exponents of class groups of real quadratic fields

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International Journal of Number Theory Vol. 4, No. 4 (2008) 597–611 c World Scientific Publishing Company

Int. J. Number Theory 2008.04:597-611. Downloaded from www.worldscientific.com by THE HARISH-CHANDRA RESEARCH INSTITUTE (HRI) on 05/12/15. For personal use only.

EXPONENTS OF CLASS GROUPS OF REAL QUADRATIC FIELDS

KALYAN CHAKRABORTY∗ , FLORIAN LUCA† and ANIRBAN MUKHOPADHYAY‡ ∗Harish-Chandra Research Institute Chhatnag Road, Jhusi, Allahabad 211 019, India †Mathematical Institute UNAM, Ap. Postal 61-3 (Xangari), CP 58089 Morelia, Michoac´ an, M´ exico ‡Institute

of Mathematical Sciences, CIT Campus Taramani, Chennai 600113, India ∗[email protected] †[email protected] ‡[email protected]

Received 6 June 2006 Accepted 27 April 2007 In this paper, we show that the number of real quadratic fields K of discriminant ∆K < x whose class group has an element of order g (with g even) is ≥ x1/g /5 if x > x0 , uniformly for positive integers g ≤ (log log x)/(8 log log log x). We also apply the result to find real quadratic number fields whose class numbers have many prime factors. Keywords: Class group; real quadratic fields. Mathematics Subject Classification 2000: 11R58; 11R29.

1. Introduction Murty [11] showed that if g is an odd positive integer, then for every > 0 the number of real quadratic fields with discriminant ≤ x and whose class group contains an element of order g is x1/2g− . Murty’s argument also gives that number of such real quadratic fields is x1/2g− when g ≡ 2 (mod 4), but when 4 | g his argument only gives a lower bound of the shape x1/4g− . Yu [16] improved on Murty’s result to x1/g− when g is odd, and Luca [10] improved it to x1/g / log x when g is even. When g = 3, Chakraborty and Murty [6] showed that the number of such fields is x5/6 and this was recently improved by Byeon and Koh [5] to x7/8 . In case of g = 5, 7, Byeon [4] proved that there are x1/2 such fields.

597

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K. Chakraborty, F. Luca & A. Mukhopadhyay

In all these estimates, the implied constants might depend on both g and . In this paper, we improve upon Luca’s estimate from [10]. More precisely, we remove the logarithmic factor, give an explicit expression for the implied constant, and show that the resulting inequality is true uniformly in some range of g versus x once x is sufficiently large. In case of imaginary quadratic fields, Ankeny and Chowla [1] proved the existence of infinitely many imaginary quadratic fields whose class number is divisible by a given positive integer g. Even though they did not mention it, their method demonstrates the existence of x1/2 such fields whose discriminant is < x in absolute value. Murty [11] improved upon this result and showed the existence of x1/2+1/g such imaginary quadratic fields. Soundararajan [15] improved Murty’s lower bound for the imaginary quadratic case. He showed that if 4 | g, then there exist x1/2+2/g− such fields, and if 4 | (g−2), then their number is x1/2+3/(g+2)− . Here, > 0 can be arbitrary.√ To fix ideas, let K = Q( d) be a real quadratic field, where d > 1 is a square free positive integer. Write CK for the class group of K. Let x be a large real number and define, for a fixed even integer g > 1, Ng (x) = #{d ≤ x : µ2 (d) = 1 and CK has an element of order g}. Here, we prove the following theorem. Theorem 1. There exists an absolute constant x0 such that if x > x0 , then the inequality x1/g 5 holds for all even positive integers 2 ≤ g ≤ (log log x)/(8 log log log x). Ng (x) ≥

Our proof follows Luca’s arguments from [10] but uses several explicit estimates from Diophantine equations, such as upper bounds for the size of the integer solutions to hyperelliptic Diophantine equations due to Baker [2] and Bugeaud [3], as well as various estimates from analytic number theory such as the Brun sieve and the Chebotarev Density theorem. In particular, the constant x0 appearing in the statement of our Theorem 1 arises as a byproduct of all such estimates. While presumably computable, we leave it as a challenge to the interested reader to find a value suitable for this constant x0 . 2. Preliminary Results We start with some preliminary results. For any odd positive integer m, let √ √ (X + X 2 + 1)m + (X − X 2 + 1)m ∈ Z[X]. Pm (X) = 2 For example, P1 (X) = X and P3 (X) = 4X 3 + 3X, etc. It is well known that Pmn (X) = Pm (Pn (X)) holds for all odd positive integers m and n. It is also well

