On some extension of Gauss' work and applications (II)
aa r X i v : . [ m a t h . N T ] O c t On some extension of Gauss’ work and applications (II)
Ick Sun Eum, Ho Yun Jung, Ja Kyung Koo and Dong Hwa Shin
Abstract
Let K be an imaginary quadratic field of discriminant d K , and let n be a nontrivial integralideal of K in which N is the smallest positive integer. Let Q N ( d K ) be the set of primitivepositive definite binary quadratic forms of discriminant d K whose leading coefficients arerelatively prime to N . We adopt an equivalence relation ∼ n on Q N ( d K ) so that the set ofequivalence classes Q N ( d K ) / ∼ n can be regarded as a group isomorphic to the ray class groupof K modulo n . We further present an explicit isomorphism of Q N ( d K ) / ∼ n onto Gal( K n /K )in terms of Fricke invariants, where K n is the ray class field of K modulo n . This would becertain extension of the classical composition theory of binary quadratic forms, originatedand developed by Gauss and Dirichlet. For a negative integer D such that D ≡ Q ( D ) be the set of primitive positivedefinite binary quadratic forms Q ( x, y ) = ax + bxy + cy of discriminant b − ac = D . Then,the modular group SL ( Z ) (or PSL ( Z )) naturally acts on Q ( D ) and gives rise to the properequivalence ∼ as Q ∼ Q ′ ⇐⇒ Q = Q ′ γ = Q ′ γ " xy for some γ ∈ SL ( Z ) . It was Gauss who first established a systematic theory of composition for binary quadratic formsin his monumental work
Disquisitiones arithmeticae published in 1801 ([10]). More precisely,the set of equivalence classes C ( D ) = Q ( D ) / ∼ becomes an abelian group under the directcomposition, called the form class group of discriminant D . Besides, in 2004 some elegant andenlightening progress was achieved in higher composition theory by Bhargava ([1], [2] and [3]).See also [5]. Mathematics Subject Classification . Primary 11E16; Secondary 11F03, 11G15, 11R37.
Key words and phrases . binary quadratic forms, class field theory, complex multiplication, modular functions.The first author was supported by the Dongguk University Research Fund of 2018 and the National ResearchFoundation of Korea (NRF) grant funded by the Korea government (MSIT) (2017R1C1B5017567). The second(corresponding) author was supported by the National Research Foundation of Korea (NRF) grant funded by theKorea Government (MIST) (2016R1A5A1008055 and 2017R1C1B2010652). The fourth author was supported bythe Hankuk University of Foreign Studies Research Fund of 2018 and the National Research Foundation of Korea(NRF) grant funded by the Korea government (MSIT) (2017R1A2B1006578). K = Q ( √ D ) and O be the order of discriminant D in the imaginary quadratic field K .Let I ( O ) be the group of proper fractional O -ideals, and let P ( O ) be its subgroup of principal O -ideals. For each Q = ax + bxy + cy ∈ Q ( D ), let ω Q be the zero of Q ( x,
1) in the complexupper half-plane H , namely, ω Q = − b + √ D a . Then, it is well known that the form class group C ( D ) is isomorphic to the O -ideal class group C ( O ) = I ( O ) /P ( O ) through the isomorphism C ( D ) ∼ → C ( O ) , [ Q ] [[ ω Q , Z ω Q + Z ] . (1)As noted in [6, Theorem 7.7] it is not necessary to prove that C ( D ) is a group. It suffices toshow that the mapping stated in (1) is a well-defined bijection, through which the O -ideal classgroup C ( O ) equips C ( D ) with a group structure, namely, the Dirichlet composition ([6, (3.7)]and [7]).On the other hand, if H O denotes the ring class field of order O , then it is generated by the j -invariant j ( O ) by the theory of complex multiplication. Furthermore, there is an isomorphism C ( O ) ∼ → Gal( H O /K ) , [ a ] ( j ( O ) j ( a )) (2)([6, Theorem 11.1 and Corollary 11.37]). Thus we attain the isomorphism C ( D ) ∼ → Gal( H O /K ) , [ Q ] ( j ( O ) j ([ − ω Q , n of the maximal order O K of K , let C ( n ) be the ray class group modulo n , namely, C ( n ) = I K ( n ) /P K, ( n )where I K ( n ) is the group of fractional ideals of K relatively prime to n and P K, ( n ) is itssubgroup given by P K, ( n ) = { ν O K | ν ∈ K ∗ such that ν ≡ ∗ n ) } . The ray class field modulo n , denoted by K n , is defined to be a unique abelian extension of K with Galois group isomorphic to C ( n ) in which every ramified prime ideal divides n . One canrefer to [12, Chapter V] or [18] for the class field theory.For a positive integer N , let Q N ( d K ) = { ax + bxy + cy ∈ Q ( d K ) | gcd( N, a ) = 1 } , and let Γ ( N ) be the congruence subgroup of SL ( Z ) defined byΓ ( N ) = ( γ ∈ SL ( Z ) | γ ≡ " ∗ (mod N M ( Z )) ) . M ( Z ) means the Z -module of 2 × Z . One can define an equivalencerelation ∼ Γ ( N ) on Q N ( d K ) by Q ∼ Γ ( N ) Q ′ ⇐⇒ Q = Q ′ γ for some γ ∈ Γ ( N ) . (3)Recently, Jung, Koo and Shin showed that if n = N O K , then the mapping φ N : Q N ( d K ) / ∼ Γ ( N ) → C ( n ) , [ Q ] [[ ω Q , Q N ( d K ) / ∼ Γ ( N ) can be regarded as a group isomorphic to C ( n ). Let F , N ( Q ) be the field of meromorphic modu-lar functions for Γ ( N ) with rational Fourier coefficients. By using the Shimura reciprocity law,they further provided the isomorphism Q N ( d K ) / ∼ Γ ( N ) ∼ → Gal( K n /K ) , [ Q ] ( h ( τ K ) h ( − ω Q ) | h ( τ ) ∈ F , N ( Q ) is finite at τ K )where τ K is the CM-point associated with the principal form of discriminant d K .In this paper, for an arbitrary nontrivial ideal n of O K where N is the least positive integer,we shall define an equivalence relation ∼ n on Q N ( d K ) so that the canonical mapping φ n : Q N ( d K ) / ∼ n → C ( n ) (5)[ Q ] [[ ω Q , ∼ n turns outto be induced from a congruence subgroup of level N (Proposition 2.8). Thus we may consider Q N ( d K ) / ∼ n as a group isomorphic to C ( n ) whose group operation will be described in § × × extended form class group Q N ( d K ) / ∼ n onto Gal( K n /K ) in terms of Fricke invariants (Theorem 5.3). To this end, it isnecessary to utilize the concept of canonical bases for nontrivial integral ideals (Proposition 2.3).We hope that this result, as some extension of Gauss’s work, would enrich the classical theoryof complex multiplication for the imaginary quadratic case. Throughout this paper, we let K be an imaginary quadratic field of discriminant d K with ring ofintegers O K . Furthermore, let n be a nontrivial ideal of O K and N be the least positive integerin n . In this section, by using an equivalence relation induced from a congruence subgroup ofSL ( Z ) of level N , we shall construct an extended form class group isomorphic to the ray classgroup C ( n ).First, the following lemma explains why we have to use Q N ( d K ) instead of the whole of Q ( d K ). Lemma . If Q = ax + bxy + cy ∈ Q ( d K ) , then the lattice [ ω Q ,
1] (= Z ω Q + Z ) is afractional ideal of K with N K/ Q ([ ω Q , /a . roof. See [6, Theorem 7.7 (i)] and [8, Lemma 2.3 (iii)].Define an equivalence relation ∼ n on the set Q N ( d K ) by Q ∼ n Q ′ ⇐⇒ [[ ω Q , ω Q ′ , C ( n ) . Note that Q ∼ n Q ′ implies Q ∼ m Q ′ for every nontrivial ideal m of O K dividing n , since there isa canonical homomorphism C ( n ) → C ( m ). Theorem . The set of equivalence classes Q N ( d K ) / ∼ n can be regarded as a group iso-morphic to C ( n ) . Proof.
By the definition of ∼ n , the mapping φ n : Q N ( d K ) / ∼ n → C ( n ) stated in (5) is awell-defined injection. Let ∼ Γ ( N ) be the equivalence relation on Q N ( d K ) given in (3), and let N = N O K . For Q, Q ′ ∈ Q N ( d K ), we deduce that Q ∼ Γ ( N ) Q ′ ⇐⇒ [[ ω Q , ω Q ′ , N )because the mapping φ N described in (4) is bijective ⇐⇒ Q ∼ N Q ′ by the definition of ∼ N . Thus ∼ Γ ( N ) is the same as ∼ N , from which we obtain the following commutative diagram: Q N ( d K ) / ∼ Γ ( N ) ∼ φ N / / (cid:15) (cid:15) (cid:15) (cid:15) C ( N ) canonical (cid:15) (cid:15) (cid:15) (cid:15) Q N ( d K ) / ∼ n φ n / / C ( n )Figure 1: A commutative diagram for surjectivity of φ n It then follows that φ n is surjective, and hence it is bijective. Therefore one can consider Q N ( d K ) / ∼ n as a group isomorphic to C ( n ) via the bijection φ n , as desired.Let Q = x + b K xy + c K y be the principal form in C ( d K ) and τ K = ω Q , namely, Q = x − d K y if d K ≡ ,x + xy + 1 − d K y if d K ≡ τ K = − b K + √ d K √ d K d K ≡ , − √ d K d K ≡ . Then we have O K = [ τ K ,
1] ([6, (5.14)]). 4 roposition . If a is a nontrivial ideal of O K , then there is a unique triple of integers ( a, b, c ) satisfying a = [ aτ K + b, c ] and < a ≤ c, ≤ b < c, a | c, a | b. (6) Here, c is the smallest positive integer belonging to a . Proof.
See [4, Theorem 6.9].
Remark . (i) We call such a Z -basis { aτ K + b, c } the canonical basis for a .(ii) Conversely, if ( a, b, c ) is a triple of integers satisfying (6) and c | N K/ Q ( aτ K + b ), then { ( aτ K + b ) ν + cν | ν , ν ∈ O K } is a nontrivial ideal of O K whose canonical basis is { aτ K + b, c } ([4, Proposition 6.12 andTheorem 6.15]).Let { ξ , ξ } be the canonical basis for n , that is, " ξ ξ = " a a N τ K (7)for some integers a and a satisfying0 < a ≤ N, ≤ a < N, a | N, a | a by Proposition 2.3.For an integer a relatively prime to N , we denote by e a any integer satisfying a e a ≡ N ).Furthermore, for α = " p qr s ∈ GL ( R ) with det( α ) > τ ∈ H , let j ( α, τ ) = rτ + s . Lemma . Let Q = ax + bxy + cy , Q ′ ∈ Q N ( d K ) . Then, Q ∼ n Q ′ if and only if there isa matrix α = " p qr s in SL ( Z ) such that Q = Q ′ α and r ≡ a ) and s ≡ e a (cid:18) a a − b K − b (cid:19) r (mod N ) . (8) Proof.
