On The Number of Isomorphism Classes of Cm Elliptic Curves Defined Over a Number Field

ON THE NUMBER OF ISOMORPHISM CLASSES OF CM ELLIPTIC CURVES DEFINED OVER A NUMBER FIELD Abstract. The theory of complex multiplication has proven to be an essential tool in number theory, mainly due to the connections with class field theory developed by Kronecker, Weber, Fricke, Hasse, Deuring, and Shimura, among others. Certain important results have been shown first in the case of complex multiplication. Thus, it is a natural question to find all the isomorphism classes of elliptic curves with complex multiplication defined over a fixed number field, for which these important results hold. In this article, we prove an upper bound on the number of isomorphism classes of CM elliptic curves defined over a number field of a fixed odd degree N , in terms of the prime factorization of N . 1. Introduction The theory of complex multiplication has proven to be an essential tool in number theory, mainly due to the connections with class field theory developed by Kronecker, Weber, Fricke, Hasse, Deuring, and Shimura, among others. Certain important results have been shown first in the case of complex multiplication, such as the analytic continuation to the entire complex plane for the L-function of an elliptic curve (see [Sil94], Ch. III, §10), some cases of the Birch and Swinnerton-Dyer conjecture (see [CW77], [Rub99]) or the main conjectures of Iwasawa theory for elliptic curves (see for example [PR04]). Thus, it is a natural question to find all the isomorphism classes of elliptic curves with complex multiplication defined over a fixed number field, for which these important results hold. In this article, we concentrate on finding an upper bound on the number of isomorphism classes of CM elliptic curves defined over a number field of a fixed degree, and concentrate in the case of odd degree. It is well known that there are only 13 isomorphism classes of elliptic curves defined over Q with complex multiplication ([Sil09], Appendix A, §3), namely the curves with j-invariant in the list: {0, 24 33 53 , −215 · 3 · 53 , 26 33 , 23 33 113 , −33 53 , 33 53 173 , 26 53 , − 215 , −215 33 , −218 33 53 , −215 33 53 113 , −218 33 53 233 293 }. However, the number of CM j-invariants varies wildly depending on the choice of field of definition, even in the case of quadratic number fields (see Table 1). For a number field L, we will write Σ(L) for the set of all CM j-invariants defined over L, but not defined over Q, so that the total number of CM j-invariants defined over L is 13 + #Σ(L). It is known that Σ(L) is a finite set, for any number field L. In this article, we show the following simple bound for #Σ(L) when the degree of L is odd. Theorem 1.1. Let L be a number field of odd degree. Then, #Σ(L) ≤ 2 log3 ([L : Q]). In particular, the number of distinct CM j-invariants defined over L is bounded by 13 + 2 log3 ([L : Q]). Remark 1.2. The simple bound √ given in Theorem 1.1 is essentially sharp. The bound is trivially sharp when L = Q. Moreover, let K = Q( −3), and for any fixed e ≥ 1, let Oe be an order of OK with conductor f = 2 · 3e . Let Ee be an elliptic curve with CM by O, and define Le = Q(j(Ee )). Then, [Le : Q] = 3e , and it follows from Theorem 1.3 that #Σ(L) = 2e − 1 = 2 log3 ([Le : Q]) − 1, which is just one unit below the bound of Theorem 1.1. Theorem 1.1 is a consequence of more refined bounds (Theorems 1.3 and 1.4; see Remark 1.5) which we discuss below after we provide some computational data (our calculations have been performed using Sage [S+ 14]; see also [Wat04] for some of the algorithms that Sage uses). For instance, we will show in Section 2 that, for a fixed √ quadratic number field L = Q( d), the number of elements in Σ(L) is given as in Table 1 (see Remark 2.15). In particular, Σ(L) = ∅ for all imaginary quadratic fields L, and in fact Σ(L) = ∅ for all but the 14 distinct real quadratic fields that appear in the table. Given a fixed integer N ≥ 2, we write O(N ) for the set of all orders of class number N in some imaginary quadratic field, and Σ(N ) for the set of CM j-invariants j(E) such 1 2 d 2 3 5 6 7 13 17 21 29 33 37 41 61 89 else √ #Σ(Q( d)) 8 4 18 2 2 6 4 2 2 2 2 2 2 2 0 Table 1. The number of non-rational CM j-invariants defined over each quadratic field. that Q(j(E))/Q is an extension of degree N (notice that #Σ(N ) = N · #O(N )). Finally, we write L(N ) for the set of all non-isomorphic fields L = Q(j(E)), where E is an elliptic curve with CM by an order of class number N (thus, #O(N ) ≥ #L(N ) for all N ), and we write L(N ) for the set of all number fields of degree N . Table 1 shows that #Σ(2) = 58 (so #O(2) = 29), and #L(2) = 14. Notice that if L/Q is quadratic and contains a CM j-invariant j(E) not defined over Q, then L = Q(j(E)), and so L ∈ L(2). For similar reasons, if N is prime, and L is a number field of degree N that contains a j-invariant j(E) of class number N , then L ∈ L(N ), i.e., L = Q(j(E)). We record the sizes of O(N ), Σ(N ), L(N ) in Table 2, for N = 1, . . . , 11, as well as the maximum number of elements of Σ(L) for L ∈ L(N ), and also the maximum of Σ(L) over all number fields L of degree N over Q (and not just those fields of the form Q(j(E))). We will explain how the data in Table 2 was collected in Section 2, Remark 2.15. N 2 3 4 5 6 7 8 9 10 11 #O(N ) 29 25 84 29 101 38 208 55 123 46 #Σ(N ) 58 75 336 145 606 266 1644 495 1230 506 #L(N ) 14 23 72 25 96 32 202 50 114 42 max{#Σ(L(N ))} 18 2 42 2 22 2 84 2 22 2 max{#Σ(L(N )))} 18 2 42 2 33 2 84 4 33 2 Table 2. The number of (a) orders of class number N ; (b) CM j-invariants j(E) such that Q(j(E)) is of degree N ; (c) fields Q(j(E)) as in (b), up to isomorphism; (d) maximum #Σ(L) for L as in (c); and maximum #Σ(L) for any L of degree N . In Table 2 and in the rest of the article, we use two abbreviations: max{#Σ(L(N ))} := max{#Σ(L) : L ∈ L(N )} = max{#Σ(L) : [L : Q] = N, L = Q(j(E)) for some CM j(E)}, max{#Σ(L(N ))} := max{#Σ(L) : L ∈ L(N )} = max{#Σ(L) : [L : Q] = N }. In this paper we give upper bounds for max{#Σ(L(N ))} and max{#Σ(L(N ))} when N is odd. Our bounds are sharp, in the sense that we exhibit examples for arbitrarily large N that attain the bound (see Examples 5.4, 5.5, and 5.9). Our first intermediary result describes the number of CM j-invariants found in an extension of the form L = Q(j(E)), where E is itself an elliptic curve with complex multiplication by an order O in an imaginary quadratic field K (i.e., L ∈ L(N ), where N is the class number of O). Theorem 1.3. Let j(E) be a j-invariant with CM by an order O in an imaginary quadratic Pfield K of conductor f, and j(E) 6∈ Q. Let L = Q(j(E)), and suppose that [L : Q] = N > 1 is odd. Let σ0 (f) = d|f d0 be the number of positive divisors of f. Then, the number of j-invariants with CM defined over L is 13 + σ0 (f) − J(K) if f is even, or if 2 does not split completely in K, and 13 + σ0 (2f) − J(K) otherwise, where J(K) is the number of rational j-invariants of curves with CM by an order of K, i.e., dK −3 −4 −7 −8 −11 −19 −43 −67 −163 else J(K ) 3 2 2 1 1 1 1 1 1 0, where dK is the discriminant of K. Our second and main result provides a bound for the number of CM j-invariants defined over any number field of odd degree N , in terms of the factorization of N . If we know the list of CM imaginary quadratic fields ON THE NUMBER OF ISOMORPHISM CLASSES OF CM ELLIPTIC CURVES DEFINED OVER A NUMBER FIELD 3 that intervene in Σ(L), and the prime factorization of their class numbers, then