# Class fields generated by coordinates of elliptic curves

• Ho Yun Jung , Ja Kyung Koo and Dong Hwa Shin
From the journal Open Mathematics

## Abstract

Let K be an imaginary quadratic field different from Q ( 1 ) and Q ( 3 ) . For a nontrivial integral ideal m of K , let K m be the ray class field modulo m . By using some inequalities on special values of modular functions, we show that a single x -coordinate of a certain elliptic curve generates K m over K .

MSC 2010: 11F03; 11G15; 11R37

## 1 Introduction

Let K be an imaginary quadratic field with ring of integers O K . Let E be the elliptic curve with complex multiplication by O K given by the Weierstrass equation:

E : y 2 = 4 x 3 g 2 x g 3 with g 2 = g 2 ( O K ) and g 3 = g 3 ( O K ) .

For z C , let [ z ] denote the coset z + O K in C / O K . Then the map

φ K : C / O K E ( C ) ( P 2 ( C ) ) [ z ] [ ( z ; O K ) : ( z ; O K ) : 1 ] ,

where is the Weierstrass -function relative to O K , is a complex analytic isomorphism of complex Lie groups ([1, Proposition 3.6 in Chapter VI]). Corresponding to E , we consider the Weber function h K : E ( C ) P 1 ( C ) given by

h K ( x , y ) = g 2 g 3 Δ x if j ( E ) 0 , 1,728 , g 2 2 Δ x 2 if j ( E ) = 1,728 , g 3 Δ x 3 if j ( E ) = 0 ,

where Δ = g 2 3 27 g 3 2 ( 0 ) and j ( E ) = j ( O K ) is the j -invariant of E . For a nontrivial ideal m of O K , by K m we mean the ray class field of K modulo m . In particular, K O K is the Hilbert class field H K of K . Then we obtain by the theory of complex multiplication that H K = K ( j ( E ) ) and

K m = H K ( h K ( x , y ) ) for some m -torsion point ( x , y ) on E

if m is proper [2, Chapter 10]. In a letter to Hecke concerning Kronecker’s Jugendtraum (= Hilbert 12th problem), Hasse asked whether every abelian extension of K can be generated only by a single value of the Weber function h K over K [3, p. 91] and Sugawara first gave partial answers to this question [4,5]. Recently, Jung et al. [6] proved that if m = N O K and N { 2 , 3 , 4 , 6 } , then

(1) K m = K h K φ K 1 N or K m = K h K φ K 2 N .

Koo et al. [7] further showed by utilizing the second Kronecker’s limit formula that (1) holds for N = 5 and N 7 . Besides, it is worth noting that Ramachandra [8] constructed a complicated primitive generator of K m over K by using special values of the product of high powers of the discriminant Δ function and Siegel functions, which is beautiful in theory.

Now, we assume that K is different from Q ( 1 ) and Q ( 3 ) , and so g 2 g 3 0 and j ( O K ) 0 , 1,728 ([9, p. 200]). Let { E K , n } n Z 0 be the family of elliptic curves isomorphic to E given by the affine models

(2) E K , n : y 2 = 4 x 3 J K ( J K 1 ) 27 K 2 n x J K ( J K 1 ) 2 2 7 2 K 3 n ,

where

J K = 1 1,728 j ( O K ) and K = J K 2 ( J K 1 ) 3 .

Then for each n Z 0 we have a parametrization

C / O K E K , n ( C ) ( P 2 ( C ) ) [ z ] [ x K , n ( z ) : y K , n ( z ) : 1 ] ,

with

x K , n ( z ) = K n g 2 g 3 Δ ( z ; O K ) and y K , n ( z ) = K n g 2 g 3 Δ 3 ( z ; O K ) .

Here we note that

(3) x K , n ( z ) = K n h K ( φ K ( [ z ] ) ) ( z C ) .

