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Normal integral bases and strict ray class groups modulo 4
Journal of Number Theory 101 (2003) 131–137
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Normal integral bases and strict ray class groups modulo 4 Fuminori Kawamoto Department of Mathematics, Faculty of Science, Gakushuin University, 1-5-1 Mejiro Toshima-ku, Tokyo 171-8588, Japan Received 25 February 2002; revised 15 March 2002 Communicated by D. Goss
Abstract Let F be a number field. We construct three tamely ramified quadratic extensions Ki =F ð1pip3Þ which are ramified at most at some given set of finite primes, such that K3 CK1 K2 ; both K1 =F and K2 =F have normal integral bases, but K3 =F has no normal integral basis. Since Hilbert–Speiser’s theorem yields that every finite and tamely ramified abelian extension over the field of rational numbers has a normal integral basis, it seems that this example is interesting (cf. [5] J. Number Theory 79 (1999) 164; Theorem 2). As we shall explain below, the previous papers (Acta Arith. 106 (2) (2003) 171–181; Abh. Math. Sem. Univ. Hamburg 72 (2002) 217–233) motivated the construction. We prove that if the class number of F is bigger than 1; or the strict ray class group Cl4l0 of F modulo 4 has an element of order X3; then there exist infinitely many triplets ðK1 ; K2 ; K3 Þ of such fields. r 2003 Elsevier Science (USA). All rights reserved. Keywords: Normal integral basis; Ray class group
1. Introduction Let F be a number field, that is, a finite extension of the field Q of rational numbers, and oF the ring of integers in F : For a finite abelian extension K=F with Galois group G ¼ GalðK=F Þ; if there exists some x in oK such that fsðxÞgsAG is a free oF -basis of oK ; then we say that K=F has a normal integral basis (abbreviated NIB). Such an element x is called a generator of NIB of K=F : Let m be the conductor of E-mail address: [email protected]. 0022-314X/03/$ - see front matter r 2003 Elsevier Science (USA). All rights reserved. doi:10.1016/S0022-314X(03)00010-6
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K=F and F ðmÞ denote the ray class field of F mod m: We asked in Kawamoto and Odai [9] whether K=F has an NIB. If K=F has an NIB, then it is tamely ramified by E. Noether’s result; therefore, m is a square-free integral divisor (i.e., the finite component of m is square-free). Let l0 ¼ l0 ðF Þ denote the product of all real primes of F : In the case where F is a totally real number field and m j l0 ; using Brinkhuis [1, Corollary 2.10] (or [2, Corollary 2.1]) and Childs [3, Theorem B], we proved in [9] the following: (I) There exists a maximal subextension Lm =F of F ðmÞ=F which has an NIB; (II) Lm =F is an elementary abelian 2-extension that is obtained by adjoining the square root of some units of F ; and for any subextension K=F of F ðmÞ=F ; we have that K=F has an NIB if and only if K is contained in Lm (here we recall that if L=K=F is a tower of abelian extensions and L=F has an NIB, then K=F also has an NIB). Furthermore, the field Lm and a generator of NIB of Lm =F were determined for all real quadratic fields and all cyclic cubic fields in [9], and for an infinite family of totally real number fields of general degree in [7]. Consequently, for m[l0 ; it might be natural to ask whether there exists a maximal subextension of F ðmÞ=F which has an NIB. The purpose of this paper is to show that such a maximal subextension does not always exist. Let A4l0 denote the group of all fractional ideals of oF relatively prime to 4; and S4l0 the subgroup of A4l0 of all principal ideals xoF such that xAF and x 1 mod 4l0 ; that is, x 1 mod 4 and x is totally positive. We put Cl4l0 :¼ A4l0 =S4l0 ; and let hF be the class number of F ; that is, the order of Cl ¼ A1 =S1 : We shall prove the following in Section 2,
Theorem 1. Let F be a number field and l0 the product of all real primes of F : Suppose that hF 41; or the ray class group Cl4l0 of F mod 4l0 is not an elementary abelian 2group. Then there exist infinitely many triplets ðp1 ; p2 ; p3 Þ of prime ideals of oF such that for all integral divisors m of F with p1 p2 p3 j m (i.e., m[l0 Þ; there does not exist a maximal subextension of F ðmÞ=F which has an NIB. In order to show Theorem 1, we actually construct three quadratic subextensions K1 ; K2 and K3 of F ðp1 p2 p3 Þ=F such that K3 CK1 K2 ; both K1 =F and K2 =F have NIB, but K3 =F has no NIB (Proposition 3). In Section 3, we give some numerical examples of these fields when F is a quadratic field.
