Exercise 21, § 4

Let $g$ be a Lie algebra and $\sigma$ an automorphism of $g$. (a) Suppose first that $K$ is algebraically closed; for all $\lambda \in K$ let $L_\lambda$ be the set of $x \in g$ which are annihilated by a power of $\sigma - \lambda I$; $g$ is the direct sum of the $L_\lambda$. Show that $[L_\lambda, L_\mu] \subset L_{\lambda \mu}$. (Note that

(b) With an arbitrary field $K$, suppose that none of the eigenvalues of $\sigma$ (in an algebraically closed extension of $K$) is a root of unity. Show that $g$ is nilpotent. (Reduce it to the case where $K$ is algebraically closed. If $\lambda_1, \ldots, \lambda_m$ are the distinct eigenvalues of $\sigma$, the $\lambda_i \lambda_j, \lambda_i \lambda_j^2, \lambda_i \lambda_j^3, \ldots$ are not all eigenvalues and hence $\mathrm{ad}_g x$ is nilpotent for $x \in L_{\lambda_i}$. Conclude using Exercise 11 applied to the set of $\mathrm{ad}_g x$, where $x$ runs through the union of the $L_{\lambda_i}$.) (c) With an arbitrary field $K$, suppose that $\sigma^q = I$, where $q$ is a prime number and that no eigenvalue of $\sigma$ is equal to 1. Show that $g$ is nilpotent. (Same method as in (b), observing that, if $g \neq \{0\}$, $g$ is not equal to the characteristic of $K$ and that for every ordered pair of eigenvalues $\lambda_i, \lambda_j$ of $\sigma$ there exists an integer $k$ such that $\lambda_i \lambda_j^k = 1$.)