The Galois group of the finite field's algebraic closure is not countable I'm trying to prove that the group $\operatorname{Gal}(\bar F_p /F_p)$ is not countable. 
My idea is to show that in the sequence 
$F_p\leq F_{p^2}\leq F_{p^4} \leq \dots \leq F_{p^{2^n}} \leq\dots $  
one can extend any automophism into to 2 distinct automorphisms of the extension. This would imply by induction:
$|\bar F_p|\geq |\cup \{F_{p^{2^n}} : n \in \mathbb{N}\}|=2^ {\aleph_0}>\aleph_0$
In any extension I can always take some element of the extension and observe its minimal polynomial. Its degree is at least 2 and hence there are 2 distinct automorphisms that each send that element to another root. Can I know for sure that there are such automorphisms that also preserve any automorphism in the previous extension? If so, then I think my proof is valid. 
I'd be glad also for more ideas for proofs
 A: What you seem to be doing could be formalized as follows. Consider a fixed sequence of bits $\overline{b}=(b_0,b_1,\ldots)\in\{0,1\}^{\Bbb{N}}$. Define the corresponding sequence of exponents
$s_k=\sum_{i=0}^kb_i2^i$ for all $k$. Define an automorphism $\sigma_{\overline{b}}$ on the union
$$
E=\bigcup_{i=0}^\infty F_{p^{2^i}}
$$
as follows. Let $\sigma_{\overline{b}}(z)=z^{p^{s_k}}$ for all $z$ in $F_{p^{2^i}}$. This is actually a well defined automorphism of $E$, because if (actually when) an element $z\in E$ belongs to two (or more) fields, say $F_{p^{2^j}}$ and $F_{p^{2^\ell}}$ with $j<\ell$, then
for all $r\ge j$ we have $z^{p^{2^r}}=z$, so consequently $z^{p^{s_\ell}}=z^{p^{s_j}}$.
If two sequences $\overline{b}$ and $\overline{b}'$ are distinct, say $b_\ell\neq b'_\ell$, then there will be a corresponding difference in the automorphisms $\sigma_{\overline{b}}$ and $\sigma_{\overline{b}'}$ in their respective restrictions to the field $F_{p^{2^{\ell+1}}}$ because $s_{\ell+1}\neq s'_{\ell+1}$. 
Therefore you get $2^{\aleph_0}$ distinct automorphisms of $E$. You can then invoke a general result telling that all of them can be lifted to automorphisms of $\overline{F}_p$ to get your claim.