2. The proof of Theorem 1 For the proof of Theorem 1, we utilize the following lemma (cf. [13, Section 3]).
Lemma 2. Let K=F be a quadratic extension of number fields, and p1 ; y; pu ðuX0Þ all prime ideals of oF which are ramified in K: Then K=F has an NIB if and only if there pffiffiffi exists some d in oF such that K ¼ F ð d Þ; d 1 mod 4; and doF ¼ p1 ?pu :
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Furthermore, if the condition is satisfied then fð1 þ basis of oK :
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pffiffiffi pffiffiffi d Þ=2; ð1 d Þ=2g is a free oF -
For aAA4l0 ; we denote by ½a the ideal class of a in Cl4l0 : A similar argument as in the proof of Theorem 4 of [12] implies: Proposition 3. Let pi ð1pip3Þ be three distinct prime ideals of oF with pi [2 that satisfy ½p1 ¼ ½p2 ¼ ½p3 1 in Cl4l0 : So, let di ði ¼ 1; 2Þ be integers in oF such that pffiffiffiffiffi pffiffiffiffiffi di 1 mod 4l0 ; d1 oF ¼ p1 p3 ; and d2 oF ¼ p2 p3 : We put K1 :¼ F ð d1 Þ; K2 :¼ F ð d2 Þ; pffiffiffiffiffiffiffiffiffi and K3 :¼ F ð d1 d2 Þ: Suppose that p3 is not a principal ideal of oF ; or ½p3 is an element of order X3: Then, both quadratic subextensions K1 =F and K2 =F of F ðp1 p2 p3 Þ=F have NIB, but K3 =F has no NIB. Furthermore, K3 =F has a relative integral basis (abbreviated RIB) if and only if p3 is principal. Proof. Let i ¼ 1; or 2: Since di oF is not square, Ki =F is a quadratic extension. From the ramification theory in Kummer extensions, pi and p3 are only prime ideals of oF which are ramified in Ki : Hence, Ki =F has an NIB by Lemma 2. The condition that di 1 mod l0 implies that all real primes of F are unramified in Ki : As both pi and p3 are tamely ramified in Ki ; it is a subfield of F ðpi p3 Þ: Since p1 is ramified in K1 ; but unramified in K2 ; we have K1 -K2 ¼ F : Therefore, K1 K2 =F is an abelian extension of type ð2; 2Þ; and K3 =F is the unique quadratic subextension of K1 K2 =F such that K3 aKi ði ¼ 1; 2Þ: As d1 d2 oF ¼ p1 p2 p23 and d1 d2 1 mod 4l0 ; the ramification theory in Kummer extensions yields that p1 and p2 are only primes of oF which are ramified in K3 : Assume that K3 =F has an NIB. Then it follows from Lemma 2 that there exists pffiffiffi some d in oF such that K3 ¼ F ð d Þ; d 1 mod 4 and doF ¼ p1 p2 : Since all real primes of F are unramified in K3 ; we have d 1 mod l0 : Hence, p1 p2 is in S4l0 : On the other hand, we have p1 p2 p23 AS4l0 : Therefore we obtain ½p3 2 ¼ 1: Also, as pffiffiffi pffiffiffiffiffiffiffiffiffi F ð d1 d2 Þ ¼ F ð d Þ; there exists some a in F such that d1 d2 ¼ da2 : Hence, p3 ¼ aoF : Thus, if p3 is not principal, or ½p3 is an element of order X3 then K3 =F has no NIB. pffiffiffi For brevity, we put b :¼ d1 d2 and x :¼ ð b bÞ=2: Since p1 and p2 are tamely ramified in K3 =F ; we have dK3 =F ¼ p1 p2 ; where dK3 =F is the discriminant of K3 =F : If p3 ¼ aoF with some a in oF ; then we have dK3 =F ¼ ðb=a2 ÞoF : The trace and norm of x=a in K3 =F are equal to b=a and ðb=a2 Þ ððb 1Þ=4Þ; respectively, which are in oF : Hence, x=a is an integer. Also, we obtain dK3 =F ¼ dK3 =F ð1; x=aÞoF ; where dK3 =F ð1; x=aÞ denotes the discriminant of 1; x=a in K3 =F : Consequently, f1; x=ag is a free oF -basis of oK3 : Note that x is an integer, as b 1 mod 4: Conversely, if fo1 ; o2 g is a free oF -basis of oK3 ; then there exists a matrix A in M2 ðoF Þ such that ð1; xÞ ¼ ðo1 ; o2 ÞA: Therefore, we have p1 p2 p23 ¼ boF ¼ dK3 =F ð1; xÞoF ¼ dK3 =F ðdet AÞ2 ; so that p3 ¼ ðdet AÞoF : Hence, p3 is principal. This proves our proposition. &
