Kummer generators and lambda invariants

3-adic L-functions

Let χ be the quadratic character attached to K 0 and let L 3 (s, χ) be the 3-adic Lfunction for χ. There is a power series f (T ) = f (T, χ) = a 0 + a 1 T + · · · ∈ Z 3 [[T ]] such that L 3 (s, χ) = f ((1 + 3) s − 1).

Therefore,

where ω is the character for Q( √ −3) and D is the discriminant of K 0 . Also, λ is the smallest index i such that a i ≡ 0 (mod 3). In particular, λ ≥ 2 if and only if a 0 ≡ a 1 ≡ 0 (mod 3).

Assume from now on that 3 ramifies in K 0 . Then 3|D and χ(3) = 0. Also, since 3|D, we have Norm(ε 0 ) = +1. The above equations yield

Proposition 1. (a) Suppose that 3 splits in F 0 and 3 ∤ h + . Then λ ≥ 2 ⇐⇒ log 3 ε 0 ≡ 0 (mod 9).

(b) Suppose that 3 splits in F 0 and 3 | h + . Then λ ≥ 2.

(c) Suppose that 3 is inert in F 0 and that 3|h − . Then

Proof. If 3 splits, then χω −1 (3) = 1, so a 0 = 0. If 3 is inert, then χω −1 (3) = −1, so a 0 = 2h − . Our assumptions imply that a 0 ≡ 0 (mod 3). The condition in part (c) is clearly equivalent to a 1 ≡ 0 (mod 3). It follows from the 3-adic class number formula that (log 3 ε 0 )/ √ D ∈ Q 3 . Therefore, log 3 ε 0 ≡ 0 (mod 9) is equivalent to log 3 ε 0 ≡ 0 (mod 9 √ D). Hence, the condition in (a) is equivalent to a 1 ≡ 0 (mod 3).

If d ≡ 2 (mod 3), write ε 2 0 = 1 + 3x + y √ 3d with 2x, 2y ∈ Z. Then

. Therefore, the condition in (b) implies that a 1 ≡ 0 (mod 3).

Remark. For a slightly different version of the proposition, see [6, Thm. 1].

Preliminaries

where we identify the groups for arbitrary i with those for i = 0. Let

σg g y y y y y y y y

n denote the subgroup of A n on which σ acts by inversion.

group L × /(L × ) 3 σ of exponent 3. Therefore, the image is trivial, which yields the surjectivity in part (a). Now suppose that x ∈ K × is a cube in L × . Then x 2 is the norm of a cube, hence is a cube. This implies that x is also a cube, so the map in part (a) in injective. The proof of (b) is similar.

. Since K 1 /K 0 is Galois, the polynomial has all three roots in K 1 , hence K 1 contains the cube roots of unity. Contradiction.

Proof. This follows from the previous two lemmas.

Structure of units

Our goal is to understand the Hilbert 3-class field of F 1 and to lay the groundwork for the proof of Theorem 2 in Section 6. The main part of the proof relies on an analysis of the Kummer generators of unramified extensions of L 1 . In our situation, these Kummer generators arise from units in K 1 .

Let E denote the units of K 1 . By the Dirichlet unit theorem, E/E 3 ≃ (Z/3Z) 5 . We have

where the groups (E/E 3 ) 1±g are the ± eigenspaces for the action of g. Lemma 3, or the fact that 1 + g is the norm from K 1 to B 1 , shows that (E/E 3 ) 1+g is represented by the units of B 1 and has dimension 2 over F 3 . Therefore (E/E 3 ) 1−g has dimension 3.

Let

is Galois equivariant for the actions of Gal(L 1 /Q) on M , A, and µ 3 . Since g acts as −1 on µ 3 and acts as −1 on M , it acts as +1 on A. In particular, any lift of g to an element of Gal( H 3 /F 1 ) commutes with Gal( H 3 /L 1 ). Since Gal( H 3 /F 1 ) is therefore abelian of order 2 × 27, we can lift g to the unique element of order 2 in Gal( H 3 /F 1 ). We continue to call this element g. The fixed field of g restricted to H 3 is a field H 3 that is Galois over F 1 . Of course, if H 3 /L 1 is unramified, then this is simply a consequence of the fact that the Hilbert 3-class field of F 1 lifts to the minus part of the Hilbert 3-class field of L 1 .

and we can rewrite the Kummer pairing as

In many ways, it is more natural to state the results over F 1 . But, for the proofs, it is often necessary to work in the larger field L 1 in order to use Kummer generators.

Since τ 3 = 1, we find that (1 − τ ) 3 kills M . There are two filtrations:

denotes the kernel of (1 − τ ) j (cf. [5]). Moreover,

There are also two filtrations on A:

This and similar facts yield nondegenerate pairings

Proof. Regard τ as a linear transformation of M . The characteristic polynomial of τ is T 3 − 1 = (T − 1) 3 . The equivalence of (a), (b), (c), (d) follows easily from a consideration of the three possibilities for the Jordan canonical form of τ . The equivalence of (e), (f), (g), and (h) follows similarly. The equivalence of (b) and (g) follows from the duality between M i and A/A i .

