Proof of Cartan's solvability criterion

Question

The following is a key lemma in popular proofs of Cartan's solvability criterion for Lie algebras over a field of characteristic zero.

Given $a\subseteq b\subseteq \mathrm{End}(V)$, let $m=\{x\in\mathrm{End}(V):[x,b]\subseteq a\}$. Then any element of $m\cap m^\perp$ is nilpotent (where $m^\perp$ is the orthogonal space to $m$ with respect to the trace form on $\mathrm{End}(V)$).

Standard proofs use the fact that if $x\in m$, and ad(y)=p(ad(x)) for a polynomial $p$ without constant term, then $y\in m$. Hence if $x\in m\cap m^\perp$, then $xy$ has trace zero for any such polynomial $p$. In all the texts I've seen, the argument then gets more complicated, using Jordan decomposition and field-theoretic tricks to show that $x$ is nilpotent.

However, if $p(t)=t^{2k-1}$, then $xp(x)=x^{2k}$, so if this has trace zero for all positive integers $k$, then $x^2$ is nilpotent (since the field has characteristic zero), hence $x$ is nilpotent as well.

Question: is this argument correct, and if so, is there a reference for it?

Edit: this argument confused ad p(x) with p(ad x)!

Answer 1

Why is it true that with $x\in \mathfrak{m}$ also $p(x)\in \mathfrak{m}$ for all polynomials $p(x)$ without constant term ? The proof I know just uses the Jordan decomposition $x=x_s+x_n$, and shows for the eigenvalues $\lambda_i$ of $x_i$, that $P(ad x_s)=ad (y)$ for a certain interpolation polynomial $P$ without constant term, and $y$ a certain matrix depending on $\lambda_i$. Then basically $0=tr (xy)=f(\lambda_1)^2+\cdots +f(\lambda_r)^2$, with linear form $f$ defined on the span of the $\lambda_i$ over $\mathbb{Q}$ gives $f=0$, hence all $\lambda_i=0$, so that $x=x_s+x_n=x_n$ is nilpotent.