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Remark 2.1. If p3 is principal and ½p3 is an element of order X3; then Proposition 3 implies that K3 =F has an RIB, but no NIB. Corollary 4. Let the assumption be as in Proposition 3. Then, for all integral divisors m of F with p1 p2 p3 j m; there does not exist a maximal subextension of F ðmÞ=F which has an NIB. Proof. Let K1 ; K2 and K3 be the fields as in Proposition 3. Since K1 K2 is a subfield of F ðp1 p2 p3 Þ; it is also a subfield of F ðmÞ: Assume that there exists a maximal subextension L=F of F ðmÞ=F which has an NIB. Then, ðK1 K2 -LÞ=F is a maximal subextension of K1 K2 =F which has an NIB. For i ¼ 1; 2; since Ki =F has an NIB, we have Ki CK1 K2 -L: Therefore, we obtain K3 CK1 K2 CK1 K2 -L: Hence, K3 =F has an NIB. This is a contradiction. Our corollary is proved. &
Proof of Theorem 1. If hF 41 (resp., Cl4l0 is not an elementary abelian 2-group), then there is some ideal a of oF that is not principal (resp. some element ½a of order X3 in Cl4l0 ). By the Dirichlet density theorem, there exist infinitely many triplets ðp1 ; p2 ; p3 Þ of prime ideals of oF not dividing 2; such that both p1 and p2 are in ½a ; and p3 is in ½a 1 : Take such a triplet ðp1 ; p2 ; p3 Þ: Then we have ½p1 ¼ ½p2 ¼ ½p3 1 ; and p3 is not principal (resp., ½p3 is of order X3). Hence, Corollary 4 implies Theorem 1. &
3. Numerical examples In this section we consider whether Cl4l0 is an elementary abelian 2-group. There exists a natural exact sequence of abelian groups: 1-ðSl0 -P4l0 Þ=S4l0 -Cl4l0 -A4l0 =ðSl0 -P4l0 Þ-1; where P4l0 denotes the subgroup of A4l0 of all principal ideals relatively prime to 4: The last group is naturally isomorphic to the strict ideal class group Cl þ ¼ Al0 =Sl0 of F by the strong approximation theorem (cf. [4, Proposition 1.2.8]). For a ring R; we denote by R the group of units in R; and put G :¼ ðoF =4oF Þ : Since each residue class in G is represented by a totally positive element x of oF with ðx; 2Þ ¼ 1 from the above theorem, we can define a natural group homomorphism by f : G-ðSl0 -P4l0 Þ=S4l0 ;
½x /½xoF ;
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where ½x denotes the residue class of x in G: If El0 is the subgroup of all totally positive units in o F ; and we put H :¼ ðEl0 þ 4oF Þ=4oF ; then we have Ker f ¼ H: Also, it is easy to see that f is surjective. Consequently, G=H is naturally isomorphic to ðSl0 -P4l0 Þ=S4l0 : We obtain the following exact sequence: 1-G=H-Cl4l0 -Cl þ -1: Hence, if either Cl þ ; or G=H is not an elementary abelian 2-group, then we see that Cl4l0 is not an elementary abelian 2-group. In particular, if hF is divisible by an odd prime, then Cl4l0 is not an elementary abelian 2-group. Example 3.1. If F ¼ Q then we have El0 ¼ f1g; so that H ¼ f1g: Since G is a cyclic group of order 2 and Cl þ ¼ f1g; we see that Cl4l0 is a cyclic group of order 2: Therefore, the assumption of Theorem 1 is not satisfied. Horie and Horie [6, Corollary 3] proved that there are only finitely many imaginary abelian fields of 2-power degrees with ideal class groups Cl þ of exponents p2 (cf. [6, Introduction] and [11]). Consequently, in the case where F is an imaginary abelian field of 2-power degree, we see that Cl4l0 is not an elementary abelian 2-group (and hF 41 holds) except for finitely many fields of this kind. pffiffiffiffi From now on, let F ¼ Qð mÞ be a quadratic field, where m ða0; 1Þ is a squarefree integer, and we shall give some numerical examples of fields K1 ; K2 and K3 which are stated in Proposition 3. We assume that there exists an element ½x H of order X3 in G=H such that x is totally positive. (When F is a quadratic field, we can give a necessary and sufficient condition for G=H to be not an elementary abelian 2-group, which will be described elsewhere.) Since G=H is naturally isomorphic to ðSl0 -P4l0 Þ=S4l0 ; we can choose some triplets ðp1 oF ; p2 oF ; p3 oF Þ of principal prime ideals in Sl0 -P4l0 ; such that each pi ð1pip3Þ is totally positive, both ½p1 and ½p2 are in ½x H; and ½p3 is in ½x 1 H: (The infinitude of such triplets follows from the