Let m be a proper nontrivial ideal of O K in such a way that K m properly contains H K . Let ω be an element of K so that [ ω ] = ω + O K is a generator of the O K -module m 1 O K / O K . In this article, we shall prove the following three assertions (Theorems 4.4, 5.2, and 6.4):

1. We have K m = K ( x K , n ( ω ) , y K , n ( ω ) 2 ) for every n Z 0 .

2. We obtain K m = K ( x K , n ( ω ) ) for every n Z 0 satisfying

n 13 24 π d K + 6 ln 229 76 N m 5 2 π d K ln 877,383 1 6 ,

where d K is the discriminant of K and N m is the least positive integer in m .

3. If m = N O K for an integer N ( 2 ) whose prime factors are all inert in K , then K m = K ( x K , n ( ω ) ) for every n Z 0 .

To this end, we shall make use of some inequalities on special values of the elliptic modular function and Siegel functions (Lemmas 4.1 and 5.1), rather than using L -function arguments adopted in [7,8,10].

Finally, we hope to utilize the aforementioned results (i), (ii), and (iii) to investigate the images of (higher dimensional) Galois representations attached to elliptic curves with complex multiplication.

## 2 Fricke and Siegel functions

In this preliminary section, we recall the definitions and basic properties of Fricke and Siegel functions.

Let H be the complex upper half-plane, that is, H = { τ C Im ( τ ) > 0 } . Let j be the elliptic modular function on H given by

j ( τ ) = j ( [ τ , 1 ] ) ( τ H ) ,

where [ τ , 1 ] stands for the lattice Z τ + Z in C and j ( [ τ , 1 ] ) is the j -invariant of an elliptic curve isomorphic to C / [ τ , 1 ] . Define the function J on H by

J ( τ ) = 1 1,728 j ( τ ) ( τ H ) .

Furthermore, for v = v 1 v 2 M 1 , 2 ( Q ) M 1 , 2 ( Z ) we define the Fricke function f v on H by

f v ( τ ) = 2 7 3 5 g 2 ( τ ) g 3 ( τ ) Δ ( τ ) ( v 1 τ + v 2 ; [ τ , 1 ] ) ( τ H ) ,

where g 2 ( τ ) = g 2 ( [ τ , 1 ] ) , g 3 ( τ ) = g 3 ( [ τ , 1 ] ) , and Δ ( τ ) = Δ ( [ τ , 1 ] ) . Note that for u , v M 1 , 2 ( Q ) M 1 , 2 ( Z )

(4) f u = f v u v or v ( mod M 1 , 2 ( Z ) )

([9, Lemma 10.4]). For a positive integer N , let N be the field given by

N = Q ( j ) if N = 1 , 1 f v v 1 N M 1 , 2 ( Z ) M 1 , 2 ( Z ) if N 2 .

Then, N is a Galois extension of 1 whose Galois group is isomorphic to GL 2 ( Z / N Z ) / I 2 ([11, Theorem 6.6]). It coincides with the field of meromorphic modular functions of level N whose Fourier coefficients belong to the N th cyclotomic field ([11, Proposition 6.9]).

## Proposition 2.1

If N 2 , v 1 N M 1 , 2 ( Z ) M 1 , 2 ( Z ) , and γ GL 2 ( Z / N Z ) / I 2 , then

f v γ = f v γ .

Moreover, if γ SL 2 ( Z / N Z ) / I 2 , then

f v γ = f v α ,

where α is any element of SL 2 ( Z ) (acting on H as fractional linear transformation) whose image in SL 2 ( Z / N Z ) / I 2 is γ .

## Proof

See [2, Theorem 3 in Chapter 6] or [11, Theorem 6.6].□

For v = v 1 v 2 M 1 , 2 ( Q ) M 1 , 2 ( Z ) , the Siegel function g v on H is given by the infinite product expansion

(5) g v ( τ ) = e π i v 2 ( v 1 1 ) q τ 1 2 ( v 1 2 v 1 + 1 6 ) ( 1 q z ) n = 1 ( 1 q τ n q z ) ( 1 q τ n q z 1 ) ( τ H ) ,

where q τ = e 2 π i τ and q z = e 2 π i z with z = v 1 τ + v 2 . Observe that g v has neither a zero nor a pole on H .

