torsion:
$[[x,y],z]=0$ means that any commutator $[x,y]$ is in the center $Z(R)$.
$[x,[y,zw]]=[x,[y,z]]w+z[x,[y,w]]+[x,z][y,w]+[y,z][x,w]$ implies that $$[x,z][y,w]+[y,z][x,w]=0,$$ in particular, when $R$ is free of $2$-torsion, $$[x,z][x,w]=0.$$
Now consider the ideal $I_a$ of $R$ generated by the elements $[a,x]$ for fixed $a\in R$ and all $x\in R$. Elements of $I_a$ are sums of the form $y[a,x]z$ (where we can have "$y=1,z=1$"), and so by items 1. and 2. we get $I_a^2=0$; in particular $I_a$ is a nil ideal, so $I_a=0$, hence $[a,R]=0$ and $a\in Z(R)$. Since this is true for all $a\in R$ we get $R=Z(R)$.
Note that the result is true also more in general for semiprime rings, i.e., rings without nonzero nilpotent ideals.