Assume that Q ∼ n Q ′ . Since Q ∼ O K Q ′ , we induce from the isomorphism given in(1) that Q = Q ′ γ for some γ ∈ SL ( Z ) . (9)And, we obtain by the definition of ∼ n that[ ω Q ,
1] = λ [ ω Q ′ ,
1] for some λ ∈ K ∗ such that λ ≡ ∗ n ) (10)= λ [ γ ( ω Q ) ,
1] by (9)= λj ( γ, ω Q ) [ ω Q , . λ = ζj ( γ, ω Q ) for some ζ ∈ O ∗ K (11)and get by (10) [ ω Q ,
1] = ζj ( γ, ω Q )[ ω Q ′ , . We then attain ζj ( γ, ω Q ) " ω Q ′ = α " ω Q for some α = " p qr s ∈ SL ( Z ) , (12)from which ω Q ′ = ζj ( γ, ω Q ) ω Q ′ ζj ( γ, ω Q ) = α ( ω Q )and rω Q + s = ζj ( γ, ω Q ) = λ ≡ ∗ n )by (11). Since aω Q ∈ O K , we deduce r ( aω Q ) + as − a ∈ n = [ a τ K + a , N ] , and hence r (cid:18) τ K + b K − b (cid:19) + as − a = ( a τ K + a ) e + N f for some e, f ∈ Z . This yields ( r − a e ) τ K + (cid:18) b K − b r + as − a − a e − N f (cid:19) = 0and so r = a e and as = a + a e + N f − b K − b r. Therefore, we achieve Q = Q ′ α and (8).Conversely, assume that there is α = " p qr s ∈ SL ( Z ) satisfying Q = Q ′ α and (8). We seethat aj ( α, ω Q ) − a = r ( aω Q ) + as − a = a e ( aω Q ) + (cid:18) a a − b K − b (cid:19) a e + aN f for some e, f ∈ Z by (8)= ( a τ K + a ) e + N ( af ) ∈ n . This implies by the fact gcd(
N, a ) = 1 that j ( α, ω Q ) ≡ ∗ n ). Since[ ω Q ′ ,
1] = [ α ( ω Q ) ,
1] = 1 j ( α, ω Q ) [ ω Q , , we get [[ ω Q ′ , ω Q , C ( n ), and hence Q ∼ n Q ′ .6 emark . The congruences (8) can be rewritten as h r s i " b K − b ) / a N/a − a /a ≡ h i " b K − b ) / a N/a − a /a (mod N M , ( Z ))where M , ( Z ) is the Z -module of 1 × Z .LetΓ n = (" c c c c ∈ SL ( Z ) | c ≡ N ) , c ≡ Na ) , c ≡ a ) , c ≡ N ) ) which is a congruence subgroup of level N . Let S = " −
11 0 and T = " . Lemma . Let α = " p qr s ∈ SL ( Z ) such that r ≡ a ) and s ≡ kr (mod N ) for some k ∈ Z . (13) Then there is a pair of integers ( m, n ) so that T n αT m ∈ Γ n . Proof.
If we set m = − k , then αT m = " p qr s − k = " p q ′ r s ′ with q ′ = − pk + q and s ′ = − rk + s. Observe by (13) that s ′ ≡ N ) . (14)Furthermore, if we let n be an integer satisfying ns ′ ≡ − q ′ (mod Na ) , (15)then we have T n αT m = " n p q ′ r s ′ = " p + nr q ′ + ns ′ r s ′ with q ′ + ns ′ ≡ Na ) and p + nr ≡ N )by (13), (14), (15) and the fact det( T n αT m ) = 1. This proves the lemma.Now, we shall show that the equivalence relation ∼ n on Q N ( d K ) is essentially induced fromthe congruence subgroup Γ n . Proposition . Let Q = ax + bxy + cy , Q ′ ∈ Q N ( d K ) . Then, Q ∼ n Q ′ if and only if Q = Q ′ T u γT v for some γ = " c c c c ∈ Γ n and ( u, v ) ∈ Z satisfying c v + c ≡ e a (cid:18) a a − b K − b (cid:19) c (mod N ) . roof. Assume that Q ∼ n Q ′ . By Lemma 2.5, there is α = " p qr s ∈ SL ( Z ) such that Q = Q ′ α and r ≡ a ) and s ≡ e a (cid:18) a a − b K − b (cid:19) r (mod N ) . (16)It then follows from Lemma 2.7 that there is ( m, n ) ∈ Z with T n αT m ∈ Γ n . If we set γ = T n αT m , u = − n and v = − m , then we establish Q = Q ′ α = Q ′ T u γT v . And, if we let γ = " c c c c , then we see that α = " p qr s = T u γT v = " u c c c c v = " c + uc ( c + uc ) v + c + uc c c v + c . Thus we attain r = c , and so c v + c = s ≡ e a (cid:18) a a − b K − b (cid:19) c (mod N )by (16).Conversely, assume that there are γ = " c c c c ∈ Γ n and ( u, v ) ∈ Z such that Q = Q ′ T u γT v and c v + c ≡ e a (cid:18) a a − b K − b (cid:19) c (mod N ) . Since γ ∈ Γ n and T u γT v = " c + uc ( c + uc ) v + c + uc c c v + c , we get c ≡ a ). Therefore, we derive by Lemma 2.5 that Q ∼ n Q ′ . Remark . In particular, if n = N O K , then we obtain a = N , a = 0 and so Γ n = Γ ( N ).Since T ∈ Γ ( N ), we conclude by Proposition 