## Proposition 2.2

Let N be an integer such that N 2 , and let v 1 N M 1 , 2 ( Z ) M 1 , 2 ( Z ) .

1. If u 1 N M 1 , 2 ( Z ) M 1 , 2 ( Z ) such that u v or v ( mod M 1 , 2 ( Z ) ) , then g u 12 N = g v 12 N .

2. The function g v 12 N belongs to N and satisfies

( g v 12 N ) γ = g v γ 12 N ( γ GL 2 ( Z / N Z ) / I 2 Gal ( N / 1 ) ) .

## Proof

1. See [12, Theorem 1.1 in Chapter 2 and p. 29].

2. See [12, Theorem 1.2 and Proposition 1.3 in Chapter 2].□

## Lemma 2.3

Let u , v M 1 , 2 ( Q ) M 1 , 2 ( Z ) such that u v and v ( mod M 1 , 2 ( Z ) ) . Then we have

( f u f v ) 6 = J 2 ( J 1 ) 3 3 9 g u + v 6 g u v 6 g u 12 g v 12 .

## Proof

See [12, p. 51].□

## 3 Extended form class groups

In this section, we review some necessary consequences of the theory of complex multiplication, and introduce extended form class groups which might be an extension of Gauss’ form class group.

Let K be an imaginary quadratic field of discriminant d K . For a positive integer N , let Q N ( d K ) be the set of primitive positive definite binary quadratic forms of discriminant d K whose leading coefficients are relatively prime to N , that is,

Q N ( d K ) = Q x y = a Q x 2 + b Q x y + c Q y 2 Z [ x , y ] gcd ( a Q , b Q , c Q ) = 1 , gcd ( a Q , N ) = 1 , a Q > 0 , b Q 2 4 a Q c Q = d K .

The congruence subgroup

Γ 1 ( N ) = γ SL 2 ( Z ) γ 1 0 1 ( mod N M 2 ( Z ) )

defines an equivalence relation N on the set Q N ( d K ) as

Q N Q Q = Q γ x y for some γ Γ 1 ( N ) .

Let

C N ( d K ) = Q N ( d K ) / N

be the set of equivalence classes. For each Q = a Q x 2 + b Q x y + c Q y 2 Q N ( d K ) , let [ Q ] N be its class in C N ( d K ) , and let

τ Q = b Q + d K 2 a Q ,

which is the zero of the quadratic polynomial Q ( x , 1 ) lying in H . For a nontrivial ideal m of O K , let us denote by Cl ( m ) the ray class group modulo m , namely, Cl ( m ) = I K ( m ) / P K , 1 ( m ) , where I K ( m ) is the group of fractional ideals of K relatively prime to m and P K , 1 ( m ) is the subgroup of P K ( m ) (the subgroup of I K ( m ) consisting of principal fractional ideals) defined by

P K , 1 ( m ) = ν O K ν O K { 0 } such that ν 1 ( mod m ) .

Then, when m = N O K , the map

C N ( d K ) Cl ( N O K ) [ Q ] N [ [ τ Q , 1 ] ] = [ Z τ Q + Z ]

is a well-defined bijection, through which we may regard C N ( d K ) as a group isomorphic to Cl ( N O K ) ([13, Theorem 2.9] or [14, Theorem 2.5 and Proposition 5.3]). The identity element of C N ( d K ) is the class [ Q pr ] N of the principal form

Q pr x 2 + b K x y + c K y 2 = x 2 d K 4 y 2 if d K 0 ( mod 4 ) , x 2 + x y + 1 d K 4 y 2 if d K 1 ( mod 4 ) .

We call this group C N ( d K ) the extended form class group of discriminant d K and level N .

In particular, C 1 ( d K ) is the classical form class group of discriminant d K , originated and developed by Gauss [15] and Dirichlet [16]. A form Q = a Q x 2 + b Q x y + c Q y 2 in Q 1 ( d K ) is said to be reduced if

a Q < b Q a Q < c Q or 0 b Q a Q = c Q .

