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- Publisher
- de Gruyter
- Copyright
- © 2022 Ho Yun Jung et al., published by De Gruyter
- ISSN
- 2391-5455
- eISSN
- 2391-5455
- DOI
- 10.1515/math-2022-0502
- Publisher site
- See Article on Publisher Site
Abstract
1IntroductionLet KKbe an imaginary quadratic field with ring of integers OK{{\mathcal{O}}}_{K}. Let EEbe the elliptic curve with complex multiplication by OK{{\mathcal{O}}}_{K}given by the Weierstrass equation: E:y2=4x3−g2x−g3withg2=g2(OK)andg3=g3(OK).E:{y}^{2}=4{x}^{3}-{g}_{2}x-{g}_{3}\hspace{1.0em}\hspace{0.1em}\text{with}\hspace{0.1em}\hspace{0.33em}{g}_{2}={g}_{2}\left({{\mathcal{O}}}_{K})\hspace{1em}\hspace{0.1em}\text{and}\hspace{0.1em}\hspace{1em}{g}_{3}={g}_{3}\left({{\mathcal{O}}}_{K}).For z∈Cz\in {\mathbb{C}}, let [z]\left[z]denote the coset z+OKz+{{\mathcal{O}}}_{K}in C/OK{\mathbb{C}}\hspace{0.1em}\text{/}\hspace{0.1em}{{\mathcal{O}}}_{K}. Then the map φK:C/OK→E(C)(⊂P2(C))[z]↦[℘(z;OK):℘′(z;OK):1],\begin{array}{rcl}{\varphi }_{K}:{\mathbb{C}}\hspace{0.1em}\text{/}\hspace{0.1em}{{\mathcal{O}}}_{K}& \to & E\left({\mathbb{C}})\hspace{1.0em}\left(\subset {{\mathbb{P}}}^{2}\left({\mathbb{C}}))\\ {[}z]& \mapsto & \left[\wp \left(z;\hspace{0.33em}{{\mathcal{O}}}_{K}):\wp ^{\prime} \left(z;\hspace{0.33em}{{\mathcal{O}}}_{K}):1],\end{array}where ℘\wp is the Weierstrass ℘\wp -function relative to OK{{\mathcal{O}}}_{K}, is a complex analytic isomorphism of complex Lie groups ([1, Proposition 3.6 in Chapter VI]). Corresponding to EE, we consider the Weber function hK:E(C)→P1(C){{\mathfrak{h}}}_{K}:E\left({\mathbb{C}})\to {{\mathbb{P}}}^{1}\left({\mathbb{C}})given by hK(x,y)=g2g3Δxifj(E)≠0,1,728,g22Δx2ifj(E)=1,728,g3Δx3ifj(E)=0,{{\mathfrak{h}}}_{K}\left(x,y)=\left\{\begin{array}{ll}\frac{{g}_{2}{g}_{3}}{\Delta }x& \hspace{0.1em}\text{if}\hspace{0.1em}\hspace{0.33em}j\left(E)\ne 0,\hspace{0.1em}\text{1,728}\hspace{0.1em},\\ \frac{{g}_{2}^{2}}{\Delta }{x}^{2}& \hspace{0.1em}\text{if}\hspace{0.1em}\hspace{0.33em}j\left(E)=\hspace{0.1em}\text{1,728}\hspace{0.1em},\\ \frac{{g}_{3}}{\Delta }{x}^{3}& \hspace{0.1em}\text{if}\hspace{0.1em}\hspace{0.33em}j\left(E)=0,\end{array}\right.where Δ=g23−27g32\Delta ={g}_{2}^{3}-27{g}_{3}^{2}(≠0\ne 0) and j(E)=j(OK)j\left(E)=j\left({{\mathcal{O}}}_{K})is the jj-invariant of EE. For a nontrivial ideal m{\mathfrak{m}}of OK{{\mathcal{O}}}_{K}, by Km{K}_{{\mathfrak{m}}}we mean the ray class field of KKmodulo m{\mathfrak{m}}. In particular, KOK{K}_{{{\mathcal{O}}}_{K}}is the Hilbert class field HK{H}_{K}of KK. Then we obtain by the theory of complex multiplication that HK=K(j(E)){H}_{K}=K\left(j\left(E))and Km=HK(hK(x,y))for somem-torsion point(x,y)onE{K}_{{\mathfrak{m}}}={H}_{K}({{\mathfrak{h}}}_{K}\left(x,y))\hspace{0.33em}\hspace{0.1em}\text{for some}\hspace{0.1em}\hspace{0.33em}{\mathfrak{m}}\hspace{0.1em}\text{-torsion point}\hspace{0.1em}\hspace{0.33em}\left(x,y)\hspace{0.33em}\hspace{0.1em}\text{on}\hspace{0.1em}\hspace{0.33em}Eif m{\mathfrak{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 KKcan be generated only by a single value of the Weber function hK{{\mathfrak{h}}}_{K}over KK[3, p. 91] and Sugawara first gave partial answers to this question [4,5]. Recently, Jung et al. [6] proved that if m=NOK{\mathfrak{m}}=N{{\mathcal{O}}}_{K}and N∈{2,3,4,6}N\in \left\{2,3,4,6\right\}, then (1)Km=KhKφK1NorKm=KhKφK2N.{K}_{{\mathfrak{m}}}=K\left({{\mathfrak{h}}}_{K}\left({\varphi }_{K}\left(\left[\frac{1}{N}\right]\right)\right)\right)\hspace{1.0em}\hspace{0.1em}\text{or}\hspace{0.1em}\hspace{1.0em}{K}_{{\mathfrak{m}}}=K\left({{\mathfrak{h}}}_{K}\left({\varphi }_{K}\left(\left[\frac{2}{N}\right]\right)\right)\right).Koo et al. [7] further showed by utilizing the second Kronecker’s limit formula that (1) holds for N=5N=5and N≥7N\ge 7. Besides, it is worth noting that Ramachandra [8] constructed a complicated primitive generator of Km{K}_{{\mathfrak{m}}}over KKby using special values of the product of high powers of the discriminant Δ\Delta function and Siegel functions, which is beautiful in theory.Now, we assume that KKis different from Q(−1){\mathbb{Q}}\left(\sqrt{-1})and Q(−3){\mathbb{Q}}\left(\sqrt{-3}), and so g2g3≠0{g}_{2}{g}_{3}\ne 0and j(OK)≠0,1,728j\left({{\mathcal{O}}}_{K})\ne 0,\hspace{0.1em}\text{1,728}\hspace{0.1em}([9, p. 200]). Let {EK,n}n∈Z≥0{\left\{{E}_{K,n}\right\}}_{n\in {{\mathbb{Z}}}_{\ge 0}}be the family of elliptic curves isomorphic to EEgiven by the affine models (2)EK,n:y2=4x3−JK(JK−1)27ℓK2nx−JK(JK−1)2272ℓK3n,{E}_{K,n}:{y}^{2}=4{x}^{3}-\frac{{J}_{K}\left({J}_{K}-1)}{27}{{\ell }_{K}}^{2n}x-\frac{{J}_{K}{\left({J}_{K}-1)}^{2}}{2{7}^{2}}\hspace{0.33em}{{\ell }_{K}}^{3n},where JK=11,728j(OK)andℓK=JK2(JK−1)3.{J}_{K}=\frac{1}{\hspace{0.1em}\text{1,728}\hspace{0.1em}}j\left({{\mathcal{O}}}_{K})\hspace{1.0em}\hspace{0.1em}\text{and}\hspace{0.1em}\hspace{1.0em}{\ell }_{K}={J}_{K}^{2}{\left({J}_{K}-1)}^{3}.Then for each n∈Z≥0n\in {{\mathbb{Z}}}_{\ge 0}we have a parametrization C/OK→EK,n(C)(⊂P2(C))[z]↦[xK,n(z):yK,n(z):1],\begin{array}{rcl}{\mathbb{C}}\hspace{0.1em}\text{/}\hspace{0.1em}{{\mathcal{O}}}_{K}& \to & {E}_{K,n}\left({\mathbb{C}})\hspace{1.0em}\left(\subset {{\mathbb{P}}}^{2}\left({\mathbb{C}}))\\ {[}z]& \mapsto & {[}{x}_{K,n}\left(z):{y}_{K,n}\left(z):1],\end{array}with xK,n(z)=ℓKng2g3Δ℘(z;OK)andyK,n(z)=ℓKng2g3Δ3℘′(z;OK).{x}_{K,n}\left(z)={{\ell }_{K}}^{n}\frac{{g}_{2}{g}_{3}}{\Delta }\wp \left(z;\hspace{0.33em}{{\mathcal{O}}}_{K})\hspace{1.0em}\hspace{0.1em}\text{and}\hspace{0.1em}\hspace{1.0em}{y}_{K,n}\left(z)=\sqrt{{\left({{\ell }_{K}}^{n}\frac{{g}_{2}{g}_{3}}{\Delta }\right)}^{3}}\wp ^{\prime} \left(z;\hspace{0.33em}{{\mathcal{O}}}_{K}).Here we note that (3)xK,n(z)=ℓKnhK(φK([z]))(z∈C).{x}_{K,n}\left(z)={{\ell }_{K}}^{n}{{\mathfrak{h}}}_{K}({\varphi }_{K}(\left[z]))\hspace{1.0em}\left(z\in {\mathbb{C}}).Let m{\mathfrak{m}}be a proper nontrivial ideal of OK{{\mathcal{O}}}_{K}in such a way that Km{K}_{{\mathfrak{m}}}properly contains HK{H}_{K}. Let ω\omega be an element of KKso that [ω]=ω+OK\left[\omega ]=\omega +{{\mathcal{O}}}_{K}is a generator of the OK{{\mathcal{O}}}_{K}-module m−1OK/OK{{\mathfrak{m}}}^{-1}{{\mathcal{O}}}_{K}\hspace{0.1em}\text{/}\hspace{0.1em}{{\mathcal{O}}}_{K}. In this article, we shall prove the following three assertions (Theorems 4.4, 5.2, and 6.4): (i)We have Km=K(xK,n(ω),yK,n(ω)2){K}_{{\mathfrak{m}}}=K({x}_{K,n}\left(\omega ),{y}_{K,n}{\left(\omega )}^{2})for every n∈Z≥0n\in {{\mathbb{Z}}}_{\ge 0}.(ii)We obtain Km=K(xK,n(ω)){K}_{{\mathfrak{m}}}=K({x}_{K,n}\left(\omega ))for every n∈Z≥0n\in {{\mathbb{Z}}}_{\ge 0}satisfying n≥1324π∣dK∣+6ln22976Nm52π∣dK∣−ln877,383−16,n\ge \frac{\frac{13}{24}\pi \sqrt{| {d}_{K}| }+6\mathrm{ln}\left(\frac{229}{76}{N}_{{\mathfrak{m}}}\right)}{\frac{5}{2}\pi \sqrt{| {d}_{K}| }-\mathrm{ln}\hspace{0.1em}\text{877,383}\hspace{0.1em}}-\frac{1}{6},where dK{d}_{K}is the discriminant of KKand Nm{N}_{{\mathfrak{m}}}is the least positive integer in m{\mathfrak{m}}.(iii)If m=NOK{\mathfrak{m}}=N{{\mathcal{O}}}_{K}for an integer NN(≥2\ge 2) whose prime factors are all inert in KK, then Km=K(xK,n(ω)){K}_{{\mathfrak{m}}}=K({x}_{K,n}\left(\omega ))for every n∈Z≥0n\in {{\mathbb{Z}}}_{\ge 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 LL-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.2Fricke and Siegel functionsIn this preliminary section, we recall the definitions and basic properties of Fricke and Siegel functions.Let H{\mathbb{H}}be the complex upper half-plane, that is, H={τ∈C∣Im(τ)>0}{\mathbb{H}}=\left\{\tau \in {\mathbb{C}}| {\rm{Im}}\left(\tau )\gt 0\right\}. Let jjbe the elliptic modular function on H{\mathbb{H}}given by j(τ)=j([τ,1])(τ∈H),j\left(\tau )=j\left(\left[\tau ,1])\hspace{1.0em}\left(\tau \in {\mathbb{H}}),where [τ,1]\left[\tau ,1]stands for the lattice Zτ+Z{\mathbb{Z}}\tau +{\mathbb{Z}}in C{\mathbb{C}}and j([τ,1])j\left(\left[\tau ,1])is the jj-invariant of an elliptic curve isomorphic to C/[τ,1]{\mathbb{C}}\hspace{0.1em}\text{/}\hspace{0.1em}\left[\tau ,1]. Define the function JJon H{\mathbb{H}}by J(τ)=11,728j(τ)(τ∈H).J\left(\tau )=\frac{1}{\hspace{0.1em}\text{1,728}\hspace{0.1em}}j\left(\tau )\hspace{1.0em}\left(\tau \in {\mathbb{H}}).Furthermore, for v=v1v2∈M1,2(Q)⧹M1,2(Z){\bf{v}}=\left[\begin{array}{cc}{v}_{1}& {v}_{2}\end{array}\right]\in {M}_{1,2}\left({\mathbb{Q}})\setminus {M}_{1,2}\left({\mathbb{Z}})we define the Fricke function fv{f}_{{\bf{v}}}on H{\mathbb{H}}by fv(τ)=−2735g2(τ)g3(τ)Δ(τ)℘(v1τ+v2;[τ,1])(τ∈H),{f}_{{\bf{v}}}\left(\tau )=-{2}^{7}{3}^{5}\frac{{g}_{2}\left(\tau ){g}_{3}\left(\tau )}{\Delta \left(\tau )}\wp \left({v}_{1}\tau +{v}_{2};\hspace{0.33em}\left[\tau ,1])\hspace{1.0em}\left(\tau \in {\mathbb{H}}),where g2(τ)=g2([τ,1]){g}_{2}\left(\tau )={g}_{2}\left(\left[\tau ,1]), g3(τ)=g3([τ,1]){g}_{3}\left(\tau )={g}_{3}\left(\left[\tau ,1]), and Δ(τ)=Δ([τ,1])\Delta \left(\tau )=\Delta \left(\left[\tau ,1]). Note that for u,v∈M1,2(Q)⧹M1,2(Z){\bf{u}},{\bf{v}}\in {M}_{1,2}\left({\mathbb{Q}})\setminus {M}_{1,2}\left({\mathbb{Z}})(4)fu=fv⇔u≡vor−v(modM1,2(Z)){f}_{{\bf{u}}}={f}_{{\bf{v}}}\hspace{0.33em}\iff \hspace{0.33em}{\bf{u}}\equiv {\bf{v}}\hspace{0.33em}\hspace{0.1em}\text{or}\hspace{0.1em}\hspace{0.33em}-{\bf{v}}\hspace{0.33em}\left(\hspace{0.1em}\text{mod}\hspace{0.1em}\hspace{0.33em}{M}_{1,2}\left({\mathbb{Z}}))([9, Lemma 10.4]). For a positive integer NN, let ℱN{{\mathcal{ {\mathcal F} }}}_{N}be the field given by ℱN=Q(j)ifN=1,ℱ1fv∣v∈1NM1,2(Z)⧹M1,2(Z)ifN≥2.