2 B) (d 1 =p 3 )=(d 2 =p 3 ) = +1, (d 1 =p 2 )=;1. The C 4 -factorization corresponding to the nontrivial 4-part of Cl 2 (k) isd = d 1 d 2 d 3 in case

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1 IMAGINARY QUADRATIC FIELDS k WIT Cl 2 (k) ' (2 2 m ) AND RANK Cl 2 (k 1 )=2 E. BENJAMIN, F. LEMMERMEYER, C. SNYDER Abstract. Let k be an imaginary quadratic number eld and k 1 the ilbert 2-class eld of k. We give acharacterization of those k with Cl 2 (k) ' (2 2 m ) such that Cl 2 (k 1 ) has 2 generators. 1. Introduction Let k be an algebraic number eld with Cl 2 (k), the Sylow 2-subgroup of its ideal class group, Cl(k). Denote by k 1 the ilbert 2-class eld of k (in the wide sense). Also let k n (for n a nonnegative integer) be dened inductively as: k 0 = k and k n+1 =(k n ) 1. Then k 0 k 1 k 2 k n is called the 2-class eld tower of k. If n is the minimal integer such that k n = k n+1, then n is called the length of the tower. If no such n exists, then the tower is said to be of innite length. At present there is no known decision procedure to determine whether or not the (2-)class eld tower of a given eld k is innite. owever, it is known by group theoretic results (see [2]) that if rank Cl 2 (k 1 ) 2, then the tower is nite, in fact of length at most 3. (ere the rank means minimal number of generators.) On the other hand, until now (seetable 1 and the penultimate paragraph of this introduction) all examples in the mathematical literature of imaginary quadratic elds with rank Cl 2 (k 1 ) 3 (let us mention in particular Schmithals [13]) have innite 2-class eld tower. Nevertheless, if weareinterested in developing a decision procedure for determining if the 2-class eld tower of a eld is innite, then a good starting point would be to nd a procedure for sieving out those elds with rank Cl 2 (k 1 ) 2. We have already started this program for imaginary quadratic number elds k. In [1] we classied all imaginary quadratic elds whose 2-class eld k 1 has cyclic 2-class group. In this paper we determine when Cl 2 (k 1 ) has rank 2 for imaginary quadratic elds k with Cl 2 (k) oftype (2 2 m ). (The notation (2 2 m ) means the direct sum of a group of order 2 and a cyclic group of order 2 m.) The group theoretic results mentioned above alsoshow that such elds have 2-class eld tower of length 2. From a classication of imaginary quadratic number elds k with Cl 2 (k) ' (2 2 m ) and our results from [1] we see that it suces to consider discriminants d = d 1 d 2 d 3 with prime discriminants d 1 d 2 > 0, d 3 < 0 such that exactly one of the (d i =p j ) equals ;1 (weletp j denote the prime dividing d j ) thus there are only two cases: A) (d 1 =p 2 )=(d 1 =p 3 ) = +1, (d 2 =p 3 )=; Mathematics Subject Classication. Primary 11R37. 1

2 2 B) (d 1 =p 3 )=(d 2 =p 3 ) = +1, (d 1 =p 2 )=;1. The C 4 -factorization corresponding to the nontrivial 4-part of Cl 2 (k) isd = d 1 d 2 d 3 in case A) and d = d 1 d 2 d 3 in case B). Note that, by our results from [1], some of these elds have cyclic Cl 2 (k 1 ) however, we do not exclude them right from the start since there is no extra work involved and since it provides a welcome check on our earlier work. The main result of the paper is that rank Cl 2 (k 1 )=2only occurs for elds of type B) more precisely, we prove the following Theorem 1. Let k be acomplex quadratic number eld with Cl 2 (k) ' (2 2 m ),and let k 1 be its 2-class eld. Then rank Cl 2 (k 1 )=2if and only if disc k = d 1 d 2 d 3 is the product of three prime discriminants d 1 d 2 > 0 and ;4 6= d 3 < 0 such that (d 1 =p 3 )=(d 2 =p 3 )=+1, (d 1 =p 2 )=;1, and h 2 (K) =2, where K is a nonnormal quartic subeld of one of the two unramied cyclic quartic extensions of k such that Q( p d 1 d 2 ) K. This result is the rst step in the classication of imaginary quadratic number elds k with rank Cl 2 (k 1 ) = 2 it remains to solve these problems for elds with rank Cl 2 (k) = 3 and those with Cl 2 (k) (4 4) since we know that rank Cl 2 (k 1 ) 5 whenever rank Cl 2 (k) 4 (using Schur multipliers as in [1]). As a demonstration of the utility of our results, we give intable 1 below a list of the rst 12 imaginary quadratic elds k, arranged by decreasing value of their discriminants, with rank Cl 2 (k) = 2 and noncyclic Cl 2 (k 1 ). Table 1 disc k factors Cl 2 (k) type f Cl 2 (K) r Cl 2 (k gen ) ;1015 ; (2 8) A x 4 ; 22x (4) 3 (2 2 8) ;1240 ; (2 4) B x 4 ; 6x 2 ; 31 (2) 2 (2 2 8) ;1443 ; (2 4) B x 4 ; 86x 2 ; 75 (2) 2 (2 2 8) ;1595 ; (2 8) A x 4 +26x (4) 3 (2 2 8) ;1615 ; (2 4) B x 4 +26x 2 ; 171 (2) 2 (2 2 8) ;1624 ; (2 8) B x 4 ; 30x 2 ; 7 (2) 2 (2 2 8) ;1780 ; (2 4) A x 4 +6x (4) 3 (2 2 4) ;2035 ; (2 4) B x 4 ; 54x 2 ; 11 (2 2) 3 (2 2 16) ;2067 ; (2 4) A x 4 + x (2 2) 3 (2 2 4) ;2072 ; (2 8) B x 4 +34x 2 ; 7 (2 2) 3 (2 2 8) ;2379 ; (4 4) 3 (4 4 8) ;2392 ; (2 4) B x 4 +18x 2 ; 23 (2) 2 (2 2 8) ere f denotes a generating polynomial for a eld K as in Theorem 1, r denotes the rank of Cl 2 (k 1 ). The cases where r = 3 follow from our theorem combined with Blackburn's upper bound for the number of generators of derived groups (it implies that nite 2-groups G with G=G 0 ' (2 4) satisfy rank G 0 3), see [3]. In order to verify that Cl 2 (k 1 ) has rank at least 3 for k = Q( p ;2379 ) it is sucient to show that its genus class eld k gen has class group (4 4 8): in fact, Cl 2 (k 1 ) then contains a quotient of(4 4 8) by (2 2) ' Gal(k 1 =k gen ), and the claim follows.

