The Recursive Structures of Manin Symbols over \(\mathbb{Q}\), Cusps and Elliptic Points on \(X_0(N)\)

1. Introduction

In his seminal monograph [5], G. Shimura defined a complete set $D(N)$ of representatives for the projective line ${\mathbb{P}}^1(\mathbb{Z}/N\mathbb{Z})$ over $\mathbb{Z}/N\mathbb{Z}$ to be all couples $\{c,d\}$ of positive integers satisfying

$$(\ast )(c,d)=1,d|N,1⩽c⩽N/d(orcinanysetofrepresentativesfor\mathbb{Z}modulo(N/d)),$$

where $(c,d)$ denote the greatest common divisor of integers c and d.

Let $[x]$ to be the greatest integer less than or equal to x. For two integers $a,b$, define

$${[\frac{a}{b}]}^{\prime }=\left\{\begin{array}{c}\frac{a}{b}-1\mathrm{if}b|a,\hfill \\ [\frac{a}{b}]\mathrm{otherwise},\hfill \end{array}\right.$$

Then $1⩽a-b{[\frac{a}{b}]}^{\prime }⩽b$. In this paper, we define

$$\begin{array}{c}\hfill D(N)=\{(c,d):c,d\in \mathbb{Z},c,d⩾1,c|N,(c,d)=1\mathrm{and}\\ \hfill (c,d-\frac{N}{c}({[\frac{cd}{N}]}^{\prime }-n))⩾2\mathrm{for}0⩽n<{[\frac{cd}{N}]}^{\prime }\}.\end{array}$$

(1)

We then establish a bijection between $D(M)\times D(N)$ and $D(MN)$ for $(M,N)=1$ in Section 2. This result gives a recursive algorithm to construct the projective line ${\mathbb{P}}^1(\mathbb{Z}/N\mathbb{Z})$ over $\mathbb{Z}/N\mathbb{Z}$.

Let $\Pi (N)=\{[\delta ;amod(\delta ,N{\delta }^{-1})]:a,\delta \in \mathbb{Z},\delta ⩾1,\delta |N,1⩽a⩽(\delta ,N{\delta }^{-1})\}$. In [2], Ju. I. Manin proved that there exists a bijection between $\Pi (N)$ and the set of cusps on $X_0(N)$. Based on Manin’s result and Cremona’s characterization(See Proposition 3), we identify $\Pi (N)$ with

$$\begin{array}{c}\hfill C(N)=\{(c,d):c,d\in \mathbb{Z},1⩽c⩽N,c|N,(c,d)=1\mathrm{and}\\ \hfill (c,d-(c,N{c}^{-1}){[\frac{d}{(c,N{c}^{-1})}]}^{\prime }+\frac{Nn}{c})⩾2\mathrm{for}0⩽n<\frac{c(c,N{c}^{-1})}{N}{[\frac{d}{(c,N{c}^{-1})}]}^{\prime }\},\end{array}$$

(2)

which is a subset of $D(N)$. In Section 3, we establish a bijection between $C(N_1{N}_2)$ and $C(N_1)\times C(N_2)$ for $(N_1,N_2)=1$. This result gives a recursive algorithm to construct the complete set of representatives of ${\Gamma }_0(N)$-inequivalent cusps.

Define

$$\begin{array}{cc}\hfill E_2(N)& =\{(1,d):(1,d)\in D(N),1+d^2\equiv 0(\mathrm{mod}N)\},\hfill \\ \hfill E_3(N)& =\{(1,d):(1,d)\in D(N),1-d+d^2\equiv 0(\mathrm{mod}N)\}.\hfill \end{array}$$

(3)

Then there exist bijections between $E_2(N),E_3(N)$ and complete sets of representatives of ${\Gamma }_0(N)$-inequivalent elliptic points of order 2, 3, respectively. In Section 4, we establish bijections between $E_2(N_1{N}_2)$ and $E_2(N_1)\times E_2(N_2),E_3(N_1{N}_2)$ and $E_3(N_1)\times E_3(N_2)$, for $(N_1,N_2)=1$. These results give a recursive algorithm for constructing the complete set $E_3(N)$ and $E_2(N)$of ${\Gamma }_0(N)$-inequivalent elliptic points of order 2, 3.

The elements in ${\mathbb{P}}^1(\mathbb{Z}/N\mathbb{Z})$ are called Manin symbols and there exists a bijection between the set of right cosets of ${\Gamma }_0(N)$ in $SL(2,\mathbb{Z})$ and ${\mathbb{P}}^1(\mathbb{Z}/N\mathbb{Z})$. The important steps in the modular elliptic algorithm are to construct the complete set $D(N)$ of representatives for the projective line ${\mathbb{P}}^1(\mathbb{Z}/N\mathbb{Z})$ and the complete set $C(N)$ of representatives of ${\Gamma }_0(N)$-inequivalent cusps. The recursive structure of $D(N),C(N),E_2(N)$ and $E_3(N)$ may give rise to a more efficient modular elliptic algorithm.

2. The recursive structure of Manin symbols over $\mathbb{Q}$

We firstly give some necessary notations and facts, for details, See [1].

Definition 1.

