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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. –  Gerry Myerson Mar 30 '12 at 0:33

1 Answer 1

Assume that $\sigma\in Gal(N/F)$ is of finite order $m$. Let $x\in N$. Let $k=[F(x):F]$. Let $K=GF(q^{km})$ be the unique extension of $F$ of degree $km$. Because $K$ is Galois over $F$, the restriction of $\sigma$ to $K$ is an automorphism of $K$ (there are also other simple ways of seeing that $\sigma(K)=K$). Let $L\subseteq K$ be the fixed field of $\sigma\vert_K$. The order of the restriction $\sigma\vert_K$ must be a factor of $m$, so $[K:L]\mid m$. Therefore $GF(q^k)\subset L$. Because $x\in GF(q^k)$, we may infer that $\sigma(x)=x$. As $x$ was an arbitrary element of $N$, we have shown that $\sigma$ is the identity mapping.

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Why the fact that the order of the restriction $\sigma|_{K}$ is a factor of $m$ implies $[K:L]|m$? It is hard to understand for me. –  Guillermo Dec 3 '14 at 8:30
@Guillermo: The extension $K/L$ is Galois, so $$[K:L]=|Gal(K/L)|=|\langle \sigma\vert_K\rangle|$$ is the order of $\sigma\vert_K$. –  Jyrki Lahtonen Dec 3 '14 at 10:17

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