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I need help proving that the polynomial $f(x)=(x-\alpha)(x-\alpha^p)\cdots(x-\alpha^{p^{n-1}})$ is in $F_p[x]$ if $\alpha\in F_{p^n}$.

This assertion is trivial for $\alpha$ already in $F_p$. So we assume it's not. We know this polynomial is degree $n$ and we know all of its roots are contained in $F_{p^n}$. We also have the result that $x^{p^n}-x$ is the product $\prod_{\alpha\in F_{p^n}}(x-\alpha)$. Thus we know that if all the $\alpha^{p^i}$ are distinct then $f(x)|x^{p^n}-x$.

My rough thoughts so far: We have a polynomial $f(x)$ of degree $n$ and we know its roots are all the powers of some element $\alpha$. Thus when we adjoin the root $\alpha$ to $F_p$ we have that the polynomial will split completely. I'm not sure where to go from here though.

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What is the question? – JSchlather Nov 26 '12 at 22:03
Sorry, I'll edit to make it more clear. – Steven-Owen Nov 26 '12 at 22:06
Hint: Show that if $f_i$ denotes the coefficient of $x^i$ in $f(x)$, then $f_i^p = f_i$ and so $f_i \in \mathbb F_p$. Look at $[f(x)]^p$ and use $(a-b)^p = a^p - b^p$ which holds in a field of characteristic $p$. – Dilip Sarwate Nov 26 '12 at 22:23
Look at this question for ideas on how to proceed. – Dilip Sarwate Nov 26 '12 at 22:31
up vote 4 down vote accepted

Based on Dilip's comments, I think the answer is much simpler than anticipated.

Consider $f(x)^p$ we have $\prod_{i=0}^{n-1}(x-\alpha^{p^i})^p$ and this gives $\prod_{i=0}^{n-1}(x^p-(\alpha^{p^i})^p)=\prod_{i=0}^{n-1}(x^p-(\alpha^{p^i+1}))=\prod_{i=1}^{n}(x^p-(\alpha^{p^i}))$. But $\alpha^{p^n}=\alpha$ so we are back to $\prod_{i=0}^{n-1}(x^p-(\alpha^{p^i}))=f(x^p)$. Let $f_i$ be the coefficient of the $x^i$ term in $f(x)$. We have $f(x)^p=\left(\sum_{i=0}^n f_ix^i\right)^p=\sum_{i=0}^n f_i^p(x^i)^p=\sum_{i=0}^n f_i^p(x^p)^i$ but we know this equals $f(x^p)=\sum_{i=0}^n f_i(x^p)^i$. Thus matching terms we get $f_i=f_i^p$ which implies that the $f_i$ are all in $F_p$ and thus $f(x)\in F_p[x]$

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+1 Good! I am glad you found the solution for yourself. – Dilip Sarwate Nov 27 '12 at 2:25

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