3 IMAGINARY QUADRATIC FIELDS 3 We mention one last feature gleaned from the table. It follows from conditional Odlyzko bounds (assuming the Generalized Riemann ypothesis) that those quadratic elds with rank Cl 2 (k 1 ) 3 and discriminant0>d>;2000 have nite class eld tower unconditional proofs are not known. ence, conditionally, we conclude that those k with discriminants ;1015 ;1595 and ;1780 have nite (2-)class eld tower even though rank Cl 2 (k 1 ) 3. Of course, it would be interesting to determine the length of their towers. The structure of this paper is as follows: we use results from group theory developed in Section 2 to pull down the condition rank Cl 2 (k 1 ) = 2 from the eld k 1 with degree 2 m+2 to a subeld L of k 1 with degree 8. Using the arithmetic of dihedral elds from Section 4 we then go down to the eld K of degree 4 occurring in Theorem Group Theoretic Preliminaries Let G be a group. If x y 2 G, then we let [x y] = x ;1 y ;1 xy denote the commutator of x and y. If A and B are nonempty subsets of G, then [A B] denotes the subgroup of G generated by the set f[a b] :a 2 A b 2 Bg. The lower central series fg j g of G is dened inductively by: G 1 = G and G j+1 = [G G j ] for j 1. The derived series fg (n) g is dened inductively by: G (0) = G and G (n+1) =[G (n) G (n) ] for n 0. Notice that G (1) = G 2 =[G G] the commutator subgroup, G 0,ofG. Throughout this section, we assume that G is a nite, nonmetacyclic, 2-group such that its abelianization G ab = G=G 0 is of type (2 2 m ) for some positive integer m (necessarily 2). Let G = ha bi, where a 2 b 2m 1modG 2 (actually modg 3 since G is nonmetacyclic, cf. [1]) c 2 =[a b] and c j+1 =[b c j ] for j 2. Lemma 1. Let G be as above (but not necessarily metabelian). Suppose that d(g 0 ) = n where d(g 0 ) denotes the minimal number of generators of the derived group G 0 = G 2 of G. Then moreover, G 0 = hc 2 c 3 c n+1 i G 2 =G 2 2 'hc 2 G 2 2ihc n+1 G 2 2i: Proof. By the Burnside Basis Theorem, d(g 2 )=d(g 2 =(G)), where (G) is the Frattini subgroup of G, i.e. the intersection of all maximal subgroups of G, see [5]. But in the case of a 2-group, (G) =G 2, see [8]. By Blackburn, [3], since G=G 2 2 has elementary derived group, we know that G 2 =G 2 2 ' hc 2 G 2 2ihc n+1 G 2 2i. Again, by the Burnside Basis Theorem, G 2 = hc 2 c n+1 i: Lemma 2. Let G be as above. Moreover, assume G is metabelian. Let be a maximal subgroup of G such that =G 0 is cyclic, and denote the index (G 0 : 0 ) by 2. Then G 0 contains an element of order 2. Proof. Without loss of generality, let = hb G 0 i. Notice that G 0 = hc 2 c 3 i and by our presentation of, 0 = hc 3 c 4 i. Thus, G 0 = 0 = hc 2 0 i. But since (G 0 : 0 )=2, the order of c 2 is 2. This establishes the lemma. Lemma 3. Let G be as above and again assume G is metabelian. Let be a maximal subgroup of G such that =G 0 is cyclic, and assume that (G 0 : 0 ) 0mod4. If d(g 0 )=2, then G 2 = hc 2 c 3 i and G j = hc2 2j;2 c3 2j;3 i for j>2.

