Nilpotent Lie algebras are closed under extensions

Question

Problem: Let $L$ be a Lie algebra and $K$ an ideal such that $L/K$ is nilpotent and such that $ad(x)|_K$ is nilpotent for all $x \in L$. Prove that $L$ is nilpotent.

By Engel's Theorem, I know that $K$ is nilpotent, which implies that $ad(K)$ is also nilpotent. So, there exists an integer $m$ such that $ad(K)^{(m)} = \{ 0 \}$, where $ad(K)^{(m)}= [ad(K), ad(K)^{(m-1)}]= \{ 0 \}$, (the $m$-th term of the lower central series of $ad(K)$). Now, because I know that an extension of a nilpotent algebra is nilpotent only if it is central, I would like to relate $K$ to $Z(L)$, using information on $ad(K)^{(m)}$, if that makes any sense. Is this a reasonable way to tackle the problem? If so, any hints to proceed?

Answer 1

The answer is yes provided that $L$ is finite-dimensional. Denote by $x^*$ a class in $L/K$, $x\in L$. Since $L/K$ is nilpotent, there exists $m$ such that $ad_{x^*}^r=0$. That is, for all $z\in L$ we have $ad_x^r(z)\in K$. But $(ad_x)|_K$ is nilpotent, hence there exists $s$ such that $0=ad_x^s(ad_x^r(z))=(ad^{r+s}_x)(z)$ for all $z\in L$. By Engel's theorem $L$ is nilpotent.