All the standard questions of genus theory for an odd prime can be asked about A and M . For instance, let H i be the fixed field of A i . Then H 1 is the "genus subfield" of H 3 ; namely, H 1 is the maximal abelian extension of F 0 contained in H 3 . Equivalently, H 1 is maximal with τ acting trivially on Gal(H 1 /F 1 ). Thus we can call A/A 1 the "genus group" for A.

We can also define M/M 1 to be the "genus unit group," as it is the maximal quotient of M with trivial action by τ . Similarly, we can call A 1 and M 1 the ambiguous subgroups. Then the nondegenerate pairing

says that the ambiguous unit group pairs nontrivially with the genus group. A similar statement can be made for

Thus, ambiguous unit classes are Kummer generators for the genus field H 1 . In general, the duality between M i and A/A i says that M i is the group of Kummer generators for the fixed field of A i . Suppose now that dim M i = i for all i. Then there are units u 1 , u 2 , and u 3 such that u i ∈ M i but u i ∈ M i−1 , and {u 1 , u 2 , u 3 } is a basis for M . The fixed field of A i is then

We can extend the definition of genus and ambiguous groups. We call A i /A i+1 and M i /M i+1 the ith higher genus group and we call A i+1 /A i and M i+1 /M i the ith higher ambiguous groups. These are all maximal quotients with trivial τ action. There are also nondegenerate pairings

See [5], section 2 for a use of some of these groups.

Capitulation

Let A(K i ) be the Sylow 3-subgroup of the class group of K i . There is a natural homomorphism A(K 0 ) → A(K 1 ) that maps the class of an ideal of K 0 to the class of the ideal generated by that ideal in K 1 .

Then ε 1−τ = γ 3 for some γ ∈ E. Taking the norm to K 0 yields Norm(γ) 3 = 1. Since ζ 3 ∈ K 0 , we have Norm(γ) = 1. By Hilbert's Theorem 90, γ = η 1−τ for some η ∈ K 1 . Then α = ε/η 3 is fixed by τ , so it lies in K 0 . Let J = (η −1 ). Then J 3 = (α) = (τ α) = (τ J) 3 , so J is fixed by τ (alternatively, J 1−τ = (γ) = (1)). Therefore, J = Ip a 1 for some integer a, where I comes from K 0 and p 1 is the prime of K 1 above 3. This implies that

It is straightforward to check that the ideal class of I 1−g is independent of the various choices made and depends only on ε mod cubes. Therefore, ψ is well-defined. If ε = ε 0 , then we may take γ = η = 1, so α = ε 0 and J and I 1−g are trivial. Therefore, ε 0 ∈ Ker(ψ).

Suppose ψ(ε) = 1. Then I 1−g = (δ) for some δ ∈ K 0 . Then

0 mod cubes. Since ε ∈ M = (E/E 3 ) 1−g , we have ε 1−g ≡ ε 2 mod cubes. Therefore, ε 2 , and hence ε, is in ε 0 mod cubes. Consequently, the kernel of ψ is ε 0 . Now let B be an ideal in Ker(A(K 0 ) → A(K 1 )). Let B = (y) with y ∈ K 1 . Then y 1−τ ∈ E. Since the norm from K 1 to K 0 of a principal ideal is principal,

. But the ideal class of B g is the ideal class of B −1 , and the ideal class of B has order 3 since it capitulates, so ψ(u 1−g ) = B. Therefore, ψ is surjective.

Proof. Let I be an ideal of K 0 that becomes principal in K 1 . Say I = (y) with y ∈ K 1 . Then y 1−τ ∈ E. It is easy to see that the map ρ : I → (τ → y 1−τ ) gives an injective homomorphism from Ker(A(K 0 ) → A(K 1 )) to H 1 (K 1 /K 0 , E) (the image is the group of locally trivial cohomology classes; see [9]).

The Herbrand quotient is given by

, the norms of units are either ± ε 0 or ± ε 3 0 . Therefore, | H 0 | = 1 or 3. Therefore, |H 1 | = 3 or 9. Let p 1 be the prime of K 1 above 3. Since B 1 has class number 1 and p 2 1 is the prime of B 1 above 3, we have p 2 1 = (β) for some β ∈ B 1 . The cocycle τ → β 1−τ gives a cohomology class in H 1 (K 1 /K 0 , E). We claim that it is not in the image of ρ. If it is, then there is an ideal I of K 0 with I = (y) in K 1 and such that

This contradicts the fact that y/β ∈ K 0 . Therefore, the cocycle is not in the image of ρ. Consequently, the image of ρ has order 1 or 3. Since ρ is injective, this completes the proof.

Proposition 4. τ does not act trivially on

or 2. It has dimension 1 if and only if the map

Proof. This follows immediately from the preceding lemma and proposition.

The following is what we need in subsequent sections.

Theorem 1. Exactly one of the following cases holds:

The map from the class group of K 0 to the class group of K 1 is injective.

(c) The norm map from the units of K 1 to the units of K 0 is surjective.

The kernel of the map from the class group of K 0 to the class group of K 1 has order 3.

(c) The cokernel of the norm map from the units of K 1 to the units of K 0 is has order 3.

