Proposition 3

Let M be a finite-dimensional vector space over K and $g$ a semi-simple subalgebra of $\mathfrak{gl}(M)$. Then $g$ contains the semi-simple and nilpotent components of its elements.

If $K_1$ is an extension of K, the Killing form of $g_{(K_1)}$ is the extension to $g_{(K_1)}$ of that of $g$ (§ 3, no. 8) and hence is non-degenerate; therefore $g_{(K_1)}$ is semi-simple. It therefore suffices to prove Proposition 3 when the base field is algebraically closed, which we shall henceforth assume to be the case.

For every subspace N of M, let $g_N$ be the subalgebra of $\mathfrak{gl}(M)$ consisting of the elements which leave N stable and whose restriction to N has trace zero. As $g = \mathscr{D}g$, $g \subset g_N$ if N is stable under $g$. Then let $g^*$ be the intersection of the normalizer of $g$ in $\mathfrak{gl}(M)$ and the algebras $g_N$ where N runs through the set of subspaces of M which are stable under g. As the semi-simple (resp. nilpotent) component s (resp. n) of $x \in gl(M)$ is a polynomial in x with no constant term and ad s (resp. ad n) is the semi-simple (resp. nilpotent) part of ad x ($§ 5$, no. 4, Lemma 2), clearly $x \in g^*$ implies $s \in g^*$ and $n \in g^*$; it therefore suffices to show that $g^* = g$. Since g is a semi-simple ideal of $g^*$, $g^* = a \times g$ (no. 1, Corollary 1 to Proposition 1). Let $a \in a$ and let N be a subspace which is minimal among the non-zero subspaces of M which are stable under g. The restriction of a to N is a scalar multiple of the identity by Burnside’s Theorem, has trace zero by construction and hence is zero since K is of characteristic 0. As M is the direct sum of subspaces such as N, it follows that $a = 0$ and hence $g^* = g$.