4 4 Proof. Assume that d(g 0 )=2. By Lemma 1, G 2 = hc 2 c 3 i and hence c 4 2hc 2 c 3 i. Write c 4 = c x 2c y 3 where x y are positive integers. Without loss of generality, let = hb c 2 c 3 i and write (G 0 : 0 )=2 for some 2. Since c 3 c wehave, c x 2 1mod0. By the proof of Lemma 2, this implies that x 0mod2. Write x = 2 x 1 for some positive integer x 1. On the other hand, since c 4 c 2 x G 4, we see that c y 3 1modG 4. If y were odd, then c 3 2 G 4. This, however, implies that G 2 = hc 2 i, contrary to our assumptions. Thus y is even, say y =2y 1. From all of this we see that c 4 = c 2 x 1 2 c 2y1 3. Consequently, by induction we have c j 2 hc 2j;2 2 c 2j;3 3 i for all j 4. Since G j = hc2 2j;2 c3 2j;3 c 2 j;1 c j c j+1 i, cf. [1], we obtain the lemma. Let us translate the above into the eld-theoretic language. Let k be an imaginary quadratic number eld of type A) or B)(see the Introduction), and let M=k be one of the two quadratic subextensions of k 1 =k over which k 1 is cyclic. If h 2 (M) =2 m+ and Cl 2 (k) =(2 2 m ), then Lemma 2 implies that Cl 2 (k 1 )contains an element of order 2. Table 2 contains the relevant information for the elds occurring in Table 1. An application of the class number formula to M=Q (see e.g. Proposition 3 below) shows immediately that h 2 (M) =2 m+, where 2 is the class number of the quadratic subeld Q( p d i d j )ofm, where (d i =p j ) = +1 in particular, we always have 2, and the assumption (G 0 : 0 ) 4isalways satised for the elds that we consider. Table 2 M 1 Cl 2 (M 1 ) M 2 Cl 2 (M 2 ) Q( p 5 p ;7 29 ) (2 16) Q( p 5 29 p ;7) (2 16) Q( p 2 p ;5 31 ) (4 4) Q( p 5 p ;2 31 ) (2 16) Q( p 13 p ;3 37 ) (2 16) Q( p 37 p ;3 13 ) (2 16) Q( p ;11 p 5 29 ) (2 16) Q( p 29 p ;5 11 ) (2 16) Q( p 5 p ;17 19 ) (4 4) Q( p 17 p ;5 19 ) (2 16) Q( p 29 p ;2 7) (2 16) Q( p 2 p ;7 29 ) (2 16) Q( p 5 89 p ;1) (4 4) Q( p 5 p ;89 ) (2 8) Q( p 37 p ;5 11 ) (4 4) Q( p 5 p ;37 11 ) (2 32) Q( p 53 p ;3 13 ) (4 4) Q( p p ;3) (2 2 4) Q( p 37 p ;2 7) (2 16) Q( p 2 p ;7 37 ) (2 16) Q( p 13 p ;2 23 ) (4 4) Q( p 2 p ;13 23 ) (2 16) We now use the above results to prove the following useful proposition. Proposition 1. Let G be a nonmetacyclic 2-group such that G=G 0 ' (2 2 m ) (hence m>1). Let and K be the two maximal subgroups of G such that =G 0 and K=G 0 are cyclic. Moreover, assume that (G 0 : 0 ) 0mod4. Finally, assume

5 IMAGINARY QUADRATIC FIELDS 5 that N isasubgroup of index 4 in G not contained in or K Then 8 (N : N 0 < ) : : = 2 m if d(g 0 )=1 = 2 m+1 if d(g 0 )=2 2 m+2 if d(g 0 ) 3 Proof. Without loss of generality we assume that G is metabelian. Let G = ha bi, where a 2 b 2m 1modG 3. Also let = hb G 0 i and K = hab G 0 i (without loss of generality). Then N = hab 2 G 0 i or N = ha b 4 G 0 i. Suppose that N = hab 2 G 0 i. First assume d(g 0 )=1.Then G 0 = hc 2 i and thus N 0 = h[ab 2 c 2 ]i. But [ab 2 c 2 ]= c for some 4 2 G 4 = hc 4 2i (cf. Lemma 1 of [1]). ence, N 0 = hc 2 2i, and so (G 0 : N 0 )=2. Since (N : G 0 )=2 m;1,we get (N : N 0 )=2 m as desired. Next, assume that d(g 0 ) = 2. Then N = hab 2 c 2 c 3 i by Lemma 1. Notice that [ab 2 c 2 ] = c and [ab 2 c 3 ] = c where j 2 G j for j = 4 5. ence N 0 = hc c N 3 i and so hc c 2 3 5iN 0. But then N 0 G 5 hc 4 2 c2 3 i = G 4 by Lemma 3. Therefore, by [5], N 0 G 4. But notice that N 3 G 4. Thus N 0 = hc 2 2 c 2 3i and so (G 0 : N 0 ) = 4 which in turn implies that (N : N 0 )=2 m+1, as desired. Finally, assume d(g 0 ) 3. Then d(g 0 =G 5 )=3. Moreover there exists an exact sequence N=N 0 ;! (N=G 5 )=(N=G 5 ) 0 ;! 1 and thus #N ab #(N=G 5 ) ab. ence it suces to prove the result for G 5 = 1 which wenow assume. N = hab 2 c 2 c 3 c 4 i and so, arguing as above, we have N 0 = hc c c N 3 i = hc c 2 3 N 3 i, where j 2 G j. But N 3 = h[ab 2 c ]i = hc 4 2 i. Therefore, N0 = hc c 2 3 i. From this we see that (G0 : N 0 ) = 8 and thus (N : N 0 )=2 m+2 as desired. Now suppose that N = ha b 4 G 0 i. Then the proof is essentially the same as above once we noticethat[a b 4 ] c 3 2 c 2 ;4 mod G 5. This establishes the proposition. 3. Number Theoretic Preliminaries Proposition 2. Let K=k be a quadratic extension, and assume that the class number of k, h(k), isodd. If K has an unramied cyclic extension M of order 4, then M=k is normal and Gal(M=k) ' D 4. Proof. Redei and Reichardt [12] proved this for k = Q the general case is analogous. We shall make extensive use of the class number formula for extensions of type (2 2): Proposition 3. Let K=k be a normal quartic extension with Galois group of type (2 2), and let k j (j =1 2 3) denote the quadratic subextensions. Then (1) h(k) =2 d;;2; q(k)h(k 1 )h(k 2 )h(k 3 )=h(k) 2 where q(k) =(E K : E 1 E 2 E 3 ) denotes the unit index of K=k (E j is the unit group of k j ), d is the number of innite primes in k that ramify in K=k, is the Z-rank of the unit group E k of k, and =0except when K k( p E k ),where =1. Proof. See [10].

