Simultaneous triangularizability over a commutative ring 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.
 A: 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.