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known that if d > 1 is a square free integer, such that the Pell equation


U 2 − dV 2 = −1

(1)

has at least one solution in positive integers (U, V ), then it has infinitely many such positive integer solutions. Further, denoting by (U1 , V1 ) the smallest positive integer solution of Eq. (1), then all the other solutions are of the form (Um , Vm ), for some odd positive integer m, where Um = Pm (U1 ). Let g be a positive integer. For any positive integer n, we may write n2g + 1 = 2 dv , where d is square free. It is clear that d > 1. Hence, in light of the above comments, ng = Um = Pm (U1 ) for some positive integer m, where (Um , Vm ) is the mth solution of the Pell equation (1) corresponding to the square free number d. We may ask the following question: Can we say something about those positive integers n such that for some g ≥ 1, the index m appearing above is > 1? Our first result here shows — a little bit more than — that the set of such positive integers is of asymptotic density zero. It will play an important role in establishing our main result Theorem 1. For any subset of positive integers A and any real number x ≥ 1, we write A(x) = A ∩ [1, x]. Lemma 1. Let A be the set of those positive integers n such that there exists a positive integer g ≥ 1 with the property that if we write n2g + 1 = dv 2 , with some square free integer d, then the pair (U, V ) = (ng , v) is not the minimal solution of the Pell equation U 2 − dV 2 = −1. Then the estimate #A(x) =

x1/3 + O(x1/5 ) 22/3

(2)

holds as x → ∞. Proof. Assume that n ∈ A(x). Then ng = Pm (u) holds with some positive integers g, m and u. Since Pmn (X) = Pm (Pn (X)) holds in Z[X] for all odd positive integers m and n, it follows that we may assume that m = p is a prime. If u = 1, we then get that √ √ (1 + 2)p + (1 − 2)p g . n = Pp (1) = 2 When g = 1, we then get that n = Pp (1) ≤ x, therefore p log x. Hence, there are at most O(log x) integers n ∈ A(x) for which g = u = 1. If g > 1, we then get that Pp (1) is a perfect power. From known results concerning perfect powers in binary recurrent sequences (see [14]), we get that the number of such exponents p is O(1). From now on, we assume that u > 1. In this case, √ √ (u + u2 + 1)p + (u − u2 + 1)p (2u + 1)p (2u)p − 1 < < . (3) 2 2 2 Assume again that g = 1. Then n = Pp (u) for some odd prime p and some positive integer u. If is clear that if u is such that Pp (u) ≤ x, then n = Pp (u) ∈ A(x). If

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p = 3, we get that 4u3 + 3u ≤ x, and the number of such positive integers u is 2−2/3 x1/3 + O(1). If p ≥ 5, then from estimates (3), we get that (2u)p ≤ 2x + 1, therefore u x1/p and p log x. Hence, the number of n ∈ A(x) for which g = 1 is x1/3 x1/3 1/5 1/7 + O(x + (log x)x ) = + O(x1/5 ). (4) 22/3 22/3 From now on, we assume that g > 1. It is well known that the roots of Pp (X) are zj = i sin((2j + 1)π/p), j ∈ {0, 1, . . . , p − 1}. In particular, Pp (X) has no double roots. Hence, from known results about perfect power values of polynomials (see again [14]), we deduce that for any fixed p ≥ 3, the equation Pp (u) = ng has only finitely many positive integer solutions (u, n, g). From now on, we assume that p > 100. Now note that u | Pp (u). Further, it is known that gcd(u, Pp (u)/u)|p, and that if this greatest common divisor is p, then p Pp (u)/u. Hence, from the equation ng = Pp (u), we deduce that either u = ng1 ,

Pp (u)/u = ng2 ,

and n1 n2 = n,

or u = pg−1 ng1

Pp (u)/u = png2 ,

and pn1 n2 = n.

