Georges Gras
email: g.mn.gras@wanadoo.fr

NO RIEMANN-HURWITZ FORMULA FOR THE p-RANKS OF RELATIVE CLASS GROUPS

Keywords: 1991 Mathematics Subject Classification. Primary 11R, Secondary ,11R29 11R37, 11S15, 11S20 ideal class groups, class field theory, ambiguous classes, p-ramification, Riemann-Hurwitz formula, Kida's formula

We disprove, by means of numerical examples, the existence of a Riemann-Hurwitz formula for the p-ranks of relative class groups in a pramified p-extension K/k of number fields of CM-type containing µp. In the cyclic case of degree p, under some assumptions on the p-class group of k, we prove some properties of the Galois structure of the p-class group of K; but we have found, through numerical experimentation, that some theoretical group structures do not exist in this particular situation, and we justify this fact. Then we show, in this context, that Kida's formula on lambda invariants is valid for the p-ranks if and only if the p-class group of K is reduced to the group of ambiguous classes, which is of course not always the case.

December 14, 2015

Generalities

In 1980, Y. Kida [Ki] proved an analogue of the Riemann-Hurwitz formula for the minus part of the Iwasawa λ-invariant of the cyclotomic Z p -extension k ∞ of an algebraic number field k of CM-type with maximal totally real subfield k + :

λ -(K) -1 = [K ∞ : k ∞ ] (λ -(k) -1) + v∤p (e v (K + ∞ /k + ∞ ) -1),
where K/k is a finite p-extension of CM-fields containing the group µ p of pth roots of unity and where e v denotes the ramification index of the place v. When K/k is p-ramified (i.e., unramified outside p) and such that K ∩ k ∞ = k the formula reduces to:

λ -(K) -1 = [K : k] (λ -(k) -1).
Many generalizations where given as in [JaMa], [JaMi], [Sch], among many others. An interesting question is to ask if such a formula can be valid for the p-ranks of the relative ideal class groups in a p-extension K/k (e.g. K/k cyclic of degree p).

In a work, using Iwasawa theory and published in 1996 by K. Wingberg [W], such a formula is proposed (Theorem 2.1, Corollary 2.2) in a very general framwork and applied to the case of a p-ramified p-extension K/k of CM-fields containing µ p .

As many people remarked, we can be astonished by a result which is not really "arithmetical" since many of our class groups investigations (as in [START_REF] Gras | Classes généralisées invariantes[END_REF], [START_REF] Gras | Sur les ℓ-classes d'idéaux dans les extensions cycliques relatives de degré premier ℓ[END_REF]) show that such a "regularity" only happens at infinity (Iwasawa theory). The proof of Kida's formula given by W. Sinnott [Sin], using p-adic L-functions, is probably the most appropriate to see the transition from one aspect to the other.

Indeed, to find a relation between the p-ranks (for instance in a cyclic extension K/k of degree p) depends for G := Gal(K/k) =: σ on non obvious structures of finite Z p [G]-modules M provided with an arithmetical norm N K/k and a transfert map j K/k (with j K/k • N K/k = 1 + σ + . . . + σ p-1 ), where the filtration of the M i := {h ∈ M, h (σ-1) i = 1} plays an important non-algebraic role because all the orders #(M i+1 /M i ) depend on arithmetical local normic computations by means of formulas, given in [START_REF] Gras | Classes généralisées invariantes[END_REF], similar to that of the case i = 0 of ambiguous ideal classes (see Section 3).

To be more convincing, we have given numerical computations and we shall see that it is not difficult to conjecture that there are infinitely many counterexamples to a formula, for the relative p-ranks, such as r - K -1 = p (r - k -1) wich may be true in some cases.

A numerical counterexample

Using PARI (from [P]), we give a program which can be used by the reader to compute easily for the case p = 3 in a biquadratique field containing a primitive 3th root of unity j, and such that its 3-class group is of order 3.

