Nilpotent ideal is in nullspace for Killing form

Statement
Suppose $$F$$ is a field, $$L$$ is a Lie algebra over $$F$$, $$\kappa$$ is the fact about::Killing form on $$L$$, and $$A$$ is a fact about::nilpotent ideal of $$L$$. Then, for $$x \in A, y \in L$$, $$\! \kappa(x,y) = 0$$.

Weaker facts

 * Abelian ideal is in nullspace for Killing form

Other related facts

 * Killing form on ideal equals restriction of Killing form
 * Cartan's first criterion
 * Cartan's second criterion

Facts used

 * 1) uses::Lower central series members are derivation-invariant
 * 2) uses::Derivation-invariant subring of ideal implies ideal

Proof
Given: A field $$F$$, a Lie algebra $$L$$ over $$F$$, a class $$c$$-nilpotent ideal $$A$$ of $$L$$. $$\kappa$$ is the Killing form on $$L$$.

To prove: For $$x \in A, y \in L$$, $$\kappa(x,y) = 0$$.

Proof: Consider:

$$\operatorname{ad}(x) \circ (\operatorname{ad}(y) \circ \operatorname{ad}(x))^c$$.

The right-most $$\operatorname{ad}(x)$$ sends any element of $$L$$ inside $$A$$, since $$A$$ is an ideal. $$\operatorname{ad}(y)$$ again sends this inside $$A$$, since $$A$$ is an ideal.

The next $$\operatorname{ad}(x)$$ now sends the element inside $$[A,A]$$. Since $$[A,A]$$ is a derivation-invariant subring of $$A$$ (fact (1)) which is an ideal in $$L$$, $$[A,A]$$ is an ideal in $$L$$ (fact (2)). So $$\operatorname{ad}(y)$$ sends the element within $$[A,A]$$.

Inductively, after $$d$$ steps, the element is in $$[\dots [A,A],A],\dots,A]$$ with $$d$$ $$A$$s. Applying $$\operatorname{ad}(x)$$ sends it to $$[\dots [A,A], \dots A]$$ with $$d+1$$ $$A$$s. This is a derivation-invariant subring of $$A$$ which is an ideal of $$L$$, so it is an ideal of $$L$$. So $$\operatorname{ad}(y)$$ preserves it.

Thus, $$(\operatorname{ad}(y) \circ \operatorname{ad}(x))$$ is in $$A_c = [ \dots [A,A],\dots, A]$$ where $$A$$ is repeated $$c$$ times. Applying $$\operatorname{ad}(x)$$ to this sends it to $$A_{c+1}$$, which is zero since $$A$$ has class $$c$$. Thus:

$$\operatorname{ad}(x) \circ (\operatorname{ad}(y) \circ \operatorname{ad}(x))^c = 0$$

From this, we get that $$(\operatorname{ad}(x) \circ \operatorname{ad}(y))^{c+1} = 0$$. Thus, $$\operatorname{ad}(x)\circ \operatorname{ad}(y)$$ is nilpotent, so it has trace zero, so $$\kappa(x,y) = 0$$.