= Solution
If $L$ is abelian and nonzero, $L'=0<L$. Otherwise choose $x\notin Z=Z(L)$ such that $Z+Fx$ is an ideal. For each $a\in L$ write
$$
[a,x]=\lambda(a)x+z(a),\qquad z(a)\in Z.
$$
The coefficient $\lambda(a)$ is uniquely defined and linear. The <Jacobi identity> gives
$$
[[a,b],x]=\lambda(b)z(a)-\lambda(a)z(b)\in Z,
$$
so $\lambda([a,b])=0$. If $\lambda\ne0$, then $L'\subseteq\ker\lambda<L$.
If $\lambda=0$, the same equation says $[[a,b],x]=0$ for all $a,b$. Hence $L'\subseteq C_L(x)$. Since $x$ is not central, its centralizer is proper. Thus in both cases
$$
\boxed{L'<L.}
$$
No algebraic closure or characteristic restriction was used.
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