{{\mathcal{ {\mathcal F} }}}_{N}=\left\{\begin{array}{ll}{\mathbb{Q}}\left(j)& \hspace{0.1em}\text{if}\hspace{0.1em}\hspace{0.33em}N=1,\\ {{\mathcal{ {\mathcal F} }}}_{1}\left({f}_{{\bf{v}}}| {\bf{v}}\in \frac{1}{N}{M}_{1,2}\left({\mathbb{Z}})\setminus {M}_{1,2}\left({\mathbb{Z}})\right)& \hspace{0.1em}\text{if}\hspace{0.1em}\hspace{0.33em}N\ge 2.\end{array}\right.Then, ℱN{{\mathcal{ {\mathcal F} }}}_{N}is a Galois extension of ℱ1{{\mathcal{ {\mathcal F} }}}_{1}whose Galois group is isomorphic to GL2(Z/NZ)/⟨−I2⟩{{\rm{GL}}}_{2}\left({\mathbb{Z}}\hspace{0.1em}\text{/}\hspace{0.1em}N{\mathbb{Z}})\hspace{0.1em}\text{/}\hspace{0.1em}\langle -{I}_{2}\rangle ([11, Theorem 6.6]). It coincides with the field of meromorphic modular functions of level NNwhose Fourier coefficients belong to the NNth cyclotomic field ([11, Proposition 6.9]).Proposition 2.1If N≥2N\ge 2, v∈1NM1,2(Z)⧹M1,2(Z){\bf{v}}\in \frac{1}{N}{M}_{1,2}\left({\mathbb{Z}})\setminus {M}_{1,2}\left({\mathbb{Z}}), and γ∈GL2(Z/NZ)/⟨−I2⟩\gamma \in {{\rm{GL}}}_{2}\left({\mathbb{Z}}\hspace{0.1em}\text{/}\hspace{0.1em}N{\mathbb{Z}})\hspace{0.1em}\text{/}\hspace{0.1em}\langle -{I}_{2}\rangle , thenfvγ=fvγ.{f}_{{\bf{v}}}^{\gamma }={f}_{{\bf{v}}\gamma }.Moreover, if γ∈SL2(Z/NZ)/⟨−I2⟩\gamma \in {{\rm{SL}}}_{2}\left({\mathbb{Z}}\hspace{0.1em}\text{/}\hspace{0.1em}N{\mathbb{Z}})\hspace{0.1em}\text{/}\hspace{0.1em}\langle -{I}_{2}\rangle , thenfvγ=fv∘α,{f}_{{\bf{v}}}^{\gamma }={f}_{{\bf{v}}}\circ \alpha ,where α\alpha is any element of SL2(Z){{\rm{SL}}}_{2}\left({\mathbb{Z}})(acting on H{\mathbb{H}}as fractional linear transformation) whose image in SL2(Z/NZ)/⟨−I2⟩{{\rm{SL}}}_{2}\left({\mathbb{Z}}\hspace{0.1em}\text{/}\hspace{0.1em}N{\mathbb{Z}})\hspace{0.1em}\text{/}\hspace{0.1em}\langle -{I}_{2}\rangle is γ\gamma .ProofSee [2, Theorem 3 in Chapter 6] or [11, Theorem 6.6].□For v=v1v2∈M1,2(Q)⧹M1,2(Z){\bf{v}}=\left[\begin{array}{cc}{v}_{1}& {v}_{2}\end{array}\right]\in {M}_{1,2}\left({\mathbb{Q}})\setminus {M}_{1,2}\left({\mathbb{Z}}), the Siegel function gv{g}_{{\bf{v}}}on H{\mathbb{H}}is given by the infinite product expansion (5)gv(τ)=−eπiv2(v1−1)qτ12(v12−v1+16)(1−qz)∏n=1∞(1−qτnqz)(1−qτnqz−1)(τ∈H),{g}_{{\bf{v}}}\left(\tau )=-{e}^{\pi {\rm{i}}{v}_{2}\left({v}_{1}-1)}{q}_{\tau }^{\frac{1}{2}\left({v}_{1}^{2}-{v}_{1}+\frac{1}{6})}\left(1-{q}_{z})\mathop{\prod }\limits_{n=1}^{\infty }\left(1-{q}_{\tau }^{n}{q}_{z})\left(1-{q}_{\tau }^{n}{q}_{z}^{-1})\hspace{1.0em}\left(\tau \in {\mathbb{H}}),where qτ=e2πiτ{q}_{\tau }={e}^{2\pi {\rm{i}}\tau }and qz=e2πiz{q}_{z}={e}^{2\pi {\rm{i}}z}with z=v1τ+v2z={v}_{1}\tau +{v}_{2}. Observe that gv{g}_{{\bf{v}}}has neither a zero nor a pole on H{\mathbb{H}}.Proposition 2.2Let N be an integer such that N≥2N\ge 2, and let v∈1NM1,2(Z)⧹M1,2(Z){\bf{v}}\in \frac{1}{N}{M}_{1,2}\left({\mathbb{Z}})\setminus {M}_{1,2}\left({\mathbb{Z}}). (i)If u∈1NM1,2(Z)⧹M1,2(Z){\bf{u}}\in \frac{1}{N}{M}_{1,2}\left({\mathbb{Z}})\setminus {M}_{1,2}\left({\mathbb{Z}})such that u≡vor−v(modM1,2(Z)){\bf{u}}\equiv {\bf{v}}\hspace{0.33em}{or}\hspace{0.33em}-{\bf{v}}\hspace{0.33em}\left(\hspace{0.1em}\text{mod}\hspace{0.1em}\hspace{0.33em}{M}_{1,2}\left({\mathbb{Z}})), then gu12N=gv12N{g}_{{\bf{u}}}^{12N}={g}_{{\bf{v}}}^{12N}.(ii)The function gv12N{g}_{{\bf{v}}}^{12N}belongs to ℱN{{\mathcal{ {\mathcal F} }}}_{N}and satisfies(gv12N)γ=gvγ12N(γ∈GL2(Z/NZ)/⟨−I2⟩≃Gal(ℱN/ℱ1)).{({g}_{{\bf{v}}}^{12N})}^{\gamma }={g}_{{\bf{v}}\gamma }^{12N}\hspace{1.0em}\left(\gamma \in {{\rm{GL}}}_{2}\left({\mathbb{Z}}\hspace{0.1em}\text{/}\hspace{0.1em}N{\mathbb{Z}})\hspace{0.1em}\text{/}\hspace{0.1em}\langle -{I}_{2}\rangle \simeq {\rm{Gal}}\left({{\mathcal{ {\mathcal F} }}}_{N}\hspace{0.1em}\text{/}\hspace{0.1em}{{\mathcal{ {\mathcal F} }}}_{1})).Proof(i)See [12, Theorem 1.1 in Chapter 2 and p. 29].(ii)See [12, Theorem 1.2 and Proposition 1.3 in Chapter 2].□Lemma 2.3Let u,v∈M1,2(Q)⧹M1,2(Z){\bf{u}},{\bf{v}}\in {M}_{1,2}\left({\mathbb{Q}})\setminus {M}_{1,2}\left({\mathbb{Z}})such that u≢vand−v(modM1,2(Z)){\bf{u}}\not\equiv {\bf{v}}\hspace{0.33em}{and}\hspace{0.33em}-{\bf{v}}\hspace{0.33em}\left(\hspace{0.1em}\text{mod}\hspace{0.1em}\hspace{0.33em}{M}_{1,2}\left({\mathbb{Z}})). Then we have(fu−fv)6=J2(J−1)339gu+v6gu−v6gu12gv12.{({f}_{{\bf{u}}}-{f}_{{\bf{v}}})}^{6}=\frac{{J}^{2}{\left(J-1)}^{3}}{{3}^{9}}\frac{{g}_{{\bf{u}}+{\bf{v}}}^{6}{g}_{{\bf{u}}-{\bf{v}}}^{6}}{{g}_{{\bf{u}}}^{12}{g}_{{\bf{v}}}^{12}}.ProofSee [12, p. 51].□3Extended form class groupsIn 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 KKbe an imaginary quadratic field of discriminant dK{d}_{K}. For a positive integer NN, let QN(dK){{\mathcal{Q}}}_{N}\left({d}_{K})be the set of primitive positive definite binary quadratic forms of discriminant dK{d}_{K}whose leading coefficients are relatively prime to NN, that is, QN(dK)=Qxy=aQx2+bQxy+cQy2∈Z[x,y]gcd(aQ,bQ,cQ)=1,gcd(aQ,N)=1,aQ>0,bQ2−4aQcQ=dK.{{\mathcal{Q}}}_{N}\left({d}_{K})=\left\{Q\left(\left[\begin{array}{c}x\\ y\end{array}\right]\right)={a}_{Q}{x}^{2}+{b}_{Q}xy+{c}_{Q}{y}^{2}\in {\mathbb{Z}}\left[x,y]\left|\begin{array}{l}\gcd \left({a}_{Q},{b}_{Q},{c}_{Q})=1,\\ \gcd \left({a}_{Q},N)=1,\\ {a}_{Q}\gt 0,\\ {b}_{Q}^{2}-4{a}_{Q}{c}_{Q}={d}_{K}\\ \end{array}\right.\right\}.The congruence subgroup Γ1(N)=γ∈SL2(Z)∣γ≡1∗01(modNM2(Z)){\Gamma }_{1}\left(N)=\left\{\gamma \in {{\rm{SL}}}_{2}\left({\mathbb{Z}})| \gamma \equiv \left[\begin{array}{cc}1& \ast \\ 0& 1\end{array}\right]\hspace{0.33em}\left(\hspace{0.1em}\text{mod}\hspace{0.1em}\hspace{0.33em}N{M}_{2}\left({\mathbb{Z}}))\right\}defines an equivalence relation ∼N{ \sim }_{N}on the set QN(dK){{\mathcal{Q}}}_{N}\left({d}_{K})as Q∼NQ′⇔Q′=Qγxyfor someγ∈Γ1(N).Q{ \sim }_{N}Q^{\prime} \hspace{0.33em}\iff \hspace{0.33em}Q^{\prime} =Q\left(\gamma \left[\begin{array}{c}x\\ y\end{array}\right]\right)\hspace{0.33em}\hspace{0.1em}\text{for some}\hspace{0.1em}\hspace{0.33em}\gamma \in {\Gamma }_{1}\left(N).Let CN(dK)=QN(dK)/∼N{{\rm{C}}}_{N}\left({d}_{K})={{\mathcal{Q}}}_{N}\left({d}_{K})\hspace{0.1em}\text{/}\hspace{0.1em}{ \sim }_{N}be the set of equivalence classes. For each Q=aQx2+bQxy+cQy2∈QN(dK)Q={a}_{Q}{x}^{2}+{b}_{Q}xy+{c}_{Q}{y}^{2}\in {{\mathcal{Q}}}_{N}\left({d}_{K}), let [Q]N{\left[Q]}_{N}be its class in CN(dK){{\rm{C}}}_{N}\left({d}_{K}), and let τQ=−bQ+dK2aQ,{\tau }_{Q}=\frac{-{b}_{Q}+\sqrt{{d}_{K}}}{2{a}_{Q}},which is the zero of the quadratic polynomial Q(x,1)Q\left(x,1)lying in H{\mathbb{H}}. For a nontrivial ideal m{\mathfrak{m}}of OK{{\mathcal{O}}}_{K}, let us denote by Cl(m){\rm{Cl}}\left({\mathfrak{m}})the ray class group modulo m{\mathfrak{m}}, namely, Cl(m)=IK(m)/PK,1(m){\rm{Cl}}\left({\mathfrak{m}})={I}_{K}\left({\mathfrak{m}})\hspace{0.1em}\text{/}\hspace{0.1em}{P}_{K,1}\left({\mathfrak{m}}), where IK(m){I}_{K}\left({\mathfrak{m}})is the group of fractional ideals of KKrelatively prime to m{\mathfrak{m}}and PK,1(m){P}_{K,1}\left({\mathfrak{m}})is the subgroup of PK(m){P}_{K}\left({\mathfrak{m}})(the subgroup of IK(m){I}_{K}\left({\mathfrak{m}})consisting of principal fractional ideals) defined by PK,1(m)=⟨νOK∣ν∈OK⧹{0}such thatν≡1(modm)⟩.