Assume first that n1 = 1. Then, since u > 1, we have that u = pg−1 , and png2 = Pp (u)/u. Hence, xg ≥ ng = Pp (u) ≥ up /2 = pp(g−1) /2 ≥ pp(g−1)/2 ≥ ppg/4 , therefore g log x pg log p, giving p log x/ log log x. Next, assume that n1 > 1. Then log u ≥ g log n1 , while p log u − log 2 = log(up /2) < log(Pp (u)/u) ≤ log(ng2 p) ≤ g log n2 + log p, therefore p−1≤

log n2 + (log p)/g p log u − log 2 ≤ . log u log n1

(5)

Since g ≥ 2 and n2 ≤ x/n1 , it follows, from (5), that (p − 1) log n1 ≤ log x − log n1 + (log p)/2, giving np1 ≤ p1/2 x.

(6)

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Further, since n1 ≥ 2, g ≥ 2 and n2 ≤ x, we have p−1≤

log n2 + (log p)/2 ≤ 2 log x + log p. log 2


Since log p < p/2 − 1 when p > 100, we get that p ≤ 4 log x. Thus, in both cases when n1 = 1 or n1 > 1, we have that p ≤ 4 log x when x is sufficiently large. We also record, for future use, that if x is large then 4 log x < x2 , therefore estimate (6) implies that n1 ≤ x2/p .

(7)

We now deal with bounding g. It is known that 2Pp (X) = Qp (2X), where Qp (X) ∈ Z[X] is a monic polynomial. A quick way to prove this fact is to first notice, by comparing leading terms, that Qp (X) ∈ Q[X] is monic, and next to notice that the roots of Qp (X) 2zj = 2i sin((2j + 1)π/p) = e(2j+1)i/p − e−(2j+1)i/p ,

j = 0, . . . , p − 1,

are all algebraic integers; thus, Qp (X) ∈ Q[X] is, in fact, a polynomial with integer coefficients. Hence, the equation Pp (u) = ng is equivalent to Qp (2u) = 2ng . At this stage, we record a result of Bugeaud from [3]. Lemma 2. Let f (X) = X d + a1 X d−1 + · · · + ad ∈ Z[X] be a polynomial of degree d ≥ 2 with integer coefficients without multiple roots. Assume that a = 0 and u are integers such that f (u) = av m . Then either m ≤ 2d log(2H + 3) or m ≤ 215(d+6) d7d |D|3/2 (log |D|)3d (log(3|a|))2 log log(27|a|), where D is the discriminant of f and H = max{|a1 |, . . . , |ad |} is the naive height of f . We apply Lemma 2 to bound the number g in terms of x. For this, we need upper bounds for the parameters H and |D| associated to the polynomial Pp (X). Since Pp (X) has only positive coefficients, it follows that √ p (1 + 2)p . H ≤1+ ai = Pp (1) < 2 i=1 As for the discriminant D of Pp (X), note that D=

p−1

Pp (zj ),

j=0

where again zj = i sin((2j + 1)π/p), j = 0, . . . , p − 1, are the roots of Pp (X). Since p ((X + X 2 + 1)p − (X − X 2 + 1)p ), Pp (X) = √ (8) 2 X2 + 1 one checks easily that ±p for j = 0, . . . , p − 1. Pp (zj ) = cos((2j + 1)π/p)

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Since |cos((2j + 1)π/p)| = |sin((p − 2(2j + 1))π/(2p))| ≥ sin(π/(2p)) ≥ 1/p for all j = 0, . . . , p − 1, and p ≥ 3, we get that |D| ≤ p2p .


Thus, from Lemma 2, we conclude that either √ g ≤ 2p log((1 + 2)p + 3) p2 , or g ≤ 215(p+6) p7p p3p (2p log p)3p (log 6)2 log log 54. Hence, there exists an absolute constant c1 > 0 such that g ≤ exp(c1 p log p). (9) We define y = log x/ log log x. If p ≤ y, then log p < (log log x)/2 and the above inequality (9) immediately implies that (10) g < exp(c2 log x log log x), where c2 = c1 /2. We now look at the case when p > y. Estimate (7) implies that n1 ≤ x2/y = exp(2 log x log log x).

