Math 254B, UC Berkeley, Spring 2002 (Kedlaya) The Principal Ideal Theorem

Reference. Milne, section V.3 (but you won’t find the proofs I’ve omitted there either); Neukirch, section VI.7 (see also IV.5); Lang, Algebraic Number Theory, section XI.5.

For a change, we’re going to prove something, if only assuming the Artin reciprocity law which we haven’t proved. Or rather, we’re going to sketch a proof that you will 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 in the old days the word “capitulate” meant “to become principal”. Etymology left to the reader.)

Theorem 1 (Principal ideal theorem). Let L be the Hilbert class field of the number field K. Then every ideal of K becomes principal in L.

(Warning: this does not mean that L has class number 1!) Example: if K = ℚ(\sqrt{−5}), then L = ℚ(\sqrt{−5}, \sqrt{−1}), and the nonprincipal ideal class of K is represented by (2, 1+\sqrt{−5}), which is generated by 1 + \sqrt{−1} in L.

The idea is that given Artin reciprocity, this reduces to a question in group theory. Namely, 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 I_{L} → Gal(M/L) is trivial. So what we need is to find a map V such that

\xymatrix{Cl(K) \ar[r] \ar[d] & Gal(L/K) \ar[d]^{V} \\ Cl(L) \ar[r] & Gal(M/L)}

commutes, then show that V is the zero map. (The horizontal arrows are the Artin maps.) Regarding the relationship between K, L and M:

(a) M is Galois over K (because its image under any element of Gal(\overline{K}/K) is still an abelian extension of L unramified at all finite and infinite places, so is contained in M) and unramified everywhere (since M/L and L/K are unramified).

(b) L is the maximal subextension of M/K which is abelian over K (since any abelian subextension is unramified over K, so contained in L).

Given a finite group G, let G^{ab} denote the maximal abelian quotient of G; that is, G^{ab} is the quotient of G by its commutator subgroup G′. Then (b) implies that Gal(M/L) is the commutator subgroup of Gal(M/K) and Gal(M/K)^{ab} = Gal(L/K).

Before returning to the principal ideal theorem, we need to do a bit of group theory. Let G be a finite group and H a subgroup (but not necessarily normal!). Let g_{1}, …, g_{n} be left coset representatives of H in G: that is, G = g_{1}H ∪ ⋯ ∪ g_{n}H. For g ∈ G, put ϕ(g) = g_{i} if g ∈ g_{i}H (i.e., g_{i}^{−1}g ∈ H). Put

V (g) = ∏_{i=1}^{n} ϕ(gg_{i})^{−1}(gg_{i});

then V (g) always lands in H. In particular, it induces a map V : G → H^{ab}.

Theorem 2. The map V : G → H^{ab} is a homomorphism, does not depend on the choice of the g_{i}, and induces a homomorphism G^{ab} → H^{ab} (i.e., kills commutators in G).

The map G^{ab} → H^{ab} is called the transfer map (in German, “Verlagerung”, hence the V).

Now let’s return to that diagram:

\xymatrix{Cl(K) \ar[r] \ar[d] & Gal(L/K) = Gal(M/K)^{ab} \ar[d]^{V} \\ Cl(L) \ar[r] & Gal(M/L) = Gal(M/L)^{ab}}

and show that the transfer map V does actually make this diagram commute; it’s enough to check this when we stick a prime 𝔭 of K in at the top left. For consistency of notation, put G = Gal(M/K) and H = Gal(M/L), so that G/H = Gal(L/K). Choose a prime 𝔮 of L over 𝔭 and a prime 𝔯 of M over 𝔮, let G_{𝔯} ⊆ G be the decomposition group of 𝔯 over K (i.e., the set of automorphisms mapping 𝔯 to itself) and let g ∈ G_{𝔯} be the Frobenius of 𝔯. (Note: since G is not abelian, g depends on the choice of 𝔯, not just on 𝔮. I.e., there’s no Artin map into G.)

Let 𝔮_{1}, …, 𝔮_{r} be the primes of L above K; then the image of 𝔭 in L is ∏_{i} 𝔮_{i}, and the image of that product under the Artin map is ∏_{i} Frob_{M/L}(𝔮_{i}); we’re trying to show this equals V (g) = ∏_{i} ϕ(gg_{j})^{−1}gg_{j}.

We do this by carefully picking the coset representatives g_{i} in the definition of V. Namely, decompose G as a union of double cosets G_{𝔯}τ_{i}H. Then the primes of L above 𝔭 correspond to these double cosets, where the double coset G_{𝔯}τ_{i}H corresponds to L ∩ 𝔯^{τ_{i}}. Choose the left coset representatives of H by writing G_{𝔯}τ_{i}H = τ_{i}H ∪ gτ_{i}H ∪ ⋯ ∪ g^{m−1}τ_{i}H for each i, then using g_{ij} = g^{j}τ_{i} as the left coset representatives to define ϕ and V. (Here m is the order of Frob_{L/K}(𝔭).) Thus the equality V (g) = ∏_{i} Frob_{M/L}(𝔮_{i}) follows from the following lemma.

Lemma 3. If L ∩ 𝔯^{τ_{i}} = 𝔮_{i}, then Frob_{M/L}(𝔮_{i}) = ∏_{j=0}^{m−1} ϕ(gg_{ij})^{−1}gg_{ij}.

Thus the principal ideal theorem now follows from the following fact.

Theorem 4. Let G be a finite group and H its commutator subgroup. Then the transfer map V : G^{ab} → H^{ab} is zero.

Exercises.

1. Prove Theorem 2. (Hint: one approach to proving independence from choices is to change one g_{i} at a time. Also, show that ϕ(gg_{1}), …, ϕ(gg_{n}) are a permutation of g_{1}, …, g_{n}.)

2. Prove Lemma 3.

3. Let ℤ[G] be the group algebra of G (formal linear combinations ∑_{g∈G} n_{g}[g] with n_{g} ∈ ℤ, multiplied by putting [g][h] = [gh]) and let I_{G} be the ideal of sums ∑ n_{g}[g] with ∑ n_{g} = 0. Show that the following diagram commutes:

\xymatrix{G/G′ \ar[r]^{V} \ar[d]^{δ} & G′/G′′ \ar[d]^{δ} \\ I_{G}/I_{G}^{2} \ar[r]|{S} & (I_{G′} + I_{G}I_{G′})/I_{G}I_{G′}}

where the maps δ are induced by g ↦ [g] − 1 and S is given by S(x) = x ∑_{h}[h], for h running over a set of left coset representatives of G/H. Also, show that each δ is an isomorphism. (Hint: do the second part first. For more hints, see Neukirch, Lemma VI.7.7; but try it first before peeking!)

4. Prove Theorem 4. (Hint: quotient by the commutator subgroup of H to reduce to the case where H is abelian. Pick generators g_{1}, …, g_{n} of G, then write down an integer matrix m_{ij} of determinant G/H such that ∏_{i=1}^{n} g_{i}^{m_{ij}} ∈ H for all j. Now apply δ to get relations in ℤ[G]/ℤ[G]I_{G′}. For more hints, see Neukirch, Theorem VI.7.6; again, try first before peeking.)