(a) 

$D_2(N)=\{(c,d):c,d\in \mathbb{Z},(c,d,N)=1\}$;

(b) 

$\forall (c_1,d_1),(c_2,d_2)\in D_2(N),$ define $(c_1,d_1)\sim (c_2,d_2)$ if $c_1{d}_2\equiv d_1{c}_2(modN)$,

then ∼ is an equivalence relation on $D_2(N)$;

(c) 

$\forall (c,d)\in D_2(N)$, define $(c:d)=\left\{(c^{\prime },d^{\prime }):(c^{\prime },d^{\prime })\in D_2(N),(c^{\prime },d^{\prime })\sim (c,d)\right\}$;

(d) 

$\mathcal{D}(N)=D_2(N)/\sim =\{(c:d):(c,d)\in D_0(N)\}$;

(e) 

$D_1(N)=\left\{(c,d):c,d\in \mathbb{Z},c,d⩾1,c|N,(c,d,\frac{N}{c})=1,cd⩽N\right\}$;

(f) 

$D(N)$ is defined in (1);

(g) 

$\mu (N)$, $v_{\infty }(N)$, $v_2(N)$ and $v_3(N)$ are the numbers of elements in $D(N),C(N),E_2(N)$ and $E_3(N)$, respectively.

Lemma 1.

Let $c,d,h\in \mathbb{Z}$, $(c,d,h)=1$,$c,d⩾1$ and $d⩽h$ then there exists an integer k such that $(c,d+hk)=1$ and $0⩽k<c$.

Proof. 

If $c=1$, take $k=0$ then $(c,d+hk)=1$. Thus let $c⩾2$ in the following. Let $c=p_1^{{\alpha }_1}\cdots p_s^{{\alpha }_s}$ be the standard factorization of c. The proof is by induction on the numbers of distinct prime divisors in c. Suppose that $c=p_1^{{\alpha }_1}$. Assume that $(p_1^{{\alpha }_1},d)⩾2$ and $(p_1^{{\alpha }_1},d+h)⩾2$ then $p_1|d$ and $p_1|d+h$. Thus $p_1|d$ and $p_1|h$, this contradicts with $(c,d,h)=1$ and hence $(c,d+hk)=1$ for some $0⩽k⩽1<c$.

Let $c_1=p_1^{{\alpha }_1}\cdots p_{s-1}^{{\alpha }_{s-1}}$. By the induction hypothesis, there exists an integer $k_1$ such that $(c_1,d+h{k}_1)=1$ and $0⩽k_1<c_1$. Then $(c_1,d+h{k}_1+h{c}_1)=1$. Assume that $(p_s^{{\alpha }_s},d+h{k}_1)⩾2$ and $(p_s^{{\alpha }_s},d+h{k}_1+h{c}_1)⩾2$ then $p_s|d+h{k}_1$ and $p_s|d+h{k}_1+h{c}_1$. Thus $p_s|h{c}_1$ and hence $p_s|h$ by $(p_s,c_1)=1$. Therefore $p_s|d$. This contradicts with $(c,d,h)=1$ and hence $(c,d+h{k}_1)=1$ or $(c,d+h{k}_1+h{c}_1)=1$. Take $k=k_1$ or$k=c_1+k_1$, then $(c,d+hk)=1$ for some $0⩽k_1⩽k⩽c_1+k_1<2c_1⩽c$. This completes the proof by the induction principal. □

Corollary 1.

Let $a,b,c\in \mathbb{Z}$, $(a,b,c)=1$ then the equation $ax+by+cyz=1$ has solutions in $\mathbb{Z}$.

Lemma 2.

There exists a bijection between $D(N)$ and $D_1(N)$.

Proof. 

Let $(c,d)\in D(N)$. Define $d_n=d-\frac{N}{c}{[\frac{cd}{N}]}^{\prime }+\frac{Nn}{c}$ for all $n\in \mathbb{Z}$. Then $1⩽d_0⩽\frac{N}{c}$ and $(c,d_0,\frac{N}{c})=1$ by $(c,d)=1$. Thus $(c,d_0)\in D_1(N)$. Define $\mathsf{\Phi }:D(N)\to D_1(N)$ by sending $(c,d)$ to $(c,d_0)$.

Let $(u,v)\in D(N)$ such that $\mathsf{\Phi }(c,d)=\mathsf{\Phi }(u,v)$. Define $v_n=v-\frac{N}{u}{[\frac{uv}{N}]}^{\prime }+\frac{Nn}{u}$ for all $n\in \mathbb{Z}$. Then $c=u$ and $d_0=v_0$. Thus $d_n=v_n$ for all $n\in \mathbb{Z}$. Let $e={[\frac{cd}{N}]}^{\prime }$ and $w={[\frac{cv}{N}]}^{\prime }$. Then $d=d_e$ and $v=v_w$. Suppose that $e<w$ then $(c,d_e)=1$ by $(c,d)=1$ but $(c,d_e)⩾2$ by $(c,v)\in D(N)$ and $d_e=v_e$, a contradiction and thus $e⩾w$. $e⩽w$ holds by a similar proof and thus $e=w$ and $(c,d)=(u,v)$. Therefore $\mathsf{\Phi }$ is an injection from $D(N)$ to $D_1(N)$.

