Math 254B, UC Berkeley, Spring 2002 (Kedlaya) Cohomology of the Ideles 2: The “Second Inequality”

Reference. Milne VII.5; Neukirch VI.4.

We proved last time that for L/K a cyclic extension of number fields, the Herbrand quotient h(C_{L}) of the idele class group of L is equal to [L : K]. This time we’ll prove the following.

Theorem 1. Let L/K be a Galois extension of number fields, with Galois group G. Then

(a) the group I_{K}/(K^{∗} Norm_{L/K} I_{L}) is finite of order at most [L : K];

(b) the group H^{1}(G, C_{L}) is trivial;

(c) the group H^{2}(G, C_{L}) is finite of order at most [L : K].

By the first inequality, for L/K cyclic, these three are equivalent and all imply that H^{2}(G, C_{L}) has order exactly [L : K]. That would suffice to prove the “class field axiom” in Neukirch’s setup.

There are two basic ways to prove this result: an analytic proof and an algebraic proof. Although the analytic proof is somewhat afield of what we have been doing (it requires some properties of zeta functions that we haven’t worked out), it’s somewhat simpler overall than the algebraic proof. So we’ll sketch it first before proceeding to the algebraic proof.

The analytic proof.

For the analytic proof, we need to recast the Second Inequality back into classical, idealtheoretic language. Let L/K be a finite Galois extension and 𝔪 a formal product of places of K. Back when we defined generalized ideal class groups, we defined the group I_{𝔪} of fractional ideals of K coprime to 𝔪 and P_{𝔪} the group of principal ideals admitting a generator α such that α ≡ 1 (mod 𝔭^{e}) if the prime power 𝔭^{e} occurs in 𝔪 for a finite prime 𝔭, and τ(α) > 0 if τ is the real embedding corresponding to a real place in 𝔪. Also, let J_{𝔪} be the group of fractional ideals of L coprime to 𝔪. Then the Second Inequality states that

#I_{𝔪}/P_{𝔪} Norm_{L/K} J_{𝔪} ≤ [L : K].

Note that we don’t have to assume 𝔪 is divisible by the ramified primes of L/K.

We’ll need the following special case of the Chebotarev density theorem, which fortunately we can prove without already having all of class field theory.

Proposition 2. Let L be a finite extension of K and let M/K be its Galois closure. Then the set S of prime ideals of K that split completely in L has Dirichlet density 1/[M : K].

Proof. A prime of K splits completely in L if and only if it splits completely in M, so we may assume L = M is Galois. Recall that the set T of unramfied primes 𝔮 of L of absolute degree 1 has Dirichlet density 1; each such prime lies over an unramified prime 𝔭 of K of absolute degree 1 which splits completely in L.

Now recall how the Dirichlet density works: the set T having Dirichlet density 1 means that

∑_{𝔮∈T} \frac{1}{Norm(𝔮)^{s}} ∼ \frac{1}{s − 1} s ↘ 1

(s approaching 1 from above, that is). If we group the primes in T by which prime of S they lie over, then we get

[L : K] ∑_{𝔭∈T} \frac{1}{Norm(𝔭)^{s}} ∼ \frac{1}{s − 1}.

That is, the Dirichlet density of S is 1/[L : K].

Now for the inequality. For χ : I_{𝔪}/P_{𝔪} → ℂ^{∗} a character, we defined the L-function

L(s, χ) = ∏_{𝔭|̸𝔪} \frac{1}{1 − χ(𝔭) Norm(𝔭)^{−s}} .

We’ll use some properties of this function that were on a previous homework. For starters,

log L(s, 1) ∼ log ζ_{K}(s) ∼ log \frac{1}{s − 1} s ↘ 1,

while if χ is not the trivial character, L(s, χ) is holomorphic at s = 1. If L(s, χ) = (s − 1)^{m(χ)}g(s) where g is holomorphic and nonvanishing at s = 1, then m(χ) ≥ 0, and

log L(s, χ) ∼ m(χ) log(s − 1) = −m(χ) log \frac{1}{s − 1}.

