Separable elements in a finite characteristic field Let $k$ be a field of characteristic $p\ne 0$ , $K$ be a finite extension of $k$ and $\alpha \in K$. 
How do I prove that either:


*

*$\exists n \in \mathbb{N}. {\alpha}^{p^n}\in k$ 

*$\exists m \in \mathbb{N}.\alpha^{p^m}\notin k$ and $\alpha^{p^m}$ is separable over $k$

 A: If $\alpha$ is separable over $k$, then we can take either $n=0$ (if $\alpha\in k$), or $m=0$ (if $\alpha\notin k$). So we may assume that $\alpha$ is not separable over $k$.
Recall that if $k$ is a field of characteristic $p\gt 0$, and $f(x)\in k[x]$ is irreducible, then $f$ is not separable if and only if it can be expressed as a polynomial in $x^p$; indeed, $f$ is not separable if and only if $f$ has repeated roots; if and only if $f$ is not relatively prime with its formal derivative $f'$; if and only if (since we are assuming $f$ is irreducible) $f'=0$; if and only if $f$ is a polynomial in $x^p$.
So let $f(x)$ be the minimal polynomial of $\alpha$ over $k$. Then $f(x) = g(x^p)$ for some polynomial $g(x)\in k[x]$, since $\alpha$ is not separable. Let $\ell$ be the largest integer such that $f(x) = h(x^{p^{\ell}})$ for some $h(x)\in k[x]$. In particular, $h(x)$ is separable, since it must be irreducible (as $f$ is irreducible), and cannot be a polynomial in $x^p$ by the definition of $\ell$. If $h(x)$ is of degree one, then $\alpha^{p^{\ell}}\in k$, so take $n=\ell$. If $h$ is not linear, then let $m=\ell$; then $\beta=\alpha^{p^m}$ is a root of $h(x)$, does not lie in $k$, and is separable over $k$. 