(d) There are units ε 1 , ε 2 ∈ K 1 such that {ε 0 , ε 1 , ε 2 } is a basis for M and such that ε 1−τ 2 = ε 1 , and ε 1+τ +τ 2 2 = 1.

Proof. Proposition 4 says that exactly one of (i)(a) and (ii)(a) holds. Proposition 2 implies that (a) is equivalent to (b) in both cases.

In particular, there are units u 1 and u 2 such that

(Note that we do not need to multiply u 1 by a cube since we can simply modify it by a cube, if necessary.) This implies that

Therefore, (i)(a) implies (i)(b), (i)(c), and (i)(d). Now assume (ii)(a). Since M [1 − τ ] has dimension greater than 1, Lemma 6 implies that (1 − τ ) 2 annihilates M . Let ε ∈ M . Then

Since the norm maps E 1+g to powers of ε 1+g 0 = 1, the map 1 + τ + τ 2 annihilates E/E 3 = (E/E 3 ) 1+g ⊕ M . Therefore, the norm from units of K 1 to units of K 0 is not surjective. Since ε 3 0 is in the image, the cokernel has order 3, which is (ii)(c).

Again assuming (ii)(a), we have units u 1 , u 2 , u 3 that form a basis of M and such that u 1−τ

, which means that the norm is surjective. Since (ii)(a) implies (ii)(c), which implies that the norm is not surjective, we have a contradiction.

Write ε 0 ≡ u a 1 u b 2 u c 3 mod cubes. Applying 1 − τ yields 1 ≡ u c 2 mod cubes. Therefore, c ≡ 0 (mod 3). By the claim, we cannot also have a ≡ 0 (mod 3). It follows that we may replace u 1 by ε 0 and obtain a basis of M satisfying the same relations as u 1 , u 2 , u 3 . We therefore assume that u 1 = ε 0 .

Since the image of the norm map has index 3, we have Norm(u 2 ) = ±(ε 0 ) 3k for some k, and by changing the sign of u 3 if necessary we may assume that ± = +. Then Norm(u 3 /ε k 0 ) = 1. Let

These are the desired units.

Continuing with the analogy with genus theory, we can ask if every ambiguous unit class in M 1 = M [1 − τ ] contains an ambiguous unit. The group of ambiguous units is generated by ε 0 . In case (i), where there is no capitulation from K 0 to K 1 , the unit ε 0 generates the group of ambiguous unit classes, so every ambiguous unit class contains an ambiguous unit. However, in case (ii), when there is capitulation from K 0 to K 1 , the dimension of M 1 is 2. It is curious to note that in this case, since ε 0 generates a subgroup of dimension 1, there are ambiguous unit classes that do not contain an ambiguous unit.

Subfields

. Then g acts as −1 on the Kummer generators of H i /L 1 and acts as −1 on µ 3 , so g acts trivially on Gal( H i /L 1 ). Therefore, H i /F 1 is abelian and has a unique element of order 2, which we call g. Let H i be the fixed field of g. Then H i /F 1 is Galois with group (Z/3Z) i .

We can actually say a lot more: the extensions H i /L 1 are lifts of extensions H ′ i /F 0 . We do not know any applications, but the proof shows that the result is closely related to the τ -structure of M .

Proof. We need the following result.

Lemma 7. Let p be a prime and let N 1 /N 0 be a Galois extension of fields of characteristic not p. Assume that Gal(N 1 /N 0 ) contains an element τ of order p. Let ε ∈ N × 1 be such that N 1 (ζ p , ε 1/p )/N 0 is Galois. Then ε τ −1 = β p for some β ∈ N 1 . Moreover, τ has an extension to an element of Gal(N 1 (ζ p , ε 1/p )/N 0 ) of order p if and only if β 1+τ +···+τ p−1 = 1.

Proof. Take any extension τ of τ to N 1 (ζ p , ε 1/p ) that is trivial on N 0 (ζ p ). Note that such extensions exist since τ must be trivial on N 0 (ζ p ) ∩ N 1 because its degree over N 0 is prime to p.

Let N ′ 0 ⊆ N 1 be the fixed field of τ . Then N 1 (ζ p , ε 1/p )/N ′ 0 (ζ p ) is Galois of order p or p 2 and hence abelian. Thus τ commutes with a generator of Gal(N 1 (ζ p , ε 1/p )/N 1 (ζ p )), so it follows (by a straightforward calculation or by the Kummer pairing) that ε τ −1 = β p for some β ∈ N 1 .

We have τ (ε 1/p ) = ε 1/p βζ for some (possibly trivial) pth root of unity ζ. An easy calculation shows that

Therefore, if β 1+τ +···+τ p−1 = 1, then τ has order p. Conversely, if τ has an extension with order p, then this extension is trivial on N 0 (ζ p ). Therefore, the above calculation shows that β 1+τ +···+τ p−1 = 1.