2.8 that ∼ n is the same as ∼ Γ ( N ) given in (3),which recovers the previous results of [8] and [14]. Now, we shall modify the classical Gauss-Dirichlet composition in order to achieve the composi-tion law on the extended form class group Q N ( d K ) / ∼ n . Although this composition law can beexplained directly from Figure 1 and [8, Remark 2.10] for the case where n = N O K , we wouldlike to include this section for completeness.Let Q = ax + bxy + cy , Q ′ = a ′ x + b ′ xy + c ′ y ∈ Q N ( d K ). By Theorem 2.2 we must have[ Q ][ Q ′ ] in Q N ( d K ) / ∼ n = φ − n (cid:0) φ n ([ Q ]) φ n ([ Q ′ ]) (cid:1) = φ − n (cid:0) [[ ω Q , ω Q ′ , (cid:1) . a = [ ω Q , ω Q ′ , a − = 1N K/ Q ( a ) a = aa ′ a = [ − aω Q , a ][ − a ′ ω Q ′ , a ′ ] , which shows that a − is an integral ideal in the ray class ( φ n ([ Q ]) φ n ([ Q ′ ])) − .Now, one can take a matrix γ in SL ( Z ) so that the new quadratic form Q ′′ = a ′′ x + b ′′ xy + c ′′ y = Q ′ γ (17)satisfies gcd( a, a ′′ , ( b + b ′′ ) /
2) = 1 ([6, Lemmas 2.3 and 2.25]). We then obtain[ ω Q , ω Q ′′ ,
1] = (cid:20) − B + √ d K aa ′′ , (cid:21) (18)where B is a unique integer modulo 2 aa ′′ satisfying B ≡ b (mod 2 a ) , B ≡ b ′′ (mod 2 a ′′ ) and B ≡ d K (mod 4 aa ′′ )([6, Lemma 3.2 and (7.13)]). Furthermore, we know by (17) and (18) that a = [ ω Q , γ ( ω Q ′′ ) ,
1] = 1 j ( γ, ω Q ′′ ) [ ω Q , ω Q ′′ ,
1] = 1 j ( γ, ω Q ′′ ) (cid:20) − B + √ d K aa ′′ , (cid:21) . (19)Let ν , ν ∈ K ∗ such that a = [ ν , ν ] and ν = ν ν ∈ H . By (19) we may just take ν = − B + √ d K aa ′′ j ( γ, ω Q ′′ ) , ν = 1 j ( γ, ω Q ′′ ) and ν = − B + √ d K aa ′′ . Since a − is an integral ideal of K , we see 1 ∈ a and so1 = uν + vν for some u, v ∈ Z . (20)Here, one can readily check gcd( N, u, v ) = 1 because a − is relatively prime to n . Take a matrix σ = " ∗ ∗ u ′ v ′ in SL ( Z ) such that σ ≡ " ∗ ∗ u v (mod N M ( Z )) , (21)which is possible by the surjectivity of the reduction SL ( Z ) → SL ( Z /N Z ) ([20, Lemma 1.38]).If we set e ω = σ ( ν ), then we deduce that[ e ω,
1] = [ σ ( ν ) ,
1] = 1 u ′ ν + v ′ [ ν,
1] = 1 u ′ ν + v ′ ν [ ν , ν ] = 1 u ′ ν + v ′ ν a . (22)Observe by (20) and (21) that u ′ ν + v ′ ν − u ′ ν + v ′ ν − ( uν + vν ) = ( u ′ − u ) ν + ( v ′ − v ) ν ∈ N a ⊆ na , u ′ ν + v ′ ν ≡ ∗ n ) . This implies by (22) [[ e ω, a ] in C ( n ) . Finally, if we let e Q be the quadratic form in Q N ( d K ) satisfying ω e Q = e ω = σ ( ν ), then weattain [ Q ][ Q ′ ] = [ e Q ] . More precisely, we have [ Q ][ Q ′ ] = "(cid:18) aa ′′ x + Bxy + B − d K aa ′′ y (cid:19) σ − . From now on, we further assume that n is a proper ideal of O K and so N ≥
2. In this section,we shall show that K n is in fact a specialization over K of a certain modular function field byutilizing Fricke invariants.For a lattice Λ in C , let g (Λ) = 60 X λ ∈ Λ −{ } λ , g (Λ) = 140 X λ ∈ Λ −{ } λ , ∆(Λ) = g (Λ) − g (Λ) (23)and j (Λ) = 1728 g (Λ) ∆(Λ) . (24)And, let ℘ ( z ; Λ) be the Weierstrass ℘ -function relative to Λ given by ℘ ( z ; Λ) = 1 z + X λ ∈ Λ −{ } (cid:26) z − λ ) − λ (cid:27) ( z ∈ C ) . (25)For a fractional ideal a of K , the Weber function h : C / a → P ( C ) is defined by h ( z ; a ) = g ( a ) ∆( a ) ℘ ( z ; a ) if K = Q ( √− ,g ( a )∆( a ) ℘ ( z ; a ) if K = Q ( √− ,g ( a ) g ( a )∆( a ) ℘ ( z ; a ) otherwise . Then, it follows from (23) and (25) that h ( νz ; ν a ) = h ( z ; a ) for any ν ∈ K ∗ . (26)As a consequence of the main theorem of complex multiplication, we get the following resultdue to Hasse ([11]). 10 roposition . If κ is a generator of the O K -module n − / O K , then K n = H K ( h ( κ ; O K )) = K ( j ( O K ) , h ( κ ; O K )) . Proof.