we can significantly improve the bound on the number of CM j-invariants defined over L, but this finer data is not required, and we also obtain a bound that only depends on the factorization of N . Theorem 1.4. Let L/Q be a number field of odd degree N = pe11 · · · perr , and let K1 , . . . , Kt be the list of imaginary quadratic fields such that there is j(E) ∈ Σ(L) where E has CM by an order of Ki , for some i = 1, . . . , t. Further, let hi be the class number of Ki , and suppose that hi > 1 for i = 1, . . . , s and hi = 1 for i = s + 1, . . . , t. Then, ! r s X X #Σ(L) ≤ 2s + 2 ej − fi,j . j=1 f f where hi = p1i,1 · · · pri,r . In particular, #Σ(L) ≤ 2 i=1 Pr j=1 ej . Remark 1.5. Let L be a number field of odd degree N = pe11 · · · perr . Theorem 1.4 shows that #Σ(L) ≤ Pr P 2 j=1 ej . Since pj ≥ 3, it is clear that the quantity ej would be maximized if r = 1, p1 = 3, and e1 = log3 (N ). Thus, r X #Σ(L) ≤ 2 ej ≤ 2 log3 (N ), j=1 which shows Theorem 1.1. The article is organized as follows. In Section 2 we include a number of well-known results on class numbers of orders in imaginary quadratic fields, and then we specialize those to the case of odd class number. In Section 3, we study the number of CM j-invariants defined over a field extension L ∈ L(N ), for some fixed N , i.e., L = Q(j(E)) for some CM curve E. In particular, we prove Theorem 1.3 (see Theorem 3.4). In Section 4 we study the intersection of ring class fields, and we use these results in Section 5 in order to prove Theorem 1.4. We also provide several explicit examples, in order to demonstrate that our bounds are, in fact, sharp in the sense that there are number fields of arbitrarily high odd degree where the bound is an equality (see Examples 5.4, 5.5, and 5.9). 2. Preliminaries In this section we collect a number of results on orders in imaginary quadratic fields, and their class numbers. Throughout the paper K will be an imaginary quadratic field with ring of integers OK , and the class number of OK will be denoted by hK . The discriminant of OK will be denoted by dK . If O is an order of OK , then we denote its class number by h(O). The basic theory of complex multiplication is summarized in the following result. Theorem 2.1 ([Sil94], Ch.2, Theorems 4.3 and 6.1; [Cox89], Theorem 11.1). Let K be an imaginary quadratic field with ring of integers OK and let E be an elliptic curve with CM by an order O of OK of conductor f. Then: (1) (2) (3) (4) The j-invariant of E, j(E), is an algebraic integer. The field L = K(j(E)) is the ring class field the order O. [Q(j(E)) : Q] = [K(j(E)) : K] = h(O), where h(O) is the class number of O. Let {E1 , . . . , Eh } be a complete set of representatives of isomorphism classes of elliptic curves with CM by O. Then {j(E1 ), . . . , j(Eh )} is a complete set of Gal(K/K) conjugates of j(E). We shall need a formula for the class number of an arbitrary order in terms of its conductor, and the class number of the maximal order. Such a formula is given in the next theorem. Theorem 2.2 ([Cox89], Theorem 7.24). Let O be the order of conductor f in an imaginary quadratic field K. Then, the class number of O is an integer multiple of hK , and it satisfies    Y dK hK · f 1 h(O) = × · 1− · , × p p [OK : O ] p|f where   · p is the Kronecker symbol. 4 We remind the reader that for an odd prime p, the Kronecker symbol is just the Legendre symbol, and if p = 2, we have the formula     if 2|dK , 0 dK = 1 if dK ≡ 1 mod 8,  2  −1 if dK ≡ 5 mod 8. In this note we are specially interested in orders with odd class number. The class number formula of Theorem 2.2 shows that, for our purposes, we only need to consider imaginary quadratic fields whose ring of integers has odd class number (we will show this carefully, and improve the characterization in Theorem 2.8 below). Genus theory will tell us that we should only consider those with prime discriminant. We remind the reader that the genus field of K is the maximal unramified extension of K which is an abelian extension of Q. Theorem 2.3 ([Cox89], Theorem 6.1). Let K be an imaginary quadratic field of discriminant dK , let µ be the number of primes dividingp dK , and let p1 , . . . , pr be the odd primes dividing dK . Set p∗i = (−1)(pi −1)/2 pi . Then, √ the genus field of K is K( p∗1 , . . . , p∗r ). In particular, the class number hK is divisible by 2µ−1 . √ Corollary 2.4. If K/Q is an imaginary quadratic extension with odd class number, then K = Q( −d) where d = 1 or 2, or a prime q ≡ 3 mod 4. Proof. By Theorem 2.3, if hK is odd, then the discriminant dK has precisely one prime divisor. In particular d = 1, 2, or an odd prime q. Moreover, since dK = −d or −4d, according to the class of −d modulo 4, it follows that if d = q is an odd prime, then q ≡ 3 mod 4.  √ Theorem 2.5. Let K = Q( −d) be an imaginary quadratic number field with d 6= 1 or 2, and let O be the order √ of OK of conductor f. Then h(O) is odd if and only if hK is odd, K = Q( −q) for some prime q ≡ 3 mod 4, n m and f = 2 q , where n ∈ {0, 1} and m ≥ 0. √ Proof. Let us first assume that K = Q( −q) for some prime q ≡ 3 mod 4, hK is odd, and f = 2n q m for some n ∈ {0, 1} and m ≥ 0. Under these hypotheses dK = −q. Since we are assuming that K 6= Q(i), we know that × × × [OK : O× ] = 1 or 3, so in either case [OK : O× ] is odd. Thus, it suffices to check that [OK : O× ]h(O) is odd. Applying Theorem 2.2 we have that       Y Y  dK 1 −q 1 × × n m [OK : O ]h(O) = hK · f · 1− = hK · 2 · q 1− p p p p p|f p|2n q m      1  hK · q m 1 − −q if n = 0, and q q =    

 hK · 2 · q m 1 − −q 1 1 − −q 1 if n = 1 2 2 q q  hK · q m if n = 0, and =
h · 2 · q m 2±1 if n = 1 K 2  hK · q m if n = 0, and = h · q m (2 ± 1) if n = 1. K Therefore, h(O) is also odd, by the assumption that hK is odd and the fact that q ≡ 3 mod 4. For the converse, let us assume that h(O) is odd. By Theorem 2.2, h(O) is a multiple of hK , so hK must be √ odd. By Corollary 2.4, and since we are assuming d 6= 1 or 2, we have K = Q( −q) for some prime q ≡ 3 mod 4, × × and so dK = −q. Since h(O) is assumed to be odd, and [OK : O× ] is odd (d 6= 1 or 2), then [OK : O× ]h(O) is odd. By Theorem 2.2 we have that    Y −q 1 × × [OK : O ]h(O) = hK · f · 1− . p p p|f It is clear that the 2-adic valuation of the denominator of 2|f, but if 4|f, then the 2-adic valuation of × [OK × Q  p|f 1−  −q p   1 p is at most 1, and this occurs when : O ]h(O) would be at least 1, and so the number would be odd. ON THE NUMBER OF ISOMORPHISM CLASSES OF CM ELLIPTIC CURVES DEFINED OVER A NUMBER FIELD 5 We conclude that  the  2-adic valuation of f is 0 or 1. Moreover, suppose p is a prime not equal to 2 or q and such −q that p|f. Then p = ±1, which implies that     −q 1 p±1 1− = , p p p × and so the 2-adic valuation of this factor is positive, since p is odd. Hence, the 2-adic valuation of [OK : O× ]h(O) satisfies × ν2 ([OK : O× ]h(O)) ≥ max{ν2 (f) − 1, 0} + #{p : p|f, p 6= 2, q}. It follows that if h(O) is odd, then the only primes dividing f may be 2 or q, but no others. Hence, f = 2n q m , with n ∈ {0, 1} and m ≥ 0 as claimed.  √ Next, we find the orders of Q(i) and Q( −2) with odd class number. We remark first that the class number of the order O of Q(i) of conductor 2 is, by Theorem 2.2, given by         hK · f Y 1·2 −4 1 dK 1 h(O) = × = 1 − = 1, 1 − p p 2 2 2 [OK : O× ] p|f and so h(O) = 1 is odd. Indeed, there are elliptic curves defined over Q with CM by O, namely those with j = 23 33 113 . Lemma 2.6. Let O be an order of Q(i) of conductor f > 2. Then h(O) is even. × Proof. Notice that if f > 2, then [OK : O× ] = 2. Applying Theorem 2.2, we see that         hK · f Y dK 1 fY −4 1 h(O) = × 1− = 1− . p p 2 p p [OK : O× ] p|f p|f We will break this lemma into three cases, according to whether f is a power of 2, or it has an odd prime divisor ≡ ±1 mod 4. Case 1: f > 2 is a power of 2 (in particular, f is divisible by 4). Let us write f = 2n for some n > 1. Applying Theorem 2.2 we obtain h(O) =         1 · 2n Y −4 1 −4 1 1− = 2n−1 1 − = 2n−1 . 