2.1. Definition and assumptions. Consider the following diagram:

Q k + = Q( √ 3.d ) k -= Q( √ -d) Q(j) k = k + (j) K + K = k( 3 √ α ) [K : k] = 3 s
We recall, in this particular context, the hypothesis of the statement of Corollary 2.2 (ii) of [W] and we shall suppose these conditions satisfied in all the sequel. For any field F , let Cℓ F be its 3-class group and let Cℓ ± F be its two usual components when F is a CM-field. We denote by H F the 3-Hilbert class field of F .

Hypothesis 2.1. (i) d > 0, d squarefree, d ≡ 0 (mod 3), (ii) p = 3 does not split in k/Q (hence d ≡ 1 (mod 3)), (iii) K + /k + is a 3-ramified cubic cyclic extension and K = K + (j), (iv)

K + is not contained in the cyclotomic Z 3 -extension of k + , (v) Cℓ k + = 1 & Cℓ k -≃ Z/3Z.
From (v), the ambiguous class number formula implies Cℓ K + = 1 (Lemma 2.2). Starting from a p-ramified cubic cyclic extension K + /k + , the associated Kummer extension K/k is defined by 3 √ α such that α ∈ k × \ k ×3 , (α) = a 3 for an ideal a of k, and α s+1 ∈ k ×3 where s ∈ Gal(K/K + ) is the complex conjugation (usual decomposition criterion of a Kummer extension over a subfield).

If the 3-rank of the 3-class group Cℓ - k ≃ Cℓ k -is r -, the 3-rank of the Galois group of the maximal Abelian 3-ramified 3-extension of k + is r -+ 1 when 3 does not split in k (see [START_REF] Gras | Class Field Theory: from theory to practice[END_REF], Proposition III.4.2.2 for the general statement). From (iv), necessarily r -≥ 1. Here we suppose #Cℓ k -= 3, hence r -+ 1 = 2 (this is also equivalent to the non-nullity of the 3-torsion subgroup T 3 , of the above Galois group, whose order is in our context #T 3 ∼ log 3 (ε)/ √ 3.d where ε is the fundamental unit of k + , see [START_REF] Gras | Class Field Theory: from theory to practice[END_REF], Remark III.2.6.5 (i)). We shall precise the choice of the extension K + /k + (i.e., α) as follows:

Let α in k -be such that (α) = a 3 (a must be a non-principal ideal since E k -is trivial); this defines a canonical cubic cyclic extension K + of k + and the numbers α j, α j 2 define the two other cubic cyclic extensions

K + 1 , K + 2 of k + , distinct from the first step of the cyclotomic Z 3 -extension K + 0 of k + defined by j. Lemma 2.2. (i) We have #Cℓ K + = 1 & #Cℓ G K = #Cℓ k = 3, where G = Gal(K/k). (ii) We have #Cℓ H k = 1.
Proof. (i) Using the Chevalley's formula in K + /k + (see e.g. [START_REF] Gras | Class Field Theory: from theory to practice[END_REF], Lemma II.6.1.2) with a trivial 3-class group for k + , the formula reduces to

#Cℓ G K + = 3 3 . (E k + : E k + ∩ N K + /k + K +× ) = 1,
since the product of ramification indices is equal to 3 (Cℓ H k = 1 implies that K + /k + is necessarily totally ramified at the single prime above 3 of k + ).

The same formula in

K/k is #Cℓ G K = #Cℓ k × 3 3 . (E k : E k ∩ N K/k K × ) with E k = ε, j ,
where ε is the fundamental unit of k + ; but, as for K + /k + , there is by assumption a single place of k ramified in K/k, thus, using the product formula, the Hasse norm theorem shows that all these units are local norms everywhere hence global norms.

