REMARKS ON QUADRATIC FIELDS WITH ... - Project Euclid

TAIWANESE JOURNAL OF MATHEMATICS Vol. 19, No. 5, pp. 1387-1399, October 2015 DOI: 10.11650/tjm.19.2015.5081 This paper is available online at http://journal.taiwanmathsoc.org.tw

REMARKS ON QUADRATIC FIELDS WITH NONCYCLIC IDEAL CLASS GROUPS Kwang-Seob Kim Abstract. Let n be an integer. Then, it is well known that there are infinitely many imaginary quadratic fields with an ideal class group having a subgroup isomorphic to Z/nZ × Z/nZ. Less is known for real quadratic fields, other than the cases that n = 3, 5, or 7, due to Craig [3] and Mestre [4, 5]. In this article, we will prove that there exist infinitely many real quadratic number fields with the ideal class group having a subgroup isomorphic to Z/nZ × Z/nZ In addition, we will prove that there exist infinitely many imaginary quadratic number fields with the ideal class group having a subgroup isomorphic to Z/nZ × Z/nZ × Z/nZ.

1. INTRODUCTION The divisibility properties of class numbers are very important for understanding the structure of ideal class groups of number fields. Numerous results about the divisibility of class numbers of quadratic fields have been given by many authors (for more details see [1, 2, 6, 7, 8, 9, 10, 11, 12]). Through such works, it has been shown that, for any integer n, there exist infinitely many imaginary (resp. real) quadratic number fields whose ideal class numbers are multiples of n. Furthermore, it has been proven that there exist infinitely many imaginary (resp. real) quadratic number fields with the property that the ideal class group has a cyclic subgroup of order n. However, this does not necessarily mean that the ideal class group has an arbitrary abelian group of order n as a subgroup. We present our conjecture, as follows. Conjecture. For any finite abelian group G, there exist infinitely many quadratic fields K such that the ideal class group of K contains a subgroup isomorphic to G. In order to prove our conjecture, it suffices to show that there exist infinitely many quadratic fields K such that the ideal class group of K contains a subgroup isomorphic to (Z/nZ)m , where m and n are arbitrary integers. However, even this statement is not close to having been proven so far. The best known quantitative result is as follows. Received July 24, 2014, accepted April 9, 2015. Communicated by Yu-Ru Liu. 2010 Mathematics Subject Classification: Primary 11R29; Secondary 11R11. Key words and phrases: Ideal class group, Real quadratic fields.

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Theorem 1.1. (Theorem 2 of [11]). For any integer n ≥ 1, there are infinitely many imaginary quadratic fields with the ideal class group having a subgroup isomorphic to Z/nZ × Z/nZ. For real quadratic fields, less has been shown, except due to Craig [3] and Mestre [4, 5] for the cases that n = 3, 5, or 7. In this article, we will prove that there exist infinitely many real quadratic number fields with an ideal class group having a subgroup isomorphic to Z/nZ × Z/nZ, for every natural number n. We approach this by modifying the method used in [11]. At the same time, we will also show that there exist infinitely many imaginary quadratic number fields with the ideal class group having a subgroup isomorphic to Z/nZ × Z/nZ × Z/nZ. A sketch of our method is as follows. We will construct a quadratic number field that has three ideal classes, [a], [a ], and [a], and also satisfies some local conditions on its discriminant K. In the case that D < 0, they are of order n and independent. In the case that D > 0, neither of them may be of order n, due to the existence of non-trivial units, but the subgroup [a], [a], [a] contains Z/nZ × Z/nZ. Next, we show that an infinite quantity of such fields exist in either case. 2. REVIEW OF YAMAMOTO’S PAPER [11] Since our proof is based on Yamamoto’s paper [11], we will review the method used there. Let n ≥ 3 be a natural number. We will fix this definition of n throughout this article. Let K be a quadratic number field with discriminant D, where we assume D = −3 or −4 in order to simplify our argument in the following. Then, let σ be the nontrivial automorphism of K over Q. Define by a fundamental unit of K if D > 0, = 1 if D < −4. Lemma 2.1. (Lemma 1 of [11]). Let x, y, z be a solution in Z of the Diophantine equation (2.1)

X 2 − Y 2 D = 4Z n

satisfying (x, z) = 1. Then, there exists an (integral ideal) a in K such that √ (a) x+y2 D = an , (b) a and aσ are relatively prime, where (α) denotes the principal ideal in K generated by an element α of K. Let p be a prime factor of n. Take another prime number with (2.2)

