For a Lie algebra $\mathcal{sl}(n, \mathbb{C})$, it is well known that the Killing form $\kappa(A,B) = 2n tr(AB)$ for all $A, B \in \mathcal{sl}(n, \mathbb{C})$.
Now let $F=\mathbb{F}_p$ be a finite field. Is there any relation, like above, between the Killing form $\kappa$ and the trace $tr$ over $\mathcal{sl}(n, \mathbb{F}_p)$? Perhaps we can conclude if $tr(AB) = 0$, then $\kappa(A,B) = 0$.
In general we do not have $\kappa(A,B)=0$ because of $tr(AB)=0$. For example, the trace form for $\mathfrak{sl}_p(\mathbb{F}_p)$ for $p>2$ is non-degenerate, but nevertheless the Killing form is identically zero. So in general it is not true that the Killing form of a simple modular Lie algebra is a non-zero multiple of the trace form.
For the Lie algebra $\mathbb{sl}_n(K)$ we have $$ \kappa(x,y)=2n\cdot tr(xy) $$ over an arbitrary field $K$. This follows from a direct computation using a basis of $\mathbb{gl}_n(K)$, which gives $\kappa(x,y)=2n tr(xy)-2tr(x)tr(y)$ on this basis. Since $\mathbb{sl}_n(K)$ is an ideal, we obtain $\kappa(x,y)=2n\cdot tr(xy)$ as the restriction for trace zero matrices.