Dirichlet density theorem.) Then we obtain integers d1 ¼ p1 p3 and d2 ¼ p2 p3 ; pffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffi satisfying the property as in Proposition 3: K1 ¼ F ð p1 p3 Þ; K2 ¼ F ð p2 p3 Þ and pffiffiffiffiffiffiffiffiffiffi K3 ¼ F ð p1 p2 Þ: Hence, both K1 =F and K2 =F have NIB; K3 =F has an RIB, but no NIB. pffiffiffiffi We treat the case where m 5 mod 8 by way of example. Let o :¼ ð1 þ mÞ=2 and M :¼ ðm 1Þ=4: The canonical basis for oF is given by f1; og: If we put a :¼ ½ 1 ; b :¼ ½1 þ 2o and g :¼ ½M þ o then G ¼ /aS /bS /gS is an abelian group of type ð2; 2; 3Þ (cf. Lemma 6.6 and its proof in [8]). Let s be the generator of GalðF =QÞ: Since s naturally acts on G; H and ðSl0 -P4l0 Þ=S4l0 ; the map f becomes a /sS-homomorphism. Also, sðaÞ ¼ a; sðbÞ ¼ ab and sðgÞ ¼ g2 : The following numerical examples are calculated by using UBASIC [10]. Example 3.2. Let m ¼ 11: Then, M ¼ 3 1 mod 4; hF ¼ 1; so that Cl þ is a trivial group. For p ¼ 5; 37; 89; 97; 137; y; the prime p is decomposed in F ; if we put p ¼ 1 þ o; 5 3o; 7 þ 5o; 7 3o; 5 7o; y for each p; then ½p ¼ g; and poF is
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a prime ideal of oF lying above p: Since El0 ¼ o F ¼ f71g; we have H ¼ /aS; therefore, the element gH is of order 3: We take such distinct prime elements p1 and p2 ; and put p3 :¼ sðp1 Þ: Then, ½p1 H ¼ ½p2 H ¼ gH: As gsðgÞAH; we obtain ½p3 H ¼ sðgÞH ¼ g 1 H (of course, we may choose a prime element p3 which is in g 1 ). If m ¼ 35 ¼ 5 7 then M ¼ 9 3 mod 4; H ¼ /aS; Cl þ is a cyclic group of order 2: For p ¼ 389; 449; 541; 1061; 1201; yðp ¼ 19 þ o; 1 7o; 23 3o; 7 11o; 35 3o; yÞ; we have ½p ¼ g: Then the same thing as above is true. Example 3.3. Let m ¼ 37: Then, M ¼ 9 1 mod 4; the fundamental unit e ¼ 5 þ 2o is not totally positive; hF ¼ 1; so that Cl þ is a trivial group. For p ¼ 41; 53; 73; 101; 181; y; the prime p is decomposed in F ; if we put p ¼ 79 þ 31o; 19 þ 7o; 11 o; 191 þ 75o; y for each p; then p is totally positive, ½p ¼ ag; and poF is a prime ideal of oF lying above p: Since El0 ¼ /e2 S and e2 1 mod 4; we have H ¼ f1g; therefore, the element agH is of order 6: We take such distinct prime elements p1 and p2 ; and put p3 :¼ sðp1 Þ: Then, ½p1 H ¼ ½p2 H ¼ agH: As agsðagÞAH; we obtain ½p3 H ¼ sðagÞH ¼ ðagÞ 1 H: If m ¼ 485 ¼ 5 97 then M ¼ 121 1 mod 4; e ¼ 21 þ 2o is not totally positive; hF ¼ 2; so that H ¼ f1g and Cl þ is a cyclic group of order 2: For p ¼ 89; 241; 389; 809; 829; yðp ¼ 15 o; 35 þ 3o; 151 13o; 31 o; 119 þ 11o; yÞ; we have ½p ¼ ag: Then the same thing as above is true.
Acknowledgments The author is grateful to Humio Ichimura for his kind suggestions.
References [1] J. Brinkhuis, Unramified abelian extensions of CM-fields and their Galois module structure, Bull. London Math. Soc. 24 (1992) 236–242. [2] J. Brinkhuis, On the Galois module structure over CM-fields, Manuscripta Math. 75 (1992) 333–347. [3] L. Childs, The group of unramified Kummer extensions of prime degree, Proc. London Math. Soc. 35 (3) (1977) 407–422. [4] H. Cohen, Advanced Topics in Computational Number Theory, in: Graduate Texts in Mathematics, Vol. 193, Springer, New York/Berlin, 2000. [5] C. Greither, D.R. Replogle, K. Rubin, A. Srivastav, Swan modules and Hilbert–Speiser number fields, J. Number Theory 79 (1999) 164–173. [6] K. Horie, M. Horie, CM-fields and exponents of their ideal class groups, Acta Arith. 55 (2) (1990) 157–170. [7] H. Ichimura, F. Kawamoto, An infinite family of totally real number fields, Acta Arith. 106 (2) (2003) 171–181. [8] F. Kawamoto, On quadratic subextensions of ray class fields of quadratic fields mod p; J. Number Theory 86 (2001) 1–38. [9] F. Kawamoto, Y. Odai, Normal integral bases of N-ramified abelian extensions of totally real number fields, Abh. Math. Sem. Univ. Hamburg 72 (2002) 217–233.