2 p p 2 2 n p|2 Thus, h(O) is even. Case 2: The conductor of O is divisible by an odd prime q ≡ 1 mod 4. Let us write f = f0 · q and again apply Theorem 2.2 to calculate    Y          0 Y f 1 −4 1 q−1 −4 1  . h(O) = 1− 1− = · f 1−  2 q p p 2 p p  0 0 p|f p|f p= 6 q    Y q−1 −4 1 0 Since q ≡ 1 mod 4, we have that ≡ 0 mod 2, and since f 1− is an integer, we see that 2 p p 0 p|f p6=q h(O) is even. Case 3: The conductor of O is divisible by an odd prime q ≡ 3 mod 4. Let us write f = f0 · q. Then,         0 Y q+1 −4 1  . h(O) = · f 1−   2 p p 0 p|f p= 6 q 6    Y  q+1 −4 1 0 Since q ≡ 3 mod 4, we have that ≡ 0 mod 2, and again f 1− is an integer, we conclude 2 p p 0 p|f p= 6 q that h(O) is even.  √ Lemma 2.7. Let O be an order of Q( −2) of conductor f > 1. Then h(O) is even. Proof. Applying Theorem 2.2, we see that        Y hK · f Y dK 1 −8 1 h(O) = × 1− =f 1− . p p p p [OK : O] p|f p|f We consider two cases, according to whether f is a power of 2. Case 1: f is a power of 2. Let us write f = 2n , with n > 0. Then,     −8 1 h(O) = 2n 1 − = 2n . 2 2 Thus, h(O) is even. Case 2: f is divisible by an odd prime q. Let us write f = q · f0 . Then,        Y    Y −8 1 −8 1 −8 1 h(O) = f 1− =q 1− · f0 1− p p q q p p 0 p|f p|f p6=q  0 = (q ± 1)  f Y p|f0  1−    −8 1  . p p  p6=q Since q is an odd prime, we know that q ± 1 is even, and since f0 Q  p|f0 1−  −8 p   1 p is an integer, we conclude p6=q that h(O) is even.  Putting together Theorem 2.5, Lemma 2.6, and Lemma 2.7, we show a formula for the class number of orders, in the case when the class number is odd. Theorem 2.8. Let K be an imaginary quadratic field, let O be an order of conductor f, and suppose that h(O) is odd. Then, exactly one of the following occurs: (1) If K = Q(i), √ then O = OK or f = 2, and h(O) = 1. (2) If K = Q( −2), then = 1. √ O = OK and h(O) √ (3) If K 6= Q(i) or Q( −2), then K = Q( −q) for some prime q ≡ 3 mod 4, and f = 2n q m , where n ∈ {0, 1} and m ≥ 0. Thus, √ (a) If K = Q( −3), and O = OK , then h(O) = 1. Otherwise, if f > 1 with f = 2n 3m , then ( 3m−1 if f = 3m , h(O) = 3m if f = 2 · 3m . √ √ √ (b) If K = Q( −q) 6= Q(i), Q( −2), Q( −3), then  m  if f = q m , hK · q h(O) = hK · q m if f = 2 · q m , q ≡ 7 mod 8,   m 3 · hK · q if f = 2 · q m , q ≡ 3 mod 8. A direct consequence of the formulas for odd class numbers is that, for a fixed imaginary quadratic field, there are only finitely many orders of odd class number below a given bound. We record this as a corollary. Corollary 2.9. Let K be an imaginary quadratic field, and let N ≥ 1 be fixed. Then, (1) There are only finitely many orders of K with odd class number ≤ N . ON THE NUMBER OF ISOMORPHISM CLASSES OF CM ELLIPTIC CURVES DEFINED OVER A NUMBER FIELD 7 (2) There are only finitely many isomorphism classes of elliptic curves E with CM by K, such that [Q(j(E)) : Q] is odd and ≤ N . Proof. Part (1) follows from Theorem 2.8, because the class number grows as the conductor grows. Part (2) follows from part (1) and the fact that there only finitely many elliptic curves with CM by a given order, by Theorem 2.1.  Remark 2.10. Of course, Corollary 2.9 is true much more generally (and not just for odd class number). It is well known that there are only finitely many imaginary quadratic fields with class number less or equal than a given bound N (see [Gol85]) and, together with Theorem 2.2, this shows that there are only finitely many orders of class number ≤ N , for any fixed N ≥ 1. Corollary 2.11. Let K be an imaginary quadratic field and suppose that E1 and E2 have CM by orders O1 and O2 of OK with conductors f1 ≤ f2 respectively, such that h(O1 ) and h(O2 ) are odd. Let fg = gcd(f1 , f2 ) and fl = lcm(f1 , f2 ), and let Og and Ol be, respectively, orders of OK with conductors fg and fl . Then: h(Og ) = gcd(h(O1 ), h(O2 )), and h(Ol ) = lcm(h(O1 ), h(O2 )). √ Proof. Throughout the proof, we will use the formulas of Theorem 2.8. If K = Q(i) or Q( −2), then the√theorem is clear √ because all the class numbers involved are equal to 1, so let us assume that K 6= Q(i) or Q( −2). If K = Q( −3) and O1 = OK (i.e., trivially. Thus, √ f1 = 1), then Og = OK and Ol = O2 , and the formulas ×hold × × we may assume that if K√= Q( −3), then neither √ O1 or O2 are OK , and therefore λ = [OK : O1 ] = [OK : O2× ] n1 m 1 n2 m2 is equal to 3 if K = Q( −3) and to 1 if K 6= Q( −3). In particular, if f1 = 2 q and f2 = 2 q , then fg = gcd(f1 , f2 ) = 2min{n1 ,n2 } 3min{m1 ,m2 } and so ( 3 · hK · q min{m1 ,m2 } /λ if n1 = n2 = 1 and q ≡ 3 mod 8, gcd(h(O1 ), h(O2 )) = hK · q min{m1 ,m2 } /λ otherwise, = h(Og ). Similarly, fl = lcm(f1 , f2 ) = 2max{n1 ,n2 } 3max{m1 ,m2 } and so ( 3 · hK · q max{m1 ,m2 } /λ lcm(h(O1 ), h(O2 )) = hK · q max{m1 ,m2 } /λ if n1 or n2 = 1 and q ≡ 3 mod 8, otherwise, = h(Ol ). Thus, the proof is complete.  We will also need a lemma about conductors in abelian extensions of imaginary quadratic fields. Lemma 2.12. Let O be an order of conductor f in an imaginary quadratic field K, and let L be the ring class field attached to O. Then: (1) Let O0 be another order in K, with conductor f0 , and ring class field L0 . Then, O ⊆ O0 if and only if f0 |f. Moreover, f0 |f implies that L0 ⊆ L. (2) Let L and L0 be ring class fields attached to O and O0 respectively, and let C = C(L/K) and C 0 = C(L0 /K) be the conductors of the abelian extensions L/K and L0 /K. If L0 ⊆ L, then C 0 is a divisor of C. (3) The relationship between the conductors f of O and C = C(L/K) is given by  √ OK if f = 2 or 3, and K = Q( −3),    O if f = 2, and K = Q(i), K C(L/K) =  (f/2)OK if f is even and f/2 is odd, and 2 splits completely in K,    fOK otherwise. In particular, C(L/K) = (f/k)OK with k = 1, 2, or 3. Proof. Parts (1), and (3) are shown, respectively in Exercise 9.19, and Exercise 9.20 in [Cox89]. It remains to justify (2), i.e., if L/K and L0 /K are abelian extensions such that K ⊆ L0 ⊆ L, then C 0 = C(L0 /K) is a divisor of C = C(L/K). In order to show this, we recall that the conductor of an abelian extension F/K is the greatest common divisor of all moduli m such that F ⊆ Km , where Km is the ray class field modulo m (see Exercise 8.6 in [Cox89]). In particular L ⊆ KC , and if L0 ⊆ L, then L0 ⊆ KC as well. From this it follows that C 0 = C(L0 /K) is a divisor of C.  8 The following result is a corollary of Theorem 2.2 and Lemma 2.12. Corollary 2.13. Let O be the order of conductor f in an imaginary quadratic field K, and let O ⊆ O0 be a suborder of conductor f0 . Then, h(O0 ) is a divisor of h(O). Proof. If f0 = 1, i.e., if O0 = OK , then hK divides h(O) by Theorem 2.2. Otherwise, assume that f0 > 1 (and so × × f > 1 as well because f0 needs to be a divisor of f for√O ⊆ O0 to hold). In this case, [OK : O× ] = [OK : O0× ], and the index equals 2 if K = Q(i), equals 3 if K = Q( −3), and 1 otherwise. Now the divisibility of class numbers 0 is clear from for h(O) in Theorem 2.2,  the  formula   since  f |f (by  Lemma 2.12) and since the same factors in Q  Q dK dK 1 1 · p appear in the product p|f 1 − p · p .  