Odd and Even Elliptic Curves with Complex Multiplication (2024)

YuriG.ZarhinDepartment of Mathematics, Pennsylvania State University,University Park, PA 16802, USAzarhin@math.psu.edu

Abstract.

We call an order $O$ in a quadratic field $K$ odd (resp. even) if its discriminant is an odd (resp.even) integer.We call an elliptic curve $E$ over ${\mathbb{C}}$ with CM odd (resp. even) if its endomorphism ring $\mathrm{End}(E)$ is an odd (resp. even)order in the imaginary quadratic field $\mathrm{End}(E)\otimes{\mathbb{Q}}$ .

Suppose that $j(E)\in{\mathbb{R}}$ and let us consider the set $\mathcal{J}({\mathbb{R}},E)$ of all $j(E^{\prime})$ where $E^{\prime}$ is any elliptic curve that enjoys the following properties.

•
$E^{\prime}$ isisogenous to $E$ ;
•
$j(E^{\prime})\in{\mathbb{R}}$ ;
•
$E^{\prime}$ has the same parity as $E$ .

We prove that the closure of $\mathcal{J}({\mathbb{R}},E)$ in ${\mathbb{R}}$ is the closed semi-infinite interval $(-\infty,1728]$ (resp. the whole ${\mathbb{R}}$ ) if $E$ is odd (resp. even).

This paper was inspired by a questionof Jean-Louis Colliot-Thélène and Alena Pirutka about the distribution of $j$ -invariants of certain elliptic curves of CM type.

1. Orders in quadratic fields

As usual ${\mathbb{Z}},{\mathbb{Q}},{\mathbb{R}},{\mathbb{C}}$ stand for the ring of integers and the fields of rational, real and complex numbers respectively.If $n$ is a positive integer then we write ${\mathbb{Z}}[1/n]$ for the subring of ${\mathbb{Q}}$ generated by ${\mathbb{Z}}$ and $1/n$ . More generally,If $F$ is a field of characteristic zero and $A$ is a subring of $F$ (with $1$ ) then we write $A[1/n]$ for the subringof $F$ generated by $A$ and $1/n$ ; clearly, $A[1/n]$ is a ${\mathbb{Z}}[1/n]$ -subalgebra of $F$ . We write ${\mathbb{Z}}_{(2)}$ for the subring of ${\mathbb{Q}}$ that consists of all fractions $\frac{a}{n}$ where $a\in{\mathbb{Z}}$ and $n$ an odd integer.

Let $K$ be a quadratic field, $O_{K}$ its ring of integers, and

\mathrm{tr}=\mathrm{tr}_{K}:K\to{\mathbb{Q}}

be the corresponding trace map attached to the field extension $K/{\mathbb{Q}}$ .We have

2{\mathbb{Z}}=\mathrm{tr}({\mathbb{Z}})\subset\mathrm{tr}(O_{K})\subset{%\mathbb{Z}}.

Let $O$ be an order in $K$ and let a positive integer $\mathfrak{f}=\mathfrak{f}_{O}$ be the conductor of $O$ . This means that

O={\mathbb{Z}}+\mathfrak{f}\cdot O_{K}\subset O_{K}

and $\mathfrak{f}$ coincides with the index $[O_{K}:O]$ of the additive subgroup $O$ in $O_{K}$ [2, Ch. 2, Sect. 7, Th.1].Since ${\mathbb{Z}}\subset O$ and $\mathrm{tr}({\mathbb{Z}})=2{\mathbb{Z}}$ ,

\mathrm{tr}(O)={\mathbb{Z}}\quad\text{or}\ 2{\mathbb{Z}}.

(1)

Recall that the discriminant $\mathrm{discr}(O)\in{\mathbb{Z}}$ of $O$ is the discriminant of the symmetric bilinear form

O\times O\to{\mathbb{Z}},\ a,b\mapsto\mathrm{tr}(ab).

By definition, the discriminant $\mathrm{discr}(K)$ of the field $K$ is $\mathrm{discr}(O_{K})$ . It is well known [2, Ch. 2, Sect. 7, Th.1] that

\mathrm{discr}(O)=\mathfrak{f}_{O}^{2}\cdot\mathrm{discr}(O_{K})=\mathfrak{f}_%{O}^{2}\cdot\mathrm{discr}(K).

(2)

Definition 1.1 (Definition-Lemma).

An order $O$ is called odd if it enjoys the following equivalent properties.

(i)
$\mathrm{tr}(O)={\mathbb{Z}}$ .
(ii)
$\mathrm{discr}(O)$ is an odd integer.
(iii)
Both $\mathrm{discr}(K)$ and the conductor $\mathfrak{f}_{O}$ are odd integers.
(iv)
There exists an integer $D\equiv 1(\bmod 4)$ that is not a square such that
$\frac{1+\sqrt{D}}{2}\in O.$

Definition 1.2 (Definition-Lemma).

An order $O$ in a quadratic field $K$ is called even if it enjoys the following equivalent properties.

(i)
$\mathrm{tr}(O)=2{\mathbb{Z}}$ .
(ii)
$\mathrm{discr}(O)$ is an even integer.
(iii)
Either $\mathrm{discr}(K)$ or the conductor $\mathfrak{f}_{O}$ is an even integer.

Clearly, every order $O$ is either odd or even. We prove Lemmas 1.1 and 1.2 in Section 5.

Definition 1.3.