See [16, Corollary to Theorem 7 in Chapter 10].Let M , ( Q ) be the set of 1 × Q . For i = 1 , , v = h v v i ∈ M , ( Q ) − M , ( Z ), the i th Fricke function f ( i ) v ( τ ) is defined on H by f ( i ) v ( τ ) = g (Λ τ ) g (Λ τ )∆(Λ τ ) ℘ ( v τ + v ; Λ τ ) if i = 1 ,g (Λ τ ) ∆(Λ τ ) ℘ ( v τ + v ; Λ τ ) if i = 2 ,g (Λ τ )∆(Λ τ ) ℘ ( v τ + v ; Λ τ ) if i = 3where Λ τ = [ τ, f (2) v ( τ ) = 120736 f (1) v ( τ ) j ( τ ) − f (3) v ( τ ) = 1373248 f (1) v ( τ ) j ( τ )( j ( τ ) − . (27)Let F = Q ( j ( τ )) and F N = Q (cid:18) j ( τ ) , f ( i ) v ( τ ) | i = 1 , , v ∈ N M , ( Z ) − M , ( Z ) (cid:19) where j ( τ ) = j (Λ τ ). Then it is well known that F N coincides with the field of meromorphicmodular functions of level N whose Fourier coefficients belong to the N th cyclotomic field ([20,Proposition 6.9 (i)]). Proposition . (i) The field F N is a Galois extension of F whose Galois group isisomorphic to GL ( Z /N Z ) / {± I } and satisfies f ( i ) v ( τ ) γ = f ( i ) v γ ( τ ) ( i = 1 , , and γ ∈ GL ( Z /N Z ) / {± I } ) . (28)(ii) If f ( τ ) ∈ F N and γ ∈ SL ( Z /N Z ) / {± I } , then f ( τ ) γ = f ( e γ ( τ )) where e γ is any element of SL ( Z ) which maps to γ through the reduction SL ( Z ) → SL ( Z /N Z ) / {± I } . Proof. (i) See [20, Theorem 6.6].(ii) See [20, (6.1.3)].Now, consider the index set V N = n v = h v v i ∈ M , ( Q ) | N is the smallest positive integer so that N v ∈ M , ( Z ) o . Kubert and Lang first defined in [15, § Fricke family of level N to be a family { h v ( τ ) } v ∈ V N of functions in F N satisfying 11F1) h u ( τ ) = h v ( τ ) if u ≡ ± v (mod M , ( Z )),(F2) h v ( τ ) γ = h v γ for all γ ∈ GL ( Z /N Z ) / {± I } ≃ Gal( F N / F ).For example, for each i = 1 , ,
3, the family { f ( i ) v ( τ ) } v ∈ V N is a Fricke family of level N byProposition 4.2 (i). If we let Fr( N ) be the set of all Fricke families of level N , then it is naturalfor us to regard it as a field under the operations { h v ( τ ) } v + { k v ( τ ) } v = { ( h v + k v )( τ ) } v , { h v ( τ ) } v · { k v ( τ ) } v = { ( h v k v )( τ ) } v . And, let F N ( Q ) be the field of meromorphic modular functions for the congruence subgroupΓ ( N ) = ( γ ∈ SL ( Z ) | γ ≡ " ∗ (mod N M ( Z )) ) with rational Fourier coefficients. Then it was shown by Eum and Shin that F N ( Q ) = Q (cid:16) j ( τ ) , f ( i )[ 1 /N ( τ ) | i = 1 , , (cid:17) and Fr( N ) is indeed isomorphic to F N ( Q ) through the mapFr( N ) ∼ → F N ( Q ) , { h v ( τ ) } v h [ 1 /N ( τ ) (29)([9, Theorem 4.3 and Proposition 6.1] and (27)).Let { h v ( τ ) } v ∈ Fr( N ) and C ∈ C ( n ). Take any integral ideal c in the class C , and let ω , ω ∈ K ∗ such that nc − = [ ω , ω ] and ω = ω ω ∈ H . Since N ∈ nc − , we get N = rω + sω for some r, s ∈ Z . Now, we define the
Fricke invariant h ( C ) by h ( C ) = h [ r/N s/N ] ( ω )if it is finite. Proposition . With the notations as above, h ( C ) depends only on the class C , not onthe choice of c , ω and ω . It belongs to K n and satisfies h ( C ) σ n ( C ′ ) = h ( CC ′ ) ( C ′ ∈ C ( n )) , where σ n is the Artin reciprocity map for modulus n . Proof.
See [15, Theorem 1.1 in Chapter 11].
Remark . (i) The Fricke invariant h ( C ) is a generalization of the Siegel-Ramachandrainvariant ([19] and [21]). 12ii) When C is the identity class of C ( n ), one can take c = O K and so nc − = n = [ ξ , ξ ] and N = 0 · ξ + 1 · ξ . Thus we have h ( C ) = h [ 0 1 /N ] ( ξ ) where ξ = ξ ξ = h [ 1 /N ( τ ) S | τ = ξ by (F1) and (F2) with S = " −
11 0 = h [ 1 /N ( S ( ξ )) by Proposition 4.2 (ii)= h [ 1 /N ( − /ξ ) . (iii) In particular, consider the Fricke family { f ( i ) v ( τ ) } v ∈ V N for i = 1 , , f ( i ) ( C ) = f ( i )[ 0 1 /N ] ( ξ ) by (ii)= g ([ ξ , ξ ]) g ([ ξ , ξ ])∆([ ξ , ξ ]) ℘ (cid:18) ξ N ; [ ξ , ξ ] (cid:19) if i = 1 ,g ([ ξ , ξ ]) ∆([ ξ , ξ ]) ℘ (cid:18) ξ N ; [ ξ , ξ ] (cid:19) if i = 2 ,g ([ ξ , ξ ])∆([ ξ , ξ ]) ℘ (cid:18) ξ N ; [ ξ , ξ ] (cid:19) if i = 3 , by (26)= h (1; n ) becase ξ = N. Proposition . If i = |O ∗ K | / , then K n is generated by f ( i ) ( C ) over H K for any C ∈ C ( n ) . Proof.
