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Let $f(X) \in \mathbb F_p[X]$ irreducible with $p$ prime and assume $\exists \alpha \in \mathbb F_{p^n}: f(\alpha) = 0$ where $n \geq 1$. I then have to prove that $f$ splits over $\mathbb F_{p^n}$.

Some general thoughts are that $\mathbb F_{p^n}$ is the splittingfield of $X^{p^n}-X \in \mathbb F_p[X]$. I can look at $\mathbb F_p(\alpha)$ which must be a subfield of $\mathbb F_{p^n}$ since $\alpha \in \mathbb F_{p^n}$. I further have that $f$ divides $X^{p^n}-X$ since $f$ is irreducible and $\alpha$ is a zero of both polynomials.

I am quite confused. Help please :)

Some more considerations: Assume $f(\beta) = 0$ Then with $f| X^{p^n}-X$ is get $f(X)g(X) = X^{p^n}-X$ so $\beta^{p^n}-\beta = 0$ s.t. $\beta \in \mathbb F_{p^n}$ and thus $f$ splits in $\mathbb F_{p^n}$ ?

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marked as duplicate by DonAntonio, Amzoti, Jyrki Lahtonen, Daniel Rust, O.L. Jun 22 '13 at 20:07

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Do you know what a normal extension is? If so, what definition are you using? –  Zev Chonoles Jun 22 '13 at 18:22
    
Nope. This term is not introduced yet. So I hope that there is some other solution. –  André Jun 22 '13 at 18:24
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It seems like you've done most of the work. If I have two polynomials $f, g \in k[X]$ with $f \mid g$ and $g$ splits over $k$, then does $f$ split as well? We ought to be able to prove that the answer is yes. [Unique factorization!] –  TTS Jun 22 '13 at 18:28
    
Possibly the oldest version this is a duplicate of? –  Jyrki Lahtonen Jun 22 '13 at 19:53
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2 Answers 2

You've done all but wrap the whole thing up:

$$f(x)\mid\left(x^{p^n}-x\right)\iff x^{p^n}-x=f(x)g(x)\implies$$

every root of $\,f\,$ is also a root of $\,x^{p^n}-x\,$ and thus...

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Another way of reaching the conclusion would be to observe that if $\alpha$ is a root of $f(x)$, then so are $\alpha^p$, $\alpha^{p^2}$,$\ldots$, $\alpha^{p^{n-1}}$ and $\alpha^{p^n}=\alpha$. As $f(x)$ is irreducible and the product $$(x-\alpha)(x-\alpha^p)\cdots(x-\alpha^{p^{n-1}})$$ is in $\mathbb{F}_p[x]$, these must be all the roots of $f(x)$. Hence they are in the field $\mathbb{F}_p[\alpha]$.

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This answer also has to be a duplicate, but I couldn't find it in five minutes of searching, so retyping it instead. –  Jyrki Lahtonen Jun 22 '13 at 19:35
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