In our situation, the lemma directly implies that τ ∈ Gal(L 1 /L 0 ) yields an element of Gal( H 1 /L 0 ) of order 3. But we need to be more explicit. Since Gal( H 1 /L 0 (ε 1/3 0 )) restricts isomorphically to Gal(L 1 /L 0 ), we choose the element that restricts to τ and continue to call this new element τ . The fixed field of τ is L 0 (ε 1/3 0 ). Now assume that we are in Case (i) of Theorem 1. Apply the lemma to the extension H 1 /L 0 with ε = ε 1 . We have ε τ −1

The lemma therefore yields an element τ ∈ Gal( H 2 /L 0 ) of order 3. Finally, apply the lemma to

We obtain τ ∈ Gal( H 3 /L 0 ) of order 3. Now assume that we are in Case (ii) of Theorem 1. We have

and (ε τ 2 ) 1+τ +τ 2 = 1. The lemma yields τ ∈ Gal( H 2 /L 0 ) of order 3. Finally, apply the lemma to H 2 /L 0 and ε 2 . We have ε τ −1

Therefore, in both cases, we obtain τ ∈ Gal( H 3 /L 0 ). Since g is the unique element of order 2 in the normal subgroup Gal( H 3 /F 1 ) of Gal( H 3 /F 0 ), we must have τ gτ −1 = g, so g and τ commute. Therefore, τ g has order 6 in Gal( H 3 /F 0 ). The fixed field of τ g is the desired field H ′ 3 . If we restrict τ g to H 2 , its fixed field yields H ′ 2 and the restriction to H 1 yields H ′ 1 .

Kummer generators

The goal of this section is to prove the following.

. Let A 1 be the 3-Sylow subgroup of the ideal class group of F 1 . (a) There are units ε 1 , ε 2 ∈ K 1 such that

and ε 1+τ +τ 2 2 = ε 0 .

(b) The 3-rank of A 1 is at most 3. (c) The 3-rank of A 1 is at least 1 if and only if

The Kummer pairing

where µ 3 is the group of cube roots of unity, is Galois equivariant. Since 3 ∤ h(K 1 ), we see that σ acts on A 1 / A 3 1 as −1. Also, σ acts on µ 3 as −1. Therefore, σ acts on V as +1. By Lemma 3, we may assume that V ⊂ K × 1 /(K × 1 ) 3 . By assumption, 3 is unramified in F 0 /Q. Since the extension L 1 /K 1 is the lift of the extension F 0 /Q, it is unramified at 3. Since it is the lift of Q( √ −3)/Q, it ramifies at most at 3 and the archimedean primes. Therefore, L 1 /K 1 is unramified at all finite primes. Let b ∈ V . We also denote the corresponding element of K × 1 by b. Therefore, σ(b) = b. Since H (3) /L 1 is unramified, we must have (b) = I 3 for some ideal I of L 1 . Since σ(I) = I, and since L 1 /K 1 is unramified at all finite primes, I comes from an ideal of K 1 , so (b) = I 3 as ideals of K 1 . Since 3 ∤ h(K 1 ) (because 3 ∤ h + , and there is exactly one ramified prime and it is totally ramified), the ideal I is principal, so b = εα 3 for some α ∈ K 1 and some unit ε of K 1 . We have shown that V is represented by units of K 1 .

Since the 3-parts of the class groups of L 1 and F 1 are isomorphic, g acts as +1 on

We are assuming that 3 ∤ h + , so part (i)(b) of Theorem 1 holds. Therefore, ε 0 , ε 1 , ε 2 exist and M ≃ F 3 [T ]/(T 3 ), where 1 − τ ↔ T . The only subspaces stable under the action of T are 0, (T 2 ), (T ), and the whole space. This means that V is one of 0, ε 0 , ε 0 , ε 1 , or ε 0 , ε 1 , ε 2 . Parts (b), (c), (d), (e) of the theorem follow immediately.

It is interesting to note that although Gal(H 3 /F 1 ) ≃ (Z/3Z) 3 , which is very much non-cyclic, the τ -structure forces it to behave like a cyclic extension of degree 27 in that the inertia subgroup is restricted to be one of four different distinguished subgroups of orders 1, 3, 9, 27. Namely, there are subgroups

and the inertia group and the decomposition group must be among these. Correspondingly, there is a tower of fields

The splitting of primes above 3, if it occurs, happens at the bottom of this tower and the ramification (possibly trivial) happens at the top.

Kummer generators in the split case

Let r be the 3-rank of the class group of K 0 . By Scholz's theorem, the 3-rank of A 0 is r or r + 1, so λ ≥ r by Lemma 2. In fact, we can do better. Proof. Let π = 1 − ζ 9 . An extension L 1 (α 1/3 )/L 1 , with α ∈ L × 1 , is everywhere unramified if and only if (α) is the cube of an ideal and α is congruent to a cube mod π 9 ([10, Exercise 9.3]). Therefore, we need to show that ε 0 and γ i are cubes mod π 9 .

We need a few preliminary calculations. We have

Substituting this expression for the −3 in the right-hand side yields −3 ≡ π 6 + π 9 (mod π 10 ).

Substituting this into the expression for √ −3 yields √ −3 ≡ π 3 − π 6 − π 7 + π 8 (mod π 10 ). Also, ζ 3 ≡ 1 − π 3 + π 7 − π 8 (mod π 10 ).

