Maximal extension almost everywhere unramified and totally split at one place

Question

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.

Answer 1

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".