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[10] Y. Kida, UBASIC calculator, Version 8.8e. [11] S. Louboutin, Determination of all non-quadratic imaginary cyclic number fields of 2-power degrees with ideal class groups of exponent p2; Math. Comp. 64 (209) (1995) 323–340. [12] H. Mann, On integral bases, Proc. Amer. Math. Soc. 9 (1958) 167–172. [13] R. Massy, Bases normales d’entiers relatives quadratiques, J. Number Theory 38 (1991) 216–239.
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Normal integral bases and strict ray class groups modulo 4 Fuminori Kawamoto Department of Mathematics, Faculty of Science, Gakushuin University, 1-5-1 Mejiro Toshima-ku, Tokyo 171-8588, Japan Received 25 February 2002; revised 15 March 2002 Communicated by D. Goss
Abstract Let F be a number field. We construct three tamely ramified quadratic extensions Ki =F ð1pip3Þ which are ramified at most at some given set of finite primes, such that K3 CK1 K2 ; both K1 =F and K2 =F have normal integral bases, but K3 =F has no normal integral basis. Since Hilbert–Speiser’s theorem yields that every finite and tamely ramified abelian extension over the field of rational numbers has a normal integral basis, it seems that this example is interesting (cf. [5] J. Number Theory 79 (1999) 164; Theorem 2). As we shall explain below, the previous papers (Acta Arith. 106 (2) (2003) 171–181; Abh. Math. Sem. Univ. Hamburg 72 (2002) 217–233) motivated the construction. We prove that if the class number of F is bigger than 1; or the strict ray class group Cl4l0 of F modulo 4 has an element of order X3; then there exist infinitely many triplets ðK1 ; K2 ; K3 Þ of such fields. r 2003 Elsevier Science (USA). All rights reserved. Keywords: Normal integral basis; Ray class group
1. Introduction Let F be a number field, that is, a finite extension of the field Q of rational numbers, and oF the ring of integers in F : For a finite abelian extension K=F with Galois group G ¼ GalðK=F Þ; if there exists some x in oK such that fsðxÞgsAG is a free oF -basis of oK ; then we say that K=F has a normal integral basis (abbreviated NIB). Such an element x is called a generator of NIB of K=F : Let m be the conductor of E-mail address: [email protected]. 0022-314X/03/$ - see front matter r 2003 Elsevier Science (USA). All rights reserved. doi:10.1016/S0022-314X(03)00010-6
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F. Kawamoto / Journal of Number Theory 101 (2003) 131–137
K=F and F ðmÞ denote the ray class field of F mod m: We asked in Kawamoto and Odai [9] whether K=F has an NIB. If K=F has an NIB, then it is tamely ramified by E. Noether’s result; therefore, m is a square-free integral divisor (i.e., the finite component of m is square-free). Let l0 ¼ l0 ðF Þ denote the product of all real primes of F : In the case where F is a totally real number field and m j l0 ; using Brinkhuis [1, Corollary 2.10] (or [2, Corollary 2.1]) and Childs [3, Theorem B], we proved in [9] the following: (I) There exists a maximal subextension Lm =F of F ðmÞ=F which has an NIB; (II) Lm =F is an elementary abelian 2-extension that is obtained by adjoining the square root of some units of F ; and for any subextension K=F of F ðmÞ=F ; we have that K=F has an NIB if and only if K is contained in Lm (here we recall that if L=K=F is a tower of abelian extensions and L=F has an NIB, then K=F also has an NIB). Furthermore, the field Lm and a generator of NIB of Lm =F were determined for all real quadratic fields and all cyclic cubic fields in [9], and for an infinite family of totally real number fields of general degree in [7]. Consequently, for m[l0 ; it might be natural to ask whether there exists a maximal subextension of F ðmÞ=F which has an NIB. The purpose of this paper is to show that such a maximal subextension does not always exist. Let A4l0 denote the group of all fractional ideals of oF relatively prime to 4; and S4l0 the subgroup of A4l0 of all principal ideals xoF such that xAF and x 1 mod 4l0 ; that is, x 1 mod 4 and x is totally positive. We put Cl4l0 :¼ A4l0 =S4l0 ; and let hF be the class number of F ; that is, the order of Cl ¼ A1 =S1 : We shall prove the following in Section 2,
Theorem 1. Let F be a number field and l0 the product of all real primes of F : Suppose that hF 41; or the ray class group Cl4l0 of F mod 4l0 is not an elementary abelian 2group. Then there exist infinitely many triplets ðp1 ; p2 ; p3 Þ of prime ideals of oF such that for all integral divisors m of F with p1 p2 p3 j m (i.e., m[l0 Þ; there does not exist a maximal subextension of F ðmÞ=F which has an NIB. In order to show Theorem 1, we actually construct three quadratic subextensions K1 ; K2 and K3 of F ðp1 p2 p3 Þ=F such that K3 CK1 K2 ; both K1 =F and K2 =F have NIB, but K3 =F has no NIB (Proposition 3). In Section 3, we give some numerical examples of these fields when F is a quadratic field.
