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Notes on class field theory

Kiran S. Kedlaya

Section 2.3 The principal ideal theorem

Reference.

[36], section V.3 (but you won’t find the proofs I’ve omitted there either); [37], section VI.7 (see also IV.5); [33], 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 2.3.7 (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 \(K = \QQ(\sqrt{-5})\text{,}\) then \(L = \QQ(\sqrt{-5}, \sqrt{-1})\text{,}\) and the nonprincipal ideal class of \(K\) is represented by \((2, 1+\sqrt{-5})\text{,}\) which is generated by \(1+\sqrt{-1}\) in \(L\text{.}\)

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 \(L\text{;}\) then an ideal of \(L\) is principal if and only if its image under the Artin map \(J_L \to \Gal(M/L)\) is trivial. So our first step will be to give a purely group-theoretic description of the map \(V: \Gal(L/K) \to \Gal(M/L)\) corresponding to the extension homomorphism \(\Cl(K) \to \Cl(L)\) (i.e., making the diagram in Figure 2.3.3 commute, in which the horizontal arrows are Artin maps).
Figure 2.3.3.
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 \(\Gal(\overline{K}/K)\) is still an unramified abelian extension of \(L\) and so is contained in \(M\text{.}\)
  2. The maximal subextension of \(M/K\) which is abelian over \(K\) is equal to \(L\text{.}\)

Subsection Translation into group theory

Definition 2.3.4.

Given a finite group \(G\text{,}\) let \(G^{\ab}\) denote the maximal abelian quotient of \(G\text{;}\) that is, \(G^{\ab}\) is the quotient of \(G\) by its commutator subgroup \(G'\text{.}\) Then the previous discussion implies that \(\Gal(M/L)\) is the commutator subgroup of \(\Gal(M/K)\) and \(\Gal(M/K)^{\ab} = \Gal(L/K)\text{.}\) We may thus relabel Figure 2.3.3 as in Figure 2.3.5.
Figure 2.3.5.
We now give the purely group-theoretic interpretation of the map \(V\) in Figure 2.3.5.

Definition 2.3.6.

Let \(G\) be a finite group and \(H\) a (not necessarily normal) subgroup. Let \(g_1, \dots, g_n\) be left coset representatives of \(H\) in \(G\text{:}\) that is, \(G = g_1H \cup \cdots \cup g_nH\text{.}\) For \(g \in G\text{,}\) put \(\phi(g) = g_i\) if \(g \in g_iH\) (i.e., \(g_i^{-1}g \in H\)). Put
\begin{equation*} V(g) = \prod_{i=1}^n \phi(gg_i)^{-1}(gg_i); \end{equation*}
then \(V(g)\) always lands in \(H\text{.}\)
Now consider what happens when we compose the map \(g \mapsto V(g): G \to H\) (which is not necessarily a homomorphism) with the projection \(H \to H^{\ab}\text{.}\) It will follow from Theorem 2.3.7 that the resulting map \(G \to H^{\ab}\) is a homomorphism which factors through \(G^{\ab}\text{.}\) The induced map \(V: G^{\ab} \to H^{\ab}\) is called the transfer map (in German “Verlagerung”, hence the use of the letter \(V\) in the notation).

Proof.

Remark 2.3.8.

