Class Groups and Quadratic Diophantine Equations∗ R.A. Mollin
Abstract First, we consider the Diophantine equation of the form x2 −Dy 2 = 16, and provide a criterion√for the central norm in the simple continued fraction expansion of D to be 16 in association with norm 32. This involves congruence conditions on the fundamental unit of the underlying real quadratic order. Then we provide necessary and sufficient conditions for the class group of complex quadratic fields to have a cyclic subgroup of order n > 1, with generator of a certain form.
1
Introduction
In [20] we provided a criterion for the central √ norm (defined below) to be 2 in the simple continued fraction expansion of D, which √ involved congruence conditions on the fundamental unit of the order Z[ D]. In Section 3 of this paper we provide a criterion for the central norm to be 16 in terms of similar congruence criteria related to two of the underlying real quadratic orders. This is a generalized Eisenstein criterion in the sense that the equation x2 − Dy 2 = ±4 was studied by Eisenstein in 1844 in terms of criteria for its solvability. We provided criteria for the solvability of this equation in terms of continued fractions and results of Lagrange in [21] concerning congruence √ conditions on the coefficients of the fundamental unit of Z[ D]. The result herein extends this by providing a criterion for the central norm to be 16 in terms of such congruences. ∗
Mathematics Subject Classification 2000: Primary: 11D09, 11R11, 11A55. Secondary: 11R29. Key words and phrases: Quadratic Diophantine equations, Continued Fractions, Central Norms, Class group, Generator.
2 There have been numerous results in the literature that provide sufficient conditions for the class number of complex quadratic fields to be divisible by a given integer, such as [1], [4]–[13], [24], and [27] but relatively few with necessary and sufficient conditions such as [26] (wherein only the prime case is considered). In section 4, we provide necessary and sufficient conditions in a wide variety of cases which yields complex quadratic fields whose class groups have cyclic subgroups of order n > 1, with generator of a specific form. We conclude with several conjectures pertaining to class number of complex quadratic fields and related quadratic polynomials.
2
Notation and Preliminaries
In Section √ 3, we will be concerned with the simple continued fraction expansions of D, where D is an integer that is not a perfect square. We denote this expansion by, √ D = q0 ; q1 , q2 , . . . , q−1 , 2q0 , √ √ √ where = ( D) is the period length, q0 = D (the √ floor of D), and q1 q2 , . . . , q−1 is a palindrome. The jth convergent of D for j ≥ 0 is given by, Aj = q0 ; q1 , q2 , . . . , qj , Bj where Aj = qj Aj−1 + Aj−2 ,
(1)
Bj = qj Bj−1 + Bj−2 ,
(2)
with A−2 √ = 0, A−1 = 1, B−2 = 1, B−1 = 0. The complete quotients are given by, (Pj + D)/Qj , where P0 = 0, Q0 = 1, and for j ≥ 1, Pj+1 = qj Qj − Pj , √ Pj + D , qj = Qj and 2 D = Pj+1 + Qj Qj+1 .
(3)
3 We will also need the following facts (which can be found in most introductory texts in number theory, such as [16]. Also, see [14] for a more advanced exposition). 2 A2j−1 − Bj−1 D = (−1)j Qj .
(4)
2 A2−1 − B−1 D = (−1) .
(5)
In particular, When is even, P/2 = P/2+1 , so by Equation (3), Q/2 | 2P/2 , where Q/2 is called the central norm, (via Equation (4)), where Q/2 | 2D.
(6)
In general, the values Qj are called the principal norms, √ since they are the norms of the principal reduced ideals in the order Z[ D],√due to the association between the simple continued fraction expansion of D and the infrastructure of the underlying real quadratic order (see [14] for instance). We will be considering Diophantine equations x2 − Dy 2 = 1, 4. The fundamental solution of such an equation means the (unique) least positive integers (x, y) = (x0 , y0 ) satisfying it. In the following (which we need in the next section), and all subsequent results, the notation for the Aj , Bj , Qj and so forth apply √ to the abovedeveloped notation for the continued fraction expansion of D. Theorem √ 1 Let D be a positive integer that is not a perfect square. Then = ( D) is even if and only if one of the following two conditions occurs. 1. There exists a factorization D = ab with 1 < a < b such that the following equation has an integral solution (x, y). ax2 − by 2 = ±1.
