Can we find a Galois extension with Galois group $\mathbf{Z}$?

Question

Let $F$ be a finite field, $\bar{F}$ its algebraic closure. A standard result in infinite Galois theory shows that $\mathbf{Z}$(corresponds to the free cyclic subgroup $\left \langle \sigma \right \rangle$ in Gal($\bar{F}$/$F$) that is generated by the relative Frobenius homomorphism) is not the Galois group of Gal($\bar{F}$/$F$) but dense in it with Krull topology. So I want to know whether there is a Galois extension with the Galois group be exactly $\mathbf{Z}$. I think the answer is false, but I don't know how to prove that...

Answer 1

No. A Galois group will always be profinite, i.e. isomorphic to the inverse limit of its finite index subgroup. A cardinality argument shows $\mathbb{Z}$ cannot be profinite.