Let $(c,d_0)\in D_1(N)$. By Lemma 1, there exists an integer k such that $(c,d_0+\frac{Nk}{c})=1$ and $0⩽k⩽c-1$. Let $0⩽k_0⩽k$ such that $(c,d_0+\frac{N{k}_0}{c})=1$ and $(c,d_0+\frac{Nn}{c})=1$ for all $0⩽n<k_0$. Define $d=d_0+\frac{N{k}_0}{c}$. Then $(c,d)\in D(N)$ and $\mathsf{\Phi }((c,d))=(c,d_0)$. Therefore $\mathsf{\Phi }$ is a surjection from $D(N)$ to $D_1(N)$. □

Lemma 3.

There exists a bijection between $\mathcal{D}(N)$ and $D(N)$, i.e., $D(N)$ is a complete system of the representatives of elements of $\mathcal{D}(N)$.

Proof. 

Define $\mathsf{\Phi }:D(N)\to \mathcal{D}(N)$ by the natural map, i.e., $\mathsf{\Phi }((c,d))=(c:d)$.

Let $(c:d)\in \mathcal{D}(N)$. Then $(c,d,N)=1$. Define $c_1=(c,N)$, $d_0$ to be the unique solution of the congruence equation $\frac{c}{c_1}x\equiv d(mod\frac{N}{c_1})$ such that $1⩽d_0⩽\frac{N}{c_1}$. Then there exists an integer y such that $\frac{c}{c_1}{d}_0+\frac{N}{c_1}y=d$. Assume that there exists a prime p such that $p|(c_1,d_0,\frac{N}{c_1})$. Then $p|d$ and $p|(c,N)$, this contradicts with $(c,d,N)=1$ and thus $(c_1,d_0,\frac{N}{c_1})=1$. Hence $(c_1,d_0)\in D_1(N)$. Then there exists the unique $(c_1,d_1)\in D(N)$ which corresponds to $(c_1,d_0)$. Hence $(c_1,d_1)\in (c:d)$, i.e., $\mathsf{\Phi }((c_1,d_1))=(c:d)$.

Assume that$(c_1,d_1),(c_2,d_2)\in D(N)$ such that $\mathsf{\Phi }((c_1,d_1))=\mathsf{\Phi }((c_2,d_2))$. Then

$(c_1:d_1)=(c_2:d_2)$ and thus there exists an integerk such that $c_1{d}_2-c_2{d}_1=Nk$. Thus $c_1|c_2{d}_1$ by $c_1|N$ and $c_2|c_1{d}_2$ by $c_2|N$. Hence $c_1|c_2$ by $(c_1,d_1)=1$ and $c_2|c_1$ by $(c_2,d_2)=1$. Therefore $c_1=c_2$ and $d_1=d_2$ by $d_1\equiv d_2(mod\frac{N}{c_1})$ and the definition of $D(N)$. Thus $\mathsf{\Phi }$ is a bijection between $\mathcal{D}(N)$ and $D(N)$. This completes the proof. □

Theorem 1.

Let $M,N\in \mathbb{Z},M,N⩾1$, $(M,N)=1$. Then there exists a bijection between $D(M)\times D(N)$ and $D(MN)$.

Proof. 

Let $(a,b)\in D(M)$ and $(c,d)\in D(N)$. Assume that there exists a prime p such that $p|(ac,bN+dM-\frac{MN}{ac}{[\frac{ac(bN+dM)}{MN}]}^{\prime },\frac{MN}{ac})$. Then $p|ac,p|\frac{M}{a}\frac{N}{c}$ and

$$p|bN+dM-\frac{MN}{ac}{[\frac{ac(bN+dM)}{MN}]}^{\prime }.$$

Then $p|a,p|\frac{M}{a}$ or $p|c,p|\frac{N}{c}$ by $(M,N)=1$,$a|M,c|N$. If $p|a,p|\frac{M}{a}$ then $p|bN$ and thus $p|N$ by $(a,b)=1$, which contradicts with $(M,N)=1$. The case of $p|c,p|\frac{N}{c}$ is tackled by a similar way. Therefore $(ac,bN+dM-\frac{MN}{ac}{[\frac{ac(bN+dM)}{MN}]}^{\prime },\frac{MN}{ac})=1$ and

$$(ac,bN+dM-\frac{MN}{ac}{[\frac{ac(bN+dM)}{MN}]}^{\prime })\in D_1(MN).$$

Define $e=ac$, $f=bN+dM-\frac{MN}{ac}({[\frac{ac(bN+dM)}{MN}]}^{\prime }-k)$ for some k such that

$$(ac,bN+dM-\frac{MN}{ac}({[\frac{ac(bN+dM)}{MN}]}^{\prime }-n))⩾2$$

for all $0⩽n<k$. Then $(e,f)\in D(MN)$. Define $\mathsf{\Phi }:D(M)\times D(N)\to D(MN)$ by sending $((a,b),(c,d))$ to $(e,f)$.