(7)
Furthermore, in this case, each of the following holds, where (x, y) = (r, s) is the fundamental solution of Equation (7). (a) Q/2 = a.
4 (b) A/2−1 = ra and B/2−1 = s. (c) A−1 = r2 a + s2 b and B−1 = 2rs. (d) r2 a − s2 b = (−1)/2 . 2. There exists a factorization D = ab with 1 ≤ a < b such that the following equation has an integral solution (x, y) with xy odd. ax2 − by 2 = ±2
(8)
Moreover, in this case each of the following holds, where (x, y) = (r, s) is the fundamental solution of Equation (8). (a) Q/2 = 2a. (b) A/2−1 = ra and B/2−1 = s. (c) 2A−1 = r2 a + s2 b and B−1 = rs. (d) r2 a − s2 b = 2(−1)/2 . Proof. All of this is proved in [19].
✷
We will require the following dual results, which are our original generalizations of results of Lagrange that inspired the work herein. They are proved in [20]. Theorem 2 If (x0 , y0 ) is the fundamental solution of x2 − Dy 2 = 1,
(9)
where D > 2 is not a perfect square, then the following are equivalent. 1. x0 ≡ 1 (mod D). √ 2. If = ( D), then ≡ 0 (mod 4), and Q/2 = 2. 3. There is a solution to the Diophantine equation x2 − Dy 2 = 2.
(10)
5 Theorem 3 If (x0 , y0 ) is the fundamental solution of x2 − Dy 2 = 1,
(11)
where D > 2 is not a perfect square, then the following are equivalent. 1. x0 ≡ −1 (mod D). √ 2. If = ( D), then ≡ 2 (mod 4), and Q/2 = 2. 3. There is a solution to the Diophantine equation x2 − Dy 2 = −2.
(12)
There is also the following result on central norms that we proved in [22]. Theorem 4 √Suppose that D = 4d c, where c is not a perfect square, c is odd, √ d ≥ 1, = ( D), and = ( c). If is even, then Q/2 = 4d if and only if √ √ A/2−1 + B/2−1 c = A −1 + B −1 c, (13) d 2 √ √ in the simple continued fraction expansions of D, respectively c. Moreover, when this occurs, ≡ /2 (mod 2). Lastly, we will require the following in the next section. Theorem √ 5 Let D > 1 be an integer that is not a perfect square and suppose that = ( D) is even. Then each of the following holds. 2 Q/2 A−1 = A2/2−1 + B/2−1 D,
(14)
Q/2 B−1 = 2A/2−1 B/2−1 ,
(15)
and Proof. This is a consequence of [18, Lemma 3.3, p. 323].
