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Suppose, $F$ is a field of char $p>0$. Let $\alpha$ be algebraic over $F$ with minimal polynomial $f$. If $f$ is not separable, it can be shown that all roots (say we have $r$ of them) of $f$ have the same multiplicity, say $m$. Then, there exists a separable polynomial $h$ and an integer $n$ such that $h(x^{p^n})=f(x)$. Now, I want to show, for every root $\beta$ of $f$, $\beta ^{p^n}$ is a root of $h$ and every root of $h$ arises this way. The first part is quite clear, but I am unable to see the second part. I should be able to show this by a counting argument and comparing degrees if I could prove that $m=p^n$, but I am unable to see this either. I tried equating degrees of extensions involved and using the fact that the separable degree of $F(\alpha ^{p^n})$ over $F$ is equal to the degree of this extension. However, I am still unable to show this. Any hints?

PS: Feel free to edit the title if it's not very descriptive.

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Hint: You know that $h$ is separable, so we can write $$ h(x)=(x-\gamma_1)(x-\gamma_1)\cdots (x-\gamma_k) $$ for some positive integer $k$ and some (pairwise) distinct elements $\gamma_1,\ldots,\gamma_k$ from some extension field of $F$. Then $$ f(x)=h(x^{p^n})=(x^{p^n}-\gamma_1)(x^{p^n}-\gamma_2)\cdots(x^{p^n}-\gamma_k). $$ Can you see the relation between zeros of the two polynomials now? In both directions? Any element of any extension of $F$ has a $p^{\rm th}$ root in some extension field.

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Thanks for the hint. This is also something I had tried before. In the other direction, the trouble for me, which $p^n$-th root of $\gamma_i$ will be a root of $f$ since there $p^n$ such candidates. –  B M Sep 21 '11 at 18:06
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@BM: Welcome to the magic of characteristic $p$: there are no nontrivial $p$-power roots of unity, so for any $x$ in a field of characteristic $p$, there is at most one $y$ such that $y^{p^n} = x$. –  Pete L. Clark Sep 21 '11 at 18:18
    
@B M: No! In a field of char $p$ a $p$th root is always unique! If $\alpha^p=\beta$ in char $p$, then $$(x-\alpha)^p=x^p-\alpha^p=x^p-\beta,$$ so the root automatically has multiplicity $p$. –  Jyrki Lahtonen Sep 21 '11 at 18:20
    
And furthermore, if you have in, say, char 0 a factorization like $$f(x)=(x^{p^n}-\gamma_1)\cdots(x^{p^n}-\gamma_k),$$ then ALL the $p^n$-th roots of ALL the $\gamma_i$:s would be zeros of $f$. –  Jyrki Lahtonen Sep 21 '11 at 18:23
    
Thanks, Jyrk and Pete. I have seen this argument several times, but somehow failed to reconcile it with this proof. –  B M Sep 21 '11 at 23:33
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