6 6 Another important result is the ambiguous class number formula. For cyclic extensions K=k, let Am(K=k) denote the group of ideal classes in K xed by Gal(K=k), i.e. the ambiguous ideal class group of K, and Am 2 its 2-Sylow subgroup. Proposition 4. Let K=k be a cyclic extension of prime degree p then the number of ambiguous ideal classes is given by #Am(K=k) =h(k) p t;1 (E : ) where t is the number of primes (including those at 1) ofk that ramify in K=k, E is the unit group of k, and is its subgroup consisting of norms of elements from K. Moreover, Cl p (K) is trivial if and only if p - #Am(K=k). Proof. See Lang [9, part II] for the formula. For a proof of the second assertion (see e.g. Moriya [11]), note that Am(K=k) is dened by the exact sequence 1 ;;;;! Am(K=k) ;;;;! Cl(K) ;;;;! Cl(K) 1; ;;;;! 1 where generates Gal(K=k). Taking p-parts we see that p - # Am(K=k) is equivalent tocl p (K) =Cl p (K) 1;. By induction we get Cl p (K) =Cl p (K) (1;)p but since (1 ; ) p 0modp in the group ring Z[G], this implies Cl p (K) Cl p (K) p. But then Cl p (K) must be trivial. We make one further remark concerning the ambiguous class number formula that will be useful below. If the class number h(k) is odd, then it is known that #Am 2 (K=k) =2 r where r = rank Cl 2 (K). We also need a result essentially due to G. Gras [4]: Proposition 5. Let K=k be a quadratic extension of number elds and assume that h 2 (k) =#Am 2 (K=k) =2. Then K=k is ramied and ( (2 2) or Z=2n Z (n 3) if # Cl 2 (K) K=k =1 ' Z=2 n Z (n 1) if # K=k =2 where K=k denotes the set of ideal classes of k that become principal (capitulate) in K. Proof. We rst notice that K=k is ramied. If the extension were unramied, then K would be the 2-class eld of k, and since Cl 2 (k) is cyclic, it would follow that Cl 2 (K) = 1, contrary to assumption. Before we start with the rest of the proof, we cite the results of Gras that we need (we could also give a slightly longer direct proof without referring to his results). Let K=k be a cyclic extension of prime power order p r, and let be a generator of G = Gal(K=k). For any p-group M on which G acts we put M i = fm 2 M : m (1;)i = 1g. Moreover, let be the algebraic norm, that is, exponentiation by :::+ pr ;1. Then [4, Cor. 4.3] reads Lemma 4. Suppose that M =1 let n be the smallest positive integer such that M n = M and write n = a(p ; 1) + b with integers a 0 and 0 b p ; 2. If # M i+1 =M i = p for i =0 1 ::: n; 1, then M ' (Z=p a+1 Z) b (Z=p a Z) p;1;b. We claim that if K=k = 2, then M = Cl 2 (K) satises the assumptions of Lemma 4: in fact, let j = j k!k denote the transfer of ideal classes. Then c 1+ = j(n K=k c) for any ideal class c 2 Cl 2 (K), hence M = j(cl 2 (k)) =1. Moreover,

7 IMAGINARY QUADRATIC FIELDS 7 M 1 = Am 2 (K=k) in our case, hence M 1 =M 0 has order 2. Since the orders of M i+1 =M i decrease towards 1 as i grows (Gras [4, Prop. 4.1.ii)]), we conclude that # M i+1 =M i = 2 for all i<n. Since a = n and b = 0 when p = 2, Lemma 4 now implies that Cl 2 (K) ' Z=2 n Z, that is, the 2-class group is cyclic. The second result of Gras that we need is [4, Prop. 4.3] Lemma 5. Suppose that M 6= 1 but assume the other conditions in Lemma 4. Then n 2 and 8 >< (Z=p 2 Z) (Z=pZ) n;2 if n<p M ' (Z=pZ) >: p or (Z=p 2 Z) (Z=pZ) n;2 if n = p (Z=p a+1 Z) b (Z=p a Z) p;1;b if n>p: If K=k = 1, then this lemma shows that Cl 2 (K) is either cyclic of order 4 or of type (2 2). (Notice that the hypothesis of the lemma is satised since K=k is ramied implying that the norm N K=k :Cl 2 (K) ;! Cl 2 (k) isonto and so the argument above this lemma applies.) It remains to show that the case Cl 2 (K) ' Z=4Z cannot occur here. Now assume that Cl 2 (K) = hci ' Z=4Z since K=k is ramied, the norm N K=k :Cl 2 (K) ;! Cl 2 (k) isonto, and using K=k = 1 once more we nd C 1+ = c, where c is the nontrivial ideal class from Cl 2 (k). On the other hand, c 2 Cl 2 (k) still has order 2 in Cl 2 (K), hence we must also have C 2 = C 1+. But this implies that C = C, i.e.that each ideal class in K is ambiguous, contradicting our assumption that # Am 2 (K=k) =2. 4. Arithmetic of some Dihedral Extensions In this section we study the arithmetic of some dihedral extensions L=Q, that is, normal extensions L of Q with Galois group Gal(L=Q) ' D 4, the dihedral group of order 8. ence D 4 may bepresented as h j 2 = 4 =1 = ;1 i. Now consider the following diagrams (Galois correspondence): h1i X XXXXX h 2 i hi h 2 i h 3 i hi h 2 i hi h i h 2 i L X XXXXX K 0 1 K 1 K K 2 K 0 2 k 1 k k 2 Q In this situation, we let q 1 =(E L : E 1 E 0 1 E K) and q 2 =(E L : E 2 E 0 2 E K) denote the unit indices of the bicyclic extensions L=k 1 and L=k 2, where E i and E 0 i are the unit groups in K i and K 0 i respectively. Finally, let i denote the kernel of the transfer of ideal classes j ki!k i :Cl 2 (k i ) ;! Cl 2 (K i )fori =1 2. The following remark will be used several times: if K 1 = k 1 ( p ) for some 2 k 1, then k 2 = Q( p a ), where a = 0 is the norm of. To see this, let = p then =, since 2 K 1. Clearly 1+ = p a 2 K and hence xed by 2. Furthermore, ( 1+ ) = +2 = =( ) = = (1+)2 = 1+ implying that p a 2 k 2. Finally notice that p a 62 Q, since otherwise p 0 = p a= p 2 K1 implying that K 1 =Q is normal, which is not the case.