{P}_{K,1}\left({\mathfrak{m}})=\langle \nu {{\mathcal{O}}}_{K}| \nu \in {{\mathcal{O}}}_{K}\setminus \left\{0\right\}\hspace{1em}\hspace{0.1em}\text{such that}\hspace{0.1em}\hspace{0.33em}\nu \equiv 1\hspace{0.33em}\left(\hspace{0.1em}\text{mod}\hspace{0.1em}\hspace{0.33em}{\mathfrak{m}})\rangle .Then, when m=NOK{\mathfrak{m}}=N{{\mathcal{O}}}_{K}, the map CN(dK)→Cl(NOK)[Q]N↦[[τQ,1]]=[ZτQ+Z]\hspace{1em}\begin{array}{lcl}{{\rm{C}}}_{N}\left({d}_{K})& \to & {\rm{Cl}}\left(N{{\mathcal{O}}}_{K})\\ {{[}Q]}_{N}& \mapsto & \left[\left[{\tau }_{Q},1]]=\left[{\mathbb{Z}}{\tau }_{Q}+{\mathbb{Z}}]\end{array}is a well-defined bijection, through which we may regard CN(dK){{\rm{C}}}_{N}\left({d}_{K})as a group isomorphic to Cl(NOK){\rm{Cl}}\left(N{{\mathcal{O}}}_{K})([13, Theorem 2.9] or [14, Theorem 2.5 and Proposition 5.3]). The identity element of CN(dK){{\rm{C}}}_{N}\left({d}_{K})is the class [Qpr]N{\left[{Q}_{{\rm{pr}}}]}_{N}of the principal form Qpr≔x2+bKxy+cKy2=x2−dK4y2ifdK≡0(mod4),x2+xy+1−dK4y2ifdK≡1(mod4).{Q}_{{\rm{pr}}}:= {x}^{2}+{b}_{K}xy+{c}_{K}{y}^{2}=\left\{\begin{array}{ll}{x}^{2}-\frac{{d}_{K}}{4}{y}^{2}& \hspace{0.1em}\text{if}\hspace{0.1em}\hspace{0.33em}{d}_{K}\equiv 0\hspace{0.33em}\left(\hspace{0.1em}\text{mod}\hspace{0.1em}\hspace{0.33em}4),\\ {x}^{2}+xy+\frac{1-{d}_{K}}{4}{y}^{2}& \hspace{0.1em}\text{if}\hspace{0.1em}\hspace{0.33em}{d}_{K}\equiv 1\hspace{0.33em}\left(\hspace{0.1em}\text{mod}\hspace{0.1em}\hspace{0.33em}4).\end{array}\right.We call this group CN(dK){{\rm{C}}}_{N}\left({d}_{K})the extended form class group of discriminant dK{d}_{K}and level NN.In particular, C1(dK){{\rm{C}}}_{1}\left({d}_{K})is the classical form class group of discriminant dK{d}_{K}, originated and developed by Gauss [15] and Dirichlet [16]. A form Q=aQx2+bQxy+cQy2Q={a}_{Q}{x}^{2}+{b}_{Q}xy+{c}_{Q}{y}^{2}in Q1(dK){{\mathcal{Q}}}_{1}\left({d}_{K})is said to be reduced if −aQ<bQ≤aQ<cQor0≤bQ≤aQ=cQ.-{a}_{Q}\lt {b}_{Q}\le {a}_{Q}\lt {c}_{Q}\hspace{1.0em}\hspace{0.1em}\text{or}\hspace{0.1em}\hspace{1.0em}0\le {b}_{Q}\le {a}_{Q}={c}_{Q}.This condition yields (6)aQ≤∣dK∣3.{a}_{Q}\le \sqrt{\frac{| {d}_{K}| }{3}}.If we let Q1,Q2,…,Qh{Q}_{1},{Q}_{2},\ldots ,{Q}_{h}be all the reduced forms of discriminant dK{d}_{K}, then we have h=∣C1(dK)∣h=| {{\rm{C}}}_{1}\left({d}_{K})| and (7)C1(dK)={[Q1]1,[Q2]1,…,[Qh]1}{{\rm{C}}}_{1}\left({d}_{K})=\{{\left[{Q}_{1}]}_{1},{\left[{Q}_{2}]}_{1},\ldots ,{\left[{Q}_{h}]}_{1}\}([9, Theorem 2.8]). Set τK=dK2ifdK≡0(mod4),−1+dK2ifdK≡1(mod4),{\tau }_{K}=\left\{\begin{array}{ll}\frac{\sqrt{{d}_{K}}}{2}& \hspace{0.1em}\text{if}\hspace{0.1em}\hspace{0.33em}{d}_{K}\equiv 0\hspace{0.33em}\left(\hspace{0.1em}\text{mod}\hspace{0.1em}\hspace{0.33em}4),\\ \frac{-1+\sqrt{{d}_{K}}}{2}& \hspace{0.1em}\text{if}\hspace{0.1em}\hspace{0.33em}{d}_{K}\equiv 1\hspace{0.33em}\left(\hspace{0.1em}\text{mod}\hspace{0.1em}\hspace{0.33em}4),\end{array}\right.and then τK=τQpr{\tau }_{K}={\tau }_{{Q}_{{\rm{pr}}}}and OK=[τK,1]{{\mathcal{O}}}_{K}={[}{\tau }_{K},1]. By the theory of complex multiplication, we obtain the following results.Proposition 3.1Let K be an imaginary quadratic field and m{\mathfrak{m}}be a nontrivial ideal of OK{{\mathcal{O}}}_{K}. (i)If m=OK{\mathfrak{m}}={{\mathcal{O}}}_{K}, then we obtainKm=HK=K(j(τK)).{K}_{{\mathfrak{m}}}={H}_{K}=K\left(j\left({\tau }_{K})).Furthermore, if Qi{Q}_{i}(i=1,2,…,h=∣C1(dK)∣i=1,2,\ldots ,h=| {{\rm{C}}}_{1}\left({d}_{K})| ) are reduced forms of discriminant dK{d}_{K}, then the singular values j(τQi)j\left({\tau }_{{Q}_{i}})are distinct (Galois) conjugates of j(τK)j\left({\tau }_{K})over KK.(ii)If m≠OK{\mathfrak{m}}\ne {{\mathcal{O}}}_{K}, then we haveKm=HK(hK(φK([ω]))){K}_{{\mathfrak{m}}}={H}_{K}({{\mathfrak{h}}}_{K}\left({\varphi }_{K}\left(\left[\omega ])))for any element ω\omega of KKfor which [ω]=ω+OK\left[\omega ]=\omega +{{\mathcal{O}}}_{K}is a generator of the OK{{\mathcal{O}}}_{K}-module m−1OK/OK{{\mathfrak{m}}}^{-1}{{\mathcal{O}}}_{K}\hspace{0.1em}\text{/}\hspace{0.1em}{{\mathcal{O}}}_{K}. All such hK(φK([ω])){{\mathfrak{h}}}_{K}\left({\varphi }_{K}\left(\left[\omega ]))are conjugate over HK{H}_{K}. More precisely, if ξi{\xi }_{i}(i=1,2,…,[Km:HK]i=1,2,\ldots ,\left[{K}_{{\mathfrak{m}}}:{H}_{K}]) are nonzero elements of OK{{\mathcal{O}}}_{K}such that{(ξi)PK,1(m)∣i=1,2,…,[Km:HK]}=PK(m)/PK,1(m)(≃Gal(Km/HK)),\{\left({\xi }_{i}){P}_{K,1}\left({\mathfrak{m}})| i=1,2,\ldots ,\left[{K}_{{\mathfrak{m}}}:{H}_{K}]\}={P}_{K}\left({\mathfrak{m}})\hspace{0.1em}\text{/}\hspace{0.1em}{P}_{K,1}\left({\mathfrak{m}})\left(\simeq {\rm{Gal}}\left({K}_{{\mathfrak{m}}}\hspace{0.1em}\text{/}\hspace{0.1em}{H}_{K})),then hK(φK([ξiω])){{\mathfrak{h}}}_{K}\left({\varphi }_{K}\left(\left[{\xi }_{i}\omega ]))are all distinct conjugates of hK(φK([ω])){{\mathfrak{h}}}_{K}\left({\varphi }_{K}\left(\left[\omega ]))over HK{H}_{K}.Proof(i)See [2, Theorem 1 in Chapter 10] and [9, Theorem 7.7 (ii)].(ii)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 3.2Let KKbe an imaginary quadratic field, N be a positive integer, and K(N){K}_{\left(N)}be the ray class field of K modulo the ideal (N)\left(N). Then the mapCN(dK)→Gal(K(N)/K)[Q]N↦f(τK)↦faQ(bQ−bK)/201(τQ)∣f∈ℱNisfiniteatτK\begin{array}{rcl}{{\rm{C}}}_{N}\left({d}_{K})& \to & {\rm{Gal}}\left({K}_{\left(N)}\hspace{0.1em}\text{/}\hspace{0.1em}K)\\ {{[}Q]}_{N}& \mapsto & \left(f\left({\tau }_{K})\mapsto {f}^{\left[\begin{array}{cc}{a}_{Q}& \left({b}_{Q}-{b}_{K})\text{/}2\\ 0& 1\end{array}\right]}\left({\tau }_{Q})| f\in {{\mathcal{ {\mathcal F} }}}_{N}\hspace{0.33em}{is}\hspace{0.33em}finite\hspace{0.33em}at\hspace{0.33em}{\tau }_{K}\right)\end{array}is a well-defined isomorphism.ProofSee [13, Remark 3.3 and Theorem 3.10].□Remark 3.3If MMand NNare positive integers such that M∣NM\hspace{0.33em}| \hspace{0.33em}N, then the natural map CN(dK)→CM(dK)[Q]N↦[Q]M\begin{array}{rcl}{{\rm{C}}}_{N}\left({d}_{K})& \to & {{\rm{C}}}_{M}\left({d}_{K})\\ {{[}Q]}_{N}& \mapsto & {\left[Q]}_{M}\end{array}is a surjective homomorphism ([13, Remark 2.10 (i)]).4Some applications of inequality on singular values of jjLet KKbe an imaginary quadratic field of discriminant dK{d}_{K}. By using inequality argument on singular values of jjdeveloped in [6], we show that coordinates of elliptic curves in the family {EK,n}n∈Z≥0{\left\{{E}_{K,n}\right\}}_{n\in {{\mathbb{Z}}}_{\ge 0}}described in (2) can be used in order to generate the ray class fields of KK.Let hK{h}_{K}denote the class number of KK, i.e., hK=∣C1(dK)∣=[HK:K]{h}_{K}=| {{\rm{C}}}_{1}\left({d}_{K})| =\left[{H}_{K}:K]. It is well known that hK=1⇔dK=−3,−4,−7,−8,−11,−19,−43,−67,−163{h}_{K}=1\hspace{0.33em}\iff \hspace{0.33em}{d}_{K}=-3,-4,-7,-8,-11,-19,-43,-67,-163([9, Theorem 12.34]). So, if hK≥2{h}_{K}\ge 2, then we have dK≤−15{d}_{K}\le -15.Lemma 4.1If hK≥2{h}_{K}\ge 2and dK≤−20{d}_{K}\le -20, then we achieve(8)J(τQ)2(J(τQ)−1)3J(τK)2(J(τK)−1)3<877,383∣qτK∣52(<1)\left|\hspace{-0.33em},\frac{J{\left({\tau }_{Q})}^{2}{\left(J\left({\tau }_{Q})-1)}^{3}}{J{\left({\tau }_{K})}^{2}{\left(J\left({\tau }_{K})-1)}^{3}}\hspace{-0.33em}\right|\lt \hspace{0.1em}\text{877,383}\hspace{0.1em}\hspace{0.33em}{| {q}_{{\tau }_{K}}| }^{\tfrac{5}{2}}\left(\lt 1)for all nonprincipal reduced forms Q of discriminant dK{d}_{K}.ProofSee [6, Lemma 6.3 (ii)].□Remark 4.2If dK=−15{d}_{K}=-15, then we obtain C1(dK)={[Q1]1,[Q2]1}{{\rm{C}}}_{1}\left({d}_{K})=\left\{{\left[{Q}_{1}]}_{1},{\left[{Q}_{2}]}_{1}\right\}with Q1=x2+xy+4y2andQ2=2x2+xy+2y2.{Q}_{1}={x}^{2}+xy+4{y}^{2}\hspace{1.0em}\hspace{0.1em}\text{and}\hspace{0.1em}\hspace{1.0em}{Q}_{2}=2{x}^{2}+xy+2{y}^{2}.Moreover, we have j(τK)=j(τQ1)=−52,515−85,9951+52andj(τQ2)=−52,515−85,9951−52j\left({\tau }_{K})=j\left({\tau }_{{Q}_{1}})=-\hspace{0.1em}\text{52,515}-\text{85,995}\hspace{0.1em}\frac{1+\sqrt{5}}{2}\hspace{1.0em}\hspace{0.1em}\text{and}\hspace{0.1em}\hspace{1.0em}j\left({\tau }_{{Q}_{2}})=-\hspace{0.1em}\text{52,515}-\text{85,995}\hspace{0.1em}\hspace{0.33em}\frac{1-\sqrt{5}}{2}([1, Example 6.2.2]). One can check that inequality (8) also holds true.Lemma 4.3Let K be an imaginary quadratic field other than Q(−1){\mathbb{Q}}\left(\sqrt{-1})and Q(−3){\mathbb{Q}}\left(\sqrt{-3}). Then we attainHK=K(ℓKn)foreveryn∈Z>0.