For the case where K is different from Q ( √−
1) and Q ( √− K = Q ( √−
1) or Q ( √− K is 1, we have n = ν O K forsome ν ∈ K ∗ and so n − = ν − O K . Thus we obtain by Proposition 4.1, (26) and Remark 4.4(iii) that K n = H K (cid:0) h ( ν − ; O K ) (cid:1) = H K ( h (1; ν O K )) = H K ( h (1; n )) = H K (cid:16) f ( i ) ( C ) (cid:17) . Since K n is an abelian extension of K , we further achieve K n = H K (cid:16) f ( i ) ( C ) σ n ( C ) (cid:17) = H K (cid:16) f ( i ) ( C C ) (cid:17) by Proposition 4.3= H K (cid:16) f ( i ) ( C ) (cid:17) . Corollary . We have K n = K (cid:0) h ( − /ξ ) | h ( τ ) ∈ F N ( Q ) is finite at − /ξ (cid:1) with ξ = ξ ξ = a τ K + a N . roof. Since { j ( τ ) } v ∈ V N is also a Fricke family of level N and j ( C ) = j ( O K ), we get byPropositions 4.3 and 4.5 that K n = K ( h ( C ) | { h v ( τ ) } v ∈ Fr( N ) such that h ( C ) is finite) . (30)We then deduce that K n = K (cid:0) h [ 1 /N ( − /ξ ) | { h v ( τ ) } v ∈ Fr( N ) with h [ 1 /N ( τ ) finite at − /ξ (cid:1) by Remark 4.4 (ii)= K (cid:0) h ( − /ξ ) | h ( τ ) ∈ F N ( Q ) is finite at − /ξ (cid:1) by the isomorphism given in (29) . By the class field theory we know that C ( n ) is isomorphic to Gal( K n /K ) through the Artinmap for modulus n . Thus, the extended form class group Q N ( d K ) / ∼ n is also isomorphic toGal( K n /K ). In this section, we shall present an explicit isomorphism of Q N ( d K ) / ∼ n ontoGal( K n /K ) in terms of Fricke invariants.For each Q = ax + bxy + cy ∈ Q N ( d K ) there is a unique integer d Q satisfying d Q ≡ − a (cid:18) b + b K (cid:19) + a (mod N ) , d Q ≡ a ) , ≤ d Q + a (cid:18) b + b K (cid:19) < N a (31)by the fact gcd( N, a ) = 1 and the Chinese remainder theorem.
Lemma . Let a be a nontrivial ideal of O K . Let { ν , ν } be a Z -basis for a such that ν /ν ∈ H , and so " ν ν = A " τ K for some A ∈ M ( Z ) such that det( A ) > . (32) Then we have det( A ) = N K/ Q ( a ) . Proof.
We derive from (32) " ν ν ν ν = A " τ K τ K . By taking determinant and squaring, we getdisc K/ Q ( a ) = det( A ) d K . Moreover, since disc K/ Q ( a ) = N K/ Q ( a ) d K ([17, Proposition 13 in Chapter III]) and det( A ) > A ) = N K/ Q ( a ). Lemma . If Q = ax + bxy + cy ∈ Q N ( d K ) and m = [ − aω Q , a ] , then we have nm = [ ξ ′ , ξ ′ ] where " ξ ′ ξ ′ = " a d Q N a − aω Q . roof. The smallest positive integer in m = [ − aω Q , a ] is a . Note by the fact gcd( N, a ) = 1and Lemma 2.1 that the ideals n and m are relatively prime. Hence the smallest positive integerin nm = n ∩ m is lcm( N, a ) =
N a . Now, if { ξ ′ , ξ ′ } is the canonical basis for nm , then " ξ ′ ξ ′ = B " τ K with B = " b b N a (33)for some unique pair of integers ( b , b ) satisfying0 < b ≤ N a, ≤ b < N a, b | N a, b | b . Since det( B ) = N K/ Q ( nm ) by (33) and Lemma 5.1= N K/ Q ( n )N K/ Q ( m )= ( a N ) a by (7), Lemmas 2.1 and 5.1 , we have b = a . On the other hand, we attain n = [ a τ K + a , N ] = (cid:20) a ( − aω Q ) − a (cid:18) b + b K (cid:19) + a , N (cid:21) . (34)Observe that nm = n ∩ m ∋ ξ ′ − ( a ( − aω Q ) + d Q ) because a ( − aω Q ) + d Q ∈ n ∩ m by (31) and (34)= a τ K + b − ( a ( − aω Q ) + d Q ) by (33) and the fact b = a = a (cid:18) − aω Q − b K + b (cid:19) + b − ( a ( − aω Q ) + d Q )= b − d Q − a (cid:18) b K + b (cid:19) . Here, since 0 ≤ b , d Q + a ( b + b K ) / < N a and N a is the smallest positive integer in nm , wemust have b = d Q + a ( b + b K ) /
2. Thus we obtain by (33) that " ξ ′ ξ ′ = " a d Q + a ( b K + b ) / N a τ K = " a d Q N a τ K + ( b K + b ) / = " a d Q N a − aω Q . Theorem . We have an isomorphism Q N ( d K ) / ∼ n → Gal( K n /K )[ Q ] = [ ax + bxy + cy ] h ( − /ξ ) h e aS " a d Q /a N ( − ω Q ) ! | h ( τ ) ∈ F N ( Q ) is finite at − /ξ ! . Proof.