≡ 1 (mod 2p),

so that −1 is a p-th power residue mod . Suppose that we have a solution x, y, z of the equation (2.1) satisfying

Remarks on Quadratic Fields with Noncyclic Ideal Class Groups

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(i) (x, z) = 1, (ii) |z, (iii) x is a p-th power non-residue mod . It can be easily deduced that (2.3)

D

= 1,

where the left side is the Kronecker symbol. By the decomposition law of primes we have = τ τ σ , where τ and τ σ are distinct conjugate prime ideals in K. Now, we set √ x+y D , α= 2 so we have τ τ σ | (α)(ασ ). Therefore, we may assume that τ | (ασ ), but τ (α), because (α) and (ασ ) are relatively prime, by the above lemma. Then, we have the following: Lemma 2.2. (Lemma 2 of [11]). If is a p-th power residue mod τ , then the ideal (α) is not the p-th power of any principal ideal in K. Let n = pe11 pe22 · · · pess be the prime decomposition of n. For each i (1 ≤ i ≤ s), we fix a prime number i , satisfying (2.4)

i ≡ 1 (mod 2pi).

Suppose that we have a solution x, y, z of the equation (2.1), satisfying (i) (x, z) = 1, (ii) i |z, for i = 1, 2, . . ., s, (iii) x is a pi -th power non-residue mod i for i = 1, 2, . . ., s. Then, set

√ x+y D . α= 2 From Lemma 2.1, we have that (α) = an with an ideal a ∈ K, and every i is decomposed in K as i = τi τiσ . Assume that τi | (ασ ). Denote by [a] the ideal class containing α. Then, we have [a]n = [(α)] = 1. Proposition 2.3. Let notations and assumptions be as above. If is a pi -th power residue mod τi for every i (1 ≤ i ≤ s), then the order of [a] is equal to n.

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Proof. Assume that [a]m = 1 for some m (1 ≤ m < n). It is obvious that m is a divisor of n, so there exists at least one prime divisor pi of n such that mpi | n. Then, [a]n/pi = 1. Therefore, there exists an integer β in K such that an/pi = (β). Then, (α) = an = (β)pi . However, this is impossible from Lemma 2.2. Therefore, we have [a]m = 1 for m = 1, 2, . . . , n − 1. It follows that the order [a] is equal to n. Remark 2.4. In the case that D < −4, we do not require the condition on from Lemma 2.2 and Proposition 2.3, since = 1. 3. CRUCIAL PROPOSITION Take three systems of prime numbers {i }, {i }, and {i }, each satisfying the condition (2.4). Moreover, assume that i , i , and i are pairwise distinct for every i (1 ≤ i ≤ s). Our aim in this section is to prove the following proposition, which plays a crucial role in our derivation. Proposition 3.1. Let x, z, x , z , x , and z constitute a non-trivial solution of the Diophantine equation (3.1)

X 2 − 4Z n = X 2 − 4Z n = X 2 − 4Z n,

satisfying: (i) (x, z) = (x , z ) = (x , z ) = 1, (ii) i |z, |z and |z , (iii) x (resp. x , x ) is a pi -th power non-residue mod i (resp. i , i ), (iv) (x + x )/2 and (x + x )/2 are pi -th power residues mod i , (v) (x + x )/2 and (x + x )/2 are pi -th power residues mod i , (vi) (x + x )/2 and (x + x )/2 are pi -th power residues mod i , for every i (1 ≤ i ≤ s). Then, the ideal class group of the field K = Q( x2 − 4z n ) has a subgroup N , such that N

Z/nZ × Z/nZ × Z/nZ if D < −4, Z/nZ × Z/nZ

where D is the discriminant of K.

if D > 0,


Proof.

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From equation 3.1, we set x2 − 4z n = x2 − 4z n = x2 − 4z n = y 2 D,

(3.2)

for some y ∈ Z. Thus, we have (3.3)

x2 − y 2 D = 4z n ,

(3.4)

x2 − y 2 D = 4z n

and x2 − y 2 D = 4z n .

(3.5)

Therefore, we obtain three solutions (x, y, z), (x , y, z ), and (x , y, z ) of the Diophantine equation (2.1). It follows from Lemma 2.1 that there are ideals a, a , and a in K, such that (α) = an , (α ) = an , and (α ) = an, where √ √ √ x + y D x + y D x + y D ,α = and α = . α= 2 2 2 Let τi , τi , and τi (1 ≤ i ≤ s) be the prime ideals in K, such that i = τi τiσ τi | (ασ ), i = τi τiσ τi | (ασ ), i = τi τiσ τi | (ασ ).