p|f0 1 − p The result that follows characterizes extensions of K contained in a ring class field. Theorem 2.14 ([Cox89], Theorem 9.18). Let K be an imaginary quadratic field. Then, an abelian extension H of K is generalized dihedral over Q if and only if L is contained in a ring class field of K. We finish this section with a remark about the computation of the data in Tables 1 and 2 Remark 2.15. The data in Tables 1 and 2 was gathered using SAGE version 6.1.2 ([S+ 14]), as follows. Given a positive integer N , the command cm orders(N) returns a√complete list of orders of class number N , i.e., a list of all the ordered pairs (−D, f) such that the order O of Q( −D) of conductor f has class number h(O) = N . The length of this list is exactly #O(N ), while Σ(N ) = N · #O(N ). Once we had a complete list of the orders of class number N , we used hilbert class polynomials(-D*f2 ) to find the minimal polynomial of the j-invariants of the elliptic curves with CM by O. With a minimal polynomial for j(E), we can define the field L = Q(j(E)). To compute #L(N ), we simply compute how many of these fields are non-isomorphic for a fixed N . Last, we used the command cm j invariants(L) that returns a list of all of the j-invariants of CM elliptic curves that are defined over the number field L. Thus, in order to order to compute # max{#Σ(L(N )))}, we simply compute the length of the list returned by cm j invariants(L), for every L ∈ L(N ) and take the maximum value. To compute # max{#Σ(L(N )))} we do the same, but this time we take the maximum over all fields of the form L = Q(j1 , ..., jr ) such that [L : Q] = N , and each ji for i = 1, . . . , r, is a j-invariant with CM by an order of class number dividing N . For instance, when N = 2, the command cm orders(2) returns the list (−3, 7), (−3, 5), (−3, 4), (−4, 5), (−4, 4), (−4, 3), (−7, 4), (−8, 3), (−8, 2), (−11, 3), (−15, 2), (−15, 1), (−20, 1), (−24, 1), (−35, 1), (−40, 1), (−51, 1), (−52, 1), (−88, 1), (−91, 1), (−115, 1), (−123, 1), (−148, 1), (−187, 1), (−232, 1), (−235, 1), (−267, 1), (−403, 1), (−427, 1). Then, the command hilbert class polynomials(-D*f2 ) for each pair in the list gave us minimal polynomials, which in turn allow us to define all the number fields L in L(2). Finally, we used cm j invariants(L), for each L in L(2) to gather the data in Table 1. 3. The case when L = Q(j(E)) In this section we consider the special case of a number field obtained by adjoining the j-invariant of a single elliptic curve with complex multiplication. Lemma 3.1. Let j(E) be a j-invariant with complex multiplication by an order O in an imaginary quadratic field K of conductor f, such that the class number of O is n > 1. Then: (1) If n > 1 is odd, then the field H = K(j(E)) contains a unique quadratic subfield, namely K. (2) Suppose that K 6⊆ Q(j(E)), and assume the extension K(j(E))/K is not 2-elementary abelian. Then Q(j(E))/Q is not Galois. Proof. Let j(E) be a j-invariant with complex multiplication by an order O in an imaginary quadratic field K of conductor f, such that the class number of O is odd > 1. Then, by the theory of complex multiplication the field H = K(j(E)) is the ring class field of O (Theorem 11.1 in [Cox89]). In particular, H/Q is Galois (Lemma 9.3 in [Cox89]), and Gal(H/Q) has order 2n. Since n is odd, H/Q contains a unique quadratic subfield, namely K (note that if this was not the case, then there would be two distinct quadratic fields inside H, and therefore Gal(H/Q) would be divisible by 4). This shows (1). ON THE NUMBER OF ISOMORPHISM CLASSES OF CM ELLIPTIC CURVES DEFINED OVER A NUMBER FIELD 9 For (2), let us assume that K 6⊆ Q(j(E)), and assume the extension H = K(j(E))/K is not 2-elementary abelian. In particular, H/Q(j(E)) is quadratic and Gal(H/Q(j(E))) ∼ = Gal(K/Q). Lemma 9.3 of [Cox89] says that Gal(H/Q) ∼ = Gal(H/K) o Gal(H/Q(j(E))) is a semi-direct product, such that the non-trivial element τ ∈ Gal(H/Q(j(E))) acts by conjugation on Gal(H/K) and sends σ to σ −1 , i.e., τ στ −1 = σ −1 . Now, Gal(Q(j(E))/Q) is Galois if and only if Gal(H/Q(j(E))) is normal in Gal(H/Q), and Proposition 11 in Section 5.5 of [DF04] says that Gal(H/Q(j(E))) is normal in H/Q if and only if the semi-direct action is trivial, i.e., σ = σ −1 for all σ ∈ Gal(H/K), which means that every non-trivial element of Gal(H/K) has order 2. However, by assumption, Gal(K(j(E))/K) is not 2-elementary abelian and therefore it contains an element σ such that σ 6= σ −1 . Hence, Q(j(E))/Q cannot be Galois.  Lemma 3.2. Let j and j 0 be distinct j-invariants with complex multiplication by an order O of conductor f, within the maximal order OK of an imaginary quadratic field K, such that h(O) is odd and > 1. Then, Q(j, j 0 ) = K(j) = K(j 0 ). Proof. Since j and j 0 have CM by the same order, they are Galois conjugates, by Theorem 2.1. By Lemma 3.1, neither Q(j) nor Q(j 0 ) are Galois extensions of Q, but K(j)/Q is Galois (Theorem 11.1 in [Cox89]), and therefore Q(j) ⊆ Q(j, j 0 ) ⊆ K(j). Since K(j)/Q(j) is quadratic, either Q(j) = Q(j, j 0 ) or K(j) = Q(j, j 0 ). Suppose for a contradiction that Q(j) = Q(j, j 0 ), i.e., Q(j) = Q(j 0 ), where the last equality follows from the fact that their degrees are equal. Let G = Gal(K(j)/Q). By Lemma 9.3 of [Cox89] we can write N = Gal(K(j)/K) and hτ i = Gal(K(j)/Q(j)), so that Gal(K(j)/Q) ∼ = N o hτ i. Since h(O) is odd, it follows that the order of N is odd, and hτ i is a 2-Sylow subgroup of G = Gal(K(j)/Q) that fixes Q(j). Since j 0 is a conjugate of j, there exists some g ∈ G such that j 0 = g(j). In particular, j 0 is fixed by τ , because j 0 ∈ Q(j), but j 0 is also fixed by τ 0 = gτ g −1 because τ 0 (j 0 ) = gτ g −1 (j 0 ) = gτ g −1 (g(j)) = gτ (j) = g(j) = j 0 , where we used the fact that τ fixes j. Hence, Q(j) is the fixed field of hτ i and Q(j) = Q(j 0 ) is the fixed field of hgτ g −1 i, so we conclude that hτ i = hgτ g −1 i, and so gτ g −1 = τ , i.e., g and τ commute (recall that τ acts on g by conjugation and τ gτ −1 = g −1 ). We distinguish two cases. Since G = N o hτ i, either g ∈ N , or g = nτ , for some n ∈ N : (1) If g ∈ N commutes with τ , i.e., gτ = τ g, then g = τ gτ −1 = g −1 , so g 2 = 1. Since N has odd order, g = 1. (2) If g = nτ for some n ∈ N , and g commutes with τ , then n = (nτ )τ = τ (nτ ), hence n = τ nτ = τ nτ −1 = n−1 , so n = 1 as before, and so g = nτ = τ . In either case we find that g = 1 or τ , i.e., g ∈ hτ i, which fixed j, and so j 0 = g(j) = j, contradicting the fact that j and j 0 are distinct. Hence, we have reached a contradiction, and we must have Q(j) 6= Q(j, j 0 ), which implies Q(j, j 0 ) = K(j) as desired.  