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Sorry for the non-specific title, but this would be hard to fit in one line.

In "Groups as Galois Groups", theorem 10.27 says that there's some absolutely irreducible component of $\mathcal{H}_r^{in}(G)$ (if you don't know what this is, don't worry about it), call it $\mathcal{H}$, defined over $\mathbb{Q}$, such that it has an $\mathbb{R}$-rational point, and for every $p$ it has a $\mathbb{Q}_p$-rational point. It goes on to say that if we had a local-global principle here, then we would get a $\mathbb{Q}$-rational point.

It seems that Moret Bailly would do: http://people.math.jussieu.fr/~harris/SatoTate/notes/MoretBailly.pdf if you pick $S_1=\{ \infty \}$ and $S_2=$ all the finite places.

What am I missing?

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  • $\begingroup$ The two sets $S_i$ should be finite if I remember correctly. Have you looked at the original paper where Moret-Bailly proved his result? $\endgroup$
    – BCnrd
    Commented Nov 29, 2010 at 18:29
  • $\begingroup$ No, but that would make sense. $\endgroup$
    – Makhalan Duff
    Commented Nov 29, 2010 at 18:35
  • $\begingroup$ If this is the usual Moret-Bailly result that these modularity lifting theorem guys apply, then all it says is that there is a K-point for some number field K in which all primes in a given finite set split. It certainly can't control the degree of K. $\endgroup$
    – Kevin Buzzard
    Commented Nov 29, 2010 at 18:41
  • $\begingroup$ Well, if S_2 is the set of all finite places, and you have a number field unramified over all of S_2, then that number field must equal Q. My guess is Brian's right. $\endgroup$
    – Makhalan Duff
    Commented Nov 29, 2010 at 18:46
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    $\begingroup$ Yes definitely the $S_i$ must be finite! The theorem is completely false otherwise---just take a standard counterexample to the Hasse principle like $3x^3+4y^3+5z^3=0$. $\endgroup$
    – Kevin Buzzard
    Commented Nov 29, 2010 at 18:55

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