Let Q = ax + bxy + cy ∈ Q N ( d K ) and C = [[ ω Q , C ( n ). Since a ϕ ( N ) ≡ N ), where ϕ is the Euler totient function, one can take c = a ϕ ( N ) [ ω Q ,
1] as an integralideal in C . We get by Lemma 2.1 that cc = N K/ Q ( c ) O K = a ϕ ( N ) N K/ Q ([ ω Q , O K = a ϕ ( N ) − O K , c − = c a − ϕ ( N )+1 O K = a − ϕ ( N )+1 [ − ω Q , . Thus we establish by Lemma 5.2 that nc − = a − ϕ ( N ) [ a τ K + a , N ][ − aω Q , a ] = a − ϕ ( N ) [ a ( − aω Q ) + d Q , N a ] (35)and N = 0 · { a − ϕ ( N ) ( a ( − aω Q ) + d Q ) } + a ϕ ( N ) − ( a − ϕ ( N ) N a ) . (36)Now, let h ( τ ) be an element of F N ( Q ) which is finite at − /ξ , and let { h v ( τ ) } v be the Frickefamily of level N induced from h ( τ ) via the isomorphism mentioned in (29). Then we achievethat h ( − /ξ ) σ n ( C ) = h ( C ) σ n ( C ) by Remark 4.4 (ii)= h ( C C ) by Proposition 4.3= h ( C )= h [ a ϕ ( N ) − /N ] (cid:18) a ( − aω Q ) + d Q N a (cid:19) by (35) and (36)= h [ 0 e a/N ] " a a d Q N a ( − ω Q ) ! by (F1)= h [ 1 /N e aS " a d Q /a N ( − ω Q ) ! since " a a d Q N a and " a d Q /a N yield the same fractional linear transformation on H = h e aS [ 1 /N " a d Q /a N ( − ω Q ) ! by Proposition 4.2 (i)= h e aS " a d Q /a N ( − ω Q ) ! due to the isomorphism in (29) . This, together with Corollary 4.6, completes the proof.
Remark . In particular, if n = N O K , then we derive that h e aS " a d Q /a N ( − ω Q ) ! = h e aS " N N ( − ω Q ) ! = h e aS ( − ω Q ) . In this last section, we shall explain how to find elements of Q N ( d K ) / ∼ n and give a couple ofconcrete examples.For each Q = ax + bxy + cy ∈ Q N ( d K ), let V Q = nh u v i ∈ M , ( Z ) | ( uω Q + v ) O K ∈ P K ( N O K ) o . Then, it is straightforward to show V Q = nh u v i ∈ M , ( Z ) | gcd( N, Q ( v, − u )) = 1 o U K = { ( m, n ) ∈ Z | mτ K + n ∈ O ∗ K } = {± (0 , } if K = Q ( √− , Q ( √− , {± (0 , , ± (1 , } if K = Q ( √− , {± (0 , , ± (1 , , ± (1 , } if K = Q ( √− Lemma . Let Q = ax + bxy + cy ∈ Q N ( d K ) and h r s i , h u v i ∈ V Q . Then, [( rω Q + s ) O K ] = [( uω Q + v ) O K ] in C ( n ) if and only if h r s i " b K − b ) / a N/a − a /a ≡ h u v i " b K − b ) / a − mb K + n − mc K m n N/a − a /a (mod N M , ( Z )) for some ( m, n ) ∈ U K . (37) Proof.
We deduce that[( rω Q + s ) O K ] = [( uω Q + v ) O K ] in C ( n ) ⇐⇒ rω Q + suω Q + v O K ∈ P K, ( n ) ⇐⇒ rω Q + suω Q + v ≡ ∗ ζ (mod n ) for some ζ ∈ O ∗ K ⇐⇒ r ( aω Q ) + as ≡ ζ { u ( aω Q ) + av } (mod n ) since gcd( N, a ) = 1 and aω Q ∈ O K ⇐⇒ r (cid:18) τ K + b K − b (cid:19) + as − ( mτ K + n ) (cid:26) u (cid:18) τ K + b K − b (cid:19) + av (cid:27) ∈ n for some ( m, n ) ∈ U K ⇐⇒ a (cid:18) r + mu b K + b − mav − nu (cid:19) ξ + 1 N (cid:26)(cid:18) − a a (cid:19) (cid:18) r + mu b K + b − mav − nu (cid:19) + r b K − b as + muc K − nu b K − b − nav (cid:27) ξ ∈ n = [ ξ , ξ ]because τ K + b K τ K + c K = 0 and " τ K = " a a N − " ξ ξ ⇐⇒ r ≡ (cid:18) − m b K + b n (cid:19) u + mav (mod a ) , (cid:18) − a a + b K − b (cid:19) r + as ≡ (cid:26) a a (cid:18) m b K + b − n (cid:19) − mc K + n b K − b (cid:27) u + (cid:18) − ma a a + na (cid:19) v (mod N ) . This proves the lemma.Now, define an equivalence relation ∼ Q on the set V Q as follows: Let h r s i , h u v i ∈ V Q .Then, h r s i ∼ Q h u v i if and only if they satisfy the congruence relation stated in (37).17 roposition . There is an algorithm to find all elements of the extended form classgroup Q N ( d K ) / ∼ n . Proof.
Let Q , Q , . . . , Q h be all of the reduced forms in Q ( d K ). One can take α i ∈ SL ( Z )so that Q ′ i = Q α i i belongs to Q N ( d K ) ( i = 1 , , . . . , h ) ([6, Lemmas 2.3 and 2.25]). Observe bythe isomorphism given in (1) that C ( O K ) = n [[ ω Q ′ , , [[ ω Q ′ , , . . . , [[ ω Q ′ h , o . (38)On the other hand, since the canonical homomorphism C ( N O K ) → C ( n ) is surjective, we have P K ( n ) /P K, ( n ) ≃ P K ( N O K ) / ( P K ( N O K ) ∩ P K, ( n )) , from which it follows by Lemma 6.1 that P K ( n ) /P K, ( n ) = n [( uω Q ′ i + v ) O K ] | hh u v ii ∈ V Q ′ i / ∼ Q ′ i o for each i = 1 , , . . . , h. (39)Now that the canonical homomorphism π n : C ( n ) → C ( O K ) is a surjection with Ker( π n ) = P K ( n ) /P K, ( n ), we see by (38) and (39) that C ( n ) = (" uω Q ′ i + v [ ω Q ′ i , | i = 1 , , . . . , h and hh u v ii ∈ V Q ′ i / ∼ Q ′ i ) = nhh γ i, [[ u v ]] ( ω Q ′ i ) , ii | i = 1 , , . . . , h and hh u v ii ∈ V Q ′ i / ∼ Q ′ i o where γ i, [[ u v ]] is an element of SL ( Z ) satisfying γ i, [[ u v ]] ≡ " ∗ ∗ u v (mod N M ( Z ))= ("" ω Q ′ γ − i, [[ u v ]] i , | i = 1 , , . . . , h and hh u v ii ∈ V Q ′ i / ∼ Q ′ i ) . Therefore we attain Q N ( d K ) / ∼ n = (cid:26)(cid:20) Q ′ γ − i, [[ u v ]] i (cid:21) | i = 1 , , . . . , h and hh u v ii ∈ V Q ′ i / ∼ Q ′ i (cid:27) . By using Remark 2.4, Lemmas 2.5, 6.1 and Proposition 6.2, we present the following exam-ples.
