 Solution

If $L$ is abelian and nonzero, $L^{\prime }=0<L$. Otherwise choose $x\notin Z=Z(L)$ such that $Z+Fx$ is an ideal. For each $a\in L$ writeThe coefficient $\lambda (a)$ is uniquely defined and linear. The Jacobi identity givesso $\lambda ([a,b])=0$. If $\lambda \ne 0$, then $L^{\prime }\subseteq \ker \lambda <L$.

If $\lambda =0$, the same equation says $[[a,b],x]=0$ for all $a,b$. Hence $L^{\prime }\subseteq C_L(x)$. Since $x$ is not central, its centralizer is proper. Thus in both casesNo algebraic closure or characteristic restriction was used.