 Solution

Let $E=F({\alpha }^p)\subseteq F(\alpha )$. If $\alpha$ is a separable algebraic element over $F$, its minimal polynomial over $E$ divides the original separable minimal polynomial. It also divides $T^p-{\alpha }^p=(T-\alpha {)}^p$, which has only one distinct root in any extension. Being separable with only one distinct root, that minimal polynomial has degree one. Hence $\alpha \in E$ and $F(\alpha )=F({\alpha }^p)$.

Conversely, if the degree-$n$ minimal polynomial $m$ over $F$ is inseparable, part (b) writes it as $m(T)=g(T^p)$ with $\mathrm{deg}g=n/p$. Then $g({\alpha }^p)=0$, soThe fields cannot be equal. This proves the equivalencethe separability criterion using a pth power.