THE CLASS NUMBER ONE PROBLEM FOR THE NORMAL CM

TRANSACTIONS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 359, Number 10, October 2007, Pages 5057–5089 S 0002-9947(07)04219-5 Article electronically published on April 16, 2007

THE CLASS NUMBER ONE PROBLEM FOR THE NORMAL CM-FIELDS OF DEGREE 32 SUN-MI PARK, HEE-SUN YANG, AND SOUN-HI KWON

Abstract. We prove that there are exactly six normal CM-fields of degree 32 with class number one. Five of them are composita of two normal CM-fields of degree 16 with the same maximal totally real octic field.

1. Introduction All the normal CM-fields of degree less than 32 with class number one are known. In this paper we will prove the following: Theorem 1. (1) There is one and only one normal CM-field of degree 32 with class number one which is not a compositum of two normal CM-subfields of degree 16 √ same maximal real subfield: the narrow Hilbert class √ with the field of Q( 5 · 29, 2). (2) There are exactly five normal CM-fields of degree 32 with relative class number one which are composita of two normal CM-fields of degree 16 with the √ √ √ 2 + 2, 3 + 3, −1), same maximal real subfield : Q( √ √ √ √ Q( 2, 17 + 4 17, −(5 + 17)/2, −(17 + 3 17)/2, α) with α8 + √ √ √ √ 4 17α6 √+ 85α + 136α2 + 68 = 0, Q(θ, 2, −1), Q(θ, 3, −1), and √ Q(θ, 7, −1) with θ 8 + 10θ 6 + 25θ 4 + 20θ 2 + 5 = 0. Moreover, those five fields have class number one. The structure of this paper is as follows. In Section 2 we review well-known results on relative class numbers of CM-fields which will be used later in this paper. According to [LOO, Lemma 2 (ii)] a totally imaginary normal number field is a CMfield if and only if the complex conjugation lies in the center of its Galois group. A normal CM-field of degree 32 is a compositum of two normal CM-fields of degree 16 with the same maximal totally real octic field if and only if the center of its Galois group is not cyclic, and there are 44 non-abelian groups of order 32: 29 out of them have non-cyclic centers. We will study all of these 44 groups. In Section 4 we focus on such composita. Section 3 is devoted to the normal CM-fields of degree 32 that are not such composita of two normal CM-fields of degree 16 with the same maximal totally real octic field.

Received by the editors May 6, 2004 and, in revised form, September 30, 2005. 2000 Mathematics Subject Classification. Primary 11R29; Secondary 11R21. c 2007 American Mathematical Society Reverts to public domain 28 years from publication

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SUN-MI PARK, HEE-SUN YANG, AND SOUN-HI KWON

2. Preliminaries Throughout this paper we use the following notation. For a number field F , let OF , dF , ζF , hF , Hil(F ), Hilnar (F ), and wF ≥ 2 be the ring of algebraic integers, the absolute value of discriminant, the Dedekind zeta function, the class number, the Hilbert class field, the narrow Hilbert class field of F , and the number of roots of unity in F , respectively. Let κF denote the residue of ζF (s) at s = 1. When F is an abelian number field, let G(F ) denote its genus field. For a finite abelian extension K of a number field k let F(K/k) be the finite part of the conductor of the extension K/k. When F is a totally real number field of degree n over Q, we denote by ∞ the product of its n real infinite primes. When F is a CM-field, we will denote by F + , h− F , and QF ∈ {1, 2}, the maximal totally real subfield, the relative class number, and the Hasse unit index of F , respectively. For a Galois extension K/k we denote by G(K/k) its Galois group. For a subgroup H of G(K/k), the fixed field of H is denoted by H . Let Dn be the dihedral group of order n, Cn the cyclic group of order n, and Qn the dicyclic group of order n. (Q8 is usually said to be the quaternion group of order 8.) For a group H let C(H) be the center of H, D(H) the commutator subgroup of H, and let H ab = H/D(H). Theorem 2. (1) ([LO2, Proposition 2]) If a CM-field N = N1 N2 is a compositum of two CM-fields N1 and N2 with the same maximal totally real subfield N1+ = N2+ , then (A)

h− N =

wN QN h− h− , QN1 QN2 wN1 wN2 N1 N2

− − and h− N1 hN2 divides 4hN . In particular, if N1 and N2 are isomorphic, then

(B)

h− N

QN = 2

h− N1 QN1

2 .

(2) ([Ok, Theorem 1] and [Lem1, Proposition 1 (d)]) Let M1 ⊆ M2 be two − CM-fields. Then h− M1 |4hM2 , and QM1 wM1 divides QM2 wM2 . (3) ([Lou2, Proposition 6]) Let M be a CM-field and t the number of ramified − t prime ideals in M/M + . Then 2t−1 divides h− M and 2 divides hM if QM = − 2. In particular, if hM = 2QM , then at most two prime ideals are ramified in M/M + . √ (4) ([Lem1, Theorem 1 (i)]) Let N = N + ( −δ) with δ ∈ N + totally positive. Assume that wN ≡ 2 mod 4. We have three mutually exclusive cases: (a) If δON + is not the square of any ideal of ON + , then QN = 1. In particular, if there is a prime ideal p lying above an odd prime p which is ramified in the extension N/N + , then QN = 1. (b) If δON + = a2 for some principal ideal a of ON + , then QN = 2. In particular, if hN + is odd and if N/N + is unramified at all finite primes, then QN = 2. (c) If δON + = a2 for some non-principal ideal a of ON + , then QN = 1. (5) If a CM-field N has a subfield M which is a CM-field with QM = 2, then √ QN = 2 unless N = N + ( −1).


NORMAL CM-FIELDS OF DEGREE 32

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(6) ([W]) Let K be a CM-field of degree 2n over Q. Then Q dK κK w K K . h− K = (2π)n dK + κK + (7) ([Lou10, Theorem 1]) For any totally imaginary number field K of degree ≥ 12 and root discriminant ≥ 2π 2 we have κK ≥ (1 − β)/(2e) if ζK (β) ≤ 0 for some β ∈ [1 − (2/ log dK ), 1), and κK ≥ 2/(e log dK ) if ζK (1 − 2/ log dK ) ≤ 0. (8) Let K range over the abelian extensions of degree m unramified at all the infinite places of a given k. Then there exists a computable constant µk ≥ 0 such that 1 m−1 m κk . κK ≤ ( log(dK /dm k ) + 2µk ) 2(m − 1) κm If K/k is unramified at all the places, then κK ≤ µm−1 k . Moreover, if k k is a real abelian field of degree n ≥ 2, then µk κk ≤

logn dk . − 1)n−1

2n+1 (n

(9) ([LO2, Proposition 15]) Let F (2) denote the compositum of all the quadratic subfields of a normal number field F of 2-power degree. Then ζF /ζF (2) is an entire function and (ζF /ζF (2) )(s) ≥ 0 for s ∈ (0, 1). √ √ √ (10) Let M = Q( p1 , p2 , p3 ), where the pi ’s are the distinct primes ≡/ 3 √ mod 4. Set k = Q( p1 p2 p3 ). Then log7 dk . 24 36 has a real zero α ∈ ( 21 , 1), then we also have κM ≤

Moreover, if ζM

1−α log8 dk . 26 36 (11) Let k be a real abelian number field of degree n ≥ 2. Then log d n−1 k . κk ≤ 2(n − 1) κM ≤

If k is a real quadratic number field and if ζk has a real zero β in ( 21 , 1), then 1−β (C) κk ≤ log2 dk . 8 √ Proof. (5) (See [Ok, Lemma 14]) QM =√2 implies M = M + ( −η) with a totally positive unit η in M + . Thus N = N + ( −η), so QN = 2 unless η = α2 for some α ∈ N +. (8) The first assertion and the second one are Corollary 2 in [Lou5]. According to [R1] (see also [R2]) for any primitive even character χ of conduct f we have |L(1, χ)| ≤ (log f )/2. Hence, the proofs of [Lou5, Theorem 11] or [Lou9, Theorem 9] now yield the third assertion.


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(10) is derived from [LO2, Proposition 16] and [R1, Corollary 1]. (11) is derived from [Lou4, Corollaries 5A and 7B] and [R1, Corollary 1].

3. The fields N which are not composita The aim of this section is to prove point (1) of Theorem 1. Throughout this section N denotes a non-abelian normal CM-field of degree 32 which is not a compositum of two normal CM-subfields of degree 16 with the same maximal real subfield. Then the center of its Galois group G(N/Q) must be cyclic. According to [TW] there are 15 non-abelian groups of order 32 whose centers are cyclic: G32/17 , G32/22 , G32/26 , G32/31 , G32/32 , and G32/i for 42 ≤ i ≤ 51. For each of those 15 groups G we will study whether there exists a CM-field N of class number one with Galois group G. In Subsection 3.1 we will show that if G(N/Q) is neither of type G32/26 nor G32/31 , then hN > 1. In Subsection 3.2 we determine all CM-fields of class number one such that their Galois groups are of type G32/26 . In Subsection 3.3 we prove that if G(N/Q) is of type G32/31 , then hN > 1. 3.1. The fields N with G(N/Q) = G32/26 , G32/31 . In this subsection we will show that if G(N/Q) = G32/26 , G32/31 , then hN ≥ 2. Lemma 1 ([LO2, Proposition 4]). Let N be a non-abelian normal CM-field of degree 2n = 2r ≥ 8. Assume that h− N is odd and N is cyclic of degree n over a quadratic subfield k. Then k is real and N is a dihedral CM-field. Proposition 1. If G(N/Q) has an element of order 16, then h− N ≥ 2. In particular, if G(N/Q) ∈ {G32/22 , G32/50 , G32/51 }, then h− is even, and if G(N/Q) = N G32/49 (= D32 ), then h− ≥ 2. N Proof. Follows from Lemma 1 and [Lef, Theorem 1.1].

Proposition 2. If G(N + /Q) is isomorphic to C2 × C2 × C2 × C2 , then hN is even. In particular, if G(N/Q) ∈ {G32/42 , G32/43 }, then hN is even. Proof. Suppose the contrary. Since hN is odd and N is a non-abelian number field, √ √ √ √ G(N + ) = N + . Hence, N + = Q( p1 , p2 , p3 , p4 ) where pi ’s are primes with √ + ≡ / 3 mod 4. Then N is the Hilbert 2-class field of k = Q( p1 p2 p3 p4 ) and the 2-class field tower of k has length one. According to [BLS, Corollary 2] the 2-rank of cl(k) is ≤ 2 which contradicts the fact that G(N + /k) cl(k) C2 ×C2 ×C2 . Proposition 3. If G(N/Q) ∈ {G32/44 , G32/46 }, then h− N is even. Proof. We will show that N = M16.1 M16.2 is a compositum of two isomorphic + + = M16.2 , and that one of these two fields non-normal CM-subfields with M16.1 say, M16.1 = M8.1 M8.2 , is a compositum of two isomorphic non-normal octic CMh−

M16.1 2 QN + + = M8.2 . Hence, h− ) and h− subfields with M8.1 N = 2 ( QM M16.1 = 16.1

− QM16.1 hM8.1 2 ( QM ) , 2 8.1 8

by (B), which implies that h− N is always even. Write G32/44 = a, b, c : a = 2 2 −1 b = c = 1, ab = ba , ac = ca5 , bc = cb and G32/46 = a, b, c, d : a2 = b2 = c2 = d4 = 1, bd = dab, cd = dabc, ab = ba, ac = ca, ad = da, bc = cb. Note that C(G32/44 ) = a4 and C(G32/46 ) = a. When G(N/Q) = G32/44 , set M16.1 = c , M16.2 = a4 c , M8.1 = b, c , and M8.2 = a6 ba−6 , c . We have + + + + a4 c = aca−1 , M16.1 = M16.2 = a4 , c , a6 ba−6 , c = a6 b, ca−6 , M8.1 = M8.2 =



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a4 , b, c . When G(N/Q) = G32/46 , set M16.1 = b , M16.2 = dbd−1 , M8.1 = + +

b, c , and M8.2 = b, ac . Then ac = d2 cd−2 , M16.1 = M16.2 = a, b , and + + M8.1 = M8.2 = a, b, c . Lemma 2 (Chebotarev’s Monodromy Theorem). Let k ⊆ K be two number fields such that K/Q and k/Q are normal, and K/k is unramified at all the finite places. Let m be the least positive integer such that g m = 1 for every g ∈ G(k/Q). Then the Galois group G(K/Q) is generated by the elements σ which are not contained in the subgroup G(K/k) and at the same time satisfy σ m = 1.

Proof. See [Lem2, Corollary 1]. Proposition 4. If G(N/Q) ∈ {G32/45 , G32/47 , G32/48 }, then

h− N 5

is even.

Proof. Write G32/45 = a, b, c | a8 = b2 = 1, c2 = a4 , ab = ba , ac = ca−1 , bc = cb, G32/47 = a, b, c | a8 = b2 = c2 = 1, ab = ba5 , ac = cba, bc = cb, and G32/48 = a, b, c | a8 = b2 = 1, c2 = a4 , ab = ba5 , ac = cba, bc = cb. Note that the centers of these three groups are a4 , so N + = a4 . Assume that G(N/Q) ∈ {G32/45 , G32/47 , G32/48 }. Since N is a compositum of two isomorphic CM-subfields b and aba−1 with the same maximal real subfield a4 , b , accord+ ing to Theorem 2, points (1) and (3), if h− is unramified at N is odd, then N/N 2 2 all the finite places, hence so is N/ a . Note that a is a normal subgroup of G(N/Q). We will show that N/ a2 cannot be unramified at all the finite places. (1) If G(N/Q) = G32/45 , then G(N/Q)/ a2 C2 × C2 × C2 . We observe that the subgroup generated by the elements of order 2 in G(N/Q) is b, acb, cab which is a group of order 16 isomorphic to the group of type 16/8 designated in [TW] (or named G9 in [JL]). According to Lemma 2, N/ a2 cannot be unramified at all the finite places and h− N is even. (2) If G(N/Q) = G32/47 or G32/48 , then G(N/Q)/ a2 D8 . We observe that the subgroup generated by the elements of order 2 or 4 in G(N/Q) is a2 , b, c which is isomorphic to D8 × C2 or Q8 × C2 according as G(N/Q) = G32/47 or G32/48 . By the same argument as in (1), it follows that h− N is even. Lemma 3. Let K ⊂ K ⊂ L be the number fields such that L/K is a Galois extension. Let K ab (K ab , respectively) be the maximal extension of K (K , resp.) contained in L which is abelian over K (K , resp.). Let m and m be the conductors of K ab /K and K ab /K , respectively, Im the group of fractional ideals of K coprime to m, and let Im be that of K coprime to m . Then the following diagram is commutative: V er G(L/K)ab −→ G(L/K )ab ↑ ↑ incl. Im −→ Im Here, Ver means the transfer map. Proof. See the functorial properties of the reciprocity map in [S, Ch. XI §3]. (See also [Mar] or [N, Ch. II and IV].) Proposition 5. If G(N/Q) ∈ {G32/17 , G32/32 }, then h− N is even. Proof. Write G32/17 = a, b, c | a8 = b2 = c2 = 1, ab = ba, ac = ca, bc = ca4 b and G32/32 = a, b | a8 = 1, a4 = b4 , ab = ba−1 . Note that C(G32/17 ) = a and C(G32/32 ) = b2 .