Assume that $\mathsf{\Phi }((a,b),(c,d))=\mathsf{\Phi }((a_1,b_1),(c_1,d_1))$ for some $(a,b),(a_1,b_1)\in D(M)$ and $(c,d),(c_1,d_1)\in D(N)$. Then

$$(ac,bN+dM-\frac{MN}{ac}({[\frac{ac(bN+dM)}{MN}]}^{\prime }-k))$$

$$=(a_1{c}_1,b_{1}N+d_{1}M-\frac{MN}{a_1{c}_1}({[\frac{a_1{c}_1(b_{1}N+d_{1}M)}{MN}]}^{\prime }-k_1)).$$

Thus $ac=a_1{c}_1$ and

$$bN+dM-\frac{MN}{ac}({[\frac{ac(bN+dM)}{MN}]}^{\prime }-k)$$

$$=b_{1}N+d_{1}M-\frac{MN}{a_1{c}_1}({[\frac{a_1{c}_1(b_{1}N+d_{1}M)}{MN}]}^{\prime }-k_1).$$

Hence $a=a_1$, $c=c_1$ by $(M,N)=1$,$a|M,a_1|M,c|N,c_1|N$. Therefore

$$bN+dM-\frac{MN}{ac}({[\frac{ac(bN+dM)}{MN}]}^{\prime }-k)$$

$$=b_{1}N+d_{1}M-\frac{MN}{ac}({[\frac{ac(b_{1}N+d_{1}M)}{MN}]}^{\prime }-k_1).$$

Thus $d\equiv d_1(mod\frac{N}{c})$ and $b\equiv b_1(mod\frac{M}{a})$by $(M,N)=1$. Hence $b=b_1$, $d=d_1$. Then $((a,b),(c,d))=((a_1,b_1),(c_1,d_1))$.

Let $(e,f)\in D(MN)$. Then $e|MN$, $(e,f)=1and(e,f-\frac{MN}{e}({[\frac{ef}{MN}]}^{\prime }-n))⩾2for0⩽n<{[\frac{ef}{MN}]}^{\prime }$. Let $a=(e,M)$, $c=(e,N)$, then $e=ac$,$a|M$ and $c|N$. Let $x_0,y_0,z_0$ be a particular solution of the equation

$$Nx+My+\frac{MN}{ac}z=f$$

(4)

then $x=\frac{M}{a}X+x_0,y=\frac{N}{c}Y+y_0,z=-cX-aY+z_0$ are solutions of $(4)$ for all integers $X,Y$. Take $b_1=x_0-\frac{M}{a}{[\frac{a{x}_0}{M}]}^{\prime },d_1=y_0-\frac{N}{c}{[\frac{c{y}_0}{N}]}^{\prime }$, then

$$N{b}_1+M{d}_1+\frac{MN}{ac}(c{[\frac{a{x}_0}{M}]}^{\prime }+a{[\frac{c{y}_0}{N}]}^{\prime }+z_0)=f,1⩽b_1⩽\frac{M}{a},1⩽d_1⩽\frac{N}{c}.$$

Then $(a,b_1,\frac{M}{a})=1$ by $a|M,(e,f)=1$ and $(c,d_1,\frac{N}{c})=1$ by $c|N,(e,f)=1$. Hence

$(a,b_1)\in D_1(M),(c,d_1)\in D_1(N)$. Let $(a,b)\in D(M)$ and $(c,d)\in D(N)$ which correspond to $(a,b_1)$ and $(c,d_1)$, respectively. Then $b=b_1+\frac{M}{a}{k}_1$ and $d=d_1+\frac{N}{c}{k}_2$ for some $k_1,k_2$. Then $Nb+Md+\frac{MN}{ac}(c{[\frac{a{x}_0}{M}]}^{\prime }+a{[\frac{c{y}_0}{N}]}^{\prime }-c{k}_1-a{k}_2+z_0)=f$. Then $(e,f)=\mathsf{\Phi }((a,b),(c,d))$.

Thus $\mathsf{\Phi }$ is a bijection between $D(M)\times D(N)$ and $D(MN)$. □

Proposition 1.

Let p be a prime and l a positive integer. Then

$$(a)D(p^l)=\{(1,d):1⩽d⩽p^l\}\cup \left\{(p^l,1)\right\}\cup$$

$$\{(p^{\alpha },kp+d):1⩽\alpha ⩽l-1,1⩽d⩽p-1,0⩽k⩽p^{l-\alpha -1}-1\};$$

$$(b)\mu (p^l)=p^l(1+\frac{1}{p});$$

$$(c)\mu (N)=N{\prod }_{p|N}\left(1+\left(\frac{1}{p}\right)\right).$$

Proof. (c) is immediately from (b) and Theorem 1. □

Algorithm 1.

(1)

Construct D($p^l)$ by Proposition 1(a);

(2)

Given $D(M)$ and $D(N)$ for $(M,N)=1$, $D(MN)$is constructed as follows. For all $(a,b)\in D(M)$,$(c,d)\in D(N)$, define $e=ac$, $f=bN+dM-\frac{MN}{ac}({[\frac{ac(bN+dM)}{MN}]}^{\prime }-k)$ for some $k\in \mathbb{Z}$ such that $(e,f)=1$ and $(ac,bN+dM-\frac{MN}{ac}({[\frac{ac(bN+dM)}{MN}]}^{\prime }-n))⩾2$ for all $0⩽n<k$. Then $(e,f)\in D(MN)$ and all elements in $D(MN)$ are constructed if all pairs in $D(M)\times D(N)$ are processed.