(11)

Further, the constant term ap−1 of Pp (u)/u is p. This can be noticed by observing that this constant term is Pp (t) = Pp (s) = p (cf. formula (8)). ap−1 = lim t→0 t s=0 Since u | Pp (u)/u − ap−1 , we get that ng1 | ng2 − p, or pg−1 ng1 | png2 − p, according to whether u = ng1 or pg−1 ng1 . Assume first that n1 = 1. Then pg−2 | ng2 − 1. Using a (trivial) lower bound for linear forms in p-adic logarithms, we get that log g log n2 + g − 2 ≤ ordp (ng2 − 1) ≤ (p − 1) log p log p < p log x + log g < 4(log x)2 + log g. This shows that g (log x)2 in this case. Hence, inequality (10) holds in this case as well if x is large. Assume now that n1 > 1. Then ng1 | ng2 − δ, where δ ∈ {1, p}. Let q be the smallest prime factor of n1 . Applying a linear form in q-adic logarithms, we get that q log n2 log p log g n1 log x log log x log g, g ≤ ordq (ng2 − δ)

log q which together with inequality (11) implies easily that g < exp(3 log x log log x).

(12)

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Comparing inequalities (10) and (12), we conclude that inequality g < exp(c3 log x log log x)

603

(13)

holds for all sufficiently large x with c3 = max{c2 , 3}. Note now that if p, g and n1 are fixed, then either u = ng1 , n2 = (Pp (u)/u)1/g or u = pg−1 ng1 , n2 = (Pp (u)/pu)1/g . In conclusion, if p, g and n1 are fixed, then n can take at most two values. Since p > 100, it follows from estimate (7) that n1 < x1/50 . Since we also have p < 4 log x, it follows, from the above observation and inequality (13), that the number of possible n ∈ A(x) for which p > 100 is ≤ 8(log x)x1/50 exp(c3 log x log log x) < x1/49 for large x, which completes the proof of the lemma. While unnecessary for the purpose of the present paper, we make a few comments about the set A. As we have seen, whenever ng is of the form Pp (u) with some odd prime p and positive integer u, we get that n ∈ A with g = 1. Furthermore, all elements n from A with g = 1 arise in this way for some prime number p and positive integer u (not necessarily uniquely determined). Let E be the subset of A consisting of all the positive integers n in A for which the corresponding g is > 1. We may ask what is the behavior of E(x) as x → ∞. Our next result addresses this question. Recall that the ABC conjecture is the statement that for every ε > 0 there exists a constant C(ε) depending on ε such that whenever A, B and C are nonzero coprime integers with A + B = C, the inequality 1+ε  p max{|A|, |B|, |C|} ≤ C(ε)  p | ABC

holds. Lemma 3. (i) The estimate #E(x) ≤ x(2+o(1))ε(x) holds as x → ∞, where ε(x) =

10 log log log log log x 3 log log log log x

1/3 .

(ii) The ABC conjecture implies that #E(x) = O(1). Proof. (i) We recall the following result of Baker from 1964 (see [2]). Lemma 4. Let f (x) = a0 X d + a1 X d−1 + · · · + ad ∈ Z[X] be a polynomial of degree d with integer coefficients. Let m ≥ 2 be fixed and let (x, y) be an integer solution to the equation y m = f (x).

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(i) If m ≥ 3 and f (X) has at least two simple roots, then 2

max{|x|, |y|} ≤ exp(exp((5m)d (d10d H)d )). (ii) If m = 2 and f (X) has at least three simple roots, then 2

max{|x|, |y|} ≤ exp(exp(exp((d10d H)d ))). Here, H = max{|a0 |, . . . , |ad |, e}. Int. J. Number Theory 2008.04:597-611. Downloaded from www.worldscientific.com by THE HARISH-CHANDRA RESEARCH INSTITUTE (HRI) on 05/12/15. For personal use only.

