p-class Groups of Cyclic Number Fields of Odd Prime Degree

Transcription

1 International Journal of Algebra, Vol. 10, 2016, no. 9, HIKARI Ltd, p-class Groups of Cyclic Number Fields of Odd Prime Degree Jose Valter Lopes Nunes Departamento de Matemática, Universidade Federal do Ceará Fortaleza, CE, Brazil J. Carmelo Interlando Department of Mathematics and Statistics San Diego State University, San Diego, CA, USA Trajano Pires da Nóbrega Neto Departamento de Matemática Universidade Estadual Paulista São José do Rio Preto, SP, Brazil José Othon Dantas Lopes Departamento de Matemática Universidade Federal do Ceará, Fortaleza, CE, Brazil Copyright c 2016 Jose Valter Lopes Nunes et al. This article is distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Abstract Let L/Q be a cyclic extension of degree p where p is an odd unramified prime in L/Q, and let p 1,..., p s be the rational primes that are ramified in L/Q. A new method of proof for showing the existence of subgroups in H(L) p, the p-part of the class group of L, is introduced. It is purely arithmetical in the sense that it is solely based on the pth residuacity of p i modulo p j, p i p j. Mathematics Subject Classification: 11R18, 11R20, 11R37, 11A15

2 430 Jose Valter Lopes Nunes et al. Keywords: cyclic extensions, class groups, power residues 1 Introduction and Background Class groups of number fields play a central role in class field theory due to their connections with unit groups [9], unramified extensions, and other number field invariants. Determining the structure of the class group H(F ) of a number field F can be a formidable task and it is one of the main problems in computational number theory. The p-part, or the p-sylow subgroup, of H(F ) is relevant for example in Iwasawa theory, elliptic curves, and for a long time it was of interest to Fermat s last theorem due to Kummer s famous criterion, namely, if p H(Q(ζ p )) then x p + y p = z p has no nontrivial integer solutions [12]; as customary, for any integer k 3, ζ k C denotes a primitive kth root of unity and Q(ζ k ) the kth cyclotomic field. Throughout the paper, L/Q denotes a cyclic extension of degree p where p is an odd unramified prime in L/Q. One has H(L) = H(L) p H(L) p where H(L) p is the p-part of H(L) and H(L) p the prime-to-p part of H(L). The cardinality of H(L), that is, the class number of L, is denoted by h(l). Relevant results needed for the rest of the work are in order. First, the conductor of L, that is, the smallest positive integer n such that L is contained in the nth cyclotomic field K = Q(ζ n ), is of the form n = p 1 p s where p 1,..., p s are distinct odd primes with p i 1 (mod p), i = 1,..., s; furthermore, the discriminant of L, Disc(L), equals n p 1, see [4, p. 186]. Hence, p 1,..., p s are precisely the rational primes that are ramified in L/Q. Since each one of them ramifies completely, one has p i O L = p p p i where p pi is the prime ideal of O L lying above p i, i = 1,..., s. If θ is a generator of Gal(L/Q) and t = Tr K/L (ζ n ) then L can be expressed as L = Q(t) and {t, θ(t),..., θ p 1 (t)} is an integral basis for L, see [8, p. 166]. In [6], using class field theory, Gerth III proved that when the rational primes that ramify in L/Q satisfy certain pth power residuacity conditions, H p (L) = (Z/pZ) s 1. The contribution of the present paper is a self-contained and new proof that, under the same conditions, H p (L) contains a subgroup isomorphic to (Z/pZ) r for r = 1,..., s. Furthermore, each subgroup is given explicitly, and several examples are provided to illustrate the result. Before proceeding, we provide a brief history of the subject and introduce further notation. By a theorem of Leopoldt [7], it is known that p h(l) if and only if exactly one rational prime ramifies in L/Q. Conner and Hurrelbrink [3] proved that if p h(l) then exactly two rational primes ramify in L/Q. Conversely, by [3, Theorem 26.9], if exactly two primes p 1 and p 2 ramify and p p 1, p 2, then p h(l) if and only if either p 1 is not a pth power modulo p 2 or p 2 is not a pth power modulo p 1. Recall that a rational integer a is a pth residue modulo a prime q if a is not divisible by q and the congruence x p a