So #Cℓ G K = 3 since #Cℓ k = #Cℓ k -= 3. (ii) We have #Cℓ G ′ H k = #Cℓ k 3 . (E k : E k ∩ N H k /k H × k )
= 1 where

G ′ := Gal(H k /k).
Corollary 2.3. We have

Cℓ K = Cℓ - K since Cℓ + K ≃ Cℓ K + = 1. 2.2. Numerical data. We have a first example with d = 211 ≡ 1 (mod 3) and α = 17+ √ -211 2
where (α) = p 3 5 for a non-principal prime ideal dividing 5 in k -. The class number of k + is 1 and that of k -is 3, which is coherent with the fact that the fundamental unit ε = 440772247 + 17519124

√ 3.211 of k + is 3-primary (indeed, ε ≡ 1 + 3. √ 3.211 (mod 9)), which implies that H k -is given via k( 3 √ ε)/k which is decomposed by means of an unramified cubic cyclic extension of k -.
So all the five conditions (i) to (v) are fulfilled.

The PARI program (see Section 5) gives in "component(H, 5)" the class number and the structure of the whole class group of K; the program needs an irreducible polynomial defining K; it is given by "P = polcompositum(

x 2 + x + 1, Q)" where Q = x 6 -17 x 3 + 5 3 is the irreducible polynomial of 3 √ α over Q (the general formula is Q = x 6 -Tr k -/Q (α) x 3 + N k -/Q (α))
; one obtains: P = x 12 -6 x 11 + 21 x 10 -84 x 9 + 243 x 8 -432 x 7 + 1037 x 6 -1896 x 5 -204 x 4 -966 x 3 + 5949 x 2 + 4905 x + 11881.

Conclusion. The program gives

Cℓ K ≃ Z/9 Z×Z/3 Z for a class number equal to 27. This yields the 3-rank R -= 2 of Cℓ - K when the 3-rank r -of Cℓ - k is equal to 1, which is incompatible with the formula

R --1 = 3 × (r --1).
But, this "Riemann-Hurwitz formula" is valid if and only if Cℓ K = Cℓ G K ≃ Z/3 Z (no exceptional 3-classes). Such a case is also very frequent (see § 5.1).

Some structural results

Denote by M a finite Z p [Γ]-module, where we assume that Γ is an Abelian Galois group of the form G × g, where G = Gal(K/k) =: σ is cyclic of order p and g ≃ Gal(k/k 0 ) (of order prime to p), where k 0 is a suitable subfield of k (so that K = k K 0 with K 0 := K g ). The existence of g allows us to take isotypic components of M (as the ±-components when the fields are of CM-type). In our example, g = s , k 0 = k + and K 0 = K + .

We have M/M p = M/M 1+σ+...+σ p-1 -Ω where Ω = u (σ -1) p-1 for an inversible element u of the group algebra Z p [G] ([Gr3], Proposition 4.1); in our case p = 3, Ω = σ 2 (σ -1) 2 . We shall use:

ω := σ (σ -1), such that ω 2 ≡ 3 (mod ν), where ν := 1 + σ + σ 2 .
For our purpose we shall have M = Cℓ - K (we refere to [START_REF] Gras | Sur les ℓ-classes d'idéaux dans les extensions cycliques relatives de degré premier ℓ[END_REF], Chap. IV, A, § 2). By class field theory, when K/k is totally ramified at the unique p | 3, the arithmetical norm

N K/k : Cℓ - K -→ Cℓ - k is surjective.
Another important fact for the structure of Cℓ K , in our particular context, is that the class of order 3 of k capitulates in K because the equality (α) = a 3 becomes ( 3 √ α) = (a) K in K (the transfert map j K/k : Cℓ - k -→ Cℓ - K is not injective). This has the following tricky consequence:

N K/k (Cℓ - K ) = Cℓ - k & (Cℓ - K ) ν = 1.
Return to the general case M = Cℓ K for any prime p and suppose #M G = p. Put M i := {h ∈ M, h (σ-1) i = 1}, i ≥ 0, and let n be the least integer i such that M i = M .