3. The recursive structure of cusps

In order to describe the cusps on $X_0(N)$, Ju. I. Manin in [2] introduced the set $\Pi (N)$, which consists of pairs of the form $[\delta ;amod(\delta ,N{\delta }^{-1})]$. Here $\delta$ runs through all positive divisors of N, and the second coordinate of the pair runs through any invertible class of residues modulo the greatest common divisor of $\delta$ and $N{\delta }^{-1}$. If $(\delta ,N{\delta }^{-1})=1$ we sometimes put simply 1 in place of the second coordinate.

Proposition 2.

Let $\delta |N,u,v\in \mathbb{Z}$; $(u,v\delta )=(v,N{\delta }^{-1})=1$. The map $\mathbb{Q}\cup \{i\infty \}\to \Pi (N)$ of the form $\frac{u}{v\delta }\mapsto [\delta ;uvmod(\delta ,N{\delta }^{-1})]$ gives an isomorphism of the set of cusps on $X_0(N)$ with $\Pi (N)$.

Proof. 

See Proposition 2.2 in [2]. □

In [1], J. E. Cremona gives the following characterization of cusps of $X_0(N)$.

Proposition 3.

For $j=1,2$ let ${\alpha }_j=p_j/q_j$ be cusps written in lowest terms. The following are equivalent:

(a) 

${\alpha }_2=M({\alpha }_1)$ for some $M\in {\Gamma }_0(N)$;

(b) 

$q_2\equiv u{q}_1(modN)$ and $u{p}_2\equiv p_1(mod(q_1,N))$, with $(u,N)=1$;

(c) 

$s_1{q}_2\equiv s_2{q}_1(mod(q_1{q}_2,N))$, where $s_j$ satisfies $p_j{s}_j\equiv 1(mod{q}_j)$.

Proof. 

See Proposition 2.2.3 in [1]. □

Definition 2.

$$(a)C_1(N)=\{(c,d):c,d\in \mathbb{Z},1⩽c⩽N,c|N,1⩽d⩽(c,N{c}^{-1}),(c,d,N{c}^{-1})=1\},$$

$$(b)C(N)isdefinedin(2).$$

Lemma 4.

There exists a bijection between $C_1(N)$ and $C(N)$.

Proof. 

It holds by $C_1(N)\subseteq D_1(N)$ , $C(N)\subseteq D(N)$ and Lemma 2. □

Lemma 5.

There exists a bijection between ${\Gamma }_0(N)\setminus \mathbb{Q}\cup \{i\infty \}$ and $C_1(N)$.

Proof. 

Let ${\gamma }_i=\left(\begin{array}{cc}{a}_i\hfill & b_i\hfill \\ c_i\hfill & d_i\hfill \end{array}\right),{\gamma }_j=\left(\begin{array}{cc}{a}_j\hfill & b_j\hfill \\ c_j\hfill & d_j\hfill \end{array}\right)\in {SL}_2(\mathbb{Z})$ such that $(c_i,d_i),(c_j,d_j)\in D(N)$ for $1⩽i<j⩽\mu (N)$ then ${SL}_2(\mathbb{Z})={\Gamma }_0(N){\gamma }_1\cup \cdots \cup {\Gamma }_0(N){\gamma }_{\mu (N)}$ and ${\Gamma }_0(N){\gamma }_i\ne {\Gamma }_0(N){\gamma }_j$. $\forall c,a\in \mathbb{Z},(c,a)=1,c⩾1$, let $\left(\begin{array}{cc}a\hfill & b\hfill \\ c\hfill & d\hfill \end{array}\right)\in {SL}_2(\mathbb{Z})$ for some $a,b$. Then there exists $\gamma \in {\Gamma }_0(N),1⩽i⩽\mu (N)$ such that $\left(\begin{array}{cc}a\hfill & b\hfill \\ c\hfill & d\hfill \end{array}\right)=\gamma {\gamma }_i$. Thus $\left(\begin{array}{cc}a\hfill & b\hfill \\ c\hfill & d\hfill \end{array}\right)(\infty )=\gamma {\gamma }_i(\infty )$, $\gamma (\frac{a_i}{c_i})=\frac{a}{c}$ and ${\Gamma }_0(N)\frac{a}{c}={\Gamma }_0(N)\frac{a_i}{c_i}$ . Then ${\Gamma }_0(N)\setminus \mathbb{Q}\cup \{i\infty \}=\{{\Gamma }_0(N)\frac{a_i}{c_i}:1⩽i⩽\mu \}$. Define $\mathsf{\Phi }:{\Gamma }_0(N)\setminus \mathbb{Q}\cup \{i\infty \}\to C_1(N)$ by

$${\Gamma }_0(N)\frac{a}{c}\mapsto (c_i,d_i-(c_i,c_i^{-1}N){\left[d_i{(c_i,c_i^{-1}N)}^{-1}\right]}^{\prime }),{\Gamma }_0(N)·i\infty \mapsto (N,1).$$