Let a, b, c, . . . be integers. Using the above congruences, we obtain

This is the general form of a cube that is congruent to 1 mod π. Let γ ∈ K × 0 be such that (γ) = I 3 for some ideal I of K 0 . By multiplying I by a principal ideal, if necessary, we may assume that I is prime to 3. Therefore, by multiplying γ by a cube, we may assume that γ is prime to 3. Write

It is easy to see that u, v do not have 3 in their denominators, so they may be regarded as elements of Z 3 . Since we are assuming that 3 splits in

Hence v √ −d is congruent mod 3 to an integer ℓ, and we can regard γ as an element of Z 3 [ζ 3 ]. Moreover, γ 2 ≡ 1 (mod √ −3). Therefore, u = 1 + 3k for some k, and

Since (γ) = I 3 , we have Norm(γ) = ±Norm(I) 3 , and therefore Norm(γ) 2 is a 6-th power, hence is congruent to 1 mod 9. Therefore,

It follows that ℓ 2 ≡ k (mod 3) and

(1 + ℓπ − (ℓ + k)π 2 ) 3 ≡ γ 2 (mod π 9 Z 3 [ζ 9 ]), so γ 2 is a cube mod π 9 . Since this calculation is valid for the completions at each of the two primes above 3, we have proved that γ 2 is globally a cube mod π 9 . This implies that L 1 (γ 1/3 )/L 1 is everywhere unramified. Now suppose that ε a0 0 γ a1 1 · · · γ ar r = β 3 for some β ∈ L 1 . By Lemmas 3 and 4, we may assume that β ∈ K 0 . Therefore, I a1 1 · · · I ar r = (β) in K 0 , which implies that a i ≡ 0 (mod 3) for all i. Consequently, ε a0 0 is a cube in K 0 , so a 0 ≡ 0 (mod 3). Therefore, ε 0 , γ 1 , . . . , γ r are independent mod cubes in L 1 , so

. . , γ 1/3 r )/L 1 is unramified of degree 3 r+1 . Since σ acts trivially on these Kummer generators, the Galois equivariance of the Kummer pairing implies that σ acts by inversion on the Galois group of this extension. Therefore, rank(A 1 ) = rank( A − 1 ) ≥ r + 1.

Corollary 1. Let d ≡ 2 (mod 3) and let r be the 3-rank of the class group of

Proof. This follows immediately from Theorem 3 and Lemma 2.

In particular, the corollary implies that if 3 | h + then λ ≥ 2, as was shown in Proposition 1 using 3-adic L-functions.

When 3 splits in F 0 , we know that λ ≥ 1. Correspondingly, in this case, Theorem 3 shows that we can always produce an explicit unramified 3-extension of L 1 using ε 1/3 0 . This gives one explanation of the fact that λ ≥ 1. Note that we obtain an unramified extension of L 1 but not necessarily of L 0 . The latter case happens when ε 0 ≡ ±1 (mod 3 √ D).

Congruences

We continue to restrict to the case that d ≡ 2 (mod 3), so 3 splits in F 0 . We complete everything at one of the two primes of L 1 above 3. The completions of K 0 and L 0 yield Q 3 (ζ 3 ), and the completions of K 1 and L 1 yield Q 3 (ζ 9 ). The element τ can be taken to be σ 4 ∈ (Z/9Z) × ≃ Gal(Q 3 (ζ 9 )/Q 3 ) and g becomes σ −1 , where σ i (ζ 9 ) = ζ i 9 . Recall that π = 1 − ζ 9 . Lemma 8. Let ε be a π-adic unit in Q 3 (ζ 9 ) such that σ −1 (ε) = ε −1 times a cube. Then ε ≡ ±ζ a 9 (mod π 5 )

for some integer a, and ε 3 ≡ ±ζ a 3 (mod π 11 ).

Proof. As a preliminary result, note that the cube of a π-adic unit is congruent to ±1 (mod π 3 ) (proof: compute the cube of a+bπ +· · · ). Also, ζ a 9 = (1−π) a ≡ 1−aπ (mod π 2 ). Therefore, if ±ε = 1+aπ+· · · , then ±εζ a 9 ≡ (1+aπ)(1−aπ) ≡ 1 (mod π 2 ).

Since

it follows that σ −1 (π j ) ≡ (−1) j π j (mod π j+1 ). Suppose that σ −1 (ε) ≡ ε −1 mod cubes. Write ±εζ a 9 = 1 + bπ 2 + · · · . Since σ −1 (ζ 9 ) = ζ −1 9 , we have

Therefore, b ≡ 0 (mod 3), so ±εζ a 9 = 1 + cπ 3 + · · · .

Since ζ 3 ≡ 1 − π 3 (mod π 4 ), we have ±εζ a 9 ζ c 3 ≡ 1 (mod π 4 ).

Suppose now that log 3 ε 0 ≡ 0 (mod π 10 ). Then a 5 ≡ 0 (mod 3). Therefore,

≡ (1 + a 6 π 6 + a 7 π 7 + a 8 π 8 + a 9 π 9 ) 1−τ ≡ 1 − a 7 π 9 (mod π 10 ).

