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If $K$ is a field extension of $F$ and if $\alpha\in K$ is not separable over $F$, show that $\alpha^{p^m}$ is separable over $F$ for some $m\geq 0$, where $p = $char$(F)$.

I know that $x^p-\alpha^p=(x-\alpha)^p$ and that $x^{p^m}-\alpha^{p^m}=(x-\alpha)(x^{p^m-1}+x^{p^m-2}\alpha+...+x\alpha^{p^m-2}+\alpha^{p^m-1})$, but I do not know how to use this and the fact that $\alpha$ is not separable over P, could anyone help me please? Thank you very much.

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I assume you want $K$ algebraic over $F$.

Consider the minimum polynomial $f$ of $\alpha$. Write it as $f(x)=a_0+a_1x+a_2x^2+\cdots +a_dx^d$. If $a_j\ne0$ for some $j$ with $p\nmid j$ then $f'$ is nonzero and $\alpha$ is separable. Otherwise $f(x)=a_0+a_p x^p+a_{2p}x^{2p}+\cdots=g(x^p)$ where $g$ is now the minimum polynomial of $\alpha^p$. If $g'$ is nonzero, then $\alpha^p$ is separable, otherwise $f(x)=h(x^{p^2})$ etc. This process will eventually conclude...

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  • $\begingroup$ Why does $a_d=0$? $\endgroup$ – J. Doe May 9 at 13:06

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