✷
In Section 4, will be dealing with orders in quadratic fields for which the following notation will be used. If D = 1 is a squarefree integer, then set ∆ = D if D ≡ 1 (mod 4), and ∆ = 4D otherwise. D is often called a
6 fundamental radicand with discriminant ∆. The maximal order √ √ √ or ring of integers of Q( D) is denoted by O∆ . If α, β ∈ Q( ∆) = Q( D), then we denote a Z-module generated by them as [α, β] = {αx + βy : x, y ∈ Z}, In particular √
O∆ = [1, ω∆ ],
where ω∆ = (σ − 1 + D)/σ with σ = 1 if D ≡ 1 (mod 4) and σ = 2 otherwise. We also need to distinguish Z-modules in O∆ that are ideals. In what follows, N (α) denotes the norm of an element of O∆ , namely N (α) = α · α where α is the algebraic conjugate of α. Theorem 6 Let ∆ be a discriminant, and let I be a nonzero Z-submodule of O∆ . Then I has a representation in the form I = [a, b + cω∆ ], where a, c are positive integers and b is a nonnegative integer with b < a. Also, I is an ideal in O∆ , called an O∆ -ideal, if and only if this representation satisfies c | a, c | b, and ac | N (b + cω∆ ). If c = 1, we say that a nonzero ideal I is a primitive O∆ -ideal. Moreover, if I is a primitive O∆ -ideal, then a is the least positive rational integer in I denoted by N (I) = a, called the norm of I. Proof. See [14, pp. 9–30]
✷
An O∆ -ideal is called reduced if there does not exist any nonzero element α ∈ I such that |α| < N (I) where |α|2 = αα = N (α). In the following, the importance of reduced ideals is highlighted. Theorem 7 Suppose that ∆ < 0 is a fundamental discriminant and I = [N (I), α] is a primitive ideal O∆ . Then each of the following holds. 1. |T r(α)| ≤ N (I) and |T r(α)| is unique in the sense that if I = [N (I), α] = [N (I), β] with |T r(α)| ≤ N (I) and |T r(β)| ≤ N (I), then |T r(α)| = |T r(β)|. 2. If I is a primitive O∆ -ideal and I = [N (I), α] with |T r(α)| ≤ N (I), then I is a reduced ideal if and only if |α| ≥ N (I).
7 3. If I is a primitive ideal of O∆ , and N (I) < ideal. 4. If I is a reduced ideal of O∆ , then N (I)
16 is an odd positive integer that is not a perfect square, and let = ( D). Also, assume that (x0 , y0 ) is the fundamental solution of x2 − Dy 2 = 16 with gcd(x, y) = 1. Then the following are equivalent 1. x0 /4 ≡ ±1 (mod c).
(16)
8 2. ≡ 0 (mod 4), Q/2 = 16, and the there is a solution to the Diophantine equation X 2 − DY 2 = ±32 with gcd(X, Y ) = 1, (17) where the ± signs correspond to those in part 1. Proof. First we assume that part 1 holds. If x0 /2d ≡ −1 (mod c), then by √ Theorem 3, = ( c) ≡ 2 (mod 4), Q /2 = 2 with A2 /2−1 − B2 /2−1 c = −2.
(18)
Moreover, if x0 ≡ 0 (mod 16), then 16(x0 /16)2 −y02 c = 1, so part 1 of Theorem 1 tells us that Q/2 = 16. Now suppose that x0 ≡ 0 (mod 16). If x0 /4 is odd, then from Equation (16), we must have c ≡ 0 (mod 8), a contradiction, so x0 /8 is an odd integer. Therefore, 8(x0 /8)2 −y02 2c = 2 and part 2 of Theorem 1 tells us that Q/2 = 16. Therefore, we may invoke Theorem 4 to conclude that ≡ 2 ≡ 0 (mod 4), and √ √ A/2−1 + B/2−1 c = A −1 + B −1 c, 4 and Theorem 5 also tells us that √ 2 √ A /2−1 + B /2−1 c , A −1 + B −1 c = 2 so we have, √ √ 2 A/2−1 + B/2−1 D = 2 A /2−1 + B /2−1 c . √ √ 3 It follows that 2 A /2−1 + B /2−1 c = X + Y D is a primitive element with