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Notes on class field theory

Section 7.4 The existence theorem

Reference.

[37] VII.6, VII.9, [38] VI.4, VI.6.
With the reciprocity isomorphism in hand, we now prove the existence theorem in its idelic formulation (see Theorem 6.4.4). As a reminder, deducing the ideal-theoretic existence theorem (Theorem 2.2.8) will also require local-global compatibility (Proposition 7.5.6), as described in Remark 6.4.10.
As in the proof of the local existence theorem (Theorem 4.3.18), having access to the reciprocity law, even in abstract form, reduces the task of proving the existence theorem to the “topological” assertion that every open subgroup of CL of finite index contains a norm subgroup. For this, we can essentially rerun the Kummer-theoretic argument from the local case. Since this argument is agnostic as to where the reciprocity map came from, it also applies in the approach that does not use abstract class field theory (Section 7.6).
After proving the existence theorem, we give the closely related algebraic proof of the Second Inequality (Theorem 7.2.9).

Subsection A base case for the existence theorem

As in the proof of the local existence theorem (Theorem 4.3.18), the key to the proof of Theorem 7.4.8 is showing that for any given number field K, we can find finite extensions L/K for which the groups NormL/KCL can be made arbitrarily small. In preparation for an inductive proof, we establish a key base case using Kummer theory.

Proof.

Let J be the preimage of U under the projection IKCK, so that J is open in IK of index p. Then J contains a subgroup of the form
V:=vS{1}×vSoKv
for some finite set S of places of K containing the infinite places, which by Corollary 6.2.10 we may choose large enough so that IK=KIK,S. The group J must also contain IKp, and hence WS.
We continue with a lemma that allows to detect whether certain elements of a number field are p-th powers based on whether this happens locally. This amounts to a carefully chosen special case of the Grunwald-Wang theorem (Remark 7.2.12).

Proof.

It is clear that
(oK,S)pWSK.
To prove the reverse inclusion, note that for any yWSK, if we set L=K(y1/p), then every place vS is split in L and every place vS is unramified in L, yielding
NormL/KIL,S=IK,S.
Since IK=KIK,S, this implies NormL/KCL=CK. By the First Inequality (Theorem 7.1.2), this implies L=K and so y(K)p.
This will in turn enable us to compute the norm group for a certain compositum of Kummer extensions.

Proof.

By Corollary 6.2.11 and the assumption that K contains a primitive p-th root of unity, the group oK,S/(oK,S)p is finite of order ps, yielding [L:K]=ps. By local reciprocity (Lemma 4.3.13), we have KWS/KNormL/KCL; to prove equality, it will suffice to show that these subgroups have the same finite index in CK.
Consider now the exact sequence
1oK,SoK,SWSIK,SWSCKKWS/K1.
By Lemma 7.4.2, the group on the left has order [oK,S:(oK,S)p]=ps. By Lemma 7.4.4, the group in the middle has order p2s. We thus have
[CK:KWS/K]=ps=[L:K]=[CK:NormL/KCL]
where the last equality follows from the global reciprocity isomorphism (Theorem 6.4.3).
Here is the local calculation used in the proof of Lemma 7.4.3.

Proof.

We separate cases as follows.
  1. If v is a real place, then p=2, p2|p|v=2, and
    Kv/(Kv)p=R/(R)2=R/R+Z/2Z.
  2. If v is a complex place, then p2|p|v=1 according to our conventions (Definition 6.1.11), and Kv/(Kv)p is trivial because C is p-divisible.
  3. If v is a finite place not lying above p, then p2|p|v=p2 and Kv/(Kv)p is generated by ζp and a uniformizer of Kv.
  4. If v is a finite place above p, then |p|v=pn for some positive integer n, so p2|p|v=pn+2. Since KvoKv×Z, it suffices to check that [oKv:(oKv)p]=pn+1. For this, see Exercise 1.
We finally put everything together to get a key special case of the existence theorem.