Let H be a subgroup of I_{𝔪} containing P_{𝔪}. By finite Fourier analysis, or orthogonality of characters,

∑_{χ:I_{𝔪}/H→ℂ^{∗}} log L(s, χ) ∼ #(I_{𝔪}/H) ∑_{𝔭∈H} \frac{1}{Norm(𝔭)^{−s}} .

We conclude that the set of primes in H has Dirichlet density

\frac{1 − ∑_{χ≠1} m(χ)}{#(I_{𝔪}/H)};

this is 1/#(I_{𝔪}/H) if the m(χ) are all zero, and 0 otherwise.

We apply this with H = P_{𝔪} Norm_{L/K} J_{𝔪}. This in particular includes every prime of K that splits completely, since such a prime is the norm of any prime of L lying over it. Thus the set of primes in H has Dirichlet density, on one hand, is at least 1/[L : K]. On the other hand, this set has density either zero or 1/#(I_{𝔪}/H). We conclude #I_{𝔪}/H ≤ [L : K], as desired.

The algebraic proof.

We now proceed to the algebraic proof of the Second Inequality. To prove the theorem in general, we can very quickly reduce to the case of L/K not just cyclic, but cyclic of prime order. The reductions are similar to those we used to compute H^{2}(L^{∗}) in the local case. First of all, if we have the theorem for all solvable groups, then if G is general and H is a Sylow p-subgroup of G, then for any G-module M,

Res : H_{T}^{r} (G, M) → H_{T}^{r} (H, M)

is injective on p-primary components (because Cor ∘ Res is multiplication by [G : H]), so we can deduce the desired result. Thus it suffices to consider L/K solvable. In that case, starting from the cyclic of prime order case we can induct using the inflation-restriction exact sequence: if K′/K is a subextension and H = Gal(L/K′), then

0 → H^{i}(G/H, C_{K′}) → H^{i}(G, C_{L}) → H^{i}(H, C_{L})

Upshot: we need only consider L/K cyclic of prime order p.

One more reduction to make things simpler: we reduce to the case where K contains a p-th root of unity. Let K′ = K(ζ_{p}) and L′ = L(ζ_{p}); then K′ and L are linearly disjoint over K (since their degrees are coprime), so [L′ : K′] = [L : K] = p and the Galois groups of L/K and L′/K′ are canonically isomorphic. It suffices to check that the homomorphism

H_{T}^{0}(Gal(L/K), C_{L}) → H_{T}^{0}(Gal(L′/K′), C_{L′})

induced by the inclusion C_{L} → C_{L′} is injective. These groups are both killed by multiplication by p, since for x ∈ C_{K}, Norm_{L/K} x = x^{p}. Thus multiplication by d = [K′ : K], which divides p − 1, is an isomorphism on these groups. If x ∈ C_{K} maps to the identity in H_{T}^{0}(Gal(L′/K′), C_{L′}), we can choose a representative of the same class as x in H_{T}^{0}(Gal(L/K), C_{L}) of the form y^{d}; then y also maps to the identity in H_{T}^{0}(Gal(L′/K′), C_{L′}). That is, y = Norm_{L′/K′}(z′) for some z′ ∈ C_{L′}, and

y^{d} = Norm_{K′/K}(y) = Norm_{L′/K}(z′) ∈ Norm_{L/K} C_{L}.

Thus x ∈ Norm_{L/K} C_{L}, so the homomorphism is injective.

The key case.

To sum up: it suffices to prove the theorem for K containing a p-th root of unity ζ_{p} and L/K cyclic of order p. We now address this case.

As in the proof of the First Inequality, we will use a set S of places of K containing the infinite places, the primes that ramify in L, and enough additional primes so that I_{K} = I_{K,S}K^{∗}. Again, we put K_{S} = I_{K,S} ∩ K^{∗}. Also we write s = #S.