3 p-class groups of cyclic number fields 431 (mod q) has a solution [5, Section III]. By Euler s criterion, a is a pth residue modulo q if and only if a (q 1)/p 1 (mod q) [10, Chapter VII]. By an abuse of notation, but for convenience, we will write (a q) p = 1 or (a q) p = 1 according to whether a is a pth residue modulo q or not. In regards to more recent work, an algorithm for computing H(F ) p with F/Q Abelian was proposed in [1]. However, in contrast to the present paper, the assumption in the former is that p [F : Q]. The main result, namely, Theorem 2, is presented in the next section along with auxiliary results and examples. The Conclusion is presented in Section 3. 2 pth Power Residuacity and the Structure of H(L) p All of L/Q, p, n, p 1,..., p s, and p p1,..., p ps remain exactly as defined in the previous section. Given any fractional ideal a of O L, we use [a] to denote the equivalence class of a in H(L). Lemma 1. Let i be any integer in {1,..., s}. If p pi is principal then (p i p j ) p = 1 for all j in {1,..., s} \ {i}. Proof. Let p pi = (α i ) for some α i O L. Without loss of generality, we may assume N L/Q (α i ) = p i. For all j {1,..., s} \ {i}, since α i / p pj, we have α i c j (mod p pj ) for some c j {1,..., p j 1}. Thus, whence c p j p i (mod p j ). p 1 p i = N L/Q (α i ) = θ k (α i ) c p j (mod p pj ), k=0 Corollary 1. If (p i p j ) p = 1 for any i, j in {1,..., s} with i j then [p pi ] = (Z/pZ), and consequently H(L) p is non-trivial. Example 1. Let p = 3, p 1 = 7, p 2 = 13 and L a subfield of Q(ζ 91 ) with [L : Q] = 3 and conductor 91. By Theorem 26.9 in [3] (as stated in the Introduction), H(L) 3 = 3. From Corollary 1, p 7 is non-principal for (7 13) 3 = 1. Thus, H(L) 3 = [p 7 ]. Table 1 lists H(L) 3 where L is a subfield of Q(ζ n ) with [L : Q] = 3 and conductor n = p 1 p 2 with p 1, p 2 in {7, 13, 19, 31, 37, 43}, and p 1 p 2. In all cases, H(L) 3 = 3. Example 2. Let p = 3, p 1 = 7, p 2 = 13, p 3 = 19, n = p 1 p 2 p 3 = 1729, and L a subfield of Q(ζ n ) of conductor n. Then p 7, p 13, and p 19 are all non-principal

4 432 Jose Valter Lopes Nunes et al. n p 1 p 2 (p 1 p 2 ) 3 (p 2 p 1 ) 3 H(L) [p 7 ] [p 19 ] [p 7 ] [p 7 ] [p 7 ] [p 13 ] [p 13 ] [p 13 ] [p 13 ] [p 19 ] [p 19 ] [p 19 ] [p 37 ] [p 31 ] [p 37 ] Table 1: H(L) 3 for certain cubic fields of conductor p 1 p 2 for (7 13) 3 = (13 19) 3 = (19 7) 3 = 1. Thus, 3 divides h(l). If we can show that [p 7 ] [p 13 ], we can immediately conclude that H(L) 3 has a subgroup isomorphic to (Z/3Z) 2 and h(l) is a multiple of 3 2. This will be done after we state and prove a corollary to the next theorem. Theorem 1. [11, p. 184] Let Q be a principal fractional ideal of O L and q 1,..., q r prime ideals of O L. Then Q can be written as the quotient (α)/(β) of two integral principal ideals such that none of the ideals q i, i = 1,..., r divides both (α) and (β). Corollary 2. Let a = p a 1 p i1... p au p iu and b = p b 1 pj1... p bv p jv where 1 r < i 1 < < i u s and 1 r < j 1 < j v s. If [a] = [b] then ( u ) pa l i l v pb l p k = 1 for k = 1,..., r. j l p ( u ) ( v ) 1. Proof. Consider the ideal I = pa l p il pb l p jl By hypothesis, I = ((a)/(b)) for some a, b O L, b 0. By Theorem 1, we may assume that neither a nor b belongs to any of the ideals p p1,..., p pr. On the other hand, taking the (ideal) norm N( ) on both sides of u pa l p il = ((a)/(b)) v pb l p jl, we obtain u v p a l i l = N((a)/(b)) p b l j l