norm −8, where X = 2A3 /2−1 + 6A /2−1 B2 /2−1 c and Y = (3A2 /2−1 B /2−1 + B3 /2−1 c)/2, which are both integers since A /2−1 B /2−1 is odd. This completes the case where x0 /4 ≡ −1 (mod c). If x0 /4 ≡ 1 (mod c), then we may invoke Theorem 2 to argue in a similar fashion to the above. Thus, we have shown that part 1 implies part 2. Assume part 2 holds. Then the solvability of Equation (17) implies the solvability of the (X/4)2 − Y 2 c = ±2. Then using the solvability of Equation (16), we may invoke Theorems 2 and 3 to get that x0 /4 ≡ ±1 (mod c), which secures the result. ✷
9 Example 1 If D = 16 · 23 = 16c = 368, then = 4, Q/2 = 16, x0 /4 = 24 = A/2−1 /4 ≡ 1 (mod c), and the fundamental solution of X 2 − DY 2 = 32 is (X, Y ) = (940, 49). If D = 16 · 107, then = 12, Q/2 = 16, x0 /4 = 962 ≡ −1 (mod c), and the fundamental solution of X 2 − DY 2 = −32 is (X, Y ) = (238700, 5769)
Remark 1 Note that when D > 1024, the solution of Equation (17) means √ that Qj = 32 for some j in the simple continued fraction expansion of D, where j is odd when there is a minus sign and j is even when there is a plus sign. this is a consequence of the theory of reduced ideals that is expounded in [14] for instance. A direct consequence of the above discussion is the following. Theorem 11 Suppose that D = 16c > 1024 where c is is a positive nonsquare integer, and (x0 , y0 ) is the fundamental solution of Equation (16). Then x0 /4 ≡ ±1 (mod c) if and only if Q/2 = 16 and Qj = 32 for some j.
4
Cyclic Subgroups
Although we state the following for fundamental discriminants, the result can be extrapolated to arbitrary quadratic orders using the ideas in [14, Section 1.5, pp. 23–30]. Theorem 12 Let D < −19 be a squarefree integer with discriminant ∆. Then C∆ has a cyclic subgroup of order n > 1 generated by either √ √ I = [N, (a + D)/2] or I = [N, a + D] if and only if one of the following holds: 1. There exist integers integers a ≥ 1, N > 1 such that D = a2 − N n where a = N n/2 , when n is odd and a = N n/2 − 1 in any case. 2. There exist integers a ≥ 1, N > 1 such that D = a2 − 4N n where a is odd and a = 2N n/2 , when n is odd, and a = 2N n/2 − 1 when n is even.
10 √ √ Proof. If part 1 holds, we may set I√= [N, a + D]. If N (a + D)√< N (ω∆ )2 , then by Theorem 8, I n = [N n , a + D] is principal since N (a + D) = N n . Moreover, by Theorem 8, I j is not principal for any j < n. It remains to show, for this case, that n | h∆ . Let g = gcd(n, h∆ ). Then there exist integers u and v such that g = un + h∆ v. Thus, I g ∼ I nu I h∆ v ∼ (I n )u (I h∆ )u ∼ 1. Hence, g = n,√so n | h∆ . If N (a + D) > N (ω∆ )2 , then N n ≥ (1 − D)2 /16, from which it follows that a2 +4a−1 ≥ N n . Since D = a2 −N n , it follows that N n/2 > a > N n/2 −2. When n is even this implies a = N n/2 − 1 and when n is odd it implies that a = N n/2 or a = N n/2 − 1 all √ of which2 contradict the hypothesis of the theorem. Similarly, if N (a + D) > D , then it is deduced that N n/2 > a > N n/2 − 1, so n is odd and a = N n/2 , contradicting the hypothesis. √ If part 2 holds, then we set I = [N, (a+ D)/2] and argue in an analogous fashion to that of part 1 to get the desired conclusion. Conversely, √ if C∆ has a cyclic subgroup of order n > 1 generated by