# Lifting the Frobenius to the absolute Galois group of the $p$-adics

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 $\bar{\mathbb{Q}}_p$?

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The third term in your exact sequence is just $\hat Z$, so there's a splitting if $Gal(\bar\mbb{Q}_p/\mbb{Q}_p)$ is complete (which it is, as it's topologically finitely generated); picking such a lift amounts to picking a lift of a topological generator of $\hat Z$. As JSE points out, there are tons of lifts, so your second question is mal-formed. – Daniel Litt Oct 5 '12 at 2:23
Welcome to MathOverflow. In the past I've enjoyed reading the discussion that your questions have inspired at mathstackexchange. – Jonah Sinick Oct 5 '12 at 6:59
For what it's worth, I've seen several papers which use the existence of such a lift of Frobenius, which is referred to as a "Lubin--Tate splitting". This is important in Ivan Fesenko's approach to nonabelian local class field theory. – David Loeffler Oct 5 '12 at 11:31

## 2 Answers

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.

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I'm not sure I've understood your question. The second map is surjective, so there is certainly an element of Gal(Qpbar/Qp) which projects to Frobenius. In fact there are lots of them. How would such a lift act on individual elements of Qpbar? It depends which lift it is.

Is that what you were asking, or did I miss the point?

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Dear Jordan, the surjectivity of the second map by itself does not imply splitting. The sequence splits if and only if there is a subgroup of $Gal(\bar{\mathbb{Q}}_p/\mathbb{Q}_p)$ that isomorphically maps onto the rightmost term, which in turn is equivalent to $Gal(\bar{\mathbb{Q}}_p/\mathbb{Q}_p)$ being a semidirect product of the left term by a subgroup isomorphic to the right term. – Alex B. Oct 5 '12 at 10:18