5 p-class groups of cyclic number fields 433 For each 1 k r, there exists c k {1,..., p k 1} such that a/b c k (mod p pk ), whence θ j (a/b) c k (mod p pk ) for j = 0,..., p 1. From N((a)/(b)) = N L/Q (a/b), it follows that p 1 N L/Q (a/b) = θ j (a/b) c p k j=0 and the corollary is proved. (i) ( 13 7 (mod p pk ), that is, u p a l i l c p k v p b l j l (mod p k ), Example 3 (Example 2, cont d). We have already seen that the ideals p 7, p 13, and p 19 are non-principal. Now, 19 )3 = (10 19) 3 = 1, so by Corollary 2, [p 7 ] [p 13 ]. (ii) ( ) 19 = (16 19) 3 = 1, so by Corollary 2, [p 7 ] [p 13 ] 2. 3 In conclusion, h(l) is a multiple of 3 2. Moreover, [p 7 ] [p 13 ] is a subgroup of H(L) p isomorphic to (Z/3Z) 2. The next theorem is the main result of the present work. When its hypotheses are satisfied, it immediately provides a method for determining subgroups of H(L) p. Theorem 2. Notation as before, let p 1,..., p r, 2 r s, be such that (p 1 p k ) p = = (p k 2 p k ) p = 1 and (p k 1 p k ) p = 1 for k = 2,..., r. Then [p p1 ],..., [p pr 1 ] = (Z/pZ) r 1. Proof. We will use induction on r. The case r = 2 can be handled by Corollary 1, so we will treat the case 2 < r < s. Assuming that (p 1 p k ) p = = (p k 2 p k ) p = 1 and (p k 1 p k ) p = 1 for k = 2,..., r + 1, we will show that [p p1 ],..., [p pr 1 ], [p pr ] = (Z/pZ) r. (1) By Corollary 1, [p pr ] = (Z/pZ), so the isomorphism in (1) can be justified by showing that [p pr ] [p p1 ],..., [p pr 1 ]. By way of contradiction, suppose [p pr ] = [p p1 ] e1 [p pr 1 ] e r 1 for some choice of non-negative integers e 1,..., e r 1, that is, [p pr ] = [p e 1 p 1 p e r 1 N ( p e 1 p 1 p e r 1 p pr p r 1 p r 1 ]. By Corollary 2, ) r 1 c p r+1 (mod p pr+1 ), that is, i=1 p e i i c p r+1 p r (mod p r+1 ) for some c r+1 {1,..., p r+1 1}. Since (p i p r+1 ) p = 1 for i = 1,..., r 1, the latter congruence implies (p r p r+1 ) p = 1, a contradiction. Thus, [p pr ] [p p1 ],..., [p pr 1 ], as desired.

6 434 Jose Valter Lopes Nunes et al. Example 4. Let p = 3, p 1 = 7, p 2 = 13, p 3 = 19, p 4 = 223, p 5 = 373, n = p 1 p 5 = , and L a subfield of Q(ζ n ) of conductor n. We have: (i) (p 1 p 2 ) 3 = 1; (ii) (p 1 p 3 ) 3 = 1, (p 2 p 3 ) = 1; (iii) (p 1 p 4 ) 3 = (p 2 p 4 ) 3 = 1, (p 3 p 4 ) 3 = 1; (iv) (p 1 p 5 ) 3 = (p 2 p 5 ) 3 = (p 3 p 5 ) 3 = 1, (p 4 p 5 ) 3 = 1. By Theorem 2, H(L) 3 has a subgroup generated by [p 7 ], [p 13 ], [p 19 ], and [p 223 ]. Hence, h(l) is a multiple of 3 4. Example 5. Let p = 5, p 1 = 11, p 2 = 31, p 3 = 61, p 4 = 191, p 5 = 541, n = p 1 p 5 = , and L a subfield of Q(ζ n ) of conductor n. We have: (i) (p 1 p 2 ) 5 = 1; (ii) (p 1 p 3 ) 5 = 1, (p 2 p 3 ) = 1; (iii) (p 1 p 4 ) 5 = (p 2 p 4 ) 5 = 1, (p 3 p 4 ) 5 = 1; (iv) (p 1 p 5 ) 5 = (p 2 p 5 ) 5 = (p 3 p 5 ) 5 = 1, (p 4 p 5 ) 5 = 1. By Theorem 2, H(L) 5 has a subgroup generated by [p 11 ], [p 31 ], [p 61 ], and [p 191 ]. Hence, h(l) is a multiple of Conclusion A method for explicitly determining subgroups of the p-part of the class group of L where L/Q is a cyclic extension of degree p and conductor n = p 1 p s with p i 1 (mod p), i = 1,..., s, was introduced. Its effectiveness relies on the assumption that p i is a pth residue modulo p j for j = 1,..., i 2, but not a pth residue modulo p i 1 for i = 2,..., r. Despite this limitation, the method allowed us to easily tackle several cases that may have been otherwise computationally challenging even for state-of-the-art algorithms [2]. Therefore, the technique could conceivably be applied prior to those algorithms, hence reducing their computational costs to a reasonable degree. The actual limitation of the presented technique is unknown to us, so we leave this as a topic for future research.

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