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Let $F$ be a field with $\mathbb{char}(F) = p$. Prove that if $E/F$ is a finite extension with $p \nmid [E:F]$, then $E/F$ is a separable extension.

I've no idea how to prove this, I'm looking for some insight to lead me in the right direction.

I've considered a proof by contradiction, supposing $E/F$ is not separable so there is some $\alpha \in F$ such that the minimal polynomial of $\alpha$ is not separable, ie. $p(x) = (x - \alpha)^2g(x)$, but I can't see how to derive a contradiction from this.

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  • $\begingroup$ Prove the contrapositive: if $E/F$ isn't separable, then $p$ divides the degree. $\endgroup$ – Qiaochu Yuan Feb 22 '16 at 1:07
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Let $x\in E$, consider $E_x$ the subfield generated by $1,x,x^2,...$. Its rank is the degree of the minimal polynomial $P$ of $x$, since $[E:F]=[E:E_x][E_x:F]$, we deduce that $[E_x:F]$ is prime with $p$. We know that $E_x=F[X]/(P)$, this implies that $P$ is separable, since if $P$ was not separable, $P(X)=Q(X^p)$ https://en.wikipedia.org/wiki/Separable_polynomial#Separable_field_extensions a fact which implies that $p$ divides the degree of $P$.

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