2. The proof of Theorem 1 For the proof of Theorem 1, we utilize the following lemma (cf. [13, Section 3]).
Lemma 2. Let K=F be a quadratic extension of number fields, and p1 ; y; pu ðuX0Þ all prime ideals of oF which are ramified in K: Then K=F has an NIB if and only if there pffiffiffi exists some d in oF such that K ¼ F ð d Þ; d 1 mod 4; and doF ¼ p1 ?pu :
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Furthermore, if the condition is satisfied then fð1 þ basis of oK :
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pffiffiffi pffiffiffi d Þ=2; ð1 d Þ=2g is a free oF -
For aAA4l0 ; we denote by ½a the ideal class of a in Cl4l0 : A similar argument as in the proof of Theorem 4 of [12] implies: Proposition 3. Let pi ð1pip3Þ be three distinct prime ideals of oF with pi [2 that satisfy ½p1 ¼ ½p2 ¼ ½p3 1 in Cl4l0 : So, let di ði ¼ 1; 2Þ be integers in oF such that pffiffiffiffiffi pffiffiffiffiffi di 1 mod 4l0 ; d1 oF ¼ p1 p3 ; and d2 oF ¼ p2 p3 : We put K1 :¼ F ð d1 Þ; K2 :¼ F ð d2 Þ; pffiffiffiffiffiffiffiffiffi and K3 :¼ F ð d1 d2 Þ: Suppose that p3 is not a principal ideal of oF ; or ½p3 is an element of order X3: Then, both quadratic subextensions K1 =F and K2 =F of F ðp1 p2 p3 Þ=F have NIB, but K3 =F has no NIB. Furthermore, K3 =F has a relative integral basis (abbreviated RIB) if and only if p3 is principal. Proof. Let i ¼ 1; or 2: Since di oF is not square, Ki =F is a quadratic extension. From the ramification theory in Kummer extensions, pi and p3 are only prime ideals of oF which are ramified in Ki : Hence, Ki =F has an NIB by Lemma 2. The condition that di 1 mod l0 implies that all real primes of F are unramified in Ki : As both pi and p3 are tamely ramified in Ki ; it is a subfield of F ðpi p3 Þ: Since p1 is ramified in K1 ; but unramified in K2 ; we have K1 -K2 ¼ F : Therefore, K1 K2 =F is an abelian extension of type ð2; 2Þ; and K3 =F is the unique quadratic subextension of K1 K2 =F such that K3 aKi ði ¼ 1; 2Þ: As d1 d2 oF ¼ p1 p2 p23 and d1 d2 1 mod 4l0 ; the ramification theory in Kummer extensions yields that p1 and p2 are only primes of oF which are ramified in K3 : Assume that K3 =F has an NIB. Then it follows from Lemma 2 that there exists pffiffiffi some d in oF such that K3 ¼ F ð d Þ; d 1 mod 4 and doF ¼ p1 p2 : Since all real primes of F are unramified in K3 ; we have d 1 mod l0 : Hence, p1 p2 is in S4l0 : On the other hand, we have p1 p2 p23 AS4l0 : Therefore we obtain ½p3 2 ¼ 1: Also, as pffiffiffi pffiffiffiffiffiffiffiffiffi F ð d1 d2 Þ ¼ F ð d Þ; there exists some a in F such that d1 d2 ¼ da2 : Hence, p3 ¼ aoF : Thus, if p3 is not principal, or ½p3 is an element of order X3 then K3 =F has no NIB. pffiffiffi For brevity, we put b :¼ d1 d2 and x :¼ ð b bÞ=2: Since p1 and p2 are tamely ramified in K3 =F ; we have dK3 =F ¼ p1 p2 ; where dK3 =F is the discriminant of K3 =F : If p3 ¼ aoF with some a in oF ; then we have dK3 =F ¼ ðb=a2 ÞoF : The trace and norm of x=a in K3 =F are equal to b=a and ðb=a2 Þ ððb 1Þ=4Þ; respectively, which are in oF : Hence, x=a is an integer. Also, we obtain dK3 =F ¼ dK3 =F ð1; x=aÞoF ; where dK3 =F ð1; x=aÞ denotes the discriminant of 1; x=a in K3 =F : Consequently, f1; x=ag is a free oF -basis of oK3 : Note that x is an integer, as b 1 mod 4: Conversely, if fo1 ; o2 g is a free oF -basis of oK3 ; then there exists a matrix A in M2 ðoF Þ such that ð1; xÞ ¼ ðo1 ; o2 ÞA: Therefore, we have p1 p2 p23 ¼ boF ¼ dK3 =F ð1; xÞoF ¼ dK3 =F ðdet AÞ2 ; so that p3 ¼ ðdet AÞoF : Hence, p3 is principal. This proves our proposition. &
