Imaginary Quadratic Fields with Class Group Exponent 5

Imaginary Quadratic Fields with Class Group Exponent 5 D.R. Heath-Brown Mathematical Institute, Oxford Abstract We show that there are finitely many imaginary quadratic number fields for which the class group has exponent 5. Indeed there are finitely many with exponent at most 6. The proof is based on a method of Pierce [13]. The problem is reduced to one of counting integral points on a certain affine surface. This is tackled using the author’s “square-sieve”, in conjunction with estimates for exponential sums. The latter are derived using the q-analogue of van der Corput’s method. Mathematics Subject Classification (2000): 11R29 (11D45, 11L07, 11N36, 11LR29)

Let −d be a negative fundamental discriminant and let E(−d) denote the ex√ ponent of the ideal class group of the imaginary quadratic field Q( −d). Under the Generalized Riemann Hypothesis one has E(−d)  (log d)/(log log d), see Boyd and Kisilevsky [3], and Weinberger [15]. However without any unproved hypothesis it is not even known that E(−d) → ∞. The problem of Euler’s “Numeri Idonei” is essentially equivalent to the case E(−d) = 2, and here it has long been known that there are finitely many admissible values of d. Both Boyd and Kisilevsky loc. cit., and Weinberger loc. cit., showed that there are finitely many fundamental discriminants for which E(−d) = 3. It should be remarked that none of the proofs, either for E(−d) = 2 or for E(−d) = 3, are effective. In this paper we shall handle the case in which E(−d) = 5. Theorem 1 There is an (ineffective) constant d5 such that E(−d) 6= 5 for every fundamental discriminant −d with d > d5 . While this is our principal result we also record the following easy observation. Theorem 2 Let r be a non-negative integer, and let E = 2r or E = 3.2r . Then there is an (ineffective) constant dE such that E(−d) 6= E for every fundamental discriminant −d with d > dE . It follows from these two results that E(−d) ≥ 7 for all large enough d. We begin by proving Theorem 2. If E(−d) = E with E as above, then the class group takes the form C (3) × C2r × C2a × . . . × C2b

1

where C (3) is the 3-part of the class group and a, . . . , b ≤ r. By Gauss’ theorem on the number of genera the number of 2-group factors above is ω(n)−1, whence #(C2r × C2a × . . . × C2b ) ≤ 2rω(d) . Moreover, from the work of either Helfgott and Venkatesh [7], or Pierce [12], we know that there exists a positive constant δ such that #(C (3) )  d1/2−δ . (In fact Helfgott and Venkatesh show that one may take δ = 0.058, although we do not need this information.) Thus if E(−d) = E we have h(−d)  d1/2−δ 2rω(d) . Since rω(d) = o(log d), this conflicts with Siegel’s (ineffective) lower bound h(−d) ε d1/2−ε when d is large enough. The theorem then follows. Indeed the argument shows that if f (d) is any function for which
f (d) = o (log d)(log log d)−1 then there are finitely many negative fundamental discriminants −d for which E(−d) = 2r or 3.2r and r ≤ f (d). We now describe the initial attack on Theorem 1. Our starting point is a result of Linnik and Vinogradov [11]. √ This shows that if ε > 0 is given, then the number of integral ideals A of Q( −d) having norm at most Z is asymptotically ZL(1, χ−d ) as soon as Z√≥ d1/4+ε . Here χ−d is the quadratic character corresponding to the field Q( −d). Since L(1, χ−d ) ε d−ε/2 (ineffectively), by Siegel’s Theorem, there are at least d1/4 such ideals, if d is large enough. If E(−d) = 5 then A5 is principal√for each integral ideal A. We then write A5 = (α) and observe that 2α = x + y −d with x, y ∈ Z, whence 4N (A)5 = x2 + dy 2 . Since x and y determine A this allows us to conclude as follows. Lemma 1 Let ε > 0 be given. Suppose that −d is a negative fundamental discriminant with E(−d) = 5. Then if Z := d1/4+ε the equation x2 + dy 2 = 4z 5 ,

x, y ∈ Z, z ∈ N, z ≤ Z

(1)

has at least d1/4 solutions when d is large enough. We shall write (1) in the form 4z 5 − dy 2 = x2 and note that we necessarily have |y| ≤ Y with Y := 2d1/8+3ε , say. We are thus led to consider the general question of estimating from above the number of integer points on affine surfaces of the type F (y, z) = x2 , subject to constraints |y| ≤ Y and |z| ≤ Z. When F is an irreducible binary form of even degree 2k, the number of solutions is Oε,k ((Y Z)1/2+ε ||F ||ε ) by Theorem 3 of Broberg [2]. (Here we use the notation ||F || to denote the maximum of the moduli of the coefficients of F .) If such a result were available in our case it would yield a bound Oε (d3/16+3ε ) for sufficiently small positive , which would be enough to complete the proof of Theorem 1. However Broberg’s result is based on the fact that the equation F (y, z) = x2 describes a curve in weighted projective space, under his assumption on F , while we have no such interpretation of the variety (1). Broberg’s bound is one of a family of such estimates originating from the work of Bombieri and Pila [1], and of the author [5]. For affine surfaces the author has shown in Theorem 15 of [4], that all integral points on an absolutely 2