We fix the prime p and look at the equation ng = Pp (u)

(14)

with g ≥ 2 and n ≤ x. By estimate (9), we have that g ≤ G(p) := exp(c1 p log p). Assume first that g ≥ 3. Since p ≥ 3 and Pp (X) has only simple roots, we may apply Lemma 4 to infer upper bounds for the integer solutions (n, (14) √ u)p of Eq. p when p and g are fixed. By Lemma 4 and the fact that H ≤ (1 + 2) < e , for a fixed g we get that the largest solution of Eq. (14) satisfies 2

n ≤ exp(exp((5g)p (p10p H)p )) ≤ exp(exp(exp(20p3 log p)))

(15)

if p is large. If g = 2, we then get, by the same Lemma 4 2

n ≤ exp(exp(exp((p10p ep )p ))).

(16)

Let κ(p) be the function appearing in the right-hand side of inequality (16). Imposing that κ(p) ≤ x1/ log log x and solving for p, we get that 2

(p10p ep )p ≤ log(log(log x/log log x)). One checks easily that the largest solution p = P (x) of the above equation satisfies P (x) = (1 + o(1))(ε(x))−1 . Thus, if we write E1 (x) for the subset of E(x) consisting of those n for which g ≥ 2 and p ≤ P (x), then #E1 (x) ≤ π(P )H(P )x1/ log log x ≤ x2/ log log x

(17)

for large values of x. Let E2 (x) = E(x)\E1 (x). Assume that n corresponds to some prime p > P . Then inequality (7) tells us that n1 ≤ x2/P ≤ x(2+o(1))ε(x) . Since p ≤ 4 log x, g satisfies inequality (13), and n is determined in at most two ways by n1 , p and g, we get that the number of such possibilities for n is (18) #E2 (x) ≤ 2π(4 log x) exp(c3 log x log log x)x(2+o(1))ε(x) = x(2+o(1))ε(x) . Since #E(x) ≤ #E1 (x) + #E2 (x), the conclusion of (i) follows from inequalities (17) and (18). (ii) Applying the ABC conjecture to the equation n2g − du2 = −1 with ε = 1/10, we get n2g (nud)11/10 .

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605

Since u ng /d1/2 , we get n2g n11(g+1)/10 d11/20 , therefore


d n2(9g−11)/11 . Let (X1 , Y1 ) be the minimal solution of the equation X 2 − dY 2 = −1. It is clear that X1 d1/2 n(9g−11)/11 . Thus, if ng is the m’th solution (here, m > 1 is odd but not necessarily prime), then ng = Pm (X1 ) X1m n(9g−11)m/11 , giving 9m

11m

ng( 11 −1)(1− g(9m−11) ) 1. Note that g ≥ 2 and 9m/11 − 1 ≥ 27/11 − 1 > 1. Further, if m ≥ 5 then 1−

11m 11m 7m − 22 13 ≥1− = ≥ g(9m − 11) 18m − 22 18m − 22 68

while if m = 3 but g ≥ 3, we then have 1−

11 5 11m ≥1− = . g(9m − 11) 16 16

Thus, n 1 if either m ≥ 5 or g ≥ 3, and since the equation n2 = P3 (u) = 4u3 + 3u has only finitely many integer solutions (n, u), it follows that there are only finitely many possibilities for n when (g, m) = (2, 3) also, which completes the proof of part (ii). We now fix a positive integer g > 1. Let Qg (X) = X g − 2 ∈ Z[X]. We write Cg = {n : Qg (X) (mod p) is irreducible for some prime p | n}. Our next result addresses the number of elements in Cg (x) as x → ∞. Lemma 5. The set Cg (x) contains all n ≤ x with at most o(x) exceptions uniformly when 1 ≤ g < (log log x)/(7 log log log x). Proof. We put G = log log x/(7 log log log x). Let g ≤ G be a positive integer. Let Pg be the set of primes p such that Qg (X) is irreducible modulo p. For a positive real number x, let πg (x) = #Pg (x). It is known that if g is fixed, then πg (x) ∼ π(x)/g as x → ∞. A uniform version of this result appears in the work of Lagarias and Odlyzko [9] (see also Pappalardi’s papers [12, 13]), where it is shown that

x π(x) √ +O πg (x) = g exp(A log x/g)

