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 $x\in m$, then so is $p(x)$and ad(y)=p(ad(x)) for anya polynomial $p$ without constant term, then $y\in m$. Hence if $x\in m\cap m^\perp$, then $xp(x)$$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.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.
QuestionQuestion: is this argument correct, and if so, is there a reference for it?
Edit: is this argument correct, and if so, is there a reference for it?confused ad p(x) with p(ad x)!