Suppose $R$ is a Artin ring and $x\in R$. Is there necessarily a commutative subring of $R$ which is also Artin and contains $x$? What if I want it to contain finitely many commuting elements $x_1,\ldots,x_n\in R$? Here, all rings are noncommutative unital, and by Artin I mean both left and right Artin.
This is true of a few standard examples. If $R$ is a finite-dimensional algebra over a field (the motivating case for me to ask, since this is a common trick) or a finite ring, this is trivial. If $R$ is a division ring, then the subfield generated by the $x_i$ works; wrapping this in Artin-Wedderburn proves the general semisimple case. What if $R$ is not semisimple? A candidate may be the localization of the subring generated by $x_1,\ldots,x_n$ at the set of elements which are units in $R$, or by Zorn a maximal commutative subring containing the $x_i$, or something similar?
A difficulty is it doesn't necessarily hold that there is a commutative subring $S$ which satisfies the property that if $I\subsetneq J$ are ideals of $S$ then $RIR\subsetneq RJR$, for example if $R$ is a matrix algebra over a field $k$ and $k[x]$ is not a field. I do not know off the top of my head whether there is an obvious counterexample to the one-sided version of this.