{H}_{K}=K\left({{\ell }_{K}}^{n})\hspace{1.0em}{for}\hspace{0.33em}every\hspace{0.33em}n\in {{\mathbb{Z}}}_{\gt 0}.ProofIf hK=1{h}_{K}=1, then the assertion is obvious because HK=K{H}_{K}=K.Now, assume that hK≥2{h}_{K}\ge 2. Let σ\sigma be an element of Gal(HK/K){\rm{Gal}}\left({H}_{K}\hspace{0.1em}\text{/}\hspace{0.1em}K), which fixes ℓKn{{\ell }_{K}}^{n}. Then we find by Proposition 3.1 (i) that 1=(ℓKn)σℓKn={J(τK)2(J(τK)−1)3}σJ(τK)2(J(τK)−1)3n=J(τQ)2(J(τQ)−1)3J(τK)2(J(τK)−1)3n1=\left|\hspace{-0.33em},\frac{{\left({{\ell }_{K}}^{n})}^{\sigma }}{{{\ell }_{K}}^{n}}\hspace{-0.33em}\right|={\left|,\frac{{\left\{J{\left({\tau }_{K})}^{2}{\left(J\left({\tau }_{K})-1)}^{3}\right\}}^{\sigma }}{J{\left({\tau }_{K})}^{2}{\left(J\left({\tau }_{K})-1)}^{3}}\right|}^{n}={\left|,\frac{J{\left({\tau }_{Q})}^{2}{\left(J\left({\tau }_{Q})-1)}^{3}}{J{\left({\tau }_{K})}^{2}{\left(J\left({\tau }_{K})-1)}^{3}}\right|}^{n}for some reduced form QQof discriminant dK{d}_{K}. Thus, QQmust be Qpr{Q}_{{\rm{pr}}}by Lemma 4.1 and Remark 4.2, and hence σ\sigma is the identity element of Gal(HK/K){\rm{Gal}}\left({H}_{K}\hspace{0.1em}\text{/}\hspace{0.1em}K)again by Proposition 3.1 (i). This observation implies by the Galois theory that HK{H}_{K}is generated by ℓKn{{\ell }_{K}}^{n}over KK.□Theorem 4.4Let K be an imaginary quadratic field other than Q(−1){\mathbb{Q}}\left(\sqrt{-1})and Q(−3){\mathbb{Q}}\left(\sqrt{-3}), and let m{\mathfrak{m}}be a nontrivial proper integral ideal of OK{{\mathcal{O}}}_{K}. Let ω\omega be an element of K such that ω+OK\omega +{{\mathcal{O}}}_{K}is a generator of the OK{{\mathcal{O}}}_{K}-module m−1/OK{{\mathfrak{m}}}^{-1}\hspace{0.1em}\text{/}\hspace{0.1em}{{\mathcal{O}}}_{K}. If Km{K}_{{\mathfrak{m}}}properly contains HK{H}_{K}, then we haveKm=K(xK,n(ω),yK,n(ω)2)foreveryn∈Z≥0.{K}_{{\mathfrak{m}}}=K({x}_{K,n}\left(\omega ),{y}_{K,n}{\left(\omega )}^{2})\hspace{1.0em}{for}\hspace{0.33em}{every}\hspace{0.33em}n\in {{\mathbb{Z}}}_{\ge 0}.ProofFor simplicity, let X=xK,n(ω)=ℓKnhK(φK([ω]))andY=yK,n(ω).X={x}_{K,n}\left(\omega )={{\ell }_{K}}^{n}{{\mathfrak{h}}}_{K}\left({\varphi }_{K}\left(\left[\omega ]))\hspace{1.0em}\hspace{0.1em}\text{and}\hspace{0.1em}\hspace{1.0em}Y={y}_{K,n}\left(\omega ).Set L=K(X,Y2)L=K\left(X,{Y}^{2})which is a subfield of Km{K}_{{\mathfrak{m}}}by Proposition 3.1 and the Weierstrass equation for EK,n{E}_{K,n}stated in (2).Suppose on the contrary that Km≠L{K}_{{\mathfrak{m}}}\ne L. Then there is a nonidentity element σ\sigma of Gal(Km/K){\rm{Gal}}\left({K}_{{\mathfrak{m}}}\hspace{0.1em}\text{/}\hspace{0.1em}K), which leaves both XXand Y2{Y}^{2}fixed. Note further that (9)σ∉Gal(Km/HK)\sigma \notin {\rm{Gal}}\left({K}_{{\mathfrak{m}}}\hspace{0.1em}\text{/}\hspace{0.1em}{H}_{K})because Km=HK(X){K}_{{\mathfrak{m}}}={H}_{K}\left(X)by Proposition 3.1 (ii). By applying σ\sigma on both sides of the equality Y2=4X3−AX−BwithA=JK(JK−1)27ℓK2nandB=JK(JK−1)2272ℓK3n,{Y}^{2}=4{X}^{3}-AX-B\hspace{1.0em}\hspace{0.1em}\text{with}\hspace{0.1em}\hspace{0.33em}A=\frac{{J}_{K}\left({J}_{K}-1)}{27}{{\ell }_{K}}^{2n}\hspace{1.0em}\hspace{0.1em}\text{and}\hspace{0.1em}\hspace{1.0em}B=\frac{{J}_{K}{\left({J}_{K}-1)}^{2}}{2{7}^{2}}{{\ell }_{K}}^{3n},we obtain Y2=4X3−AσX−Bσ.{Y}^{2}=4{X}^{3}-{A}^{\sigma }X-{B}^{\sigma }.It then follows that (10)(Aσ−A)X=B−Bσ.\left({A}^{\sigma }-A)X=B-{B}^{\sigma }.Since AB=ℓK5n+1273AB=\frac{{{\ell }_{K}}^{5n+1}}{2{7}^{3}}generates HK{H}_{K}over KKby Lemma 4.3, we deduce by (9) and (10) that Aσ−A≠0{A}^{\sigma }-A\ne 0and X=−Bσ−BAσ−A∈HK.X=-\frac{{B}^{\sigma }-B}{{A}^{\sigma }-A}\in {H}_{K}.Then we obtain HK=HK(X)=Km,{H}_{K}={H}_{K}\left(X)={K}_{{\mathfrak{m}}},which contradicts the hypothesis Km⊋HK{K}_{{\mathfrak{m}}}\hspace{0.33em} \supsetneq \hspace{0.33em}{H}_{K}.Therefore, we conclude that □Km=L=K(xK,n(ω),yK,n(ω)2).{K}_{{\mathfrak{m}}}=L=K({x}_{K,n}\left(\omega ),{y}_{K,n}{\left(\omega )}^{2}).Proposition 4.5Let KKbe an imaginary quadratic field other than Q(−1){\mathbb{Q}}\left(\sqrt{-1})and Q(−3){\mathbb{Q}}\left(\sqrt{-3}), and let m{\mathfrak{m}}be a nontrivial proper integral ideal of OK{{\mathcal{O}}}_{K}. Let ω\omega be an element of K such that ω+OK\omega +{{\mathcal{O}}}_{K}is a generator of the OK{{\mathcal{O}}}_{K}-module m−1/OK{{\mathfrak{m}}}^{-1}\hspace{0.1em}\text{/}\hspace{0.1em}{{\mathcal{O}}}_{K}. Then we haveKm=K(xK,n(ω))forsufficientlylargen∈Z≥0.{K}_{{\mathfrak{m}}}=K({x}_{K,n}\left(\omega ))\hspace{1.0em}{for}\hspace{0.33em}{sufficiently}\hspace{0.33em}{large}\hspace{0.33em}n\in {{\mathbb{Z}}}_{\ge 0}.ProofNote that ℓK=J(τK)2(J(τK)−1)3{\ell }_{K}=J{\left({\tau }_{K})}^{2}{\left(J\left({\tau }_{K})-1)}^{3}is nonzero because KKis different from Q(−1){\mathbb{Q}}\left(\sqrt{-1})and Q(−3){\mathbb{Q}}\left(\sqrt{-3}). There are two possible cases: hK=1{h}_{K}=1or hK≥2{h}_{K}\ge 2. Case 1.If hK=1{h}_{K}=1(and so HK=K{H}_{K}=K), then for any n∈Z≥0n\in {{\mathbb{Z}}}_{\ge 0}K(xK,n(ω))=HK(ℓKnhK(φK([ω])))by (3)=HK(hK(φK([ω])))by Proposition 3.1 (i)=Kmby Proposition 3.1 (ii).\begin{array}{rcl}K({x}_{K,n}\left(\omega ))& =& {H}_{K}({{\ell }_{K}}^{n}{{\mathfrak{h}}}_{K}\left({\varphi }_{K}\left(\left[\omega ])))\hspace{1.0em}\hspace{0.1em}\text{by (3)}\hspace{0.1em}\\ & =& {H}_{K}({{\mathfrak{h}}}_{K}\left({\varphi }_{K}\left(\left[\omega ])))\hspace{1.0em}\hspace{0.1em}\text{by Proposition 3.1 (i)}\hspace{0.1em}\\ & =& {K}_{{\mathfrak{m}}}\hspace{1.0em}\hspace{0.1em}\text{by Proposition 3.1 (ii)}\hspace{0.1em}.\end{array}Case 2.Consider the case where hK≥2{h}_{K}\ge 2. Let Gal(HK/K)={σ1=id,σ2,…,σhK}{\rm{Gal}}\left({H}_{K}\hspace{0.1em}\text{/}\hspace{0.1em}K)=\left\{{\sigma }_{1}={\rm{id}},{\sigma }_{2},\ldots ,{\sigma }_{{h}_{K}}\right\}and d=[Km:HK]d=\left[{K}_{{\mathfrak{m}}}:{H}_{K}]. Observe by Proposition 3.1 (i) that for each i=1,2,…,hKi=1,2,\ldots ,{h}_{K}there is a unique reduced form Qi{Q}_{i}of discriminant dK{d}_{K}, and so J(τK)σi=J(τQi)J{\left({\tau }_{K})}^{{\sigma }_{i}}=J\left({\tau }_{{Q}_{i}}). By Lemma 4.1 and Remark 4.2 we can take a sufficiently large positive integer mmso that ℓKσiℓKmd=J(τQi)2(J(τQi)−1)3J(τK)2(J(τK)−1)3md<NKm/HK(hK(φK([ω])))NKm/HK(hK(φK([ω])))σifor alli=2,3,…,hK.{\left|,\frac{{{\ell }_{K}}^{{\sigma }_{i}}}{{\ell }_{K}}\right|}^{md}={\left|,\frac{J{\left({\tau }_{{Q}_{i}})}^{2}{\left(J\left({\tau }_{{Q}_{i}})-1)}^{3}}{J{\left({\tau }_{K})}^{2}{\left(J\left({\tau }_{K})-1)}^{3}}\right|}^{md}\lt \left|\hspace{-0.33em},\frac{{{\rm{N}}}_{{K}_{{\mathfrak{m}}}\text{/}{H}_{K}}({{\mathfrak{h}}}_{K}\left({\varphi }_{K}\left(\left[\omega ])))}{{{\rm{N}}}_{{K}_{{\mathfrak{m}}}\text{/}{H}_{K}}{({{\mathfrak{h}}}_{K}\left({\varphi }_{K}\left(\left[\omega ])))}^{{\sigma }_{i}}}\hspace{-0.33em}\right|\hspace{1.0em}\hspace{0.1em}\text{for all}\hspace{0.1em}\hspace{0.33em}i=2,3,\ldots ,{h}_{K}.We then see by (3) and Proposition 3.1 (i) that if n∈Z≥0n\in {{\mathbb{Z}}}_{\ge 0}satisfies n≥mn\ge mand 2≤i≤hK2\le i\le {h}_{K}, then NKm/HK(xK,n(ω))σiNKm/HK(xK,n(ω))=ℓKσiℓKndNKm/HK(hK(φK([ω])))σiNKm/HK(hK(φK([ω])))<1.\left|\hspace{-0.33em},\frac{{{\rm{N}}}_{{K}_{{\mathfrak{m}}}\text{/}{H}_{K}}{\left({x}_{K,n}\left(\omega ))}^{{\sigma }_{i}}}{{{\rm{N}}}_{{K}_{{\mathfrak{m}}}\text{/}{H}_{K}}\left({x}_{K,n}\left(\omega ))}\hspace{-0.33em}\right|={\left|,\frac{{{\ell }_{K}}^{{\sigma }_{i}}}{{\ell }_{K}}\right|}^{nd}\left|\hspace{-0.33em},\frac{{{\rm{N}}}_{{K}_{{\mathfrak{m}}}\text{/}{H}_{K}}{({{\mathfrak{h}}}_{K}\left({\varphi }_{K}\left(\left[\omega ])))}^{{\sigma }_{i}}}{{{\rm{N}}}_{{K}_{{\mathfrak{m}}}\text{/}{H}_{K}}({{\mathfrak{h}}}_{K}\left({\varphi }_{K}\left(\left[\omega ])))}\hspace{-0.33em}\right|\lt 1.This observation implies that (11)K(NKm/HK(xK,n(ω)))=HK.K({N}_{{K}_{{\mathfrak{m}}}\text{/}{H}_{K}}\left({x}_{K,n}\left(\omega )))={H}_{K}.Hence we derive that if n≥mn\ge m, then □K(xK,n(ω))=K(xK,n(ω),NKm/HK(xK,n(ω)))sinceK(xK,n(ω))(⊆Km)is an abelian extension ofK=HK(xK,n(ω))by (11)=Kmby (3) and Proposition 3.1.\begin{array}{rcl}K({x}_{K,n}\left(\omega ))& =& K({x}_{K,n}\left(\omega ),{N}_{{K}_{{\mathfrak{m}}}\text{/}{H}_{K}}\left({x}_{K,n}\left(\omega )))\hspace{0.33em}\hspace{0.1em}\text{since}\hspace{0.1em}\hspace{0.33em}K\left({x}_{K,n}\left(\omega ))\hspace{1em}\left(\subseteq {K}_{{\mathfrak{m}}})\hspace{1em}\hspace{0.1em}\text{is an abelian extension of}\hspace{0.1em}\hspace{0.33em}K\\ & =& {H}_{K}({x}_{K,n}\left(\omega ))\hspace{1.0em}\hspace{0.1em}\text{by (11)}\hspace{0.1em}\\ & =& {K}_{{\mathfrak{m}}}\hspace{1.0em}\hspace{0.1em}\text{by (3) and Proposition 3.1}.\end{array}5Generation of ray class fields by xx-coordinates of elliptic curvesBy using some interesting inequalities on special values of Siegel functions, we shall find a concrete bound of nnin Proposition 4.5 for which if nnis greater than or equal to this, then xK,n(ω){x}_{K,n}\left(\omega )generates Km{K}_{{\mathfrak{m}}}over KK.Lemma 5.1Let v∈M1,2(Q)⧹M1,2(Z){\bf{v}}\in {M}_{1,2}\left({\mathbb{Q}})\setminus {M}_{1,2}\left({\mathbb{Z}}), and let τ∈H\tau \in {\mathbb{H}}such that ∣qτ∣=∣e2πiτ∣≤e−π3| {q}_{\tau }| =| {e}^{2\pi {\rm{i}}\tau }| \le {e}^{-\pi \sqrt{3}}. (i)We have∣gv(τ)∣<2.29∣qτ∣−124.