I= [N, (a + D)/2] and by Theorem 9, we may assume that 2a/σ < N ≤ |∆|/3. By the choice of I, a2 − D < ((1 − D)/4)2 , when D ≡ 1 (mod 4), except for certain values of D > −20, which are excluded by hypothesis. Similarly, we have that a2 − D < D2 when D ≡ 1 (mod 4). Therefore, we may invoke Theorem 8 to conclude that I n ∼ 1 if and only if a2 − D = N n or a2 − D = 4N n . If D ≡ 1 (mod 4), a = N n/2 with n odd and D = a2 − N n , then N n/2 − 1 < a < N , forcing n = 2, a contradiction. If D ≡ 1 (mod 4) with these conditions, then N n/2 − 1 < a < N/2, which is impossible. We have shown that the conditions in part 1 must hold. It remains to verify those for part 2. If D ≡ 1 (mod 4) and a = 2N n/2 with n odd, then 2N n/2 − 1 < a < N , which is impossible, as is the case a = 2N n/2 − 1 with √ n even. Similar arguments hold if the group is generated by I = [N, a + D]. ✷ Example 2 If D = −307, then −307 = 62 − 73 , with 6 = 73/2 = 18, N = 7, n = 3, so by part 1 of Theorem 12, C−307 has a cyclic subgroup of order 3. In fact it is a group of order 3. Note here that −307 ≡ 1 (mod 4) and this illustrates part 1 of the theorem. There is no representation of √ the generator of the cyclic group in the form I = [N, (a + −307)/2] since
11 3 a2 + 307 = 4N has no solutions. This may be checked since we can assume that a < N √< 307/3 < 11. Thus, only part 1 applies and the generator is I = [7, 6 + −307].
Part 2 of Theorem 12 is illustrated as follows. Example 3 Let D = −1607, then D = 212 − 4 · 29 = a2 − 4N n . Since a = 21 = 2N n/2 = 45, then C∆ = C−1607 has a cyclic subgroup of order 9. In fact, C∆ is a cyclic group of order 27. To demonstrate how sharp the bounds in Theorem 12 happen to be, we present the following illustrations. Example 4 Let D = −163, for which D = 812 − 4 · 412 = a2 − 4 · N n , but a = 41 = 2N n/2 − 1 = 2 · 81 − 1 so Theorem 12 does not apply. Indeed, this is the largest discriminant of class number 1 in the solution of Gauss’s problem (see [14, Theorem 4.1.3, p. 109]). Similarly, if D = −67, then D = 332 − 4 · 172 = a2 − 4 · N n , but a = 33 = 2N n/2 − 1 = 2 · 17 − 1 so Theorem 12 does not apply; and this is the second largest discriminant of class number 1 in complex quadratic orders. Also, the third on the list is D = −43 = 212 −4·112 , with a = 21 = 2·11−1 = 2N n/2 −1. Example 4 shows how the bounds in Theorem 12 are the best possible. The following further illustrates this contention, presented in connection with other results in the literature. Example 5 If D = −341 = 27596462 − 3775 , then a = 275946 = 3775/2 = N n/2 , and 5 does not divide h∆ = h−341 = 28. Note that by [4], this value of −D ≡ 1 (mod 4) is the only such value representable in the form D = a2 −N p for an odd prime p for which p does not divide the class number. (See also [23] where some errors from [4] are corrected.) Also, for D = −71 = 212 − 29 we have that a = 21 = 29/2 − 1 = n/2 N − 1, contradicting the hypothesis of part 1 of Theorem 12. Indeed 9 does not divide h−71 = 7. However, if we view the representation of D as follows, we may employ part 2 since D = 212 − 4 · 27 = a2 − 4 · N n . Here a = 2N n/2 = 29/2 = 22. These examples portray the exact nature of the conditions in Theorem 12 and the precision of the bounds.
