Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

share|improve this question
2  
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
2  
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 2

up vote 4 down vote accepted

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.

share|improve this answer

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?

share|improve this answer
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.