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(a) Assume that G(N/Q) = G32/17 . The field N is a compositum of two isomorphic CM-subfields b and cbc−1 with the same maximal real subfield a4 , b . 4 Suppose that h− N is odd. By Theorem 2, points (1) and (3), N/ a is unramified at all the finite places and so is N/ a . Since the ring of integers Z is principal, according to Lemma 3 the transfer map V er : Gab 32/17 → a would be trivial. 4 However, V er(aD(G32/17 )) = a , which is a contradiction. Hence, h− N is always even. (b) Assume that G(N/Q) = G32/32 . The field N is a compositum of two isomorphic CM-subfields a2 b2 and a6 b2 with the same maximal real subfield

a2 b2 , a4 . (Note that a6 b2 = ba2 b2 b−1 .) Suppose that h− N is odd. By Theorem 2, points (1) and (3), N/ b4 is unramified at all the finite places and so is N/ b . 4 Hence, V er : Gab 32/32 → b would be trivial. However, V er(aD(G32/32 )) = b , which is a contradiction. Hence, h− N is always even. 3.2. The fields N with G(N/Q) = G32/26 . Write G32/26 = a, b, c | a8 = b2 = c2 = 1, ab = ba, ac = ca−1 , bc = ca4 b. Since C(G32/26 ) = a2 b C4 , the complex conjugation is a4 and G(N + /Q) D8 × C2 . We will prove the following: Proposition 6. G(N/Q) = G32/26 and hN = 1 if and only if √ √ N = Hilnar (Q( 5 · 29, 2)). We proceed as follows. First, we prove that if hN = 1, then √ √ N = Hilnar (Q( p1 p2 , p3 )), where pi ’s are primes ≡ / 3 mod 4. Second, we find a C such that if dN ≥ C, then − h− N > 1. Third, using Hecke L-functions we compute hN for the fields N with dN < C and prove Proposition 6. Lemma 4. (1) If h− N is odd, then QN = 2 and N has a unique real quadratic subfield k such that G(N/k) Q16 , and N/k is unramified at all the finite places. Moreover, dk = d1 d2 d3 is the product of three distinct discriminants of real quadratic fields and (i) (di , dj ) = 1 for i = j; (ii) ( dp21 ) = ( pd12 ) = 1 for all primes p1 | d1 and p2 | d2 ; √ (iii) G(N + /k( d3 )) = C4 and G(N + /k( dj )) = C2 × C2 for j = 1, 2. √ / 3 mod 4. (2) If hN = 1, then k = Q( d1 d√ 2 d3 ) and these di ’s are primes di ≡ √ In addition, N = Hilnar (Q( d1 d2 , d3 )). 4 −1 Proof. (1) Assume that h− , N is a compositum of N is odd. Since a b = cbc 4 two isomorphic CM-subfields b and a b with the same maximal real subfield

a4 , b . By Theorem 2, QN = 2 and N/N + is unramified at all the finite places. Since a4 is the unique element of order 2 in a, cb Q16 , N/ a, cb is unramified at all the finite places. Set k = a, cb . The remaining statements follow from [Lem2, Theorem 2]. (2) Let M = a2 . Then G(M/Q) = C2 × C2 × C2 and M is the maximal = 1, √ then √ M = normal subfield of N + which is an abelian extension of k. If hN √ / 3 mod 4 and M = Q( d1 , d2 , d3 ). Hil(k) = G(k). Hence di ’s are primes di ≡ + Let K = √ then N = Hil(K) and N = Hilnar (K). Note that √ a . If hN = 1, G(N/k( d1 )) √ G(N/k( d2 )) Q8 . Moreover, according to (1) point (iii) we have K = k( d3 ).



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From now on p1 , p2 , and p3 denote distinct primes ≡ / 3 mod 4 satisfying ( pp21 ) = √ √ √ = 1. Write K = a = Q( p1 p2 , p3 ), k = a, cb = Q( p1 p2 p3 ), and √ √ √ M = a2 = Q( p1 , p2 , p3 ). ( pp12 )

Lemma 5. Assume that hN = 1. √ (i) Q( pi ) has class number one for i = 1, 2, and 3. √ √ √ (ii) cl(Q( p1 p2 )) C4 , cl(Q( p1 p3 )) cl(Q( p2 p3 )) C2 . √ √ Proof. (i) The class number of Q( pi ) is odd and Hil(Q( pi )) lies in N . Hence √ the class number of Q( pi ) must be equal to one. (ii) Since G(N/k) = a, cb is the dicyclic group of order 16, it has two quaternion subgroups of order 8: a2 , cb and a2 , acb. Changing p1 to p2 if necessary we √ √ √ √ may assume that a2 , acb = Q( p1 p2 p3 , p1 ) and a2 , cb = Q( p1 p2 p3 , p2 ). Note that the element ab is of order 8 and (ab)2 = a2 . The fixed field of ab √ √ √ lies in a2 = Q( p1 , p2 , p2 ) and N/ ab is unramified at all the finite places. √ √ √ √ √ Hence, Q( p1 , p2 , p3 )/ ab is unramified and ab = Q( p1 p2 , p2 p3 ). Thus, √ √ √

a, b = Q( p1 p2 ), ab, acb = Q( p2 p3 ), and ab, cb = Q( p1 p3 ). Since √ √ √ + + G(N /Q( p1 p2 )) C4 ×C2 and Hil(Q( p1 p2 )) is contained in N , Hil(Q( p1 p2 )) must be the fixed field of the inertia group of a prime ideal in N + lying above p3 . Note that the inertia group is of order 2 and all intermediate fields be√ tween N + and Q( p1 p2 ) are normal extensions over Q. Because the 2-rank of √ √ cl(Q( p1 p2 )) is equal to 1, Hil(Q( p1 p2 )) is one of the two cyclic quartic ex√ √ tensions of Q( p1 p2 ) contained in N + and cl(Q( p1 p2 )) C4 . Similarly, we √ √ √ √ √ √ verify that Hil(Q( p2 p3 )) = Q( p2 , p3 ) and Hil(Q( p1 p3 )) = Q( p1 , p3 ). √ √ Since hN = hN + = 1, Hil(Q( p2 p3 )) and Hil(Q( p1 p3 )) are contained in N + . √ √ Note that G(N + /Q( p2 p3 )) ab, acb/ a4 D8 and G(N + /Q( p1 p3 )) √ √ √ √

ab, cb/ a4 D8 . The field Q( p2 , p3 ) (respectively, Q( p1 , p3 )) is the unique √ √ intermediate field between N + and Q( p2 p3 ) (respectively, Q( p1 p3 )) that is an √ √ unramified and normal over Q( p2 p3 ) (respectively, Q( p1 p3 )). It follows that √ √ cl(Q( p2 p3 )) cl(Q( p1 p3 )) C2 . √ From now on we assume that N is unramified over k = Q( p1 p2 p3 ) at all the 8 finite places and QN = 2. Hence dN = d16 k and dN + = dk . We first show that if − 5 dk ≥ 9.1 × 10 , then hN > 1. Set βN = 1 − (2/ log dN ). (i) Assume that ζN (βN ) ≤ 0. By Theorem 2, points (7), (8), and (10), κN ≥

1 23 e log dk

and κN + ≤ µM κ2M ≤

23 log15 dk . 36 77

Hence, (D)

h− N ≥

36 77 220 eπ 16

d4k log16 dk

5 and h− N > 1 if dk ≥ 9.1 × 10 . (ii) Assume that ζN (βN ) > 0. Since N (2) = M , by Theorem 2, point (9), we have ζM (βN ) > 0 and there is a positive real number β such that βN < β < 1 and


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ζM (β) = 0. Hence, ζN + (β) = ζN (β) = 0, κN ≥ (1 − β)/(2e), 2(1 − β) log16 dk . 36 77 4 and we end up with a lower bound for h− N which is 2 times better than (D). Thus − (D) is always valid and we conclude that hN > 1 if dk ≥ 9.1 × 105 . Now, there are 510 triplets (p1 , p2 , p3 ) such that dk ≤ 9.1 × 105 and p1 , p2 , and p3 satisfy Lemmas 4 and 5. For those 510 triplets (p1 , p2 , p3 ) we compute explicitly κM (see [Lou5, Section 4.2]) and improve our upper bound for κN + : κN + ≤ µM κ2M ≤

27 log8 dk . 77 as follows. If ζN (βN ) ≤ 0, then

κN + ≤ κM · µM κM ≤ κM We improve the lower bound for h− N

d4k . κM log9 dk If ζN (βN ) > 0, then there is a positive real number β ∈ (βN , 1) such that ζM (β) = ζN (β) = 0. Hence, 4 (1 − β) ≥ , κN ≥ 2e eBN (2) where log2 dF BN (2) = max F ⊂N (2) L(1, χF ) h− N ≥

(E)

77

224 eπ 16

[F :Q]=2

( dpF

) denotes the corresponding quadratic character of F . In fact, and χF (p) = ζM (β) = 0 implies that there exists a subfield F ⊂ M = N (2) with [F : Q] = 2 such that ζF (β) = 0. Hence, 1≤

1−β 1 − β log2 dF ≤ BN (2) , 8 L(1, χF ) 8

by (C). Thus (F)

h− N ≥

77

d4k

219 eπ 16

κM BN (2) log8 dk

.

For these 510 triplets (p1 , p2 , p3 ) we compute two lower bounds for h− N in (E) and (F) respectively, and then take the worse one. There are 467 out of these 510 triplets with h− remaining 43(= 510 − 467) triplets N > 1. Finally, only 19 out of the√ √ (p1 , p2 , p3 ) satisfy cl(K) C4 , where K = Q( p1 p2 , p3 ) (use [KT]). For these 19 triplets (p1 , p2 , p3 ) we construct the primitive Hecke characters χ of order 8 on the narrow class group of K and evaluate L(0, χ) using the technique developed in [Lou7] and [Lou8], where L(s, χ) is the Hecke L-function. According to [Lou8] we have 1 √ h− N = QN wN NQ( 2, i)/Q ( 24 L(0, χ)). √ √ We verify that there is√one and only one field K that has h− N = 1: K = Q( 5 · 29, 2) and L(0, χ) = 16 + 8 2. Using the package [KT] we verify that √ √ G(Hilnar (Q( 5 · 29, 2))/Q) G32/26 and hHil(Q(√5·29,√2)) = 1. This completes the proof of Proposition 6.



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3.3. The fields N with G(N/Q) = G32/31 . Write G32/31 = a, b, c | a4 = b4 = c2 = 1, ab = ba, ac = ca, bc = cab−1 . Since C(G32/31 ) = a, the complex conjugation is a2 and N + = a2 . The aim of this subsection is to prove: Proposition 7. If G(N/Q) = G32/31 , then hN > 1. Before starting the proof we observe that k1 = cb, b2 , k2 = a, b , and k3 =

a, b2 , c are the only quadratic subfields of N . Note that G(N + /k1 ) = C4 × C2 , G(N + /k2 ) = C4 × C2 , and G(N + /k3 ) = C2 × C2 × C2 . Lemma 6. If G(N/Q) = G32/31 and h− N is odd, then every prime divisor of dk3 divides dk1 and is not equal to 3 modulo 4. Such primes are the unique primes having ramification index 4 in N/Q. Proof. Since cb2 c−1 = a2 b2 , N is a compositum of two isomorphic CM-subfields

b2 and a2 b2 with the same maximal real subfield a2 , b2 . According to The+ orem 2 the fact that h− is unramified at all the finite N is odd implies that N/N places. Let L3 = ab2 , ac . Since ab2 , ac Q8 and (ab2 )2 = a2 , N/L3 is unramified at all the finite places. The only prime divisors of dN are hence those of dL3 . The subgroup ab2 , ac is normal in G32/31 with G32/31 / ab2 , ac C4 . Since

ab2 , ac ⊂ a, b2 , c and k3 ⊂ L3 , every prime divisor p of dk3 is totally ramified in / 3 mod 4 and the ramification index of p in the cyclic extension L3 /Q. Hence, p ≡ N is equal to 4. Let L1 = cb . As (cb)2 = a and L1 ⊂ N + , N/L1 is unramified at all the finite places. Hence, p is totally ramified in L1 and p | dk1 . Remark 1. The fact that if h− N is odd, then N/L3 is unramified at all the finite places will be used in the sequel. The proof of Proposition 7 consists of four parts. Part 1: We claim that if dL3 ≥ 8.7 × 1014 , then h− N > 1. Proof. Set βN = 1 − 2/ log dN . (i) Assume that ζN (βN ) ≤ 0. By Theorem 2 κN ≥

2 1 = e log dN 4e log dL3

and κN + ≤ κL3 · (µL3 κL3 )3 ≤

log15 dL3 . 218 312

Hence, h− N ≥

22 312 d2L3 eπ 16 log16 dL3

12 and h− N > 1 if dL3 ≥ 6.1 × 10 . (2) is the biquadratic (ii) Assume that ζN (βN ) > 0. Since Gab 32/31 C4 × C2 , N (2) bicyclic subfield of N containing k1 , k2 , and k3 , i.e., N = L4 = a, b2 ( a, b2 =

cb, b2 ∩ a, b ∩ a, b2 , c). According to Theorem 2, point (9), ζN (2) (βN ) > 0, which yields ζN (2) (β) = 0 for some β ∈ (βN , 1). Thus ζki (β) = 0 for some i=1, 2, or 3. 2 and κki > 1−β Hence, κN ≥ (1−β) 2e 8 log dki . By Theorem 2, point (8), we have

κN + ≤

7 1 log(dN + /d8ki )κki + 2µki κki κki , 14


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and µki κki ≤ get

1 8

log2 dki . Since d2ki ≤ dL3 and dN + /d8ki = (dL3 /(dki )2 )4 < d4L3 , we 37 57 (1 − β) log16 dL3 , 233 77 dL3 2 218 77 ≥ 7 7 16 , 3 5 eπ log16 dL3

κN + ≤ h− N

− 14 and h− N > 1 if dL3 ≥ 8.7 × 10 . (However, we can obtain N ≥ 0.978 and hN > 1 14 if dL3 ≥ 5.6 × 10 using [Lou1, Proposition A].)