irreducible affine surface are contained in a small number of curves. Specifically, let F (x, y, z) ∈ Z[x, y, z] have total degree D and consider integral points on F (x, y, z) = 0 with |x| ≤ X, |y| ≤ Y, |z| ≤ Z. Write T for the largest value of X a Y b Z c corresponding to a monomial occurring in F . Then there are at most ) ( s (log X)(log Y )(log Z) D,ε (T ||F ||)ε exp 2 log T curves, which contain all the integral points to be counted. Moreover the curves have degrees which can be bounded in terms of D and ε alone. In our case we have T ≥ X 2 , leading to a bound np o √ ε (T ||F ||)ε exp 2(log Y )(log Z) ε d 3/4+6ε on the number of curves. Unfortunately this is too large for our purposes. It therefore appears that estimates arising from the above-mentioned techniques are insufficient, and we turn instead to a method which may be traced back to the work of Hooley [9], and which the author has dubbed the “square sieve”. In its simplest form, for which the reader is referred to the author’s paper [6] for details, the method depends on an inequality of the shape #{(x, y, z) : x2 = F (y, z)}  (#P)−1

X X F (y, z) { ( )}2 . p y,z

(2)

p∈P

Here P is a suitable set of primes. On expanding the square one is led to consider incomplete character sums of the type X F (y, z) ( ). pp0 y,z

(3)

One can factor the corresponding complete sums as a product of two terms X y,z∈Fp

(

F (y, z) )ep (ay + bz) = p

X

ep (ay + bz)

x,y,z∈Fp x2 =F (y,z)

which may be estimated via a result of Hooley [8]. If we choose P to be the set of primes from a suitable interval (P, 2P ] the outcome is that #{(x, y, z) ∈ Z3 : x2 = F (y, z), |y| ≤ Y, |z| ≤ Z} ε (Y ZP −1 + P 2 )(P ||F ||)ε , providing that F (y, z) cannot be factorized as G(y, z)2 H(αy + βz). Choosing P = (Y Z)1/3 then leads to a bound Oε ((Y Z)2/3 (Y Z||F ||)ε ). For our problem, with Y = 2d1/8+3ε and Z = d1/4+ε , the bound is Oε (d1/4+3ε ), which is not quite sufficient for our purposes. When Y is appreciably smaller than Z the procedure above is inefficient. A superior version of the square-sieve for such situations has been developed by Pierce [13], who applies it to the equation 4z 3 − dy 2 = x2 , 1 ≤ z ≤ Z = d1/2 , |y| ≤ Y = 2d1/4 3

and shows that there are Oε (d27/56+ε ) solutions. √ This allows her to deduce that the 3-part h3 (−d) of the class number for Q( −d) is Oε (d27/56+ε ). The method of the previous paragraph would only give an exponent 1/2 + ε, which is trivial in this context. Pierce uses an inequality of the form (2) in which P is a set of products p = uv of primes, with u roughly of size v 2 . She completes the exponential sum (3) with respect to the longer variable z only, and applies the q-analogue of van der Corput’s method with respect to the incomplete sum over the shorter variable y. For this last step it is necessary for there to be a factor of pp0 = uvu0 v 0 having a convenient size, and this is why it is necessary to use composite values of p. For the most part Pierce’s method extends immediately to the general equation x2 = F (y, z), but we shall confine our attention to the case in which F (y, z) = 4z k − dy 2 , with a fixed prime exponent k ≥ 3. Here we prove the following bound. Lemma 2 Let η > 0 and a prime k ≥ 3 be given, and suppose that −d is a negative fundamental discriminant. Assume that Y ≤ Z ≤ Y 3 and Z ≥ dη . Then #{(x, y, z) ∈ Z × Z × N : x2 = 4z k − dy 2 , |y| ≤ Y, z ≤ Z} η,k

{Y 7/8 Z 1/2 + Y 11/14 Z 4/7 }(log Z)3 .