| {g}_{{\bf{v}}}\left(\tau )| \lt 2.29| {q}_{\tau }\hspace{-0.33em}{| }^{-\tfrac{1}{24}}.(ii)If v∈1NM1,2(Z)⧹M1,2(Z){\bf{v}}\in \frac{1}{N}{M}_{1,2}\left({\mathbb{Z}})\setminus {M}_{1,2}\left({\mathbb{Z}})for an integer N≥2N\ge 2, then we obtain∣gv(τ)∣>0.76∣qτ∣112N.| {g}_{{\bf{v}}}\left(\tau )| \gt \frac{0.76| {q}_{\tau }\hspace{-0.33em}{| }^{\tfrac{1}{12}}}{N}.ProofLet v=v1v2{\bf{v}}=\left[\begin{array}{cc}{v}_{1}& {v}_{2}\end{array}\right]and z=v1τ+v2z={v}_{1}\tau +{v}_{2}. By Proposition 2.2 (i) we may assume that 0≤v1≤120\le {v}_{1}\le \frac{1}{2}. Set s=∣qτ∣s=| {q}_{\tau }| . (i)We then derive that ∣gv(τ)∣≤∣qτ∣12(v12−v1+16)(1+∣qz∣)∏n=1∞(1+∣qτ∣n∣qz∣)(1+∣qτ∣n∣qz∣−1)by (5)=s12(v12−v1+16)(1+sv1)∏n=1∞(1+sn+v1)(1+sn−v1)≤s−124(1+1)∏n=1∞(1+sn−12)2since0≤v1≤12andv12−v1+16≥−112≤2s−124∏n=1∞e2(e−π3)n−12because1+x<exforx>0=2s−124e2∑n=1∞(e−π3)n−12=2s−124e2e−π321−e−π3<2.29s−124.\begin{array}{rcl}| {g}_{{\bf{v}}}\left(\tau )| & \le & | {q}_{\tau }\hspace{-0.25em}{| }^{\tfrac{1}{2}\left({v}_{1}^{2}-{v}_{1}+\tfrac{1}{6})}\left(1+| {q}_{z}| )\mathop{\displaystyle \prod }\limits_{n=1}^{\infty }\left(1+| {q}_{\tau }\hspace{-0.25em}{| }^{n}| {q}_{z}| )\left(1+| {q}_{\tau }\hspace{-0.25em}{| }^{n}| {q}_{z}\hspace{-0.25em}{| }^{-1})\hspace{1.0em}\hspace{0.1em}\text{by (5)}\hspace{0.1em}\\ & =& {s}^{\tfrac{1}{2}\left({v}_{1}^{2}-{v}_{1}+\tfrac{1}{6})}\left(1+{s}^{{v}_{1}})\mathop{\displaystyle \prod }\limits_{n=1}^{\infty }\left(1+{s}^{n+{v}_{1}})\left(1+{s}^{n-{v}_{1}})\\ & \le & {s}^{-\tfrac{1}{24}}\left(1+1)\mathop{\displaystyle \prod }\limits_{n=1}^{\infty }{\left(1+{s}^{n-\frac{1}{2}})}^{2}\hspace{1.0em}\hspace{0.1em}\text{since}\hspace{0.1em}0\le {v}_{1}\le \frac{1}{2}\hspace{0.33em}\hspace{0.1em}\text{and}\hspace{0.1em}\hspace{0.33em}{v}_{1}^{2}-{v}_{1}+\frac{1}{6}\ge -\frac{1}{12}\\ & \le & 2{s}^{-\tfrac{1}{24}}\mathop{\displaystyle \prod }\limits_{n=1}^{\infty }{e}^{2{\left({e}^{-\pi \sqrt{3}})}^{n-\frac{1}{2}}}\hspace{1.0em}\hspace{0.1em}\text{because}\hspace{0.1em}\hspace{0.33em}1+x\lt {e}^{x}\hspace{0.33em}\hspace{0.1em}\text{for}\hspace{0.1em}\hspace{0.33em}x\gt 0\\ & =& 2{s}^{-\tfrac{1}{24}}{e}^{2{\displaystyle \sum }_{n=1}^{\infty }{\left({e}^{-\pi \sqrt{3}})}^{n-\frac{1}{2}}}\\ & =& 2{s}^{-\tfrac{1}{24}}{e}^{\tfrac{2{e}^{-\tfrac{\pi \sqrt{3}}{2}}}{1-{e}^{-\pi \sqrt{3}}}}\\ & \lt & 2.29\hspace{0.33em}{s}^{-\tfrac{1}{24}}.\end{array}(ii)Furthermore, we see that □∣gv(τ)∣≥s12(v12−v1+16)∣1−qz∣∏n=1∞(1−sn+v1)(1−sn−v1)by (5)≥s112min∣1−e2πiN∣,1−s1N∏n=1∞(1−sn−12)2becausev12−v1+16≤16≥s112minsinπN,1−(e−π3)1N∏n=1∞e−4(e−π3)n−12since1−x>e−2xfor0<x<12>s1121Ne−4∑n=1∞(e−π3)n−12because bothsin(πx)and1−e−π3xare>xfor0<x≤12=s1121Ne−4e−π321−e−π3>0.76s112N.\begin{array}{rcl}| {g}_{{\bf{v}}}\left(\tau )| & \ge & {s}^{\tfrac{1}{2}\left({v}_{1}^{2}-{v}_{1}+\tfrac{1}{6})}| 1-{q}_{z}| \mathop{\displaystyle \prod }\limits_{n=1}^{\infty }\left(1-{s}^{n+{v}_{1}})\left(1-{s}^{n-{v}_{1}})\hspace{1.0em}\hspace{0.1em}\text{by (5)}\hspace{0.1em}\\ & \ge & {s}^{\tfrac{1}{12}}\min \left\{| 1-{e}^{\tfrac{2\pi {\rm{i}}}{N}}| ,1-{s}^{\tfrac{1}{N}}\right\}\mathop{\displaystyle \prod }\limits_{n=1}^{\infty }{\left(1-{s}^{n-\frac{1}{2}})}^{2}\hspace{1.0em}\hspace{0.1em}\text{because}\hspace{0.1em}\hspace{0.33em}{v}_{1}^{2}-{v}_{1}+\frac{1}{6}\le \frac{1}{6}\\ & \ge & {s}^{\tfrac{1}{12}}\min \left\{\sin \frac{\pi }{N},1-{\left({e}^{-\pi \sqrt{3}})}^{\tfrac{1}{N}}\right\}\mathop{\displaystyle \prod }\limits_{n=1}^{\infty }{e}^{-4{\left({e}^{-\pi \sqrt{3}})}^{n-\frac{1}{2}}}\hspace{1em}\hspace{0.1em}\text{since}\hspace{0.1em}\hspace{0.33em}1-x\gt {e}^{-2x}\hspace{0.33em}\hspace{0.1em}\text{for}\hspace{0.1em}\hspace{0.33em}0\lt x\lt \frac{1}{2}\\ & \gt & {s}^{\tfrac{1}{12}}\frac{1}{N}{e}^{-4{\displaystyle \sum }_{n=1}^{\infty }{\left({e}^{-\pi \sqrt{3}})}^{n-\frac{1}{2}}}\hspace{1.0em}\hspace{0.1em}\text{because both}\hspace{0.1em}\hspace{0.33em}\sin \left(\pi x)\hspace{0.33em}\hspace{0.1em}\text{and}\hspace{0.1em}\hspace{0.33em}1-{e}^{-\pi \sqrt{3}x}\hspace{0.33em}\hspace{0.1em}\text{are}\hspace{0.1em}\gt x\hspace{0.33em}\hspace{0.1em}\text{for}\hspace{0.1em}\hspace{0.33em}0\lt x\le \frac{1}{2}\\ & =& {s}^{\tfrac{1}{12}}\frac{1}{N}{e}^{\tfrac{-4{e}^{-\tfrac{\pi \sqrt{3}}{2}}}{1-{e}^{-\pi \sqrt{3}}}}\\ & \gt & \frac{0.76{s}^{\tfrac{1}{12}}}{N}.\end{array}Theorem 5.2Let KKbe an imaginary quadratic field other than Q(−1){\mathbb{Q}}\left(\sqrt{-1})and Q(−3){\mathbb{Q}}\left(\sqrt{-3}). Let m{\mathfrak{m}}be a proper nontrivial ideal of OK{{\mathcal{O}}}_{K}in which Nm{N}_{{\mathfrak{m}}}(≥2\ge 2) is the least positive integer. Let ω\omega be an element of K such that ω+OK\omega +{{\mathcal{O}}}_{K}is a generator of the OK{{\mathcal{O}}}_{K}-module m−1/OK{{\mathfrak{m}}}^{-1}\hspace{0.1em}\text{/}\hspace{0.1em}{{\mathcal{O}}}_{K}. If Km{K}_{{\mathfrak{m}}}properly contains HK{H}_{K}, thenKm=K(xK,n(ω)){K}_{{\mathfrak{m}}}=K\left({x}_{K,n}\left(\omega ))for every nonnegative integer n satisfying(12)n≥1324π∣dK∣+6ln22976Nm52π∣dK∣−ln877,383−16.n\ge \frac{\frac{13}{24}\pi \sqrt{| {d}_{K}| }+6\mathrm{ln}\left(\frac{229}{76}{N}_{{\mathfrak{m}}}\right)}{\frac{5}{2}\pi \sqrt{| {d}_{K}| }-\mathrm{ln}\hspace{0.1em}\text{877,383}\hspace{0.1em}}-\frac{1}{6}.ProofSince NmOK⊆m{N}_{{\mathfrak{m}}}{{\mathcal{O}}}_{K}\subseteq {\mathfrak{m}}and ω∈m−1⧹OK\omega \in {{\mathfrak{m}}}^{-1}\setminus {{\mathcal{O}}}_{K}, we have (13)Nmω=aτK+bfor somea,b∈Zsuch thatab∉NmM1,2(Z).{N}_{{\mathfrak{m}}}\omega =a{\tau }_{K}+b\hspace{1.0em}\hspace{0.1em}\text{for some}\hspace{0.1em}\hspace{0.33em}a,b\in {\mathbb{Z}}\hspace{0.33em}\hspace{0.1em}\text{such that}\hspace{0.1em}\hspace{0.33em}\left[\begin{array}{cc}a& b\end{array}\right]\notin {N}_{{\mathfrak{m}}}{M}_{1,2}\left({\mathbb{Z}}).Let nnbe a nonnegative integer satisfying (12). If hK=1{h}_{K}=1, then the assertion holds by the proof (Case 1) of Proposition 4.5.Now, we assume hK≥2{h}_{K}\ge 2. Since Km{K}_{{\mathfrak{m}}}properly contains HK{H}_{K}, one can take a nonidentity element ρ\rho of Gal(Km/HK){\rm{Gal}}\left({K}_{{\mathfrak{m}}}\hspace{0.1em}\text{/}\hspace{0.1em}{H}_{K}). Note that ρ\rho does not fix xK,n(ω){x}_{K,n}\left(\omega )due to the fact Km=HK(xK,n(ω)){K}_{{\mathfrak{m}}}={H}_{K}\left({x}_{K,n}\left(\omega ))by (3) and Proposition 3.1 (ii). Suppose on the contrary that xK,n(ω){x}_{K,n}\left(\omega )does not generate Km{K}_{{\mathfrak{m}}}over KK. Then there exists at least one nonidentity element σ\sigma in Gal(Km/K(xK,n(ω)))=Gal(HK(xK,n(ω))/K(xK,n(ω))){\rm{Gal}}\left({K}_{{\mathfrak{m}}}\hspace{0.1em}\text{/}\hspace{0.1em}K\left({x}_{K,n}\left(\omega )))={\rm{Gal}}\left({H}_{K}\left({x}_{K,n}\left(\omega ))\hspace{0.1em}\text{/}\hspace{0.1em}K\left({x}_{K,n}\left(\omega ))). Let PPbe a quadratic form in QNm(dK){{\mathcal{Q}}}_{{N}_{{\mathfrak{m}}}}\left({d}_{K})such that [P]Nm{\left[P]}_{{N}_{{\mathfrak{m}}}}maps to σ\sigma through the surjection CNm(dK)⟶∼Gal(K(Nm)/K)⟶restrictionGal(Km/K)μ↦μ∣Km\begin{array}{ccccc}{{\rm{C}}}_{{N}_{{\mathfrak{m}}}}\left({d}_{K})& \mathop{\longrightarrow }\limits^{ \sim }& {\rm{Gal}}\left({K}_{\left({N}_{{\mathfrak{m}}})}\hspace{0.1em}\text{/}\hspace{0.1em}K)& \mathop{\longrightarrow }\limits^{\hspace{0.1em}\text{restriction}\hspace{0.1em}}& {\rm{Gal}}\left({K}_{{\mathfrak{m}}}\hspace{0.1em}\text{/}\hspace{0.1em}K)\\ & & \mu & \mapsto & \mu {| }_{{K}_{{\mathfrak{m}}}}\end{array}whose first map is the isomorphism given in Proposition 3.2. It follows from (7), Proposition 3.2, and Remark 3.3 that P=Qγfor some nonpricipal reduced form Q andγ∈SL2(Z).P={Q}^{\gamma }\hspace{1.0em}\hspace{0.1em}\text{for some nonpricipal reduced form\hspace{0.5em}}Q\text{\hspace{0.5em} and}\hspace{0.1em}\hspace{0.33em}\gamma \in {{\rm{SL}}}_{2}\left({\mathbb{Z}}).Here we observe that (14)τP=γ−1(τQ)andaQ≥2.