Math 254B, UC Berkeley, Spring 2002 (Kedlaya) Local Class Field Theory: Addenda

Reference. Neukirch, IV.4-IV.6; also V.1 for the first section.

The local existence theorem. We’ve seen two ways to derive the local reciprocity law, but I somehow managed not to prove the local existence theorem. Fortunately, it’s easy. (Again, this argument only works in characteristic 0.)

Theorem 1. Let K be a finite extension of ℚ_{p}. Then every open subgroup of K^{∗} of finite index is of the form Norm_{L/K} L^{∗} for some finite (abelian) extension L of K.

Recall that Norm_{L/K} L^{∗} = Norm_{M/K} M^{∗} for M = L ∩ K^{ab} by the local reciprocity law, so we don’t have to make sure L is abelian.

Proof. Let N be an open subgroup of K^{∗} of index n. Then (K^{∗})^{n} ⊆ N, and it suffices to prove that (K^{∗})^{n} contains Norm_{L/K} L^{∗} for some L. Let ζ_{n} be a primitive n-th root of unity, let μ_{n} be the group generated by ζ_{n}, and put K_{1} = K(ζ_{n}). By Kummer theory, there is a canonical isomorphism between Hom(Gal(K_{1}^{ab}/K_{1}), μ_{n}) and K_{1}^{∗}/(K_{1}^{∗})^{n}. But the latter is finite for any local field K_{1} (exercise), so if we let L be the compositum of all ℤ/nℤ-extensions of K_{1}, then L/K_{1} is a finite extension. Now Gal(L/K_{1}) ≅ K_{1}^{∗}/ Norm_{L/K_{1}} L^{∗} is a group of exponent n, so (K_{1}^{∗})^{n} ⊆ Norm_{L/K_{1}} L^{∗}. On the other hand,

#K_{1}^{∗}/(K_{1}^{∗})^{n} = # Gal(L/K_{1}) = #K_{1}^{∗}/ Norm_{L/K_{1}} L^{∗},

so (K_{1}^{∗})^{n} = Norm_{L/K_{1}} L^{∗}. In particular, (K^{∗})^{n} contains Norm_{K_{1}/K}(K_{1}^{∗})^{n} = Norm_{L/K} L^{∗}, as desired.

More on abstract class field theory.

First of all, the claim that Gal(ℚ^{ab}/ℚ) is isomorphic to ℤ̂ is false; that group is of course ℤ̂^{∗}. But it has a natural quotient isomorphic to ℤ̂; it’s easiest to construct this factor by factor, by exhibiting a surjection ℤ_{p}^{∗} → ℤ_{p}. (Otherwise, put, ℚ(ζ_{p^{∞}})/ℚ has a subextension with Galois group ℤ_{p}.) For p ≠ 2, ℤ_{p}^{∗} can be written as the product of the group of (p − 1)- st roots of unity with the subgroup 1 + pℤ_{p}, and the latter is isomorphic to pℤ_{p} via the logarithm map. For p = 2, ℤ_{p}^{∗} can be written as the product of {±1} with the subgroup 1 + 4ℤ_{2}, which again is isomorphic to 4ℤ_{2} via the logarithm map.

Second, the proof of the multiplicativity of the reciprocity map was seriously broken. Having finally deciphered the proof in Neukirch, I’ll give a version of that instead.

Recall notation: L/K is a finite extension of finite extensions of k, H is the semigroup of g ∈ Gal(L^{unr}/K) such that d_{K}(g) ∈ ℕ (that’s positive integers, not nonnegative), and r′ : H → A_{K}/ Norm_{L/K} A_{L} is defined as follows. For g ∈ Gal(L^{unr}/K), let M be the fixed field of g, let π_{M} be a uniformizer of M, and set r′(g) = Norm_{M/K}(π_{M}).