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Remark 2.1. If p3 is principal and ½p3 is an element of order X3; then Proposition 3 implies that K3 =F has an RIB, but no NIB. Corollary 4. Let the assumption be as in Proposition 3. Then, for all integral divisors m of F with p1 p2 p3 j m; there does not exist a maximal subextension of F ðmÞ=F which has an NIB. Proof. Let K1 ; K2 and K3 be the fields as in Proposition 3. Since K1 K2 is a subfield of F ðp1 p2 p3 Þ; it is also a subfield of F ðmÞ: Assume that there exists a maximal subextension L=F of F ðmÞ=F which has an NIB. Then, ðK1 K2 -LÞ=F is a maximal subextension of K1 K2 =F which has an NIB. For i ¼ 1; 2; since Ki =F has an NIB, we have Ki CK1 K2 -L: Therefore, we obtain K3 CK1 K2 CK1 K2 -L: Hence, K3 =F has an NIB. This is a contradiction. Our corollary is proved. &
Proof of Theorem 1. If hF 41 (resp., Cl4l0 is not an elementary abelian 2-group), then there is some ideal a of oF that is not principal (resp. some element ½a of order X3 in Cl4l0 ). By the Dirichlet density theorem, there exist infinitely many triplets ðp1 ; p2 ; p3 Þ of prime ideals of oF not dividing 2; such that both p1 and p2 are in ½a ; and p3 is in ½a 1 : Take such a triplet ðp1 ; p2 ; p3 Þ: Then we have ½p1 ¼ ½p2 ¼ ½p3 1 ; and p3 is not principal (resp., ½p3 is of order X3). Hence, Corollary 4 implies Theorem 1. &
3. Numerical examples In this section we consider whether Cl4l0 is an elementary abelian 2-group. There exists a natural exact sequence of abelian groups: 1-ðSl0 -P4l0 Þ=S4l0 -Cl4l0 -A4l0 =ðSl0 -P4l0 Þ-1; where P4l0 denotes the subgroup of A4l0 of all principal ideals relatively prime to 4: The last group is naturally isomorphic to the strict ideal class group Cl þ ¼ Al0 =Sl0 of F by the strong approximation theorem (cf. [4, Proposition 1.2.8]). For a ring R; we denote by R the group of units in R; and put G :¼ ðoF =4oF Þ : Since each residue class in G is represented by a totally positive element x of oF with ðx; 2Þ ¼ 1 from the above theorem, we can define a natural group homomorphism by f : G-ðSl0 -P4l0 Þ=S4l0 ;
½x /½xoF ;
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where ½x denotes the residue class of x in G: If El0 is the subgroup of all totally positive units in o F ; and we put H :¼ ðEl0 þ 4oF Þ=4oF ; then we have Ker f ¼ H: Also, it is easy to see that f is surjective. Consequently, G=H is naturally isomorphic to ðSl0 -P4l0 Þ=S4l0 : We obtain the following exact sequence: 1-G=H-Cl4l0 -Cl þ -1: Hence, if either Cl þ ; or G=H is not an elementary abelian 2-group, then we see that Cl4l0 is not an elementary abelian 2-group. In particular, if hF is divisible by an odd prime, then Cl4l0 is not an elementary abelian 2-group. Example 3.1. If F ¼ Q then we have El0 ¼ f1g; so that H ¼ f1g: Since G is a cyclic group of order 2 and Cl þ ¼ f1g; we see that Cl4l0 is a cyclic group of order 2: Therefore, the assumption of Theorem 1 is not satisfied. Horie and Horie [6, Corollary 3] proved that there are only finitely many imaginary abelian fields of 2-power degrees with ideal class groups Cl þ of exponents p2 (cf. [6, Introduction] and [11]). Consequently, in the case where F is an imaginary abelian field of 2-power degree, we see that Cl4l0 is not an elementary abelian 2-group (and hF 41 holds) except for finitely many fields of this kind. pffiffiffiffi From now