By Proposition 3, ${\Gamma }_0(N)\frac{a_i}{c_i}={\Gamma }_0(N)\frac{a_j}{c_j}$ iff $c_i{d}_j\equiv c_j{d}_i(mod(c_i{c}_j,N))$. Then

$c_i{d}_j=c_j{d}_i+(c_i{c}_j,N)h$ for some $h\in \mathbb{Z}$. Thus $c_i=c_j$ by $c_i|N$, $c_j|N$, $(c_i,d_i)=1$ and $(c_j,d_j)=1$. Hence $c_i{d}_j\equiv c_j{d}_i(mod(c_i{c}_j,N))$ iff $d_i\equiv d_j(mod(c_i,c_i^{-1}N))$. Therefore $\mathsf{\Phi }$ is a bijection between ${\Gamma }_0(N)\setminus \mathbb{Q}\cup \{i\infty \}$ and $C_1(N)$. □

Lemma 6.

There exists a bijection between ${\Gamma }_0(N)\setminus \mathbb{Q}\cup \{i\infty \}$ and $C(N)$.

Proof. 

It is immediately from Lemma 4 and 5. □

Lemma 7.

Let $(N_1,N_2)=1$. Then there exists a bijection between $C_1(N_1{N}_2)$ and $C_1(N_1)\times C_1(N_2)$.

Proof. 

Let $(c,d)\in C_1(N_1{N}_2)$ then $c|N_1{N}_2,d⩽(c,N_1{N}_2{c}^{-1}),(d,c,N_1{N}_2{c}^{-1})=1$. Let $c_1=(c,N_1),c_2=(c,N_2)$ then $c=c_1{c}_2,(c_1,c_2)=1$ and $(d,c_1{c}_2,N_1{c}_1^{-1}{N}_2{c}_2^{-1})=1$. Thus $(d,(c_1,N_1{c}_1^{-1}))=1$, $(d,(c_2,N_2{c}_2^{-1}))=1$ by $(c,N_1{N}_2{c}^{-1})=(c_1,N_1{c}_1^{-1})(c_2,N_2{c}_2^{-1})$. Let $d_1=d-(c_1,N_1{c}_1^{-1}){[d{(c_1,N_1{c}_1^{-1})}^{-1}]}^{\prime }$ and $d_2=d-(c_2,N_2{c}_2^{-1}){[d{(c_2,N_2{c}_2^{-1})}^{-1}]}^{\prime }$ then $(d_1,(c_1,N_1{c}_1^{-1}))=1$ and $(d_2,c_2,N_2{c}_2^{-1})=1$. Thus $(c_1,d_1)\in C_1(N_1)$ and $(c_2,d_2)\in C_1(N_2)$. Define $\mathsf{\Phi }:C_1(N_1{N}_2)\to C_1(N_1)\times C_2(N_2)$ by $(c,d)\mapsto ((c_1,d_1),(c_2,d_2))$.

For any $((c_1,d_1),(c_2,d_2))\in C_1(N_1)\times C_1(N_2)$, let $c=c_1{c}_2$ there exists an integer d such that $d\equiv d_1(mod(c_1,N_1{c}_1^{-1}))$, $d\equiv d_2(mod(c_2,N_2{c}_2^{-1}))$ and

$$1⩽d⩽(c_1,N_1{c}_1^{-1})(c_2,N_2{c}_2^{-1})=(c,N_1{N}_2{c}^{-1})$$

by $((c_1,N_1{c}_1^{-1}),(c_2,N_2{c}_2^{-1}))=1$. Thus $(c,d)\in C_1(N_1{N}_2)$ and hence $\mathsf{\Phi }$ is a surjective map.

Let $\mathsf{\Phi }((c,d))=\mathsf{\Phi }((c^{\prime },d^{\prime }))$. Then $((c_1,d_1),(c_2,d_2))=((c_1^{\prime },d_1^{\prime }),(c_2^{\prime },d_2^{\prime }))$, $(c_1,d_1)=(c_1^{\prime },d_1^{\prime })$ and $(c_2,d_2)=(c_2^{\prime },d_2^{\prime })$. Thus $c_1=c_1^{\prime },c_2=c_2^{\prime },d_1=d_1^{\prime }$ and $d_2=d_2^{\prime }$. Hence $c=c_1{c}_2=c_1^{\prime }{c}_2^{\prime }=c^{\prime }$ and $d=d^{\prime }$ by $d\equiv d_1(mod(c_1,N_1{c}_1^{-1}))$, $d\equiv d_2(mod(c_2,N_2{c}_2^{-1}))$, $d^{\prime }\equiv d_1^{\prime }(mod(c_1,N_1{c}_1^{-1}))$ and $d^{\prime }\equiv d_2^{\prime }(mod(c_2,N_2{c}_2^{-1}))$. Therefore $\mathsf{\Phi }$ is an injective map. Then $\mathsf{\Phi }$ is a bijection between $C_1(N_1{N}_2)$ and $C_1(N_1)\times C_1(N_2)$. □

Theorem 2.

Let $(N_1,N_2)=1$. Then there exists a bijection between $C(N_1{N}_2)$ and $C(N_1)\times C(N_2)$.

