Lemma 1

Let $x, y\in$ End(V).

(i) Assume that V is finite dimensional. Then $x$ is triangularizable if and only if $V =\sum_{a\in k}V^a(x)$.

(ii) If there exists an integer $n$ such that (ad $x$)$^ny= 0$, each $V^a(x)$ is stable under $y$.

(iii) Assume that V is finite dimensional. If $V =\sum_{a\in k}V^a(x)$ and if each

$V^a(x)$ is stable under $y$, there exists an integer $n$ such that (ad $x$)$^ny= 0$.

Part (i) follows from Algebra, Chap. VII, §5, no. 2, Prop. 3.

Let E = End(V). Let B be the bilinear map $(u, v) \rightarrow u(v)$ from $E\times V$ to V. By the definition of ad $x$,

$x(B(u, v)) =B(u, x(v)) +$ B((ad $x$)$(u), v$)

for $x\in E,u\in E,v\in V$. Let $x$ operate on E via ad $x$. By Prop. 2 (ii), $B(E^0(x),V^a(x))\subset V^a(x)$ for all $a\in k$. If (ad $x$)$^ny= 0$, then $y\in E^0(x)$, so $y(V^a(x))\subset V^a(x)$, which proves (ii).

To prove (iii), we can replace V by $V^a(x)$ and $x$ (resp. $y$) by its restriction to $V^a(x)$. Replacing $x$ by $x-a$, we can assume that $x$ is nilpotent. Then, (ad $x$)$^{2 dim V-1}= 0$ (Chap. I, §4, no. 2), which proves (iii).