Given g_{1}, g_{2} ∈ H, set g_{3} = g_{1}g_{2}; we want to show that r′(g_{1})r′(g_{2}) = r′(g_{3}). Let M_{i} be the fixed field of g_{i}, let π_{i} be a uniformizer of π_{i}, and put ρ_{i} = r(g_{i}) = Norm_{M_{i}/K}(π_{i}); since v_{K}(ρ_{i}) = d_{K}(g_{i}), the ratio ρ_{1}ρ_{2}/ρ_{3} is a unit; we want to show that it is in Norm_{L/K}(U_{L}).

Again, the trouble with ρ_{1}ρ_{2}/ρ_{3} is that it is made up of norms from different fields, and to make progress we need to rewrite it more uniformly. Choose ϕ ∈ Gal(L^{unr}/K) with d_{K}(ϕ) = 1, and put d_{i} = d_{K}(g_{i}); then we can write g_{i} = ϕ^{d_{i}}h_{i}^{−1} for some h_{i} with d_{K}(h_{i}) = 0, that is, h_{i} ∈ Gal(L^{unr}/K^{unr}).

Proposition 2. Let M be the fixed field of some h ∈ Gal(L^{unr}/K) with d_{K}(h) = n a nonnegative integer, and suppose ϕ ∈ H satisfies d_{K}(ϕ) = 1. Then for any x ∈ A_{M},

Norm_{M/K}(x) = Norm_{L^{unr}/K^{unr}}(xx^{ϕ} ⋯ x^{ϕ^{n−1}}).

Put

σ_{i} = π_{i}π_{i}^{ϕ} ⋯ ϕ_{i}^{ϕ^{d_{i}−1}};

then if u = σ_{1}σ_{2}/σ_{3}, we have u ∈ U_{L^{unr}} and ρ_{1}ρ_{2}/ρ_{3} = Norm_{L^{unr}/K^{unr}}(u). Now

\frac{u^{ϕ}}{u} = \frac{σ_{1}^{ϕ}σ_{2}^{ϕ}σ_{3}}{σ_{1}σ_{2}σ_{3}^{ϕ}} = \frac{π_{1}^{ϕ^{d_{1}}}π_{2}^{ϕ^{d_{2}}}π_{3}}{π_{1}π_{2}π_{3}^{ϕ^{d_{3}}}} = \frac{π_{1}^{h_{1}}π_{2}^{h_{2}}π_{3}}{π_{1}π_{2}π_{3}^{h_{3}}}

since π_{i} is fixed by g_{i} = ϕ^{d_{i}}h_{i}^{−1}. The relationship among the h_{i} is ϕ^{d_{1}}h_{1}^{−1}ϕ^{d_{2}}h_{2}^{−1} = ϕ^{d_{3}}h_{3}^{−1}, that is, h_{3} = h_{2}ϕ^{−d_{2}}h_{1}ϕ^{d_{2}}.

Pick a single uniformizer π_{L} of L; then we can write π_{i} = v_{i}π_{L} for some unit v_{i} ∈ L^{unr}, and

\frac{u^{ϕ}}{u} = \frac{v_{1}^{h_{1}}v_{2}^{h_{2}}v_{3}}{v_{1}v_{2}v_{3}^{h_{3}}} \frac{π_{L}^{h_{1}}π_{L}^{h_{2}}}{π_{L}π_{L}^{h_{3}}} .

The first term is clearly a product of expressions of the form y_{i}^{g_{i}}/y_{i} for y_{i} ∈ U_{L^{unr}} and g_{i} ∈ Gal(L^{unr}/K^{unr}). On the other hand, since π_{L} and π_{L}^{h_{1}} are invariant under ϕ, we get

\frac{π_{L}^{h_{1}}π_{L}^{h_{2}}}{π_{L}π_{L}^{h_{3}}} = \frac{π_{L}^{h_{1}}π_{L}^{h_{2}}}{π_{L}^{ϕ^{d_{2}}}π_{L}^{h_{2}ϕ^{−d_{2}}h_{1}ϕ^{d_{2}}}}

= \frac{π_{L}^{h_{2}}}{π_{L}^{d_{2}}} (\frac{π_{L}^{d_{2}}}{π_{L}^{h_{2}}} )^{ϕ^{−d_{2}}h_{1}ϕ^{d_{2}}} .