The plan is to explicitly produce a subgroup of C_{K} of index [L : K] consisting of norms from C_{L}. We do this by using an auxiliary set of places T disjoint from S. For such T, we define

J = ∏_{v∈S}(K_{v}^{∗})^{n} × ∏_{v∈T} K_{v}^{∗} × ∏_{v∉S∪T} 𝔬_{K_{v}}^{∗}.

Let Δ = (L^{∗})^{p} ∩ K_{S}. We will show that:

(a) L = K(Δ^{1/p}).

(b) We can choose a set T of s − 1 primes such that Δ is the kernel of the map K_{S} → ∏_{v∈T} K_{v}^{∗}/(K_{v}^{∗})^{n}.

(c) For such a set T, if we put C_{K,S,T} = JK^{∗}/K^{∗}, then

#C_{K}/C_{K,S,T} = [L : K] = p.

(d) With the same notation, C_{K,S,T} ⊆ Norm_{L/K} C_{L}.

That will imply #C_{K}/ Norm_{L/K} C_{L} ≤ p, as desired.

We first concentrate on (a). By Kummer theory, since K contains a primitive p-th root of unity, we can write L = K(D^{1/p}) for D = (L^{∗})^{p} ∩ K^{∗}. Thus K(Δ^{1/p}) ⊆ L and since there is no room between K and L for an extension ([L : K] being prime), all that we have to check is that Δ ≠ (K_{S})^{p}. Choose a single x ∈ D such that L = K(x^{1/p}). For each v ∉ S, the extension K_{v}(x^{1/p})/K_{v} is unramified, so we can write x as a unit times a p-th power, say x = u_{v}y_{v}^{p}. If we put y_{v} = 1 for v ∈ S, we can assemble the y_{v} into an idele y, which by I_{K} = K^{∗}I_{K,S} we can rewrite as zw for z ∈ K^{∗} and w ∈ I_{K,S}. Now for v ∉ S, (x/z^{p})_{v} = u_{v}/w_{v}^{p} ∈ 𝔬_{K_{v}}^{∗}. Thus x/z^{p} ∈ (L^{∗})^{p} ∩ K_{S} ∈ Δ but x ∉ (K^{∗})^{p}. We conclude L = K(Δ^{1/p}).

Now we move to (b). Put N = K(K_{S}^{1/p}). By Kummer theory,

Gal(N/K) ≅ Hom(K_{S}/K_{S}^{p}, ℤ/pℤ).

By the generalization of Dirichlet’s units theorem to S-units, K_{S} modulo torsion is a free abelian group of rank s−1, and the torsion subgroup consists of roots of unity, so is cyclic of order divisible by p. Thus K_{S}/K_{S}^{p} ≅ (ℤ/pℤ)^{s}. Choose generators g_{1}, …, g_{s−1} of Gal(N/L); these correspond in Hom(K_{S}/K_{S}^{p}, ℤ/pℤ) to a set of homomorphisms whose common kernel is precisely Δ/K_{S}^{p}.

So to establish (b), we need to find for each g_{i} a place v_{i} such that the kernel of g_{i} is the same as the kernel of K_{S} → K_{v}^{∗}/(K_{v}^{∗})^{p}. Let N_{i} be the fixed field of g_{i}; as noted last time, it follows from the First Inequality that there are infinitely many primes of N_{i} that do not split in N. So we can choose a place w_{i} of each N_{i} such that their restrictions v_{i} to K are distinct, not contained in S, and don’t divide p.

We claim N_{i} is the maximal subextension of N/K in which v_{i} splits completely (a/k/a the “decomposition field” of v_{i}). On one hand, v_{i} does not split completely in N, so the decomposition field is no larger than N_{i}. On the other hand, the decomposition field is the fixed field of the decomposition group, which has exponent p and is cyclic (since v_{i} does not ramify in N). Thus it must have index p in N, so must be N_{i} itself.

