Nilpotency index of the nilpotent elements of non commutative rings.

Question

Let $A$ and $B$ be two nilpotent elements of a rinf $R$ with index $m$ and $n$ respectively such that $AB \neq BA$. Can we say anything anout nilpotency of $AB$? Also can we say anything about the positive integer $k$ such that $(AB)^k = 0$ if $AB$ is nilpotent.

Answer 1

$\begin{bmatrix}0&1\\ 0&0\end{bmatrix}\begin{bmatrix}0&0\\ 1&0\end{bmatrix}=\begin{bmatrix}1&0\\0&0\end{bmatrix}$ is a case where two noncommiting matrices which square to zero have a product which is idempotent, which is kind of an opposite of nilpotent.

As for the degree of nilpotency of $A$ and $B$ based on that of $AB$, it is trivial to make examples that do anything. Just consider a quotient of the free algebra in noncommuting variables $\mathbb Z\langle x,y\rangle/(x^n, (xy)^m, y^q)$ for whatever positive integers $n,m,q$ you want.