This means that ε 1 is congruent to a cube mod π 9 , so L 1 (ε

Conversely, suppose that L 1 (ε 1/3 1 )/L 1 is unramified, so ε 1 is congruent to a cube mod π 9 . Although we already know from Lemma 2 and Proposition 1 that log 3 ε 0 ≡ 0 (mod π 10 ), we prove this algebraically.

Since ζ 3 = ζ 3 9 , we have that ζ a 3 ε 1 is congruent to a cube mod π 9 . The π-adic expansion of ε 2 yields ζ a 3 ε 1 = ζ a 3 ε 1−τ 2 ≡ (1 + a 5 π 5 + · · · ) 1−τ (mod π 9 ) ≡ 1 + a 5 π 7 − a 5 π 8 .

By Lemma 8, this must be congruent to ζ c 3 mod π 9 for some c. Since ζ c 3 ≡ 1−cπ 3 (mod π 4 ), we must have ζ c 3 = 1, which implies that a 5 ≡ 0 (mod 3). Therefore, log 3 ε 0 ≡ 0 (mod π 10 ). This completes the proof. Proof. When 3 ∤ h + , this follows from the equivalence of (a) and (b) in the theorem, plus Theorem 2(d). When 3 | h + , Theorem 3 implies that rank(A 1 ) ≥ 2 and hence that λ ≥ 2.

Therefore, we can see whether or not λ ≥ 2 already at F 1 . When λ ≥ 2, we can obtain more information about Kummer generators. Proposition 6. Assume that d ≡ 2 (mod 3), that 3 ∤ h + , and that λ ≥ 2. Proof. Choose a prime of Q above 3 and work in the completion at this prime. Theorem 4 says that log 3 ε 0 ≡ 0 (mod π 15 ), so log 3 (ε 1/3 0 ) ≡ 0 (mod π 9 ). Let

Then y ≡ 1 (mod π 9 ) and log 3 y = log 3 (ε 1/3 0 ). The kernel of log 3 consists of (integral and fractional) powers of 3 times roots of unity. Therefore, y = ±ζ a 9 ε 1/3 0 for some integer a, and ε 1/3 0 ≡ ±ζ −a 9 (mod π 9 ), so ε 1/3 0 is congruent to a cube in Q 3 (ζ 27 ) mod π 9 , so the extension Q 3 (ζ 27 , ε 1/9 0 )/Q 3 (ζ 27 ) is unramified. But this is the completion of the extension L 2 (ε 1/9 0 )/L 2 at any of the primes above 3. Since this extension is unramified at all other primes because ε 0 is a unit, the extension is everywhere unramified. This proves (a) and (b).

We now prove (c). The proof of Scholz's theorem (see [10,Theorem 10.10]) yields that 3-rank of class group of F 0 = δ, where δ = 1 if L 0 (ε 1/3 0 )/L 0 is unramified and δ = 0 otherwise. The relation ±ζ a 9 ε 1/3 0 = y ≡ 1 (mod π 9 ) shows that, in any completion at a prime above 3, ε 0 ≡ ±ζ −a 3 (mod π 15 ) for some a that possibly depends on the choice of completion.

, we see that ζ a 3 is a cube of an element of L 0 mod π 9 if and only if a ≡ 0 (mod 3). However, since the only Galois conjugates of ε 0 are ε 0 and ε −1 0 , if ε 0 ≡ 1 (mod π 9 ) at one completion at a prime above 3, then this holds for all such completions, hence holds globally. In other words, if we have a ≡ 0 (mod 3) when working in one completion, then this holds in all the other completions.

Suppose now that 3|h − . Then δ = 1, which means that we must have ε 0 ≡ 1 (mod π 9 ) and a ≡ 0 (mod 3). Therefore, ε

(mod π 9 ) means that ε 1/3 0 is congruent to a cube mod π 9 . Therefore, L 1 (ε 1/9 0 )/L 1 is unramified. Conversely, suppose L 1 (ε 1/9 0 )/L 1 is unramified. Then ε 1/3 0 is congruent to a cube mod π 9 . This means that a ≡ 0 (mod 3), so ε 0 ≡ 1 (mod π 9 ). Therefore,

When 3|h − , part (c) implies that A 1 is not elementary. We can also prove this as follows. Suppose that A 1 is elementary. Since 3 ∤ h + , the rank of A 0 is at most 1. Since A 0 is a quotient of A 1 , it is also elementary, hence of order 3. Let 3 ei = |A i |. Gold [1] proves that if d ≡ 2 (mod 3) and if e 1 − e 0 ≤ 2 then λ = e 1 − e 0 . By Theorem 2, the rank of A 1 is at most 3, so e 1 ≤ 3. Therefore e 1 − 1 = e 1 − e 0 ≤ 2, so λ = e 1 − 1. But Lemma 2 says that e 1 ≤ λ, so we have a contradiction. This proves that A 1 is not elementary. Note that we did not need any assumption on λ for this argument.

Another point of view

Gross-Koblitz [3] show that if 3 splits in F 0 as pp, then

where (α) = p h − and the logarithm is taken in the p-adic completion. Therefore,

This yields the following:

Lemma 9. λ ≥ 2 if and only if log 3 α ≡ 0 (mod 9).