Lemma 3.3. Let E be an elliptic curve with complex multiplication by an order O of conductor f, within the maximal order OK of an imaginary quadratic field K, such that h(O) is odd. Let O0 be another order with conductor f0 of odd class number, and such that there is an inclusion of ring class fields K(O0 ) ⊆ K(O). Then, there exists a unique j-invariant j 0 with CM by O0 , such that any elliptic curve E 0 with j(E 0 ) = j 0 satisfies Q(j(E 0 )) ⊆ Q(j(E)). Proof. Let E, O, O0 , f, and f0 be as in the statement, and let E 00 for the moment be any elliptic curve with CM by the order of O0 of conductor f0 . Since K(j(E)) is the ring class field of O (Theorem 11.1 in [Cox89]) we have by assumption that the ring class field K(j(E 00 )) = K(O0 ) is contained in K(j(E)) = K(O). In particular, Q(j(E 00 )) ⊆ K(j(E)). By Lemma 9.3 of [Cox89] we can write N = Gal(K(j(E))/K), so that Gal(K(j(E))/Q) ∼ = N o hτ i. Since h(O) is odd, it follows that the order of N is odd, and hτ i is a 2-Sylow subgroup of G = Gal(K(j(E))/Q). Since h(O0 ) = [Q(j(E 00 )) : Q] is odd by assumption, it follows that Q(j(E 00 )) is fixed by some element τ 00 of order 2 of G. Since hτ 00 i is also a 2-Sylow subgroup, it follows that hτ i and hτ 00 i are conjugates, and so there is g ∈ G such that τ = gτ 00 g −1 . Let us define j 0 = g(j(E 00 )). Then, j 0 is fixed by τ : τ (j 0 ) = gτ 00 g −1 (j 0 ) = gτ 00 g −1 (g(j(E 00 ))) = gτ 00 (j(E 00 )) = g(j(E 00 )) = j 0 , 10 where we have used the fact that τ 00 fixes j(E 00 ). Hence, j 0 is a conjugate of j(E 00 ) that is fixed by τ . Since the fixed field of hτ i is precisely Q(j(E)), we conclude that j 0 ∈ Q(j(E)). Moreover, since j 0 is a conjugate of j(E 00 ), any elliptic curve E 0 with j(E 0 ) = j 0 has CM by the order O0 of conductor f0 as desired. It remains to show that j 0 is the unique j-invariant with CM by O0 contained in Q(j(E)). Suppose that j 00 is another j-invariant with CM by O0 which is also in Q(j(E)). Then, j 00 , j(E 0 ) are two j-invariants with CM by O0 contained in Q(j(E)), and if they are distinct, then K(j(E 0 )) ⊆ Q(j(E)), by Lemma 3.2. However, [K(j(E 0 )) : Q] is even, while [Q(j(E)) : Q] is odd, so this is a contradiction. Hence, we must have j 00 = j(E 0 ), and there is a unique such j-invariant within Q(j(E)) as claimed.  Armed with these lemmas we are now ready to prove the following theorem. Theorem 3.4. Let L be a number field of degree d, and let j(E) ∈ L be a j-invariant with complex multiplication by an order O in an imaginary quadratic field K of conductor f, and j(E) 6∈ Q. (1) The class number of O is a divisor of d = [L : Q]. (2) Let F = Q(j(E)) be a number field of odd degree n > 1, and suppose j(E 0 ) ∈ F is another j-invariant with complex multiplication, such that j(E 0 ) 6∈ Q. Then, the curve E 0 has CM by O0 also in K. Pan order 0 (3) Let F = Q(j(E)) ⊆ L, and suppose that [F : Q] = n > 1 is odd. Let σ0 (f) = d|f d be the number of positive divisors of f. Then, the number of j-invariants with CM defined over F is 13 + σ0 (f) − J(K) if f is even, or if 2 does not split completely in K, and 13 + σ0 (2f) − J(K) otherwise, where J(K) is the number of rational j-invariants of curves with CM by an order of K, i.e., dK J(K ) −3 3 −4 2 −7 2 −8 1 −11 −19 −43 −67 −163 1 1 1 1 1 else 0. Proof. Let L be a number field of degree d, and let j(E) be a j-invariant with complex multiplication by an order O in an imaginary quadratic field K of conductor f, and j(E) 6∈ Q. (1) From Theorem 2.1, we have [Q(j(E)) : Q] = h(O), and since Q(j(E)) ⊆ L, it follows that h(O) is a divisor of d = [L : Q]. (2) Let j(E 0 ) ∈ Q(j(E)) be another j-invariant with complex multiplication by an order O0 in an imaginary quadratic field K 0 , such that j(E 0 ) 6∈ Q. Now consider Q(j(E 0 )) ⊆ Q(j(E)) ⊆ K(j(E)) = H. Since H/Q is Galois, it follows that the Galois closure of Q(j(E 0 )) is contained in H. Moreover, j(E 0 ) is not in Q, so Q(j(E 0 ))/Q is non-trivial and of odd degree (because Q(j(E))/Q is of odd degree and j(E 0 ) ∈ Q(j(E))). Since Q(j(E 0 ))/Q is of odd degree, K 0 is not a subfield of Q(j(E 0 )) and K 0 (j(E 0 ))/K 0 cannot be 2elementary abelian. Thus, by Lemma 3.1, it follows that Q(j(E 0 ))/Q is not Galois. Since K 0 (j(E 0 )) is Galois over Q (by Lemma 9.3 in [Cox89]), we have K 0 (j(E 0 )) ⊆ H, and in particular, K 0 ⊆ H. But Lemma 3.1 says that H/Q contains a unique quadratic subfield, namely K, and so K = K 0 . Hence E and E 0 have complex multiplications by orders of a common imaginary quadratic field K. (3) Let F = Q(j(E)) ⊆ L, and suppose that [F : Q] = n > 1 is odd. Let j(E 0 ) be another CM j-invariant defined over F , with CM by an order O0 in an imaginary quadratic field K 0 . Now consider Q(j(E 0 )) ⊆ Q(j(E)) ⊆ K(j(E)) = H. Since H/Q is Galois, it follows that the Galois closure of Q(j(E 0 )) is contained in H. There are two possibilities. Either (a) Q(j(E 0 ))/Q is Galois. In this case, since [Q(j(E 0 )) : Q] = n0 is a divisor of n, which is odd, then n0 is also odd. Thus K 0 cannot be contained in Q(j(E 0 )) and K(j(E 0 ))/K cannot be 2-elementary abelian. Thus, by Lemma 3.1, we must have n0 = 1. Hence, j(E 0 ) ∈ Q. (b) Or Q(j(E 0 ))/Q is not Galois. Since K 0 (j(E 0 )) is a ring class field, it is Galois over Q, and therefore K 0 (j(E 0 )) ⊆ H. In particular, K 0 ⊆ H. By Lemma 3.1, it follows that K = K 0 , and O and O0 are orders within the same imaginary quadratic field K. Let L = K(j(E)) and L0 = K(j(E 0 )) be ring class fields attached to O and O0 respectively (of conductor f and f0 respectively), and let C = C(L/K) and C 0 = C(L0 /K) be the conductors of the abelian extensions L/K and L0 /K. Since L0 ⊆ L, then C 0 is a divisor of C. We distinguish several possibilities according to the choice of K. The results below follow from Lemma 2.12, part (3): (i) If K = Q(i), then either • f0 |f, so that O0 ⊆ O; or ON THE NUMBER OF ISOMORPHISM CLASSES OF CM ELLIPTIC CURVES DEFINED OVER A NUMBER FIELD 11 • f0 = 2 and f = 1, so that O0 = O = OK . Since the class number of OK is 1, this means that j(E), j(E 0 ) ∈ Q. √ (ii) If K = Q( −3), then either • f0 |f, so that O0 ⊆ O; or • f0 = 2 or 3, and f = 1, so that O0 = O = OK . Since the class number of OK is 1, this 0 means that j(E), √ j(E ) ∈ Q. (iii) If K 6= Q(i) or Q( −3), then either • f0 |f, so that O0 ⊆ O; or • f0 |2f, where f is odd, and 2 splits completely in K. Hence, in all cases, either j(E 0 ) ∈ Q, or f0 |f, or f is odd, and 2 splits completely in K, and f0 |2f. Since every conductor f0 as above corresponds to a unique order O0 with K(O0 ) ⊆ K(O), which in turn corresponds to a unique j-invariant j(E 0 ) ∈ Q(j(E)) by Lemma 3.3, it follows that the number of j(E 0 ) ∈ Q(j(E)) with j(E 0 ) 6∈ Q is precisely 13 + σ0 (f) − J(K) if f is even, or if 2 does not split completely in K, and 13 + σ0 (2f) − J(K) otherwise, as desired. Notice, the correction factor J(K) ensures that no CM elliptic curves defined over Q are double counted.  We conclude this section with a theorem about fields generated by several CM j-invariants. Theorem 3.5. Let K be an imaginary quadratic field and suppose that E1 and E2 have CM by orders O1 and O2 of OK with conductors f1 and f2 respectively. Then, if [Q(j(E1 ) : Q] and [Q(j(E2 )) : Q] are odd, then (1) There exists an elliptic curve Eg with CM by an order Og of OK with conductor fg = gcd(f1 , f2 ) such that Q(j(E1 )) ∩ Q(j(E2 )) = Q(j(Eg )), and (2) Assume that Q(j(E1 ), j(E2 )) is an extension of odd degree. Then, there exists an elliptic curve El with CM by an order Ol of OK with conductor fl = lcm(f1 , f2 ) such that Q(j(E1 ), j(E2 )) = Q(j(El )). Proof. Let fg = gcd(f1 , f2 ), and let Og be the order of K with conductor fg . Since fg divides f1 and f2 , it follows from Lemma 2.12 that there are inclusions of ring class fields K(Og ) ⊆ K(Oi ) for i = 1, 2. By assumption h(O1 ) and h(O2 ) are odd, so Lemma 3.3 implies that there is a unique j-invariant jg such that any elliptic curve Eg with j(Eg ) = jg has CM by Og , and Q(j(Eg )) is contained in both Q(j(E1 )) and Q(j(E2 )), so Q(j(Eg )) ⊆ Q(j(E1 )) ∩ Q(j(E2 )). Since the degree of Q(j(E1 )) ∩ Q(j(E2 )) divides the GCD of the degrees of Q(j(E1 )) and Q(j(E2 )), it follows that h(Og ) = [Q(j(Eg ) : Q] is a divisor of gcd(h(O1 ), h(O2 )). But Corollary 2.11 shows that h(Og ) = gcd(h(O1 ), h(O2 )), and hence Q(j(Eg )) = Q(j(E1 )) ∩ Q(j(E2 )). This completes the proof of part (1). For (2), let fl = lcm(f1 , f2 ), and let Ol be the order of K with conductor fl . Let El0 be for the moment any elliptic curve with CM by Ol . Since fi divides