TITLE: Lifting the Frobenius to the absolute Galois group of the $p$-adics
QUESTION [6 upvotes]: I tend to ask questions on mathstackexchange, because I don't feel adequate yet for mathoverflow. I had previously asked this question there (I have now deleted it), where it was quite popular, but it didn't seem like anybody knew the answer. I thought that perhaps this question should graduate to mathoverflow. Here it is, as it had originally appeared:
I've been playing around with the $p$-adics, and I wondered about the structure of their Galois group.
We have the short exact sequence:
$$1\rightarrow Gal(\bar{\mathbb{Q}}_p/\mathbb{Q}_p^{un})\rightarrow Gal(\bar{\mathbb{Q}}_p/\mathbb{Q}_p)\rightarrow Gal(\mathbb{Q}_p^{un}/\mathbb{Q}_p)\rightarrow 1$$
I as wondering if this exact sequence is split. I.e., is it true that you can lift the Frobenius automorphism of $Gal(\mathbb{Q}_p^{un}/\mathbb{Q}_p)$ to $Gal(\bar{\mathbb{Q}}_p/\mathbb{Q}_p)$? How would the lifting of this automorphism act on individual elements of $\bar{\mathbb{Q}}_p$?

REPLY [4 votes]: As others have mentioned, the exact sequence is continuously split.  Furthermore, any lift of Frobenius will act as you expect on the subfield $\mathbb{Q}_p^{nr}$, i.e., the extension you get by adjoining all prime-to-$p$ roots of unity.  Indeed, you can form a $\mathbb{Q}_p$-basis of $\mathbb{Q}_p^{nr}$ using roots of unity, and then the action is determined by $p$-th powers on the basis elements.
As far as other elements are concerned, the set of lifts is a torsor under $\operatorname{Gal}(\overline{\mathbb{Q}_p}/\mathbb{Q}_p^{nr})$, so you have a large amount of freedom.  For example, your typical $S_3$-extension whose intersection with $\mathbb{Q}_p^{nr}$ is quadratic will have three Frobenius lifts.