Theorem 1

Let g be a Lie algebra, r its radical and s its nilpotent radical. Then $s = \mathscr{D}g \cap r$.

We already know that $s \subset \mathscr{D}g \cap r$. Hence it will suffice to show that if $\rho$ is a finite-dimensional simple representation of g then $\rho(\mathscr{D}g \cap r) = \{0\}$. Let k be the least integer $\geq 0$ such that $\rho(\mathscr{D}^{k+1}r) = \{0\}$; we write $g' = \rho(g)$, $a' = \rho(D^k r)$; as $\mathscr{D}^k r$ is an ideal of g, $a'$ is an ideal of $g'$; this ideal is commutative since $\rho(\mathscr{D}^{k+1}r) = \{0\}$. If V is the space of $\rho$, $g' \subset \operatorname{gl}(V)$ and V is a simple $g'$-module. Then $\rho(\mathscr{D}g \cap \mathscr{D}^k r) \subset \mathscr{D}g' \cap a' = \{0\}$. If $k > 0$, then $\mathscr{D}^k r \subset \mathscr{D}g$ and $\rho(\mathscr{D}^k r) = \{0\}$, contrary to the definition of k. Hence $k = 0$, that is