AAA Abstract and Applied Analysis 1687-0409 1085-3375 Hindawi Publishing Corporation 570154 10.1155/2012/570154 570154 Research Article Divisibility Criteria for Class Numbers of Imaginary Quadratic Fields Whose Discriminant Has Only Two Prime Factors Pekin A. Akca Haydar Department of Mathematics Faculty of Science Istanbul University 34134 Istanbul Turkey istanbul.edu.tr 2012 9 12 2012 2012 17 10 2012 04 11 2012 2012 Copyright © 2012 A. Pekin. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We will prove a theorem providing sufficient condition for the divisibility of class numbers of certain imaginary quadratic fields by 2g, where g>1 is an integer and the discriminant of such fields has only two prime divisors.

1. Introduction

Let K=Q(D) be the quadratic fields with discriminant D and h=h(D) its class number. In the narrow sense, the class number of K is denoted by h+(D), where, if D>0, then h+(D)=2h(D) and the fundamental unit εD has norm 1, otherwise h+(D)=h(D). If the discriminant of |D| has two distinct prime divisors, then by the genus theory of Gauss the 2-class group of K is cyclic. The problem of the divisibility of class numbers for number fields has been studied by many authors. There are Hartung , Honda , Murty , Nagel , Soundararajan , Weinberger , Yamamoto , among them. Ankeny and Chowla  proved that there exists infinitely many imaginary quadratic fields each with class numbers divisible by g where g is any given rational integer. Later, Belabas and Fouvry  proved that there are infinitely many primes p such that the class number of the real quadratic field K=Q(p) is not divisible by 3. Furthermore, many authors [7, 1013] have studied the conditions for h+(D) to be divisible by 2n when the 2-class group of K is cyclic. However the criterion for h+(D) to be divisible by 2n is known for only n4 and the existence of quadratic fields with arbitrarily large cyclic 2-class groups is not known yet. Recently, Byeon and Lee  proved that there are infinitely many imaginary quadratic fields whose ideal class group has an element of order 2g and whose discriminant has only two prime divisors. In this paper, we will prove a theorem that the order of the ideal class group of certain imaginary quadratic field is divisible by 2g. Moreover, we notice that the discriminant of these fields has only different two prime divisors. Finally, we will give a table as an application to our main theorem.

2. Main Theorem

Our main theorem is the following.

Theorem 2.1.

Let D=pq be square-free integer with primes pq1(mod4). If there is a prime r1(mod8) satisfying (D/r)=1, then th(D) for at least positive integer t where t2.

In order to prove this theorem we need the following fundamental lemma and some theorems.

Lemma 2.2.

If D is of the form p·q where p and q are primes pq1(mod4), then there is a prime r1(mod8) such that (D/r)=1.

Proof.

Let a and b be quadratic nonresidues for p and q are primes such that (a/p)=-1, (b/q)=-1, where () denotes Legendre symbol and g·c·d(p,q)=1. Therefore, by Chinese Remainder Theorem, we can write wa(modp), wb(modq) for a positive integer w. Now, we consider the numbers of the form pqk+w such that pqk0+w1(mod8) for some 1k08. Since pqk0+w are distinct residues mod(8) for some 1k08, then we get pq(8n+k0)+w=8pqn+pqk0+w, n0. We assert that g·c·d(8pq,pqk0+w)=1. Really, we suppose that g·c·d(8pq,pqk0+w)=m>1, then there is a prime s such that sm, and so we have s8pq, spqk0+w. Thereby this follows that s=2, p or q. But since pqk0+w1(mod8), then s2 and sm; this is in contradiction with wa(modp), wb(modq). Therefore, g·c·d(8pq,pqk0+w)=1 holds. Thus, by the Dirichlet theorem on primes, there is a prime r satisfying r=pq(8n+k0)+w=8pqn+pqk0+w. Hence, it is seen that r1(mod8).

The following theorem is generalized by Cowles .

Theorem 2.3.

Let r, m, t be positive integers with m>1 and t>1, and let n=r2-4mt be square-free and negative. If mc is not the norm of a primitive element of OK whenever c properly divides t, then th(n).

Cowles proved this theorem by using the decomposition of the prime divisors in OK. But Mollin has emphasized in  that it contains some misprints and then he has provided the following theorem which is more useful in practise than Theorem 2.4.

Theorem 2.4.

Let n be a square-free integer of the form n=r2-4mt where r, m, and t are positive integers such that m>1 and t>1. If r24mt-1(m-1), then th(n).

Theorem 2.5.

Let n be a square-free integer, and let m>1, t>1 be integers such that

mt is the norm of a primitive element from K=Q(n),

mc is not the norm of a primitive element from K for all c properly dividing t,

if t=|m|2, then n1(mod8).

Then t divides the exponent of ψK, where ψK is the class group of K.

3. Proof of Main Theorem

Now we will provide a proof for the fundamental theorem which is more practical than all of the works above mentioned.

Proof.

From the assumption of Lemma 2.2, it follows that there is suitable prime r with r1(mod8) such that (D/r)=1. However, from the properties of the Legendre symbol, we can write (Dy2/r2)=1 for any integer y. Since (2,r)=1, then we have (Dy2/rt)=1. Therefore, there are integers x=a/2,y=b/2 such that the equation x2-Dy2=rt has a solution in integers. Hence, we can write a2-Db2=4rt, where ab(mod2). From this equation, it is seen that rt is the norm of a primitive element of OK, and, then by Theorem 2.5, t divides h(n).

We have the following results.

Corollary 3.1.

Let D be a square-free and negative integer in the form of D=n2-4r2g=p·q with n>1, g>1 are positive integers and p, q, r are primes such that pq1(mod4), r1(mod8). If r2g is the norm of a primitive element of OK, then the order of the ideal class group of K=Q(D) is 2g.

Corollary 3.2.

Let D be a square-free and negative integer in the form of D=p·q, then there exists exactly 34433 imaginary quadratic fields satisfying assertion of the main theorem.

4. Table

The above-mentioned imaginary quadratic fields K=Q(D) correspond to some values of D(5D106) which are given in Table 1. We have provided a table of the examples to illustrate the results above, using C programming language. Moreover, it is easily seen that the class numbers of imaginary quadratic fields of K=(QD) are divisible by 2g from Table 1.

D p q r h ( D )
65 5 13 17 8
1165 5 233 41 20
3341 13 257 41 72
10685 5 2137 73 116
30769 29 1061 41 112
45349 101 449 17 168
95509 149 641 17 176
97309 73 1333 89 216
102689 29 3541 73 496
125009 41 3049 17 504
18497 53 349 41 168
20453 113 181 17 116
223721 137 1633 97 496
378905 5 75781 41 592
567137 17 333613 89 640
650117 13 50009 17 848
735929 373 1973 41 1664
847085 5 169417 73 936
874589 241 3629 17 1160
875705 5 175141 41 1328
876461 53 16537 73 1584
971081 109 8909 17 1464
971413 29 33497 73 336
978809 13 75293 89 1728
987169 97 10177 17 624
999997 757 1321 17 380
Acknowledgment

This work was partially supported by the scientific research project with the number IU-YADOP 12368.