Let $E$ be an elliptic curve with complex multiplication (CM) over the field ${\mathbb{C}}$ of complex numbers.Then its endomorphism ring $\mathrm{End}(E)$ is an order in the imaginary quadratic field

K:=\mathrm{End}(E)\otimes{\mathbb{Q}}=:\mathrm{End}^{0}(E).

We say that $E$ is odd (resp. even) if the order $\mathrm{End}(E)$ is odd (resp. even).

Clearly, every elliptic curve over ${\mathbb{C}}$ with CM is either odd or even. The following assertions will be proven in Section 6.

Proposition 1.4.

Let $\phi:E_{1}\to E_{2}$ be an isogeny of complex elliptic curves with CM.Suppose that $n=\deg(\phi)$ is an odd integer.Then $E_{1}$ is odd (resp. even) if and only if $E_{2}$ is odd (resp. even).In other words, $E_{1}$ and $E_{2}$ have the same parity.

Proposition 1.5.

Suppose that $E_{1}$ and $E_{2}$ are isogenous CM elliptic curves over ${\mathbb{C}}$ . Suppose that there are elliptic curves $E_{1,{\mathbb{R}}}$ and $E_{2,{\mathbb{R}}}$ over the field ${\mathbb{R}}$ of real numbers such that there are isomorphisms of complex elliptic curves

E_{1}\cong E_{1,{\mathbb{R}}}\times_{{\mathbb{R}}}{\mathbb{C}},\quad E_{2}=E_{%2,{\mathbb{R}}}\times_{{\mathbb{R}}}{\mathbb{C}}.

Then $E_{1,{\mathbb{R}}}$ and $E_{2,{\mathbb{R}}}$ are isogenous over ${\mathbb{R}}$ .

Remark 1.6.

Let $E$ be an elliptic curve over ${\mathbb{C}}$ . It is well known [5, Ch. 3, Sect. 2, Prop. 3.7] that $j(E)\in{\mathbb{R}}$ if and only if $E$ may be defined over ${\mathbb{R}}$ , i.e.,there is an elliptic curve $E_{{\mathbb{R}}}$ over ${\mathbb{R}}$ such that $E\cong E_{{\mathbb{R}}}\times_{{\mathbb{R}}}{\mathbb{C}}$ . (See also [11, Appendix A, Prop. 1.2(b)] and [9, Ch. 4, Sect. 4.1]).)

Examples 1.7.

(e)
If $E$ is an elliptic curve $y^{2}=x^{3}-x$ then it is well known that the order $\mathrm{End}(E)={\mathbb{Z}}[\sqrt{-1}]$ has even discriminant $-4$ and therefore $E$ is even.
(o)
If $E$ is an elliptic curve $y^{2}=x^{3}-1$ then it is well known that the order $\mathrm{End}(E)={\mathbb{Z}}\left[\frac{-1+\sqrt{-3}}{2}\right]$ has odd discriminant $-3$ and therefore $E$ isodd.

2. $j$ -invariants of CM elliptic curves

In what follows, $E$ is an elliptic curve over ${\mathbb{C}}$ .The following assertion will be proven in Section 3.

Proposition 2.1.

Let us considerthe subset $J_{E,\mathrm{is}}$ of all $j(E^{\prime})$ where $E^{\prime}$ runs through the set of ellipticcurves over ${\mathbb{C}}$ such that there exists an isogeny $E\to E^{\prime}$ of odd degree.

Then $J_{E,\mathrm{is}}$ is dense in ${\mathbb{C}}$ .

Remark 2.2.

If $E$ has CM thenall $E^{\prime}$ that appear in Proposition 2.1 are elliptic curves with CMand the corresponding imaginary quadratic fields $\mathrm{End}^{0}(E^{\prime})$ are isomorphic to $\mathrm{End}^{0}(E)$ .

Notice that in light of Examples 1.7, there exist an odd CM curve and an even CM curve. Now,combining Proposition 2.1 with Proposition 1.4, we obtain the following statement.

Corollary 2.3.

Let $J^{\mathrm{odd}}({\mathbb{C}})\subset{\mathbb{C}}$ be the set of all $j(E^{\mathrm{odd}})$ where $E^{\mathrm{odd}}$ runs through the setof all odd elliptic curves over ${\mathbb{C}}$ with CM.

Let $J^{\mathrm{ev}}({\mathbb{C}})\subset{\mathbb{C}}$ be the set of all $j(E^{\mathrm{ev}})$ where $E^{\mathrm{ev}}$ runs through the setof all even elliptic curves over ${\mathbb{C}}$ with CM.

Then each of two subsets $J^{\mathrm{odd}}({\mathbb{C}})$ and $J^{\mathrm{ev}}({\mathbb{C}})$ is dense in ${\mathbb{C}}$ .

Our main result is the following assertion.

Theorem 2.4.

Suppose that $E$ is a CM curve with $j(E)\in{\mathbb{R}}$ . Let us consider the set $\mathcal{J}({\mathbb{R}},E)$ of all $j(E^{\prime})$ where $E^{\prime}$ is any elliptic curve over ${\mathbb{C}}$ that enjoys the following properties.

•
$E^{\prime}$ is isogenousto $E$ ;
•
$j(E^{\prime})\in{\mathbb{R}}$ ;
•
$E^{\prime}$ has the same parity as $E$ .

