Extensions of number fields ramified at few places Let $K$ be a number field and $S$ a finite set of places of $K$. Then there is a maximal algebraic extension $K^S / K$ unramified outside $S$.

How large do we have to take $S$ to be in order that $K^S / K$ is infinite?

For $K = \mathbf{Q}$ it's necessary and sufficient that $S$ contain at least one finite place. This is not a necessary condition for every $K$, since the Golod--Shafarevich theorem shows that there are fields such that $K^S$ is infinite when $S$ is the empty set; but is it sufficient? That is:

Does there exist a field $K$ and a finite place $v$ such that $K^{\{v\}} / K$ is finite?

There certainly exist examples such that $K^{\{v\}} / K$ contains no solvable extension of $K$, e.g. $K = \mathbf{Q}(\sqrt{3})$, $v$ the prime $1 + 2\sqrt{3}$ of norm 11.
 A: @David   Does your question mean (even in vague terms) : "Given $K$, what is the "smallest" $S$ such that $K^S/K$ is infinite ?". Because the examination of miscellaneous known cases seems to suggest the contrary of a uniform answer. Throughout, fix a prime number $p$ (assumed to be odd to avoid petty trouble) and shift the problem from $K^S$ to $K^S (p)$ = the maximal pro-$p$-extension of $K$ unramified outside $S$, with Galois group $ {G^S_K}(p)= Gal(K^S (p)/K)$ .
1) Take $S$ containing all the $p$-places of $K$. It is well known by CFT that ${G^S_K}(p)^{ab}$ is a $\mathbf Z_p$-module of rank $1+r_2+ \delta$, where $r_2$ is the number of complex places of $K$ and $\delta \in \mathbf N$ is conjecturally null. So the point is whether $S$ can be downsized or not.
2) Let us first remove some $p$-places from $S$. Consider an imaginary quadratic field $k$ s.t. $p$ splits completely in $k$, say $(p)=P.P^*$, and an elliptic curve $E$ defined over a number field $F$, with complex multiplication by the ring of integers of $k$. Suppose moreover that $E$ has good ordinary reduction at any $p$-place. Then the field $F(E_{P^{\infty}})$ obtained by adding all the $P^n$-torsion points of $E$ is a $\mathbf Z_p$-extension of $K=F(E_P)$. Thus $K^S/K$ is infinite, with  $S$={$P$}.
3) Let us be go to extremities and cut $S$ down to the empty set, so that $K^S (p)$ is the so-called "$p$-class field tower". The Golod-Safarevic theorem says that there exists a function $\gamma (n)$ s.t., for all number fields $k$ of degree $n$ with finite $p$-class field tower, $\gamma(n)$ > the minimal number of generators of the $p$-class group of $k$. One can even take $\gamma(n)=2+2\sqrt(n+1)$. Moreover, for a finite $p$-group $G$, it is known by cohomological considerations that $r(G) > \frac 14 d(G)^2$ , where $d(G)$ (resp. $r(G)$) dotes the minimal number of generators (resp. relations) of $G$. The combination of these two properties produces examples of finite as well as of infinite $p$-class field towers.
4) Another approach lies in the "tame" situation, in which $S$ contains no $p$-place, hence $K^S (p)$ contains no $\mathbf Z_p$-extension, hence ${G^S_K}(p)^{ab}$ is finite (this property is called FAB). At the beginning of this century, somewhat to the general surprise, J. Labute produced a family of Galois groups $G^S_{\mathbf Q}(p)$ called mild, of cohomological dimension $2$, hence torsion free, hence infinite. Roughly speaking, the initial terms of the relations of a mild pro-$p$-group $G$ must satisfy special combinatorial properties in order that the graded algebra built on the lower $p$-central series of $G$ has a nice convenient description in terms of the corresponding free graded algebra. Subsequent developments by A. Schmidt et. al. provide an infinite supply of examples over arbitrary number fields. One should
also remark the relevance of arithmetical FAB groups with regard to the Fontaine-Mazur conjecture on $p$-adic representations  ./.