fl , for i = 1, or 2, it follows that K(Oi ) ⊆ K(Ol ), and therefore Lemma 3.3 implies that there are elliptic curves E10 and E20 with CM by O1 and O2 respectively, such that j(E10 ), j(E20 ) ∈ Q(j(El0 )). Next, from part (1) of this theorem we know that there exists an elliptic curve Eg0 such that Q(j(E10 )) ∩ Q(j(E20 )) = Q(j(Eg0 )). Thus, K(j(E10 )) ∩ K(j(E20 )) = K(j(Eg0 )) and by our previous remarks it follows that K(j(E10 ), j(E20 )) ⊆ K(j(El0 )). Next we remark that K(j(E10 ), j(E20 ))/Q is Galois (because each K(j(Ei0 ))/Q is Galois) and compute [K(j(E10 ), j(E20 )) : Q] = [K(j(E10 )) : Q] · [K(j(E20 )) : Q] [K(j(E10 )) : Q] · [K(j(E20 )) : Q] = 0 0 [K(j(E1 )) ∩ K(j(E2 )) : Q] [K(j(Eg0 )) : Q] 2h(O1 ) · 2h(O2 ) 2h(O1 ) · h(O2 ) = 2h(Og ) gcd(h(O1 ), h(O2 )) = 2lcm(h(O1 ), h(O2 )) = 2h(Ol ) = [K(j(El0 )) : Q], = where we have used Corollary 2.11 twice in the last two strings of equalities. Thus, we conclude that we have an equality K(j(E10 ), j(E20 )) = K(j(El0 )). From Galois theory, we know that Gal(K(j(El0 ))/K) is isomorphic to 12 the subgroup of Gal(K(j(E10 ))/K) × Gal(K(j(E20 ))/K) given by n o n (σ1 , σ2 ) : σ1 K(j(E 0 ))∩K(j(E 0 )) = σ2 K(j(E 0 ))∩K(j(E 0 )) = (σ1 , σ2 ) : σ1 1 2 1 2 K(j(Eg0 )) = σ2 o K(j(Eg0 )) . Theorem 2.1 implies that j(E1 ) and j(E10 ) are roots of the same polynomial, and so there exists an element σ1 ∈ Gal(K(j(E10 ))/K) such that σ1 (j(E10 )) = j(E1 ). Similarly, there is a σ2 ∈ Gal(K(j(E20 ))/K) such that σ2 (j(E20 )) = j(E2 ). Notice that for i = 1 or 2, the element σi sends j(Eg0 ) to some conjugate σi (j(Eg0 )) of j(Eg ), but σi (j(Eg0 )) ∈ σi (Q(j(Ei0 ))) = Q(j(Ei )), and therefore σi (j(Eg0 )) ∈ Q(j(E1 )) ∩ Q(j(E2 )). Since j(Eg ) is the unique j-invariant with CM by Og in Q(j(E1 )) and Q(j(E2 )) (by Lemma 3.3), it also follows that j(Eg ) is the unique such j-invariant in Q(j(E1 )) ∩ Q(j(E2 )). Thus, it must be that σ1 (j(Eg0 )) = j(Eg ) = σ2 (j(Eg0 )). Hence σ1 K(j(Eg0 )) = σ2 K(j(Eg0 )) . Thus there is an element σ ∈ Gal(K(j(El0 )/K) corresponding to the pair (σ1 , σ2 ), such that σ(j(E10 )) = j(E1 ) and σ(j(E20 )) = j(E2 ). Letting El = σ(El0 ) we have j(E1 ), j(E2 ) ∈ Q(j(El )) by construction, and El has CM by the order of OK of conductor fl . Since K(j(El )) = K(j(E1 ), j(E2 )) and Q(j(E1 ), j(E2 )) is of odd degree by assumption, it must be that Q(j(El )) = Q(j(E1 ), j(E2 )), as desired. This concludes the proof of (2).  4. Intersections of Ring Class Fields of different Quadratic Imaginary Fields In this section we study the intersection of ring class fields attached to distinct imaginary quadratic fields. Lemma 4.1. Let K1 , . . . , Kn be distinct imaginary quadratic fields of odd class number, and let K be the compositum K1 · · · Kn . Then, the only imaginary quadratic subfields of K of odd class number are K1 , . . . , Kn . √ Proof. By Corollary 2.4, each Ki is of the form Q( −qi ) where Qn qi = 1 or a prime, with qi 6= qj for i 6= j. In particular, every quadratic subfield of the compositum K = i=1 Ki is of the form v  u n uY F = Q t (−qi )ti  i=1 where each ti = 0 or 1. But, if ti = 1 for two different indices, then either F is real quadratic, or two different primes divide the discriminant and Theorem 2.3 would imply that the class number of F is even. Therefore, if F ⊆ K is an imaginary quadratic field of odd class number, then ti = 1 for a unique index i, and so F = Ki as desired.  Lemma 4.2. Let K1 , . . . , Kn be distinct imaginary quadratic fields, and let Hi be a ring class field of Ki , for each i = 1, . . . , n, such that [Hi : Ki ] = ni is odd. Let H be the compositum of all Hi . Then, any quadratic subfield F ⊆ H is contained in the compositum K of all Ki , for i = 1, . . . , n. Proof. Since the fields Ki are distinct, the compositum K has Q degree 2n over Q. And since [Hi : Ki ] = hi is n n odd and Hi /Q Galois, it follows that [H : Q] is a divisor of 2 ( i=1 hi ), and therefore the power of 2 dividing n [H : Q] is precisely 2 . If F ⊆ H was a quadratic field not contained in K, then F K/Q would be of degree 2n+1 , and since F K ⊆ H, it would follow that 2n+1 divides [H : Q], and a contradiction arises.  Lemma √4.3. Let E1 , . . . , En be elliptic curves with CM by orders Oi of distinct imaginary quadratic fields Ki = Q( −di ). Assume that the extension Q(j(E1 ), . . . , j(En )) is of odd degree. Then, if E is a CM elliptic curve defined over Q(j(E1 ), . . . , j(En )), then E has CM by an order of class number 1 or an order of Ki , for some i = 1, . . . , n. Proof. Let K be the compositum√of all Ki , for i = 1, . . . , n. Suppose that E has CM by an order O of the imaginary quadratic field F = Q( −d) and that [Q(j(E)) : Q] = m. If the class number of O is 1, then j(E) is defined over Q, so let us assume that m > 1. Since every Q(j(Ei )) is of odd degree, we have that Q(j(E1 ), . . . , j(En ))/Q is of odd degree, and since j(E) ∈ Q(j(E1 ), . . . , j(En )), it follows that [Q(j(E)) : Q] = m > 1 is also odd. In particular, the Galois closure of Q(j(E)) is the field F (j(E)) by Lemma 3.1. Also, we know that Ki (j(Ei )) is Galois, and so K(j(E1 ), . . . , j(En ))/Q is also Galois, and so F (j(E)) ⊆ K(j(E1 ), . . . , j(En )). It follows that F ⊆ K, by Lemma 4.2. But F is an imaginary quadratic field of odd class number contained in K, and Lemma 4.1 implies that F = Ki for some i = 1, . . . , n.  ON THE NUMBER OF ISOMORPHISM CLASSES OF CM ELLIPTIC CURVES DEFINED OVER A NUMBER FIELD 13 Theorem 4.4. Let K1 , . . . , Kn be distinct imaginary quadratic fields of odd class number. LetQH1 , . . . , Hn be n ring class fields of K1 , . . . , Kn respectively, such that [Hi : Ki ] = ni are all odd. Let H = i=2 Hi . Then H1 ∩ H = Q. Proof. Let F = H1 ∩ H. Since each Hi is a ring class field, we know that H1 /Q and H/Q are both Galois extensions, and therefore their intersection, F , is also Galois over Q. Next, we know that since [H1 : Q] = 2n1 with n1 odd, it must be that H1 contains a unique quadratic extension (Lemma 3.1); namely, K1 , and in particular, F ⊆ H1 contains at most one quadratic extension. Qn But if K1 ⊆ F , then K1 ⊆ H, and Lemma 4.2 would imply that K1 is contained in the compositum K = i=2 Ki . By Lemma 4.1, and since all Ki are distinct of odd class number, K1 cannot be contained in K. It follows that F has no quadratic subextension, and in particular F is of odd degree. Since K1 ∩ F = Q we get that Gal(F K1 /K1 ) ∼ = Gal(F/(F ∩ K1 )) = Gal(F/Q), by Galois theory, but we know that F K1 /K1 is an abelian extension since it is a subextension of H1 /K1 . Therefore, we also have that F/Q is an abelian extension. Now, Galois theory implies that Gal(F K1 /Q) ∼ = Gal(K1 /Q) × Gal(F/Q), which is clearly abelian. From Theorem 2.14 we know that F K1 /K1 is contained in a ring class field if and only if F K1 /Q is generalized dihedral, but all nontrivial generalized dihedral groups are nonabelian. Thus it must be that F K1 /K1 is a trivial extension and since K1 6⊆ F it follows that F/Q is trivial and F = Q.  