Part 2: We claim that (G)

h− N =

− − QN hK8.1 hK8.2 2 , 8 QK8.1 QK8.2

where K8.1 = b and K8.2 = a2 b . Proof. Since G(N/k2 ) = a, b C4 × C4 , there are four octic CM-subfields of N containing k2 : K8.1 = b , K8.2 = a2 b , L8.1 = a3 b and L8.2 = ab . Note that + + + cbc−1 = ab3 and ca2 bc−1 = (ab)−1 . Then K8.1 = K8.2 = a2 , b and L+ 8.1 = L8.2 =

a2 , ab . Let χ and ψ be two primitive quartic Hecke characters associated with the cyclic quartic extensions K8.1 /k2 and L8.1 /k2 , respectively. Then χψ 2 and χ2 ψ are associated with the extensions K8.2 /k2 and L8.2 /k2 , respectively. According to (17) in [Lou8] we have 1 1 − 2 h− K8.1 = QK8.1 wK8.1 NQ(i)/Q ( 2 L(0, χ)), hK8.2 = QK8.2 wK8.2 NQ(i)/Q ( 2 L(0, χψ )), 2 2 1 1 − 2 h− L8.1 = QL8.1 wL8.1 NQ(i)/Q ( 2 L(0, ψ)), hL8.2 = QL8.2 wL8.2 NQ(i)/Q ( 2 L(0, χ ψ)), 2 2 − − − h− K8.1 = hL8.1 , hK8.2 = hL8.2 and h− N = QN wN NQ(i)/Q

1 2 2 L(0, χ)L(0, χψ )L(0, ψ)L(0, χ ψ) , 28 = wK8.i = wL8.i = 2 for i=1, 2.

which proves (G). Note that wN

Part 3: Suppose that hN = 1. We claim the following: / 3 mod 4 such that (i) hk2 = hk3 = 1. There exist two distinct primes p, q≡ √ √ √ k1 = Q( pq), k2 = Q( q), and k3 = Q( p). (More precisely, it will be verified in the proof of point (iv) below that p = 2 or p ≡ 1 mod 8.) The ramification indices of p and q in N/Q are 4 and 2, respectively. (ii) Without loss of generality we may assume that hK8.1 = 1, h− K8.2 = hK8.2 = 2, and QK8.1 = QK8.2 = 1. In addition, p and q are the only ramified primes in N/Q. √ √ √ (iii) Set L4 = a, b2 . Then L4 = Q( p, q), hL4 = 2, cl(Q( pq)) = C4 , ( pq ) = √ ( pq ) = 1, and Nk1 /Q (1 ) = −1, where 1 is the fundamental unit of k1 = Q( pq). (iv) Let p and p be the two prime ideals in k2 lying above p. Then F(K8.1 /k2 ) = 3 p and F(K8.2 /k2 ) = pp if p is odd, F(K8.1 /k2 ) = p4 and F(K8.2 /k2 ) = p4 p if p=2. Before proving Part 3 we recall the following result. Proposition 8 ([Mas, Corollaries 2.2 and 2.3]). Let E/F be an extension of number fields. Then hF divides [E : F ]hE . Moreover, if no non-trivial abelian subextension of E/F is unramified over F , then hF divides hE .



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Proof. (i) To begin with we will prove that G(k2 ) = k2 and G(k3 ) = k3 . Note that there are five intermediate fields between N and k2 which are normal over Q:

a2 , a , a2 , b2 , ab2 , and a, b2 . Moreover, G32/31 / a2 G6 in [JL] (or type 16/9 in [TW]), G32/31 / a D8 , G32/31 / a2 , b2 D8 , G32/31 / ab2 C2 × C4 , and G32/31 / a, b2 C2 × C2 . Since hN = 1, G(k2 ) ⊂ N and G(k2 ) = a, b2 or G(k2 ) = k2 . Since a, b2 contains k1 , k2 , and k3 , G(k2 ) = a, b2 . Otherwise, every prime divisor p of dk3 divides dk2 and at the same time divides dk1 by Lemma 6. Thus, p is totally ramified in a, b2 /Q. It is contrary to the fact that a, b2 /k2 √ is unramified. Hence, k2 = G(k2 ), hk2 = 1, and k2 = Q( q) for some prime q ≡ /3 √ / 3 mod 4. mod 4. Similarly, hk3 = 1 and k3 = Q( p) for some prime p ≡ (ii) Let P be a prime ideal of N dividing p and G0 (P) its inertia group. Since N/N + is unramified at all the finite places, a2 ∈ / G0 (P) and G0 (P) must be one of the four cyclic subgroups of order 4 of G(N/k2 ) = a, b: ab, ab3 , b, a2 b. Changing P to P = cP, or K8.1 to K8.2 if necessary we may assume that K8.1 = G0 (P) . According to Proposition 8 hN16.1 = hK8.1 = 1, and hK8.2 |2, where + N16.1 = b2 . Then G0 (P ) = cG0 (P)c−1 = ab3 . Since G(N/K8.1 ) = a2 , b + and G(N/K8.1 ) ∩ G0 (P ) = a2 b2 , the prime ideal P ∩ OK8.2 and the prime ideal + . By Proposition 8 h− P∩OK8.2 are ramified in K8.2 /K8.2 K8.2 = hK8.2 = 2. From (G) it follows that QK8.1 = QK8.2 = 1. For the second assertion, we suppose that there exists a prime r with (r, pq) = 1 which is ramified in N/Q. Its ramification index is equal to 2. There are at least two prime ideals lying above r in N . Let R be a prime ideal lying above r such that its inertia group G0 (R) is equal to a2 b2 . Since +

a2 b2 ∩ G(N/K8.1 ) = {1}, R is unramified in N/K8.1 . Since G0 (R) ⊂ G(N/K8.1 ) + 2 2 and G0 (P ) ∩ G(N/K8.1 ) = a b , R ∩ OK8.1 and P ∩ OK8.1 are ramified in + . It follows that 2 divides h− K8.1 /K8.1 K8.1 , which contradicts hK8.1 = 1. Hence, p and q are the only prime divisors of dN . (iii) Since G(N + /k1 ) cb, b2 / a2 C4 × C2 , there are three octic subfields of N + containing k1 : a , a2 , b2 , and ab2 . Since N/N + is unramified at all the finite places, so are N/ a and N/ ab2 . Let B = a2 , b2 . Since a2 , b2 G and cb, b2 / a2 , b2 C4 , G(B/k1 ) is a cyclic group of order 4. We claim that B = Hil(k1 ). For every prime ideal lying above p in N + , its inertia group in the extension N + /k1 is of order 2. Hence, it must be G(N + /B). Since the ramification index of q in N/Q is equal to 2, the prime ideal lying above q is unramified in N + /k1 . By (ii) p and q are the only ramified primes in N/Q. Hence, B/k1 is unramified at all the places. From Proposition 8 it follows that hB = 1, B = Hil(k1 ), and √ √ the unique intermediate field L4 = a, b2 = Q( p, q) between B and k1 has class number 2. According to [BLS, Proposition 1] ( pq ) = ( pq ) = 1. Since hN = 1, Hil(k1 ) = Hilnar (k1 ). Namely, Nk1 /Q (1 ) = −1. (iv) Using the discriminant-conductor formula for the abelian extension N/k2 we can get the finite parts of the conductors F(χ)(= F(K8.1 /k2 )) and F(χψ 2 )(= F(K8.2 /k2 )) from dN (= d8L3 ). Let us determine dL3 . If p and q are odd, then dL3 = p3 q 2 . We claim that if p is odd, then p ≡ 1 mod 8. Let ab2 = T . Then N/T is unramified at all the finite places. Since ab2 G32/31 and G32/31 / ab2 C4 ×C2 , T has two cyclic quartic subfields L3 = ab2 , ac and L5 = ab2 , c . The inertia group of a prime ideal lying above q in T must be the Galois group G(T /L5 ). Hence L5 is a totally real field of conductor p if p is odd, 24 otherwise. Thus, p ≡ 1 mod 8 if p is odd. We have dL3 = 211 q 2 if p=2 or dL3 = p3 26 if q=2. Note that the case that dL3 = p3 24 cannot occur.


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Part 4: There are 5447 pairs of (p, q) which satisfy dL3 < 8.7 × 1014 , p = 2 or p ≡ 1 mod 8, q ≡ / 3 mod 4, hQ(√p) = hQ(√q) = 1, ( pq ) = ( pq ) = 1, Nk1 /Q (1 ) = −1, and hQ(√pq) = 4. For these 5447 pairs of (p, q) we compute explicitly κL4 . Using the κ

log4 d

better upper bound κN + ≤ ( L6 4 log(dN + /d4L4 ) + 24 33L4 )3 κL4 we improve the lower bound for h− N obtained in Part 1 as in the proof of Proposition 6. If ζN (βN ) ≤ 0, then (H)

h− N ≥

d2L3 1 . 216 eπ 16 log dL3 ( κL6 4 log (dN + /d4L4 ) + (log4 dL4 )/(24 33 ))3 κL4

If ζN (βN ) > 0, then (I)

h− N ≥

d2L3 1 . 212 eπ 16 BN (2) ( κL6 4 log (dN + /d4L4 ) + (log4 dL4 )/(24 33 ))3 κL4

For all of these 5447 pairs of (p, q) we compute the lower bounds in (H) and (I) respectively, and then take the smaller one. There are 5236 out of these 5447 pairs with h− N > 1. For the remaining 211(= 5447 − 5236) pairs of (p, q) we compute and h− h− K8.1 K8.2 . We examine whether there exist Hecke characters χ of order 4 with conductor p∞ if p ≡ 1 mod 8, p4 ∞ otherwise. There are 103 pairs of (p, q) that satisfy χ(2 ) = 1 and χ(−1) = 1, where 2 denotes the fundamental unit of k2 . For these 103 pairs of (p, q) we compute L(0, χ) by using the technique developed in [Lou8] to obtain that h− K8.1 = 1 if and only if (p, q) ∈ {(17, 53), (89, 5), (137, 2)}. For these three fields we compute h− K8.2 and verify that there is no pair of (K8.1 , K8.2 ) − , h ) = (1, 2). Our computational results are given in the table such that (h− K8.1 K8.2 below. (p, q) L(0, χ) h− L(0, χψ 2 ) h− h− K8.1 K8.2 N (17,53) 2 + 2i 1 −8 − 4i 10 52 (89,5) 2 + 2i 1 −24 − 4i 74 372 (137,2) −2 − 2i 1 −8 + 36i 170 852 It follows that there is no field N with hN = 1, as claimed. Now, the proof of point (1) of Theorem 1 is complete.

4. The composita N = M16.1 M16.2 The aim of this section is to prove point (2) of Theorem 1. In this section N = M16.1 M16.2 is a normal CM-field of degree 32 which is a compositum of two normal CM-fields, M16.1 and M16.2 , of degree 16 over Q with the same maximal + + = M16.2 . In order to determine all such fields N real octic subfield M8+ := M16.1 − with hN = 1 we divide our computations into 4 cases according to the Galois group G(M8+ /Q) ∈ {D8 , Q8 , C4 × C2 , C2 × C2 × C2 }. It is known that if L is an imaginary abelian number field with relative class number one, then [L : Q] ≤ 24 ([CK]). If + h− N = 1, then G(M8 /Q) cannot be isomorphic to C8 . Otherwise, M16.1 , M16.2 , and N are imaginary abelian number fields and h− N > 1. For an abelian number field ∗ K we write G(K/Q) = Cm × Cn when K = Lk is a compositum of an imaginary cyclic number field L of degree m over Q and a real cyclic number field k of degree n with L ∩ k = Q.



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Proposition 9 ([JL] and [TW]). There are 9 non-abelian groups of order 16 : G1 G2 G3

= a, b : a8 = b2 = 1, b−1 ab = a−1 = D16 For 1 ≤ i ≤ 3 = a, b : a8 = 1, b2 = a4 , b−1 ab = a−1 = Q16 C(Gi ) = a4 , Gi /a4 = D8 8 2 −1 3 = a, b : a = b = 1, b ab = a D(Gi ) = a2 , Gi /D(Gi ) = C2 × C2 G4 = a, b, z : a4 = z, b2 = z 2 = 1, b−1 ab = az C(G4 ) = a2 , G4 /a2 = C2 × C2 , G4 /a4 = C4 × C2 , D(G4 ) = a4 G5 = a, b, z : a4 = 1, b2 = z, z 2 = 1, b−1 ab = az C(G5 ) = a2 , b2 , G5 /a2 , b2 = C2 × C2 , G5 /a2 = D8 G5 /b2 = C4 × C2 , G5 /a2 b2 = Q8 , D(G5 ) = b2 G6 = a, b, z : a4 = 1, b2 = z 2 = 1, b−1 ab = az, az = za, bz = zb C(G6 ) = a2 , z, G6 /a2 , z = C2 × C2 , G6 /a2 = D8 G6 /z = C4 × C2 , G6 /a2 z = D8 , D(G6 ) = z G7 = a, b, z : a4 = b2 = z 2 = 1, b−1 ab = a−1 , az = za, bz = zb = D8 × C2 C(G7 ) = a2 , z, G7 /a2 , z = C2 × C2 G7 /a2 = C2 × C2 × C2 , G7 /z = D8 , G7 /a2 z = D8 , D(G7 ) = a2 G8 = a, b, z : a4 = z 2 = 1, a2 = b2 , b−1 ab = a−1 , az = za, bz = zb = Q8 × C2 C(G8 ) = a2 , z, G8 /a2 , z = C2 × C2 G8 /a2 = C2 × C2 × C2 , G8 /z = Q8 , G8 /a2 z = Q8 , D(G8 ) = a2 G9 = a, b, z : a2 = b2 = z 4 = 1, b−1 ab = az 2 , az = za, bz = zb C(G9 ) = z, G9 /z = C2 × C2 , G9 /z 2 = C2 × C2 × C2 , D(G9 ) = z 2

Theorem 3 ([LO2] and [Lou3]). There are precisely 13 non-abelian normal CMfields M of degree 16 with relative class number one: five with G(M/Q) = D16 , one with G(M/Q) = G6 and G(M + /Q) = C4 × C2 , one with G(M/Q) = G6 and G(M + /Q) = D8 , one with G(M/Q) = G7 and G(M + /Q) = C2 × C2 × C2 , three with G(M/Q) = G7 and G(M + /Q) = D8 , one with G(M/Q) = G8 and G(M + /Q) = Q8 , and one with G(M/Q) = G9 . Proof. In [LO2] and [Lou3] the fields with relative class number one are known except for the fields which are composita of an abelian octic CM-field and a dihedral octic CM-field. There are four non-abelian CM-fields of degree 16 with class number one which are composita of an abelian octic CM-field and a dihedral octic CM-field with the same maximal totally real subfield ([Lou3]). Using [CK] and [YK] we verify that those are the only such fields with relative class number one. Proposition 10. Let N = M16.1 M16.2 be a compositum of two normal CM -fields + + of degree 16 with the same totally real octic subfield M16.1 = M16.2 . Assume that M16.1 = M8.1 M8.2 is a compositum of two isomorphic octic CM-fields with the + + = M8.2 . If h− same totally real quartic subfield M8.1 N = 1 and wM16.1 = 2, then − − − hM16.1 = 1 or hM16.2 = 1. In particular, if hN = 1 and G(M16.1 /Q) ∈ {G1 , G9 }, − − then h− M16.1 = 1 or hM16.2 = 1, if hN = 1 and G(M16.1 /Q) ∈ {G3 , G4 }, then − h− M16.1 = 2QM16.1 and hM16.2 = 1. Proof. By Theorem 2 (1) we have (J)

h− N =

h− wN − QN M16.1 hM16.2 QM16.2 wM16.2 QM16.1 wM16.1

and h− M16.1

QM16.1 = 2

h− M8.1 QM8.1

2 .

