Proposition 5

We retain the notations of Prop. 4, and denote by $\mathfrak{p}$ the ideal of $\mathbf{S}(\mathfrak{a}^*)$ generated by the homogeneous elements of I of degree $\geq 1$. Let $x\in \mathfrak{a}$. Then $x$ is nilpotent if and only if $f(x) = 0$ for all $x\in \mathfrak{p}.^3$

Extending the base field if necessary, we can assume that $\mathfrak{a}=\mathfrak{g}$ is splittable. Assume that $x$ is nilpotent. For any finite dimensional linear representation $\rho$ of $\mathfrak{g}$, and any integer $n\geq 1$, we have Tr($\rho (x)^n$) $= 0$, so $f(x) = 0$ for all homogeneous $f\in I(\mathfrak{g}^*)$ of degree $\geq 1$ (Th. 1 (ii)), and hence $f(x) = 0$ for all $f\in \mathfrak{p}$. Conversely, if $f(x) = 0$ for all $f\in \mathfrak{p}$, then Tr((ad $x$)$^n$) $= 0$ for all $n\geq 1$ (Th. 1 (ii)), so $x$ is nilpotent.