This condition yields

(6) a Q d K 3 .

If we let Q 1 , Q 2 , , Q h be all the reduced forms of discriminant d K , then we have h = C 1 ( d K ) and

(7) C 1 ( d K ) = { [ Q 1 ] 1 , [ Q 2 ] 1 , , [ Q h ] 1 }

([9, Theorem 2.8]). Set

τ K = d K 2 if d K 0 ( mod 4 ) , 1 + d K 2 if d K 1 ( mod 4 ) ,

and then τ K = τ Q pr and O K = [ τ K , 1 ] . By the theory of complex multiplication, we obtain the following results.

## Proposition 3.1

Let K be an imaginary quadratic field and m be a nontrivial ideal of O K .

1. If m = O K , then we obtain

K m = H K = K ( j ( τ K ) ) .

Furthermore, if Q i ( i = 1 , 2 , , h = C 1 ( d K ) ) are reduced forms of discriminant d K , then the singular values j ( τ Q i ) are distinct (Galois) conjugates of j ( τ K ) over K .

2. If m O K , then we have

K m = H K ( h K ( φ K ( [ ω ] ) ) )

for any element ω of K for which [ ω ] = ω + O K is a generator of the O K -module m 1 O K / O K . All such h K ( φ K ( [ ω ] ) ) are conjugate over H K . More precisely, if ξ i ( i = 1 , 2 , , [ K m : H K ] ) are nonzero elements of O K such that

{ ( ξ i ) P K , 1 ( m ) i = 1 , 2 , , [ K m : H K ] } = P K ( m ) / P K , 1 ( m ) ( Gal ( K m / H K ) ) ,

then h K ( φ K ( [ ξ i ω ] ) ) are all distinct conjugates of h K ( φ K ( [ ω ] ) ) over H K .

## Proof

1. See [2, Theorem 1 in Chapter 10] and [9, Theorem 7.7 (ii)].

2. See [2, Theorem 7 and its Corollary in Chapter 10].□

By modifying Shimura’s reciprocity law ([11, Theorem 6.31, Propositions 6.33 and 6.34]), Eum et al. established the following proposition.

## Proposition 3.2

Let K be an imaginary quadratic field, N be a positive integer, and K ( N ) be the ray class field of K modulo the ideal ( N ) . Then the map

C N ( d K ) Gal ( K ( N ) / K ) [ Q ] N f ( τ K ) f a Q ( b Q b K ) / 2 0 1 ( τ Q ) f N is f i n i t e a t τ K

is a well-defined isomorphism.

## Proof

See [13, Remark 3.3 and Theorem 3.10].□

## Remark 3.3

If M and N are positive integers such that M N , then the natural map

C N ( d K ) C M ( d K ) [ Q ] N [ Q ] M

is a surjective homomorphism ([13, Remark 2.10 (i)]).

## 4 Some applications of inequality on singular values of j

Let K be an imaginary quadratic field of discriminant d K . By using inequality argument on singular values of j developed in [6], we show that coordinates of elliptic curves in the family { E K , n } n Z 0 described in (2) can be used in order to generate the ray class fields of K .

Let h K denote the class number of K , i.e., h K = C 1 ( d K ) = [ H K : K ] . It is well known that

h K = 1 d K = 3 , 4 , 7 , 8 , 11 , 19 , 43 , 67 , 163

([9, Theorem 12.34]). So, if h K 2 , then we have d K 15 .

## Lemma 4.1

If h K 2 and d K 20 , then we achieve

(8) J ( τ Q ) 2 ( J ( τ Q ) 1 ) 3 J ( τ K ) 2 ( J ( τ K ) 1 ) 3 < 877,383 q τ K 5 2 ( < 1 )

for all nonprincipal reduced forms Q of discriminant d K .

## Proof

See [6, Lemma 6.3 (ii)].□

## Remark 4.2

If d K = 15 , then we obtain C 1 ( d K ) = { [ Q 1 ] 1 , [ Q 2 ] 1 } with

Q 1 = x 2 + x y + 4 y 2 and Q 2 = 2 x 2 + x y + 2 y 2 .

