Remark

(1) Let $s$ be the nilpotent radical of $g$. As every decreasing sequence of vector subspaces of $g$ is stationary, there exists a finite number of finite-dimensional simple representations of $g$ whose kernels have intersection $s$.

The direct sum of these representations is semi-simple and has kernel s. It follows that the set of kernels of finite-dimensional semi-simple representations of g has a least element, namely s.

(2) By Proposition 4 (c) of § 4, no. 3, s is also the intersection of the largest nilpotency ideals of the finite-dimensional representations of g. In particular, s is contained in the largest nilpotent ideal of g and is therefore a nilpotent ideal of g.

(3) Every linear form $\lambda$ on g which is zero on $\mathscr{D}g$ is a simple representation (with space K) of g, whence $\lambda(s) = \{0\}$. It follows that $s \subset \mathscr{D}g$. On the other hand, s is contained in the radical r of g by Remark 2. We shall prove that $s = r \cap \mathscr{D}g$.