Proposition 4.5. Let n ≥ 1, and suppose that Ei , for i = 1, . . . , n, is an elliptic curve with CM by an order Oi of an imaginary quadratic field Ki of odd class number, with Ki 6= Kj for i 6= j, and such that j(Ei ) 6∈ Q. Further, assume that Q(j(E1 ), . . . , j(En ))/Q is an extension of odd degree. Then: [Q(j(E1 ), . . . , j(En )) : Q] = n Y [Q(j(Ei )) : Q]. i=1 Proof. We will prove the formula using induction on n. The statement is trivial for n = 1, so let us assume that it is true for n = k − 1, for some k ≥ 2, and let Ei , for i = 1, . . . , k, be elliptic curves as in the statement of the theorem. Let K be the compositum of K2 , . . . , Kk , and let K0 = K1 K be the compositum of K1 , . . . , Kk . It follows from Lemma 4.1 that [K0 : Q] = 2k and [K : Q] = 2k−1 . Consider the following diagram H0 = K0 (j(E1 ), . . . , j(Ek )) H = K(j(E2 ), . . . , j(Ek )) H1 = K1 (j(E1 )) Q and notice that every field in the diagram is Galois over Q, because they are compositums of ring class fields. Since H = K(j(E2 ), . . . , j(Ek )) is the compositum of all ring class fields Hi = Ki (j(Ei )) for i = 2, . . . , k, and since all the imaginary quadratic fields are distinct, it follows from Theorem 4.4 that H1 and H are disjoint, and therefore [K0 (j(E1 ), . . . , j(Ek )) : Q] = [K1 (j(E1 )) : Q] · [H : Q]. By the induction hypothesis, since [K : Q] = 2k−1 and K/Q is Galois, and Q(j(E2 ), . . . , j(Ek )) is of odd degree by assumption, it follows that [H : Q] = 2k−1 · [Q(j(E2 ), . . . , j(Ek )) : Q] = 2k−1 k Y [Q(j(Ei )) : Q]. i=2 14 Hence, if we write H0 = K0 (j(E1 ), . . . , j(Ek )), we have 2k · [Q(j(E1 ), . . . , j(Ek )) : Q] = [H0 : Q] = [K0 (j(E1 ), . . . , j(Ek )) : Q] = [K1 (j(E1 )) : Q] · [H : Q] ! k Y k−1 = (2 · [Q(j(E1 )) : Q]) · 2 [Q(j(Ei )) : Q] i=2 = 2k · k Y [Q(j(Ei )) : Q], i=1 Qk which shows [Q(j(E1 ), . . . , j(Ek )) : Q] = i=1 [Q(j(Ei )) : Q], as desired, as this completes the proof of the induction step, and thus the statement is true for all n ≥ 1.  Theorem 4.6. Suppose that Ei , for i = 1, 2, 3, . . . , n, is an elliptic curve with CM by an order Oi of an imaginary quadratic field Ki of odd class number. Further suppose that the extension Q(j(E1 ), . . . , j(En ))/Q is of odd degree, and Ki 6= Kj if i 6= j. Then, (1) Q(j(E1 )) ∩ Q(j(E2 ), . . . , j(En )) = Q, and (2) If E is an elliptic curve with CM by O an order of an imaginary quadratic field K such that j(E) ∈ Q(j(E1 ), j(E2 ), . . . , j(En )), then K = Ki and j(E) ∈ Q(j(Ei )), for some i = 1, 2, 3, . . . , n. Proof. Let K be the compositum of K2 through Kn . Consider Q(j(E1 )) ⊆ K(j(E1 )) = H1 and Q(j(E2 ), . . . , j(En )) ⊆ K(j(E2 ), . . . , j(En )) = H. By Theorem 4.4, we have H1 ∩ H = Q, and therefore Q(j(E1 )) ∩ Q(j(E2 ), . . . , j(En )) = Q. This shows part (1). Suppose that E is an elliptic curve with CM such that j(E) ∈ Q(j(E1 ), . . . , j(En )). From Lemma 4.3 we know that E must have CM by an order of class number 1 (and thus j(E) ∈ Q) or an order of Ki , for some i = 1, . . . , n. Without loss of generality assume that E has CM by O ⊂ K1 . Now, by Proposition 4.5 we have n Y [Q(j(E1 ), . . . , j(En )) : Q] = [Q(j(Ei )) : Q]. i=1 Since E and E1 have CM by orders of odd class number of the same imaginary quadratic field, and Q(j(E), j(E1 )) is contained in an extension of odd degree by assumption, Theorem 3.5 implies that there is an elliptic curve El with CM by an order of K1 such that Q(j(E), j(E1 )) = Q(j(El )). Thus, Q(j(E1 ), j(E2 ), . . . , j(En )) = Q(j(E), j(E1 ), j(E2 ), . . . , j(En )) = Q(j(El ), j(E2 ), . . . , j(En )) and so, using Proposition 4.5 once again, we obtain [Q(j(E1 ), . . . , j(En )) : Q] = [Q(j(El ), . . . , j(En )) : Q] = [Q(j(El )) : Q] · n Y [Q(j(Ei )) : Q]. i=2 It follows that [Q(j(E1 )) : Q] = [Q(j(El )) : Q], and since Q(j(E1 )) ⊆ Q(j(E), j(E1 )) = Q(j(El )), we conclude that Q(j(E1 )) = Q(j(El )) = Q(j(E), j(E1 )) and so j(E) ∈ Q(j(E1 )) as desired.  5. A Sharp Upper Bound We are finally ready to find an upper bound on the number of elliptic curves with CM defined over a number L of odd degree over Q. In order to simplify the situation further, we introduce the following notation. Definition 5.1. Given a number field L/Q and an imaginary quadratic field K, let Σ(L, K) be the set jinvariants of elliptic curves defined over L with CM by an order of OK that are not defined over Q. We define the field of definition of Σ(L, K) as Q(Σ(L, K)) = Q({j : j ∈ Σ(L, K)}). Also, let Σ(L) be the set of j-invariants of elliptic curves defined over L with CM that are not defined over Q. The field of definition of Σ(L) will be denoted by Q(Σ(L)) = Q({j : j ∈ Σ(L)}). We begin by showing that Σ(L, K) is a finite set. Lemma 5.2. If L is a number field of odd degree over Q, then (1) Σ(L, K) is a finite set, and (2) There exists an elliptic curve E with CM by an order O of OK such that Q(Σ(L, K)) = Q(j(E)). ON THE NUMBER OF ISOMORPHISM CLASSES OF CM ELLIPTIC CURVES DEFINED OVER A NUMBER FIELD 15 Proof. If E is an elliptic curve with CM by K and j(E) ∈ L, then [Q(j(E)) : Q] is a divisor of N = [L : Q]. Since L/Q is assumed to have odd degree, it follows that [Q(j(E)) : Q] is odd and ≤ N . Hence, Corollary 2.9 implies that there are only finitely many possibilities for j(E). Hence Σ(L, K) is finite. This shows (1). Now, part (2) follows by induction on the size of Σ(L, K) by Theorem 3.5.  It is worth pointing out that the previous finiteness result (Lemma 5.2, part (1)) is also true in even degree (see Remark 2.10). Proposition 5.3. Let L/Q be a number field of odd degree and let K be an imaginary quadratic field such that Σ(L, K) 6= ∅. Further, suppose that d = [Q(Σ(L, K)) : Q] = pe11 · · · perr , and hK = pf11 · · · pfrr , for some distinct primes pi , and some ei , fi ≥ 0. Then, r X (1) If hK = 1, we have #Σ(L, K) ≤ 2 ei , and i=1 ! r r X X (2) If hK > 1, then #Σ(L, K) ≤ 2 1 + (ei − fi ) ≤ 2 ei . i=1 In all cases, #Σ(L, K) ≤ 2 r X i=1 ei . i=1 Proof. Notice first that the assumption that L/Q is an odd degree extension guarantees that d is also odd since Q(Σ(L, K)) is a subfield of L. From Lemma 5.2, we know that there is an elliptic curve with CM by an order O of OK such that Q(Σ(L, K)) = Q(j(E)). Suppose that the conductor of O is f. Therefore, d = [Q(Σ(L, K)) : Q] = [Q(j(E)) : Q] = h(O) and so it must be that hK divides d and so hK√ must be odd. In particular, 0 ≤ fi ≤ ei for i = 1, . . . , r. Also, Lemmas 2.6 and 2.7 show that K 6= Q(i) or Q( −2) because Σ(L, K) = ∅ in those cases. √ Combining all of this we get that K = Q( −q) and f = 2n q m , with q ≡ 3 mod 4, n ∈ {0, 1}, and m ≥ 0. Further, we have that dK = −q. Assume that hK 6= 1, so that at least one exponent fi > 0 in the prime factorization of hK . Applying Theorem 3.4, and noting that J(K) = 0 because we are assuming hK 6= 1, we can see that ( σ0 (f) if n = 1, or 2 does not split completely in K, #Σ(L, K) = σ0 (2f) otherwise (in particular n = 0), ( σ0 (2q m ) if n = 1, or 2 does not split completely in K, = m σ0 (2 · q ) otherwise (in particular n = 0), ≤ 2(m + 1). By Theorem 2.8 we have that   if f is odd, hK · f e1 er p1 · · · pr = d = [Q(Σ(L, K)) : Q] = h(O) = hK · 2f if f is even and dK ≡ 1 mod 8,   hK · 3·f if f is even and dK ≡ 5 mod 8. 