By Theorem 2 (2) it follows that QN wN /(QM16.2 wM16.2 ) is an integer. If h− N =1 and − − − = 1, then h /Q =2. Hence, h /(Q w )=1 and h h− M M M 16.1 16.1 16.1 M16.1 M16.1 M16.1 M16.2 =1. Now, we claim that if G(M16.1 /Q) ∈ {G1 , G3 , G4 , G9 }, then G(M16.1 /Q) has two elements α and β of order 2 such that β = γαγ −1 for some γ ∈ G(M16.1 /Q), α, β


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Table 1. Here k2 is the unique quadratic real subfield of M such that G(M + /k2 ) = C4 . N r. 1 2 3 4 5 6 7 8 9

M √ Hil+ (Q( 514)) √ + Hil (Q( 505)) √ Hil+ (Q( 905)) √ Hil+ (Q( 689)) √ Hil+ (Q( 793)) + √ M ( −1) √ M + ( −1) + √ M ( −3) √ M + ( −(13 + 2 13))

+ M √ √ Q( 257, 17 + 4 2) √ √ Q( 101, 2 5) 11 +√ √ Q( 181, (27 + 5)/2) √ √ Q( 53, (15 + 13)/2) √ √ Q( 61, (19 + 3 13)/2) √ √ Q( 2, 3 + 3) √ √ Q( 11, 15 + 8 3) √ √ Q( 2, (15 + 3 17)/2) √ √ Q( 17, (143 + 31 13)/2)

k √2 Q( 514) √ Q( 505) √ Q( 905) √ Q( 689) √ Q( 793) √ Q( 2) √ Q( 11) √ Q( 34) √ Q( 17)

G(M/Q) D16 D16

QM 2 2

hM 3 1

D16

2

1

D16

2

1

D16

2

1

G7 G7

2 2

1 1

G7

2

1

G6

2

1

contains the complex conjugation τ , and D(G(M16.1 /Q)) contains τ . The fact that τ ∈ D(G(M16.1 /Q)) ensures that wM16.1 = 2. It is sufficient to take α and β as follows (with the notation of Proposition 9):

α β τ

G1 G3 b b a−2 ba2 = a4 b a−2 ba2 = a4 b a4 a4

G4 b a−1 ba = a4 b a4

G9 a b−1 ab = az 2 z2

4.1. In this subsection we assume that G(M8+ /Q) = D8 . Then we have G(M16.1 /Q), G(M16.2 /Q) ∈ {D16 , Q16 , G3 , G5 , G6 , G7 }. In Table 1 we list the nine normal CM+ fields M of degree 16 such that h− M = 1 and G(M /Q) = D8 (see [LO2, Theorem 10] and [Lou3, Theorem 3]). Using [KT] we verify the following which will be used in the proofs of several propositions. √ √ Lemma 7. (1) If k = Q( 2, 3 + 3), 2Ok = I 8 and 3Ok = J 4 , then there are exactly four totally imaginary quadratic extensions K of k such that K is a subfield of the ray class field of k with modulus I 17 J ∞ and K is normal over Q: Hilnar (k) with Galois group G7 , one of conductor I 10 J ∞ with Galois group G , and two of conductor I 14 J ∞ with Galois group G3 . √ 5 √ (2) If k = Q( 11, 15 + 8 3) and 3Ok = I 4 , then the ray class field of k with modulus I∞ contains one and only one quadratic extension K of k which is normal over Q: Hilnar (k) with Galois group G7 . √ √ (3) If k = Q( 2, (15 + 3 17)/2) and 3Ok = J1 J2 , then there is no quadratic extension of with√conductor J1 J2 ∞. √ k√ 2 2 I17 , (4) If k = Q( 2, 5, 17), 2Ok = I22 I22 , 5Ok = I52 I52 and 17Ok = I17 then there is no totally imaginary quadratic extension K of k such that } and K/Q is normal with G(K/Q) = G9 , F(K/k) ∈ {I2e I2e , I5 I5 , I17 I17 where e = √ 2, 4,√or √ 5. 2 2 I37 , (5) If k = Q( 2, 5, 37), 2Ok = I22 I22 , 5Ok = I52 I52 and 37Ok = I37 then there is no totally imaginary quadratic extension K of k such that } and K/Q is normal with G(K/Q) = G9 , F(K/k) ∈ {I2e I2e , I5 I5 , I37 I37 where e = 2, 4, or 5.



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√ (6) If k = Q( 17 + 4 17), 2Ok = p2 p2 and 17Ok = p417 , then there is one and only one (up to isomorphism) quadratic extension K of k such that the conductor of K/k is of the form pe2 p17 ∞ with e = 2 or 3 and the normal closure N of K has G(N/Q) = G4 : F(K/k) = p32 p17 and K = Q(θ) with 4 + 136θ 2 + 68 = 0. θ 8 + 17θ 6 + 85θ √ √ (7) If k = Q( 5, (65 + 65)/2), 5Ok = P45 and 13Ok = P413 , then there is one and only one normal CM-field K containing k with G(K/Q) = G4 and F(K/k) = P5 P13 . √ (8) If k = Q( (5 − 5)/2), 2Ok = p22 and 5Ok = p45 , then there is one and only one (up to isomorphism) quadratic extension K of k such that the conductor of K/k is of the form pe2 p5 ∞ with e ∈ {2, 4, 5} and the normal closure N of K satisfies G(N/Q) = G4 : F(K/k) = p42 p5 and K = Q(θ) 6 4 2 with θ 8 + 10θ +√25θ + 20θ + 5 = 0. (9) If k = Q( 5 + 5), 2Ok = p22 and 5Ok = p45 , then there is no quadratic extension K of k such that the conductor of K/k is of the form pe2 p5 ∞ with e ∈ {2, 4, 5} the normal closure N of K satisfies G(N/Q) = G4 . and √ (10) If k = Q( 2 + 2) and 2Ok = p42 , then there is no quadratic extension K of k such that the conductor of K/k is of the form pe2 (3)g ∞ and the normal closure N of K satisfies G(N/Q) = G4 , where e ∈ {2, 4, 6, 8, 9} and g ∈ {0, 1}. √ (11) If k = Q( 3(2 + 2)), 2Ok = p42 and 3Ok = p23 , then there is one and only one (up to isomorphism) quadratic extension K of k such that the conductor of K/k is of the form pe2 p3 ∞ and the normal closure N of K satisfies G(N/Q) = G4 , where e ∈ {2, 4, 6, 8, 9}: this extension K/k is of conductor p92 p3 ∞. Proposition 11. If any one of G(M16.i /Q) for i = 1, 2 is isomorphic to G2 or G3 , then h− N > 1. Proof. (i) Assume that G(M16.1 /Q) = G2 (= Q16 ). According to [Lou6, Theo− − − rem 3] 2|h− M16.1 and QM16.1 = 1. By (J) if hN = 1, then hM16.1 = 2, hM16.2 = 1, and M16.2 must be one of the nine fields M in Table 1. Hence, we have √ √ √ M16.2 = Q( −1, 2, 3 + 3). Indeed, for the eight other M ’s in Table 1 at least + + four prime ideals are ramified in M + /Q (here M + = M16.1 = M16.2 ), and since + + in M /M16.1 and G(M16.1 /Q) = Q16 , these four prime ideals of M16.1 are ramified 16.1 √ √ − + 23 |hM16.1 . Now, (2) = I 8 and (3) = J 4 in M16.1 = Q( 2, 3 + 3). We need to √ √ verify whether there exists a normal CM-field M16.1 containing Q( 2, 3 + 3) √ √ √ such that G(M16.1 /Q) = Q16 and h− 3 + 3)/Q( 2) is M16.1 = 2. Since Q( 2, + ) = I e J for some 2 ≤ e ≤ 17. (See [He, Theorem 119] and cyclic, F(M16.1 /M16.1 [Co, Theorem 10.2.9].) According to Lemma 7, point (1), there is no such field M16.1 . We conclude that if any one of G(M16.i /Q) for i = 1, 2 is isomorphic to G2 , then h− N > 1. − (ii) Assume that G(M16.1 /Q) = G3 . By Proposition 10 if h− N = 1, then hM16.1 = 2QM16.1 , h− M16.2 = 1 and M16.2 must be one of the nine fields in Table 1. For a given octic dihedral field M + in Table 1 we will construct M16.1 containing M + such that G(M16.1 /Q) = G3 . Note that the subgroup a2 , ab = Q8 and a2 , b = D8 .


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√ √ √ 2 (ii-a) Suppose that M √ √16.1 ⊃ Q( 257, 17 + 4 2). Then a , ab = Q( 257) or Q( 2). If G(M16.1 /Q( 2)) = Q8 , then the four prime lying above 257 are ideals √ √ ramified in the quadratic extension M16.1 /Q( 257, 17 + 4 2), whence 23 |h− M16.1 . √ If G(M16.1 /Q( 257)) = Q8 , then the four prime ideals lying above 2 are ramified √ √ in the quadratic extension M16.1 /Q( 257, 17 + 4 2), whence 23 |hM − . By the 16.1 same argument we verify that if there exists a normal CM-field M16.1 such that + is one of eight fields M + in Table 1 except for the G(M16.1 /Q) = G3 and M16.1 − field numbered 6, then 4|hM16.1 . (ii-b) We take the field number 6 in Table 1 as M16.2 and assume that G(M16.1 /Q) √ √ + = G3 and M16.1 = Q( 2, 3 + 3). Since h− M16.1 = 2QM16.1 , at most two prime + + and F(M16.1 /M16.1 ) = I e J for some 2 ≤ e ≤ 17, ideals are ramified in M16.1 /M16.1 where I and J are as in (i). From Lemma 7, point (1), there are two such fields √ √ √ √ F(M16.1 /Q( 2, 3 + 3)) = F(M16.1 /Q( 2, √ 3 + 3)) = M16.1 and M16.1 : √ √ I 14 J , G(M16.1 /Q( √ 3)) = D8 , G(M16.1 /Q(− 6)) = Q8−, G(M16.1 /Q( 6)) = D8 , and G(M16.1 /Q( 3)) = Q8 . To determine hM16.1 and hM we compute the rela16.1 tive class numbers of their non-normal octic CM-subfields using the technique de by √ + veloped in [Lou7]: the quadratic extension M8.1 of Q( 3 + 3)(= ) with con M8.1 √ − + 9 ) ductor p3 p2 ∞ for hM16.1 and the quadratic extension M8.1 of Q( 3 + 6)(= M8.1 − 8 with conductor p3 p2 ∞ for hM , where p3 , p3 are the unique prime ideal lying 16.1 + + and M8.1 respectively, and p2 , p2 are those above 2. We obtain above 3 in M8.1 − = 6. By point (2) of Theorem 2 it follows that neither h− h− M8.1 = hM8.1 M16.1 nor − − hM divides 4, and hN > 1. 16.1

In the following Propositions 12-15 we prove that if any one of G(M16.i /Q) for i = 1, 2 is isomorphic to D16 , then h− N ≥ 2. Lemma 8. Let N = M16.1 M16.2 be a compositum of two dihedral CM-fields M16.1 + + − and M16.2 of degree 16 with M16.1 = M16.2 . Assume that h− M16.1 = 1. Then hN = 1 − if and only if hM16.2 = 2. − Proof. Since there is no pair of (M16.1 , M16.2 ) such that h− M16.1 = hM16.2 = 1 and + + = M16.2 (see [Lef] and [Lou6]), it is sufficient to show that QM16.2 must be M16.1 − equal to 1 in order to prove that h− N = 1 implies hM16.2 = 2. According to [LO2, Theorem10] M16.1 must be one of the fields numbered from 1 to 5 in Table 1. For these five fields we have QM16.1 = 2, hM + is odd, and h+ = 2hM + ≡ 2 mod + M16.1 16.1 16.1 √ + + + 4, where hM + is the narrow class number of M16.1 . Hence M16.1 = M16.1 ( −), 16.1

+ . There is only one such unit up where = 1 is a totally positive unit of M16.1 = 2. By Theorem 2 (4) we would have M16.2 = to square. Suppose that Q M 16.2 √ + M16.2 ( −) = M16.1 . This is a contradiction. Now, assume that h− M16.2 = 2. It = 1. follows that wN = QN = 2 and h− N

Proposition 12. If N = M16.1 M16.2 is a compositum of two dihedral CM-fields of + + = M16.2 , then h− degree 16 with M16.1 N ≥ 2. Proof. By Proposition 10 and Lemma 8 it is sufficient to show that there is no pair − of (M16.1 , M16.2 ) such that h− M16.1 = 1 and hM16.2 = 2. To do this we will prove + that there is no dihedral CM-field M16 of degree 16 with h− M16 = 2 such that M16 is



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equal to any one of the five octic subfields M + of M numbered from 1 to 5 in Table 1. We proceed as follows. First, we fix the real octic field M + and find a constant + + C0 such that if h− M16 ≤ 2 and M16 = M , then dM16 ≤ C0 . Second, we make a list + of dihedral CM-fields M16 such that M16 = M + , dM16 ≤ C0 , and exactly two prime + . Third, we compute the relative class numbers of ideals are ramified in M16 /M16 these dihedral fields M16 . + and k2 as in Table 1. We (i) Let M4 be the only quartic normal subfield of M16 have 3 (ζM16 /ζM4 )(s) = |L(s, χi , M16 /k2 )|2 ≥ 0, i=1

where χ is any one of four octic characters of degree one associated with the cyclic extension M16 /k2 . For all five subfields M4 in M + in Table 1 we verify that ζM4 (s) ≤ 0 for s ∈ (0, 1), whence ζM16 (s) ≤ 0 for s ∈ (0, 1). Now, we apply Theorem 2, points (7) and (8), and get κM16 ≥ 2/(elogdM16 )

(K) and

κM + ≤ κk2 (µk2 κk2 )3 ≤

(L)

16

log7 dk2 . 210

+ . Using (K) and (L) Let D be the relative discriminant of the extension M16 /M16 + we obtain in Table 2 an upper bound C such that NM /Q (D)1/8 ≤ C if h− M16 ≤ 2. 16 + + = 2 implies that N (F(M /M )) has only one (ii) The fact that h− 16 16 M16 M /Q 16

+ prime divisor p, and this p is not ramified in M16 : pOM + = pp , pOM16 = P2 P2 , 16 + . Indeed, the where P and P are the unique ramified prime ideals in M16 /M16 + primes ramified in k2 split completely in M16 = Hil(k2 ). Hence, if h− M16 = 2, + . Let L be any then the prime divisors of dM + cannot be ramified in M16 /M16 16 + + such that G(M16 /L) = C2 × C2 . Then one of two quadratic subfields of M16 G(M16 /L) = D8 . Let K4 be any one of the two non-normal quartic subfields + of M16 containing L and let K8 be any one of the two non-normal octic CMsubfields of M16 containing K4 . (Hence K4 = K8+ .) Let us determine F(K8 /K4 ). In a normal extension K/k for a prime ideal Q of K we denote by G−1 (Q) and G0 (Q) the decomposition group and the inertia group of Q, respectively. Since + G−1 (p) = G−1 (p ) = G(M16 /k2 ) and G0 (p) = G0 (p ) = {id}, pOK4 remains prime. + = G0 (P ) = G(M16 /M16 ), Since G−1 (P) = G−1 (P ) = G(M16 /k2 ) and G0 (P) √ pOK4 is ramified in√K8 /K4 . Let us write k2 = Q( rs) with r, s two distinct primes and L = Q( r). In K4 , sOK4 = a2 bb with N (a) = N (b) = N (b ) = s. Exchanging b and b if necessary we may assume that b is ramified in K8 /K4 .