Notice that we improve on the bound Y Z 1/2 whenever Y ≤ Z ≤ Y 3−η . When Y = 2d1/8+3ε and Z = d1/4+ε we obtain a bound  d1/4−1/112+4ε . This plainly suffices to deduce Theorem 1 from Lemma 1. Similarly, with Y = 2d1/4 and Z = d1/2 we may recover Pierce’s bound for h3 (−d). Rather than repeat Pierce’s argument in its entirety, we merely detail the necessary changes. We begin by writing Y = N and Z = L, so as to accord with Pierce’s notation. We write U0 = Qα , V0 = Q1−α , so that Pierce’s primes 0 u and v satisfy U0  u  U0 and V0  v  V0 respectively. We shall require that U0  dη/8 , which suffices to ensure that u takes  U0 (log Q)−1 possible prime values u - d. Similarly we shall require that V0  dη/8 . In our setting the bound in (2.6) of [13] becomes LN Q−1 (log Q)2 . The analogue of Pierce’s Lemma 3.2 is the bound |S(d, p; t, z)| ≤ kp1/2 ,

(4)

which follows from Theorems 2C’ and 2G of Schmidt [14]. Similarly in place of Pierce’s Lemma 3.4 we have the bound |W (d, p; h, s, t)| ≤ 4k(k − 1)p3/2 for p ≥ 2k − 1 and p - (h, s), which follows from parts (1), (2), (3) and (4b) of Katz [10, Theorem 1.3]. On the other hand, if p|(h, s) it follows from (4) that |W (d, p; h, s, t)| ≤ k 2 p2 . Thus Pierce’s Lemma 3.7 extends to a bound |W (d, p; h, s, t)| ≤ 4k(k − 1)p3/2 (p, h, s)1/2

4

in general. We may note that the terms d(r0 ) in [13, (3.7)] are at most 4, since r0 is a product of two primes. It then follows that Proposition 3.1 of Pierce [13] becomes −1/2

C(d, f, g)  (Q−2 L + 1){N QU0

1/2

+ N 1/2 Q2 U0−1 + N 1/2 QU0 }(log Q)2 . (5)

We should remark that in the proof of (5) there is a tacit assumption that N  V02 , corresponding to the requirement that r1 ≤ N in [13, §3]. We have regarded the exponent k as fixed, so that the implied constant above is allowed to depend on k. We follow this convention for the remainder of this paper. In a completely analogous way the bounds in [13, Proposition 4.1] become 3/4

+ N 1/2 U0

3/4

+ N 1/2 V0

C(d, u, u0 )  (U0−2 L + 1){N U0

3/2

+ N 1/2 U0 }(log Q)2

5/4

3/2

+ N 1/2 V0

for N  U0 and C(d, v, v 0 )  (V0−2 L + 1){N V0

5/4

5/4

}(log Q)2

5/4

for N  V0 respectively. The terms N 1/2 U0 and N 1/2 V0 are clearly redundant. Finally, we can extend Lemma 6.1 of [13] to state that |Y (d, p; v, x0 , z, h)| ≤ kp1/2 for p - dv, as an immediate corollary of (4). This enables us to estimate Pierce’s error terms in our more general setting, providing that we choose our parameters with U0 , V0  L, so as to ensure that (in Pierce’s notation) K  1. We then find that the terms “A−2 |E(U)|” and “A−2 |E(V)|” are  {LN U0−1 V0−2 + N U0 V0−1 }(log Q)2 and  {LN V0−1 U0−2 + N V0 U0−1 }(log Q)2 respectively, providing that U0 , V0  L. We may now assemble all these estimates to deduce that #{(x, y, z) ∈ Z × Z × N : x2 = 4z k − dy 2 , |y| ≤ Y, z ≤ Z}  L.T. + M.S. + P.S.1 + P.S.2 + E1 + E2 ,

(6)

with L.T. = LN Q−1 (log Q)2 , M.S. =

−1/2

(Q−2 L + 1){N QU0

3/4

+ N 1/2 U0 }(log Q)3 ,

3/4

+ N 1/2 V0

P.S.1

= V0−1 (U0−2 L + 1){N U0

P.S.2

= U0−1 (V0−2 L + 1){N V0

E1

1/2

+ N 1/2 Q2 U0−1 + N 1/2 QU0 }(log Q)2 , 3/2

3/2

}(log Q)3 ,

= {LN U0−1 V0−2 + N U0 V0−1 }(log Q)2

and E2 = {LN V0−1 U0−2 + N V0 U0−1 }(log Q)2 , subject to the conditions V02  N,

max(U0 , V0 )  min(L, N ) and U0 , V0  dη/8 . 5

(7)

We wish to choose U0 and V0 to optimize our estimate, subject to the condition that Q = U0 V0 . In order to simplify matters we content ourselves with minimizing the term −1/2