8 8 Recall that a quadratic extension K = k( p ) is called essentially ramied if O k is not an ideal square. This denition is independent of the choice of. Proposition 6. Let L=Q be a non-cm totally complex dihedral extension not containing p ;1, and assume that L=K 1 and L=K 2 are essentially ramied. If the fundamental unit of the real quadratic subeld of K has norm ;1, thenq 1 q 2 =2. Proof. Notice rst that k cannot be real (in fact, K is not totally real by assumption, and since L=k is a cyclic quartic extension, no innite prime can ramify in K=k) thus exactly one of k 1, k 2 is real, and the other is complex. Multiplying the class number formulas, Proposition 3, for L=k 1 and L=k 2 (note that = 0 since both L=K 1 and L=K 2 are essentially ramied) we nd that 2q 1 q 2 is a square. If we can prove that q 1 q 2 2, then 2q 1 q 2 is a square between 2 and 8, which implies that we must have 2q 1 q 2 = 4 and q 1 q 2 = 2 as claimed. We start by remarking that if becomes a square in L, where isarootofunity in L, then so does one of. This follows from the fact that the only non-trivial roots of unity that can be in L are the sixth roots of unity h 6 i, and here 6 = ; 2 3. Now we prove that q 1 2 under the assumptions we made the claim q 2 2will then follow by symmetry. Assume rst that k 1 is real and let " be the fundamental unit of k 1. We claim that p " 62 L. Suppose otherwise then k 1 ( p " ) is one of K 1, K 0 1 or K. If k 1 ( p " )=K 1, then K 0 1 = k 1 ( p "0 )andk = k1 ( p ""0 ). (ere and below x 0 = x.) This however cannot occur since by assumption "" 0 = ;1 implying that p ;1 2 L, a contradiction. Similarly, if k 1 ( p " )=K, then again p ;1 2 L. Thus p " 62 L, and E 1 = h;1 " i for some unit 2 E 1. Suppose that p u 2 L for some unit u 2 k1. Then L = K 1 ( p u ), contradicting our assumption that L=K 1 is essentially ramied. The same argument shows that p u0 62 L, hence either E L = h " 0 i and q 1 =1orE L = h " p u0 i for some unit u 2 k1 and q 1 =2. ere is a root of unity generating the torsion subgroup W L of E L. Next consider the case where k 1 is complex, and let " denote the fundamental unit of k 2. Then " stays fundamental in L by the argument above. Let be a fundamental unit in K 1. If became a square in L, then clearly L=K 1 could not be essentially ramied. Thus if we have q 1 4, then " = 2 is a square in L. Applying to this relation we nd that ;1 ="" 0 isasquareinl, contradicting the assumption that L does not contain p ;1. Proposition 7. Suppose that q 2 = 1. Then K 2 =k 2 is essentially ramied if and only if 2 =1 if K 2 =k 2 is not essentially ramied, then 2 = h[b]i, where K 2 = k 2 ( p ) and () =b 2. Proof. First notice that if K 2 =k 2 is not essentially ramied, then 2 6=1: in fact, in this case we have () = b 2, and if we had 2 = 1, then b would have to be principal, say b =(). This implies that = " 2 for some unit " 2 k 2, which in view of q 2 = 1 implies that " must be a square. But then would be a square, and this is impossible. Conversely, suppose 2 6=1. Let a be a nonprincipal ideal in k 2 of absolute norm a, and assume that a =() in K 2. Then 1;2 = for some unit 2 E 2, and similarly ;3 = 0, where 0 is a unit in E 0 2. But then 0 = 1+;2 ; 3 2 = N L=k = N L=k a = a 2 2 = 1 in L, where 2 = means equal up to a square in L. Thus 0 is a square in L, so our assumption that q 2 = 1 implies that 0

9 IMAGINARY QUADRATIC FIELDS 9 must be a square in k 2. The same argument show that = 0 is a square in k 2, hence we nd 2 k 2. Thus 1;2 is xed by 2 and so := 2 2 k 2. This gives K 2 = k 2 ( p ), hence K 2 =k 2 is not essentially ramied, and moreover, a b. From now on assume that k is one of the imaginary quadratic elds of type A) or B) as explained in the Introduction. Let k 1 = Q( p d 1 )andk 2 = Q( p d 2 d 3 ) in case A), and k 1 = Q( p d 3 )andk 2 = Q( p d 1 d 2 ) in case B). Then there exist two unramied cyclic quartic extensions of k which are D 4 over Q (see Proposition 2). Let us say a few words about their construction. Consider e.g. case B) by Redei's theory (see [12]), the C 4 -factorization d = d 1 d 2 d 3 implies that unramied cyclic quartic extensions of k = Q( p d ) are constructed by choosing a \primitive" solution (x y z) of d 1 d 2 X 2 + d 3 Y 2 = Z 2 and putting L = k( p d 1 d 2 p ) with = z+x p d 1 d 2 (primitive here means that should not be divisible by rational integers) the other unramied cyclic quartic extension is then el = k( p d 1 d 2 p d 1 ). Since 4 =(x p d 1 d 2 + y p d 3 + z) 2 for = 1 2 (z + yp d 3 ), we also have L = k( p d 3 p ) etc. If d 3 = ;4, then it is easy to see that we may choose as the fundamental unit of k 2 if d 3 6= ;4, then genus theory says that a) the class number h of k 2 is twice an odd number u and b) the prime ideal p 3 above d 3 in k 2 is in the principal genus, so p u 3 =( 3 ) is principal. Again it can be checked that = 3 for a suitable choice of the sign. Example. Consider the case d = ; here 3 = (3+2 p 10 ), and the positive sign is correct since 3+2 p 10 (1+ p 10 ) 2 mod 4 is primary. The minimal polynomial of p 3 is f(x) =x 4 ; 6x 2 ; 31: compare Table 1. The elds K 2 = k 2 ( p ) and e K 2 = k 2 ( p d 2 ) will play a dominant role in the proof below they are both contained in M = F ( p ) for F = k 2 ( p d 2 ), and it is the ambiguous class group Am(M=F) that contains the information we are interested in. Lemma 6. The eld F has odd class number (even in the strict sense), and we have # Am(M=F) j 2. In particular, Cl 2 (M) is cyclic (though possibly trivial). Proof. The class group in the strict sense of k 2 is cyclic of order 2 by Redei's theory [12] (since (d 2 =p 3 )=(d 3 =p 2 )=;1 in case A) and (d 1 =p 2 )=(d 2 =p 1 )=;1 incase B)). Since F is the ilbert class eld of k 2 in the strict sense, its class number in the strict sense is odd. Next we apply the ambiguous class number formula. In case A), F is complex, and exactly the two primes above d 3 ramify in M=F. Note that M = F ( p ) with primary of norm d 3 y 2 there are four primes above d 3 in F, and exactly two of them divide to an odd power, so t = 2 by the decomposition law in quadratic Kummer extensions. By Proposition 4 and the remarks following it, #Am 2 (M=F) =2=(E : ) 2, and Cl 2 (M) is cyclic. In case B), however, F is real since 2 k 2 has norm d 3 y 2 < 0, it has mixed signature, hence there are exactly two innite primes that ramify in M=F. As in case A), there are two nite primes above d 3 that ramify in M=F, so we get #Am 2 (M=F) =8=(E : ). Since F has odd class number in the strict sense, F has units of independent signs. This implies that the group of units that are positive at the two ramied innite primes has Z-rank 2, i.e. (E : ) 4by consideration of the innite primes alone. In particular, # Am 2 (M=F) 2 in case B).