Since h− M16 =

QM16 2 h− M16

h−

− ( QKK8 )2 , if h− M16 = 2, then hK8 /QK8 = 2 and QM16 = 1. It 8

follows that if = 2, then the unique ramified prime ideals in K8 /K4 are (p) √ √ and b, and QK8 = 1. Note that p is inert in Q( r) and in Q( s). Hence, if 1/8 + = p ≤ NM + /Q (D)1/8 ≤ C and ( pr ) = ( ps ) = −1. h− M16 = 2, then NM16 /Q (pp ) 16 According to Hecke’s Theorem F(K8 /K4 ) = (p)b for p = 2, F(K8 /K4 ) = (2)e b with e = 2 or 3 (indeed, we will see below e = 3) in the case that p = 2 and + M16 is the real subfield M + of the fields numbered from 2 to 5 in Table 1. (See Theorem 10.2.9 in [Co] and Theorem 119 in [He].) We may assume that s is odd.


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Table 2. M+ √ √ Q( 257, 17 + 4 2) √ √ Q( 101, 11 + 2 5) √ √ Q( 181, (27 + 5)/2) √ √ Q( 53, (15 + 13)/2)

K = K8+ 4 √ Q(17 + 4 2) √ Q( 2 5) 11 +√ Q( (27 + 5)/2)

C

p

25 25

5 2

22

2 17

23

2 5

Q(

2

Q(

√ √ Q( 61, (19 + 3 13)/2) 23

√ (15 + 13)/2)

√ (19 + 3 13)/2)

F(K8 /K4 ) h− K8 (5)b (2)3 b

274 34

(2)3 b (17)b

146 658

(2)3 b (5)b

82 34

(2)3 b

50

+ In the case that M16 is the real subfield of the first field M in Table 1 we take √ √ L = Q( 2), K4 = Q( 17 + 4 2) and s=257. Furthermore, we can get rid of p with p ≡ 3 mod 4. In fact, the narrow class number h+ 4 of K4 is odd. There √ h+ 4 = (α) and K4 ( −α) is exists a totally positive element α in K4 such that b contained √ in Hilnar (k2 ). Since s is odd and b is the unique prime ideal ramified in K4 ( −α)/K4 , the equation x2 ≡ −α mod p22 is solvable for each prime p2 lying √ above 2. We can write K8 = K4 ( −αp). Hence, if p = 2, then F(K8 /K4 ) = (2)3 b. If p = 2, then the prime ideal(s) lying above 2 is(are) not ramified in K8 /K4 and / 3 mod 4. Suppose that p ≡ 3 mod x2 ≡ −αp mod p22 is solvable. It follows that p ≡ 4. Then x2 ≡ −1 mod p22 would be solvable. This implies that the prime ideal(s) √ lying above 2 is(are) not ramified in K4 ( −1)/K4 . This contradicts the fact that h+ 4 is odd. (iii) For these p ≡/ 3 mod 4 such that p ≤ C and ( pr ) = ( ps ) = −1 we have √ computed the relative class numbers of K8 = K4 ( −αp) by using the technique developed in [Lou7]. In all there are nine such fields K8 and we have verified that − − h− K8 > 2, which permits us to conclude hM16 > 2 and hN ≥ 2. We give our computational results in Table 2.

Proposition 13. If G(M16.1 /Q) = D16 and G(M16.2 /Q) = G5 , then h− N ≥ 2. Proof. Since h− M16.2 ≥ 2, it follows from Proposition 10 and Theorem 3 that if there = 1, then h− exists a field N = M16.1 M16.2 with h− M16.1 = 1. Moreover, the unique √ N √ + + = M16.2 is contained in the octic quartic normal subfield Q( m1 , m2 ) of M16.1 √ √ quaternion subfield M8.2 of M16.2 . Since the subfields Q( m1 , m2 ) of the fields m1 m2 in Table 1 satisfy ( m2 ) = ( m1 ) = +1, there are at least four ramified prime ideals √ √ − in M8.2 /Q( m1 , m2 ), whence 23 |h− M8.2 and hN > 1. This is a contradiction. Proposition 14. If G(M16.1 /Q) = D16 and G(M16.2 /Q) = G6 , then h− N ≥ 2. Proof. Let M16.2 be a compositum of two normal octic CM-fields M8.1 and M8.2 + + such that M8.1 = M8.2 , G(M8.1 /Q) = D8 , and G(M8.2 /Q) = C4∗ × C2 . According to Proposition 10 if there exists a field N = M16.1 M16.2 of h− N = 1, then − = 1 or h = 1. First, we verify that there is no pair of fields either h− M16.1 M16.2



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− (M16.1 , M16.2 ) such that h− 1, 2, or 4. Indeed, we verM16.1 = 1 and hM16.2 = √ √ ify that there is no real bicyclic biquadratic field Q( m1 , m2 ) which is a sub− field of a dihedral CM-field M16.1 with hM16.1 = 1 and at the same time is contained in the intersection of an octic dihedral CM-field M8.1 with h− M8.1 = 1, 2, or 4 and an imaginary abelian octic field M8.2 with G(M8.2 /Q) = C4∗ × C2 and h− M8.2 = 1, 2, or 4 ([YK] and [CK]). Second, we will show that there is no pair of 1, 2, or 4 and h− = 1. The unique fields (M16.1 , M16.2 ) such that h− M16.1 = M 16.2 √ √ √ field M16.2 with h− −(9 + 13)/2, −(13 + 2 13)). Set M16.2 = 1 is Q( 17, √ √ √ + M8+ = M16.2 = Q( 17, (13 + 2 13)(9 + 13)/2). Suppose that there exists 4 4 a field M16.1 containing M8+ with G(M16.1 √ /Q) = D16 . We have (13) = I1 I2 + + 2 2 and (17) = J1 J2 in M8 . Since M8 /Q( 17) is cyclic, I1 and I2 are ramified in + M16.1 /M8+ . We claim that J1 and J2 are ramified in M16.1 /M√ 8 . Suppose that J1 and J2 remain prime in M16.1 . Then G−1 (J1 ) = G(M16.1 /Q( 13)) = D8 , G0 (J1 ) is of order 2, and G−1 (J1 )/G0 (J1 ) is cyclic. In D8 there is no normal subgroup of order 2 such that its factor group is cyclic 1 and

of order 4. Suppose now that J√ J2 split in M16.1 /M8+ . Write(17) = 4i=1 Pi 2 in M16.1 , (17) = pp in Q( 13), √ √ 2 and (17) = PP in L4 = Q( (13 + 2 13)(9 + 13)/2). Assume that p|P1 . The √ decomposition field of P1 in the extension M16.1 /Q( 13) must be one of the three √ √ √ √ fields L4 , Q( 13, 17) or L4 = Q( (13 − 2 13)(9 − 13)/2). Note that p is in√ √ √ √ √ ert in L4 /Q( 13) and ramified in Q( 13, 14)/Q( 13) and L4 /Q( 13). This is a contradiction. Consequently, if there is a CM-field M16.1 containing M8+ with G(M16.1 /Q) = D16 , then 23 |h− M16.1 . This completes the proof.

Proposition 15. If G(M16.1 /Q) = D16 and G(M16.2 /Q) = G7 , then h− N ≥ 2. Proof. Let M16.2 = M8.1 M8.2 be a compositum of two normal octic CM-fields M8.1 + + and M8.2 such that M8.1 = M8.2 , G(M8.1 /Q) = D8 , and G(M8.2 /Q) = C2∗ ×C2 ×C2 . − − According to Proposition 10 if h− N = 1, then hM16.1 = 1 or hM16.2 = 1. − = 1, then the biCase (i) : Assume that h− M16.1 = 1. According to [LO2] if hM √16.1 √ + quadratic bicyclic subfield K of M16.1 satisfies K ∈ A := {Q( m1 , m2 ); (m1 , m2 ) ∈ {(2, 257), (5, (13, 53), (13, 61)}}. Using [YK] we verify that M8.1 = 101), (5, 181), √ √ √ Q( 5, 101, −(33 + 6 5)), of which h− M8.1 = 2, is the only dihedral octic CM+ − field M8.1 such that M8.1 ∈ A and hM8.1 divides 4. The compositum of M8.1 and √ √ √ √ √ Q( 11 + 2 5, 101) contains the subfield Q( 5, 101, −3) whose relative class − number is equal to 30. Hence, we have shown that if h− M16.1 = 1, then hN ≥ 2. − Case (ii): Assume that hM16.2 = 1. According to [Lou3] there are three such fields M16.2 : the fields numbered 6, 7, and 8 in Table 1. For these three fields + = M16.2 we examine whether there exists a normal CM-field M16.1 such that M16.1 + M16.2 , G(M16.1 /Q) = D16 , and at most two prime ideals are ramified in the exten+ . sion M16.1 /M16.1 (ii-a) Assume that M16.2 is the field numbered 6 in Table 1. In the proof of Proposition 11, part (i), and Lemma 7, point (1), we verified that there is no √ √ + dihedral CM-field M16.1 of degree 16 containing M16.2 = L8 := Q( 2, 3 + 3) such that the only ramified prime ideals in M16.1 /L8 lie above 2 or 3. Hence if


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there is a dihedral CM-field M16.1 of degree 16 containing L8 , then at least four prime ideals are ramified in the extension M16.1 /L8 , whence 8 | h− M16.1 . Because if a prime p with p = 2, 3 divides dM16.1 , then there are at least two prime ideals lying above p which are ramified in M16.1 /L8 . there is a dihedral (ii-b) Assume that M16.2 is the field numbered 7 in Table 1. If √ √ + CM-field M16.1 of degree 16 containing M16.2 = L8 := Q( 11, 15 + 8 √3), then the prime ideal lying above 3 is ramified in the cyclic extension M16.1 /Q( 11) and − QM16.1 = 1. In this case h− N = 1 implies that hM16.1 = 2, F(M16.1 /L8 ) = I, where I is the unique prime ideal lying above 3 in L8 . According to Lemma 7, point (2), there is no dihedral CM-field M16.1 of degree 16 containing L8 with F(M16.1 /L8 ) = I. (ii-c) Assume that M16.2 is the field numbered 8 in Table 1. Suppose that there is a dihedral CM-field M16.1 of degree 16 containing √ √ + M16.2 = L8 = Q( 2, (15 + 3 17)/2). Since two prime ideals J1 and J2 lying above 3 are ramified in√ the √ √ √ extension L8 /Q( 2, 17), they are totally ramified in the extension M16.1 /Q( 2, 17), which − implies QM16.1 = 1. Hence, if h− N = 1, then hM16.1 = 2 and F(M16.1 /L8 ) = J1 J2 . According to Lemma 7, point (3), there is no normal CM-field M16.1 containing L8 with F(M16.1 /L8 ) = J1 J2 . The proof is complete. In Propositions 16-18 below we deal with the composita N = M16.1 M16.2 such that G(M16.i /Q) ∈ {G5 , G6 , G7 } for i = 1, 2 . Lemma 9. Let L be an octic quaternion CM-field. (1) ([Lou2, Corollary 10]) h√− L ≡ 2 mod 4 if and only if L is a pure quaternion √ + CM-field with L = Q( 2, q), where q ≡ 3 mod 8 is prime. + (2) If h√− L ≡ 4√ mod 8, then L is a pure quaternion extension of Q and L = Q( 2p1 , p2 ), where p1 ≡ 3 mod 4 and p2 ≡ 3 mod 8 are prime. Proof. (2) Assume that h− L ≡ 4 mod 8. Counting the number of ramified prime ideals in L it can be easily shown that L is a pure quaternion extension of Q. First we prove that the number of prime divisors of dL is equal to 3 and for each prime divisor p of dL there is only one prime ideal lying above p in L. Suppose the√number √ that √ √ of prime divisors of dL is equal to 2. Then L+ is of the form Q( 2, p), Q( 2, q), √ √ or Q( r, s), where p, q, r, and s are odd primes with q ≡ 3 mod √ 8, p ≡ 1 mod √ 8, r, s ≡ 1 mod 4, and ( rs ) = +1 (see [K, 2.2 Proposition]). If L+ = Q( 2, q), then √ √ √ √ − + + ≡ 0 mod 8. h− L ≡ 2 mod 4. If either L = Q( 2, p) or L = Q( r, s), then hL √ √ + Second, we claim that 2 must be ramified. Otherwise we have L = Q( p1 , p2 p3 ), where p1 , p2 , and p3 are odd primes with p1 ≡ 1 mod 4 and p2 p3 ≡ 1 mod 4. Since √ √ p2 p3 Q( p1 , p2 p3 ) is embedded in an octic quaternion field, ( −1 p1 ) = ( p1 ) = +1. It √ follows that p1 splits in Q( p2 p3 ), two prime ideals lying above p1 in L+ are ramified √ √ + in L/L+ , whence h− L ≡ 0 mod 8. Write L = Q( m1 , m2 ), where m1 and m2 are square free positive integers. Third, we verify that 2 is totally ramified. Suppose √ the contrary. Say 2 is not ramified in Q( m2 ). Then either (m1 , m2 ) = (2, p1 p2 ) or (m1 , m2 ) = (2p, q), where p1 , p2 , p, and q are odd primes satisfying p1 p2 ≡ 1 2p q 2 −1 2 −1 −1 mod 4, ( −1 p1 ) = ( p1 ), ( p2 ) = ( p2 ), q ≡ 1 mod 4, ( q ) = ( q ), and ( p ) = ( p ). √ √ Note that Q( p, q) with p ≡ 3 mod 4 and q ≡ 1 mod 4 cannot be embedded



