International Journal of Algebra, Vol. 10, 2016, no. 9, 429 - 435
HIKARI Ltd, www.m-hikari.com
http://dx.doi.org/10.12988/ija.2016.6753
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
c 2016 Jose Valter Lopes Nunes et al. This article is distributed under the
Copyright 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 p1 , . . . , ps 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 pi modulo pj , pi 6= pj .
Mathematics Subject Classification: 11R18, 11R20, 11R37, 11A15
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(ζp ))| then xp + y p = z p has no nontrivial integer solutions
[12]; as customary, for any integer k ≥ 3, ζk ∈ C denotes a primitive kth root
of unity and Q(ζk ) 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)6=p where
H(L)p is the p-part of H(L) and H(L)6=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(ζn ), is of the form n = p1 · · · ps where p1 , . . . , ps
are distinct odd primes with pi ≡ 1 (mod p), i = 1, . . . , s; furthermore, the
discriminant of L, Disc(L), equals np−1 , see [4, p. 186]. Hence, p1 , . . . , ps are
precisely the rational primes that are ramified in L/Q. Since each one of them
ramifies completely, one has pi OL = pppi where ppi is the prime ideal of OL
lying above pi , i = 1, . . . , s. If θ is a generator of Gal(L/Q) and t = TrK/L (ζn )
then L can be expressed as L = Q(t) and {t, θ(t), . . . , θp−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,
Hp (L) ∼
= (Z/pZ)s−1 . The contribution of the present paper is a self-contained
and new proof that, under the same conditions, Hp (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 p1 and p2 ramify and
p 6= p1 , p2 , then p || h(L) if and only if either p1 is not a pth power modulo p2
or p2 is not a pth power modulo p1 . Recall that a rational integer a is a pth
residue modulo a prime q if a is not divisible by q and the congruence xp ≡ a
431
p-Class groups of cyclic number fields
(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, p1 , . . . , ps , and pp1 , . . . , pps remain exactly as defined in the
previous section. Given any fractional ideal a of OL , we use [a] to denote the
equivalence class of a in H(L).
Lemma 1. Let i be any integer in {1, . . . , s}. If ppi is principal then (pi |pj )p =
1 for all j in {1, . . . , s} \ {i}.
Proof. Let ppi = (αi ) for some αi ∈ OL . Without loss of generality, we may
assume NL/Q (αi ) = pi . For all j ∈ {1, . . . , s} \ {i}, since αi ∈
/ ppj , we have
αi ≡ cj (mod ppj ) for some cj ∈ {1, . . . , pj − 1}. Thus,
pi = NL/Q (αi ) =
p−1
Y
θk (αi ) ≡ cpj
(mod ppj ),
k=0
whence cpj ≡ pi (mod pj ).
Corollary 1. If (pi |pj )p = −1 for any i, j in {1, . . . , s} with i 6= j then
h[ppi ]i ∼
= (Z/pZ), and consequently H(L)p is non-trivial.
Example 1. Let p = 3, p1 = 7, p2 = 13 and L a subfield of Q(ζ91 ) 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, p7 is non-principal for (7|13)3 =
−1. Thus, H(L)3 = h[p7 ]ii. Table 1 lists H(L)3 where L is a subfield of Q(ζn )
with [L : Q] = 3 and conductor n = p1 p2 with p1 , p2 in {7, 13, 19, 31, 37, 43},
and p1 6= p2 . In all cases, |H(L)3 | = 3.
Example 2. Let p = 3, p1 = 7, p2 = 13, p3 = 19, n = p1 p2 p3 = 1729, and L
a subfield of Q(ζn ) of conductor n. Then p7 , p13 , and p19 are all non-principal
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Jose Valter Lopes Nunes et al.
n
91
133
217
259
301
247
403
481
559
589
703
817
1147
1333
1591
p2 (p1 |p2 )3 (p2 |p1 )3 H(L)3
13
−1
1 h[p7 ]i
19
1
−1 h[p19 ]i
31
−1
−1 h[p7 ]i
37
−1
−1 h[p7 ]i
43
−1
1 h[p7 ]i
19
−1
−1 h[p13 ]i
31
−1
1 h[p13 ]i
37
−1
−1 h[p13 ]i
43
−1
−1 h[p13 ]i
31
−1
1 h[p19 ]i
37
−1
1 h[p19 ]i
43
−1
−1 h[p19 ]i
37
1
−1 h[p37 ]i
43
−1
−1 h[p31 ]i
43
−1
1 h[p37 ]i
p1
7
7
7
7
7
13
13
13
13
19
19
19
31
31
37
Table 1: H(L)3 for certain cubic fields of conductor p1 p2
for (7|13)3 = (13|19)3 = (19|7)3 = −1. Thus, 3 divides h(L). If we can show
that h[p7 ]i =
6 h[p13 ]i, we can immediately conclude that H(L)3 has a subgroup
isomorphic to (Z/3Z)2 and h(L) is a multiple of 32 . 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 OL and
q1 , . . . , qr prime ideals of OL . Then Q can be written as the quotient (α)/(β)
of two integral principal ideals such that none of the ideals qi , i = 1, . . . , r
divides both (α) and (β).
Corollary 2. Let a = papi11 . . . papiuu and b = pbp1j1 . . . pbpvjv where 1 ≤ r < i1 <
· · · < iu ≤ s and 1 ≤ r < j1 < · · · jv ≤ s. If [a] = [b] then
Qu a` !
`=1 pi` p
= 1 for k = 1, . . . , r.
