Lemma 1

Let V be a finite-dimensional vector space over K, g a subalgebra of gl(V) such that V is a simple g-module and a a commutative ideal of g. Then $a \cap \mathscr{D}g = \{0\}$. Let S be the subalgebra of $\mathscr{L}(V)$ generated by 1 and a.

If b is an ideal of g contained in a such that $\operatorname{Tr} bs = 0$ for all $b \in b$ and all $s \in S$, then in particular, by definition of S, $\operatorname{Tr}(b^n) = 0$ for every integer $n > 0$ and hence b is nilpotent (Algebra, Chapter VII, § 5, no. 5, Corollary 4 to Proposition 13); as the elements of b are all nilpotent, $b = \{0\}$ (§ 4, no. 3, Lemma 2). We first apply this to the ideal $[g, a]$ of g. If $x \in g, a \in a, s \in S$, then $\operatorname{Tr}[x, a]s = \operatorname{Tr}(xas - axs) = \operatorname{Tr} x(as - sa) = 0$ since $as = sa$; hence $[g, a] = \{0\}$. Hence the elements of g commute with those of a and hence also with those of S. If $x, y$ belong to g and $s \in S$, then

since $ys = sy$; then taking b to be the ideal $\mathscr{D}g \cap a$, it follows that $\mathscr{D}g \cap a = \{0\}$.