{\tau }_{P}={\gamma }^{-1}\left({\tau }_{Q})\hspace{1.0em}\hspace{0.1em}\text{and}\hspace{0.1em}\hspace{1.0em}{a}_{Q}\ge 2.Then we deduce that 1=(xK,n(ω)−xK,n(ω)ρ)σxK,n(ω)−xK,n(ω)ρbecause σ is the identity on K(xK,n(ω)) which containsxK,n(ω)ρ=(ℓKnfu(τK)−(ℓKnfu(τK))ρ)σℓKnfu(τK)−(ℓKnfu(τK))ρwithu=aNmbNmby (3), (13), and the definitions of hK,φK and a Fricke function={J(τK)2n(J(τK)−1)3n(fu(τK)−fv(τK))}σJ(τK)2n(J(τK)−1)3n(fu(τK)−fv(τK))for somev∈1NmM1,2(Z)⧹M1,2(Z)such thatu≢vand−v(modM1,2(Z))by Proposition 3.1(ii) and (4) sinceρ∈Gal(Km/HK)⧹{idKm}=J(τP)2(J(τP)−1)3J(τK)2(J(τK)−1)3nfu′(τP)−fv′(τP)fu(τK)−fv(τK)for someu′,v′∈1NmM1,2(Z)⧹M1,2(Z)such thatu′≢v′and−v′(modM1,2(Z))by Proposition 3.2=J(γ−1(τQ))2(J(γ−1(τQ))−1)3J(τK)2(J(τK)−1)3nfu′(γ−1(τQ))−fv′(γ−1(τQ))fu(τK)−fv(τK)by (14)=J(τQ)2(J(τQ)−1)3J(τK)2(J(τK)−1)3nfu″(τQ)−fv″(τQ)fu(τK)−fv(τK)withu″=u′γ−1andv″=v′γ−1by the fact J∈ℱ1 and Proposition 2.1=J(τQ)2(J(τQ)−1)3J(τK)2(J(τK)−1)3n+16gu″+v″(τQ)gu″−v″(τQ)gu(τK)2gv(τK)2gu+v(τK)gu−v(τK)gu″(τQ)2gv″(τQ)2by Lemma 2.3<877,383∣qτK∣52n+162.296∣qτQ∣−112∣qτK∣−160.76Nm6∣qτK∣16∣qτQ∣13by Lemmas 4.1, 5.1, and Remark 4.2because∣qτK∣=e−π∣dK∣≤e−π15and∣qτQ∣=e−π∣dK∣aQ≤e−π3by (6)=877,383∣qτK∣52n+1622976Nm6∣qτQ∣−512∣qτK∣−13≤877,383∣qτK∣52n+1622976Nm6∣qτK∣−524∣qτK∣−13since∣qτK∣−1>1and∣qτQ∣−1=(∣qτK∣−1)1aQ≤(∣qτK∣−1)12by (14)=877,383e−52π∣dK∣n+1622976Nm6e1324π∣dK∣.\begin{array}{rcl}1& =& \left|\hspace{-0.33em},\frac{{({x}_{K,n}\left(\omega )-{x}_{K,n}{\left(\omega )}^{\rho })}^{\sigma }}{{x}_{K,n}\left(\omega )-{x}_{K,n}{\left(\omega )}^{\rho }}\hspace{-0.33em}\right|\\ & & \hspace{0.1em}\text{because\hspace{0.5em}}\sigma \text{\hspace{0.5em}is the identity on\hspace{0.5em}}\hspace{0.1em}K\left({x}_{K,n}\left(\omega ))\hspace{0.1em}\text{\hspace{0.5em}which contains}\hspace{0.1em}\hspace{0.33em}{x}_{K,n}{\left(\omega )}^{\rho }\\ & =& \left|\hspace{-0.33em},\frac{{({{\ell }_{K}}^{n}{f}_{{\bf{u}}}\left({\tau }_{K})-{\left({{\ell }_{K}}^{n}{f}_{{\bf{u}}}\left({\tau }_{K}))}^{\rho })}^{\sigma }}{{{\ell }_{K}}^{n}{f}_{{\bf{u}}}\left({\tau }_{K})-{\left({{\ell }_{K}}^{n}{f}_{{\bf{u}}}\left({\tau }_{K}))}^{\rho }}\hspace{-0.33em}\right|\hspace{1.0em}\hspace{0.1em}\text{with}\hspace{0.33em}{\bf{u}}=\left[\begin{array}{cc}\frac{a}{{N}_{{\mathfrak{m}}}}& \frac{b}{{N}_{{\mathfrak{m}}}}\end{array}\right]\\ & & \hspace{0.1em}\text{by (3), (13), and the definitions of\hspace{0.5em}}\hspace{0.1em}{{\mathfrak{h}}}_{K},{\varphi }_{K}\hspace{0.1em}\text{\hspace{0.5em}and a Fricke function}\hspace{0.1em}\\ & =& \left|\hspace{-0.33em},\frac{{\left\{J{\left({\tau }_{K})}^{2n}{\left(J\left({\tau }_{K})-1)}^{3n}({f}_{{\bf{u}}}\left({\tau }_{K})-{f}_{{\bf{v}}}\left({\tau }_{K}))\right\}}^{\sigma }}{J{\left({\tau }_{K})}^{2n}{\left(J\left({\tau }_{K})-1)}^{3n}({f}_{{\bf{u}}}\left({\tau }_{K})-{f}_{{\bf{v}}}\left({\tau }_{K}))}\hspace{-0.33em}\right|\\ & & \hspace{0.1em}\text{for some}\hspace{0.1em}\hspace{0.33em}{\bf{v}}\in \frac{1}{{N}_{{\mathfrak{m}}}}{M}_{1,2}\left({\mathbb{Z}})\setminus {M}_{1,2}\left({\mathbb{Z}})\hspace{0.33em}\hspace{0.1em}\text{such that}\hspace{0.1em}\hspace{0.33em}{\bf{u}}\not\equiv {\bf{v}}\hspace{0.33em}\hspace{0.1em}\text{and}\hspace{0.1em}\hspace{0.33em}-{\bf{v}}\hspace{0.33em}\left(\hspace{0.1em}\text{mod}\hspace{0.1em}\hspace{0.33em}{M}_{1,2}\left({\mathbb{Z}}))\\ & & \hspace{0.1em}\text{by Proposition 3.1(ii) and (4) since}\hspace{0.1em}\hspace{0.33em}\rho \in {\rm{Gal}}\left({K}_{{\mathfrak{m}}}\hspace{0.1em}\text{/}\hspace{0.1em}{H}_{K})\setminus \left\{{{\rm{id}}}_{{K}_{{\mathfrak{m}}}}\right\}\\ & =& {\left|,\frac{J{\left({\tau }_{P})}^{2}{\left(J\left({\tau }_{P})-1)}^{3}}{J{\left({\tau }_{K})}^{2}{\left(J\left({\tau }_{K})-1)}^{3}}\right|}^{n}\left|\hspace{-0.33em},\frac{{f}_{{\bf{u}}^{\prime} }\left({\tau }_{P})-{f}_{{\bf{v}}^{\prime} }\left({\tau }_{P})}{{f}_{{\bf{u}}}\left({\tau }_{K})-{f}_{{\bf{v}}}\left({\tau }_{K})}\hspace{-0.33em}\right|\\ & & \hspace{0.1em}\text{for some}\hspace{0.1em}\hspace{0.33em}{\bf{u}}^{\prime} ,{\bf{v}}^{\prime} \in \frac{1}{{N}_{{\mathfrak{m}}}}{M}_{1,2}\left({\mathbb{Z}})\setminus {M}_{1,2}\left({\mathbb{Z}})\hspace{0.33em}\hspace{0.1em}\text{such that}\hspace{0.1em}\hspace{0.33em}{\bf{u}}^{\prime} \not\equiv {\bf{v}}^{\prime} \hspace{0.33em}\hspace{0.1em}\text{and}\hspace{0.1em}\hspace{0.33em}-{\bf{v}}^{\prime} \hspace{0.33em}\left(\hspace{0.1em}\text{mod}\hspace{0.1em}\hspace{0.33em}{M}_{1,2}\left({\mathbb{Z}}))\\ & & \hspace{0.1em}\text{by Proposition 3.2}\hspace{0.1em}\\ & =& {\left|,\frac{J{\left({\gamma }^{-1}\left({\tau }_{Q}))}^{2}{\left(J\left({\gamma }^{-1}\left({\tau }_{Q}))-1)}^{3}}{J{\left({\tau }_{K})}^{2}{\left(J\left({\tau }_{K})-1)}^{3}}\right|}^{n}\left|\hspace{-0.33em},\frac{{f}_{{\bf{u}}^{\prime} }\left({\gamma }^{-1}\left({\tau }_{Q}))-{f}_{{\bf{v}}^{\prime} }\left({\gamma }^{-1}\left({\tau }_{Q}))}{{f}_{{\bf{u}}}\left({\tau }_{K})-{f}_{{\bf{v}}}\left({\tau }_{K})}\hspace{-0.33em}\right|\hspace{1.0em}\hspace{0.1em}\text{by (14)}\hspace{0.1em}\\ & =& {\left|,\frac{J{\left({\tau }_{Q})}^{2}{\left(J\left({\tau }_{Q})-1)}^{3}}{J{\left({\tau }_{K})}^{2}{\left(J\left({\tau }_{K})-1)}^{3}}\right|}^{n}\left|\hspace{-0.33em},\frac{{f}_{{{\bf{u}}}^{^{\prime\prime} }}\left({\tau }_{Q})-{f}_{{{\bf{v}}}^{^{\prime\prime} }}\left({\tau }_{Q})}{{f}_{{\bf{u}}}\left({\tau }_{K})-{f}_{{\bf{v}}}\left({\tau }_{K})}\hspace{-0.33em}\right|\hspace{1.0em}\hspace{0.1em}\text{with}\hspace{0.1em}\hspace{0.33em}{{\bf{u}}}^{^{\prime\prime} }={\bf{u}}^{\prime} {\gamma }^{-1}\hspace{0.33em}\hspace{0.1em}\text{and}\hspace{0.1em}\hspace{0.33em}{{\bf{v}}}^{^{\prime\prime} }={\bf{v}}^{\prime} {\gamma }^{-1}\\ & & \hspace{0.1em}\text{by the fact\hspace{0.5em}}\hspace{0.1em}J\in {{\mathcal{ {\mathcal F} }}}_{1}\hspace{0.1em}\text{\hspace{0.5em}and Proposition 2.1}\hspace{0.1em}\\ & =& {\left|,\frac{J{\left({\tau }_{Q})}^{2}{\left(J\left({\tau }_{Q})-1)}^{3}}{J{\left({\tau }_{K})}^{2}{\left(J\left({\tau }_{K})-1)}^{3}}\right|}^{n+\tfrac{1}{6}}\left|\hspace{-0.33em},\frac{{g}_{{{\bf{u}}}^{^{\prime\prime} }+{{\bf{v}}}^{^{\prime\prime} }}\left({\tau }_{Q}){g}_{{{\bf{u}}}^{^{\prime\prime} }-{{\bf{v}}}^{^{\prime\prime} }}\left({\tau }_{Q}){g}_{{\bf{u}}}{\left({\tau }_{K})}^{2}{g}_{{\bf{v}}}{\left({\tau }_{K})}^{2}}{{g}_{{\bf{u}}+{\bf{v}}}\left({\tau }_{K}){g}_{{\bf{u}}-{\bf{v}}}\left({\tau }_{K}){g}_{{{\bf{u}}}^{^{\prime\prime} }}{\left({\tau }_{Q})}^{2}{g}_{{{\bf{v}}}^{^{\prime\prime} }}{\left({\tau }_{Q})}^{2}}\hspace{-0.33em}\right|\hspace{1.0em}\hspace{0.1em}\text{by Lemma 2.3}\hspace{0.1em}\\ & \lt & {\left(\text{877,383}| {q}_{{\tau }_{K}}{| }^{\frac{5}{2}}\right)}^{n+\tfrac{1}{6}}\frac{2.2{9}^{6}| {q}_{{\tau }_{Q}}\hspace{-0.25em}{| }^{-\tfrac{1}{12}}| {q}_{{\tau }_{K}}\hspace{-0.25em}{| }^{-\tfrac{1}{6}}}{{\left(\frac{0.76}{{N}_{{\mathfrak{m}}}}\right)}^{6}| {q}_{{\tau }_{K}}\hspace{-0.25em}{| }^{\tfrac{1}{6}}| {q}_{{\tau }_{Q}}\hspace{-0.25em}{| }^{\tfrac{1}{3}}}\hspace{1.0em}\hspace{0.1em}\text{by Lemmas 4.1, 5.1, and Remark 4.2}\hspace{0.1em}\\ & & \hspace{0.1em}\text{because}\hspace{0.1em}\hspace{0.33em}| {q}_{{\tau }_{K}}| ={e}^{-\pi \sqrt{| {d}_{K}| }}\le {e}^{-\pi \sqrt{15}}\hspace{0.33em}\hspace{0.1em}\text{and}\hspace{0.1em}\hspace{0.33em}| {q}_{{\tau }_{Q}}| ={e}^{-\tfrac{\pi \sqrt{| {d}_{K}| }}{{a}_{Q}}}\le {e}^{-\pi \sqrt{3}}\hspace{0.1em}\text{by (6)}\hspace{0.1em}\\ & =& {\left(\text{877,383}| {q}_{{\tau }_{K}}{| }^{\frac{5}{2}}\right)}^{n+\tfrac{1}{6}}{\left(\frac{229}{76}{N}_{{\mathfrak{m}}}\right)}^{6}| {q}_{{\tau }_{Q}}\hspace{-0.25em}{| }^{-\tfrac{5}{12}}| {q}_{{\tau }_{K}}\hspace{-0.25em}{| }^{-\tfrac{1}{3}}\\ & \le & {\left(\text{877,383}| {q}_{{\tau }_{K}}{| }^{\frac{5}{2}}\right)}^{n+\tfrac{1}{6}}{\left(\frac{229}{76}{N}_{{\mathfrak{m}}}\right)}^{6}| {q}_{{\tau }_{K}}\hspace{-0.25em}{| }^{-\tfrac{5}{24}}| {q}_{{\tau }_{K}}\hspace{-0.25em}{| }^{-\tfrac{1}{3}}\\ & & \hspace{0.1em}\text{since}\hspace{0.1em}\hspace{0.33em}| {q}_{{\tau }_{K}}{| }^{-1}\gt 1\hspace{0.33em}\hspace{0.1em}\text{and}\hspace{0.1em}\hspace{0.33em}| {q}_{{\tau }_{Q}}{| }^{-1}={(| {q}_{{\tau }_{K}}{| }^{-1})}^{\tfrac{1}{{a}_{Q}}}\le {(| {q}_{{\tau }_{K}}{| }^{-1})}^{\tfrac{1}{2}}\hspace{0.33em}\hspace{0.1em}\text{by (14)}\hspace{0.1em}\\ & =& {\left(\text{877,383}{e}^{-\frac{5}{2}\pi \sqrt{| {d}_{K}| }}\right)}^{n+\tfrac{1}{6}}{\left(\frac{229}{76}{N}_{{\mathfrak{m}}}\right)}^{6}{e}^{\tfrac{13}{24}\pi \sqrt{| {d}_{K}| }}.\end{array}Now, by taking logarithm we obtain the inequality 0<n+16ln877,383−52π∣dK∣+6ln22976Nm+1324π∣dK∣0\lt \left(n+\frac{1}{6}\right)\left(\mathrm{ln}\hspace{0.1em}\text{877,383}\hspace{0.1em}-\frac{5}{2}\pi \sqrt{| {d}_{K}| }\right)+6\mathrm{ln}\left(\frac{229}{76}{N}_{{\mathfrak{m}}}\right)+\frac{13}{24}\pi \sqrt{| {d}_{K}| }with ln877,383−52π∣dK∣≤ln877,383−52π15<0.