= Separability criterion using a pth power
{title2=$\alpha\text{ separable over }F\iff F(\alpha)=F(\alpha^p)$}
For an <algebraic element> $\alpha$ over a <field> $F$ of characteristic $p>0$, suppose first that $\alpha$ is a <separable algebraic element>. Its <minimal polynomial> over $E=F(\alpha^p)$ divides its original <separable polynomial> and also $T^p-\alpha^p=(T-\alpha)^p$. It is therefore separable with only one distinct root, so has degree one and $\alpha\in E$. Conversely, if the irreducible <minimal polynomial> $m$ is inseparable, its <formal derivative> vanishes and $m(T)=g(T^p)$ for some $g$ of degree $\deg(m)/p$. Thus $[F(\alpha^p):F]\leq\deg(m)/p<[F(\alpha):F]$, proving the converse.
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