In lieu of establishing Theorem 2.3.7 directly, one can derive it from properties of homology of finite groups. See Exercise 2.
Setting aside the proof of Theorem 2.3.7 for the moment, let’s see that this does indeed give the correct map in Figure 2.3.5 when we take \(G = \Gal(M/K)\) and \(H = \Gal(M/L)\text{,}\) so that \(G/H = \Gal(L/K)\text{.}\) This amounts to computing what happens when we apply all of the maps starting with a prime \(\gothp\) of \(K\) at the top left of the diagram.
Choose a prime \(\gothq\) of \(L\) over \(\gothp\) and a prime \(\gothr\) of \(M\) over \(\gothq\text{,}\) let \(G_{\gothr} \subseteq G\) be the decomposition group of \(\gothr\) over \(K\) (i.e., the stabilizer of \(\gothr\) under the action of \(G\) on the primes above \(\gothp\)), and let \(g \in G_{\gothr}\) be the Frobenius of \(\gothr\text{.}\) Keep in mind that since \(G\) is not abelian, \(g\) depends on the choice of \(\gothr\text{,}\) not just on \(\gothq\text{;}\) that is, there’s no Artin map into \(G\text{.}\)
Let \(\gothq_1, \dots, \gothq_r\) be the primes of \(L\) above \(\gothp\text{;}\) then the image of \(\gothp\) in \(L\) is \(\prod_i \gothq_i\text{,}\) and the image of that product under the Artin map is \(\prod_i \Frob_{M/L}(\gothq_i)\text{.}\) To show that this equals \(V(g)\text{,}\) we make a careful choice of the coset representatives \(g_i\) in the definition of \(V\text{.}\) Namely, decompose \(G\) as a union of double cosets \(G_{\gothr} \tau_i H\text{.}\) Then the primes of \(L\) above \(\gothp\) correspond to these double cosets, where the double coset \(G_{\gothr} \tau_i H\) corresponds to \(L \cap \gothr^{\tau_i}\text{.}\) Let \(m\) be the order of \(\Frob_{L/K}(\gothp)\) and write \(G_{\gothr} \tau_i H = \tau_iH \cup g\tau_i H \cup \cdots \cup g^{m-1}\tau_i H\) for each \(i\text{;}\) we then use the elements \(g_{ij} = g^j \tau_i\) as the left coset representatives to define \(\phi\) and \(V\text{.}\) Thus the equality \(V(g) = \prod_i \Frob_{M/L}(\gothq_i)\) 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 \(\gothm\text{,}\) and \(\gothm \gotho_L\) 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 \(\gothm\)), then the induced map \(\Cl^{\gothm}(K) \to \Cl^{\gothm}(L)\) 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. In fact, it is an open problem to show that every number field \(K\) admits an extension which has class number 1.
One approach to constructing such an extension would be to consider the class field tower over \(K\text{,}\) in which \(K_0 = K\) and for each positive integer \(i\text{,}\) \(K_i\) is the Hilbert class field of \(K_{i-1}\text{.}\) However, Golod and Shafarevich showed that in certain cases this sequence grows without bound; for example, this holds if \(K\) is an imaginary quadratic field in which at least six distinct primes of \(\QQ\) ramify. In particular, in such cases \(K\) admits an infinite unramified extension. (See [4] for more discussion.)

Remark 2.3.13.

Let \(K\) be a number field. Let \(M\) be the \(\gotho_K\)-submodule of \(K[x]\) consisting of integer-valued polynomials, meaning those that map \(\gotho_K\) 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 [58], 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 [35], 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.

Hint.
One approach to proving independence from choices is to change one \(g_i\) at a time. Also, notice that \(\phi(gg_1), \dots, \phi(gg_n)\) are a permutation of \(g_1, \dots, g_n\text{.}\)

3.

Let \(H \subseteq G\) be an inclusion of finite groups. Let \(G'\) and \(H'\) be the commutator subgroups of \(G\) and \(H\text{.}\) Let \(\ZZ[G]\) be the group algebra of \(G\text{,}\) i.e., the (noncommutative) ring of formal linear combinations \(\sum_{g \in G} n_g [g]\) with \(n_g \in \ZZ\text{,}\) multiplied by putting \([g][h] = [gh]\text{.}\) Let \(I_G \subset \ZZ[G]\) be the ideal of sums \(\sum n_g[g]\) with \(\sum n_g = 0\) (called the augmentation ideal; see Section 3.3). Let
\begin{equation*} \delta: H/H' \to (I_{H}+I_GI_{H})/I_GI_{H} \end{equation*}
be the homomorphism taking the class of \(h\) to the class of \([h]-1\text{.}\) Prove that \(\delta\) is an isomorphism.
Hint.
Show that the elements
\begin{equation*} [g]([h]-1) \qquad \mbox{for \(g \in \{g_1,\dots,g_n\}, h \in H\)} \end{equation*}
form a basis of \(I_H + I_G I_H\) as a \(\ZZ\)-module. For more clues, see [37], Lemma VI.7.7.

4.

With notation as in Exercise 3, prove that the diagram in Figure 2.3.14 commutes, where \(S\) is given by \(S(x) = x ([g_1] + \cdots + [g_n])\text{.}\)
Figure 2.3.14.

5.

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 \(\ZZ/e_1 \ZZ \times \cdots \times \ZZ/e_m \ZZ\text{.}\) Let \(f_i\) be an element of \(G\) lifting a generator of \(\ZZ/e_i \ZZ\) and put \(h_i = f_i^{-e_i} \in H\text{;}\) then \(0 = \delta(f_i^{e_i} h_i)\text{,}\) which can be rewritten as \(\delta(f_i) \mu_i\) for some \(\mu_i \in \ZZ[G]\) congruent to \(e_i\) modulo \(I_G\text{.}\) Now check that
\begin{equation*} n \mu_1 \cdots \mu_m \equiv [g_1] + \cdots + [g_n] \pmod{I_H \ZZ[G]}\text{.} \end{equation*}
For more details, see [37], Theorem VI.7.6.
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