I'm currently studying Lie algebras and i ran into the Killing form, defined for a Lie algebra $\mathfrak{g}$ over a field $k$ by: $\mathfrak{g}\times\mathfrak{g} \rightarrow k, (x,y) \mapsto tr(ad(x)ad(y))$ . This map is useful because if $k$ is algebraically closed and its characteristic is zero, the Killing form is non degenerate iff $\mathfrak{g}$ is semisimple.
On the other hand, given a field extension of finite degree $K/k$, one defines the trace map $K \rightarrow k$ which sends $x\in K$ to $tr(r_x)$, where $r_x \in End_k(K)$ is the multiplication by $x$. Now, $K/k$ is separable iff the bilinear symmetric form $K \times K \rightarrow k, (x,y) \mapsto tr(r_xr_y)$ is non degenerate.
My question is: is there a reason why these two definitions look the same and give such similar results?
Edit. Maybe one way to proceed could be to search a Lie algebra whose Killing form is related to the trace map of $K/k$. The map $r : K \rightarrow End_k(K), x \mapsto r_x$ sends $K$ into $GL(n,k)$, so one may consider the Zariski closure of the image $r(K)$; this will be an algebraic group with an associated Lie algebra. Now, everything is constructed from $K/k$, so I'd expect some relations. Could this be the case?