10 10 Next we derive some relations between the class groups of K 2 and e K 2 these relations will allow us to use each of them as our eld K in Theorem 1. Proposition 8. Let L and e L be the two unramied cyclic quartic extensions of k, and let K 2 and e K 2 be two quadratic extensions of k 2 in L and e L,respectively, which are not normal over Q. a) We have 4 j h(k 2 ) if and only if 4 j h( e K 2 ) b) If 4 j h(k 2 ), then one of Cl 2 (K 2 ) or Cl 2 ( e K 2 ) has type (2 2), whereas the other is cyclic of order 4. Proof. Notice that the prime dividing disc(k 1 ) splits in k 2. Throughout this proof, let p be one of the primes of k 2 dividing disc(k 1 ). If we write K 2 = k 2 ( p ) for some 2 k 2, then e K 2 = k 2 ( p d 2 ). In fact, K 2 and e K 2 are the only extensions F=k 2 of k 2 with the properties 1. F=k 2 is a quadratic extension unramied outside p 2. kf=k is a cyclic extension. Therefore it suces to observe that if k 2 ( p ) has these properties, then so does k 2 ( p d 2 ). But this is elementary. In particular, the compositum M = K 2 e K2 = k 2 ( p d 2 p ) is an extension of type (2 2) over k 2 with subextensions K 2, e K2 and F = k 2 ( p d 2 ). Clearly F is the unramied quadratic extension of k 2, so both M=K 2 and M= e K 2 are unramied. If K 2 had 2-class number 2, then M would have odd class number, and M would also be the 2-class eld of e K 2. Thus 2 k h(k 2 ) implies that 2 k h( e K 2 ). This proves part a) of the proposition. Before we goon,wegive a asse diagram for the elds occurring in this proof: Now assume that 4 j h(k 2 ). M K 2 ek 2 k 2 N F 1 F 2 F Since Cl 2 (M) is cyclic by Lemma 6, there is a unique quadratic unramied extension N=M, and the uniqueness implies at once that N=k 2 is normal. ence G = Gal(N=k 2 ) is a group of order 8 containing a subgroup of type (2 2) ' Gal(N=F): in fact, if Gal(N=F) were cyclic, then the primes ramifying in M=F would also ramify in N=M contradicting the fact that N=M is unramied. There are three groups satisfying these conditions: G =(2 4), G =(2 2 2) and G = D 4. We claim that G is non-abelian once we have proved this, it follows that exactly one of the groups Gal(N=K 2 ) and Gal(N= e K 2 ) is cyclic, and that the other is not, which is what we want toprove. So assume that G is abelian. Then M=F is ramied at two nite primes q and q 0 of F dividing p (in k 2 ) if F 1 and F 2 denote the quadratic subextensions of N=F dierent from M then F 1 =F and F 2 =F must be ramied at a nite prime (since F has odd class number in the strict sense: see Lemma 6) since both F 1 and F 2 are normal (even abelian) over k 2, ramication at q implies ramication at the conjugated ideal q 0. ence both q and q 0 ramify in F 1 =F and F 2 =F, and since they

11 IMAGINARY QUADRATIC FIELDS 11 also ramify in M=F, they must ramify completely in N=F, again contradicting the fact that N=M is unramied. We have proved that Cl 2 (K 2 ) and Cl 2 ( e K 2 ) contain subgroups of type (4) and (2 2), respectively. Now we wish to apply Proposition 5. But we have to compute #Am 2 ( e K 2 =k 2 ). Since the class number of e K 2 is even, it is sucient toshow that #Am 2 ( e K 2 =k 2 ) 2. In case A), there is exactly one ramied prime (it divides d 1 ), hence #Am 2 ( e K 2 =k 2 ) = 2=(E : ) 2. In case B), there are two ramied primes (one is innite, the other divides d 3 ), hence #Am 2 ( e K 2 =k 2 ) =4=(E : ) but ;1 is not a norm residue at the ramied innite prime, hence (E : ) 2 and #Am 2 ( e K 2 =k 2 ) 2 as claimed. Now Proposition 5 implies that Cl 2 (K 2 ) is cyclic of order 4, and that Cl 2 ( e K 2 ) ' (2 2). This concludes our proof. Proposition 9. Assume that k is one of the imaginary quadratic elds of type A) or B) as explained in the Introduction. Then there exist two unramied cyclic quartic extensions of k. Let L be one of them, and write k 1 = Q( p d 1 ) and k 2 = Q( p d 2 d 3 ) in case A), and k 1 = Q( p d 3 ) and k 2 = Q( p d 1 d 2 ) in case B). Then h 2 (L) = 1 4 h 2(k)h 2 (K 1 )h 2 (K 2 ) unless possibly when d 3 = ;4 in case B). Proof. Observe that = 0 in case A) and B) Kuroda's class number formulas for L=k 1 and L=k 2 gives in case A) and h 2 (L) = q 1h 2 (K 1 ) 2 h 2 (K) 2h 2 (k 1 ) 2 h 2 (L) = q 1h 2 (K 1 ) 2 h 2 (K) 4h 2 (k 1 ) 2 = q 2h 2 (K 2 ) 2 h 2 (K) 4h 2 (k 2 ) 2 = q 2h 2 (K 2 ) 2 h 2 (K) 2h 2 (k 2 ) 2 in case B). Multiplying them together and plugging in the class number formula for K=Q yields h 2 (L) 2 = q 1 q 2 h 2 (K 1 ) 2 h 2 (K 2 ) 2 h 2 (k) 2 8 h 2 (k 1 ) 2 h 2 (k 2 ) 2 : Now h 2 (k 1 )=1,h 2 (k 2 )=2andq 1 q 2 =2(by Proposition 6), and taking the square root we nd h 2 (L) = 1 4 h 2(k)h 2 (K 1 )h 2 (K 2 ) as claimed. 