Let Ri (resp. Ri and Ri ) be the set of all integers in K that are a pi -th power residue mod τi (resp. τi and τi ). Since α ≡ x(mod τi ), α ≡

α≡

x + x (mod τi ), 2

x + x (mod τi ), 2 α ≡ x (mod τi ),

x + x (mod τi ), 2 x + x (mod τi ), α ≡ 2

α≡

and x + x x + x (mod τi ), α ≡ (mod τi ), α ≡ x (mod τi ), 2 2 it follows form the conditions (iii)-(vi) of Proposition 3.1 that α ≡

(3.6)

α∈ / Ri ,

α ∈ Ri ,

α ∈ Ri ,

(3.7)

α ∈ Ri ,

α ∈ / Ri,

α ∈ Ri ,

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and α ∈ Ri ,

(3.8)

α ∈ Ri,

α ∈ / Ri ,

for every i (1 ≤ i ≤ s). 3.1. The case with D < −4. It follows from Proposition 2.3 that the ideal classes [a], [a ], and [a ] have the same order n. Suppose that the following equation holds for m, m , m > 0:

[a]m[a ]m [a ]m = 1.

(3.9)

Then, there exists a number β ∈ K such that

)

am a(m ) a(m

(3.10)

= (β).

Taking the n-th power of both sides of (3.10), we obtain

)

αm α(m ) α(m

(3.11)

= ±β n .

Define di by pdi i (m, m, m), and ei by pei i n (1 ≤ i ≤ s). We claim that di ≥ ei for all i. Suppose that di < ei holds for some i, and set (3.12)

m = pdi m0 , m = pdi m0 , m = pdi m0 , n = pdi n0 ,

where pi | n0 . It follows from (3.11) that (3.13)

αm0 α(m0 ) α(m0 ) = ±β n0 ,

as K contains no root of 1 other than ±1. Since α(m0 ) ∈ Ri , α(m0 ) ∈ Ri, and / Ri , so we have that pi | m0 . ±β n0 ∈ Ri , we have that αm0 ∈ Ri . However, α ∈ Similarly, we also have that pi | m0 and pi | m0 . Hence, and from (3.12), we have that pidi +1 | (m, m, m). This contradicts the definition of di . Therefore, we have that di ≥ ei for every i. Accordingly, we have that n | m, n | m , and n | m . Let N be the subgroup of the ideal class group generated by [a], [a ], and [a ]. Then, N is isomorphic to Z/nZ × Z/nZ × Z/nZ. 3.2. The case with D > 0. Let n = pe11 pe22 · · · pess


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be the prime decomposition of n. We will show that the ideal class group of K has a subgroup isomorphic to Z/pei Z × Z/pei Z (1 ≤ i ≤ s). Let be a fixed fundamental unit of K. Define I := {i | ∈ Ri, 1 ≤ i ≤ s}, I := {i | ∈ Ri, 1 ≤ i ≤ s}, and

I := {i | ∈ Ri , 1 ≤ i ≤ s}.

Then, let m, m , and m be the orders of the ideal classes [a], [a ], and [a], respectively of (m 2.2ei that m is a multiple | n,ei m | n, and m | n). It follows from Lemma i∈I pi . We claim that m and m are multiples of i∈I / pi . Assume that pi m | n, for some i ∈ / I. Then, there exists a number β in K such that a(n) = (α ) = (β)pi . Therefore, we have

α = ±k β pi for some k ∈ Z.

/ Ri, we obtain that pi | k. It follows that ±k β pi ∈ Ri . Since α ∈ Ri, β pi ∈ Ri, and ∈ Therefore, we see that α ∈ Ri . This is acontradiction. It follows that pi m n, for ei all / I. Therefore, m is a multiple of i∈I / pi . Similarly, m is also a multiple of i∈ ei i∈I / pi . By the same reasoning, we have that

pei i m, pei i m , and pei i m . i∈I

i∈I /

In addition, we have i∈I /

i∈I /

pei i m, pei i m , and pei i m . i∈I

i∈I /

e

We claim that pj j divides at least two of m, m , and m , for each j. e

3.2.1. Case 1 - pj j m e

Without loss of generality, suppose that pj j m for some j. We know that m is a ei ei multiple of i∈I / pi , and is also a multiple of i∈I / pi . Therefore, e

pj j In other words,

i∈I /

pei i

e

and pj j

i∈I /

pei i .

e

e

pei i and pj j

pei i . pj j

i∈I

i∈I

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Kwang-Seob Kim e

This implies that pj j divides m (resp. m ), and that j is contained in I (resp. I ). e e Set n ˜ := n/pj j . Then, the ideal classes [a(ñ) ] and [a(ñ)] have the same order, pj j . Suppose that the following equation holds for m > and m > 0:

[a(ñ)]m [a(ñ) ]m = 1.

