Galois group of $\overline{F}/F$ 
Let $K$ be the algebraic closure of $F$ where $F$ is a finite field. Show that $Gal(K/F) \simeq \hat{\mathbb{Z}}$.

I know that $\hat{\mathbb{Z}} = \varprojlim \mathbb{Z}_{n}$, so its enough to show that $Gal(K/F) \simeq \varprojlim \mathbb{Z}_{n}$. I can solve the problem if $|F|=p$, but and if $|F| = p^{n}$? Can someone help me?
 A: The reasoning is almost the same as when $|F| = p$.  
Let $k$ be the field with $p$ elements with algebraic closure $\overline{k}$.  If $F$ is a finite subfield of $\overline{k}$, there is for each $n \geq 1$ a unique subfield $L$ of $\overline{k}$ containing $F$ with $\operatorname{Gal}(L/F) \cong \mathbb Z/n\mathbb Z$.  
As you vary $L$, the inverse system of these Galois groups $\operatorname{Gal}(L/F)$ coincides with the inverse system of the $\mathbb Z/n\mathbb Z$, so taking inverse limits you get a canonical isomorphism of $\operatorname{Gal}(\overline{k}/F)$ with $\hat{\mathbb Z}$ just the same as when $F$ has $p$ elements.
A: More generally, Gal$(\bar F/F)$ is the inverse limit of its finite quotients, so it suffices to pick up, if possible, a convenient inductive system of finite extensions of $F$. If $F$ is a finite field (its characteristic is irrelevant), we know that every finite extension of $F$ is cyclic , with Galois group generated by the Frobenius automorphism. Hence Gal$(\bar F/F)\cong \hat {\mathbf Z}$. Of course this doesn't work all the time : we don't know the absolute Galois group of $\mathbf Q$ !