12 Unfortunately, not all generators of class groups are of the form in Theorem 12. For instance, we have the following. √ Example 6 If D = −907, then there is not ideal I = [N, (a + D)/2] with N (I) = N 3 , even though h∆ = 3 since a2 + 907 = σ 2 N 3 √ is not solvable for 2 2 3 σ = 1, 2. However, 25 + 3 · 907 = 4 · 13 , and (25 + 3 D)/2) = I 3 is a 3 principal ideal with norm √ 13 and I is a non-principal ideal of norm √ 13. We may set I = [13, (25 + 9D)/2] which is an ideal in the order Z[ −907 · 9]. Then we must look at the√ relationship between the class numbers of this order and the maximal order Z[ −907]. The formula for this relationship is known. If h∆ is the class number of the order Z[f∆2 ∆0 ] where √ f∆ is the conductor, and if h∆0 is the class number of the maximal order Z[ ∆0 ], then ∆0 1− , (19) h∆ = h∆0 p where the product is taken over all prime divisors p of f∆ and the symbol on the right is the Kronecker symbol (see [14, pp. 25–26]). In the case of ∆0 = −907 and ∆ = −9 · 907, √ h∆ = 12 = 4 · h∆0 . Thus, the existence of the ideal I above of order 3 in Z[ ∆] implies the existence of one in the maximal order since 3 does not divide the right hand product in Equation (19). It is these kinds of considerations that have to be addressed when considering the general case. We did this for real quadratic orders in [17]. The reader may employ the ideas therein to create a similar scenario for the complex case. For the complex fields of the type satisfying Theorem 12, this is the best possible result as we have demonstrated via Examples 2–5. Example 6 is special for another reason. The value D = −907 represents the largest complex quadratic field of class number 3, which was proved in in [2] in 1998. Also, in 1972, it was shown in [3] that there are only finitely many complex quadratic fields with exponent e∆ = 3. In order to introduce some conjectures in this regard, we need the following. We introduce notation to coincide with [14] where all the background is developed for the following link with quadratic polynomials. Since we are discussing only negative discriminants, we sketch the results only for that case, but it holds for the positive case as well. Let ∆ < 0 be a fundamental discriminant and let
2 x + x + (1 − D)/4 if D = ∆ ≡ 1 (mod 4), f∆ (x) = x2 − D otherwise ,
13 called the Euler-Rabinowitsch polynomial, and for ∆ < −3, set F (∆) = max{d(f∆ (x)) : x = 0, 1, . . . , |∆|/4 − 1}, where d(y) denotes the number (counting multiplicity) of prime divisors of y. F (∆) is sometimes called the Ono invariant. This is all related to Gauss’s class number problems as outlined in [14] and [15]. (In fact, this links the results of this section to the previous section since the results of [15] provide quadratic Diophantine equations with solutions that are primes in connection with complex quadratic fields.) In particular, we gave criteria for the exponent of the class group of a complex quadratic order to be 2. Now we look at the case of exponent 3. In what follows N + 1 is the number distinct primes dividing ∆ where N ≥ 0, We now propose the following. Conjecture 1 If ∆ < 0 is a fundamental discriminant, then e∆ = 3 if and only if h∆ = 3F (∆)−2 and N = 0. This is related to another conjecture posed in [5]. Conjecture 2 (Cohen-Sonn) If ∆ < 0 is a fundamental discriminant, then h∆ = 3 if and only if N = 0 and F (∆) = 3. Classical results tell us that e∆ = 3 implies that N = 0. Also, in [5], they prove one direction of Conjecture 2, namely that h∆ = 3 implies N = 0 and F (∆) = 3. Both conjectures remain open. A closing illustration of Conjecture 1 in conjunction with Theorem 12 comes from a rare example where e∆ = 3 = h∆ (see [25]). The reader will be able to verify from existing tables (see [14, Appendix D, pp. 335–345], for instance) that the case is most often that e∆ = h∆ =3. In fact, the authors of [25, p. 72] state that the product of three cyclic groups of order 3 is not likely to occur as the class group of any imaginary quadratic field. Example 7 Let ∆ = −4027, then ∆ = 692 − 4 · 133 , so by Theorem 12, 3 | h∆ since a = 69 = 2N n/2 = 2 · 133/2 = 93. Also, F (∆) = 4 and h∆ = 9 with C∆ being a product of two cyclic groups of order 3. Acknowledgements: The author’s research is supported by NSERC Canada grant # A8484.
14
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[email protected] 