(3.14)

We know that a fundamental unit of K is a pj -th power residue mod τj and τj . Then, we can show that the subgroup generated by [a(ñ)] and [a(ñ) ] is isomorphic to e e Z/pj j Z × Z/pj j Z, in the same way as in the case where D < −4. e

3.2.2. Case 2 - pj j divides m, m and m e

e

Next, we assume that pj j divides m, m , and m . Set n ˜ := n/pj j . For convenience, we will employ the following notations: ˜ a := an˜ , a˜ := a(ñ) , and a˜ := a(ñ) . e

Then, the ideal classes [˜ a], [˜ a ], and [ã] have the same order, pj j . Without loss of generality, suppose that [ã] ∩ [ã ] = 1. Then, the subgroup e e generated by [˜ a] and [˜ a ] is isomorphic to Z/pj j Z × Z/pj j Z. We are done. r

r

Suppose that [˜ a] ∩ [˜ a ] = 1, i.e., [ã]pj = [ã ]pj , for some r (1 ≤ r < ej ). This means that r

r

[ã]pj s [ã ]pj = 1,

(3.15)

for some integer s coprime to pj . Then, there exists a number β ∈ K such that r

(ej −r)

Taking the pj (3.17)

r

ãpj s · ã(pj ) = (β).

(3.16)

-th power of both sides of (3.16), we obtain (ej −r)

±k αs α = β pj

.

/ Rj and As ej > r, we have that ±k αs ∈ Rj and ±k α ∈ Rj . However, α ∈ (ej −r)

/ Rj , so we have that pj k. Because αs ∈ Rj , α ∈ Rj , and β pj ∈ Rj , we α ∈ k have that ∈ Rj . Since k is relatively prime to pj , we have that ∈ Rj . Suppose q

q

a]pj = [ã]pj , for some q (1 ≤ q < ej ). This implies that [˜ a] ∩ [˜ a ] = 1, i.e., [˜ that

(3.18)

q

q

[ã]pj t [ã]pj = 1,

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for some integer t that is relatively prime to pj . Then, there exists a number γ ∈ K such that q

(ej −q)

Taking the pj (3.20)

q

ãpj t · ã(pj ) = (γ).

(3.19)

-th power of both sides of (3.19), we obtain (ej −q)

±l αt α = γ pj

.

Since ej > q, we have that ±l αt ∈ Rj and ±l α ∈ Rj . We know that ∈ Rj . Therefore, we obtain that α ∈ Rj . This is a contradiction. Therefore, [ã] and [ã] should only meet at the identity. It follows that the subgroup generated by [ã] and [ã] e e is isomorphic to Z/pj j Z × Z/pj j Z. We are done. e

e

In conclusion, K has a subgroup isomorphic to Z/pj j Z × Z/pj j Z for all j. That is, K has a subgroup isomorphic to Z/nZ × Z/nZ. 4. MAIN THEOREM Let us recall the prime decomposition of n: n = pe11 pe22 · · · pess . We need one more lemma before we state our main theorem. Lemma 4.1. For each prime number pi = 2, there exist infinitely many prime numbers , such that (a) ≡ 1 (mod 2pi), (b) −1 is an n-th power residue mod , (c) 2 is a pi -th power non-residue mod . Proof. Define F := Q(21/pi , ζ2n). Then, F is Galois over Q. It follows from Chebotarev’s density theorem that there exist infinitely many prime numbers , whose decomposition fields are equal to Q(ζ2n ). We can easily deduce that such primes satisfy the conditions of the lemma. The lemma below is for the case where pi = 2, for some i. Lemma 4.2. There exist infinitely many prime numbers , such that (a) ≡ 1 (mod 4), (b) the p2i ’s and −1 are n-th power residues mod , (c) p21 p22 · · · p2s + 1 is not a square mod .

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Proof.

Define 2/n 2/n F := Q p1 , p2 , ..., p2/n s , ζ2n

and

Fˆ := F

(p21 p22 · · · p2s + 1) .

