Lemma 2

Let $x$ be an endomorphism of a finite-dimensional vector space V and s (resp. n) its semi-simple (resp. nilpotent) component (cf. Algebra, Chapter VIII, § 9, no. 4, Definition 4). Let $\mathrm{ad}\, x, \mathrm{ad}\, s, \mathrm{ad}\, n$ be the respective images of $x, s, n$ in the adjoint representation of $\mathfrak{gl}(V)$. Then $\mathrm{ad}\, s$ (resp. $\mathrm{ad}\, n$) is the semi-simple (resp. nilpotent) component of $\mathrm{ad}\, x$ and is equal to a polynomial in $\mathrm{ad}\, x$ with coefficients in K and no constant term.

We know that $\mathrm{ad}\, x = \mathrm{ad}\, s + \mathrm{ad}\, n, [\mathrm{ad}\, s, \mathrm{ad}\, n] = 0$ and $\mathrm{ad}\, n$ is nilpotent ($§ 4$, Lemma 1). We show that $\mathrm{ad}\, s$ is semi-simple. It suffices to do this when K is algebraically closed (cf. Algebra, Chapter VIII, § 9, no. 2, Proposition 3). Then let $(e_i)_{1 \leq i \leq n}$ be a basis of V such that $s(e_i) = \lambda_i e_i$ ($\lambda_i \in K$). Let $(E_{ij})$ be the canonical basis of $M_n(K) = \mathfrak{gl}(V)$. By formulae (5) of $§ 1$,

and hence $\mathrm{ad}\, s$ is semi-simple. The last assertion of the lemma follows from Algebra, Chapter VIII, § 9, no. 4, Proposition 8.