Moreover, we have

j ( τ K ) = j ( τ Q 1 ) = 52,515 85,995 1 + 5 2 and j ( τ Q 2 ) = 52,515 85,995 1 5 2

([1, Example 6.2.2]). One can check that inequality (8) also holds true.

## Lemma 4.3

Let K be an imaginary quadratic field other than Q ( 1 ) and Q ( 3 ) . Then we attain

H K = K ( K n ) for e v e r y n Z > 0 .

## Proof

If h K = 1 , then the assertion is obvious because H K = K .

Now, assume that h K 2 . Let σ be an element of Gal ( H K / K ) , which fixes K n . Then we find by Proposition 3.1 (i) that

1 = ( K n ) σ K n = { J ( τ K ) 2 ( J ( τ K ) 1 ) 3 } σ J ( τ K ) 2 ( J ( τ K ) 1 ) 3 n = J ( τ Q ) 2 ( J ( τ Q ) 1 ) 3 J ( τ K ) 2 ( J ( τ K ) 1 ) 3 n

for some reduced form Q of discriminant d K . Thus, Q must be Q pr by Lemma 4.1 and Remark 4.2, and hence σ is the identity element of Gal ( H K / K ) again by Proposition 3.1 (i). This observation implies by the Galois theory that H K is generated by K n over K .□

## Theorem 4.4

Let K be an imaginary quadratic field other than Q ( 1 ) and Q ( 3 ) , and let m be a nontrivial proper integral ideal of O K . Let ω be an element of K such that ω + O K is a generator of the O K -module m 1 / O K . If K m properly contains H K , then we have

K m = K ( x K , n ( ω ) , y K , n ( ω ) 2 ) for every n Z 0 .

## Proof

For simplicity, let

X = x K , n ( ω ) = K n h K ( φ K ( [ ω ] ) ) and Y = y K , n ( ω ) .

Set L = K ( X , Y 2 ) which is a subfield of K m by Proposition 3.1 and the Weierstrass equation for E K , n stated in (2).

Suppose on the contrary that K m L . Then there is a nonidentity element σ of Gal ( K m / K ) , which leaves both X and Y 2 fixed. Note further that

(9) σ Gal ( K m / H K )

because K m = H K ( X ) by Proposition 3.1 (ii). By applying σ on both sides of the equality

Y 2 = 4 X 3 A X B with A = J K ( J K 1 ) 27 K 2 n and B = J K ( J K 1 ) 2 2 7 2 K 3 n ,

we obtain

Y 2 = 4 X 3 A σ X B σ .

It then follows that

(10) ( A σ A ) X = B B σ .

Since

A B = K 5 n + 1 2 7 3

generates H K over K by Lemma 4.3, we deduce by (9) and (10) that A σ A 0 and

X = B σ B A σ A H K .

Then we obtain

H K = H K ( X ) = K m ,

which contradicts the hypothesis K m H K .

Therefore, we conclude that

K m = L = K ( x K , n ( ω ) , y K , n ( ω ) 2 ) .

## Proposition 4.5

Let K be an imaginary quadratic field other than Q ( 1 ) and Q ( 3 ) , and let m be a nontrivial proper integral ideal of O K . Let ω be an element of K such that ω + O K is a generator of the O K -module m 1 / O K . Then we have

K m = K ( x K , n ( ω ) ) for sufficiently large n Z 0 .

## Proof

Note that K = J ( τ K ) 2 ( J ( τ K ) 1 ) 3 is nonzero because K is different from Q ( 1 ) and Q ( 3 ) . There are two possible cases: h K = 1 or h K 2 .

1. If h K = 1 (and so H K = K ), then for any n Z 0

K ( x K , n ( ω ) ) = H K ( K n h K ( φ K ( [ ω ] ) ) ) by (3) = H K ( h K ( φ K ( [ ω ] ) ) ) by Proposition 3.1 (i) = K m by Proposition 3.1 (ii) .