Then the closure of $\mathcal{J}({\mathbb{R}},E)$ in ${\mathbb{R}}$ is the closed semi-infinite interval $(-\infty,1728]$ (resp. the whole ${\mathbb{R}}$ ) if $E$ is odd (resp. even).

We prove Theorem 2.4 in Section 4. Our proof is based on auxiliary results about isogenies of complex elliptic curves with CM that will bediscussed in Section 3.

Remark 2.5.

It follows from Remark 1.6 and Proposition 1.5 that Theorem 2.4.is equivalent to the following statement.

Theorem 2.6.

Let $E_{{\mathbb{R}}}$ be an elliptic curve over ${\mathbb{R}}$ such that its complexification $E=E_{{\mathbb{R}}}\times_{{\mathbb{R}}}{\mathbb{C}}$ is of CM type.Let us consider the set $\mathcal{J}_{0}({\mathbb{R}},E_{{\mathbb{R}}})$ of all $j(E_{R}^{\prime})$ where $E^{\prime}_{{\mathbb{R}}}$ is any elliptic curve over ${\mathbb{R}}$ thatenjoys the following properties.

•
$E^{\prime}_{{\mathbb{R}}}$ is ${\mathbb{R}}$ -isogenous to $E_{{\mathbb{R}}}$ ;
•
The complexification $E^{\prime}:=E^{\prime}_{{\mathbb{R}}}\times_{{\mathbb{R}}}{\mathbb{C}}$ of $E_{{\mathbb{R}}}$ has the same parity as $E$ .

Then the closure of $\mathcal{J}_{0}({\mathbb{R}},E_{{\mathbb{R}}})$ in ${\mathbb{R}}$ is the closed semi-infinite interval $(-\infty,1728]$ (resp. the whole ${\mathbb{R}}$ ) if $E$ is odd (resp. even).

The paper is organized as follows. In Section 3 we discuss auxiliary resultsabout isogenies of complex elliptic curves. We prove our main results in Section 4,using properties of odd and even orders and parity preservation under isogenies of odd degrees that will be provenin Section 5 and 5 respectively. In Section 7 we use the techniques developed in Section3, in order to describe explicitly $2$ -torsion in the class group of proper $O$ -ideals where $O$ is an odd corderin an imaginary quadratic field. In Section 8 we “classify” CM elliptic curve, whose $j$ -invariant is real andthe endomorphism ring is a given odd order.

3. The upper half-plane and elliptic curves

Definition 3.1.

If $\omega_{1}$ and $\omega_{2}$ are nonzero complex numbers such that $\omega_{1}/\omega_{2}\not\in{\mathbb{R}}$ then we write $[\omega_{1},\omega_{2}]$ for the discrete lattice ${\mathbb{Z}}\omega_{1}+{\mathbb{Z}}\omega_{2}$ of rank $2$ in ${\mathbb{C}}$ .

Let

\mathfrak{H}:=\{x+\mathbf{i}y\mid x,y\in\mathbb{R};\ y>0\}\subset{\mathbb{C}}

be the upper half-plane. If $\tau\in\mathfrak{H}$ then we write $\mathcal{E}_{\tau}$ for the complex elliptic curvesuch that the complex tori $\mathcal{E}_{\tau}({\mathbb{C}})$ and ${\mathbb{C}}/\Lambda_{\tau}$ are isomorphic. Here

\Lambda_{\tau}:=[\tau,1]={\mathbb{Z}}\tau+{\mathbb{Z}}.

See [12, Ch. V, p. 408–411] for a Weierstrass equation of $\mathcal{E}_{\tau}$ . (In the notation of [12, Ch. V], our $\mathcal{E}_{\tau}$ is $E_{q}$ with $q=\exp(2\pi\mathbf{i}\tau)$ .)

Remark 3.2.

If $\tau_{1},\tau_{2}\in\mathfrak{H}$ then the group $\mathrm{Hom}(\mathcal{E}_{\tau_{1}},\mathcal{E}_{\tau_{2}})$ of hom*omorphisms from $\mathcal{E}_{\tau_{1}},\mathcal{E}_{\tau_{2}}$ can be be canonically identified with

\{u\in{\mathbb{C}}\mid u(\Lambda_{\tau_{1}})\subset\Lambda_{\tau_{2}}\}\subset%{\mathbb{C}}.

(3)

Each $u$ that satisfies (3) corresponds to the hom*omorphism

\phi_{u}:\mathcal{E}_{\tau_{1}}\to\mathcal{E}_{\tau_{2}}

of elliptic curves such that the corresponding action of $\phi_{u}$ on the complex points is as follows.

\mathcal{E}_{\tau_{1}}({\mathbb{C}})={\mathbb{C}}/\Lambda_{\tau_{1}}\to{%\mathbb{C}}/\Lambda_{\tau_{2}}=\mathcal{E}_{\tau_{2}}({\mathbb{C}}),\quad z+%\Lambda_{\tau_{1}}\mapsto uz+\Lambda_{\tau_{2}}.

If $u\neq 0$ then $\phi_{u}$ is an isogeny,whose degree coincides with the index $[\Lambda_{\tau_{2}}:u(\Lambda_{\tau_{1}})]$ of the subgroup $u(\Lambda_{\tau_{1}})$ in $\Lambda_{\tau_{2}}$ [6, p. 9]. In particular, if

\Lambda_{\tau_{2}}=u(\Lambda_{\tau_{1}})

(i.e., the index is $1$ ), $\phi_{u}$ is an isomorphism of elliptic curves $\mathcal{E}_{\tau_{1}}$ and $\mathcal{E}_{\tau_{2}}$ .