Proof.

Subsection Proof of the existence theorem

Building on the base case offered by Lemma 7.4.5, we now finish the proof of the existence theorem.

Proof.

Take K=K(ζp). Let U be the inverse image of U in CK. By Theorem 6.4.3 [CK:NormK/KCK]=[K:K] is coprime to p; consequently, [CK:U]=p. By Lemma 7.4.5, there exists a finite extension L/K such that NormL/KCLU; then NormL/KCLNormK/KUU.

Proof.

We proceed by induction on the index [CK:U], with Lemma 7.4.6 as the base case. Otherwise, choose an intermediate subgroup V between U and CK. By the induction hypothesis, V contains N:=NormL/KCL for some finite extension L of K. Then
[N:(UN)]=[UN:U][V:U].
Let W be the subgroup of CL consisting of those x whose norms lie in U. Then
[CL:W][N:UN][V:U],
so by the induction hypothesis W contains NormM/LCM for some finite extension M/L. Thus U contains NormM/KCM, as desired.

Proof.

For any finite abelian extension L/K, NormL/KCL is a subgroup of CK which is open (by Lemma 6.4.1) of index [L:K] (by Theorem 6.4.3). Moreover, by Corollary 5.3.13, the correspondence LNormL/KCL is injective.
Conversely, let U be an open subgroup of CK of finite index. By Lemma 7.4.7, there exists a finite extension L1/K such that NormL1/KCL1U. By the adelic norm limitation theorem (Theorem 6.4.5), we also have NormL1/KCL1=NormL2/KCL2U for L2/K the maximal abelian subextension of L1/K. By Theorem 6.4.3 again, we have an isomorphism Gal(L2/K)CK/NormL2/KCL2, via which the subgroup U/NormL2/KCL2 corresponds to a subgroup H of Gal(L2/K). Taking L to be the fixed field of H, we deduce that NormL/KCL=U as desired.

Remark 7.4.9.

As with the proof of the local existence theorem, the proof of Theorem 7.4.8 is constructive in principle but not in practice: it involves constructing some extension much larger than the desired abelian extension, then invoking the norm limitation theorem to get down to an abelian extension. We remind the reader that there is no easy fix known for this (Remark 2.2.10).

Subsection An algebraic approach to the Second Inequality

Drawing inspiration from the calculation of norm groups given in Lemma 7.4.3, we now explain how to use similar ideas to give an algebraic proof of the Second Inequality. Again, the key case is where L/K is a cyclic extension of number fields of prime degree p and ζpK. To modify the calculation from Lemma 7.4.3 to compute the norm group of a single Kummer extension, we use a second set of places.

Proof.

By Kummer theory (Theorem 1.2.6), we can choose a finite set S of places of K containing all infinite places, all places that ramify in L, and all places above p so that L=K(Δ1/p) for Δ=oK,S(L)p. This remains true after enlarging S, so by Corollary 6.2.10 we can further ensure that IK=IK,SK.
Put N:=K((oK,S)1/p). By Kummer theory again,
Gal(N/K)Hom(oK,S/(oK,S)p,Z/pZ).
By Corollary 6.2.11, oK,S/(oK,S)p(Z/pZ)s. Choose generators g1,,gs1 of Gal(N/L); these correspond in Hom(oK,S/(oK,S)p,Z/pZ) to a set of homomorphisms whose common kernel is precisely Δ/(oK,S)p. We thus need to find, for each gi, a place vi such that the kernel of gi is the same as the kernel of oK,SKvi/(Kvi)p; we can then take T={v1,,vs1}.
Let Ni be the fixed field of gi; by Corollary 7.1.15 (which we deduced from the First Inequality) or Proposition 7.2.3 (which was part of the analytic proof of the Second Inequality), there are infinitely many primes of Ni that do not split in N. So we can choose a place wi of each Ni such that their restrictions vi to K are distinct, not contained in S, and don’t divide p.
We claim Ni is the maximal subextension of N/K in which vi splits completely (i.e., the decomposition field of vi). On one hand, vi does not split completely in N, so the decomposition field is no larger than Ni. On the other hand, the decomposition field is the fixed field of the decomposition group, which has exponent p and is cyclic (since vi does not ramify in N). Thus it must have index p in N, so must be Ni itself.
Thus L=Ni is the maximal subextension of N in which all of the vi split completely. We conclude that for xoK,S, x belongs to Δ iff Kvi(x1/p)=Kvi for all i, which occurs iff xKvip. That is, Δ is precisely the kernel of the map oK,SiKvi/(Kvi)p. This proves the claim.
We have the following modified version of Lemma 7.4.2.

