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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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    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 $.