Proposition 4

Let $\mathfrak{g}$ be a Lie algebra, $l$ the rank of $\mathfrak{g},c$ the nilpotency class of the Cartan subalgebras of $\mathfrak{g}$, and $x\in \mathfrak{g}$. There exists an $l$-dimensional subalgebra of $\mathfrak{g}$ whose nilpotency class is $\leq c$ and which contains $x$.

Let T be an indeterminate. Let $k'=k(T)$ and $\mathfrak{g}'=\mathfrak{g}\otimes_kk'$. If $\mathfrak{h}$ is a Cartan subalgebra of $\mathfrak{g},\mathfrak{h}\otimes_kk'$ is a Cartan subalgebra of $\mathfrak{g}'$, hence the rank of $\mathfrak{g}'$ is $l$ and the nilpotency class of the Cartan subalgebras of $\mathfrak{g}'$ is $c$.

Choose a regular element $y$ of $\mathfrak{g}$. With the notations of §2, no. 2, we have $a_l(y)\not= 0$. Denote also by $a_l$ the polynomial function on $\mathfrak{g}'$ that extends $a_l$. Then the element $a_l(x+ Ty)$ of $k[T]$ has dominant coefficient $a_l(y)$. In particular, $x+ Ty$ is regular in $\mathfrak{g}'$. Let $\mathfrak{h}'$ be the nilspace of ad($x+ Ty$) in $\mathfrak{g}'$. Then dim $\mathfrak{h}'=l$ and the nilpotency class of $\mathfrak{h}'$ is $c$. Put $\mathfrak{k}=\mathfrak{h}'\cap (\mathfrak{g}\otimes_kk[T])$; then $\mathfrak{k}\otimes_{k[T]}k(T) =\mathfrak{h}'$.

Let $\varphi$ be the homomorphism from $k[T]$ to $k$ such that $\varphi (T) = 0$, and let $\psi$ be the homomorphism $1\otimes \varphi$ from $\mathfrak{g}\otimes_kk[T]$ to $\mathfrak{g}$. Then $\psi (\mathfrak{k})$ is a subalgebra of $\mathfrak{g}$ whose nilpotency class is $\leq c$ and which contains $\psi (x+ Ty) =x$.

In the free $k$[T]-module $\mathfrak{g}\otimes_kk[T],\mathfrak{k}$ is a submodule of rank $l$, and $(\mathfrak{g}\otimes_kk[T])/\mathfrak{k}$ is torsion free, so the submodule $\mathfrak{k}$ is a direct summand of $\mathfrak{g}\otimes_kk[T] ($Algebra, Chap. VII, §4, no. 2, Th. 1). Hence dim$_k\psi (\mathfrak{k}) =l$, which completes the proof.