Moreover, the 3-adic analytic class number formula implies that

so we obtain the interesting relation

Note that α is a 3-unit that lies in the + component for the action of g and the − component for the action of σ on 3-units mod powers of 3. The units ε i that we worked with in previous sections are in the components of the parities opposite from α. The fact that Theorem 4 is based on the 3-adic L-function at both 0 and 1 means that it is natural that it involves both F 0 and K 0 , which are related to each other through reflection theorems, hence through the relations between Kummer generators and class groups. In the present situation, everything involves only the 3-adic L-function at 0, hence involves only F 0 , not K 0 . We prove the following analogue of Theorem 4. It is interesting to note that the higher genus groups in the proof of Theorem 1 are replaced by "higher ambiguous groups" of ideal classes, namely ideal classes that are annihilated by some power of 1 − τ .

Theorem 5. The following are equivalent: (a) λ ≥ 2 (b) the 3-rank of A 1 is greater than or equal to 2 (c) α is a norm for F 1 /F 0 .

Proof. Lemma 2 shows that (b) implies (a).

Assume (a). Lemma 9 implies that log 3 α ≡ 0 (mod 9), where the logarithm is taken in the p-adic completion of F 0 . Let γ = exp((1/3) log 3 (α)). Then log 3 (γ 3 ) = log 3 (α), so γ 3 and α differ by a root of unity in the completion of F 0 at p. Since 3 splits in F 0 , this completion is Q 3 , whose only roots of unity are ±1. Therefore, α is a local cube, hence a local norm at p for F 1 /F 0 . Since F 1 /F 0 is unramified at all primes not above 3, α is a local norm for all places except possibly p. By the product formula for the norm residue symbol, α is a norm also at p. Since F 1 /F 0 is cyclic, Hasse's norm theorem says that α is a global norm from F 1 . (A similar argument appears in [1].) Assume (c). Write Norm F1/F0 (β) = α. Then the ideal norm of (β) is (α) = p h − . Let P be the prime above p, so Norm F1/F0 (P) = p. Therefore,

Let ord(B) denote the order of an ideal B in the class group of F 1 . Since the natural map from the class group of F 0 to the class group of F 1 is injective (see [10,Proposition 13.26]), if B is an ideal of F 0 , then its order in the class group of F 0 equals its order in the class group of F 1 , so we need to consider only the order in F 1 .

Lemma 10. ord(P) = 3ord(p).

Proof. (cf. [2, proof of Proposition 2]) If 3|ord(P), then the lemma follows immediately from the fact that P 3 = p. So we need to show that 3|ord(P).

Suppose P b is principal for some b > 0. Then (P/P 1 ) b = (γ) for some γ ∈ F 1 . Let δ = γ/γ. Then (δ) = (P/P 1 ) 2b .

Since this ideal is fixed by τ , it follows that δ 1−τ is a unit. Since it has absolute value 1 at all embeddings into C, it is a root of unity, hence ±1. Since the norm from F 1 to F 0 of δ 1−τ is 1, we must have δ 1−τ = +1. Therefore, δ ∈ F 0 . Since 2b = v P (δ) = 3v p (δ), we have 3|b. Therefore, 3|ord(P).

Case 1. Suppose 3 n |ord(p). Then 3 n+1 |ord(P). Therefore, 3|c. It follows that the subgroup of the ideal class group generated by P and J has order at least 3 times the order of the subgroup generated by P, hence has order at least 3 n+2 . Therefore, |A 1 | ≥ 9|A 0 |. The following lemma implies that A 1 cannot be cyclic (cf. [8]).

where m ≥ 1 by Lemma 10. Then τ ∈ Aut(A 1 ) has order 3, hence corresponds to multiplication by 1+3 m−1 x for some x. Therefore, 1 + τ + τ 2 ≡ 3 (mod 3 m ). The map 1 + τ + τ 2 is the endomorphism of A 1 given by the norm followed by the natural map from A 0 to A 1 . Since F 1 /F 0 is totally ramified, the norm map from A 1 to A 0 is surjective. Since A 0 lies in the minus component with respect to complex conjugation, the map A 0 → A 1 is injective. It follows that so A 0 ≃ 3Z/3 m Z. Case 2. Suppose 3 n ∤ ord(p). Then P h − is principal, so the subgroup of A 1 generated by A 0 and P has exponent at most h − .

Lemma 12. Every ideal class of A 1 that is fixed by τ contains an ideal of the form IP a , with I an ideal from F 0 .

Proof. Let B represent a fixed ideal class, so B 1−τ = (γ) for some γ ∈ F 1 . Then the norm from F 1 to F 0 of γ/γ is a unit and has absolute value 1 at all places, hence is a root of unity. The only roots of unity in F 1 are ±1, so Norm(γ 2 /γ 2 ) = 1, which means that γ 2 /γ 2 = δ 1−τ for some δ ∈ F 1 . Therefore, B 2 /δ is fixed by τ , so it is of the form P a I with I an ideal from F 0 . Since the class of B has 3-power order, the class of B also contains an ideal of this form.

The order of the subgroup of the class group of F 1 that is fixed by τ is 3h − (see [4,Ia,Satz 13]). Therefore, this subgroup has order greater than its exponent, so it is noncyclic. It follows that A(F 1 ) is noncyclic.