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uniformly when g < B(log x)1/8 , where A and B are some fixed positive constants. We put y = exp((log log x)10 ) and we use the above result to find a lower bound for the sum 1 x dπg (t) Σg (x) = = . p t y


y≤p≤x p∈Pg

Note that since we already have g ≤ G, it follows that g < B(log y)1/8 for all positive integers g in our range provided that x is large. Hence, by partial summation, for large x we have x dπg (t) Σg (x) = t y x πg (t) πg (t) t=x + dt = t t=y t2 y

x 1 π(t) +O dt + O(1) = gt2 t exp(A(log t)1/2 /g) y

x log x 1 = log log t + O(1) + O g exp(A(log y)1/2 /g) t=y 1 (log log x − 10 log log log x) + O(1), g where in the last estimate above we used the fact that the estimate log x = exp(−A(log log x)5 /g + log log x) exp(A(log y)1/2 /g) =

(19)

< exp(−7(log log x)4 + log log x) = o(1) holds as x → ∞ in our range for g. We are now ready to prove Lemma 5. We let x be large, assume that g ≤ G and let Bg (x) be the complement of Cg (x) in [1, x]. We may certainly assume that g > 1, otherwise there is nothing to prove. Hence, Bg (x) is contained in the set of n ≤ x which have no prime factor p ∈ Pg ∩ [y, x]. By Brun’s method (see, for example, [8, Theorem 2.3]), the number of such numbers does not exceed

1 1− #Bg (x) x p y≤p≤x p∈Pg

   1 1    + O

x exp−  p p2   



y≤p≤x p∈Pg

p≥2

x(log log x)5 x(log log x)10/g

. 1/g (log x) (log x)1/g

(20)

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We thus get

#Bg (x)

g≤G

x(log log x)5 G x ,

log log x (log x)1/G


which shows that ∩g≤G Cg (x) contains all positive integers n ≤ x with at most O(x/ log log x) exceptions, which is even a slightly stronger result than the one aimed for. 3. The Proof of Theorem 1 We follow the method of proof from [10]. We let x be a large positive real number and g ≤ (log log x)/(16 log log log x) be a positive integer. We set z = (x/5)1/2g and put Dg for the set of all odd integers 1 < n ≤ y in Cg (y) which are not in A. Since (2g) < (log log x)/(8 log log log x), it follows that for large values of x, the inequality (2g) < (log log y)/(7 log log log y) holds. Thus, by Lemmas 1 and 5, we conclude that #Dg ≥ z/2.1 − 1 > x1/2g /5 if x > x0 . We shall use the numbers n ∈ Dg to create real quadratic fields having an element of order 2g in their class group. For each n ∈ Dg , we write n2g + 1 = dv 2 , for some positive integers d and v x/5 + 1 < x/4. As n > 1 is odd and with d squarefree. Clearly, d ≤ n2g + 1 < √ 2g is even, we get that 2 d. Let K = Q[ d]. We note that the discriminant of K is 4d < x. Further, since n ∈ A, it follows that the pair (U, V ) = (ng , v) is the smallest solution in positive integers of the equation U 2 − dV 2 = −1. In particular, √ g ζ = n + v d is the fundamental unit in OK . Thus, the fields K obtained as above when n ∈ Cg are all distinct. We are now left with showing that CK has an element of exact order 2g. For this, we note the factorization √ √ n2g = 1 − dv 2 = (1 − v d)(1 + v d) √ √ √ in Z[ d]. Since d is even, the two principal ideals [1 + v d] and [1 − dv] are coprime. The above factorization now leads to the conclusion that each of the ideals above must be a perfect 2gth power. Hence, √ I 2g = [1 + v d] for some ideal I in OK . This shows that the order of I in the group CK divides 2g. It remains to prove that this order is exactly 2g. Suppose on the contrary that the order √of I is less than 2g. Then there exists a prime r | 2g and an element α = a + b d ∈ OK such that ideal-wise √ (21) [αr ] = [1 + v d]. Here, a and b are nonzero integers with gcd(a, db) = 1. Identifying generators in (21) and using the fact that ζ is the fundamental unit OK , we get the following Diophantine equation √ (22) 1 + v d = αr ζ s , ε ∈ {±1}, s ∈ Z.