\mathrm{ln}\hspace{0.1em}\text{877,383}-\frac{5}{2}\pi \sqrt{| {d}_{K}| }\le \mathrm{ln}\text{877,383}\hspace{0.1em}-\frac{5}{2}\pi \sqrt{15}\lt 0.But this contradicts (12). Therefore, we conclude that Km=K(xK,n(ω)){K}_{{\mathfrak{m}}}=K({x}_{K,n}\left(\omega )).□Example 5.3Let K=Q(−5)K={\mathbb{Q}}\left(\sqrt{-5}), and so dK=−20{d}_{K}=-20. Note that 2 is ramified in Ksince2∣dK,13 is inert in Kdue todK13=−1,23 splits completely in KbecausedK23=1\left\{\begin{array}{ll}\hspace{0.1em}\text{2 is ramified in\hspace{0.5em}}K\text{}\hspace{0.1em}& \hspace{0.1em}\text{since}\hspace{0.1em}\hspace{0.33em}2\hspace{0.33em}| \hspace{0.33em}{d}_{K},\\ \hspace{0.1em}\text{13 is inert in\hspace{0.5em}}K\text{}\hspace{0.1em}& \hspace{0.1em}\text{due to}\hspace{0.1em}\hspace{0.33em}\left(\frac{{d}_{K}}{13}\right)=-1,\\ \hspace{0.1em}\text{23 splits completely in\hspace{0.5em}}K\text{}\hspace{0.1em}& \hspace{0.1em}\text{because}\hspace{0.1em}\hspace{0.33em}\left(\frac{{d}_{K}}{23}\right)=1\end{array}\right.([9, Proposition 5.16]). Let m=p1p2p3{\mathfrak{m}}={{\mathfrak{p}}}_{1}{{\mathfrak{p}}}_{2}{{\mathfrak{p}}}_{3}be the product of three prime ideals p1=[1+−5,2],p2=[13−5,13],andp3=[15+−5,23]{{\mathfrak{p}}}_{1}=\left[1+\sqrt{-5},2],\hspace{1.0em}{{\mathfrak{p}}}_{2}=\left[13\sqrt{-5},13],\hspace{1.0em}\hspace{0.1em}\text{and}\hspace{0.1em}\hspace{1.0em}{{\mathfrak{p}}}_{3}=\left[15+\sqrt{-5},23]of OK{{\mathcal{O}}}_{K}satisfying 2OK=p122{{\mathcal{O}}}_{K}={{\mathfrak{p}}}_{1}^{2}, 13OK=p213{{\mathcal{O}}}_{K}={{\mathfrak{p}}}_{2}, and 23OK=p3p¯323{{\mathcal{O}}}_{K}={{\mathfrak{p}}}_{3}{\overline{{\mathfrak{p}}}}_{3}. In this case, by checking the degree formula for [Km:HK]\left[{K}_{{\mathfrak{m}}}:{H}_{K}]we see that Km{K}_{{\mathfrak{m}}}properly contains HK{H}_{K}. Since m=p1p2p3⊃p12p2p3p¯3=(2⋅13⋅23)OK,{\mathfrak{m}}={{\mathfrak{p}}}_{1}{{\mathfrak{p}}}_{2}{{\mathfrak{p}}}_{3}\supset {{\mathfrak{p}}}_{1}^{2}{{\mathfrak{p}}}_{2}{{\mathfrak{p}}}_{3}{\overline{{\mathfrak{p}}}}_{3}=\left(2\cdot 13\cdot 23){{\mathcal{O}}}_{K},we obtain Nm=2⋅13⋅23=598{N}_{{\mathfrak{m}}}=2\cdot 13\cdot 23=598, and hence one can estimate 1324π∣dK∣+6ln22976Nm52π∣dK∣−ln877,383−16=1324π20+6ln22976⋅59852π20−ln877,383−16≈2.286282.\frac{\frac{13}{24}\pi \sqrt{| {d}_{K}| }+6\mathrm{ln}\left(\frac{229}{76}{N}_{{\mathfrak{m}}}\right)}{\frac{5}{2}\pi \sqrt{| {d}_{K}| }-\mathrm{ln}\hspace{0.1em}\text{877,383}\hspace{0.1em}}-\frac{1}{6}=\frac{\frac{13}{24}\pi \sqrt{20}+6\mathrm{ln}\left(\frac{229}{76}\cdot 598\right)}{\frac{5}{2}\pi \sqrt{20}-\mathrm{ln}\hspace{0.1em}\text{877,383}\hspace{0.1em}}-\frac{1}{6}\approx 2.286282.If ω\omega is an element of KKsuch that ω+OK\omega +{{\mathcal{O}}}_{K}is a generator of the OK{{\mathcal{O}}}_{K}-module m−1/OK{{\mathfrak{m}}}^{-1}\hspace{0.1em}\text{/}\hspace{0.1em}{{\mathcal{O}}}_{K}, then we obtain by Theorem 5.2 that Km=K(xK,n(ω))for alln≥3.{K}_{{\mathfrak{m}}}=K\left({x}_{K,n}\left(\omega ))\hspace{1.0em}\hspace{0.1em}\text{for all}\hspace{0.1em}\hspace{0.33em}n\ge 3.Remark 5.4At this stage, we conjecture that Theorem 5.2 may hold for every n∈Z≥0n\in {{\mathbb{Z}}}_{\ge 0}.6Ray class fields of special moduliLet KKbe an imaginary quadratic field other than Q(−1){\mathbb{Q}}\left(\sqrt{-1})and Q(−3){\mathbb{Q}}\left(\sqrt{-3}), and let NN(≥2\ge 2) be an integer whose prime factors are all inert in KK. In this last section, we consider the special case where m=NOK{\mathfrak{m}}=N{{\mathcal{O}}}_{K}and show that Theorem 5.2 is also true for every n∈Z≥0n\in {{\mathbb{Z}}}_{\ge 0}.Lemma 6.1Let f∈ℱ1f\in {{\mathcal{ {\mathcal F} }}}_{1}. If f has neither a zero nor a pole on H{\mathbb{H}}, then it is a nonzero rational number.ProofSee [2, Theorem 2 in Chapter 5] and [17, Lemma 2.1].□For an integer N≥2N\ge 2, let SN=st∈M1,2(Z)∣0≤s,t<Nandgcd(N,s,t)=1.{S}_{N}=\left\{\left[\begin{array}{cc}s& t\end{array}\right]\in {M}_{1,2}\left({\mathbb{Z}})| 0\le s,t\lt N\hspace{0.33em}\hspace{0.1em}\text{and}\hspace{0.1em}\hspace{0.33em}\gcd \left(N,s,t)=1\right\}.We define an equivalence relation ≡N{\equiv }_{N}on the set SN{S}_{N}as follows: for u,v∈SN{\bf{u}},{\bf{v}}\in {S}_{N}u≡Nv⇔u≡vor−v(modNM1,2(Z)).{\bf{u}}{\equiv }_{N}{\bf{v}}\hspace{1.0em}\iff \hspace{1.0em}{\bf{u}}\equiv {\bf{v}}\hspace{0.33em}\text{or}\hspace{0.33em}-{\bf{v}}\hspace{0.33em}\left(\hspace{0.1em}\text{mod}\hspace{0.1em}\hspace{0.33em}N{M}_{1,2}\left({\mathbb{Z}})).Let PN={(u,v)∣u,v∈SNsuch thatu≢Nv}andmN=∣PN∣.{P}_{N}=\{\left({\bf{u}},{\bf{v}})| {\bf{u}},{\bf{v}}\in {S}_{N}\hspace{0.33em}\hspace{0.1em}\text{such that}\hspace{0.1em}\hspace{0.33em}{\bf{u}}{\not\equiv }_{N}{\bf{v}}\}\hspace{1.0em}\hspace{0.1em}\text{and}\hspace{0.1em}\hspace{1.0em}{m}_{N}=| {P}_{N}| .Since 10,01\left[\begin{array}{cc}1& 0\end{array}\right],\left[\begin{array}{cc}0& 1\end{array}\right]represent distinct classes in SN/≡N{S}_{N}\hspace{0.1em}\text{/}\hspace{0.1em}{\equiv }_{N}, we claim mN≥2{m}_{N}\ge 2.Lemma 6.2If N is an integer such that N≥2N\ge 2, then we have∏(u,v)∈PNf1Nu−f1Nv6=k{J2(J−1)3}mNforsomek∈Q⧹{0}.\prod _{\left({\bf{u}},{\bf{v}})\in {P}_{N}}{\left({f}_{\tfrac{1}{N}{\bf{u}}}-{f}_{\tfrac{1}{N}{\bf{v}}}\right)}^{6}=k{\{{J}^{2}{\left(J-1)}^{3}\}}^{{m}_{N}}\hspace{1.0em}{for}\hspace{0.33em}some\hspace{0.33em}k\in {\mathbb{Q}}\setminus \left\{0\right\}.ProofFor ab∈M1,2(Z)\left[\begin{array}{cc}a& b\end{array}\right]\in {M}_{1,2}\left({\mathbb{Z}})with gcd(N,a,b)=1\gcd \left(N,a,b)=1, let πNab{\pi }_{N}\left(\left[\begin{array}{cc}a& b\end{array}\right]\right)denote the unique element of SN{S}_{N}satisfying πNab≡ab(modNM1,2(Z)).{\pi }_{N}\left(\left[\begin{array}{cc}a& b\end{array}\right]\right)\equiv \left[\begin{array}{cc}a& b\end{array}\right]\hspace{0.33em}\left(\hspace{0.1em}\text{mod}\hspace{0.1em}\hspace{0.33em}N{M}_{1,2}\left({\mathbb{Z}})).Let α∈M2(Z)\alpha \in {M}_{2}\left({\mathbb{Z}})with gcd(N,det(α))=1\gcd \left(N,\det \left(\alpha ))=1, and let α˜\widetilde{\alpha }be its image in GL2(Z/NZ)/⟨−I2⟩{{\rm{GL}}}_{2}\left({\mathbb{Z}}\hspace{0.1em}\text{/}\hspace{0.1em}N{\mathbb{Z}})\hspace{0.1em}\text{/}\hspace{0.1em}\langle -{I}_{2}\rangle (≃Gal(ℱN/ℱ1)\simeq {\rm{Gal}}\left({{\mathcal{ {\mathcal F} }}}_{N}\hspace{0.1em}\text{/}\hspace{0.1em}{{\mathcal{ {\mathcal F} }}}_{1})). Setting f=∏(u,v)∈PNf1Nu−f1Nv6,f=\prod _{\left({\bf{u}},{\bf{v}})\in {P}_{N}}{\left({f}_{\tfrac{1}{N}{\bf{u}}}-{f}_{\tfrac{1}{N}{\bf{v}}}\right)}^{6},we find that fα˜=∏(u,v)∈PNf1Nuα˜−f1Nvα˜6by Proposition 2.1=∏(u,v)∈PNf1NπN(uα)−f1NπN(vα)6by (4)=f{f}^{\widetilde{\alpha }}=\prod _{\left({\bf{u}},{\bf{v}})\in {P}_{N}}{\left({f}_{\tfrac{1}{N}{\bf{u}}\widetilde{\alpha }}-{f}_{\tfrac{1}{N}{\bf{v}}\widetilde{\alpha }}\right)}^{6}\hspace{1.0em}\hspace{0.1em}\text{by Proposition 2.1}\hspace{0.1em}=\prod _{\left({\bf{u}},{\bf{v}})\in {P}_{N}}{\left({f}_{\tfrac{1}{N}{\pi }_{N}\left({\bf{u}}\alpha )}-{f}_{\tfrac{1}{N}{\pi }_{N}\left({\bf{v}}\alpha )}\right)}^{6}\hspace{1.0em}\hspace{0.1em}\text{by (4)}\hspace{0.1em}=fbecause the mapping SN→SN{S}_{N}\to {S}_{N}, u↦πN(uα){\bf{u}}\mapsto {\pi }_{N}\left({\bf{u}}\alpha ), gives rise to an injection (and so, a bijection) of PN{P}_{N}into itself. This observation implies by the Galois theory that fflies in ℱ1{{\mathcal{ {\mathcal F} }}}_{1}.On the other hand, we attain f=∏(u,v)∈PNJ2(J−1)339g1N(u+v)6g1N(u−v)6g1Nu12g1Nv12by Lemma 2.3=gJ2(J−1)339mNwithg=∏(u,v)∈PNg1N(u+v)6g1N(u−v)6g1Nu12g1Nv12.\begin{array}{rcl}f& =& \displaystyle \prod _{\left({\bf{u}},{\bf{v}})\in {P}_{N}}\frac{{J}^{2}{\left(J-1)}^{3}}{{3}^{9}}\frac{{g}_{\frac{1}{N}\left({\bf{u}}+{\bf{v}})}^{6}{g}_{\frac{1}{N}\left({\bf{u}}-{\bf{v}})}^{6}}{{g}_{\frac{1}{N}{\bf{u}}}^{12}{g}_{\frac{1}{N}{\bf{v}}}^{12}}\hspace{1.0em}\hspace{0.1em}\text{by Lemma 2.3}\hspace{0.1em}\\ & =& g{\left\{\frac{{J}^{2}{\left(J-1)}^{3}}{{3}^{9}}\right\}}^{{m}_{N}}\hspace{1.0em}\hspace{0.1em}\text{with}\hspace{0.1em}\hspace{0.33em}g=\displaystyle \prod _{\left({\bf{u}},{\bf{v}})\in {P}_{N}}\frac{{g}_{\frac{1}{N}\left({\bf{u}}+{\bf{v}})}^{6}{g}_{\frac{1}{N}\left({\bf{u}}-{\bf{v}})}^{6}}{{g}_{\frac{1}{N}{\bf{u}}}^{12}{g}_{\frac{1}{N}{\bf{v}}}^{12}}.\end{array}Since ffand JJbelong to ℱ1{{\mathcal{ {\mathcal F} }}}_{1}, so does gg. Moreover, since gghas neither a zero nor a pole on H{\mathbb{H}}, it is a nonzero rational number by Lemma 6.1. Therefore, we obtain □f=k{J2(J−1)3}mNfor somek∈Q⧹{0}.f=k{\{{J}^{2}{\left(J-1)}^{3}\}}^{{m}_{N}}\hspace{1.0em}\hspace{0.1em}\text{for some}\hspace{0.1em}\hspace{0.33em}k\in {\mathbb{Q}}\setminus \left\{0\right\}.Lemma 6.3Let KKbe an imaginary quadratic field and NNbe an integer with N≥2N\ge 2. If every prime factor of N is inert in KK, then the principal ideal (sτK+t)OK\left(s{\tau }_{K}+t){{\mathcal{O}}}_{K}is relatively prime to NOKN{{\mathcal{O}}}_{K}for all s,t∈Zs,t\in {\mathbb{Z}}such that gcd(N,s,t)=1\gcd \left(N,s,t)=1.ProofWe see that NK/Q(sτK+t)=(sτK+t)(sτ¯K+t)=τKτ¯Ks2+(τK+τ¯K)st+t2=cKs2−bKst+t2.