Exercise 11, § 5

Exceptionally, we assume in this exercise that $k$ is a perfect field of characteristic $p >0$.

Let $\mathfrak{g}$ be a Lie $p$-algebra (Chap. I, §1, Exerc. 20). If $x\in \mathfrak{g}$, denote by $\langle x\rangle$ the smallest Lie $p$-subalgebra containing $x$. It is commutative and generated as a $k$-vector space by the $x^{p^i}$ where $i= 0,1, . .$.. Then $x$ is said to be nilpotent (resp. semi-simple) if the $p$-map of $\langle x\rangle$ is nilpotent (resp. bijective).

a) Show that $x$ can be decomposed uniquely in the form $x=s+n$, with $s, n\in \langle x\rangle ,s$ semi-simple and $n$ nilpotent (apply Exerc. 23 of Chap. I, §1). If $f$ is a $p$-homomorphism from $\mathfrak{g}$ to $\mathfrak{g}\mathfrak{l}$(V), $f(s)$ and $f(n)$ are the semi-simple and nilpotent components of the endomorphism $f(x)$; this applies in particular to $f=$ ad.

$b) A$ subalgebra of $\mathfrak{g}$ is said to be decomposable if it contains the semi-simple and nilpotent components of its elements. Show that, if $\mathfrak{b}$ and $\mathfrak{c}$ are vector subspaces of $\mathfrak{g}$ such that $\mathfrak{b}\subset \mathfrak{c}$, the set of $x\in \mathfrak{g}$ such that $[x,\mathfrak{c}]\subset \mathfrak{b}$ is decomposable (same proof as Prop. 3); in particular, every Cartan subalgebra of $\mathfrak{g}$ is decomposable.

c) Let $\mathfrak{t}$ be a commutative subalgebra of $\mathfrak{g}$ consisting of semi-simple elements, and maximal with this property. Let $\mathfrak{h}$ be the commutant of $\mathfrak{t}$ in $\mathfrak{g}$. Let $x\in \mathfrak{h}$ and let $x=s+n$ be its canonical decomposition; since $\mathfrak{h}$ is decomposable (cf. $b$)$),s, n\in \mathfrak{h}$. Show that the subalgebra generated by $\mathfrak{t}$ and $s$ is commutative and consists of semi-simple elements, hence coincides with $\mathfrak{t}$. Deduce that ad$_{\mathfrak{h}}x=$ ad$_{\mathfrak{h}}n$ is nilpotent, and hence that $\mathfrak{h}$ is nilpotent. Since $\mathfrak{h}=\mathfrak{g}^0(\mathfrak{h})$, $\mathfrak{h}$ is a Cartan subalgebra of $\mathfrak{g}($§2, Prop. 4).

In particular, every Lie $p$-algebra over a finite field has a Cartan subalgebra.$^9$