Thus \frac{u^{ϕ}}{u} is the product of expressions of the form y_{i}^{g_{i}}/y_{i} for y_{i} ∈ U_{L^{unr}} and g_{i} ∈ Gal(L^{unr}/K^{unr}). What we need to show is that Norm_{L^{unr}/K^{unr}}(U_{L}) ∈ Norm_{L/K}(U_{L}); that will follow from the following lemma.

Lemma 3. If x ∈ U_{L^{unr}} has the property that x^{ϕ}/x = ∏_{i} y_{i}^{g_{i}}/y_{i} for some y_{i} ∈ U_{L^{unr}} and g_{i} ∈ Gal(L^{unr}/K^{unr}), then Norm_{L^{unr}/K^{unr}}(x) belongs to Norm_{M/K} U_{M} for every finite subextension M of L^{unr}/K.

Proof. (This is Neukirch, Lemma IV.5.4.) It suffices to consider M containing x and each y_{i}, and also containing L. Let n = [M : K], and let F be the fixed field of σ = ϕ^{n}. Let G be the degree n unramified extension of G, i.e., the fixed field of σ^{n}. By the vanishing of H_{T}^{0}(Gal(G/F), U_{G}), we can write x = Norm_{G/F} (x̃) and y_{i} = Norm_{G/F} (ỹ_{i}) for some x̃, ỹ_{i} ∈ U_{G}. Put

α = \frac{x̃^{ϕ}}{x̃} ∏_{i} \frac{ỹ_{i}}{ỹ_{i}}^{g_{i}};

then Norm_{G/F} (α) = 1. By the vanishing of H_{T}^{−1}(Gal(G/F), U_{G}), that means α = w^{σ}/w for some w ∈ G. That means we can write

\frac{x̃^{ϕ}}{x̃} = \frac{(ww^{ϕ} ⋯ w^{ϕ^{n−1}})^{ϕ}}{ww^{ϕ} ⋯ w^{ϕ^{n−1}}} ∏_{i} \frac{ỹ_{i}^{g_{i}}}{ỹ_{i}} .

Apply Norm_{L^{unr}/K^{unr}} to both sides. This kills off ỹ_{i}^{g_{i}}/ỹ_{i} for each i, leaving

Norm_{L^{unr}/K^{unr}}(x̃)^{ϕ−1} = Norm_{L^{unr}/K^{unr}}(ww^{ϕ} ⋯ w^{ϕ^{n−1}})^{ϕ−1}.

Put z = Norm_{L^{unr}/K^{unr}}(x̃/(ww^{ϕ} ⋯ w^{ϕ^{n−1}})); then the previous equation reads z^{ϕ} = z, which means z ∈ U_{K}. We then have

Norm_{L^{unr}/K^{unr}}(x) = Norm_{L^{unr}/K^{unr}}(Norm_{G/F} (x̃))
= Norm_{L^{unr}/K^{unr}}(x̃x̃^{ϕ^{n}} ⋯ x̃^{ϕ^{n(n−1)}})
= Norm_{L^{unr}/K^{unr}}(zz^{ϕ^{n}} ⋯ z^{ϕ^{n(n−1)}}) Norm_{L^{unr}/K^{unr}}(ww^{ϕ} ⋯ w^{ϕ^{n^{2}−1}})
= z^{n} Norm_{L^{unr}/K^{unr}}(tt^{ϕ} ⋯ t^{ϕ^{n−1}}),

if we put t = ww^{ϕ^{n}} ⋯ w^{ϕ^{n(n−1)}}. Now applying the above proposition, and noting that trivially z^{n} = Norm_{F/K}(z), we get Norm_{L^{unr}/K^{unr}}(x) = Norm_{M/K}(z) Norm_{F/K}(t) ∈ Norm_{M/K} U_{M}, as desired.