How does class field theory help us deduce the splitting of nonprincipal prime ideals? $\newcommand\Gal{\operatorname{Gal}}$I had a general question about the significance of global class field theory.  One of the goals, as I understand, is to answer the following question: 

Given $L/K$ abelian, $g$ a divisor of $[L : K]$, describe all those primes $\mathfrak p$ of $K$ which are unramified in $L$ and split into exactly $g$ primes.

When $K = \mathbb{Q}$, I can see how to answer the question: find some number $m$, divisible only by ramified primes, for which $L \subseteq F :=\mathbb{Q}(\zeta)$, where $\zeta$ is a primitive $m$th root of unity.  
Via the mapping $x \mapsto (\zeta \mapsto \zeta^x)$, one can identify $(\mathbb{Z}/m \mathbb{Z})^{\ast}$ with $Gal(F/\mathbb{Q})$, and therefore identify $H = \Gal(F/K)$ as a subgroup of units modulo $m$.  
Then, an unramified, positive prime number $p$ splits into $g$ primes in $K$, if and only if $g$ is the smallest number for which $p^g \in H$.  Obviously such primes may be characterized by "congruence conditions" modulo $m$.
Similarly, I can sort of see that for a principal prime ideal $\pi \mathcal O_K$ of $K$, with $\pi$ being "positive" (in terms of the various embeddings $K \rightarrow \mathbb{C}$), one can embed $K$ into a cyclotomic extension $K(\zeta)$ and put congruence conditions modulo $m \mathcal O$ on how $\pi$ splits.
But what does class field theory say about the splitting of a nonprincipal prime ideal?  The thing I have heard about is how we can use "congruence conditions" to characterize how prime ideals split in abelian extension, but what do congruence conditions even mean here?
 A: This is why we define ray class groups. I follow notation from Pete Clark's notes.
Let $\mathfrak{m}$ be a modulus of $K$ -- that is to say, the formal product of an ideal $\mathfrak{m}_0$  and some
collections $\infty_1$, $\infty_2$, ..., $\infty_a$ of embeddings $K \to \mathbb{R}$. Let $I(\mathfrak{m})$ be the group of fractional ideals generated by the prime ideals not dividing $\mathfrak{m}_0$. Let $P(\mathfrak{m})$ be the subgroup of $I(\mathfrak{m})$ given by principal fractional ideal $(\alpha)$ where $\alpha \equiv 1 \bmod \mathfrak{m_0}$ and where $\infty_j(\alpha)>0$ for each $j$. The ray class group $Cl(\mathfrak{m})$ is $I(\mathfrak{m})/P(\mathfrak{m})$.
Example 1: Take $K=\mathbb{Q}$ and the modulus $m \infty$, where $\infty$ is the unique embedding $\mathbb{Q} \to \mathbb{R}$. Then $I(m \infty)$ is the subgroup of $\mathbb{Q}^{\times}/(\pm 1)$ generated by primes which are relatively prime to $m$. The subgroup $P(m \infty)$ is positive rational numbers which $1 \bmod m$. The quotient $I(m \infty)/P(m \infty)$ is $\mathbb{Z}/(m \mathbb{Z})^{\times}$.
Example 2: Take the trivial modulus. Then $I(m)$ is the fractional ideal group and $P(m)$ is the group of principal ideals. The quotient $I(m \infty)/P(m \infty)$ is the class group.
Example 3: Let $K = \mathbb{Q}(\sqrt{3})$ and consider the modulus $\infty_1 \infty_2$, where $\infty_1(a+b \sqrt{3}) = a+b \sqrt{3} \in \mathbb{R}$ and $\infty_2(a+b \sqrt{3}) = a-b \sqrt{3}$. Then $I(\infty_1 \infty_2)$ is the fractional ideal group. Since $\mathcal{O}_K$ is a PID, the fractional ideal group is $K^{\times}/\mathcal{O}_K^{\times} = K^{\times} / (\pm (2+\sqrt{3})^{\mathbb{Z}})$. The subgroup $P(\infty_1 \infty_2)$ is elements of $K^{\times} / (\pm (2+\sqrt{3})^{\mathbb{Z}})$ which have a representative $\alpha$ with $\infty_1(\alpha)$ and $\infty_2(\alpha)>0$. Since all units of $\mathcal{O}_K$ have norm $1$, we can also say that $P(\infty_1 \infty_2)$ is ideals $(\alpha)$ where $N(\alpha)>0$.
The Artin Reciprocity theorem in its weakest form says the following: For any abelian extension $L/K$, there is a modulus $\mathfrak{m}$ of $L$ such that, for any $\mathfrak{p} \in I(\mathfrak{m})$, the splitting of $\mathfrak{p}$ in $L$ is determined by its class modulo $P(\mathfrak{m})$. Returning to the examples:
Example 1 continued For $p$ a prime of $\mathbb{Z}$ with $GCD(p,m)=1$, the splitting of $p$ in $\mathbb{Q}(\zeta_m)$ is determined by $p$ modulo $m$.
Example 2 continued If $H$ is the Hilbert class field of $K$, then the splitting of a prime $\mathfrak{p}$ in $H$ is determined by the ideal class of $\mathfrak{p}$. For example, the Hilbert class field of $\mathbb{Q}(\sqrt{-5})$ is $\mathbb{Q}(\sqrt{-5}, i)$. We obtain that a prime $\mathfrak{p}$ of $\mathbb{Q}(\sqrt{-5})$ splits in $\mathbb{Q}(\sqrt{-5}, i)$ if and only if $\mathfrak{p}$ is principal.
Example 3 continued A prime $(\pi)$ of $\mathbb{Q}(\sqrt{3})$ splits in $\mathbb{Q}(\sqrt{3}, i)$ if and only if $N(\pi)>0$.