Then, Fˆ is Galois over Q. Since p21 p22 · · · p2s + 1 is not a square, and is relatively prime / F . It follows from Chebotarev’s density theorem that there to n, p21 p22 · · · p2s + 1 ∈ exist infinitely many prime numbers , whose decomposition fields are equal to F . We can easily see that such ’s satisfy the conditions of the lemma. Main theorem. For any n > 0, there exist infinitely many real (resp. imaginary) quadratic number fields K such that the ideal class group of K has a subgroup which is isomorphic to Z/nZ × Z/nZ (resp. Z/nZ × Z/nZ × Z/nZ). Proof. For each pi , fix three prime numbers i , i , and i , satisfying the three conditions from Lemma 4.1 and Lemma 4.2. We will assume that all of the i ’s, i ’s, and i are distinct. Therefore, we can find an integer ci (resp. ai , bi ) such that cni ≡ −1 mod i (resp. ani ≡ −1 mod i , bni ≡ −1 mod i ) when pi = 2. If pi is equal to 2 for some i, we can find an integer ci (resp. ai , bi ) such that cni ≡ −(p1 p2 · · · ps )2 mod i (resp. ani ≡ −(p1 p2 · · · ps )2 mod i , bni ≡ −(p1 p2 · · · ps )2 mod i ). By the Chinese remainder theorem, we can find integers A, B, and C, satisfying ⎧ ⎪ A ≡ 0, B ≡ 1, C ≡ ci (mod i ) for all i, ⎪ ⎨ (4.1) A ≡ ai , B ≡ 0, C ≡ 1 (mod i ) for all i, ⎪ ⎪ ⎩ A ≡ 1, B ≡ bi, C ≡ 0 (mod i ) for all i, and

(4.2)

⎧ ⎪ B ≡ 1, C ≡ 0 (mod q) for q ∈ SA \ {i }, ⎪ ⎨ C≡0 (mod q) for q ∈ SB \ {i}, ⎪ ⎪ ⎩ C≡1 (mod q) for q ∈ S(B n−An ) ,

where Sm denotes the set of prime factors of an integer m. (It can easily be checked that (A, B) = 1 and {i }, {i }, {i }, SA \ {i}, SB \ {i}, and S(B n−An ) are disjoint sets.) Since (A, B) = 1 and (C, B n − An ) = 1, we have (4.3)

(A, B n − C n ) = 1 and (B, C n − An ) = 1.

4.1. Case 1 - Real quadratic number fields We now set


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⎧ ⎪ z = AB, x = An + B n − C n , ⎪ ⎪ ⎨ x = −An + B n + C n , z = BC, ⎪ ⎪ ⎪ ⎩ x = An − B n + C n , z = CA.

(4.4)

Then, we have A2n + B 2n + C 2n − 2(AB)n − 2(BC)n − 2(CA)n =x2 − 4z n =x2 − 4z n

(4.5)

=x2 − 4z n , and

x≡

x ≡ x ≡

(mod i ) if pi = 2,

2

p21 p22 · · · p2s + 1 (mod i ) if pi = 2, (mod i ) if pi = 2,

2 p21 p22

···

p2s

+ 1 (mod

i )

if pi = 2,

(mod i ) if pi = 2,

2

p21 p22 · · · p2s + 1 (mod i ) if pi = 2,

z ≡ 0, (mod i ), z ≡ 0, (mod i ), z ≡ 0 (mod i ),

for all i (1 ≤ i ≤ s). In addition, we have x + x = Bn , 2

x + x = Cn, 2

x + x = An . 2

From (4.3), we also have that (x, z) = (x , z ) = (x , z ) = 1. It follows from this that x, z, x, z , x, z is a solution of the Diophantine equation (3.1), satisfying all of the conditions of Proposition 3.1 (note that pi |n, for all 1 ≤ i ≤ s). Since it holds that (4.6)

x2 − 4z n = A2n + B 2n + C 2n − 2(AB)n − 2(BC)n − 2(CA)n = C 2n − 2C n (An + B n ) + (An − B n )2 ,

and C is determined by a congruence condition, we can let√the value x2 − 4z n be positive, by choosing a suitable C. Now, we set K = Q( x2 − 4z n ). It follows from Proposition 3.1 that the ideal class group of a real quadratic number field K has Z/nZ × Z/nZ as a subgroup. The infinite property follows directly from the existence, as follows. Assume that there exist only a finite number of such K’s, and denote the set of them by K. Let k be the maximum value of the class number of such K’s (note that the class number is a finite ring of integers of number fields). Then, we can obtain (nt)2 > k, by choosing a suitable t. Let K be a real quadratic field, having a subgroup

