Let $R$ be a commutative ring with unity and $A,B\in M\_n(R)$ satisfying the property 


(\*) All elements of the two-side ideal, in $M\_n(R)$, generated by $AB-BA$, are nilpotent.


McCoy showed that, if $R$ is an algebraically closed field, then $A,B$ are simultaneously triangularizable (noted ST). Else, McCoy, again, gave a condition that is equivalent to (\*) in


A theorem on matrices over a commutative ring, Bulletin of AMS, 45 (1939) 740-744, in free access here: 
<http://www.ams.org/journals/bull/1939-45-10/S0002-9904-1939-07070-5/S0002-9904-1939-07070-5.pdf>


Unfortunately, the previous condition seems (to me) almost useless.


Then my question is: if the matrices, over a ring, $A,B$ satisfy (\*) and are separately triangularizable, then are they ST ?


Thanks in advance.



Simultaneous triangularizability over a commutative ring
When you write that all elements in the two-sided ideal $\langle AB-BA\rangle$ are nilpotent, what precisely do you mean by "nilpotent". Here is an example that I believe contradicts simultaneous diagonalizability. Let $R$ be the commutative, unital ring $\mathbb{C}[\epsilon]/\langle \epsilon^2 \rangle$, i.e., the ring of dual numbers. Let $A$ and $B$ be the following $2\times 2$ matrices with coefficients in $R$, $$A = \left[ \begin{array}{rr} 1 & \epsilon \\ 0 & 1 \end{array} \right],$$ $$B = \left[ \begin{array}{rr} 1 & 0 \\ \epsilon & 1 \end{array} \right].$$ Then the two-sided ideal is simply $\epsilon M\_2(R)$, which I consider to consist of "nilpotent" elements. Yet $A$ and $B$ are not simultaneously triangularizable in the sense that they simultaneously stabilize a complete filtration of the free $R$-module $R^{\oplus 2}$ by direct summand $R$-submodules.


**Edit.** My computation was wrong. The matrices $A$ and $B$ commute as elements in the algebra $M\_2(R)$. So, of course the ideal $\langle AB-BA \rangle$ consists of nilpotent elements; it is just $\{0\}$.


Nonetheless, there is no rank $1$, $R$-submodule of $R^{\oplus 2}$ that is a direct summand and that is stabilized by both $A$ and $B$. The point is that there is no "rigidity" for such summands as in the semisimple case. So the standard argument about simultaneous triangularizability does not apply.


