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Let $N$ be an algebraic closure of a finite field $F$. How to prove that any automorphism in $\operatorname{Gal}(N/F)$ is of infinite order?

I have shown: Letting $|F|=q$,
1) $N$ is the union of all extensions $M$ of $F$ such that $|M|=q^n$, for some $n\geq 1$.
2) $\operatorname{Gal}(N/F)$ is abelian.

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    There is, of course, one trivial counterexample.2012-03-30

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