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√ √ 2 in an octic quaternion field. If L+ = Q( 2, p1 p2 ) and if ( −1 p1 ) = ( p1 ) = 1 or √ √ − 2 −1 2 + ( −1 p2 ) = ( p2 ) = 1, then hL ≡ 0 mod 8. If L = Q( 2, p1 p2 ), ( p1 ) = ( p1 ) = −1, √ −1 2 and ( p2 ) = ( p2 ) = −1, then p1 ≡ p2 ≡ 3 mod 8, 2 splits in Q( p1 p2 ), and h− L ≡ 0 √ √ 2p − −1 + mod 8. If L = Q( 2p, q) with q ≡ 1 mod 4, then ( q ) = ( q ) = 1 and hL ≡ 0 mod 8. Finally, the fact that 2 is totally ramified implies that (m1 , m2 ) = (2p, q) q 2p −1 with q ≡ 3 mod 4, ( −1 p ) = ( p ) = −1, and ( q ) = ( q ) = −1. Thus p ≡ 3 mod 4 and q ≡ 3 mod 8. Proposition 16. (1) There is one and only one CM-field M16 of degree 16 + with G(M16 /Q) = G5 such that G(M16 /Q) and every CM-subfield L = D8 √ √ − of M16 has hL | 4 : M16 = Q( 3 + 3, −(2 + 2)) with h− M16 = 2 and QM16 = 2. + + + (2) If G(M16.1 /Q) = G(M16.2 /Q) = G5 , and M16.1 = M16.2 with G(M16.1 /Q) − = D8 , then hN ≥ 2. + + (3) If G(M16.1 /Q) = G5 , G(M16.2 /Q) = G6 , and M16.1 = M16.2 with + − G(M16.1 /Q) = D8 , then hN ≥ 2. + + (4) If G(M16.1 /Q) = G5 , G(M16.2 /Q) = G7 , and M16.1 = M16.2 with + − G(M16.1 /Q) = D8 , then there is exactly one field N with hN = 1 : √ √ √ Q( (2 + 2), 3 + 3, −1). Moreover this field has class number one. Proof. (1) The field M16 = M8.1 M8.2 is a compositum of two octic CM-fields with − + + G(M8.1 /Q) = C4∗ ×C2 and h− |4, and M8.1 = M8.2 . M8.1 |4, G(M8.2 /Q) = Q8 and hM8.2 √ √ Now, out of the 51 such M8.1 ’s (see [CK]), M8.1 = Q( 3, −(2 + 2)) is the + can be embedded in an octic quaternion field M8.2 with only one such that M8.1 √ √ √ √ − hM8.2 | 4 (use Lemma 9), and we have M8.2 = Q( 2, 3, −(2 + 2)(3 + 2)). − Since h− M8.1 = QM8.1 = 2, hM8.2 = 2, QM8.2 = 1, wM8.2 = 2, wM16 = wM8.1 , and QM16 = 2 (for QM8.1 = 2 divides QM16 ), we obtain h− M16 = 2 by (J). (2) follows immediately from (1). √ √ (3) and (4) Set M16.1 = Q( −(2 + 2), 3 + 3). The field M16.2 = L8.1 L8.2 is a compositum of two octic CM-fields L8.1 and L8.2 such that G(L8.1 /Q) = C4∗ × C2 if G(M16.2 /Q) = G6 , G(L8.1 /Q) = C2∗ × C2 × C2 if G(M16.2 /Q) = G7 , and G(L8.2 /Q) = D8 . We look for all possible N = M16.1 M16.2 with h− By [LO1] and [YK] there is only one octic dihedral CM-field L8.2 N = 1. √ √ √ √ + − such that L8.2 = Q( 2, 3) and hL8.2 | 4 : L8.2 = Q( 2, −(3 + 3)) with h− L8.2 = 2 and QL8.2 = 1. For (3) we verify that there is only one imaginary √ √ abelian number field L8.1 suchthat G(L8.1 /Q) = C4∗ × C2 , L+ 8.1 = Q( 2, 3), √ √ −(2 + 2)) ([CK]). However, the compositum and h− L8.1 | 4 : L8.1 = Q( 3, √ √ Q( −(2 + 2), −(3 + 3)) is not appropriate for us since its maximal real sub√ √ √ √ field is the octic quaternion field Q( 2, 3, (2 + 2)(3 + 3)). Consequently, there is no field N with h− N = 1. For (4) we find all imaginary √ √ abelian number = Q( 2, 3), and h− fields L8.1 such that G(L8.1 /Q) = C2∗ × C2 × C2 , L+ 8.1 L8.1 | 4.


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√ √ √ Such fields L8.1 can be written of the form L8.1 = Q( 2, 3, −m), where m = 1 or m is a prime with hQ(√−m) | 4. Otherwise, there are at least four ramified prime h−8.1 ≡ 0 mod 8. We verify that there is precisely one ideals in L8.1 /L+ 8.1 and √ L√ √ such field : L8.1 = Q( 2, 3, −1) with h− 1 and QL8.1 = 2. Set M16.2 = L8.1 = √ √ √ √ √ √ Q( 2, 3 + 3, −1) and N = M16.1 M16.2 = Q( 2 + 2, 3 + 3, −1). Since √ √ √ h− 2), −1, −3), M16.2 = 1, QM16.2 = 2, wN = wL = 48 with L = Q( −(2 + √ √ wM16.1 = wF = 2 with F = Q( 3, −(2 + 2)), and wM16.2 = wD = 24 with √ √ √ D = Q( −1, −2, −3), we have h− N = 1. (For wL , wF , and wD√seeTable√II in [He].) By using [KT] we verify that the class number of Q( 2 + 2, 3 + 3) is equal to one. Remark 2. Assume that M16.1 = M8.1 M8.2 and M16.2 = L8.1 L8.2 are two composita + + + + of two normal octic CM-field with M16.1 = M16.2 and M8.1 = M8.2 = L+ 8.1 = + L8.2 . Let N = M16.1 M16.2 , R16.1 = M8.1 L8.1 , R16.2 = M8.2 L8.2 , T16.1 = M8.1 L8.2 , + + + + = R16.1 , M16.1 = T16.1 , and T16.2 = L8.1 M8.2 . We can easily verify that M16.1 + + + + + and R16.1 = T16.1 . Since G(M16.1 /Q), G(R16.1 /Q), and G(T16.1 /Q) are the three factor groups of G(N + /Q), we can determine G(N + /Q) from these factor groups. + + + + Moreover, R16.1 = R16.2 and T16.1 = T16.2 . (a) In Proposition 16 (2) we have G(R16.1 /Q) = C4∗ × C2 × C2 , G(R16.2 /Q) = + G8 , G(T16.1 /Q) = G(T16.2 /Q) = G5 , G(T16.1 /Q) = D8 , and G(N + /Q) = G7 . (b) In Proposition 16 (3) we have G(R16.1 /Q) = C4∗ × C4∗ , G(R16.2 /Q) = G5 , + G(T16.1 /Q) = G6 , G(T16.2 /Q) = G5 , G(T16.1 /Q) = D8 , and G(N + /Q) = G6 (cf. Proposition 27 (3) below). (c) In Proposition 16 (4) we have G(R16.1 /Q) = C4∗ × C2∗ × C2 , G(R16.2 /Q) = + G5 , G(T16.1 /Q) = G5 , G(T16.2 /Q) = G8 , G(T16.1 /Q) = Q8 , and G(N + /Q) = G5 (cf. Propositions 19 (2) and 19 (4) below). The following remark will be needed in the sequel: Remark 3. (a) There are eleven pairs of octic dihedral CM-fields (L, M ) such √ √ − + that h− = M + = Q( m1 , m2 ): (m1 , m2 ) ∈ S = L | 4, hM | 4, and L {(2, 7), (2, 17), (5, 11), (5, 29), (3, 13), (21, 85), (10, 65)} ([LO1] and [YK]). (b) Assume that (m1 , m2 ) ∈ S. Let l be a positive integer and √ √ √ F = Q( m1 , m2 , −l). If h− F | 4 and all imaginary subfields of F have relative class numbers dividing 4, then (m1 , m2 , −l) ∈ T = { (2, 7, −1), (2, 7, −3), (2, 17, −1), (2, 17, −3), (5, 11, −1), (3, 13, −1)}. + Proposition 17. (1) If G(M16.1 /Q) = G(M16.2 /Q) = G6 and G(M16.1 /Q) = − D8 , then hN ≥ 2. + /Q) = D8 , then (2) If G(M16.1 /Q) = G6 , G(M16.2 /Q) = G7 , and G(M16.1 − hN ≥ 2.

Proof. Let M8.1 , M8.2 be two octic normal CM-fields such that M16.1 = M8.1 M8.2 , + + = M8.2 , G(M8.1 /Q) = C4∗ × C2 , and G(M8.2 /Q) = D8 . Let L8.1 , L8.2 be two M8.1



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+ octic normal CM-fields such that M16.2 = L8.1 L8.2 , L+ 8.1 = L8.2 , G(L8.2 /Q) = D8 , ∗ ∗ G(L8.1 /Q) = C4 × C2 or C2 × C2 × C2 , according to G(M16.2 /Q) = G6 or G7 . Then √ √ + + + M8.1 = M8.2 = L+ 8.1 = L8.2 = Q( m1 , m2 ) is the unique quartic normal subfield √ √ + of M16.1 . If h− N = 1, then Q( m1 , m2 ) is contained in two distinct octic dihedral − hence (m1 , m2 ) ∈ S. CM-fields M8.2 and L8.2 with hM8.2 | 4 and h− L8.2 | 4, √ √ (1) Assume that G(M16.2 /Q) = G6 . Then Q( m1 , m2 ) is contained in two abelian CM-fields M8.1 and L8.1 of Galois group C4∗ × C2 , and the compositum M8.1 L8.1 is an imaginary abelian number field with Galois group C4∗ × C4∗ or C4∗ × C2 × C2 . According to [CK] there is only one pair (m1 , m2 ) such that √ √ + + h− = Q( m1 , m2 ) with (m1 , m2 ) ∈ S : (m1 , m2 ) = M8.1 L8.1 | 4 and M8.1 = L8.1 √ √ √ √ (5, 29), M8.1 , L8.1 ∈ {Q( 29, −(5 + 2 5)), Q( 5, −(29 + 2 29))}, and

M8.2 , L8.2

√ √ ∈ {Q( 29, −(25 + 3 5)/2), √ √ √ √ Q( 29, −(7 + 2 5)), Q( 29, −(21 + 6 5))}.

We verify that there is no quadruple of fields (M8.1 , M8.2 , L8.1 , L8.2 ) such that the composita M8.1 M8.2 and L8.1 L8.2 have the same maximal real subfield. (2) Assume that G(M16.2 /Q) = G7 . The compositum M8.1 L8.1 is an imaginary ∗ ∗ abelian number field with √ Galois group C4 × C2 × C2 . By Remark 3 (b) we have √ √ L8.1 = Q( m1 , m2 , −l) with (m1 , m2 , −l) ∈ T. We verify that there are two quadruples of fields (M8.1 , M8.2 , L8.1 , L8.2 ) such that − − − (a) h− M8.1 , hM8.2 , hL8.1 , hL8.2 ∈ {1, 2, 4}, + + + = M8.2 = L+ (b) M8.1 8.1 = L8.2 , and + + + (c) (M8.1 M8.2 ) = (L = D8 : 8.1 L8.2 ) , G((M8.1 M8.2 ) /Q) √ √ √ √ M8.1 = Q( 7, −(2 + 2)), M8.2 = Q( 7, −(3 + 2)), √ √ √ √ √ L8.1 = Q( 2, 7, −1), L8.2 = Q( 14, −(4 + 2)); √ √ √ √ M8.1 = Q( 7, −(2 + 2)), M8.2 = Q( 14, −(4 + 2)), √ √ √ √ √ L8.1 = Q( 2, 7, −1), L8.2 = Q( 7, −(3 + 2)). − − For these two cases we verify that h− M16.1 = 4, hM16.2 = 2, and hN = 2QN ≥ 2.

Remark 4. We keep the notation in the proof of Proposition 17. Let R16.1 = M8.1 L8.1 , R16.2 = M8.2 L8.2 , T16.1 = M8.1 L8.2 , and T16.2 = L8.1 M8.2 . (a) In Proposition 17 (1) there are three possibilities : (i) If G(N + /Q) = G7 , then G(R16.1 /Q) = C4∗ × C2 × C2 , G(R16.2 /Q) = + G7 , G(T16.1 /Q) = G(T16.2 /Q) = G6 , and G(T16.1 /Q) = D8 (cf. Proposition 20 (3) below). (ii) If G(N + /Q) = G5 , then G(R16.1 /Q) = C4∗ × C4∗ , G(R16.2 /Q) = G6 , + G(T16.1 /Q) = G(T16.2 /Q) = G5 , and G(T16.1 /Q) = Q8 (cf. Propositions 19 (1) and 20 (2) below). (iii) If G(N + /Q) = G6 , then G(R16.1 /Q) = C4∗ × C4∗ , G(R16.2 /Q) = G6 , + G(T16.1 /Q) = G(T16.2 /Q) = G6 , and G(T16.1 /Q) = D8 (cf. Proposition 28 (2) below).


