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Suppose $ q = p^r$. Let $F$ be the splitting field of $X^q - X$. Let $\phi : F \to F$ be the Frobenius automorphism $\phi(x) = x^p$. Then let $F' \subseteq F$ be the fixed field of $ < \phi^r >$.

Claim: $x \in F'$ iff $\phi^r(x) = x $.

Why is the claim true? I can see one direction easily (if $\phi^r(x) = x$, then $x$ is fixed by $\phi^r$). But what about the other direction? If $x \in F'$, then $x$ is fixed by some $\phi^{kr}$. Why must it be fixed by $\phi^r$?

Thanks

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    $\begingroup$ You misunderstand the definition of "fixed field." It means that every element of the subgroup fixes an element of the subfield (whereas you seem to think it means some element fixes it). $\endgroup$ – Qiaochu Yuan Jan 19 '12 at 20:05
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    $\begingroup$ Ah, that makes sense. Thanks $\endgroup$ – Matt Jan 19 '12 at 20:19
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If $\phi^r(x) = x$, then $x$ is fixed by $\phi^r \in \langle \phi^r \rangle$, so $x \in F'$.

Conversely, if $x \in F'$, then $x$ is fixed by everything in $\langle \phi^r \rangle$, and so in particular is fixed by $\phi^r$ i.e. $\phi^r(x) = x $.

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