On the other hand, if w take $\tau_{1}=\tau_{2}=\tau$ and put

O_{\tau}:=\{u\in{\mathbb{C}}\mid u(\Lambda_{\tau})\subset\Lambda_{\tau}\},

(4)

then the map

O_{\tau}\to\mathrm{End}(\mathcal{E}_{\tau}),\ u\mapsto\phi_{u}

(5)

is a ring isomorphism.

Remark 3.3.

We have

{\mathbb{Z}}\subset O_{\tau},\quad\Lambda_{\tau}\supset O_{\tau}\cdot 1=O_{%\tau}\subset{\mathbb{C}},

i.e.,

{\mathbb{Z}}\subset O_{\tau}\subset\Lambda_{\tau}\subset{\mathbb{C}}.

(6)

It follows easily that if $\Lambda_{\tau}$ is a subring of ${\mathbb{C}}$ then $O_{\tau}=\Lambda_{\tau}$ .

Remark 3.4.

(i)

It is well known [9, Ch. 4, Sect. 4.4, Prop. 4.5] that $\mathcal{E}_{\tau}$ has CM if and only if $\tau$ is an (imaginary) quadratic irrationality,i.e., ${\mathbb{Q}}(\tau)={\mathbb{Q}}+{\mathbb{Q}}\tau$ is an imaginary quadratic field. If this is the case then $O_{\tau}$ is an order in the imaginary quadratic field ${\mathbb{Q}}(\tau)$ ; in particular,

O_{\tau}=\{u\in{\mathbb{Q}}(\tau)\mid u(\Lambda_{\tau})\subset\Lambda_{\tau}.

(7)

In addition, extending isomorphism (5) by ${\mathbb{Q}}$ -linearity,we get a canonical isomorphism of ${\mathbb{Q}}$ -algebras

{\mathbb{Q}}(\tau)=O_{\tau}\otimes{\mathbb{Q}}={\mathbb{Q}}(\tau)\to\mathrm{%End}(\mathcal{E}_{\tau})\otimes{\mathbb{Q}}=\mathrm{End}^{0}(\mathcal{E}_{\tau%}),\quad u\otimes r\mapsto\phi_{u}\otimes r

between ${\mathbb{Q}}(\tau)$ and the endomorphism algebra $\mathrm{End}^{0}(\mathcal{E}_{\tau})$ of $\mathcal{E}_{\tau}$ .

(ii)

Suppose that $\mathcal{E}_{\tau}$ has CM, i.e., ${\mathbb{Q}}(\tau)$ is an imaginary quadratic field. In light of (6),

{\mathbb{Z}}\subset O_{\tau}\subset\Lambda_{\tau}={\mathbb{Z}}\tau+{\mathbb{Z}%}\subset{\mathbb{Q}}(\tau).

(8)

It follows from Remark 3.3 that if $\Lambda_{\tau}$ is an order in ${\mathbb{Q}}(\tau)$ then

O_{\tau}=\Lambda_{\tau}.

(ii)

Let $\mathcal{E}_{\tau_{1}}$ and $\mathcal{E}_{\tau_{2}}$ be two elliptic curves with complex multiplication. Then $\mathcal{E}_{\tau_{1}}$ and $\mathcal{E}_{\tau_{2}}$ are isogenous if and only if the corresponding imaginary quadratic fields coincide, i.e.,

{\mathbb{Q}}(\tau_{1})={\mathbb{Q}}(\tau_{2})

(see [9, Ch. 4, Sect. 4.4, Prop. 4.9]).

The group $\mathrm{GL}_{2}({\mathbb{R}})^{+}$ of two-by-two real matrices with positive determinant acts transitively on $\mathfrak{H}$ by fractional-linear transformations. Namely, the continuous map

M,\tau\mapsto M(\tau)=\frac{a\tau+b}{c\tau+d}\quad\forall\tau\in\mathfrak{H},%\ M=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\mathrm{GL}_{2}({\mathbb{R}})^{+}.

defines the transitive action of $\mathrm{GL}_{2}({\mathbb{R}})^{+}$ on $\mathfrak{H}$ .(Actually, even the action of the subgroup $\mathrm{SL}(2,{\mathbb{R}})$ on $\mathfrak{H}$ is transitive.) We will mostly deal with the action of the subgroup

G:=\mathrm{GL}(2,{\mathbb{Z}}_{(2)})^{+}=\{M\in\mathrm{GL}(2,{\mathbb{Z}}_{(2)%})\mid\det(M)>0\}\subset\mathrm{GL}_{2}({\mathbb{R}})^{+}.

Lemma 3.5.

Suppose that a matrix

M=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\mathrm{GL}_{2}({\mathbb{Z}}_{(2)})

has positive determinant

\det(M)=ad-bc\in{\mathbb{Z}}_{(2)}^{*},

i.e, $M\in G$ .Then there exists an isogeny $E_{M(\tau)}\to\mathcal{E}_{\tau}$ of odd degree.

Proof.

By definition of ${\mathbb{Z}}_{(2)}$ , there are integers $\tilde{a},\tilde{b},\tilde{c},\tilde{d}$ ,and a positive odd integer $n$ such that

a=\tilde{a}/n,\ b=\tilde{b}/n,\ c=\tilde{c}/n,\ d=\tilde{d}/n.

