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

Fix a finite set of primes $S$ and an additional prime $p$. Let $K$ be the maximal extension of $\mathbb{Q}$ that is unramified outside $S$ and $\infty$ and totally split at $p$. Is the extension $K$ finite?

My intuitive guess would be no, but the simple constructions (based on class field theory) I tried so far do not prove this, at least for the ground field $\mathbb{Q}$. In contrast, for imaginary quadratic fields, there are extensions with Galois group $\mathbb{Z}_{\ell}^2$ ramified only above $\ell$, and any place not above $\ell$ splits completely in an infinite subextension, as $\mathbb{Z}_{\ell}^2$ has no procyclic subgroups of finite index.

share|improve this question
add comment

1 Answer 1

up vote 8 down vote accepted

Nope.

I'm lacking a reference in front of me at the moment (see NSW's Cohomology of Number Fields, or Gras's Class Field Theory -- I'll update with a precise reference later), but there are remarkably clean formulas for the generator and relation ranks for the Galois group of the maximal $\ell$-extension of $\mathbb{Q}$ unramified outside $S$ and completely split at $T$, for finite sets of primes $S$ and $T$. Throwing out some silly cases, these depend only on $|S|$ and $|T|$ (and, in your problem, maybe even just $|S|-|T|$). In your case, where $|T|=1$, it's just a matter of making $S$ big enough (again, a reference will say how big, but right now, I think $|S|=4$ does the trick.)

Edit to add in in a precise reference (though the above book references certainly contain the results as well): Christian Maire's "Finitude de tours et p-tours T-ramifiees moderees, S-decomposees".

share|improve this answer
add comment

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.