5. Classification In this section we apply the results obtained in the last few sections to give a proof for Theorem 1. Proof of Theorem 1. Let L be one of the two cyclic quartic unramied extensions of k, and let N be the subgroup of Gal(k 2 =k) xing L. Then N satises the assumptions of Proposition 1, thus there are only the following possibilities: d(g 0 ) h 2 (L) h 2 (K 1 )h 2 (K 2 ) 1 2 m m m+2 8

12 12 ere, the rst two columns follow from Proposition 1, the last (which we do not claim to hold if d 3 = ;4 in case B)) is a consequence of the class number formula of Proposition 9. In particular, we have d(g 0 ) 3 if one of the class numbers h 2 (K 1 ) or h 2 (K 2 )isatleast8. Therefore it suces to examine the cases h 2 (K 2 ) = 2 and h 2 (K 2 ) = 4 (recall from above that h 2 (K 2 )isalways even). We start by considering case A) it is sucient to show that h 2 (K 1 )h 2 (K 2 ) 6= 4. We now apply Proposition 5 notice that we maydosoby the proof of Proposition 8. a) If h 2 (K 2 ) = 2, then # 2 = 2 by Proposition 5, hence q 2 = 2 by Proposition 7 and then q 1 = 1 by Proposition 6. The class number formulas in the proof of Proposition 9 now give h 2 (K 1 ) = 1 and h 2 (L) =2 m. It can be shown using the ambiguous class number formula that Cl 2 (K 1 ) is trivial if and only if " 1 is a quadratic nonresidue modulo the prime ideal over d 2 in k 1 by Scholz's reciprocity law, this is equivalent to(d 1 =d 2 ) 4 (d 2 =d 1 ) 4 = 1, and this agrees with the criterion given in [1]. b) If h 2 (K 2 ) = 4, we may assume that Cl 2 (K 2 ) = (4) from Proposition 8.b). Then # 2 =2by Proposition 5, q 2 =2by Proposition 7 and q 1 =1by Proposition 6. Using the class number formula we get h 2 (K 1 )=2andh 2 (L) =2 m+2. Thus in both cases wehave h 2 (K 1 )h 2 (K 2 ) 6= 4, and by the table at the beginning of this proof this implies that rank Cl 2 (k 1 ) 6= 2 in case A). Next we consider case B) here we have to distinguish between d 3 6= ;4 (case B 1 )andd 3 = ;4 (case B 2 ). Let us start with case B 1 ). a) If h 2 (K 2 )=2,then # 2 =2,q 2 = 2 and q 1 =1asabove. The class number formula gives h 2 (K 1 ) = 2 and h 2 (L) =2 m+1. b) If Cl 2 (K 2 ) = (4) (which wemay assume without loss of generalityby Proposition 8.b)) then # 2 = 2, q 2 = 2 and q 1 = 1, again exactly as above. This implies h 2 (K 1 )=4andh 2 (L) =2 m+3. Finally, consider case B 2 ). ere we apply Kuroda's class number formula (see [10]) to L=k 1, and since h 2 (k 1 )= 1andh 2 (K 1 )=h 2 (K 0 1), we get h 2 (L) = 1 2 q 1h 2 (K 1 ) 2 h 2 (k) =2 m q 1 h 2 (K 1 ) 2. From K 2 = k 2 ( p " ) (for a suitable choice of L the other possibility is e K 2 = k 2 ( p d 2 " )), where " is the fundamental unit of k 2, we deduce that the unit ", which still is fundamental in k, becomes a square in L, and this implies that q 1 2. Moreover, we have K 1 = k 1 ( p ), where 1mod4 are prime factors of d 1 and d 2 in k 1 = Q(i), respectively. This shows that K 1 has even class number, because K 1 ( p )=K 1 is easily seen to be unramied. Thus 2 j q 1,2j h 2 (K 1 ), and so we nd that h 2 (L) is divisible by 2 m 24=2 m+3. In particular, we always have d(g 0 ) 3 in this case. This concludes the proof. The referee (whom we'd like to thank for a couple of helpful remarks) asked whether h 2 (K) = 2 and h 2 (K) > 2 innitely often. Let us show how to prove that both possibilities occur with equal density. Before we can do this, we have to study the quadratic extensions K 1 and K e 1 of k 1 more closely. We assume that d 2 = p and d 3 = r are odd primes in the following, and then sayhow to modify the arguments in the case d 2 =8ord 3 = ;8. The primes p and r split in k 1 as po 1 = pp 0 and ro 1 = rr 0. Let h denote the