on, let F ¼ Qð mÞ be a quadratic field, where m ða0; 1Þ is a squarefree integer, and we shall give some numerical examples of fields K1 ; K2 and K3 which are stated in Proposition 3. We assume that there exists an element ½x H of order X3 in G=H such that x is totally positive. (When F is a quadratic field, we can give a necessary and sufficient condition for G=H to be not an elementary abelian 2-group, which will be described elsewhere.) Since G=H is naturally isomorphic to ðSl0 -P4l0 Þ=S4l0 ; we can choose some triplets ðp1 oF ; p2 oF ; p3 oF Þ of principal prime ideals in Sl0 -P4l0 ; such that each pi ð1pip3Þ is totally positive, both ½p1 and ½p2 are in ½x H; and ½p3 is in ½x 1 H: (The infinitude of such triplets follows from the Dirichlet density theorem.) Then we obtain integers d1 ¼ p1 p3 and d2 ¼ p2 p3 ; pffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffi satisfying the property as in Proposition 3: K1 ¼ F ð p1 p3 Þ; K2 ¼ F ð p2 p3 Þ and pffiffiffiffiffiffiffiffiffiffi K3 ¼ F ð p1 p2 Þ: Hence, both K1 =F and K2 =F have NIB; K3 =F has an RIB, but no NIB. pffiffiffiffi We treat the case where m 5 mod 8 by way of example. Let o :¼ ð1 þ mÞ=2 and M :¼ ðm 1Þ=4: The canonical basis for oF is given by f1; og: If we put a :¼ ½ 1 ; b :¼ ½1 þ 2o and g :¼ ½M þ o then G ¼ /aS /bS /gS is an abelian group of type ð2; 2; 3Þ (cf. Lemma 6.6 and its proof in [8]). Let s be the generator of GalðF =QÞ: Since s naturally acts on G; H and ðSl0 -P4l0 Þ=S4l0 ; the map f becomes a /sS-homomorphism. Also, sðaÞ ¼ a; sðbÞ ¼ ab and sðgÞ ¼ g2 : The following numerical examples are calculated by using UBASIC [10]. Example 3.2. Let m ¼ 11: Then, M ¼ 3 1 mod 4; hF ¼ 1; so that Cl þ is a trivial group. For p ¼ 5; 37; 89; 97; 137; y; the prime p is decomposed in F ; if we put p ¼ 1 þ o; 5 3o; 7 þ 5o; 7 3o; 5 7o; y for each p; then ½p ¼ g; and poF is
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a prime ideal of oF lying above p: Since El0 ¼ o F ¼ f71g; we have H ¼ /aS; therefore, the element gH is of order 3: We take such distinct prime elements p1 and p2 ; and put p3 :¼ sðp1 Þ: Then, ½p1 H ¼ ½p2 H ¼ gH: As gsðgÞAH; we obtain ½p3 H ¼ sðgÞH ¼ g 1 H (of course, we may choose a prime element p3 which is in g 1 ). If m ¼ 35 ¼ 5 7 then M ¼ 9 3 mod 4; H ¼ /aS; Cl þ is a cyclic group of order 2: For p ¼ 389; 449; 541; 1061; 1201; yðp ¼ 19 þ o; 1 7o; 23 3o; 7 11o; 35 3o; yÞ; we have ½p ¼ g: Then the same thing as above is true. Example 3.3. Let m ¼ 37: Then, M ¼ 9 1 mod 4; the fundamental unit e ¼ 5 þ 2o is not totally positive; hF ¼ 1; so that Cl þ is a trivial group. For p ¼ 41; 53; 73; 101; 181; y; the prime p is decomposed in F ; if we put p ¼ 79 þ 31o; 19 þ 7o; 11 o; 191 þ 75o; y for each p; then p is totally positive, ½p ¼ ag; and poF is a prime ideal of oF lying above p: Since El0 ¼ /e2 S and e2 1 mod 4; we have H ¼ f1g; therefore, the element agH is of order 6: We take such distinct prime elements p1 and p2 ; and put p3 :¼ sðp1 Þ: Then, ½p1 H ¼ ½p2 H ¼ agH: As agsðagÞAH; we obtain ½p3 H ¼ sðagÞH ¼ ðagÞ 1 H: If m ¼ 485 ¼ 5 97 then M ¼ 121 1 mod 4; e ¼ 21 þ 2o is not totally positive; hF ¼ 2; so that H ¼ f1g and Cl þ is a cyclic group of order 2: For p ¼ 89; 241; 389; 809; 829; yðp ¼ 15 o; 35 þ 3o; 151 13o; 31 o; 119 þ 11o; yÞ; we have ½p ¼ ag: Then the same thing as above is true.
Acknowledgments The author is grateful to Humio Ichimura for his kind suggestions.
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