From the exact sequence 1 [START_REF] Gras | Sur les ℓ-classes d'idéaux dans les extensions cycliques relatives de degré premier ℓ[END_REF], Proposition 4.1 and Corollaire 4.3, assuming M ν = 1 and #(M i+1 /M i ) = p for all i < n, we obtain the following structure of Z-module:

→ M 1 = M G -→ M i+1 ω ---→ M ω i+1 ⊆ M i → 1, with #M G = p, we obtain that M ω i+1 = M i and #(M i+1 /M i ) = p for i = 0, . . . , n -1. From
M ≃ (Z/p a+1 Z) b × (Z/p a Z) p-1-b , where n = a (p -1) + b, 0 ≤ b ≤ p -2. This implies (assuming M + = 1) that the p-rank R -of M -is R -= p -1 if a ≥ 1 and R -= b if a = 0 (i.e., b = n ≤ p -2).
So, in the case r -= 1, we have the Riemann-Hurwitz formula R

--1 = p (r --1) if and only if b = n = 1 which is equivalent to M -= M -G . Otherwise, R -can take any value in [1, p -1], even if r -= 1.
This general isomorphism comes from the formula ([Gr2], Corollaire 2.8):

# M i+1 /M i G = #Cℓ k × e v [K : k] . #N K/k (M i ) . (Λ i : Λ i ∩ N K/k (K × ))
,

where N K/k (M i ) := cℓ k (N K/k (I i )) for a suitable ideal group I i such that cℓ K (I i ) = M i , and where Λ i = {x ∈ k × , (x) ∈ N K/k (I i )}.
In our case there is a single ramified place p | 3 and the elements of Λ i , being norms of ideals, are everywhere local norms except perhaps at p; so (Λ i : Λ i ∩ N K/k (K × )) = 1 under the product formula of class field theory and the Hasse norm theorem, and

# M i+1 /M i G = 3 #N K/k (M i ) .
In our numerical example with d = 211, we get necessarily a = b = 1, n = 3, giving the structure Z/9Z × Z/3Z.

For more structural results when M ν = 1, see [START_REF] Gras | Sur les ℓ-classes d'idéaux dans les extensions cycliques relatives de degré premier ℓ[END_REF], Chap. IV, § 2, Proposition 4.3 valid for any p ≥ 2. In our biquadratic case and p = 3, we obtain interesting structures for which a theoretical study should be improved. From the general PARI program we have obtained the following numerical examples:

(i) For d = 1759, for which #Cℓ k + = 1, Cℓ k -≃ Z/27Z, and α = 37 + 20 √ -d of norm 89 3 , the structure is Cℓ - K ≃ Z/27Z (i.e., Cℓ - K = (Cℓ - K ) G ). (ii) For d = 2047, for which #Cℓ k + = 1, Cℓ k -≃ Z/9Z, and α = 332 + 11 √ -d of norm 71 3 , the structure is Cℓ - K ≃ Z/9Z × Z/3Z × Z/3Z. (iii) For d = 1579, for which #Cℓ k + = 1, Cℓ k -≃ Z/9Z, and α = 1 2 (115 + 3 √ -d) of norm 19 3 , the structure is Cℓ - K ≃ Z/27Z × Z/9Z × Z/3Z. 4. The structure M ≃ Z/3 a Z × Z/3 a Z does not exist
Of course, we keep the same numerical assumptions about the 3-class groups of k + and k -(especially Cℓ k -≃ Z/3Z), the non-splitting of 3 in k/Q, and the Kummer construction of the 3-ramified cubic cyclic extension K/k. Theorem 4.1. Under all the Hypothesis 2.1, we get

Cℓ K = Cℓ - K ≃ Z/3 a+1 Z × Z/3 a Z, for some a ≥ 0. The case a = 0 is equivalent to Cℓ K = Cℓ G K .
Proof. Let n ≥ 1 be the least integer such that M n = M := Cℓ K . From the relations

M i+1 /M i ≃ Z/3Z, for 0 ≤ i ≤ n -1, we get #M = 3 n . Let h n ∈ M be such that N K/k (h n ) generates Cℓ k (equivalent to h n ∈ M n \M n-1 ). Since M ω i+1 = M i for 0 ≤ i ≤ n -1, M n is the Z p [G]-module generated by h n and for all i, 0 ≤ i ≤ n -1, h n-i := h ω i n generates M n-i . We have h ω 2 = h 3 for all h ∈ M .
Let m, 1 ≤ m ≤ n; the structure of M m is given (as for M , see Section 3) by:

