# Extension of $\mathbb Q$ which splits only at primes in $S$

Suppose that I'm given a set of rational primes $S$ with positive Dirichlet density, and a finite set of primes $R$, disjoint from $S$. Does there exist a number field $K$ that is;

• unramified outside $R$;
• splits only at primes in some subset $S'\subseteq S$?

In my situation, I have a semisimple representation $\rho$ of $G_\mathbb Q =\mathrm{Gal}(\overline{\mathbb Q}/\mathbb Q)$, unramified outside $R$, and I have information about the action of Frobenius elements $\mathrm{Frob}_p$ for $p\in S$ (let's say I know their characteristic polynomials). I'd like to know if this completely determines my representation on some open subgroup of $G_\mathbb Q$.

For many choices of $R$ and $S$ the answer is obviously no. For example, if $R$ is empty, then the answer is no, because there are no unramified extensions of $\mathbb{Q}$.
For a more interesting example, let $S$ be the set of primes $p \equiv 3 \pmod{4}$ and $R$ be any finite subset of $\{ 2, 5, 13, 17, 29, \ldots \}$. Then there is no number field that splits only at primes $S' \subseteq S$. This is because if $K/\mathbb{Q}$ is such a number field, any prime that splits in $K[i]$ must be $\equiv 1 \pmod{4}$, and $S$ doesn't contain any such primes.
• Thanks for your answer! Expecting an open subgroup was way too optimistic, but I guess the follow up question is do I know my representation on an open set? So does there exist a Galois number field $K$ and a conjugacy class $C\subset\mathrm{Gal}(K/\mathbb Q)$ such that $\{\mathrm{Frop}_p\} = C\implies p\in S$. In your example, $\mathbb Q(i)$ with the non-trivial automorphism would work. – Ariel Weiss Feb 24 '17 at 11:31
• The answer again is no. Philosophically, the reason is that there are only countably many Chebotarev sets, and uncountably many subsets of the primes with positive density. Hence, there are many special properties that Chebotarev sets must satisfy that general sets of primes do not. For example, let $S = \{ p~\text{ prime } : \text{ the smallest } k \text{ so that } p+k \text{ is prime is } \geq \log(\log(p)) \}$. Then, $S$ has natural density $1$, but doesn't contain bounded gaps. However, Chebotarev sets have bounded gaps. – Jeremy Rouse Feb 24 '17 at 19:10