Proof. 

It is immediately from Lemma 4 and 7. □

Proposition 4.

Let p be a prime and l a positive integer. Then

$$(a)C(p^l)=\{(1,1),(p^l,1)\}\cup$$

$$\{(p^{\alpha },kp+d):1⩽\alpha ⩽l-1,1⩽d⩽p-1,0⩽k⩽p^{min\{\alpha ,l-\alpha \}-1}-1\};$$

$$(b)v_{\infty }(p^l)=\left\{\begin{array}{c}(p+1)p^{\frac{l}{2}-1}if2|l,\hfill \\ 2p^{\frac{l-1}{2}}otherwise\hfill \end{array}\right.;$$

$$(c)v_{\infty }(N)={\prod }_{p|N}{v}_{\infty }(p^l).$$

Proof. (c) is immediately from (b) and Theorem 2. □

Algorithm 2.

(1)

Construct C($p^l$) by Proposition 4(a);

(2)

Let $N=N_1{N}_2$ for $(N_1,N_2)=1$. Given $C(N_1)$ and $C(N_2)$. $C(N)$is constructed as follows. For all $(c_1,d_1)\in C(N_1)$,$(c_2,d_2)\in C(N_2)$, define $c=c_1{c}_2$. Determinate $d_0$ such that

$d_0\equiv d_1(mod(c_1,N_1{c}_1^{-1}))$, $d_0\equiv d_2(mod(c_2,N_2{c}_2^{-1}))$ and

$$1⩽d_0⩽(c_1,N_1{c}_1^{-1})(c_2,N_2{c}_2^{-1}).$$

Determinate $d=d_0+\frac{Nk}{c}$ such that $(c,d)=1$ and $(c,d_0+\frac{Nn}{c})⩾2for0⩽n<k$. Then $(c,d)\in C(N_1{N}_2)$ and all elements in $C(N_1{N}_2)$ are constructed if all pairs in $C(N_1)\times C(N_2)$ are processed.

4. The recursive structure of elliptic points of $X_0(N)$

Let $\rho =\frac{-1+\sqrt{3}i}{2}$. $E_2(N)$ and $E_3(N)$ is defined in (3). Then

$$\{\gamma (i):(1,d)\in E_2(N),\gamma =\left(\begin{array}{cc}a\hfill & b\hfill \\ 1\hfill & d\hfill \end{array}\right)\in SL(2,\mathbb{Z})forsomefixeda,b\}and$$

$$\{\gamma (\rho ):(1,d)\in E_3(N),\gamma =\left(\begin{array}{cc}a\hfill & b\hfill \\ 1\hfill & d\hfill \end{array}\right)\in SL(2,\mathbb{Z})forsomefixeda,b\}$$

are complete sets of representatives of ${\Gamma }_0(N)$-inequivalent elliptic points of order 2, 3, respectively.

Theorem 3.

Let $N_1,N_2\in \mathbb{Z},N_1,N_2⩾1$ and $(N_1,N_2)=1$. Then

(a) 

there exists a bijection between $E_3(N_1)\times E_3(N_2)$ and $E_3(N_1{N}_2)$;

(b) 

there exists a bijection between $E_2(N_1)\times E_2(N_2)$ and $E_2(N_1{N}_2)$.

Proof. (a) Let $(1,d_1)\in E_3(N_1)$ and $(1,d_2)\in E_3(N_2)$. Let d be the unique integer such that $d\equiv d_1(mod{N}_1)$,$d\equiv d_2(mod{N}_2)$ and $1⩽d⩽N_1{N}_2$ then $d^2-d+1\equiv 0(mod{N}_1{N}_2)$.

Hence $(1,d)\in E_3(N_1{N}_2)$. Define

$$\mathsf{\Phi }:E_3(N_1)\times E_3(N_2)\to E_3(N_1{N}_2),((1,d_1),(1,d_2))\mapsto (1,d).$$

Then $\mathsf{\Phi }$ is a bijection between $E_3(N_1)\times E_3(N_2)$ and $E_3(N_1{N}_2)$. The proof of (b) is similar to that of (a) and omitted. □

Proposition 5.

Let $p\in \mathbb{Z}$ be a prime and $l\in \mathbb{Z},l⩾1$. Then

$$v_2(p^l)=\left\{\begin{array}{c}0ifp\equiv 3(mod4)or4|p^l,\hfill \\ 1ifp=2,\hfill \\ 2ifp\equiv 1(mod4).\hfill \end{array}\right.$$

Proof. 

Let $(1,d)\in E_2(p^l)$ then $d^2+1\equiv 0(mod{p}^l)$. Since the system of two equations $x^2+1\equiv 0(modp)$ and $2x\equiv 0(modp)$ has a common solution iff $p=2$, the number of solutions of $x^2+1\equiv 0(mod{p}^l)$ is equal to that of $x^2+1\equiv 0(modp)$ if $p\ne 2$. The cases of $p=2$ or $4|p^l$ are trivial and we then let $p⩾3$ in the following. Then $x^2+1\equiv 0(modp)$ has a solution iff $\left(\frac{-1}{p}\right)=1$ iff $p\equiv 1(mod4)$ by $\left(\frac{-1}{p}\right)={(-1)}^{\frac{p-1}{2}}$ . In addition, $x^2+1\equiv 0(modp)$ has two and only two solutions when it is solvable. This completes the proof. □

Proposition 6.