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We distinguish the following two cases: Case 1: r = 2


Without loss of generality, we may assume that s ∈ {0, 1}. Taking norms in Eq. (22), we get √ −n2g = 1 − dv 2 = NK (1 + v d) = NK ()NK (α2 )NK (ζ s ) = (−1)s NK (α)2 . From the above equation, we conclude that s is odd, and so Eq. (21) becomes √ (1 + v d)ζ = α2 . (23) Equation (23) implies (1 +

2 b 2g n2g + 1)(ng + n2g + 1) = a + n +1 . v

Equating the rational and irrational parts from Eq. (24), we get 2 b 2 2g = n2g + ng + 1, a + (n + 1) v

b 2a = ng + 1. v

(24)

(25)

From the last two equations above, we get g

2 n +1 2 2g = n2g + ng + 1; a + (n + 1) 2a 2g

n + 2ng + 1 2 2g a + (n + 1) = n2g + ng + 1; 4a2 (n2g + 1)(n2g + 2ng + 1) 2g = n + n g + 1 a2 . 4 The last equation above can be rewritten as a4 +

(n2g + 1)(ng + 1)2 = 0. a4 − n2g + ng + 1 a2 + 4

(26)

Solving Eq. (26), we get 2

(ng + 1) n2g + 1 or a2 = . 2 2 None of the above two equalities is possible as 2 is not a square and U = ng is not a solution of the Pell equation U 2 − 2V 2 = −1 (for if it were, it should be the first one which is U = n = 1, which is not allowed). a2 =

Case 2: r > 2 In this case, we have

√ 1 + v d = αr ζ s .

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We may choose α so that s is odd and negative. To see that, we start with any α satisfying the above equation for some and s, and then replace α by αζ t , where t is a large positive integer incongruent to s modulo 2. After this replacement, the exponent s will be replaced by s − rt, which is negative for large t and odd, and is replaced by (−1)rt . Replacing now α by −α, we may assume that = −1. Thus, √ (27) 1 + v d = −αr ζ −s , s > 0 is odd. Taking the norm in both sides of Eq. (27), we get √ −n2g = 1 − dv 2 = NK (1 + v d) = NK (−1)NK (αr )NK (ζ −s ) = −NK (α)r . Thus, NK (α) = n2g/r , which leads to a2 − db2 = n2g/r . We rewrite Eq. (27) as (1 +

(28)

√ dv)ζ s = −αr .

Now using relation (28), the last equation above becomes (1 + n2g + 1)(ng + n2g + 1)s = −(a + η a2 − n2g/r )r ,

(29)

where η ∈ {±1}. Identifying the rational and irrational parts from Eq. (29), we get (1 + n2g + 1)(ng + n2g + 1)s + (1 − n2g + 1)(ng − n2g + 1)s = −(a + a2 − n2g/r )r − (a − a2 − n2g/r )r . Reducing the above relation modulo q, where q | n is in Pg , we get that the left-hand side of the above relation is 2 mod q, while its right-hand side is −(2a)r = (−2a)r . Thus, (−2a)r ≡ 2

(mod q),

showing that if we put x0 = −2a, then the two polynomials X 2g/r −x0 and X 2g −2 = Q2g (X) modulo q have a root in common. Thus, Q2g (X) is not irreducible modulo q, which contradicts the choice of q. Thus, we have reached a contradiction. The theorem is therefore proved. 4. Application In this section, we point out an application of our Theorem 1 to class numbers of real quadratic fields having “many” prime factors.

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Theorem 2. For any ε > 0 there exists x0 = x0 (ε) such that if x > x0 , then there exist at least exp(8 log x log log log x/ log log x) real quadratic number fields K of discriminant < x whose class groups have exponents eK with at least (1 − ε) log log log x/ log log log log x distinct prime factors. Proof. Let ε > 0 and x be a large positive real number. We put w = (1 − ε/16) log log log x and g = p≤w p. By the Prime Number Theorem, g = e(1+o(1))w as x → ∞, therefore if x is large then g < (1 − ε)(log log x)/(8 log log log x). It is clear that g is even. By Theorem 1, the number of real quadratic number fields having discriminant < x and whose class group has exponent a multiple of g exceeds

log x 8 log x log log log x x1/g = exp − log 5 > exp 5 g log log x if x > x0 (ε). Since for each one of them we have ω(eK ) ≥ ω(g) = π(w) > (1 − ε)

log log log x log log log log x

for large enough x, the result follows. Noting that eK ≤ hK ∆K log ∆K , where hK is the class number and ∆K is the discriminant of K, Theorem 2 above guarantees the existence of a sequence of real quadratic fields K with eK → ∞ such that ω(eK ) ≥ (1 + o(1))

log log log eK log log log log eK

as eK → ∞.