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isomorphic to Z/ntZ × Z/ntZ. Then, K is also contained in K. This contradicts the maximality of k. 4.2. Case 2 - Imaginary quadratic number fields Let t be a multiple of the product of all of prime numbers in {i }, {i}, {i }, S[(B−A)n−(C−A)n ] , S[(B−A)n−(B−A)n ] , and S[(C−A)n −(C−B)n ] . From (4.3), we can check that 1 = t − A, (B − A)n − (C − A)n (4.7) = t − B, (B − A)n − (B − C)n = t − C, (C − A)n − (C − B)n . Now, we set ⎧ ⎪ z = (A − t)(B − t), x = (A − t)n + (B − t)n − (C − t)n , ⎪ ⎨ n n n (4.8) x = −(A − t) + (B − t) + (C − t) , z = (B − t)(C − t), ⎪ ⎪ ⎩ x = (A − t)n − (B − t)n + (C − t)n , z = (A − t)(C − t). Then, we also have x2 − 4z n = x2 − 4z n = x2 − 4z n ,

(4.9)

and x≡

x ≡

x ≡

(mod i ) if pi = 2,

2


2


2 p21 p22

···

p2s

+ 1 (mod

i )

if pi = 2,

z ≡ 0, (mod i ), z ≡ 0, (mod i ), z ≡ 0 (mod i ),

for all i (1 ≤ i ≤ s). In addition, we have x + x x + x x + x = (B − t)n , = (C − t)n , = (A − t)n . 2 2 2 From (4.7), we also have (x, z) = (x , z ) = (x , z ) = 1. It follows from this that x, z, x, z , x, z is a solution of the Diophantine equation (3.1), satisfying all of the conditions of Proposition 3.1 (note that pi | n for all 1 ≤ i ≤ s). Since it holds that (4.10)

x2 − 4z n = −3t2n + {terms of lower degree in t},


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2 n we can let √ the value x − 4z be negative by taking a sufficiently large t. Now, set K = Q( x2 − 4z n ). It follows from Proposition 3.1 that the ideal class group of an imaginary quadratic number field K has Z/nZ × Z/nZ × Z/nZ as a subgroup. The infinite property can be shown in the same way as for the previous case.

REFERENCES 1. N. Ankeny and S. Chowla, On the divisibility of the class numbers of quadratic fields, Pacific J. Math., 5 (1955), 321-324. 2. D. Byeon, Imaginary quadratic fields with noncyclic ideal class groups, Ramanujan J., 11(2) (2006), 159-163. 3. M. Craig, A construction for irregular discriminants, Osaka J. Math., 14(2) (1977), 365-402. 4. J. F. Mestre, Courbes elliptiques et groups de classes d’ide´ aux de certains corps quadratiques, Seminar on Number Theory, 1979-1980, Exp. No. 15, 18 pp., Univ. Bordeaux I, Talence, 1980. 5. J. F. Mestre, Groupes de classes d’ide´ aux non cyclique de corps de nombre, Seminar on Number Theory, Paris 1981-1982 (Paris, 1981/1982), Progr. Math., 38, Birkha¨ user Boston, Boston, MA, 1983, pp. 189-200. 6. S. R. Louboutin, On the divisibility of the class number of imaginary quadratic number fields, Proc. Amer. Math. Soc., 137(12) (2009), 4025-4028. 7. M. R. Murty, Exponents of class groups of quadratic fields, in: Topics in Number Theory, (University Park, PA, 1997), Mathematical Applications, Vol. 467, Kluwer Acad. Publ., Dordrecht, 1999, pp. 229-239. ¨ 8. T. Nagell, Uber die Klassenzahl imaginar quadratischer Zahko¨ rper, Abh. Math. Seminar Univ. Hambrug, 1 (1922), 140-150. 9. K. Soundararajan, Divisibility of class numbers of imaginary quadratic fields, J. London Math. Soc., 61(2) (2000), 681-690. 10. P. Weinberger, Real quadratic fields with class numbers divisible by n, J. Number Thoery, 5 (1973), 237-241. 11. Y. Yamamoto, On unramified Galois extensions of quadratic number fields, Osaka J. Math., 7 (1970), 57-76. 12. G. Yu, A note on the divisibility of class numbers of real quadratic fields, J. Number Theory, 97 (2002), 35-44. Kwang-Seob Kim School of Mathematics Korea Institute for Advanced Study (KIAS) 85 Hoegiro Seoul 130-722 Republic of Korea E-mail: [email protected]