I've solved an exercise in Humphreys that said:
Show that if Lie algebra $L$ is nilpotent, then its Killing form is zero.
I'm wondering is the opposite true? In Humphreys, we work mainly with Lie algebras over an algebraically closed field of characteristic zero, so let's assume that.
This is what I've tried:
Let's assume that Killing form is zero. That gives us solvability of $L$, so we know (as a corollary of Lie's theorem) that there is a basis of $L$ in which all matrices in $ad \ L$ are upper triangular. Then, $\kappa(x,y) = tr(ad(x) \ ad(y))$ is a scalar product of diagonals of $ad(x)$ and $ad(y)$ in that basis. If I could show that this somehow implies that for every $x \in L$ diagonal of $ad(x)$ is all zeros, it would follow all eigenvalues of $ad(x)$ are zero and this would imply that $ad(x)$ is nilpotent. Then, Engel's theorem would give us nilpotency.
But I don't know how to do that.
No, $L$ is not necessarily nilpotent: here's an example of a complex non-nilpotent Lie algebra with vanishing Killing form: take $L=\mathbb{C}e_1\oplus\mathbb{C}e_2\oplus\mathbb{C}e_3$ where $$[e_1,e_2]=e_2,\quad [e_1,e_3]=ie_3,\quad [e_2,e_3]=0.$$ (Jacobi's identity is easy to check). Since $e_2$ is an eigenvector of $\mathrm{ad}_{e_1}$ with eigenvalue $1\neq0$, $L$ is not nilpotent (Engel's Theorem). Now we have: $$[\mathrm{ad}_{e_1}]=\begin{pmatrix}0&0&0\\0&1&0\\0&0&i\end{pmatrix},\quad [\mathrm{ad}_{e_2}]=\begin{pmatrix}0&0&0\\-1&0&0\\0&0&0\end{pmatrix},\quad [\mathrm{ad}_{e_3}]=\begin{pmatrix}0&0&0\\0&0&0\\-i&0&0\end{pmatrix},$$ and it's easy to check by hand that $K$ is nil.