Let $p\in \mathbb{Z}$ be a prime and $l\in \mathbb{Z},l⩾1$. Then

$$v_3(p^l)=\left\{\begin{array}{c}0ifp\equiv 2(mod3)or9|p^l,\hfill \\ 1ifp=3,\hfill \\ 2ifp\equiv 1(mod3).\hfill \end{array}\right.$$

Proof. 

Let $(1,d)\in E_3(p^l)$ then $d^2-d+1\equiv 0(mod{p}^l)$. Since the system of two equations $x^2-x+1\equiv 0(modp)$ and $2x-1\equiv 0(modp)$ has a common solution iff $p=3$, the number of solutions of $x^2-x+1\equiv 0(mod{p}^l)$ is equal to that of $x^2-x+1\equiv 0(modp)$ if $p\ne 3$. The cases of $p=2,3$ or $9|p^l$ are trivial and we then let $p⩾5$ in the following. $x^2-x+1\equiv 0(modp)$ has a solution iff $y^2+3\equiv 0(modp)$has a solution by taking $x=\frac{y+1}{2}$ and substituting $p-y$ for y when $y\equiv 0(mod2)$. Then $x^2-x+1\equiv 0(modp)$ has a solution iff $\left(\frac{-3}{p}\right)=1$ iff $p\equiv 1(mod3)$ by

$$\left(\frac{-3}{p}\right)=\left(\frac{3}{p}\right)\left(\frac{-1}{p}\right),\left(\frac{3}{p}\right)={(-1)}^{\frac{p-1}{2}}\left(\frac{p}{3}\right),\left(\frac{-1}{p}\right)={(-1)}^{\frac{p-1}{2}}$$

and $\left(\frac{-3}{p}\right)=\left(\frac{p}{3}\right)$ . In addition, $x^2-x+1\equiv 0(modp)$ has two and only two solutions if it is solvable. This completes the proof. □

As an application of Theorem 4, we give an elementary proof of the following well-known results by Proposition 5 and 6 (See Proposition 1.43 in [5]).

Corollary 2.

$(1)v_2(N)=\left\{\begin{array}{c}0if4|N,\hfill \\ {\prod }_{p|N}\left(1+\left(\frac{-1}{p}\right)\right)otherwise.\hfill \end{array}\right.$

$(2)v_3(N)=\left\{\begin{array}{c}0if4|N,\hfill \\ {\prod }_{p|N}\left(1+\left(\frac{-3}{p}\right)\right)otherwise.\hfill \end{array}\right.$

Corollary 3.

Let $g(N)$ be the genus of modular curve $X_0(N)$. Then for any $(N_1,N_2)=1$,

$$g(N_1{N}_2)=1+\frac{\mu (N_1)\mu (N_2)}{12}-\frac{v_2(N_1)v_2(N_2)}{4}-\frac{v_3(N_1)v_3(N_2)}{3}-\frac{v_{\infty }(N_1)v_{\infty }(N_2)}{2}.$$

Proof. 

It is immediately from Theorem 1, 2, 3 and the formula for the genus of $X_0(N)$

$$g(N)=1+\frac{\mu (N)}{12}-\frac{v_2(N)}{4}-\frac{v_3(N)}{3}-\frac{v_{\infty }(N)}{2}.$$

□

Algorithm 3.

(1)

Construct $E_3$ ($p^l$) by the general method;

(2)

Let $N=N_1{N}_2$ for $(N_1,N_2)=1$. Given $E_3(N_1)$ and $E_3(N_2)$. $E_3(N)$ is constructed as follows. For all $(1,d_1)\in E_3(N_1)$,$(1,d_2)\in E_3(N_2)$, Determinate d such that

$$d\equiv d_1(mod{N}_1),d\equiv d_2(mod{N}_2)and1⩽d⩽N.$$

Then $(1,d)\in E_3(N)$ and all elements in $E_3(N)$ are constructed if all pairs in $E_3(N_1)\times E_3(N_2)$ are processed.

5. Concluding Remarks

In [6], Stein mentioned that another approach to list ${\mathbb{P}}^1(\mathbb{Z}/N\mathbb{Z})$ is to use that

$${\mathbb{P}}^1(\mathbb{Z}/N\mathbb{Z})\cong \prod _{p|N}{\mathbb{P}}^1(\mathbb{Z}/p^{v_p}\mathbb{Z}),$$

where $v_p=\mathrm{or}{\mathrm{d}}_p(N)$, and that it is relatively easy to enumerate the elements of ${\mathbb{P}}^1(\mathbb{Z}/p^n\mathbb{Z})$ for a prime power $p^n$. However, this approach had never been implemented by anyone as for as I know. Thus the results in this paper could been regarded as an explicit implementation of Stein’s ideas. The implementations of all the algorithms described in this paper have been completely written in Wolfram Language. We plan rewrite these programs in the free open source computer algebra system SAGE and integrate them into Stein’s program [3] and Walker’s program [8].