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It seems that it is conjectured, that the absolute Galois group of a number field determines already the number field up to isomorphism.

I would like to know if there is a profinite group G such that only a unique number field has G as a Galois group.

Does the topology matter?

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    $\begingroup$ Your result in the first sentence is not a conjecture but rather a theorem, due to Neukirch and Uchida. See Minhyong's answer to mathoverflow.net/questions/23711/… $\endgroup$
    – KConrad
    Commented Jul 28, 2011 at 15:53
  • $\begingroup$ Finite Galois extensions of Q are examples where their absolute Galois group is not shared (up to isom. as profinite groups) by the absolute Galois group of any other number field. That's a consequence of the Neukirch--Uchida work and the fact that a finite Galois extension of Q is not isomorphic to any other number field but itself inside a fixed algebraic closure of Q. $\endgroup$
    – KConrad
    Commented Jul 28, 2011 at 15:55
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    $\begingroup$ The recent paper Class Field Theory as a Dynamical System, G. Cornellisen (arxiv.org/abs/1107.2159), has a short exposition on the matter, and shows some very exciting recent results. $\endgroup$
    – Dror Speiser
    Commented Jul 28, 2011 at 16:05

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As mentioned by Conrad in the comments, your first statement is a theorem by Neukirch-Uchida (Neukirch-Ikeda-Iwasawa-Uchida, according to some sources).

Your second question is much more complicated, depending on how it is interpreted. As stated the answer is "yes, in fact infinitely many by Neukirch-Uchida".

But if you mean if we know a concrete profinite group $G$ such that $G\cong \mathrm{Gal}(\overline{K}/K)$ for some number field $K$, then the answer is "definitely no". But there are beautiful math behind this and related question, see for example the answers to this question of mine.

I'm not sure what you mean by "does the topology matter". I don't think there is an alternative to using the Krull topology or something equivalent, if you want a working infinite Galois theory.

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