It is now time to prove the reciprocity law, the primary result in class field theory.  I know I haven’t posted on this topic in a little while, so new readers (if they don’t already know this material) may want to review the strategy of the proof and the meaning of the Artin lemma (which is useful in reducing this to the cyclotomic case).

1. The cyclic reciprocity law

Well, I’ve already stated it before multiple times, but here it is:

Theorem 1 (Reciprocity law, cyclic case) Let {L/k} be a cyclic extension of number fields of degree {n}. Then the reciprocity law holds for {L/k}: there is an admissible cycle {\mathfrak{c}} such that the kernel of the map {I(\mathfrak{c}) \rightarrow G(L/k)} is {P_{\mathfrak{c}} N(\mathfrak{c})}, and the Artin map consequently induces an isomorphism\displaystyle J_k/k^* NJ_L \simeq I(c)/P_{\mathfrak{c}} N(\mathfrak{c}) \simeq G(L/k).


The proof of this theorem is a little sly and devious.

Recall that, for any admissible cycle {\mathfrak{c}}, we have

\displaystyle (I(\mathfrak{c}): P_{\mathfrak{c}} N(\mathfrak{c})) = n

by the conjunction of the first and second inequalities, and the Artin map {I(\mathfrak{c}) \rightarrow G(L/k)} is surjective. If we prove that the kernel of the Artin map is contained in {P_{\mathfrak{c}} N(\mathfrak{c})}, then we’ll be done by the obvious count.

This is what we shall do.

Let {\mathfrak{a}} be a fractional ideal prime to {\mathfrak{c}} which is in the kernel of the Artin map. We will prove that {\mathfrak{a} \in P_{\mathfrak{c}} N(\mathfrak{c})} in a slow, careful way. The first step is to factor {\mathfrak{a} = \prod_i \mathfrak{p}_i^{n_i}}. We will apply Artin’s lemma for each prime {\mathfrak{p}_i}, leading to a lattice

where each {\mathfrak{p}_i} splits completely in {E_i}, and {LE_i/E_i} is cyclotomic, contained in {E_i(\zeta_{m_i})} for some suitable {m_i}. We can choose the {m_i} successively so that {m_i} is not divisible by any of the primes dividing {m_j, j<i}.

Lemma 2 The field {E}, the compositum of all the {E_i}, satisfies {E \cap L =k}.

This is a fairly straightforward consequence of Galois theory. OK, cool. So we still need to do the reduction though.

1.1. The ideal {\mathfrak{d}}

We can choose {\mathfrak{d}}, an ideal of {E}, such that {(\mathfrak{d}, LE/E) = \sigma} for {\sigma \in G(L/k)} a generator. We can do this by the surjectivity of the Artin map, and moreover so that {\mathfrak{d}} is prime to the primes above {\mathfrak{c}}. Then {\mathfrak{d}_k := N^E_k \mathfrak{d}} satisfies {(\mathfrak{d}_k, L/k) = \sigma} too. This ideal {\mathfrak{d}_K} is going to be a generic bookkeeping device with which we adjust {\mathfrak{a}} to get something in {P_{\mathfrak{c}} N(\mathfrak{c})}.

1.2. Bookkeeping

Suppose {(\mathfrak{p}_i, L/k) = \sigma^{q_i}}. Then if {\mathfrak{P}_i} is a prime of {E_i} prolonging {\mathfrak{p}_i}, we have {(\mathfrak{P}_i, LE_i/E_i) = \sigma^{q_i}} as well by complete splitting. In particular, we have

\displaystyle (\mathfrak{P}_i N^{E}_{E_i} \mathfrak{d}^{-q_i}, LE_i/E_i) = 1

but since {LE_i/E_i} is cyclotomic, we have the reciprocity law, and this means we can write

\displaystyle \mathfrak{P}_i =( N^E_{E_i} \mathfrak{d})^{-q_i}( y_i) N^{LE_i}_{E_i}(\mathfrak{l}_i)

for {y_i} very close to 1 at all the primes ramified in {L/k}, and {\mathfrak{l}_i} not divisible by those primes.

So, let’s take the norm down to {k}; then

\displaystyle \mathfrak{p}_i = \mathfrak{d}_k^{-q_i} (N^{E_i}_k y_i)( N^{LE_i}_k (\mathfrak{l}_i)) .

But now {N^{E_i}_k y_i} is close to 1 at the primes of {\mathfrak{c}}, i.e. in {\mathfrak{c}}, if {\mathfrak{k}} was chosen properly (i.e. REALLY close to 1 at those primes, which is kosher in view of the reciprocity law for cyclotomic extensions). Moreover {N^{LE_i}_k(\mathfrak{l}_i)} is a norm from {L}.

The proof is now almost over. We have represented {\mathfrak{p}_i} as a power of a fixed ideal times something in {P_{\mathfrak{c}} N(\mathfrak{c})}.

1.3. Putting everything together

In particular, we have {\mathfrak{p}_i} is {\mathfrak{d}_k^{-q_i}} times some norm times something close to 1 at relevant primes. Taking the product, we see that {\mathfrak{a}} is {\mathfrak{d}_k^{-\sum n_i q_i}} times some norm times something close to 1 at relevant primes. But now {\sigma^{\sum n_i q_i} = 1} because this is equal to {(\mathfrak{a}, L/k)}, so that sum is divisible by {n}, and {\mathfrak{d}_k^{-\sum n_i q_i}} is a norm. We have proved that {\mathfrak{a}} is in the {P_{\mathfrak{c}} N(\mathfrak{c})} subgroup, and this completes the proof.

2. The general reciprocity law

Again, it wouldn’t be appropriately dramatic if I didn’t state it in a box, right?

Theorem 3 (Artin reciprocity) For a finite abelian extension {L/k} of number fields, the Artin map has a conductor {\mathfrak{c}} and so factors into an isomorphism\displaystyle J_k/k^* NJ_L \simeq I(\mathfrak{c})/P_{\mathfrak{c}} N(\mathfrak{c}) \rightarrow G(L/k).


The proof is now surprisingly easy. We can represent {L} as a compositum {L_1 \dots L_j} of cyclic extensions over {k}, by writing {1 \subset G(L/k)} as an intersection of subgroups {H_i} such that {G(L/k)/H_i} is cyclic and taking the fixed fields {L_i} of {H_i}. The compositum {L_1 \dots L_j} corresponds by Galois theory to the intersection {1} of {H_i}, and {G(L_i/k) \simeq G(L/k)/H_i} is cyclic.

Now, there exist cycles {\mathfrak{c}_1, \dots, \mathfrak{c}_j} such that each {P_{\mathfrak{c}_i}} maps to 1 in {G(L_i/k)} by the reciprocity law (and existence of conductor) for cyclic extensions. Taking the product {\mathfrak{c} = \prod \mathfrak{c}_i}, we find a conductor {\mathfrak{c}} such that the Artin map on {P_{\mathfrak{c}}} acts trivially on each {L_i}, hence on {L}.

Thus {\mathfrak{c}} is a conductor for the Artin symbol on {L/k}, so the Artin map factors through

\displaystyle I(\mathfrak{c})/P_{\mathfrak{c}} N(\mathfrak{c}) \rightarrow G(L/k)

and since, first of all, it is surjective; and second of all, the former group is of order at most that of the latter (by the second inequality), we have that the Artin map is an isomorphism.

The proof of the reciprocity law, the prime result of class field theory, is now complete.

Next time, we will take stock of how far we have come, discuss corollaries, and plan for the future.