The conditions on $\det(M)$ imply that $\tilde{a}\tilde{d}-\tilde{b}\tilde{c}$ is a positive odd integer.Then

M(\tau)=\frac{a\tau+b}{c\tau+d}=\frac{\tilde{a}\tau+\tilde{b}}{\tilde{c}\tau+%\tilde{d}}=\tilde{M}(\tau)

where the matrix

\tilde{M}=\begin{pmatrix}\tilde{a}&\tilde{b}\\\tilde{c}&\tilde{d}\end{pmatrix}.

(Since

\tilde{a}\tilde{d}-\tilde{b}\tilde{c}=\det(\tilde{M})=\frac{1}{n^{2}}\det(M)>0,

we get

\tilde{M}\in\mathrm{GL}(2,{\mathbb{Z}}_{(2)})^{+}\subset\mathrm{GL}_{2}({%\mathbb{R}})^{+}.)

We have

\Lambda_{M(\tau)}=\Lambda_{\tilde{M}(\tau)}=\left[\frac{\tilde{a}\tau+\tilde{b%}}{\tilde{c}\tau+\tilde{d}},\ 1\right].

This means that

(\tilde{c}\tau+\tilde{d})\Lambda_{M(\tau)}=[\tilde{a}\tau+\tilde{b},\ \tilde{c%}\tau+\tilde{d}]

is a subgroup in $[\tau,1]=\Lambda_{\tau}$ of odd index $\tilde{a}\tilde{d}-\tilde{b}\tilde{c}$ .This implies that if we put

u:=\tilde{c}\tau+\tilde{d}\in{\mathbb{C}},\quad k:=\tilde{a}\tilde{d}-\tilde{b%}\tilde{c}

then $u\left(\Lambda_{M(\tau)}\right)$ is a subgroup of odd index $k$ in $\Lambda_{\tau}$ . This gives us the isogeny $\phi_{u}:E_{M(\tau)}\to\mathcal{E}_{\tau}$ of odd degree $k$ . This ends the proof.∎

Proof of Proposition 2.1.

Let $B\in\mathrm{GL}(2,{\mathbb{R}})^{+}$ . The weak approximation for the field ${\mathbb{Q}}$ [1, Th. 1] with respect to places $\{\infty,\ 2\}$ implies the existence of a sequence $\{B_{n}\}$ of $2\times 2$ matrices with rational entries such that $\{B_{n}\}$ converges to $B$ in the real topology and to the identity matrix $\begin{pmatrix}1&0\\0&1\end{pmatrix}$ in the $2$ -adic topology. Removing first finite terms of the sequence, we may and will assume that for all $n$

B_{n}\in\mathrm{GL}(2,{\mathbb{Z}}_{(2)})\subset\mathrm{GL}(2,{\mathbb{Q}}),%\quad\det(B_{n})>0;

the latter inequality means that all $B_{n}\in\mathrm{GL}(2.{\mathbb{R}})^{+}$ and therefore

B_{n}\in\mathrm{GL}(2,{\mathbb{Z}}_{(2)})^{+}=G.

It follows that $B(\tau)\in\mathfrak{H}$ is the limit of the sequence

B_{n}(\tau)\in\mathfrak{H}

in the complex topology. Now the transitivity of $\mathrm{GL}(2,{\mathbb{R}})^{+}$ implies thatevery orbit $G\tau=\{M(\tau)\mid M\in G\}$ is dense in $\mathcal{H}$ in the complex topology.

Recall that the classical holomorphic (hence, continuous) modular function

j:\mathfrak{H}\to{\mathbb{C}}

takes on every complex value. It follows that for every $\tau$ the set $\{j(M(\tau))\mid M\in G\}$ is dense in ${\mathbb{C}}$ .

Let $E$ be an elliptic curve with complex multiplication. Then there exists $\tau\in H$ such thatthe elliptic curves $E$ and $\mathcal{E}_{\tau}$ are isomorphic and therefore

j(E)=j(\mathcal{E}_{\tau})=j(\tau).

In light of Lemma 3.5, if $M\in G$ then there is anisogeny $E_{M(\tau)}\to E$ of odd degree. Since

j(E_{M(\tau)})=j(M(\tau)),

the set $J_{E,\mathrm{is}}$ of complex numbers contains $\{j(M(\tau))\mid M\in G\}$ , which is dense in ${\mathbb{C}}$ . Hence, $J_{E,\mathrm{is}}$ is also dense in ${\mathbb{C}}$ ,which ends the proof.∎

Example 3.6.

Let $\tau\in\mathfrak{H}$ . Let $n,m$ be positive odd integers.Then the matrix

M=\begin{pmatrix}m&0\\0&n\end{pmatrix}

satisfies the conditions of Lemma 3.5. We have

M(\tau)=\frac{m}{n}\tau.

It follows from Lemma 3.5 that there exists an isogeny $E_{\frac{m}{n}\tau}\to\mathcal{E}_{\tau}$ of odd degree.This implies that if $\tau$ is an imaginary quadratic irrationality then, by Proposition 1.4 (to be proven in Section 6),the CM elliptic curves $E_{\frac{m}{n}\tau}$ and $\mathcal{E}_{\tau}$ have the same parity.

Example 3.7.

Let $\tau=\frac{1+\mathbf{i}y}{2}\in\mathfrak{H}$ where $y\in{\mathbb{R}}_{+}$ . Let $n,m$ be positive odd integers. Let us put

\tau_{m,n}:=\frac{1+\mathbf{i}\frac{m}{n}y}{2}\in\mathfrak{H}.