Qv
b` k
`=1 pj
p
`
Q
−1
u
v
a`
b`
Proof. Consider the ideal I =
p
p
. By hypothesis, I =
`=1 pi`
`=1 pj`
((a)/(b)) for some a, b ∈ OL , b 6= 0. By Theorem 1, we may assume that
neither a nor b belongs to any of the ideals pQ
p1 , . . . , ppr . On the other
Qv hand,
u
a`
taking the (ideal) norm N (·) on both sides of `=1 ppi = ((a)/(b)) · `=1 pbp`j ,
`
`
we obtain
u
v
Y a
Y b
pi`` = N ((a)/(b)) ·
pj``
Q
`=1
`=1
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p-Class groups of cyclic number fields
For each 1 ≤ k ≤ r, there exists ck ∈ {1, . . . , pk − 1} such that a/b ≡
ck (mod ppk ), whence θj (a/b) ≡ ck (mod ppk ) for j = 0, . . . , p − 1. From
N ((a)/(b)) = NL/Q (a/b), it follows that
NL/Q (a/b) =
p−1
Y
j
θ (a/b) ≡
cpk
(mod ppk ), that is,
j=0
u
Y
`=1
pai``
≡
v
Y
cpk ·
pbj``
(mod pk ),
`=1
and the corollary is proved.
Example 3 (Example 2, cont’d). We have already seen that the ideals p7 , p13 ,
and p19 are non-principal. Now,
19 = (10 | 19)3 = −1, so by Corollary 2, [p7 ] 6= [p13 ].
(i) 13
7
3
2 (ii) 137 19 = (16 | 19)3 = −1, so by Corollary 2, [p7 ] 6= [p13 ]2 .
3
In conclusion, h(L) is a multiple of 32 . Moreover, h[p7 ]i ⊕ h[p13 ]i 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 p1 , . . . , pr , 2 ≤ r ≤ s, be such that
(p1 |pk )p = · · · = (pk−2 |pk )p = 1 and (pk−1 |pk )p = −1
for k = 2, . . . , r. Then h[pp1 ], . . . , [ppr−1 ]i ∼
= (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
(p1 |pk )p = · · · = (pk−2 |pk )p = 1 and (pk−1 |pk )p = −1
for k = 2, . . . , r + 1, we will show that
h[pp1 ], . . . , [ppr−1 ], [ppr ]i ∼
(1)
= (Z/pZ)r .
By Corollary 1, h[ppr ]i ∼
= (Z/pZ), so the isomorphism in (1) can be justified
by showing that [ppr ] 6∈ h[pp1 ], . . . , [ppr−1 ]i. By way of contradiction, suppose
[ppr ] = [pp1 ]e1 · · · [ppr−1 ]er−1 for some choice of non-negative integers e1 , . . . , er−1 ,
that is, [ppr ] = [pep11 · · · pepr−1
]. By Corollary 2,
r−1
e1
r−1
Y
pp1 · · · pepr−1
p
r−1
≡ cr+1 (mod ppr+1 ), that is,
pei i ≡ cpr+1 ·pr (mod pr+1 )
N
ppr
i=1
for some cr+1 ∈ {1, . . . , pr+1 − 1}. Since (pi |pr+1 )p = 1 for i = 1, . . . , r − 1,
the latter congruence implies (pr |pr+1 )p = 1, a contradiction. Thus, [ppr ] 6∈
h[pp1 ], . . . , [ppr−1 ]i, as desired.
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Jose Valter Lopes Nunes et al.
Example 4. Let p = 3, p1 = 7, p2 = 13, p3 = 19, p4 = 223, p5 = 373, n =
p1 · · · p5 = 143816491, and L a subfield of Q(ζn ) of conductor n. We have:
(i) (p1 |p2 )3 = −1;
(ii) (p1 |p3 )3 = 1, (p2 |p3 ) = −1;
(iii) (p1 |p4 )3 = (p2 |p4 )3 = 1, (p3 |p4 )3 = −1;
(iv) (p1 |p5 )3 = (p2 |p5 )3 = (p3 |p5 )3 = 1, (p4 |p5 )3 = −1.
By Theorem 2, H(L)3 has a subgroup generated by [p7 ], [p13 ], [p19 ], and [p223 ].
Hence, h(L) is a multiple of 34 .
Example 5. Let p = 5, p1 = 11, p2 = 31, p3 = 61, p4 = 191, p5 = 541,
n = p1 · · · p5 = 2149388131, and L a subfield of Q(ζn ) of conductor n. We
have:
(i) (p1 |p2 )5 = −1;
(ii) (p1 |p3 )5 = 1, (p2 |p3 ) = −1;
(iii) (p1 |p4 )5 = (p2 |p4 )5 = 1, (p3 |p4 )5 = −1;
(iv) (p1 |p5 )5 = (p2 |p5 )5 = (p3 |p5 )5 = 1, (p4 |p5 )5 = −1.
By Theorem 2, H(L)5 has a subgroup generated by [p11 ], [p31 ], [p61 ], and [p191 ].
Hence, h(L) is a multiple of 54 .
3
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 = p1 · · · ps
with pi ≡ 1 (mod p), i = 1, . . . , s, was introduced. Its effectiveness relies on
the assumption that pi is a pth residue modulo pj for j = 1, . . . , i − 2, but not
a pth residue modulo pi−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.
p-Class groups of cyclic number fields
435
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Received: August 12, 2016; Published: October 7, 2016