$\operatorname{Gal}(\overline{K}/K)=\mathbb{Z}$ possible?

Question

Is it possible to have $\operatorname{Gal}(\overline{K}/K)=\mathbb{Z}$?

My question comes from the link beetween covering and field extensions. For covering the simplest example is $\operatorname{Gal}(\mathbb{R}/\mathbb{S}^1)=\mathbb{Z}$.

Maybe something like $\mathscr{M}(\mathbb{C})/\mathscr{M}(\mathbb{C}^*)$ where $\mathscr{M}(X)$ is the field of meromorphics functions on a Riemann surface?

Answer 1

You can get pretty close: there are fields whose absolute Galois groups are the profinite integers $\widehat{\mathbb{Z}}$. In particular every finite field has this property. The finite extensions of finite fields look exactly like the finite covers of $S^1$ (but the infinite extensions are more complicated).

Answer 2

No: every Galois group is profinite, and any infinite profinite group has cardinality at least $2^{\aleph_0}$.