M m ≃ (Z/3 am+1 Z) bm × (Z/3 am Z) 2-bm , m = 2 a m + b m , b m ∈ {0, 1}. (i) Case m = 2 e. Then M m ≃ Z/3 e Z × Z/3 e Z. (ii) Case m = 2 e + 1. Then M m ≃ Z/3 e+1 Z × Z/3 e Z.
With the previous notations we have, for the two cases:

M m = h m ⊕ h m-1 , as Z-module. Indeed, suppose that h = h A m = h B m-1 , A, B ∈ Z. Then h A-B ω m = 1. Since h m is anihilated by ω m and not by ω m-1 , we get A -B ω ∈ (ω m ).
In the case m = 2e, ω m ≡ 3 e (mod ν) giving in the algebra

Z 3 [G] = Z 3 [ω]
, the relations A ≡ B ≡ 0 (mod 3 e ), hence h = 1. Since #M 2e = 3 2e , the elements h m and h m-1 are independent of order 3 e .

In the case m = 2e + 1, ω m ≡ 3 e ω (mod ν) and A -B ω ≡ 0 (mod 3 e ω), giving A = 3 e A ′ and B = 3 e B ′ , then A ′ -B ′ ω ≡ 0 (mod ω Z 3 [ω]), which implies A ′ ≡ 0 (mod 3) hence the result in this case with h m of order 3 e+1 and h m-1 of order 3 e . Suppose that the structure of M is M 2e ≃ Z/3 e Z × Z/3 e Z, e ≥ 1, and let F be the subfield of

H K fixed by M 2(e-1) = M 3 ; so F is a cubic cyclic extension of KH k and Gal(F/K) ≃ Z/3Z × Z/3Z.
The 3-extension F/k is Galois: indeed, if σ is a generator of Gal(K/k), the action of σ on Gal(F/K) = Gal(H K /K)/M 3 is given, via the correspondence of class field theory, by the action of σ on h 2e := h 2e M 3 and on h 2e-1 := h 2e-1 M 3 . From

ω = σ (σ -1) = σ -2 (σ -1), we have h σ 2e = h 2e × h σ 2
2e-1 , and σ acts on h 2e-1 by h

σ 2e-1 = h σ 2e-1 M 3 ; but h σ(σ-1) 2e-1 = h 2(e-1) ∈ M 3 , so h σ-1 2e-1 ∈ M 3 and h σ 2e-1 = h 2e-1
, hence the result and the fact that F/H k is Galois. But a group of order p 2 (p prime) is Abelian and F/H k is the direct compositum of KH k and L such that L is the decomposition field at 3 giving a cyclic extension L/H k , unramified of degree 3. But we know (Lemma 2.2) that the 3-class group of H k is trivial (contradiction).

Remarks 4.2. (i) Since 3 is non-split in k/Q, the unique prime ideal P | 3 in K is principal: indeed, P 1+s = P 2 gives the square of the extension of P + | 3 in K + which is 3-principal (Lemma 2.2); so cℓ K (P) = 1. By class field theory, P splits completely in H K /K.

(ii) In the case n even for the above reasoning, an analog of the field F does not exist as Galois field over H k .

(iii) Note that the parameter a can probably take any value; we have for instance obtained the following example:

For d = 12058, for which #Cℓ k + = 2, Cℓ k -≃ Z/3Z, and α = 989 + 26 √ -d of norm 209 3 , the structure is Cℓ - K ≃ Z/3 4 Z × Z/3 3 Z. A polynomial defining K is: P = x 12 -6 x 11 +21 x 10 -4006 x 9 +17892 x 8 -35730 x 7 +22212821 x 6 -66531354 x 5 -113482743 x 4 -35777798264 x 3 +54059937672 x 2 +54106942656 x+83308554531904.

