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Notes on class field theory

Section 2.3 The principal ideal theorem

Reference.

[37], section V.3 (but you won’t find the proofs I’ve omitted there either); [38], section VI.7 (see also IV.5); [34], section XI.5.

Subsection Statement of the theorem

For a change, we’re going to prove something, although the proof will depend on the Artin reciprocity law which we haven’t proved. Or rather, we’re going to sketch a proof that you will get to fill in by doing the exercises. (Why should I have all the fun?)
The following theorem is due to Furtwängler, a student of Hilbert. (It’s also called the “capitulation” theorem, because the word “capitulate” was formerly used to mean “to become principal”. Etymology left to the reader.)

Proof.

This will follow by combining Theorem 2.3.8 (construction of the transfer homomorphism), Lemma 2.3.9 (implication that vanishing of the transfer homomorphism implies the desired result), and Theorem 2.3.10 (vanishing of the transfer homomorphism).

Example 2.3.2.

If K=Q(5), then L=Q(5,1), and the nonprincipal ideal class of K is represented by (2,1+5), which is generated by 1+1 in L.

Remark 2.3.3.

The property of making every ideal of K principal is in no way specific to the Hilbert class field. See Exercise 1.

Subsection First steps of the proof

The idea of the proof is to apply Artin reciprocity to reduce to a problem purely in finite group theory, which we then solve. To this end, let M be the Hilbert class field of L; then an ideal of L is principal if and only if its image under the Artin map JLGal(M/L) is trivial. So our first step will be to give a purely group-theoretic description of the map V:Gal(L/K)Gal(M/L) corresponding to the extension homomorphism Cl(K)Cl(L) (i.e., making the diagram in Figure 2.3.4 commute, in which the horizontal arrows are Artin maps).
Figure 2.3.4.
In order to proceed further, we must extract more information about the Galois groups in question.
  1. The extension M/K is unramified because M/L and L/K are. It is also Galois: its image under any element of Gal(K/K) is still an unramified abelian extension of L and so is contained in M.
  2. The maximal subextension of M/K which is abelian over K is equal to L.

Subsection Translation into group theory

Definition 2.3.5.

Given a finite group G, let Gab denote the maximal abelian quotient of G; that is, Gab is the quotient of G by its commutator subgroup G. Then the previous discussion implies that Gal(M/L) is the commutator subgroup of Gal(M/K) and Gal(M/K)ab=Gal(L/K). We may thus relabel Figure 2.3.4 as in Figure 2.3.6.
Figure 2.3.6.
We now give the purely group-theoretic interpretation of the map V in Figure 2.3.6.

Definition 2.3.7.

Let G be a finite group and H a (not necessarily normal) subgroup. Let g1,,gn be left coset representatives of H in G: that is, G=g1HgnH. For gG, put ϕ(g)=gi if ggiH (i.e., gi1gH). Put
V(g)=i=1nϕ(ggi)1(ggi);
then V(g) always lands in H. (Note that ϕ(gg1),,ϕ(ggn) is a permutation of g1,,gn.)
Now consider what happens when we compose the map gV(g):GH (which is not necessarily a homomorphism) with the projection HHab. It will follow from Theorem 2.3.8 that the resulting map GHab is a homomorphism which factors through Gab. The induced map V:GabHab is called the transfer map (in German “Verlagerung”, hence the use of the letter V in the notation).

Proof.

Setting aside the proof of Theorem 2.3.8 for the moment, let’s see that this does indeed give the correct map in Figure 2.3.6 when we take G=Gal(M/K) and H=Gal(M/L), so that G/H=Gal(L/K). This amounts to computing what happens when we apply all of the maps starting with a prime p of K at the top left of the diagram.
Choose a prime q of L over p and a prime r of M over q, let GrG be the decomposition group of r over K (i.e., the stabilizer of r under the action of G on the primes above p), and let gGr be the Frobenius of r. Keep in mind that since G is not abelian, g depends on the choice of r, not just on q; that is, there’s no Artin map into G.
Let q1,,qr be the primes of L above p; then the image of p in L is iqi, and the image of that product under the Artin map is iFrobM/L(qi). To show that this equals V(g), we make a careful choice of the coset representatives gi in the definition of V. Namely, decompose G as a union of double cosets GrτiH. Then the primes of L above p correspond to these double cosets, where the double coset GrτiH corresponds to Lrτi. Let m be the order of FrobL/K(p) and write GrτiH=τiHgτiHgm1τiH for each i; we then use the elements gij=gjτi as the left coset representatives to define ϕ and V. Thus the equality V(g)=iFrobM/L(qi) follows from the following lemma.

Proof.

Subsection The final group-theoretic ingredient

With this, Theorem 2.3.1 follows from the following fact.

Proof.

Remark 2.3.11.

The fact that Theorem 2.3.10 is so general means that we can easily obtain some extensions of Theorem 2.3.1. For example, it was observed by Iyanaga that if L is the ray class field of K of some modulus m, and moL is the extension of this modulus to L (that is, extend the finite part and take all places of L above the infinite places in m), then the induced map Clm(K)Clm(L) again vanishes.

