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Let $F$ be a finite field of order $p^n$ for some prime $p$ and positive integer $n$. This is well known that group of field automorphism of $F$ is cyclic and generate by the following: $\alpha:F\rightarrow F$; $\alpha(x)=x^p$, how can I find the fixed points of an arbitrary element of $Aut(F)$?

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    $\begingroup$ a) what is $\alpha^k$? b) write down what $\alpha^k(x) = x$ means. $\endgroup$ – Daniel Fischer Dec 11 '14 at 13:26
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Hints:

If $\;\overline{\Bbb F_p}\;$ is an algebraic closure of $\;\Bbb F_p\;$ , then for all

$$n\in\Bbb N\;,\;\;\Bbb F_{p^n}=\{\;\omega\in\overline{\Bbb F_p}\;:\;\;\omega^{p^n}-\omega=0\;\}$$

Now, what does the following mean?: $$x\in\Bbb F\;,\;\;x\in \text{Fix}\,(\alpha^k)\iff x^{p^k}=x\;\ldots$$

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