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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.

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  • $\begingroup$ You mean $AB - BA$? $\endgroup$
    – Qiaochu Yuan
    May 21, 2014 at 18:15
  • $\begingroup$ @ Quiaochu, yes, of course. $\endgroup$
    – loup blanc
    May 21, 2014 at 18:19

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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.

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  • $\begingroup$ @ Jason, I am in a hurry ; i'll read your post tomorrow. We say that a matrix is nilpotent iff there is $k$ s.t. $A^k=0$. In this case, we have not necessarily $A^n=0$. $\endgroup$
    – loup blanc
    May 21, 2014 at 18:35
  • $\begingroup$ @loupblanc: If that is your definition, then, indeed, every element in the ideal $\langle AB-BA\rangle$ is nilpotent for my example. $\endgroup$
    – Jason Starr
    May 21, 2014 at 18:38
  • $\begingroup$ Thanks Jason. Thus, if $A,B$ commute and are triangularizable, then they have not necessarily a common eigenvector !! $\endgroup$
    – loup blanc
    May 23, 2014 at 8:44

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