2 ( f1 p1 · · · pfrr q m if f is odd, or dK ≡ 1 mod 8, = f1 fr m 3p1 · · · pr q otherwise. Pr Pr Pr In particular, i=1 ei ≥ m + i=1 fi , and therefore m ≤ i=1 (ei − fi ). It follows that ! r X #Σ(L, K) ≤ 2(m + 1) ≤ 2 1 + (ei − fi ) . i=1 Since we are assuming hK > 1, we know that at least one index fi > 0. Thus, ! ! ! r r r r r X X X X X #Σ(L, K) ≤ 2(m + 1) ≤ 2 1 + (ei − fi ) ≤ 2 1 + ei − fi ≤ 2 1 + ei − 1 = 2 ei . i=1 i=1 i=1 i=1 i=1 This proves (2) in the case of hK > 1. If hK = 1, then we distinguish three cases according to whether q = 3, q = 7, or q ∈ D = {11, 19, 43, 67, 163}. For each of these 3 cases we also need to consider if n = 0 or 1. In all six 16 possible cases we calculate d = h(O) and the corresponding sum   q m−1 if q = 3, n = 0, 2(m − 1)         m   q if q = 3, n = 1, 2m       r q m 2m X if q = 7, n = 0, d = h(O) = , 2 ei = q m  if q = 7, n = 1, 2m   i=1     m     q if q ∈ D, n = 0, 2m      m  3q if q ∈ D, n = 1, 2(m + 1)  σ0 (f) − 3      σ0 (f) − 3    σ (2f) − 2 0 #Σ(L, K) = σ0 (f) − 2      σ0 (f) − 1    σ0 (f) − 1 if if if if if if q q q q q q of the exponents, as well as #Σ(L, K):  if q = 3, n = 0, 2m − 2 if q = 3, n = 0,      if q = 3, n = 1, 2m if q = 3, n = 1,    2m if q = 7, n = 0, if q = 7, n = 0, = 2m if q = 7, n = 1, if q = 7, n = 1,      if q ∈ D, n = 0, 2m if q ∈ D, n = 0,   if q ∈ D, n = 1, 2m + 2 if q ∈ D, n = 1,  = 3, n = 0, (m + 1) − 3 if q      = 3, n = 1, 2(m + 1) − 3 if q    2(m + 1) − 2 if q = 7, n = 0, = 2(m + 1) − 2 if q = 7, n = 1,     ∈ D, n = 0,  (m + 1) − 1 if q    ∈ D, n = 1, 2(m + 1) − 1 if q  = 3, n = 0, m−2 if      = 3, n = 1, 2m − 1 if    2m = 7, n = 0, if = 2m = 7, n = 1, if     ∈ D, n = 0,  m if    ∈ D, n = 1, 2m + 1 if q q q q q q = 3, n = 0, = 3, n = 1, = 7, n = 0, = 7, n = 1, ∈ D, n = 0, ∈ D, n = 1. Here it is worth noting that the condition Σ(L, K) 6= ∅ imposes some lower bounds on m, so that #Σ(L, K) > 0. r X In all of these cases, we have that #Σ(L, K) ≤ 2 ei , as claimed (and there is equality in certain cases!).  i=1 √ Example 5.4. Let K = Q( −7). Let E be an elliptic curve with complex multiplication by the order O of OK of conductor f = 2 · 7m . In this case hk = 1 and h(O) = 7m , so if we define L = Q(j(E)), then [L : Q] = 7m . Since Q(Σ(L, K)) ⊆ L, and Q(j(E)) ⊆ Q(Σ(L, K)), it follows that L = Q(Σ(L, K)). Thus, Proposition 5.3 says that #Σ(L, K) ≤ 2m. While Theorem 3.4 part (3) says that #Σ(L, K) = σ0 (f) − J(K) = σ0 (2 · 3m ) − 2 = 2m. Therefore, the upper bound established in Proposition 5.3 is sharp. √ Example 5.5. Let K = Q( −23). Let E be an elliptic curve with complex multiplication by the order of OK with conductor f = 2 · 23m . In this case, we have that hK = 3 and so if we define L = Q(j(E)), then [L : Q] = 3 · 23m . Since Q(Σ(L, K)) ⊆ L by definition, and Q(j(E)) ⊆ Q(Σ(L, K)), it follows that L = Q(Σ(L, K)). Thus, Proposition 5.3, part (2), says that #Σ(L, K) ≤ 2(1 + (m + 1) − 1) = 2(m + 1). While Theorem 3.4 part (3) says that #Σ(L, K) = σ0 (f) = σ0 (2 · 23m ) = 2(m + 1). Therefore, the upper bound established in Proposition 5.3 is again sharp. Remark 5.6. Examples 5.4 and 5.5 show that the bound established in Proposition 5.3 is sharp when the class number of K is 1 or if the class number of K is greater than 1. In the proof of Proposition√5.3, however, one can see that when the class number of K is 1, then the bound can be sharp only for K = Q( −7). Next, we will show that Σ(L) is a union of sets of the form Σ(L, K), for a finite number of quadratic fields K. Lemma 5.7. Given a number field L/Q of odd degree n, there exists a finite list of imaginary quadratic fields St K1 , ..., Kt of odd class number such that Σ(L) = i=1 Σ(L, Ki ) with Σ(L, Ki ) ∩ Σ(L, Kj ) = ∅ if i 6= j, and Σ(L, Ki ) 6= ∅ for each i = 1, . . . , t. b be the Galois closure of L. Let KL be the largest 2Proof. Let L be a number field of odd degree, and let L elementary abelian extension contained in L, and let K1 , . . . , Ku , for some u ≥ 1, be all the imaginary quadratic b fields of odd class number contained in KL ⊆ L. Let K be an imaginary quadratic field such that there is an elliptic curve E with CM by an order O in K, and such that j(E) ∈ L but not in Q. Since [L : Q] is odd, and Q(j(E)) ⊆ L, it follows that h(O) is odd, and b and so therefore Lemma 3.1 implies that the Galois closure of Q(j(E)) is K(j(E)). In particular, K(j(E)) ⊆ L b Since the class number h(O) is odd, and hK divides h(O), it follows that hK is odd, and therefore K K ⊆ L. is one of K1 , . . . , Ku . If we let K1 , . . . , Kt , for some t ≤ u, be the subset of fields such that Σ(L, Ki ) 6= ∅ for St i = 1, . . . , t, then we have shown that Σ(L) ⊆ i=1 Σ(L, Ki ). Clearly Σ(L, Ki ) ⊆ Σ(L) ⊆ L for each i, and so St we have an equality Σ(L) = i=1 Σ(L, Ki ). ON THE NUMBER OF ISOMORPHISM CLASSES OF CM ELLIPTIC CURVES DEFINED OVER A NUMBER FIELD 17 The disjointness Σ(L, Ki ) ∩ Σ(L, Kj ) = ∅ if i 6= j follows from the fact that an elliptic curve cannot have CM by orders of two different imaginary quadratic fields.  Theorem 5.8. Let L/Q be a number field of odd degree n = pe11 · · · perr , and let K1 , . . . , Kt be the list of those imaginary quadratic fields given by Lemma 5.7, such that K1 , . . . , Ks , for some 0 ≤ s ≤ t, are those with odd f f class number > 1. Let hi be the class number of Ki , and write hi = p1i,1 · · · pri,r , for i = 1, . . . , s. Then, ! r s r X X X #Σ(L) ≤ 2s + 2 ej − fi,j ≤ 2 ej . j=1 i=1 j=1 Moreover, if t = s = 0, then Σ(L) is empty. Proof. From Lemma 5.7 we knowSthat there is a finite list of imaginary quadratic fields of odd class number t K1 , K2 , . . . , Kt such that Σ(L) = i=1 Σ(L, Ki ) with Σ(L, Ki ) ∩ Σ(L, Kj ) = ∅ if i 6= j, and Σ(L, Ki ) 6= ∅ for each i = 1, . . . , t (in particular, if t = 0, then Σ(L) = ∅). Assume that t ≥ 1, and let K1 , . . . , Ks , for some 0 ≤ s ≤ t, be the subset of those fields with odd class number > 1. By Lemma 5.2, for each i there is an elliptic curve Ei with CM by an order in Ki such that Q(Σ(L, Ki )) = Q(j(Ei )). Moreover, by Theorem 4.6, we have that for fixed i, Q(j(Ei )) ∩ Q({j(Ek ) : 1 ≤ k ≤ t, k 6= i}) = Q, and Proposition 4.5 shows that [Q(j(E1 ), . . . , j(En )) : Q] = t Y [Q(j(Ei )) : Q]. i=1 Hence, " t [ [Q(Σ(L)) : Q] = Q ! Σ(L, Ki ) # " :Q = i=1 " = = t Y t Y # Q (Σ(L, Ki )) : Q i=1 # Q (j(Ei )) : Q = [Q(j(E1 ), . . . , j(Et )) : Q] i=1 t Y t Y i=1 i=1 [Q(j(Ei )) : Q] = e e [Q(Σ(L, Ki )) : Q]. e Let us write [Q(Σ(L, Ki )) : Q] = ni = p1i,1 p2i,2 . . . pri,r . Since Q(Σ(L)) ⊆ L, the previous equations show that Qt Pt i=1 ni divides n. Hence, ej ≥ i=1 ei,j . Now it follows from Proposition 5.3 that #Σ(L) = t X #Σ(L, Ki ) ≤ i=1 s X #Σ(L, Ki ) + #Σ(L, Ki ) i=s+1 i=1 ≤ t X s X     r t r X X X 2 2 1 + (ei,j − fi,j ) + ei,j  i=1 j=1 = 2s + 2 ≤ 2s + 2 r X t X j=1 i=1 r X i=s+1 ei,j − ej − j=1 s X j=1 ! fi,j i=1 s X ! fi,j . i=1 Moreover, by Proposition 5.3 we also know #Σ(L) = t X i=1 as desired. #Σ(L, Ki ) ≤ t X i=1  2 r X j=1  ei,j  = 2 r X t X j=1 i=1 ei,j ≤ 2 r X ej , j=1  18 √ √ √ Example 5.9. Let K1 = Q( −7), K2 = Q( −31), and K3 = Q( −47). Let Ei , for i = 1, 2, 3, be an elliptic curve with complex multiplication by an order Oi of OKi , respectively. Further, suppose that the conductor of the orders O1 , O2 , and O3 are f1 = 2 · 7m1 , f2 = 2 · 31m2 , and f3 = 2 · 47m3 with m1 , m2 , m3 ≥ 1 respectively. In this case, we have that h(O1 ) = 7m1 , h(O2 ) = 3 · 31m2 , and h(O3 ) = 5 · 47m3 , by Theorem 2.8. Let L = Q(j(E1 ), j(E2 ), j(E3 )). From the proof of Theorem 5.8, we know that [L : Q] = 3 Y [Q(j(Ei )) : Q] = 3 · 5 · 7m1 · 31m2 · 47m3 . i=1 Since Q(Σ(L)) ⊆ L and L = Q(j(E1 ), j(E2 ), j(E3 )) ⊂ Q(Σ(L)) by definition, we know that L = Q(Σ(L)) and that Σ(L) = Σ(L, K1 ) ∪ Σ(L, K2 ) ∪ Σ(L, K3 ). Thus, from Theorem 3.4 part (3), we have that #Σ(L) = #Σ(L, K1 ) + #Σ(L, K2 ) + #Σ(L, K3 ) = (σ(2 · 7m1 ) − J(K1 )) + σ0 (f2 ) + σ0 (f3 ) = (σ0 (2 · 7m1 ) − 2) + σ0 (2 · 31m2 ) + σ0 (2 · 41m3 ) = (2(m1 + 1) − 2) + 2(m2 + 1) + 2(m3 + 1) = 4 + 2m1 + 2m2 + 2m3 . Next, Theorem 5.8 says that #Σ(L) ≤ 2(1 + 1 + m1 + m2 + m3 ) = 4 + 2m1 + 2m2 + 2m3 . Therefore, the bound established in Theorem 5.8 is in fact sharp. References x2 ny 2 . [Cox89] David A. Cox. Primes of the form + A Wiley-Interscience Publication. John Wiley & Sons, Inc., New York, 1989. Fermat, class field theory and complex multiplication. [Cox12] David A. Cox. Galois theory. Pure and Applied Mathematics (Hoboken). 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