Example . Let K = Q ( √−
5) with d K = −
20, and let n = [2 τ K + 4 ,
6] with N = 6. Notethat since 2 is ramified in K and 3 splits in K , n has prime ideal factorization n = p p , where p = [ τ K + 1 ,
2] is the prime ideal of K lying above 2 and p = [ τ K + 2 ,
3] is a prime ideal of K lying above 3. We know that there are two reduced forms of discriminant − Q = x + 5 y and Q = 2 x + 2 xy + 3 y . Take Q ′ = Q and Q ′ = Q h −
11 0 i = 7 x − xy + 2 y . V Q ′ / ∼ Q ′ = nhh ii , hh iio and V Q ′ / ∼ Q ′ = nhh ii , hh iio and γ , [[ 0 1 ]] = " , γ , [[ 1 0 ]] = " −
11 0 , γ , [[ 0 1 ]] = " , γ , [[ 1 3 ]] = " . Thus we obtain Q ( − / ∼ n = (cid:26) X = (cid:20) Q ′ γ − , [[ ]] (cid:21) = [ x + 5 y ] , X = (cid:20) Q ′ γ − , [[ ]] (cid:21) = [5 x + y ] ,X = (cid:20) Q ′ γ − , [[ ]] (cid:21) = [7 x − xy + 2 y ] , X = (cid:20) Q ′ γ − , [[ ]] (cid:21) = [83 x − xy + 42 y ] (cid:27) with the following group table: X X X X X X X X X X X X X X X X X X X X X X X X Table 1: Group table of Q ( − / ∼ [2 τ K +4 , Example . Let K = Q ( √−
23) with d K = −
23, and let n = [3 τ K + 9 ,
12] with N = 12.There are three reduced forms Q = x + xy + 6 y , Q = 2 x − xy + 3 y , Q = 2 x + xy + 3 y . If we take Q ′ = Q , Q ′ = Q h − − i = 29 x − xy + 4 y , Q ′ = Q h − − i = 41 x − xy + 6 y , then we get V Q ′ i / ∼ Q ′ i = nhh ii , hh ii , hh ii , hh iio ( i = 1 , ,V Q ′ / ∼ Q ′ = nhh ii , hh ii , hh ii , hh iio and γ i, [[ 0 1 ]] = " , γ i, [[ 0 5 ]] = " , γ i, [[ 2 1 ]] = " − −
12 1 , γ i, [[ 2 7 ]] = " ( i = 1 , ,γ , [[ 0 1 ]] = " , γ , [[ 0 5 ]] = " , γ , [[ 2 3 ]] = " , γ , [[ 2 9 ]] = " . Q ( − / ∼ [3 τ K +9 , = (cid:26) Y , , = (cid:20) Q ′ γ − , [[ ]] (cid:21) = [ x + xy + 6 y ] , Y , , = (cid:20) Q ′ γ − , [[ ]] (cid:21) = [829 x − xy + 144 y ] ,Y , , = (cid:20) Q ′ γ − , [[ ]] (cid:21) = [23 x + 23 xy + 6 y ] , Y , , = (cid:20) Q ′ γ − , [[ ]] (cid:21) = [59 x − xy + 12 y ] ,Y , , = (cid:20) Q ′ γ − , [[ ]] (cid:21) = [29 x − xy + 4 y ] , Y , , = (cid:20) Q ′ γ − , [[ ]] (cid:21) = [2561 x − xy + 426 y ] ,Y , , = (cid:20) Q ′ γ − , [[ ]] (cid:21) = [403 x − xy + 54 y ] , Y , , = (cid:20) Q ′ γ − , [[ ]] (cid:21) = [2743 x − xy + 552 y ] ,Y , , = (cid:20) Q ′ γ − , [[ ]] (cid:21) = [41 x − xy + 6 y ] , Y , , = (cid:20) Q ′ γ − , [[ ]] (cid:21) = [3749 x − xy + 624 y ] ,Y , , = (cid:20) Q ′ γ − , [[ ]] (cid:21) = [127 x + 199 xy + 78 y ] , Y , , = (cid:20) Q ′ γ − , [[ ]] (cid:21) = [2467 x − xy + 468 y ] (cid:27) which is isomorphic to Z × Z × Z via the mapping Q ( − / ∼ [3 τ K +9 , → Z × Z × Z , Y a, b, c ( a, b, c ) . References [1] M. Bhargava,
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Department of Mathematics EducationDongguk University-GyeongjuGyeongju-si, Gyeongsangbuk-do 38066Republic of Korea
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Applied Algebra and Optimization Re-search CenterSungkyunkwan UniversitySuwon-si, Gyeonggi-do 16419Republic of Korea
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Department of Mathematical SciencesKAISTDaejeon 34141Republic of Korea
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Department of MathematicsHankuk University of Foreign StudiesYongin-si, Gyeonggi-do 17035Republic of Korea
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