This completes the proof that (c) implies (b).

The theorem has the interesting corollary that when 3 splits in F 0 = Q( √ −d), we can determine whether λ ≥ 2 by looking at the first level F 1 of the Z 3extension. Namely, we have λ ≥ 2 if and only if rank A 1 ≥ 2. This was also obtained from the results proved using Kummer generators in Section 8.

When 3 ∤ h + but 3 | h − , the condition that α is a local cube in the completion at p can be strengthened to saying that α is a global cube.

Proposition 7. Assume that 3 splits in F 0 , that 3 ∤ h + , and that 3 | h − . Then λ ≥ 2 if and only if α is a cube in F 0 .

Proof. Assume that λ ≥ 2. By the proof of Theorem 5, α is a cube in the completion of F 0 at p. Therefore, L 0 (α 1/3 )/L 0 is unramified at p. This implies that L 0 (α 1/3 )/L 0 is unramified at p.

αα = 3 h − (the unit must be +1 since both sides are positive), which is a cube since 3 | h − . Therefore,

It follows that both p and p are unramified in L 0 (α 1/3 )/L 0 . Since only primes above 3 can be ramified in this extension, the extension must be everywhere unramified. The fact that αα is a cube says that σ (= complex conjugation) acts by inversion on α mod cubes. Also, σ acts by inversion on µ 3 . The Galois equivariance of the Kummer pairing implies that σ acts trivially on Gal(L 0 (α 1/3 )/L 0 ), which means that this extension corresponds to part of the ideal class of K 0 . Since 3 ∤ h + , this extension must be trivial, so α is a cube in L 0 . By Lemma 4, α is a cube in F 0 . Conversely, if α is a cube in F 0 , say α = β 3 , then log 3 α = 3 log 3 β ≡ 0 (mod 9), since log 3 x ≡ 0 (mod 3) for all x ∈ Q 3 . Therefore, λ ≥ 2 by Lemma 9.

The conditions that 3 ∤ h + and 3 | h − are essential in Proposition 7. When d = 35, we have h − = h + = 2 and α = (1 + √ −35)/2. Therefore, log 3 α ≡ 0 (mod 9), so λ ≥ 2. However, α is not a cube. When d = 107, we have h − = h + = 3 and α = (1 + √ −107)/2. Therefore, log 3 α ≡ 0 (mod 9), so λ ≥ 2. Again, α is not a cube. for both primes above 3. This makes S Galois over F 0 since S is the maximal subextension of H/F 1 in which the primes above 3 split.

Since τ has order 3 and acts on the 3-group Gal(S/F 1 ), it has a non-trivial quotient on which it acts trivially. Therefore S 0 , the maximal abelian extension of F 0 contained in S, is a nontrivial extension of F 1 . Let N be a degree 3 extension of F 1 contained in S 0 . Then N/F 0 cannot be a cyclic degree 9 extension because the ramification is at the bottom. Let N ′ be the inertia field for p, one of the primes above 3 in F 0 . Then N ′ /F 0 is a cyclic degree 3 extension in which p is unramified. Since the prime above p splits in N/F 1 and p ramifies in N/N ′ , it follows that p splits in N ′ /F 0 .

Let γ be a Kummer generator for L 0 N ′ /L 0 . Since σ acts by inversion on Gal(L 0 N ′ /L 0 ) and by inversion on µ 3 , the Galois equivariance of the Kummer pairing implies that σ acts trivially on γ mod cubes. By Lemma 4, we may assume that γ ∈ K 0 . Let p 0 be the prime of K 0 above 3 and let a = v p1 (γ). Since the extension L 0 N ′ /L 0 is unramified at p, we must have a ≡ 0 (mod 3). Moreover, the extension is unramified at all primes not above 3, so we have (γ) = I 3 for some ideal I of K 0 . Since 3 ∤ h + , the ideal I is principal, which means that γ is a power of ε 0 times a cube. Therefore, N ′ L 0 = L 0 (ε 1/3 0 ), and N ′ /F 0 is the cyclic degree 3 subextension of the Hilbert class field of F 0 that lifts to this subextension of the Hilbert class field of L 0 . Since p splits in N ′ /F 0 , the image of p under the Artin map is contained in the subgroup of order h ′ /3 that fixes N ′ . Therefore, the image of p h − /3 is trivial, so p h − /3 is principal. The only units in F 0 are ±1, so p h − = (α) implies that α is a cube in F 0 , as desired.

The inert case

Suppose that 3 is inert in F 0 and that 3|h − . Then Proposition 1 says that

When 3 h − , this is a much more subtle situation than the case when 3 splits. We are not simply asking that a sufficiently high power of 3 divide log 3 ε 0 , for example. Instead, we are asking that the nonzero congruence class of h − mod 9 match the congruence class of an expression involving h + and log 3 ε 0 . In contrast to most situations, what is relevant is not simply the power of 3 that divides the numbers, but rather the congruence classes mod 3 that are obtained when the numbers are divided by suitable powers of 3 (that is, not just the "order of vanishing" but also the "coefficient of the leading term"). We hope to treat this situation in the future (but we make no promises).