2. Consider the case where h K 2 . Let Gal ( H K / K ) = { σ 1 = id , σ 2 , , σ h K } and d = [ K m : H K ] . Observe by Proposition 3.1 (i) that for each i = 1 , 2 , , h K there is a unique reduced form Q i of discriminant d K , and so J ( τ K ) σ i = J ( τ Q i ) . By Lemma 4.1 and Remark 4.2 we can take a sufficiently large positive integer m so that

K σ i K m d = J ( τ Q i ) 2 ( J ( τ Q i ) 1 ) 3 J ( τ K ) 2 ( J ( τ K ) 1 ) 3 m d < N K m / H K ( h K ( φ K ( [ ω ] ) ) ) N K m / H K ( h K ( φ K ( [ ω ] ) ) ) σ i for all i = 2 , 3 , , h K .

We then see by (3) and Proposition 3.1 (i) that if n Z 0 satisfies n m and 2 i h K , then

N K m / H K ( x K , n ( ω ) ) σ i N K m / H K ( x K , n ( ω ) ) = K σ i K n d N K m / H K ( h K ( φ K ( [ ω ] ) ) ) σ i N K m / H K ( h K ( φ K ( [ ω ] ) ) ) < 1 .

This observation implies that

(11) K ( N K m / H K ( x K , n ( ω ) ) ) = H K .

Hence we derive that if n m , then

K ( x K , n ( ω ) ) = K ( x K , n ( ω ) , N K m / H K ( x K , n ( ω ) ) ) since K ( x K , n ( ω ) ) ( K m ) is an abelian extension of K = H K ( x K , n ( ω ) ) by (11) = K m by (3) and Proposition 3.1 .

## 5 Generation of ray class fields by x -coordinates of elliptic curves

By using some interesting inequalities on special values of Siegel functions, we shall find a concrete bound of n in Proposition 4.5 for which if n is greater than or equal to this, then x K , n ( ω ) generates K m over K .

## Lemma 5.1

Let v M 1 , 2 ( Q ) M 1 , 2 ( Z ) , and let τ H such that q τ = e 2 π i τ e π 3 .

1. We have

g v ( τ ) < 2.29 q τ 1 24 .

2. If v 1 N M 1 , 2 ( Z ) M 1 , 2 ( Z ) for an integer N 2 , then we obtain

g v ( τ ) > 0.76 q τ 1 12 N .

## Proof

Let v = v 1 v 2 and z = v 1 τ + v 2 . By Proposition 2.2 (i) we may assume that 0 v 1 1 2 . Set s = q τ .

1. We then derive that

g v ( τ ) q τ 1 2 ( v 1 2 v 1 + 1 6 ) ( 1 + q z ) n = 1 ( 1 + q τ n q z ) ( 1 + q τ n q z 1 ) by (5) = s 1 2 ( v 1 2 v 1 + 1 6 ) ( 1 + s v 1 ) n = 1 ( 1 + s n + v 1 ) ( 1 + s n v 1 ) s 1 24 ( 1 + 1 ) n = 1 ( 1 + s n 1 2 ) 2 since 0 v 1 1 2 and v 1 2 v 1 + 1 6 1 12 2 s 1 24 n = 1 e 2 ( e π 3 ) n 1 2 because 1 + x < e x for x > 0 = 2 s 1 24 e 2 n = 1 ( e π 3 ) n 1 2 = 2 s 1 24 e 2 e π 3 2 1 e π 3 < 2.29 s 1 24 .

2. Furthermore, we see that

g v ( τ ) s 1 2 ( v 1 2 v 1 + 1 6 ) 1 q z n = 1 ( 1 s n + v 1 ) ( 1 s n v 1 ) by (5) s 1 12 min 1 e 2 π i N , 1 s 1 N n = 1 ( 1 s n 1 2 ) 2 because v 1 2 v 1 + 1 6 1 6 s 1 12 min sin π N , 1 ( e π 3 ) 1 N n = 1 e 4 ( e π 3 ) n 1 2 since 1 x > e 2 x for 0 < x < 1 2 > s 1 12 1 N e 4 n = 1 ( e π 3 ) n 1 2 because both sin ( π x ) and 1 e π 3 x are > x for 0 < x 1 2 = s 1 12 1 N e 4 e π 3 2 1 e π 3 > 0.76 s 1 12 N .

