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Does the Galois group of every infinite $p$-extension $K$ of a number field $k$ contain a (closed) subgroup such that the quotient group is isomorphic to $\mathbb{Z}_p$?

My feeling is "yes", but I'm not sure if we need the extension $K/k$ to be abelian. My intuition would be that the "infinite" part of $Gal(K/k)$ comes from at least one copy of $\mathbb{Z}_p$. Am I right about this?

Thank you in advance!

Update (11/21/2013): In response to hunter's answer, I want to ask, under which conditions does a $p$-extension contain the $\Bbb Z_p$-extension? Update (28/11/2013): The question above is possibly too broad and not answerable.

BIS HD
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  • @BrunoJoyal Sorry, I was mixing two different things. You are, of course, right, that the finiteness of the class number $h_k$ implies that the extension of the Hilbert class field of $k$ is finite and therefore every quotient as well. I removed the background. – BIS HD Nov 18 '13 at 13:55

1 Answers1

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No, take the compositum of all quadratic extensions of $\mathbb{Q}$.Every element of the Galois group is $2$-torsion, so this cannot possibly have a copy of $\mathbb{Z}_2$ as a quotient. If you allow only finitely many primes to ramify, then the answer becomes yes (because your Galois group is then finitely topologically generated, and now appeal to the structure theory of $p$-groups).

hunter
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  • Thank you! So if we restrict ourselves to extensions $K/k$ where $K \subset k_S$ is in the maximal unramified $p$-extension outside $S$ for a finite set of primes $S$ of $k$, then - if $K/k$ is infinite, it contains $\mathbb{Z}_p$? Have you got a reference for the last statement concerning the structure theory of $p$-groups?) – BIS HD Nov 18 '13 at 14:37
  • Now that I think about it, we need more than just group theory, since an infinite pro $p$ group can have finite abelianization. I retract my broader claim! – hunter Nov 19 '13 at 10:51
  • Could you be more detailed about your thoughts of a pro $p$ group having finite abelianization? Thanks :-) – BIS HD Nov 20 '13 at 10:43
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    When I made my claim "the answer becomes yes" I was wrongly thinking that the abelianization of an infinite pro $p$ group is still infinite, and then the claim is just that an infinite, finitely generated $\mathbb{Z}_p$ module has $\mathbb{Z}_p$ as a quotient. However, that claim was wrong (I googled to find explicit counterexamples). – hunter Nov 20 '13 at 11:14