Lemma 5

Let V be a finite dimensional vector space, A and B endomorphisms of V. Assume that A is nilpotent and that $[A,[A,B]] = 0$. Then AB is nilpotent.

Put $C = [A,B]$. Since $[A,C] = 0$,

$[A$, BC$^p] = [A,B]C^p= C^{p+1}$

for every integer $p\geq 0$. Consequently, Tr(C$^p$) $= 0$ for $p\geq 1$, which proves that C is nilpotent (Algebra, Chap. VII, §3, no. 5, Cor. 4 of Prop. 13). Now let $\overline{k}$ be an algebraic closure of $k$, and let $\lambda \in \overline{k},x\in V\otimes_k\overline{k}$ be such that AB$x=\lambda x$, $x\not= 0$. The relation $[[B,A],A] = 0$ shows that $[B,A^p] =p[B,A]A^{p-1}$ for every integer $p\geq 0$. Let $r$ be the smallest integer such that $A^rx= 0$. Then

$\lambda A^{r-1}x= A^{r-1}$AB$x= A^rBx=$ BA$^rx-[B,A^r]x=-r[B,A]A^{r-1}x$.

Since $[B,A]$ is nilpotent and since $A^{r-1}x\not= 0$, this proves that $\lambda = 0$. Thus, all the eigenvalues of AB are zero, hence the lemma.