{{\rm{N}}}_{K\text{/}{\mathbb{Q}}}\left(s{\tau }_{K}+t)=\left(s{\tau }_{K}+t)\left(s{\overline{\tau }}_{K}+t)={\tau }_{K}{\overline{\tau }}_{K}{s}^{2}+\left({\tau }_{K}+{\overline{\tau }}_{K})st+{t}^{2}={c}_{K}{s}^{2}-{b}_{K}st+{t}^{2}.Now, we claim that NK/Q(sτK+t){{\rm{N}}}_{K\text{/}{\mathbb{Q}}}\left(s{\tau }_{K}+t)is relatively prime to NN. Indeed, we have two cases: dK≡0(mod4){d}_{K}\equiv 0\hspace{0.33em}\left(\hspace{0.1em}\text{mod}\hspace{0.1em}\hspace{0.33em}4)or dK≡1(mod4){d}_{K}\equiv 1\hspace{0.33em}\left(\hspace{0.1em}\text{mod}\hspace{0.1em}\hspace{0.33em}4). Case 1. Consider the case where dK≡0(mod4){d}_{K}\equiv 0\hspace{0.33em}\left(\hspace{0.1em}\text{mod}\hspace{0.1em}\hspace{0.33em}4), and then bK=0{b}_{K}=0and cK=−dK4{c}_{K}=-\frac{{d}_{K}}{4}. Let ppbe a prime factor of NN. Since ppis inert in KK, it must be odd and satisfy dKp=−1\left(\frac{{d}_{K}}{p}\right)=-1. If NK/Q(sτK+t)=−dK4s2+t2≡0(modp),{{\rm{N}}}_{K\text{/}{\mathbb{Q}}}\left(s{\tau }_{K}+t)=-\frac{{d}_{K}}{4}{s}^{2}+{t}^{2}\equiv 0\hspace{0.33em}\left(\hspace{0.1em}\text{mod}\hspace{0.1em}\hspace{0.33em}p),then the fact dKp=−1\left(\frac{{d}_{K}}{p}\right)=-1forces us to obtain s≡0(modp)s\equiv 0\hspace{0.33em}\left(\hspace{0.1em}\text{mod}\hspace{0.1em}\hspace{0.33em}p)and so t≡0(modp)t\equiv 0\hspace{0.33em}\left(\hspace{0.1em}\text{mod}\hspace{0.1em}\hspace{0.33em}p). But this contradicts the fact gcd(N,s,t)=1\gcd \left(N,s,t)=1. Therefore, NK/Q(sτK+t){{\rm{N}}}_{K\text{/}{\mathbb{Q}}}\left(s{\tau }_{K}+t)is relatively prime to pp, and hence to NN.Case 2. Let dK≡1(mod4){d}_{K}\equiv 1\hspace{0.33em}\left(\hspace{0.1em}\text{mod}\hspace{0.1em}\hspace{0.33em}4), and so bK=1{b}_{K}=1and cK=1−dK4{c}_{K}=\frac{1-{d}_{K}}{4}. Let ppbe a prime factor of NN. Since ppis inert in KK, we derive dK≡5(mod8)ifp=2,dKp=−1ifp>2.\left\{\begin{array}{ll}{d}_{K}\equiv 5\hspace{0.33em}\left(\hspace{0.1em}\text{mod}\hspace{0.1em}\hspace{0.33em}8)& \hspace{0.1em}\text{if}\hspace{0.1em}\hspace{0.33em}p=2,\\ \left(\frac{{d}_{K}}{p}\right)=-1& \hspace{0.1em}\text{if}\hspace{0.1em}\hspace{0.33em}p\gt 2.\end{array}\right.Then we find that NK/Q(sτK+t)=1−dK4s2−st+t2≡s2+st+t2(modp)ifp=2,4′{(s−2t)2−dKs2}(modp)ifp>2,{{\rm{N}}}_{K\text{/}{\mathbb{Q}}}\left(s{\tau }_{K}+t)=\frac{1-{d}_{K}}{4}{s}^{2}-st+{t}^{2}\equiv \left\{\begin{array}{ll}{s}^{2}+st+{t}^{2}\hspace{0.33em}\left(\hspace{0.1em}\text{mod}\hspace{0.1em}\hspace{0.33em}p)& \hspace{0.1em}\text{if}\hspace{0.1em}\hspace{0.33em}p=2,\\ 4^{\prime} \{{\left(s-2t)}^{2}-{d}_{K}{s}^{2}\}\hspace{0.33em}\left(\hspace{0.1em}\text{mod}\hspace{0.1em}\hspace{0.33em}p)& \hspace{0.1em}\text{if}\hspace{0.1em}\hspace{0.33em}p\gt 2,\end{array}\right.where 4′4^{\prime} is an integer such that 4⋅4′≡1(modp)4\cdot 4^{\prime} \equiv 1\hspace{0.33em}\left(\hspace{0.1em}\text{mod}\hspace{0.1em}\hspace{0.33em}p). When p=2p=2, we see that s2+st+t2≡1(modp){s}^{2}+st+{t}^{2}\equiv 1\hspace{0.33em}\left(\hspace{0.1em}\text{mod}\hspace{0.1em}\hspace{0.33em}p)because ssand ttare not both even. When p>2p\gt 2, if NK/Q(sτK+t)≡0(modp){{\rm{N}}}_{K\text{/}{\mathbb{Q}}}\left(s{\tau }_{K}+t)\equiv 0\hspace{0.33em}\left(\hspace{0.1em}\text{mod}\hspace{0.1em}\hspace{0.33em}p), then the fact dKp=−1\left(\frac{{d}_{K}}{p}\right)=-1yields that s≡0(modp)s\equiv 0\hspace{0.33em}\left(\hspace{0.1em}\text{mod}\hspace{0.1em}\hspace{0.33em}p)and so t≡0(modp)t\equiv 0\hspace{0.33em}\left(\hspace{0.1em}\text{mod}\hspace{0.1em}\hspace{0.33em}p). But this again contradicts gcd(N,s,t)=1\gcd \left(N,s,t)=1. Hence NK/Q(sτK+t){{\rm{N}}}_{K\text{/}{\mathbb{Q}}}\left(s{\tau }_{K}+t)is relatively prime to pp, and so to NN. Therefore, the principal ideal (sτK+t)OK\left(s{\tau }_{K}+t){{\mathcal{O}}}_{K}is relatively prime to NOKN{{\mathcal{O}}}_{K}.□Theorem 6.4Let KKbe an imaginary quadratic field other than Q(−1){\mathbb{Q}}\left(\sqrt{-1})and Q(−3){\mathbb{Q}}\left(\sqrt{-3}), and let NNbe an integer such that N≥2N\ge 2. Let ω\omega be an element of KKso that ω+OK\omega +{{\mathcal{O}}}_{K}is a generator of the OK{{\mathcal{O}}}_{K}-module N−1OK/OK{N}^{-1}{{\mathcal{O}}}_{K}\hspace{0.1em}\text{/}\hspace{0.1em}{{\mathcal{O}}}_{K}. If every prime factor of NNis inert in KK, then we attainK(N)=K(xK,n(ω))foreveryn∈Z≥0.{K}_{\left(N)}=K({x}_{K,n}\left(\omega ))\hspace{1.0em}{for}\hspace{0.33em}every\hspace{0.33em}n\in {{\mathbb{Z}}}_{\ge 0}.ProofSince K(N){K}_{\left(N)}is an abelian extension of KK, K(xK,n(ω))K\left({x}_{K,n}\left(\omega ))is also an abelian extension of KKcontaining xK,n1N{x}_{K,n}\left(\frac{1}{N}\right)by Proposition 3.1 (ii). Since (sτK+t)OK∈PK(NOK)for allst∈SN\left(s{\tau }_{K}+t){{\mathcal{O}}}_{K}\in {P}_{K}\left(N{{\mathcal{O}}}_{K})\hspace{1.0em}\hspace{0.1em}\text{for all}\hspace{0.1em}\left[\begin{array}{cc}s& t\end{array}\right]\in {S}_{N}by Lemma 6.3, we obtain by Proposition 3.1 (ii) that xK,nsτK+tN∈K(xK,n(ω))for allst∈SN.{x}_{K,n}\left(\frac{s{\tau }_{K}+t}{N}\right)\in K\left({x}_{K,n}\left(\omega ))\hspace{1.0em}\hspace{0.1em}\text{for all}\hspace{0.1em}\hspace{0.33em}\left[\begin{array}{cc}s& t\end{array}\right]\in {S}_{N}.We then deduce that K(xK,n(ω))∋∏(u,v)∈PNxK,nu1τK+u2N−xK,nv1τK+v2N6withu=u1u2andv=v1v2=∏(u,v)∈PNℓKnhKφKu1τK+u2N−ℓKnhKφKv1τK+v2N6by (3)=∏(u,v)∈PNℓKn27356f1Nu(τK)−f1Nv(τK)6by the definitions of φK,hK and Fricke functions=ℓKn27356mN∏(u,v)∈PNf1Nu−f1Nv6(τK)=ℓKn27356mN[k{J2(J−1)3}mN](τK)for some k∈Q⧹{0} by Lemma 6.2=kℓK6n+1242330mN.\begin{array}{rcl}K\left({x}_{K,n}\left(\omega ))& \ni & \displaystyle \prod _{\left({\bf{u}},{\bf{v}})\in {P}_{N}}{\left({x}_{K,n}\left(\frac{{u}_{1}{\tau }_{K}+{u}_{2}}{N}\right)-{x}_{K,n}\left(\frac{{v}_{1}{\tau }_{K}+{v}_{2}}{N}\right)\right)}^{6}\hspace{1em}\hspace{0.1em}\text{with}\hspace{0.1em}\hspace{0.33em}{\bf{u}}=\left[\begin{array}{cc}{u}_{1}& {u}_{2}\end{array}\right]\hspace{0.33em}\hspace{0.1em}\text{and}\hspace{0.1em}\hspace{0.33em}{\bf{v}}=\left[\begin{array}{cc}{v}_{1}& {v}_{2}\end{array}\right]\\ & =& \displaystyle \prod _{\left({\bf{u}},{\bf{v}})\in {P}_{N}}{\left({{\ell }_{K}}^{n}{{\mathfrak{h}}}_{K}\left({\varphi }_{K}\left(\left[\frac{{u}_{1}{\tau }_{K}+{u}_{2}}{N}\right]\right)\right)-{{\ell }_{K}}^{n}{{\mathfrak{h}}}_{K}\left({\varphi }_{K}\left(\left[\frac{{v}_{1}{\tau }_{K}+{v}_{2}}{N}\right]\right)\right)\right)}^{6}\hspace{1.0em}\hspace{0.1em}\text{by (3)}\hspace{0.1em}\\ & =& \displaystyle \prod _{\left({\bf{u}},{\bf{v}})\in {P}_{N}}\left\{{\left(\frac{{{\ell }_{K}}^{n}}{{2}^{7}{3}^{5}}\right)}^{6}{\left({f}_{\tfrac{1}{N}{\bf{u}}}\left({\tau }_{K})-{f}_{\tfrac{1}{N}{\bf{v}}}\left({\tau }_{K})\right)}^{6}\right\}\\ & & \hspace{0.1em}\text{by the definitions of\hspace{0.5em}}\hspace{0.1em}{\varphi }_{K},{{\mathfrak{h}}}_{K}\hspace{0.1em}\text{\hspace{0.5em}and Fricke functions}\hspace{0.1em}\\ & =& {\left(\frac{{{\ell }_{K}}^{n}}{{2}^{7}{3}^{5}}\right)}^{6{m}_{N}}\left\{\displaystyle \prod _{\left({\bf{u}},{\bf{v}})\in {P}_{N}}{\left({f}_{\tfrac{1}{N}{\bf{u}}}-{f}_{\tfrac{1}{N}{\bf{v}}}\right)}^{6}\right\}\left({\tau }_{K})\\ & =& {\left(\frac{{{\ell }_{K}}^{n}}{{2}^{7}{3}^{5}}\right)}^{6{m}_{N}}{[}k{\{{J}^{2}{\left(J-1)}^{3}\}}^{{m}_{N}}]\left({\tau }_{K})\hspace{1.0em}\hspace{0.1em}\text{for some\hspace{0.5em}}\hspace{0.1em}k\in {\mathbb{Q}}\setminus \left\{0\right\}\hspace{0.1em}\text{\hspace{0.5em}by Lemma 6.2}\hspace{0.1em}\\ & =& k{\left(\frac{{{\ell }_{K}}^{6n+1}}{{2}^{42}{3}^{30}}\right)}^{{m}_{N}}.\end{array}Therefore, we achieve that □K(xK,n(ω))=KxK,n(ω),kℓK6n+1242330mN=HK(xK,n(ω))by Lemma 4.3=K(N)by (3) and Proposition 3.1.\begin{array}{rcl}K\left({x}_{K,n}\left(\omega ))& =& K\left({x}_{K,n}\left(\omega ),k{\left(\frac{{{\ell }_{K}}^{6n+1}}{{2}^{42}{3}^{30}}\right)}^{{m}_{N}}\right)\\ & =& {H}_{K}({x}_{K,n}\left(\omega ))\hspace{1.0em}\hspace{0.1em}\text{by Lemma 4.3}\hspace{0.1em}\\ & =& {K}_{\left(N)}\hspace{1.0em}\hspace{0.1em}\text{by (3) and Proposition 3.1}.\end{array}
Journal
Open Mathematics – de Gruyter
Published: Jan 1, 2022
Keywords: class field theory; complex multiplication; modular functions; 11F03; 11G15; 11R37