For d = 86942, for which #Cℓ k + = 4, Cℓ k -≃ Z/3Z, and α = 557 + 3 √ -d of norm 103 3 , the structure is Cℓ - K ≃ Z/3 5 Z × Z/3 4 Z. A polynomial defining K is: P = x 12 -6 x 11 +21 x 10 -2278 x 9 +10116 x 8 -20178 x 7 +3449985 x 6 -10289502 x 5 -8954865 x 4 -2399550304 x 3 + 3642928674 x 2 + 3641624304 x + 1191621124996.

In conclusion, the structure of Cℓ - K strongly depends on the hypothesis on the order of Cℓ - k and not only of its 3-rank.

  Cℓ k -= 12, #Cℓ k + = 4 P = x 12 -6 x 11 + 21 x 10 -2302 x 9 + 10224 x 8 -20394 x 7 + 2701601 x 6 -8043702 x 5 -2977323 x 4 -1568242964 x 3 + 2378397756 x 2 + 2373361968 x + 495396376336 classgroup :[600, [30, 10, 2]] 5.2. Case Cℓ K ≃ Z/9Z × Z/3Z (a = 1). -= 3, #Cℓ k + = 1 P = x 12 -6 x 11 + 21 x 10 -84 x 9 + 243 x 8 -432 x 7 + 1037 x 6 -1896 x 5 -204 x 4 -966 x 3 + 5949 x -= 6, #Cℓ k + = 2 P = x 12 -6 x 11 +21 x 10 -406 x 9 +1692 x 8 -3330 x 7 +66813 x 6 -190530 x 5 -45783 x 4 -5155600 x 3 + 8296296 x 2 + 8156544 x + 238640704 -= 24, #Cℓ k + = 4 P = x 12 -6 x 11 + 21 x 10 -1846 x 9 + 8172 x 8 -16290 x 7 + 1374653 x 6 -4075170 x 5 + 711057 x 4 --= 30, #Cℓ k + = 2 P = x 12 -6 x 11 +21 x 10 -134 x 9 +468 x 8 -882 x 7 +62369 x 6 -184542 x 5 -436569 x 4 -1322320 x Cℓ k -= 30, #Cℓ k + = 2 P = x 12 -6 x 11 + 21 x 10 -1478 x 9 + 6516 x 8 -12978 x 7 + 1302965 x 6 -3870042 x 5 -2782815 x 4 -543509224 x 3 + 830485104 x 2 + 829417344 x + 150779996416 classgroup :[19440, [270, 18, 2, 2]]d = 4567 a = 195, b = 1 #Cℓ k -= 33, #Cℓ k + = 7 P = x 12 -6 x 11 +21 x 10 -440 x 9 +1845 x 8 -3636 x 7 +63557 x 6 -179844 x 5 +66765 x 4 -3988930 x 3 + 6294021 x 2 + 6052866 x + 109286116 classgroup : [18711, [2079, 9]] 5.4. Case Cℓ K ≃ Z/81Z × Z/27Z (a = 3). = 12058 a = 989, b = 26 #Cℓ k -= 42, #Cℓ k + = 2 P = x 12 -6x 11 +21 x 10 -4006 x 9 +17892 x 8 -35730 x 7 +22212821 x 6 -66531354 x 5 -113482743 x 4 -35777798264 x 3 + 54059937672 x 2 + 54106942656 x + 83308554531904 classgroup :[30618, [1134, 27]]d = 15607 a = 534, b = 1 #Cℓ k -= 39, #Cℓ k + = 1 P =x 12 -6 x 11 + 21 x 10 -2186 x 9 + 9702 x 8 -19350 x 7 + 1764719 x 6 -5236188 x 5 + 2322777 x 4 -638361238 x 3 + 965957748 x 2 + 958427808 x + 89817692416 classgroup :[28431, [1053, 27]] Cℓ k -= 120#Cℓ k + = 4 P = x 12 -6 x 11 +21 x 10 -3430 x 9 +15300 x 8 -30546 x 7 +7597005 x 6 -22699458 x 5 -18168075 x 4 -7877764840 x 3 + 11909686236 x 2 + 11905200672 x + 5526956498704 classgroup :[174960, [3240, 54]]d = 47194 a = 293, b = 2 #Cℓ k -= 120#Cℓ k + = 4 P =x 12 -6 x 11 + 21 x 10 -1222 x 9 + 5364 x 8 -10674 x 7 + 905093 x 6 -2683338 x 5 -2064183 x 4 -313267016 x 3 + 480721224 x 2 + 480118080 x + 75097921600 classgroup :[699840, [3240, 54, 2, 2]]

