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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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    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).2012-01-19
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    Ah, that makes sense. Thanks2012-01-19

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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 $.