N QU0

1/2

+ N 1/2 Q2 U0−1 + N 1/2 QU0

which occurs in the expression for M.S.. We therefore choose U0 = max(N 1/2 , Q2/3 ), so that V0 = min(QN −1/2 , Q1/3 ). In particular we see that V0 ≤ U0 , and that the conditions (7) are satisfied when N ≤ L and N 1/2 dη/8  Q  N 3/2 . (8) Our choice of U0 , V0 then yields M.S.  (Q−2 L + 1)(N 3/4 Q + N 1/2 Q4/3 )(log Q)2 , so that M.T. + M.S.  {LN + LN 3/4 + LN 1/2 Q1/3 + N 3/4 Q2 + N 1/2 Q7/3 }Q−1 (log Q)2 . It is clear from (8) that the first term in the braces dominates the second and third terms, whence M.T. + M.S.  {LN + N 3/4 Q2 + N 1/2 Q7/3 }Q−1 (log Q)2 . Next, since V0 ≤ U0 we have −1/4

P.S.1 + P.S.2  {LN V0

1/2

+ LN 1/2 U0

7/4

+ N U0

5/2

+ N 1/2 U0 }Q−1 (log Q)3

and E1 + E2  {LN V0−1 + N U02 }Q−1 (log Q)2 , whence P.S.1 + P.S.2 + E1 + E2 −1/4

1/2

+ LN 1/2 U0

5/2

+ N 1/2 U0

+ N U02 }Q−1 (log Q)3



{LN V0



{LN Q−1/12 + LN 9/8 Q−1/4 + LN 3/4 + LN 1/2 Q1/3 + N 7/4 +N 1/2 Q5/3 + N 2 + N Q4/3 }Q−1 (log Q)3

on substituting our choices for U0 and V0 . If we assume that N ≤ L then the condition (8) ensures that each term in the braces above is dominated by LN , with the exception of the sixth and eighth terms, which are dominated by N 3/4 Q2 . We therefore deduce that #{(x, y, z) ∈ Z × Z × N : x2 = 4z k − dy 2 , |y| ≤ Y, z ≤ Z} 

{LN + N 3/4 Q2 + N 1/2 Q7/3 }Q−1 (log Q)3 .

The optimal choice of Q is then Q = min(L1/2 N 1/8 , L3/7 N 3/14 ) which satisfies (8) for N ≤ L ≤ N 3 and L ≥ dη . On recalling that Y = N and Z = L we then see that Lemma 2 follows. 6

References [1] E. Bombieri and J. Pila, The number of integral points on arcs and ovals, Duke Math. J., 59 (1989), 337-357. [2] N. Broberg, Rational points on finite covers of P1 and P2 , J. Number Theory, 101 (2003), 195-207. [3] D.W. Boyd and H. Kisilevsky, On the exponent of the ideal class groups of complex quadratic fields, Proc. Amer. Math. Soc.. 31 (1972), 433-436. [4] D.R. Heath-Brown, Counting rational points on algebraic varieties, (to appear). [5] D.R. Heath-Brown, The density of rational points on curves and surfaces, Ann. of Math. (2), 155 (2002), 553-595. [6] D.R. Heath-Brown, The square sieve and consecutive square-free numbers, Math. Ann., 266 (1984), 251-259. [7] H. Helfgott and A. Venkatesh, Integral points on elliptic curves and 3torsion in class groups, (to appear). [8] C. Hooley, On exponential sums and certain of their applications, Number theory days, 1980 (Exeter, 1980), 92-122, London Math. Soc. Lecture Note Ser., 56, (Cambridge Univ. Press, Cambridge-New York, 1982). [9] C. Hooley, On the representations of a number as the sum of four cubes.I, Proc. London Math. Soc. (3), 36 (1978), 117-140. [10] N.M. Katz, On a question of Lillian Pierce, Forum Math., to appear. [11] Yu.V. Linnik and A.I. Vinogradov, Hypoelliptic curves and the least prime quadratic residue, Dokl. Akad. Nauk SSSR, 168 (1966), 259-261. [12] L.B. Pierce, The 3-part of class numbers of quadratic fields, J. London Math. Soc. (2), 71 (2005), 579-598. [13] L.B. Pierce, A bound for the 3-part of class numbers of quadratic fields by means of the square sieve, Forum Math., to appear. [14] W.M. Schmidt, Equations over finite fields. An elementary approach, Lecture Notes in Mathematics, Vol. 536. (Springer-Verlag, Berlin-New York, 1976). [15] P.J. Weinberger, Exponents of the class groups of complex quadratic fields, Acta Arith., 22 (1973), 117-124. Author’s address Mathematical Institute, 24–29, St. Giles’, Oxford OX1 3LB England [email protected] 7