Proof.

It is again clear that
(oK,ST)pWS,TK.
To prove the reverse inclusion, it will again suffice to prove that yWS,TK, if we set L=K(y1/p), then NormL/KCL=CK; namely, Theorem 7.1.2 will then imply L=K and so y(K)p.
Since oK,SvToKv/(oKv)p is surjective, any element of IK,ST can be written as the product of an element of oK,S with an element of IK,ST which is a p-th power at each place of T. In particular, by Lemma 4.3.13 such an element is a norm from L at each place of T; we can now reprise the proof of Lemma 7.4.2, skipping over the places in T, to deduce that we have a norm from L.

Proof.

By Lemma 7.4.11, we have an exact sequence
1oK,ST(oK,ST)pIK,STWS,TCKKWS,T/K1,
with which we may compute as in Lemma 7.4.3: the left group has order p#(ST)=p2s1 by Corollary 6.2.11 while the middle group has order p2s by Lemma 7.4.4, so
[CK:KWS,T/K]=p.
Meanwhile, we can check by local reciprocity that NormL/KIL,S=IK,S (compare the proof of Lemma 7.4.11).
  • For vS, elements of (Kv)p are norms from any abelian extension of Kv of exponent p (by Lemma 4.3.13).
  • For vT, v splits in L and so Lw=Kv.
  • For vST, v is unramified in L and so NormLw/KvoLw=oKv.
Hence KWS,TNormL/KCL, completing the proof.

Proof.

For xCK, NormL/K(x)=xp; this implies that both groups in question are killed by p. In particular, multiplication by d=[K:K], which divides p1, is an isomorphism on these groups.
Suppose xCK maps to the identity in HT0(Gal(L/K),CL). We can then choose a representative of the class of x in HT0(Gal(L/K),CL) of the form yd; then y also maps to the identity in HT0(Gal(L/K),CL). That is, y=NormL/K(z) for some zCL, and
yd=NormK/K(y)=NormL/K(z)NormL/KCL.
Thus xNormL/KCL, as needed.

Proof.

As in the proof of Theorem 7.2.9, we use an induction argument to reduce to proving that for L/K cyclic,
HT0(Gal(L/K),CL)[L:K].
In fact, the same induction (considering H2 in place of HT0) allows us to further reduce to the case where [L:K]=p is prime.
Let K=K(ζp) and L=L(ζp); 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. By Lemma 7.4.12 and Lemma 7.4.13,
#HT0(Gal(L/K),CL)#HT0(Gal(L/K),CL)[L:K]=p
as desired.

Exercises Exercises

1.

Complete the proof of Lemma 7.4.4 by showing that if |p|v=pn for some positive integer n, then [oKv:(oKv)p]=pn+1. (Remember that K is a number field containing a primitive p-th root of unity.)
Hint.
Using the logarithm map, we obtain an isomorphism oKv/μKvZpn (even when p=2). See [38], Proposition II.5.7 for details.

2.

Let K be a number field. Prove that for every positive integer n, CKn is the intersection of the norm groups NormL/KCL over all abelian extensions L/K of exponent n.