Notice that $n-m$ is even and therefore $\frac{n-m}{2}\in{\mathbb{Z}}$ .Then the matrix

M=\begin{pmatrix}m&\frac{n-m}{2}\\0&n\end{pmatrix}

satisfies the conditions of Lemma 3.5. We have

M(\tau)=m\cdot\frac{1+\mathbf{i}y}{2n}+\frac{n-m}{2n}=\frac{n}{2n}+m\cdot\frac%{\mathbf{i}y}{2n}=\frac{1+\mathbf{i}\frac{m}{n}y}{2}=\tau_{m,n}.

It follows from Lemma 3.5 that there exists an isogeny $\mathcal{E}_{\tau_{m,n}}\to\mathcal{E}_{\tau}$ of odd degree.By Proposition 1.4 (to be proven in Section 6), if $\tau$ is an imaginary quadratic irrationality then the CM elliptic curves $\mathcal{E}_{\tau_{m,n}}$ and $\mathcal{E}_{\tau}$ have the same parity.

In what follows, if $D$ is a negative real number then we write $\sqrt{D}$ for $\sqrt{|D|}\cdot\mathbf{i}\in\mathfrak{H}$ .

Example 3.8.

If $\tau$ is a point in $\mathfrak{H}$ such that ${\mathbb{Q}}(\tau)$ is an imaginary quadratic field and the lattice $\Lambda_{\tau}$ is an order in ${\mathbb{Q}}(\tau)$ then $O_{\tau}=\Lambda_{\tau}$ . E.g., let $D$ be a negative integer that is congruent to $1$ modulo $4$ and

\tau=\frac{1+\sqrt{D}}{2}\in\mathfrak{H}.

Then $\Lambda_{\tau}={\mathbb{Z}}+{\mathbb{Z}}\frac{1+\sqrt{D}}{2}$ is an order in the imaginary quadratic field ${\mathbb{Q}}(\tau)={\mathbb{Q}}(\sqrt{D})$ .By Remark 3.4(ii),

O_{\tau}=\Lambda_{\tau}={\mathbb{Z}}+{\mathbb{Z}}\frac{1+\sqrt{D}}{2}.

It follows from Lemma 1.1 that $O_{\tau}$ is an odd order in the quadratic field ${\mathbb{Q}}(\sqrt{D})={\mathbb{Q}}(\tau)\cong\mathrm{End}^{0}(\mathcal{E}_{%\tau})$ . This means that the elliptic curve $\mathcal{E}_{\tau}$ is odd.

Now let $n,m$ be any positive odd integers. Let us consider the complex numbers

\tau_{m,n}(D)=\frac{1+\mathbf{i}\frac{m}{n}\sqrt{|D|}}{2}=\frac{1+\frac{m}{n}%\sqrt{D}}{2}\in\mathfrak{H}.

(9)

It follows from Example 3.7 (applied to $y=\sqrt{|D|}$ ) that all the elliptic curves $\mathcal{E}_{\tau_{n,m}(D)}$ are also odd.

Lemma 3.9 (Key Lemma).

Let $\tau\in\mathfrak{H}$ such that ${\mathbb{Q}}(\tau)$ is an imaginaryquadratic field, i.e., $\mathcal{E}_{\tau}$ is an elliptic curve with CM.

(i)
If $\mathrm{Re}(\tau)=0$ then $\mathcal{E}_{\tau}$ is even.
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(ii)

Suppose that $\mathrm{Re}(\tau)=1/2$ . Let $D$ be a negative integer such that ${\mathbb{Q}}(\tau)={\mathbb{Q}}(\sqrt{D})$ and $D\equiv 1(\bmod 4)$ .Then

\frac{1+\sqrt{D}}{2}\in O_{\tau}

(10)

if and only if there is a positive odd integer $\beta$ such that $\beta$ divides $D$ and

\tau=1/2+\frac{1}{2\beta}\sqrt{D}.

If this is the case then $\mathcal{E}_{\tau}$ is odd.

Proof.

(i). Assume that $\mathrm{Re}(\tau)=0$ . Suppose that $\mathcal{E}_{\tau}$ is odd, i.e., the order $O_{\tau}$ is odd. It followsfrom Lemma 1.1that there is a negative integer

D\equiv 1(\bmod 4)

such that

\frac{1+\sqrt{D}}{2}\in O_{\tau}\subset K={\mathbb{Q}}(\sqrt{D})={\mathbb{Q}}+%{\mathbb{Q}}\sqrt{D}.

Since $\tau\in{\mathbb{Q}}(\sqrt{D})={\mathbb{Q}}+{\mathbb{Q}}\cdot\sqrt{D}$ and $\mathrm{Re}(\tau)=0$ , there is a positive rational number $r$ such that

\tau=r\sqrt{D}.

This implies that

\Lambda_{\tau}={\mathbb{Z}}+{\mathbb{Z}}\cdot r\sqrt{D}.

It follows from the definition of $O_{\tau}$ (see (4)) that

\frac{1+\sqrt{D}}{2}\Big{(}\Lambda_{\tau}\Big{)}\subset\Lambda_{\tau}.

This implies that there are integers $\alpha,\beta\in{\mathbb{Z}}$ such that

\frac{1+\sqrt{D}}{2}\cdot 1=\alpha+\beta r\sqrt{D}.

