 Solution

In characteristic $p$, let $K={\mathbb{F}}_p(s,t)$ and $L={\mathbb{F}}_p(s^{1/p},t^{1/p})$, with $s,t$ having algebraic independence. The monomials $s^{i/p}{t}^{j/p}$, $0\le i,j<p$, form a $K$-basis, so $[L:K]=p^2$. Their independence follows by grouping polynomial monomials according to their exponents modulo $p$, after clearing rational-function denominators.

Every $\alpha \in L$ has ${\alpha }^p\in K$, so its minimal polynomial divides $T^p-{\alpha }^p$ and $[K(\alpha ):K]\le p$. Thus this finite purely inseparable extension has no primitive element. It demonstrates why the separable field extension hypothesis cannot simply be removed.