Thus L = ∩N_{i} is the maximal subextension of N in which all of the v_{i} split completely. We conclude that for x ∈ K_{S}, x belongs to Δ iff K_{v_{i}}(x^{1/p}) = K_{v_{i}} for all i, which occurs iff x ∈ K_{v_{i}}^{p}. That is, Δ is precisely the kernel of the map K_{S} → ∏_{i} K_{v_{i}}^{∗}/(K_{v_{i}}^{∗})^{p}. In fact, under this map, K_{S} actually maps to the units in K_{v_{i}}^{∗} for each i.

Next, we verify (c), using the following lemma.

Lemma 3. J ∩ K^{∗} = (K_{S∪T})^{p}.

Proof. Clearly K_{S∪T}^{p} ⊆ J ∩K^{∗}; we have to work to show the other inclusion. Take y ∈ J ∩K^{∗} and M = K(y^{1/p}). We’ll show that Norm_{M/K} C_{M} = C_{K}; by the First Inequality, this will imply M = K, so y ∈ (K^{∗})^{p} ∩ J = (K_{S∪T})^{p}.

Since I_{K} = I_{K,S}K^{∗}, it is enough to choose α ∈ I_{K,S} and show that α/x ∈ Norm_{L/K} I_{M} for some x ∈ K^{∗}. As noted above, the map

K_{S} → ∏_{v∈T} 𝔬_{K_{v}}^{∗}/(𝔬_{K_{v}}^{∗})^{p}

is surjective, and #K_{S}/Δ = p^{s−1}. That’s also the order of the product, so the map is actually an isomorphism. Thus we can find x ∈ K_{S} so that α/x has component the p-th power of a unit of K_{v} at each v ∈ T. In particular, such a component is the norm of its p-th root, so α/x is a norm at each v ∈ T. For v ∈ S, we don’t have anything to check: because y is a p-th power at v, M_{w} = K_{v}. Finally, for v ∉ S ∪ T, M_{w}/K_{v} is unramified, so any unit is a norm. Thus α/x is indeed a norm. We conclude Norm_{M/K} C_{M} = C_{K}, so M = K and y ∈ K_{S∪T}^{p}, as desired.

Given the lemma, we now have an exact sequence

1 → I_{K,S∪T} ∩ K^{∗}/J ∩ K^{∗} → I_{K,S∪T} /J → I_{K,S∪T} K^{∗}/JK^{∗} → 1.

We can rewrite I_{K,S∪T} K^{∗} as simply I_{K}. So the group on the right is precisely C_{K}/C_{K,S,T}. By the lemma, the group on the left is K_{S∪T}^{∗} /(K_{S∪T}^{∗})^{n}, which has order p^{2s−1} because K_{S∪T} is free of rank 2s − 2 plus a cyclic group of order a multiple of p. And the group in the middle is the product of K_{v}^{∗}/(K_{v}^{∗})^{p} over all v ∈ S, and each of those has order p^{2} (generated by ζ_{p} and a uniformizer of K_{v}). Adding it all up, we get #C_{K}/C_{K,S,T} = p.

Finally, to check (d), it suffices to check that J ⊆ Norm_{L/K} I_{L}, which we may check component by component. It’s automatic for the places v ∉ S ∪ T, since those places are unramified, so every unit is a norm. For places v ∈ S, any element of (K_{v}^{∗})^{p} is a norm from K_{v}(K_{v}^{1/p}) (see the proof of the local existence theorem), so also from L_{w}. Finally, for places v ∈ T, from the construction of T, we see that Δ ⊆ (K_{v}^{∗})^{p}, so L_{w} = K_{v}, and so all of K_{v}^{∗} are norms.

Aside. We get from this calculation that H_{T}^{−1}(G, C_{L}) = 1, so H_{T}^{0}(G, L^{∗}) → H_{T}^{0}(G, C_{L}) is injective. That is,

K^{∗}/ Norm_{L/K} L^{∗} → ⊕_{v}K_{v}^{∗}/ Norm_{L_{w}/K_{v}} L_{w}^{∗}

is injective. In other words, we have an interesting “local-to-global” statement, namely Hasse’s Norm Theorem: if L/K is cyclic, x ∈ K^{∗} is a norm if and only if it is locally a norm.