Taking the real parts of both sides, we get $\alpha=1/2$ , which contradicts the integrality of $\alpha$ .The obtained contradiction proves that $\mathcal{E}_{\tau}$ is even, which proves (i).

(ii). Assume that $\mathrm{Re}(\tau)=1/2$ . Recall that we are given thenegative integer $D\equiv 1(\bmod 4)$ such that

K={\mathbb{Q}}(\tau)={\mathbb{Q}}(\sqrt{D})={\mathbb{Q}}\cdot 1\oplus{\mathbb{%Q}}\cdot\sqrt{D}\subset{\mathbb{C}}.

Then there is a positive rational number $r\in{\mathbb{Q}}$ such that

\tau=\frac{1}{2}+r\sqrt{D}.

It follows that

\Lambda_{\tau}={\mathbb{Z}}\cdot 1+{\mathbb{Z}}\cdot\tau={\mathbb{Z}}+{\mathbb%{Z}}\left(\frac{1}{2}+r\sqrt{D}\right)\subset{\mathbb{C}}.

By definition of $O_{\tau}$ (see (4)), $\frac{1+\sqrt{D}}{2}\in O_{\tau}$ if and only if

\frac{1+\sqrt{D}}{2}\left(\Lambda_{\tau}\right)\subset\Lambda_{\tau}.

(11)

However, since $\{1,\ 1/2+r\sqrt{D}\}$ is a basis of the ${\mathbb{Q}}$ -vector space ${\mathbb{Q}}(\sqrt{D})$ , there are certain rational numbers $\alpha,\beta,\gamma,\delta\in{\mathbb{Q}}$ such that

\frac{1+\sqrt{D}}{2}\cdot 1=\alpha+\beta\left(\frac{1}{2}+r\sqrt{D}\right),%\quad\frac{1+\sqrt{D}}{2}\left(\frac{1}{2}+r\sqrt{D}\right)=\gamma+\delta\left%(\frac{1}{2}+r\sqrt{D}\right).

(12)

The inclusion (11) is equivalent to the condition

\alpha,\beta,\gamma,\delta\in{\mathbb{Z}}.

(13)

Opening the brackets and collecting terms in (12), we get

\frac{1+\sqrt{D}}{2}=\left(\alpha+\beta/2\right)+\beta r\sqrt{D};\quad\left(%\frac{1}{4}+rD/2\right)+\frac{1}{2}\left(r+\frac{1}{2}\right)\sqrt{D}=\left(%\gamma+\delta/2\right)+\delta r\sqrt{D}.

Taking the real parts and “imaginary” parts, we get

1/2=\alpha+\beta/2,\ 1/2=\beta r;\quad 1/4+rD/2=\gamma+\delta/2;\quad\frac{1}{%2}\left(r+1/2\right)=\delta r.

(14)

The second equality of (14) implies that

r=\frac{1}{2\beta},\quad\tau=\frac{1}{2}+\frac{1}{2\beta}\sqrt{D}.

(15)

Since $r$ is positive, $\beta$ is also positive.

It follows from first equality of (14) that $\alpha$ is an integer if and only if $\beta$ is an odd integer.This implies that if $\beta$ is not an odd integer then (13) does not hold. So, in the course of the proof we may and will assume that

\beta\in 1+2{\mathbb{Z}},\quad\alpha\in{\mathbb{Z}}.

(16)

It follows from the last equality of (14) combined with (15) that

\frac{1}{4}+\frac{1}{4\beta}=\frac{\delta}{2\beta},

which means that

\delta=\frac{\beta+1}{2}.

In light of (16),

\delta=\frac{\beta+1}{2}\in{\mathbb{Z}}.

(17)

In addition, $\delta$ is even (resp. odd) if $\beta\equiv 3(\bmod 4)$ (resp. if $\beta\equiv 1(\bmod 4)$ ). This means that

\frac{\delta}{2}\in{\mathbb{Z}}\ \text{if}\ \beta\equiv 3(\bmod 4);\quad\frac{%\delta}{2}\in\frac{1}{2}{\mathbb{Z}}\setminus{\mathbb{Z}}\ \text{if}\ \beta%\equiv 1(\bmod 4).

(18)

Now let us explore the integrality of $\gamma$ .Combining (15) with third equality of (14), we get

\frac{\beta+D}{4\beta}=\frac{1}{4}+\frac{D}{4\beta}=\gamma+\frac{\delta}{2}.

(19)

Since both $\beta$ and $D$ are odd integers, their sum $\beta+D$ is an even integer.It follows that if the odd integer $\beta$ does not divide the odd integer $D$ then $\gamma+\delta/2\not\in\frac{1}{2}{\mathbb{Z}}$ and therefore (13) does not hold. So, in the course of the proof we may and will assume that

\beta\mid D.

(20)

It follows that the even integer $\beta+D$ is divisible by $2\beta$ .

Taking into account that $D\equiv 1(\bmod 4)$ ,we conclude that if $\beta\equiv 3(\bmod 4)$ then $\beta+D$ is divisible by $4\beta$ and therefore

\gamma+\frac{\delta}{2}=\frac{\beta+D}{4\beta}\in{\mathbb{Z}}.

Combining with (18), we get

\frac{\delta}{2},\ \gamma+\frac{\delta}{2}\in{\mathbb{Z}}\quad\text{if}\ \beta%\equiv 3(\bmod 4).

