Math 254B, UC Berkeley, Spring 2002 (Kedlaya) An “Abstract” Reciprocity Map

Reference. Milne VII.5; Neukirch VI.4, but only loosely.

In this lecture, we’ll manufacture a canonical isomorphism Gal(L/K)^{ab} → C_{K}/ Norm_{L/K} C_{L} for any finite extension L/K of number fields, where C_{K} and C_{L} are the corresponding idele class groups. However, we won’t yet know it agrees with our proposed reciprocity map, which is the product of the local reciprocity maps. We’ll check that in the next lecture.

Cyclotomic extensions.

The cyclotomic extensions (extensions by roots of unity) of a number field play a role in class field theory analogous to the role played by the unramified extensions in local class field theory. This makes it essential to make an explicit study of them for use in proving the main results.

First of all, we should make a distinction that will come up repeatedly. The extension ∪_{n}ℚ(ζ_{n}) of ℚ obtained by adjoining all roots of unity has Galois group ℤ̂^{∗} = ∏_{p} ℤ_{p}^{∗}. That group has a lot of torsion, since each ℤ_{p}^{∗} contains a torsion subgroup of order p − 1 (or 2, if p = 2). If we take the fixed field for the torsion subgroup of ℤ^{∗}, we get a slightly smaller extension, which I’ll call the small cyclotomic extension of ℚ and denote ℚ^{smcy}. Its Galois group is ∏_{p} ℤ_{p} = ℤ̂. For K a number field, define K^{smcy} = Kℚ^{smcy}; then Gal(K^{smcy}/K) ≅ ℤ̂ as well, even if K contains some extra roots of unity.

The reciprocity map via abstract CFT.

First of all, we choose an isomorphism of Gal(ℚ^{smcy}/ℚ) with ℤ̂; our results are not going to depend on the choice. That gives a continuous surjection

d : Gal(\overline{ℚ}/ℚ) → Gal(ℚ^{smcy}/ℚ) ≅ ℤ̂;

if we regard ℚ^{smcy}/ℚ as the “maximal unramified extension” of ℚ, we can define the ramification index e_{L/K} and inertia degree f_{L/K} for any extension of number fields, by the rules

f_{L/K} = [L ∩ ℚ^{smcy} : K ∩ ℚ^{smcy}], e_{L/K} = \frac{[L : K]}{f_{L/K}}.

To use abstract class field theory to exhibit the reciprocity map, we need a “henselian valuation” v : C_{ℚ} → ℤ̂, i.e., a homomorphism satisfying:

(i) v(C_{ℚ}) is a subgroup Z of ℤ̂ containing ℤ with Z/nZ ≅ ℤ/nℤ for all n ∈ ℕ;

(ii) v(Norm_{K/ℚ} C_{K}) = f_{K/ℚ}Z for all finite extensions K/ℚ.

Once we have that, our calculations from the last few lectures imply that the class field axiom is satisfied: for L/K cyclic,

#H_{T}^{0}(Gal(L/K), C_{L}) = [L : K], #H_{T}^{1}(Gal(L/K), C_{L}) = 1.

So then abstract class field theory will kick in.

We can make that valuation using Artin reciprocity for ℚ(ζ_{n})/ℚ. Recall how that works: there is a canonical surjection

I_{n} → (ℤ/nℤ)^{∗} ≅ Gal(ℚ(ζ_{n})/ℚ) :

for p not dividing n, the ideal (p) goes to p ∈ (ℤ/nℤ)^{∗} and then to the automorphism ζ_{n} ↦ ζ_{n}^{p}, which indeed does act as the p-th power map modulo any prime of ℚ(ζ_{n}) above p.

That induces a homomorphism I_{ℚ} → (ℤ/nℤ)^{∗} as follows: given an idele α, pick x ∈ ℚ^{∗} so that α_{ℝ}/x > 0, and for each prime p with p^{e}|n, α/x has p-component congruent to 1 modulo p^{e}. Then map α/x to (ℤ/nℤ)^{∗} as follows:

α/x ↦ ∏_{ℓ|̸n} ℓ^{v_{ℓ}(α_{ℓ}/x)}

This gives a well-defined map: if y is an alternate choice for x, then x/y ≡ 1 (mod n) and x/y > 0, so the product on the right side is precisely x/y itself, and so is congruent to 1 in (ℤ/nℤ)^{∗}.

