Let $K/\mathbb{Q}$ be a number field. We say that a rational prime $p$ splits in $K$ if there exists a prime $\mathfrak{p}$ of $K$ above $p$ of interia degree $1$.
Is a number field $K$ uniquely determined by the set of primes which split in $K$?
A well-known application of the Chebotarev density theorem (Neukirch Cor. 13.10) says that this is true when $K/\mathbb{Q}$ is Galois (note that here a prime splits if and only if it splits completely). So I am really interested in what happens in the non-Galois case.
Note also that the answer to the analogous question for completely split primes is no; indeed a prime is completely split in $K$ if and only if it is completely split in the Galois closure of $K$.
Best Answer
See exercises (6.3) and (6.4) of Cassels-Frohlich book on algebraic number theory. In these exercises, an example of two number fields $E,E'$ with the same zeta function is given; therefore, the set of primes which split in $E,E'$ are the same.
This amounts to constructing two subgroups $H,H'$ in a finite group $G$, which are not conjugate but meet every conjugacy class of $G$ in the same number of elements.
Since the OP has asked for split (but not necessarily completely split), the fact that equality of zeta functions implies that the split primes are the same needs a small argument. Let $p$ be a prime and $f_1,\cdots, f_r$ be the residue class degrees in $E$ over $p$ (in decreasing order); simlarly $g_1,\cdots, g_s$ the residue class degrees in $E'$ over $p$ (in decreasing order). Assume $f_1\geq g_1$. The local zeta functions at $p$ are
$$(1-X^{f_1})\cdots (1-X^{f_r})=(1-X^{g_1})\cdots(1-X^{g_s})$$where $X=p^{-s}$. Since the multiplicity of the root $X=1$ is the same, we get $r=s$. The cyclotomic polynomial $\Phi _{f_1}(X)$ divides the left hand side and hence the right hand side. Therefore, by the uniqueness of irreducible factors on the LHS-RHS, $g_1=f_1$. We can thus cancel the factor $1-X^{f_1}$ on both sides and use induction on $r$ to conclude that $f_i=g_i$ for all $i$.