Prime distribution in CM field For any prescribed infinite set of primes $P\subseteq \mathbb{Z}$, is there a general heuristic to finding a field satisfying certain algebraic or geometric constraints such that these primes all split?
I am interested, in particular, in obtaining a CM field (quadratic imaginary extension of either $\mathbb{Q}$ or, more generally, of a totally real field) in which infinitely many primes in a prescribed infinite set of primes $P$ split, and for my application, I can moreover assume that $P$ has density $0$.
I do not need to write down the explicit CM field (and it also does not need to be unique), as long as I know I can find a CM field for any degree $d$ over $\mathbb{Q}$. My background is largely in geometry, so I do not know if a question of this form is routine or difficult.
 A: I'm not 100% sure I understand your question correctly, so let me first explain what I understand your problem to be. First, by a prime "splitting" in a field $K$, I understand you to mean that $p$ splits completely. Second, by "CM field," I assume you do not allow the field to be totally real (some people's definition of CM fields allows this). Finally, I understand your problem to be this: Given an infinite set $P$ of primes, does there exist a CM field $K$ such that infinitely many (but not necessarily all) primes $p$ in $P$ split in $K$. The answer to this question is that it is not always possible, so one will require more information about the particular set $P$.
Lemma: Let $K_1, K_2, \ldots K_n$ be CM fields. Then there exist infinitely many primes $p$ which do not split in any of the fields $K_i$.
Proof: A prime $p$ splits completely in a field $K$ if and only if it splits completely in the Galois closure of $K$, and thus we may assume the $K_i$ are all Galois. The Galois closure of a CM field is CM. Let $G$ denote the Galois group of the compositum of the $K_i$. By the Ceborarev Density Theorem, there exists infinitely many primes $p$ for which the Frobenius conjugacy class of $p$ is the conjugacy class of (any) complex conjugation in $G$. The image of complex conjugation in $G$ in the quotient $\mathrm{Gal}(K_i/\mathbf{Q})$ is also complex conjugation, which is non-trivial, because $K_i$ is a CM field. Hence such a prime $p$ does not split completely in $K_i$. $\square$
We now construct an infinite set $P$ of primes as follows.
Enumerate the (countably infinite) CM fields as $K_1, K_2, \ldots$. Then choose $p_n \in P$ to be any prime not in $\{p_1,\ldots,p_{n-1}\}$ that does not split in any of the fields $K_1, \ldots, K_n$. That this is possible follows from the previous lemma. By construction, for any CM field $K$, only finitely many primes in $P$ split in $K$.
You say that "I can moreover assume that $P$ has density $0$." This is a slightly strange thing to say, because this assumption makes the problem more difficult. If you assume that $P$ does have positive density, then you can find a CM field such that infinitely many primes in $P$ split in an imaginary quadratic field $K$. Indeed, for a prime $p$ not to split in any of the fields
$$\mathbf{Q}(\sqrt{-1}), \mathbf{Q}(\sqrt{-2}), \mathbf{Q}(\sqrt{-3}), \ldots \mathbf{Q}(\sqrt{-n}),$$
it must be the case that the quadratic symbol $(-k/p) = -1$ for all $k \le n$. The density of such primes tends to zero as $n$ tends to infinity. Hence, taking $n$ large enough so that the density of such primes is smaller than the density of $P$, it follows that a positive proportion of primes in $P$ splits in at least one of the fields above.
If one increases the set of CM fields to the set of all non-trivial finite extensions of $\mathbf{Q}$ (or even just all such fields which are either CM or totally real), then one can take $K$ to be either $\mathbf{Q}(\sqrt{-2})$, $\mathbf{Q}(\sqrt{-3})$, or $\mathbf{Q}(\sqrt{6})$. This is because the identity $(-2/p)(-3/p) = (6/p)$ means that every prime $p$ splits in at least one of these fields.