We now have maps I_{ℚ} → (ℤ/nℤ)^{∗} which are easily seen to be compatible, so by taking inverse limits we get I_{ℚ} → ℤ̂^{∗} ≅ Gal(ℚ^{cyc}/ℚ). We define v by piping this map through the projection Gal(ℚ^{cyc}/ℚ) → Gal(ℚ^{smcy}/ℚ) and then using our chosen isomorphism Gal(ℚ^{smcy}/ℚ) ≅ ℤ̂.

Another way to say this: I_{ℚ} can be written as ℚ^{∗} × ℝ^{∗} × ℤ̂^{∗}, and the map to ℤ̂^{∗} is just projection onto the third factor! In particular, the map factors through C_{ℚ}, and property (i) above is straightforward.

To check (ii), we need to do the same thing that we just did a bit more generally. For K now a number field, define the map

I_{n} → (ℤ/nℤ)^{∗} ⊇ Gal(K(ζ_{n})/K),

where now I_{n} is the group of fractional ideals of K coprime to (n), by sending a prime 𝔭 first to its absolute norm. We then have to check that the result is always in the image of Gal(K(ζ_{n})/K), but in fact it must be: whatever the Frobenius of 𝔭 is, it sends ζ_{n} to a power of ζ_{n} congruent to ζ_{n}^{Norm(𝔭)} modulo 𝔭. Since 𝔭 is prime to n, it’s prime to the difference between any two powers of ζ_{n}, so the Frobenius of 𝔭 must in fact send ζ_{n} to ζ_{n}^{Norm(𝔭)}. This tells us first that the map above sends I_{n} to Gal(K(ζ_{n})/K) and second that it coincides with the Artin map.

From the First Inequality, we can deduce the following handy fact:

Proposition 1. For L/K a finite abelian extension of number fields, the Artin map always surjects onto Gal(L/K).

Proof. If the Artin map only hit the subgroup H of Gal(L/K), the fixed field M of H would have the property that all but finitely many primes of M split completely in L. We’ve already seen that this contradicts the First Inequality.

In particular, the Artin map I_{n} → Gal(K(ζ_{n})/K) we wrote down above is surjective. Using that, we can verify (ii): given a prime ideal 𝔭 of K, the Artin map of K(ζ_{n})/K

applied to it gives the same element of (ℤ/nℤ)^{∗} as the Artin map of ℚ applied to Norm_{K/ℚ}(𝔭). And the Artin map of K(ζ_{n})/K surjects onto Gal(K(ζ_{n})/K), which has index [K∩ℚ(ζ_{n}) : ℚ] in (ℤ/nℤ)^{∗}. This verifies (ii).

Thus lo and behold, we get from abstract class field theory a reciprocity isomorphism for any finite extension of number fields:

r′_{L/K} : C_{K}/ Norm_{L/K} C_{L} →^{∼} Gal(L/K)^{ab}.

These are compatible in the usual way, so we get a map r′_{K} : C_{K} → Gal(K^{ab}/K). Of course, we don’t know what this map is, so we can’t yet use it to recover Artin reciprocity. (That depended on the reciprocity map being the product of the local maps.) But at least we get the norm limitation theorem: if L/K is a finite extension and M = L ∩ K^{ab}, then Norm_{L/K} C_{L} = Norm_{M/K} C_{M}.

We do know one thing about this map: for “unramified” extensions L/K (i.e., L ⊆ K^{smcy}), the “Frobenius” in Gal(L/K) maps to a “uniformizer” in C_{K}. That is, the element of Gal(L/K) coming from the element of Gal(K^{smcy}/K) which maps to 1 under d_{K} is the element of C_{K} which maps to 1 under v_{K}. But we made v_{K} simply by mapping C_{K} to Gal(K^{smcy}/K) via the Artin map and then identifying the latter with ℤ̂ by the same identification we used to make d_{K}. Upshot: the choice of that identification drops out, and the reciprocity map coincides with the Artin map that we wrote down earlier.

To do next time: check that r′_{L/K} agrees with the map that I called r_{L/K}, namely the product of the local reciprocity maps. (Remember, I need this in order to recover Artin reciprocity in general.)