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Let $k$ be a field of characteristic $p\ne 0$. Let $K$ be a finite extension of $k$. Let $\alpha \in K$. How do I prove that either ${\alpha}^{p^n}\in k$ for some $n$ or there exists an integer $m$ such that $\beta ={\alpha}^{p^m}\notin k$ and $\beta$ is separable over $k$?

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I have deleted a comment of yours, Tiez Ered, which disrupted the formatting and did not seem to be related to the question. –  Mariano Suárez-Alvarez Mar 18 '12 at 21:16
    
Mariano, it was just to pass the character limit. Thanks. –  Herband Mar 18 '12 at 21:19
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up vote 4 down vote accepted

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 $k$.

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Can you please explain why $h$ is separable? –  Herband Mar 18 '12 at 21:22
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@TiezEred: Because, by definition of $\ell$, $h(x)$ cannot be expressed as a polynomial in $x^p$. An irreducible polynomial over a field of characteristic $p\gt 0$ is inseparable if and only if $h$ has multiple roots, if and only if $h$ is not relatively prime with its formal derivative $h'$, if and only if (by irreducibility of $h$) $h'=0$, if and only if $h$ is a polynomial in $x^p$. –  Arturo Magidin Mar 18 '12 at 21:24
    
Arturo, thank you for the addition. –  Herband Mar 18 '12 at 21:30
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