Abelian 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 an fact about::abelian ideal of $$L$$. Then, for $$x \in A, y \in L$$, $$\kappa(x,y) = 0$$.

Stronger facts

 * Nilpotent 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

Proof
Given: A field $$F$$, a Lie algebra $$L$$ over $$F$$, an abelian ideal $$A$$ of $$F$$. $$\kappa$$ is the Killing form on $$L$$.

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

Proof: Consider:

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

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 left-most $$\operatorname{ad}(x)$$ then sends this element to zero, since $$A$$ is abelian.

Thus:

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

From this, we get that $$(\operatorname{ad}(x) \circ \operatorname{ad}(y))^2 = 0$$. Thus, $$\operatorname{ad}(x)\circ \operatorname{ad}(y)$$ is nilpotent, so it has trace zero, so $$\kappa(x,y) = 0$$.