	d = 970
	a = 563, b = 20 #d = 211 a = 17, b = 1 #Cℓ k 2 + 4905 x + 11881 classgroup : [27, [9, 3]] d = 214 a = 89, b = 6 #Cℓ k classgroup : [54, [18, 3]] . . . d = 4531 a = 403, b = 5 #Cℓ k d = 1141 a = 449, b = 8 #Cℓ k 487867520 x 3 + 740542656 x 2 + 735780864 x + 74927017984 classgroup : [7776, [216, 18, 2]] d = 1174 a = 21, b = 5 #Cℓ k 3 + 3316650 x 2 + 3570000 x + 885062500 classgroup : [2430, [270, 9]] . . . d = 4087 a = 357, b = 8 . d = 45517 a = 845, b = 6 #d . . #

-= 12, #Cℓ k + = 2 P = x 12 -6 x 11 + 21 x 10 -856 x 9 + 3717 x 8 -7380 x 7 + 308855 x 6 -904506 x 5 -62898 x 4 -53921936 x 3 + 83258895 x 2 + 82428357 x + 4694853361 classgroup :

[1728, [36, 12, 4]

]

d = 4639 a = 361, b = 2 #Cℓ k -= 51, #Cℓ k + = 1 P =

x 12 -6 x 11 + 21 x 10 -1494 x 9 + 6588 x 8 -13122 x 7 + 834341 x 6 -2463738 x 5 + 888141 x 4 -212657184 x 3 + 323349156 x 2 + 320016960 x + 21950200336 classgroup :

[7344, [612, 12]

] 5.3. Case Cℓ K ≃ Z/27Z × Z/9Z (a = 2).

Acknowledgments. I thank Christian Maire for telling me about difficulties with the techniques developed in [W], Thong Nguyen Quang Do and Jean-François Jaulent for similar conversations and advices about it.

Numerical results

We give explicit numerical computations of Cℓ K , for various biquadratics fields k satisfying the conditions (i) to (v) assumed in the Hypothesis 2.1.

The following PARI program gives in "component(H, 5)" the class number and the structure of the whole class group Cℓ K of K in the form

For simplicity we compute an α being an integer without non-trivial rational divisor. So (α) is the cube of an ideal if and only if N k -/Q (α) ∈ Q ×3 . Then the irreducible polynomial defining K is given by P = polcompositum(

In all the examples, we recall that the 3-class group of k + is trivial, and that of k -is of order 3 exactely. Thus, the 3-class group of K + is trivial and

, the case n = 3 yields to a = 1, b = 1, and so on. od(G,3) == 0&M od(G, 9)! = 0, print(""); print("d = ", d); print("a = ", a); print("b = ", b); print("hm = ", hm, "h = ", h); print(P ); print("classgroup : ", F ); a = 10 3 ; b = 10 2 )))))))))}

The instruction "if(Mod(G, 3)==0 & Mod(G, 9)!=0" must be adapted to the relevant needed structure (3 a+1 , 3 a ).

We give below an extract of the numerical examples we have obtained; for a complete table, please see the Section 5 of: https://www.researchgate.net/publication/286452614 5.1. Case Cℓ K ≃ Z/3Z (a = 0). This implies Cℓ K = Cℓ G K ≃ Z/3Z: