Jordan–Chevalley parts agree under the adjoint representation

Suppose $x=x_s+x_n$ is an abstract Jordan decomposition. Then ${\mathrm{ad}}_x={\mathrm{ad}}_{x_s}+{\mathrm{ad}}_{x_n}$ by linearity of $\mathrm{ad}$, the two summands commute because $[{\mathrm{ad}}_{x_s},{\mathrm{ad}}_{x_n}]={\mathrm{ad}}_{[x_s,x_n]}=0$, and by [L4] the first is semisimple while the second is nilpotent. The uniqueness assertion of [L1] therefore identifies them with the additive Jordan–Chevalley parts of ${\mathrm{ad}}_x$. This proves (i).

Now let ${\mathrm{ad}}_x=S+N$ be the additive Jordan–Chevalley decomposition of the derivation $T={\mathrm{ad}}_x$; by [L1] there are polynomials $p,q\in \mathbb{C}[t]$ with $S=p(T)$ and $N=q(T)$. We show that $S$ is a derivation and that $S$ acts on the generalized eigenspace of $T$ for $\lambda$ as multiplication by $\lambda$. The generalized eigenspaces ${\mathfrak{g}}_{\lambda }=\{y:(T-\lambda {)}^{k}y=0\text{ for some }k\}$ satisfy $\mathfrak{g}={⨁}_{\lambda }{\mathfrak{g}}_{\lambda }$ and $[{\mathfrak{g}}_{\lambda },{\mathfrak{g}}_{\mu }]\subseteq {\mathfrak{g}}_{\lambda +\mu }$: the second claim follows from the binomial expansion $(T-\lambda -\mu {)}^n[y,z]={\sum }_{i+j=n}\left(\genfrac{}{}{0px}{}{n}{i}\right)[(T-\lambda {)}^{i}y,(T-\mu {)}^{j}z]$. Each ${\mathfrak{g}}_{\lambda }$ is invariant under $T$, hence under $p(T)=S$. On ${\mathfrak{g}}_{\lambda }$ the operator $T$ equals $\lambda \cdot 1$ plus a commuting nilpotent operator, so $p(T)=p(\lambda )\cdot 1+(\text{nilpotent})$ there; since $S$ is semisimple and restricts semisimply to the invariant subspace ${\mathfrak{g}}_{\lambda }$, this forces $S{\mathrm{\mid }}_{{\mathfrak{g}}_{\lambda }}=p(\lambda )\cdot 1$. On the other hand $S-\lambda \cdot 1=(T-\lambda \cdot 1)-N$ is a difference of two commuting nilpotent operators on ${\mathfrak{g}}_{\lambda }$, hence nilpotent; comparing with the scalar operator $(p(\lambda )-\lambda )\cdot 1$ gives $p(\lambda )=\lambda$. Therefore $S{\mathrm{\mid }}_{{\mathfrak{g}}_{\lambda }}=\lambda \cdot 1$, and for $y\in {\mathfrak{g}}_{\lambda }$, $z\in {\mathfrak{g}}_{\mu }$ one has $S[y,z]=(\lambda +\mu )[y,z]=[\lambda y,z]+[y,\mu z]=[Sy,z]+[y,Sz]$ because $[y,z]\in {\mathfrak{g}}_{\lambda +\mu }$. Thus $S$ is a derivation; $T$ is a derivation by the Jacobi identity, so $N=T-S$ is a derivation too.

By [L2] there are unique $y_s,y_n\in \mathfrak{g}$ with $S={\mathrm{ad}}_{y_s}$ and $N={\mathrm{ad}}_{y_n}$. Then ${\mathrm{ad}}_{y_s+y_n}=S+N={\mathrm{ad}}_x$, so $y_s+y_n=x$ by injectivity of $\mathrm{ad}$ [L3]; also ${\mathrm{ad}}_{[y_s,y_n]}=[S,N]=0$, so $[y_s,y_n]=0$ by [L3]; and ${\mathrm{ad}}_{y_s}=S$ is semisimple while ${\mathrm{ad}}_{y_n}=N$ is nilpotent. Hence $x=y_s+y_n$ is an abstract Jordan decomposition by [L4].

If $x=u+v$ is any abstract Jordan decomposition, then ${\mathrm{ad}}_u,{\mathrm{ad}}_v$ is a commuting semisimple-plus-nilpotent decomposition of ${\mathrm{ad}}_x$ by step 1.1's computation, so uniqueness in [L1] gives ${\mathrm{ad}}_u=S={\mathrm{ad}}_{y_s}$ and ${\mathrm{ad}}_v=N={\mathrm{ad}}_{y_n}$; injectivity of $\mathrm{ad}$ [L3] gives $u=y_s$ and $v=y_n$. Hence the decomposition of (ii) is unique. If $\mathfrak{g}=0$ then $x=0=y_s=y_n$ and every assertion holds with the zero endomorphism, which is both semisimple and nilpotent; no nonempty choice is made anywhere in this argument, the Axiom of Choice being used only through the appeal to [L1].