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(b) In Proposition 17 (2) we have G(R16.1 /Q) = C4∗ × C2∗ × C2 , G(R16.2 /Q) = + G6 , G(T16.1 /Q) = G6 , G(T16.2 /Q) = G7 , G(T16.1 /Q) = D8 , and G(N + /Q) = G6 (cf. Proposition 28 (2) below). We close this subsection with the following proposition. Proposition 18. If G(M16.1 /Q) = G(M16.2 /Q) = G7 , then h− N ≥ 2. Proof. Let M16.1 = M8.1 M8.2 and M16.2 = L8.1 L8.2 such that G(M8.1 /Q) = + G(L8.1 /Q) = C2∗ × C2 × C2 , and G(M8.2 /Q) = G(L8.2 /Q) = D8 . Set M8.1 = √ √ √ √ √ + + + M8.2 = L8.1 =√L8.2 = Q( m1 , m2 ), M8.1 = Q( m1 , m2 , −m0 ), and L8.1 = √ √ Q( m1 , m2 , −l0 ). Suppose that h− N = 1. Then (m1 , m2 ) ∈ S, (m1 , m2 , −m0 ) ∈ T, and (m1 , m2 , −l0 ) ∈ T , where S and T are as in Remark 3. Hence, (m1 , m2 , √ √ −m0 , −l0 ) ∈ {(2, 7, −1, −3), (2, 17, −1, −3)}. Since h− = 8, we exclude Q( 6, −17) (m1 , m2 , −m0 , −l0 ) = (2, 17, −1, −3). There are two possible choices for quadruples (M8.1 , M8.2 , L8.1 , L8.2 ): √ √ √ √ √ (a) M8.1 = Q( 2, 7, −1), M8.2 = Q( 7, −(3 + 2)), √ √ √ √ √ L8.1 = Q( 2, 7, −3), and L8.2 = Q( 14, −(4 + 2)); √ √ √ √ √ (b) M8.1 = Q( 2, 7, −3), M8.2 = Q( 7, −(3 + 2)), √ √ √ √ √ L8.1 = Q( 2, 7, −1), and L8.2 = Q( 14, −(4 + 2)). √ √ √ √ + However, in case (a) we have M16.1 = Q( 3 + 2, 7) = Q( 3(4 + 2), 14) = √ √ √ √ + + + M16.2 , and in case (b) M16.1 = Q( 3(3 + 2), 7) = Q( (4 + 2), 14) = M16.2 . − Consequently, hN ≥ 2. Remark 5. We keep the notation in the proof of Propostion 18. Let R16.1 = M8.1 L8.1 , R16.2 = M8.2 L8.2 , T16.1 = M8.1 L8.2 , and T16.2 = L8.1 M8.2 . In Proposition 18 we have G(R16.1 /Q) = C2∗ × C2 × C2 × C2 , G(R16.2 /Q) = G7 , G(T16.1 /Q) = + G(T16.2 /Q) = G7 , G(T16.1 /Q) = D8 , and G(N + /Q) = G7 (cf. Proposition 20 (4) below). 4.2. In this subsection we assume that G(M8+ /Q) = Q8 . Then G(M16.1 /Q), G(M16.2 /Q) ∈ {G5 , G8 }. Proposition 19. (1) If G(M16.1 /Q) = G(M16.2 /Q) = G5 , then h− N ≥ 2. there is only one com(2) If G(M16.1 /Q) = G5 and G(M16.2 /Q) = G8 , then √ √ √ positum N = M16.1 M16.2 with h− = 1 : N = Q( 2 + 2, 3 + 3, −1). N (3) If G(M16.1 /Q) = G(M16.2 /Q) = G8 , then h− ≥ 2. N Proof. (1) This case is treated in Proposition 17 (1). (See Remark 4 (a) (ii).) (2) This case is dealt with in Proposition 16 (4). (See Remark 2 (c).) (3) To begin with we prove that if K16 = A1 A2 is a compositum of two octic + quaternion CM-fields A1 and A2 with the same maximal real subfield A+ 1 = A2 , + then the Galois group G(K16 /Q) is isomorphic to G8 , G(K16 /Q) = C2 ×C2 ×C2 , and + h− K16 ≥ 8. From Proposition 9 it follows that G(K16 /Q) = G8 and G(K16 /Q) = C2 × − − − C2 ×C2 . Note that hK16 = (QK16 /2)hA1 hA2 and that there is only one octic quater√ √ √ √ nion CM-field with relative class number 2: Q( 2, 3, −(2 + 2)(3 + 3) )



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Table 3. N r. 1 2 3 4 5

M16 √ √ √ √ Q( 2, 17, 5, −(5 + 2 2)) √ √ √ √ Q( 5, 29, 3, −(7 + 2 5)) √ √ √ √ Q( 3, 13, 2, −(5 + 13)/2) √ √ √ √ Q( 5, 11, 2, −(7 + 5)) √ √ √ √ Q( 21, 85, 5, −(37 + 7 21)/2)

QM16

h− M16

2

1

1

2

2

4

2

4

2

4

([Lou2]). Moreover, this is the only imaginary pure quaternion extension of Q √ √ √ √ √ √ containing Q( 2, 3). If A1 = Q( 2, 3, −(2 + 2)(3 + 3)), then there are at √ √ least four ramified prime ideals in the extension A2 /Q( 2, 3), whence 8 | h− A2 and √ √ √ √ −(2 + 2)(3 + 3) ), then 8 | h− K16 . If neither A1 nor A2 is equal to Q( 2, 3, − − , h ≥ 4 and h ≥ 8Q . Let M = M M h− K16 16.1 8.1 8.2 and M16.2 = L8.1 L8.2 with A1 A2 K16 G(M8.1 /Q) = G(L8.1 /Q) = C2∗ ×C2 ×C2 and G(M8.2 /Q) = G(L8.2 /Q) = Q8 . Then the relative class number of the compositum M8.2 L8.2 is ≥ 8, and h− N > 1. 4.3. In this subsection we assume that G(M8+ /Q) = C2 × C2 × C2 . By Proposition 9 and the proof of Proposition 19, point (3), we exclude the normal CM-fields M with G(M/Q) = G8 , hence G(M16.1 /Q), G(M16.2 /Q) ∈ {G7 , G9 , C4∗ ×C2 ×C2 , C2∗ × C2 × C2 × C2 }. We will prove that if G(M8+ /Q) = C2 × C2 × C2 , then h− N ≥ 2. For the proof of Proposition 20 we need the following remark: Remark 6. According to [LO1] and [YK] there are five normal CM-fields M16 with Galois group G(M16 /Q) = G7 such that M16 = F8.1 F8.2 is a compositum of two + + − octic dihedral CM-fields with F8.1 = F8.2 and h− F8.1 , hF8.2 ∈ {1, 2, 4}. These fields are given in Table 3. + /Q) = C2 × C2 × C2 for i = 1, 2. Proposition 20. Assume that G(M16.i (1) If G(M16.1 /Q) = G(M16.2 /Q) = G7 , then h− N ≥ 2. (2) If G(M16.1 /Q) = G7 and G(M16.2 /Q) = G9 , then h− N ≥ 2. (3) If G(M16.1 /Q) = G7 and G(M16.2 /Q) = C4∗ × C2 × C2 , then h− N ≥ 2. (4) If G(M16.1 /Q) = G7 and G(M16.2 /Q) = C2∗ × C2 × C2 × C2 , then h− N ≥ 2. + Proof. Let M16.1 = M8.1 M8.2 with G(M8.1 /Q) = G(M8.2 /Q) = D8 and M8.1 = + M8.2 . (1) By Remark 6 there is no pair of (M16.1 , M16.2 ) such that M16.1 = M8.1 M8.2 + + with M8.1 = M8.2 and G(M8.1 /Q) = G(M8.2 /Q) = D8 , M16.2 = L8.1 L8.2 with + + − − − L8.1 = L8.2 and G(L8.1 /Q) = G(L8.2 /Q) = D8 , h− M8.1 , hM8.2 , hL8.1 , hL8.2 ∈ {1, 2, 4}, + + and M16.1 = M16.2 . It follows that if G(M16.1 /Q) = G(M16.2 /Q) = G7 , then h− N > 1. (2) Suppose that h− N = 1. Then M16.1 is one of the five fields M16 in Table 3. According to [LO2] if h− M16.2 = 1, then √ √ √ √ √ √ M16.2 = Q( 2, 5, 37, −(2 2 + 3 5)(2 + 5)).


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√ √ √ + There is no field M16 in Table 3 such that M16 = Q( 2, 5, 37). So, by Propo− − 4. We will examine sition 10 if h− N = 1, then hM16.1 = 1 and hM16.2 = 2 or √ √ √ whether there exists a normal CM-field M16.2 containing √Q(√ 2,√ 5, 17) such that + G(M16.2 /Q) = G9 and h− M16.2 = 2 or 4. Set M16.2 = Q( 2, 5, 17). Let us deter+ . If there exists a prime p mine which prime ideals can be ramified in M16.2 /M16.2 different from 2, 5, and 17 such that the prime ideals lying above p are ramified in + + 2 2 2 2 , then 8 | h− the extension M16.2 /M16.2 M16.2 . In M16.2 we let (2) = I2 I2 , (5) = I5 I5 , 2 2 and (17) = I17 I17 . If h− ramified prime ideals M16.2 = 2 or 4, then the number t of √ √ √ + in M16.2 /M16.2 is ≤ 3. The first field in Table 3 is Hilnar (Q( 2, 5, 17)), hence + ) = I2e I2 e , I5 I5 , or t = 0. Since M16.2 /Q is normal, t = 2 and F(M16.2 /M16.2 I17 I17 , where e = 2, 4, or 5. According to Lemma 7, point (4), there is no such M16.2 . It follows that if G(M16.1 /Q) = G7 and G(M16.2 /Q) = G9 , then h− N > 1. (3) This case is treated in Proposition 17 (1). (See Remark 4 (a) (i).) (4) This case is treated in Proposition 19. (See Remark 5.) Proposition 21. (1) If G(M16.1 /Q) = G(M16.2 /Q) = G9 , then h− N ≥ 2. (2) If G(M16.1 /Q) = G9 and G(M16.2 /Q) = C4∗ × C2 × C2 , then h− N ≥ 2. (3) If G(M16.1 /Q) = G9 and G(M16.2 /Q) = C2∗ × C2 × C2 × C2 , then h− N ≥ 2. Proof. (1) From Proposition 10 it follows that if there exists a field N with h− N = 1, = 1. Hence, then M16.1 or M16.2 , say M16.1 , satisfies h− M16.1 √ √ √ √ √ √ M16.1 = Q( 2, 5, 37, (2 2 + 3 5)(2 − 5)) √ √ √ + ([LO2, Theorem 20]) and M16.1 = Q( 2, 5, 37). We claim that h− M16.2 > − + 2 2 2 2 2 2 I37 . 2QM16.2 , whence hN ≥ 2. In M16.1 (2) = I2 I2 , (5) = I5 I5 , and (37) = I37 + + − Suppose that there exists a field M16.2 with M16.1 = M16.2 , hM16.2 = 2QM16.2 . By + . Note Theorem 2, point (3), at most two prime ideals are ramified in M16.2 /M16.2 + that M16.1 = Hilnar (M16.1 ). Hence, there are exactly two ramified prime ideals + + and F(M16.2 /M16.1 ) = I2e I2e with e = 2, 4, or 5, I5 I5 , or I37 I37 . in M16.2 /M16.2 According to Lemma 7, point (5), there is no such M16.2 . It follows the conclusion. √ √ √ (2) Suppose that h− N = 1. Then, M16.2 = Q( −(5 + 2 5), 2, 3), the only imaginary abelian number field with Galois group C4∗ × C2 × C2 of relative class number dividing 4 (see [CK]). Since h− = 4, h− 10 and M16.1 = 1 by Proposition √ √ √M16.2 √ √ √ + + M16.1 is as in (1). Since Q( 2, 5, 37) = M16.1 = M16.2 = Q( 2, 3, 5), we have a contradiction. − (3) Suppose that h− N = 1. Since hM16.2 > 1 (see [CK]), by Proposition 10 we is as in (1) and (2). Since at most two prime have h− M16.1 = 1. Hence, M16.1 √ √ √ + + + ideals of M√ = M = Q( 2, 5, 37) are ramified in M16.2 /M16.2 , we have 16.2 √ √16.1 √ − − M16.2 = Q( 2, 5, 37, −1) and 240 = hM16.2 |(4hN ). A contradiction. 4.4. In this subsection we assume that G(M8+ /Q) = C4 × C2 . Then G(M16.1 /Q), G(M16.2 /Q) ∈ {G4 , G5 , G6 , C4∗ × C4∗ , C4∗ × C2∗ × C2∗ , C8∗ × C2 }. Proposition 22. If G(M16.1 /Q) = G4 and G(M16.2 /Q) ∈ {G4 , G5 , C8∗ × C2 }, then h− N ≥ 2. Proof. It follows from Theorem 3 and Propositions 10 and 24 below.



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Proposition 23. If G(M16.1 /Q) = G4 and G(M16.2 /Q) = G6 , then there is one and only one field N =M16.1 M16.2 with h− N = 1. Namely, N is the composi√ √ 8 6 17 + 4 17)(θ) with + 85θ 4 + 136θ 2 + 68 = 0 tum of M16.1 = Q( 2, θ + 17θ √ √ √ √ and M16.2 = Q( 2, 17, −(5 + 17)/2, −(17 + 3 17)/2). Moreover, hN = 1, √ √ N + = Q( 2, 17 + 4 17)(α) with α8 − 2α7 − 147α6 + 218α5 + 4966α4 − 9058α3 − 14325α2 + 25774α + 1361 = 0, G(N + /Q) = C8 × C2 , dN + = 224 1714 , and dN = d2N + = d2M16.1 = 248 1728 . − Proof. From Proposition 10 if h− = 1, then h− M16.1 = 2QM16.1 , hM16.2 = 1, M16.2 = N √ √ √ √ √ √ + = Q( 2, 17 + 4 17) Q( 2, 17, −(5 + 17)/2, −(17 + 3 17)/2) and M16.2 √ (see [Lou3]). Since M16.1 /Q( 2) is cyclic, the two prime ideals lying above 17 are √ √ + + ramified in M16.1 /M16.1 with M16.1 = Q( 2, 17 + 4 17). If h− M16.1 = 2QM16.1 , + . Hence, then at most two prime ideals are ramified in M16.1 /M16.1 √ √ F(M16.1 /Q( 17 + 4 17, 2)) = I17 I17 , √ √ 4 4 where (17) = I17 I 17 in Q( 2, 17 + 4 17). Let L8 be any one of two non-normal √ √ + octic CM-subfields ofM16.1 . Then L8 = Q( 17 + 4 17) or Q( 17 + 17). We √ √ + Q( 17 + 17), the cyclic claim that L+ 8 = Q( 17 + 4 17). Suppose that L8 = √ quartic field of conductor 136. Then M16.1 /Q( 17 + 4 17) is cyclic, the two prime + − ideals lying above 2 would be ramified in M16.1 /M16.1 and 8|h− M16.1 . Since hM16.1 = − QM16.1 hL8 2 ( 2 QL8 ) ,

− the fact that h− M16.1 = 2QM16.1 implies that hL8 = 2QL8 . Hence, the

4 only ramified prime ideals in L8 /L+ 8 are p2 and p17 , where (2) = p2 p2 and (17) = p17 √ + e in Q( 17 + 4 17). We have F(L8 /L8 ) = p2 p17 with e = 2 or 3. According to Lemma 7, point (6), we have e = 3 and L8 = Q(θ) with θ 8 + 17θ 6 + 85θ 4 + 136θ 2 + 68 = 0. We verify that h− Hence L8 = 2 by using the technique developed in [Lou7]. √ √ − − + hM16.1 = 2 and hN = 1. The conductor of the extension N /Q( 2, 17 + 4 17) is I17 I17 . We verify that hN + = 1 and G(N + /Q) = C8 × C2 .

