Class Field Theory

Class field theory makes sense for arbitrary number fields, but for simplicity in this section and because it is all that is needed for our application to the BSD conjecture, we assume henceforth that $K$ is a totally imaginary number field, i.e., one with no real embeddings.

Let $L/K$ be a finite abelian extension of number fields, and let $\mathfrak{a}$ be any unramified prime ideal in $\O_K$. Let $b$ be an prime of $\O_L$ over $\mathfrak{a}$ and consider the extension $k_\b= \O_L/\b$ of the finite field $k_\mathfrak{a}=\O_K/\mathfrak{a}$. There is an element $\overline{\sigma} \in \Gal (k_\b/k_\mathfrak{a})$ that acts via $q$th powering on $k_\b$, where $q=\char93 k_\mathfrak{a}$. A basic fact one proves in algebraic number theory is that there is an element $\sigma \in \Gal (L/K)$ that acts as $\overline{\sigma}$ on $\O_L/\b$; moreover, replacing $b$ by a different ideal over $\mathfrak{a}$ just changes $\sigma$ by conjugation. Since $\Gal (L/K)$ is abelian it follows that $\sigma$ is uniquely determined by $\mathfrak{a}$. The association $\mathfrak{a}\mapsto \sigma = [\mathfrak{a}, L/K]$ is called the Artin reciprocity map.

Exercise 3.10   Prove that if an unramified prime $\mathfrak{p}$ of $K$ splits completely in an abelian exension $L/K$, then $[\mathfrak{p},L/K] = 1$.

Let $c$ be an integral ideal divisible by all primes of $K$ that ramify in $L$, and let $I(\c)$ be the group of fractional ideals that are coprime to $c$. Then the reciprocity map extends to a map

$$I(\c) \to \Gal (L/K)\qquad \quad a\mapsto [\mathfrak{a},L/K]$$

Let

$$P(\c) = \{(\alpha) : \alpha \in K^*,\quad \alpha\equiv 1 \pmod{\c}\}.$$

Here $\alpha \equiv 1\pmod{\c}$ means that $\ord _\mathfrak{p}(\alpha-1) \geq \ord _p(\c)$ for each prime divisor $\mathfrak{p}\mid \c$.

Definition 3.11 (Conductor of Extension)   The conductor of an abelian extension $L/K$ is the largest (nonzero) integral ideal $\c=\c_{L/K}$ of $\O_K$ such that $[(\alpha),L/K] = 1$ for all $\alpha\in K^*$ such that $\alpha \equiv 1\pmod{\c}$.

Proposition 3.12   The conductor of $L/K$ exists.

If $\c=\c_{L/K}$ is the conductor of $L/K$ then Artin reciprocity induces a group homomorphism

$$I(\c)/P(\c) \to \Gal (L/K).$$

Definition 3.13 (Ray Class Field)   Let $c$ be a nonzero integral ideal of $\O_K$. A ray class field associated to $c$ is a finite abelian extension $K_{\c}$ of $K$ such that whenever $L/K$ is an abelian extension such that $\c_{L/K} \mid \c$, then $L\subset K_\c$.

Theorem 3.14 (Existence Theorem of Class Field Theory)   Given any nonzero integral ideal $c$ of $\O_K$ there exists a unique ray class field $K_\c$ associated to $c$, and the conductor of $K_\c$ divides $c$.

Theorem 3.15 (Reciprocity Law of Class Field Theory)   Let $L/K$ be a finite abelian extension.

1. The Artin map is a surjective homomorphism $I(\c_{L/K}) \to \Gal (L/K)$.
2. The kernel of the Artin map is $N_{L/K}(I_L) \cdot P(\c_{L/K})$, where $N_{L/K}(I_L)$ is the group of norms from $L$ to $K$ of the fractional ideals of $L$.

Definition 3.16 (Hilbert Class Field)   The Hilbert class field of a number field $K$ is the maximal unramified abelian extension of $K$.

In particular, since the Hilbert class field is unramified over $K$, we have:

Theorem 3.17   Let $K$ be a number field and let $H$ be the Hilbert class field of $K$. The Artin reciprocity map induces an isomorphism

$$\Cl (\O_K) \xrightarrow{ \cong } \Gal (H/K).$$