(30)

In the subsequent paper [7], we will prove the existence of a sequence of real quadratic number fields K whose class numbers tend to infinity and such that their class groups have exponents with even more distinct prime factors than shown by inequality (30) above. Acknowledgments Most of the early work on this paper was done during a visit by the second-named author to the Harish-Chandra Research Institute in Allahabad, India. Most of the later work was done during a visit by the last two authors to CRM, Université de Montréal, Canada. The hospitality and support of these institutions are gratefully acknowledged. F. L. thanks also Professors F. Pappalardi and I. E. Shparlinski for helpful advice. During the preparation of this paper, F. L. was also supported in part by Grants SEP-CONACyT 46755, PAPIIT IN104005, a TWAS Associateship and a Guggenheim Fellowship. References [1] N. Ankeny and S. Chowla, On the divisibility of the class numbers of quadratic fields, Pacific J. Math. 5 (1955) 321–324.

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[2] A. Baker, Bounds for the solutions of the hyperelliptic equation, Proc. Cambridge Philos. Soc. 65 (1969) 439–444. [3] Y. Bugeaud, Sur la distance entre deux puissances pures, C. R. Acad. Sci. Paris Sér. I Math. 332 (1996) 1119–1121. [4] D. Byeon, Real quadratic fields with class number divisible by 5 or 7, Manuscripta Math. 120(2) (2006) 211–215. [5] D. Byeon and E. Koh, Real quadratic fields with class number divisible by 3, Manuscripta Math. 111(2) (2003) 261–263. [6] K. Chakraborty and M. R. Murty, On the number of real quadratic fields with class number divisible by 3, Proc. Amer. Math. Soc. 131(1) (2002) 41–44. [7] K. Chakraborty, F. Luca and A. Mukhopadhyay, Class numbers with many prime factors, to appear in J. Number Theory (2008); doi: 10.1016/j.jnt.2008.03.010. [8] H. Halberstam and H.-E. Richert, Sieve Methods (Academic Press, London, 1974). [9] J. C. Lagarias and A. M. Odlyzko, Effective versions of the Chebotarev density theorem, in Algebraic Number Fields, ed. A. Fr¨ ohlich (Academic Press, New York, 1977), pp. 409–464. [10] F. Luca, A note on the divisibility of class numbers of real quadratic fields, C. R. Math. Acad. Sci. Soc. R. Canada 25(3) (2003) 71–75. [11] M. R. Murty, Exponents of class groups of quadratic fields, in Topics in Number Theory (University Park, PA, 1997), Math. Appl., Vol. 467 (Kluwer Acad. Publ., Dordrecht, 1999), pp. 229–239. [12] F. Pappalardi, On Hooley’s theorem with weights, Rend. Sem. Mat. Univ. Pol. Torino 53 (1995) 375–388. [13] F. Pappalardi, Squarefree values of the order function, New York J. Math. 9 (2003) 331–344. [14] T. N. Shorey and R. Tijdeman, Exponential Diophantine Equations, Cambridge Tracts in Mathematics, Vol. 87 (Cambridge University Press, Cambridge, 1986). [15] K. Soundararajan, Divisibility of class numbers of imaginary quadratic fields, J. London Math. Soc. 61(2) (2000) 681–690. [16] G. Yu, A note on the divisibility of class numbers of real quadratic fields, J. Number Theory 97 (2002) 35–44.

This article has been cited by:


1. Kalyan Chakraborty, Florian Luca, Anirban Mukhopadhyay. 2008. Class numbers with many prime factors. Journal of Number Theory 128, 2559-2572. [CrossRef]

exponents of class groups of real quadratic fields

exponents of class groups of real quadratic fields

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