Algebraic closure of finite field 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.
 A: 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.
A: We know that $\operatorname{Gal}(N/F) \cong \widehat{\Bbb Z} \cong \displaystyle \prod_{p~\text{prime}} \Bbb Z_p$, so it suffices to prove that $\Bbb Z_p$ is torsion-free.
Well, it follows from the fact that $\Bbb Z_p$ is a ring of characteristic zero, which follows from the fact that it is an integral domain and that the map $\Bbb Z \to \Bbb Z_p$ is injective.