The following result will be used in the sequel. Proposition 24 ([CK]). (1) There are two imaginary abelian number fields with Galois group C4∗ × C4∗ and relative class numbers equal to one. (2) There are seven imaginary abelian number fields with Galois group C4∗ × C2∗ × C2∗ and relative class numbers equal to one. (3) There is one and only one field with Galois group C8∗ ×C2 and relative class number dividing 4. These fields are compiled in Table 4. Proposition 25. If G(M16.1 /Q) = G4 and G(M16.2 /Q) = C4∗ × C4∗ , then h− N ≥ 2. − − Proof. If h− N = 1, then hM16.1 = 2QM16.1 and hM16.2 = 1. There are two fields M16.2 − with hM16.2 = 1. √ √ + = F8 := (i) Let us take M16.2 = Q( −(5 + 2 5), −(13 + 2 13)); then M16.2 √ √ Q( 5, (65 + 65)/2) (see [CK] and [G]). We claim that h− M16.1 > 2QM16.1 for


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Table 4. Here θ 8 −θ 7 +10θ 6 −11θ 5 +15θ 4 −61θ 3 +58θ 2 −47θ+103 = 0 and Q(θ) is the imaginary octic cyclic number field of conductor 51 (see [PK]). N r. 1 2 3 4 5 6 7 8 9 10

M √ √ Q( −(5 + 2 5), −(13 + 2 13)) √ √ Q( −(5 + 2 5), −(2 + 2)) √ √ √ Q( −(5 + 2 5), −1, −2) √ √ √ Q( −(5 + 2 5), −3, −1) √ √ √ Q( −(5 + 2 5), −3, −7) √ √ √ Q( −(5 + 2 5), −3, −2) √ √ √ Q( −(5 + 2 5), −1, −7) √ √ √ Q( −(2 + 2), −3, −1) √ √ √ Q( −(2 + 2), −5, −1) √ Q(θ, 3)

G(M/Q)

h− M

QM

hM +

C4∗ × C4∗

1

2

1

C4∗

1

2

1

C4∗ × C2∗ × C2∗

1

2

1

C4∗ × C2∗ × C2∗

1

2

1

C4∗ × C2∗ × C2∗

1

2

1

C4∗ × C2∗ × C2∗

1

2

1

C4∗

C2∗

1

2

1

C4∗ × C2∗ × C2∗

1

2

1

C4∗

1 4

1 1

1 1

×

×

C2∗

C4∗

×

× C2∗ × C2∗ C8∗ × C2

+ every normal √ CM-field M16.1 √ with G(M16.1 /Q) = G4 and M16.1 = F8 . Since M16.1 /Q( 5) and M16.1 /Q( 13) are cyclic, the unique prime ideal P5 lying above + + 5 in M16.1 and the unique prime ideal P13 lying above 13 in M16.1 are ramified in + M16.1 /M16.1 . Since if F(M16.1 /F8 ) = P5 P13 , then at least four prime ideals are ramified in M16.1 /F8 and 8|h− M16.1 , we assume that F(M16.1 /F8 ) = P5 P13 . Then M16.1 is equal to the field K defined in Lemma 7, point (7). LetL8 be any one √ of two non-normal octic CM-subfields of M16.1 . Then L+ 65)/2) 8 = Q( (65 + √ or Q( (65 + 7 65)/2). We verify that there is no totally imaginary quadratic √ extension of Q( (65 + 7 65)/2) of conductor p5 p13 ∞, where p5 (resp. p13 ) is √ the prime ideal lying above 5 (resp. 13) in Q( (65 + 7 65)/2). Then L+ 8 = √ + 4 4 Q( (65 + 65)/2) and F(L8 /L+ 8 ) = I5 I13 , where (5) = I5 and (13) = I13 in L8 . We verify that h− L8 = 10, whence the normal closure of L8 has relative class number 2 ) /2 = 50. This proves that h− > 2QM16.1 . (h− L8 M16.1 √ √ (ii) We now let M16.2 = Q( −(5 + 2 5), −(2 + 2)) with √ √ + = Q( 10 + 10, 5) M16.2

(see [CK] and [G]). By the similar way we verify that every normal CM-field M16.1 √ √ + with G(M16.1 /Q) = G4 and M16.1 = Q( 10 + 10, 5) has h− M16.1 > 2QM16.1 . This completes the proof. Proposition 26. There are exactly three composita N = M16.1 M16.2 such that G(M16.1 /Q) = G4 , G(M16.2 /Q) = C4∗ × C2∗ × C2∗ , and h− N =1: √ √ √ √ √ √ Q(θ, 2, −1), Q(θ, 3, −1), Q(θ, 7, −1)



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with θ 8 + 10θ 6 + 25θ 4 + 20θ 2 + 5 = 0. Moreover, these three fields have class number one. Proof. For the seven fields M16.2 numbered 3-9 in Table 4 of Proposition 24 we + such that will construct normal CM-fields M16.1 containing the given field M16.2 + at most two prime ideals are ramified in the extension M16.1 /M16.2 . Let G4 be + = a4 , G(M16.1 / a ) = G(M16.1 / ab ) = as Proposition 9. We have M16.1 4 C8 , G(M16.1 / a , b ) = C2 ×C2 , and G(M16.1 / a2 b ) = C4 . The field M16.1 has two isomorphic octic non-normal CM-subfields, M8.1 = a4 b and M8.2 = b . Then h− M16.1 = order to

− QM16.1 hM8.1 2 − ( QM ) . Therefore, h− M16.1 = 2QM16.1 implies hM8.1 = 2QM8.1 . 2 8.1 + have h− N = 1 there are at most two ramified primes in M8.1 /M8.1 .

In

√ √ √ (i) Let M16.2 = Q( −(5 + 2 5), −1, −2). Let us construct the CM-fields √ √ + + + M16.1 containing M16.2 = Q( (5 − 5)/2, 2). Then a4 , b (= M8.1 = M8.2 )= √ √ Q( (5 − 5)/2) or Q( 5 + 5). √ √ + + = M8.2 = Q( (5 − 5)/2). Then a2 b = Q( 5 + 5) (i-a) Assume that M8.1 √ + and M16.1 /Q( 5 + 5) is cyclic. In M8.1 (5) = p45 and (2) = p22 . Then + F(M8.1 /M8.1 ) = pe2 p5

with e = 2, 4, or 5, and M8.1 = Q(θ) with θ 8 + 10θ 6 + 25θ 4 + 20θ 2 + 5 = 0, by Lemma 7, point (8). We verify that h− = 2, whence h− M√ M16.1 = 2. Hence the 8.1 √ relative class number of the field N = M8.1√ ( 2, −1) is equal to 1. For convenience we write Q(θ) = F8 . Note that N + = L( 2) and hN + = 1, where L = Q(η) with η 8 − 10η 6 + 25η 4 − 20η 2 + 5 = 0. √ + + (i-b) Assume now that M = M = Q( 5 + 5). Then M16.1 / 8.1 8.2 √ + + 4 2 )= Q( (5 − 5)/2) is cyclic. In M8.1 (5) = p5 and (2) = p2 . Then F(M8.1 /M8.1 e p2 p5 with e = 2, 4, or 5, and according to Lemma 7, point (9), there is no such field M8.1 . √ √ √ (ii) Let M16.2 = Q( −(5 + 2 5), −3, −1). By a similar way as (i) we get √ √ √ − M16.1 = F8 ( 3) with h− M16.1 = 2, whence N = F8 ( 3, −1) has hN = 1. We verify that hN + = 1. √ √ √ (iii) Let M16.2 = Q( −(5 + 2 5), −3, −7). Let M16.1 be a normal CM √ √ + field with G(M16.1 /Q) = G4 containing M16.2 = Q( 3(5 + 5)/2, 21). Note that √ √ + . (5) = I1 4 I2 4 in Q( 3(5 + 5)/2, 21), I1 and I2 are ramified in M16.1 /M16.1 √ √ 2

a b = Q( 7(5 − 5)/2) or Q( 3(5 − 5)/2). √ (iii-a) If M16.1 /Q( 7(5 − 5)/2) is cyclic, then the prime ideal lying above 3 is √ √ ramified in M16.1 /Q( 3(5 + 5)/2, 21), whence QM16.1 = 1 and 4 | h− M16.1 . √ (iii-b) If M16.1 /Q( 3(5 − 5)/2) is cyclic, then the prime ideal lying above 7 is √ √ ramified in M16.1 /Q( 3(5 + 5)/2, 21), whence QM16.1 = 1 and 4 | h− M16.1 .


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√ √ √ (iv) Let M16.2 = Q( −(5 + 2 5), −3, −2). By the same argument as (iii) we prove that CM-field M16.1 with G(M16.1 /Q) = G4 containing every√ normal √ + M16.2 = Q( 3(5 + 5)/2, 6) satisfies 4 | h− M16.1 . √ √ √ (v) Let M16.2 = Q( −(5 + 2 5), −1, −7). By a similar way as (i) we obtain √ √ √ − M16.1 = F8 ( 7) with h− M16.1 = 2, whence N = F8 ( 7, −1) has hN = 1. We verify that hN + = 1. √ √ √ √ √ + = Q( 2 + 2, 3). (vi) Let M16.2 = Q( −(2 + 2), −3, −1) with M16.2 √ √ Then a2 b = Q( 3(2 + 2)) or Q( 2 + 2). √ √ (vi-a) Assume that M16.1 /Q( 3(2 + 2)) is cyclic. In Q( 2 + 2) (2) = p42 and (3) remains prime. However, according to Lemma 7, point (10), the octic subfield M8.1 of M16.1 cannot satisfy h− = 2QM8.1 . M8.1√ Then the prime ideal lying (vi-b) Assume now that M16.1 /Q( 2 + 2) is cyclic. √ √ above 3 is totally ramified in M16.1 /Q( 2 + 2). In Q( 3(2 + 2)) (2) = p42 and (3) = p23 . The field K defined in Lemma 7, point (11), is the unique (up to √ + isomorphism) octic CM-field with K = Q( 3(2 + 2)) such that at most two prime ideals are ramified in K/K + and the normal closure N of K has G(N/Q) = √ √ + G4 and N = Q( 2 + 2, 3). Set M8.1 = K. We verify that h− M8.1 = 10. √ √ √ (vii) Let M16.2 = Q( −(2 + 2), −5, −1). By a similar computation as (vi) we verify that there is no normal CM-field M16.1 containing √ √ + = Q( 2 + 2, 5) M16.2 such that G(M16.1 /Q) = G4 and h− M16.1 | 4. The proof is completed.

Proposition 27. (1) If G(M16.1 /Q) = G(M16.2 /Q) = G5 , then h− N ≥ 2. ≥ 2. (2) If G(M16.1 /Q) = G5 and G(M16.2 /Q) = G6 , then h− N (3) If G(M16.1 /Q) = G5 and G(M16.2 /Q) = C4∗ × C4∗ , then h− N ≥ 2. (4) If G(M16.1 /Q) = G5 and G(M16.2 /Q) = C4∗ × C2∗ × C2∗ , then there is one and only one N with N = M16.1 M16.2 and h− N =1: √ √ √ N = Q( 2 + 2, 3 + 3, −1). (5) If G(M16.1 /Q) = G5 and G(M16.2 /Q) = C8∗ × C2 , then h− N > 1. Proof. Let M8.1 , M8.2 , L8.1 , and L8.2 be four normal CM-fields such that M16.1 = + + + = M8.2 = L+ M8.1 M8.2 , M16.2 = L8.1 L8.2 , M8.1 8.1 = L8.2 . (1) We have G(M8.1 /Q) = G(L8.1 /Q) = D8 and G(M8.2 /Q) = G(L8.2 /Q) = Q8 . Then the compositum M8.2 L8.2 is a normal CM-field with Galois group G8 . − According to Proposition 19 (3) h− M8.2 L8.2 ≥ 8, hence hN ≥ 2. √ √ + + + + (2) Let M8.1 = M8.2 = L8.1 = L8.2 = Q( m1 , m2 ). We have G(M8.1 /Q) = − − G(L8.1 /Q) = G(L8.2 /Q) = D8 and G(M8.2 /Q) = Q8 . If h− N = 1, then hM8.1 , hL8.1 , and h− L8.2 divide 4 and (m1 , m2 ) ∈ {(2, 17), (5, 29)} by [LO1] and [YK]. If (m1 , m2 ) = (2, 17), then two prime ideals lying above 2 and two prime ideals lying above 17



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are ramified in M8.2 . Hence 8 | h− M8.2 . By the same argument if (m1 , m2 ) = (5, 29), − . It follows that h then 8 | h− M8.2 N ≥ 2 (cf. Lemma 9). (3) This case is treated in Proposition 16 (3). (See Remark 2 (b).) (4) This case is treated in Proposition 16 (4). (See Remark 2 (c).) − (5) Suppose that h− N = 1. Then hM16.2 | 4, whence M16.2 must be the field + is numbered bicyclic subfield of M16.2 √ biquadratic √ √ √ 10 in Table 4. The unique real in an octic Q( 3, 17). By [K, 2.2 Proposition] Q( 3, 17) cannot be embedded √ √ quaternion field. Hence, there is no field M16.1 containing Q( 3, 17), and we conclude that h− N > 1. Proposition 28. (1) If G(M16.1 /Q) = G(M16.2 /Q) = G6 , then h− N > 1. > 1. (2) If G(M16.1 /Q) = G6 and G(M16.2 /Q) is abelian, then h− N Proof. (1) It is sufficient to notice that there is no quadruple of octic dihedral CM-fields with the same maximal real subfield whose relative class numbers divide 4. (2) If G(M16.2 /Q) = C4∗ × C4∗ or C4∗ × C2∗ × C2 , then h− N ≥ 2 (see Remark 4). From Remark 3 (a) it follows Suppose that G(M16.2 /Q) = C8∗ × C2 and h− N = 1. √ √ + + = M16.2 that the unique biquadratic bicyclic real subfield Q( m1 , m2 ) of M16.1 − satisfies (m1 , m2 ) ∈ S where S is as in Remark 3. The fact that hM16.2 | 4 implies / S, we have a contradiction. (m1 , m2 ) = (51, 3). Since (51, 3) ∈ In conclusion the proof of point (2) of Theorem 1 is completed by Propositions 11-28. Acknowledgements All computations were performed using KASH ([KT]) and Pari-GP ([P]). S.-H. Kwon is greatly indebted to C. Fieker for helps concerning KASH. The authors wish to express their gratitude to S. Louboutin for several helpful suggestions during the preparation of this version. S.-H. Kwon was supported by the Korea Research Foundation under grant KRF-2005-C00002. References [BLS] [CK] [Co] [G] [H]

[He] [JL] [K] [KT]

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Department of Mathematics, Korea University, 136-701, Seoul, Korea E-mail address: [email protected] Department of Mathematics, Korea University, 136-701, Seoul, Korea Current address: Korea Minting and Security Printing Corporation, 54, Gwahakro, YusongGu, 305-713 Daejon, Korea E-mail address: [email protected] Department of Mathematics Education, Korea University, 136-701, Seoul, Korea E-mail address: [email protected]


THE CLASS NUMBER ONE PROBLEM FOR THE NORMAL CM

THE CLASS NUMBER ONE PROBLEM FOR THE NORMAL CM

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