13 IMAGINARY QUADRATIC FIELDS 13 odd class number of k 1 and write p h = () and r h = () for primary elements and (this is can easily be proved directly, but it is also a very special case of ilbert's rst supplementary law for quadratic reciprocity in elds K with odd class number h (see [7]): if a h = O K for an ideal p with odd norm, then can be chosen primary (i.e. congruent toa square mod 4O K ) if and only if a is primary (i.e. ["=a] = +1 for all units " 2O K, where [ = ] denotes the quadratic residue symbol in K)). Let [ = ] denote the quadratic residue symbol in k 1. Then [=][ 0 =] =[p=] =(p=r) =;1, so we maychoose the conjugates in such away that [=] =+1and[ 0 =] =[= 0 ]=;1. Put K 1 = k 1 ( p ) and e K 1 = k 1 ( p 0 ) we claim that h2 ( e K 1 ) = 2. This is equivalent to h 2 ( e L 1 ) = 1, where e L 1 = k 1 ( p p 0 ) is a quadratic unramied extension of e K 1. Put e F 1 = k 1 ( p ) and apply the ambiguous class number formula to e F 1 =k 1 and e L 1 = e F 1 : since there is only one ramied prime in each ofthese two extensions, we nd Am( e F 1 =k 1 ) = Am( e L 1 = e F 1 ) = 1 note that we have used the assumption that [= 0 ]=;1 in deducing that r 0 is inert in e F 1 =k 1. In our proof of Theorem 1 we have seen that there are the following possibilities when h 2 (K 2 ) j 4: q 2 Cl 2 (K 2 ) q 1 h 2 (K 1 ) eq 2 Cl 2 ( e K 2 ) h 2 (L) 2 (2) (2) 2 m+1 2 (4) 1 4? (2 2) 2 m+3 In order to decide whether eq 2 =1oreq 2 = 2, recall that we have h 2 (K 1 )=4 thus ek 1 must be the eld with 2-class number 2, and this implies h 2 ( L) e =2 m+2 and eq 2 =1. In particular we see that 4 j h 2 (K 2 ) if and only if 4 j h 2 (K 1 ) as long as K 1 = k 1 ( p ) with [=] = +1. The ambiguous class number formula shows that Cl 2 (K 1 ) is cyclic, thus 4 j h 2 (K 1 ) if and only if 2 j h 2 (L 1 ), where L 1 = K 1 ( p ) is the quadratic unramied extension of K 1. Applying the ambiguous class number formula to L 1 =F 1, where F 1 = k 1 ( p ), we see that 2 j h 2 (L 1 ) if and only if (E : ) = 1. Now E is generated by a root of unity (which always is a norm residue at primes dividing r 1mod4) and a fundamental unit ". Therefore (E : ) = 1 if and only if f"=r 1 g = f"=r 2 g = +1, where ro F1 = R 1 R 2 and where f= g denotes the quadratic residue symbol in F 1. Since f"=r 1 gf"=r 2 g =["=r] = +1, we have proved that 4 j h 2 (K 1 ) if and only if the prime ideal R 1 above r splits in the quadratic extension F 1 ( p " ). But if we xp and q, this happens for exactly half of the values of r satisfying (p=r) =;1, (q=r) = +1. If d 2 = 8 and p = 2, then 2O k1 = 0 22, and we have tochoose 2 h =() in such a way that k 1 ( p )=k 1 is unramied outside p. The residue symbols [=2] are dened as Kronecker symbols via the splitting of 2 in the quadratic extension k 1 ( p )=k 1. With these modifactions, the above arguments remain valid. References [1] E. Benjamin, F. Lemmermeyer, C. Snyder, Imaginary Quadratic Fields k with Cyclic Cl 2 (k 1 ), J. Number Theory 67 (1997), 229{245. [2] N. Blackburn, On Prime-Power Groups in which the Derived Group has Two Generators, Proc. Cambridge Phil. Soc. 53 (1957), 19{27.

14 14 [3] N. Blackburn, On Prime Power Groups with Two Generators, Proc. Cambridge Phil. Soc. 54 (1958), 327{337. [4] G. Gras, Sur les `-classes d'ideaux dans les extensions cycliques relatives de degre premier `, Ann. Inst. Fourier 23 (1973), 1{48. [5] P. all, A Contribution to the Theory of Groups of Prime Power Order, Proc. London Math. Soc. 36 (1933), 29{95. [6]. asse, Bericht uber neuere Untersuchungen und Probleme aus der Theorie der algebraischen Zahlkorper, Teil II: Reziprozitatsgesetz, Physica Verlag, Wurzburg [7] D. ilbert, Uber die Theorie des relativ-quadratischen Zahlkorpers, Math. Ann. 51 (1899), 1{127. [8] B. uppert, Endliche Gruppen I, Springer Verlag, eidelberg, [9] S. Lang, Cyclotomic Fields. I, II, Springer Verlag [10] F. Lemmermeyer, Kuroda's Class Number Formula, Acta Arith (1994), 245{260. [11] M. Moriya, Uber die Klassenzahl eines relativzyklischen Zahlkorpers von Primzahlgrad, Jap. J. Math. 10 (1933), 1{18. [12] L. Redei,. Reichardt, Die Anzahl der durch 4 teilbaren Invarianten der Klassengruppe eines beliebigen quadratischen Zahlkorpers, J. Reine Angew. Math. 170 (1933), 69{74. [13] B. Schmithals, Konstruktion imaginarquadratischer Korper mit unendlichem Klassenkorperturm, Arch. Math. 34 (1980), 307{312. (E. Benjamin) Mathematics Department, Unity College address: benjamin@mint.net (F. Lemmermeyer) Erwin-Rohde-Str. 19, eidelberg, Germany address: hb3@ix.urz.uni-heidelberg.de (C. Snyder) Department of Mathematics and Statistics, University of Maine, and, Research Institute of Mathematics, Orono address: snyder@gauss.umemat.maine.edu