Remark

Assume that $k$ is perfect and of characteristic 2. Let $\mathfrak{h}=\mathfrak{s}\mathfrak{l}(2, k)$, and

let M be the $\mathfrak{h}$-module $k$ (for the identity map of $\mathfrak{h}$). If $x=$ is an

arbitrary element of $\mathfrak{h}$, denote by $\lambda (x)$ the unique $\lambda \in k$ such that $\lambda^2=a^2+bc$. A calculation shows immediately that $M = M^{\lambda}(\mathfrak{h})$; on the other hand, $M_{\lambda}(\mathfrak{h}) = 0$ and $\lambda$ is neither linear nor zero on $[\mathfrak{h},\mathfrak{h}]$, even though $\mathfrak{h}$ is nilpotent.