Subsection Additional remarks

Remark 2.3.12.

One important qualification of Theorem 2.3.1 is that L need not itself have class number 1. This raises the question of whether every number field K admits a finite (not necessarily abelian) extension which has class number 1.
One approach to constructing such an extension would be to consider the class field tower over K, in which K0=K and for each positive integer i, Ki is the Hilbert class field of Ki1. If this were to stabilize at some point Ki, then the latter would be a field of class number 1.
However, Golod and Shafarevich showed that in certain cases this sequence grows without bound, in which case K admits an infinite (solvable) unramified extension. For example, if K is an imaginary quadratic field in which at least six distinct finite places of Q ramify, or a real quadratic field in which at least eight distinct finite places of Q ramify, then the 2-part of the class field tower is unbounded. See [4], Chapter IX for a proof of this statement and some further discussion; [44] for a number of additional results, including some examples of quadratic fields with prime discriminant with infinite class field towers; and [58] for more results plus a survey of recent (as of 2016) literature.
In fact, it can be shown (see Exercise 8) that the finiteness of the class field tower is equivalent to the existence of a finite extension of K of class number 1, which a priori need not be solvable or even Galois! That is, if a number field K is contained in a number field L of class number 1, then we can always ensure that L/K is Galois and solvable. In particular, by the previous paragraph there exist number fields which admit no finite extensions of class number 1.

Remark 2.3.13.

Let K be a number field. Let M be the oK-submodule of K[x] consisting of integer-valued polynomials, meaning those that map oK into itself. The field K is said to be a Pólya field if M admits a basis consisting of polynomials of pairwise distinct degrees; such a basis is called a regular basis. Any field with trivial class group is a Pólya field, but not conversely. The terminology is due to Zantema [62], who showed among other things that every cyclotomic field is a Pólya field.
Using Theorem 2.3.1, Leriche showed that the Hilbert class field of any number field is a Pólya field (see [36], Corollary 3.2). In particular, every number field can be embedded into a Pólya field via an abelian extension, whereas it is unknown whether every number field can be embedded into a field of class number one (and Remark 2.3.12 shows that solvable extensions are definitely not enough).

Exercises Exercises

1.

Let K be a number field. Prove without using class field theory that there exists a finite extension L/K such that every prime of K becomes principal in L.
Hint.
Since Cl(K) is finite, it suffices to fix a prime p and produce L such that p becomes principal in L. Again by the finiteness of Cl(K), there exists a positive integer a such that pa is principal; adjoin a suitable root of a generator.

2.

With notation as in Definition 2.3.7, prove that the map V:GHab is independent of the choice of g1,,gn.
Hint.
Independence of the order of the cosets follows because we are mapping into Hab. To check independence of the representatives of individual cosets, change one gi to gih for some hH; this changes ϕ(ggi)1ggi by a conjugation which has no effect in Hab.

3.

With notation as in Definition 2.3.7, prove that the map V:GHab is a homomorphism which factors through Gab.
Hint.
For the homomorphism, rewrite V(g)V(g) as iϕ1(ggi)ggiϕ1(ggj)ggj where j runs through the indices in a different order so that gi=ϕ(ggj). For the factorization through Gab, check directly from the formula that V(g)V(g)=V(g)V(g).

4.

Hint.
Note that ϕ(ggij)=gi(j+1) for j=0,,m2, so all of the terms in the product vanish except for τi1gmτi. We directly verify that this is the desired Frobenius element by noting that the residue class degree of q over p equals m. See also [38], Proposition IV.5.9.

5.

Let HG be an inclusion of finite groups. Let G and H be the commutator subgroups of G and H. Let Z[G] be the group algebra of G, i.e., the (noncommutative) ring of formal linear combinations gGng[g] with ngZ, multiplied according to the rule that [g][h]=[gh]. Let IGZ[G] be the ideal of sums gGng[g] with gGng=0 (called the augmentation ideal; see Section 3.3). Let
δ:H/H(IH+IGIH)/IGIH
be the homomorphism taking the class of h to the class of [h]1. Prove that δ is an isomorphism.
Hint.
Show that the elements
[g]([h]1) for g{g1,,gn},hH
form a basis of IH+IGIH as a Z-module. For more clues, see [38], Lemma VI.7.7.

7.

Hint.
Quotient by the commutator subgroup of H to reduce to the case where H is abelian. Apply the classification of finite abelian groups to write G/H as a product of cyclic groups Z/e1Z××Z/emZ. Let fi be an element of G lifting a generator of Z/eiZ and put hi=fieiH. In the notation of Exercise 5 we have 0=δ(fieihi), which can be rewritten as δ(fi)μi for some μiZ[G] congruent to ei modulo IG. Now check that
nμ1μm[g1]++[gn](modIHZ[G]).
For more details, see [38], Theorem VI.7.6.

8.

Let L/K be an extension of number fields such that Cl(L) is trivial. Prove that L contains the Hilbert class field of K. Then deduce the following corollary: the class field tower of K stabilizes if and only if K is contained in some number field of class number 1.
Hint.
Take the compositum of L with the Hilbert class field of K.