$\newcommand\Gal{\operatorname{Gal}}$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?


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 $N(\pi)>0$.


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