Do you have an example of a non-commutative ring $R$ (possibly not unital) such that $$[[x,y],z]=0 \qquad \forall x,y,z \in R$$
where $[x,y]=xy-yx$ ?
Of course this is true if $R$ is commutative, since $[x,y]=0$ for any $x,y \in R$. I tried $R = M_2(\Bbb F_p)$, but it seems painful to check whether $[[x,y],z]=0$ holds or not. Maybe there is an easier idea? Thank you very much!
The simplest example I can think of is the (non-unital) algebra of strictly upper triangular $3\times3$-matrices: $$ R=\left\{\left(\begin{array}{ccc}0&a&b\\0&0&c\\0&0&0\end{array}\right)\bigg\vert\ a,b,c\in\Bbb{R}\right\}. $$ All the commutators have $a=c=0$, but the $b$-component of $[x,y]$ is non-zero iff the $a$-component of $x$ and the $c$-component of $y$ are both non-zero. Therefore we also get $[[x,y],z]=0$ for all $x,y,z\in R$.
Unless I missed something we can actually make $R$ unital by allowing all the upper triangular matrices such that the diagonal entries are all equal. After all, adding a scalar multiple of $I_3$ is not going to change the commutators one bit.
Let $F$ be the free algbra generated by letters $a$ and $b$, endow it with its usual grading, and let $I$ be the ideal of $F$ generated by all elements of the form $[[x,y],z]$ with $x$, $y$ and $z$ in $F$. Since the iterated bracket is trilinear, it is enough to consider homogeneous elements $x$, $y$ and $z$ when generating $I$, and therefore $I$ is a homogeneous ideal. The ring $F/I$ satisfies the desired identity.
We want to show that $R$ is non trivial and non-commutative. If the three letters are homogeneous, the element $[[x,y],z]$ cannot be nonzero and have degree $\leq 2$. Indeed, that can only happen if one of $x$, $y$ or $z$ has degree zero, and then the iterated bracket is itself zero. It follows that $I$ does not contain $[a,b]$, so that $R$ is nontrivial and noncommutative.
An example is the $(2n+1)$-dimensional Heisenberg Lie Algebra $R$, together with a non-commutative bilinear product $(x,y)\mapsto x\cdot y$ satisfying $$ x\cdot y-y\cdot y=[x,y]. $$ Such a bilinear product exists (in fact, several such products, also non-commutative ones, and also given by matrix multiplication) and is called a pre-Lie algebra structure on $R$. Of course we have $[[x,y],z]=0$, since the Heisenberg Lie algebra is $2$-step nilpotent as a Lie algebra, i.e., $[L,L]\subseteq Z(L)$. Here we consider $L$ as a matrix Lie algebra (taking a faithful linear representation if necessary).