## Theorem 5.2

Let K be an imaginary quadratic field other than Q ( 1 ) and Q ( 3 ) . Let m be a proper nontrivial ideal of O K in which N m ( 2 ) is the least positive integer. Let ω be an element of K such that ω + O K is a generator of the O K -module m 1 / O K . If K m properly contains H K , then

K m = K ( x K , n ( ω ) )

for every nonnegative integer n satisfying

(12) n 13 24 π d K + 6 ln 229 76 N m 5 2 π d K ln 877,383 1 6 .

## Proof

Since N m O K m and ω m 1 O K , we have

(13) N m ω = a τ K + b for some a , b Z such that a b N m M 1 , 2 ( Z ) .

Let n be a nonnegative integer satisfying (12). If h K = 1 , then the assertion holds by the proof (Case 1) of Proposition 4.5.

Now, we assume h K 2 . Since K m properly contains H K , one can take a nonidentity element ρ of Gal ( K m / H K ) . Note that ρ does not fix x K , n ( ω ) due to the fact K m = H K ( x K , n ( ω ) ) by (3) and Proposition 3.1 (ii). Suppose on the contrary that x K , n ( ω ) does not generate K m over K . Then there exists at least one nonidentity element σ in Gal ( K m / K ( x K , n ( ω ) ) ) = Gal ( H K ( x K , n ( ω ) ) / K ( x K , n ( ω ) ) ) . Let P be a quadratic form in Q N m ( d K ) such that [ P ] N m maps to σ through the surjection

C N m ( d K ) Gal ( K ( N m ) / K ) restriction Gal ( K m / K ) μ μ K m

whose first map is the isomorphism given in Proposition 3.2. It follows from (7), Proposition 3.2, and Remark 3.3 that

P = Q γ for some nonpricipal reduced form  Q   and γ SL 2 ( Z ) .

Here we observe that

(14) τ P = γ 1 ( τ Q ) and a Q 2 .

Then we deduce that

1 = ( x K , n ( ω ) x K , n ( ω ) ρ ) σ x K , n ( ω ) x K , n ( ω ) ρ because  σ  is the identity on  K ( x K , n ( ω ) )  which contains x K , n ( ω ) ρ = ( K n f u ( τ K ) ( K n f u ( τ K ) ) ρ ) σ K n f u ( τ K ) ( K n f u ( τ K ) ) ρ with u = a N m b N m by (3), (13), and the definitions of  h K , φ K  and a Fricke function = { J ( τ K ) 2 n ( J ( τ K ) 1 ) 3 n ( f u ( τ K ) f v ( τ K ) ) } σ J ( τ K ) 2 n ( J ( τ K ) 1 ) 3 n ( f u ( τ K ) f v ( τ K ) ) for some v 1 N m M 1 , 2 ( Z ) M 1 , 2 ( Z ) such that u v and v ( mod M 1 , 2 ( Z ) ) by Proposition 3.1(ii) and (4) since ρ Gal ( K m / H K ) { id K m } = J ( τ P ) 2 ( J ( τ P ) 1 ) 3 J ( τ K ) 2 ( J ( τ K ) 1 ) 3 n f u ( τ P ) f v ( τ P ) f u ( τ K ) f v ( τ K ) for some u , v 1 N m M 1 , 2 ( Z ) M 1 , 2 ( Z ) such that u v and v ( mod M 1 , 2 ( Z ) ) by Proposition 3.2 = J ( γ 1 ( τ Q ) ) 2 ( J ( γ 1 ( τ Q ) ) 1 ) 3 J ( τ K ) 2 ( J ( τ K ) 1 ) 3 n f u ( γ 1 ( τ Q ) ) f v ( γ