This implies that

\gamma=\left(\gamma+\frac{\delta}{2}\right)-\frac{\delta}{2}\in{\mathbb{Z}}%\quad\text{if}\ \beta\equiv 3(\bmod 4).

On the other hand, suppose that $\beta\equiv 1(\bmod 4)$ . Then the integer

\frac{D}{\beta}\equiv 1(\bmod 4)

and therefore the integer $(\beta+D)/\beta\equiv 2(\bmod 4)$ ,which implies that

\frac{\beta+D}{4\beta}\in\frac{1}{2}{\mathbb{Z}}\setminus{\mathbb{Z}}.

In light of (18), $\frac{\delta}{2}\in\frac{1}{2}{\mathbb{Z}}\setminus{\mathbb{Z}}$ and therefore

\gamma=(\gamma+\delta/2)-\delta/2=\frac{\beta+D}{4\beta}-\delta/2\in{\mathbb{Z%}}\quad\text{if}\ \beta\equiv 1(\bmod 4).

This implies that all $\alpha,\beta,\gamma,\delta$ are integers if and only if $r=1/2\beta$ where $\beta$ is a positive odd integer dividing $D$ . This ends the proof of (ii), in light of (15).∎

Example 3.10.

Fix a negative integer $d$ that is congruent to $1$ modulo $4$ (e.g., we may take $d=-3$ ). If $k$ and $n$ are positive odd integers then $k^{2}n^{2}d$ is also a negative integer that iscongruent to $1$ modulo $4$ . If we put

D=k^{2}n^{2}d,\quad\beta=n^{2}

then $\beta$ is a positive odd integer dividing $D$ . It followsthat if we put

\tau(n,k):=\frac{1}{2}+\frac{1}{2\beta}\sqrt{D}=\frac{1}{2}+\frac{k\sqrt{d}}{2n}

(21)

then the elliptic curve $\mathcal{E}_{\tau(n,k)}$ is odd.

Recall that $n$ and $k$ could be any odd positive integers. Notice also that all CM elliptic curves $\mathcal{E}_{\tau(n,k)}$ are isogenous to each other, because the imaginary quadratic field

{\mathbb{Q}}(\tau(n,k))={\mathbb{Q}}(\sqrt{d})

does not depend on $n,k$ .

Corollary 3.11.

Let $d$ be a square-free negative integer such that $d\equiv 1(\bmod 4)$ .If $n$ and $m$ are positive odd integers and

\tau=\frac{1}{2}+\frac{n}{m}\sqrt{d}

then $\mathcal{E}_{\tau}$ is even.

Proof.

In light of Example 3.7 applied to $y=2\sqrt{|d|}$ , it suffices to check the case $n=m=1$ , i.e., we may (and will) assume that

\tau=1/2+\sqrt{d}.

Clearly, $K:={\mathbb{Q}}(\tau)={\mathbb{Q}}(\sqrt{d})$ . Suppose that $\mathcal{E}_{\tau}$ is odd, i.e., the order $O_{\tau}$ is odd.Let $\mathfrak{f}$ be the conductor of $O_{\tau}$ . By Lemma 1.1, $\mathfrak{f}$ is odd, hence,

\frac{1-\mathfrak{f}}{2}\in{\mathbb{Z}}.

It follows from [2, Ch. 2, Sect. 7, Th.1]that

O_{\tau}={\mathbb{Z}}+\mathfrak{f}\frac{1+\sqrt{d}}{2}{\mathbb{Z}}\ni\frac{1-%\mathfrak{f}}{2}+\mathfrak{f}\frac{1+\sqrt{d}}{2}=\frac{1+\mathfrak{f}\sqrt{d}%}{2}.

So,

\frac{1+\mathfrak{f}\sqrt{d}}{2}\in O_{\tau}.

Notice that

\frac{1+\mathfrak{f}\sqrt{d}}{2}=\frac{1+\sqrt{D}}{2}

where $D=\mathfrak{f}^{2}d$ is a negative integer that is also congruent to $1$ modulo $4$ , because $\mathfrak{f}$ is odd.By Lemma 3.9, there is an odd positive integer $\beta$ such that

\tau=1/2+\frac{1}{2\beta}\sqrt{D}=1/2+\frac{\mathfrak{f}}{2\beta}\sqrt{d}.

Since $\tau=1/2+\sqrt{d}$ , we get $\frac{\mathfrak{f}}{2\beta}=1$ , i.e., $\mathfrak{f}=2\beta$ . Since $\mathfrak{f}$ is odd, we get the desired contradiction.∎

4. $j$ -invariants: the real case

We will need the following results that either contained in [12, Ch. V, Sect. 2] (see also [7, Ch. 14, Sect. 4] and [6, Ch. 1, Sect. 4.3]) or follow readily from them.

•

Let us consider the disjoint subsets $\mathfrak{T}_{1}$ and $\mathfrak{T}_{2}$ of $\mathfrak{H}$ defined as follows.

\mathfrak{T}_{1}:=\{\mathbf{i}t\mid t\in{\mathbb{R}},\ t\geq 1\};\quad%\mathfrak{T}_{2}:=\{\frac{1}{2}+\mathbf{i}t\mid t\in{\mathbb{R}},\ t>1/2\}.

Let us put

\mathfrak{T}:=\mathfrak{T}_{1}\cup\mathfrak{T}_{2}\subset\mathfrak{H}.

If $s$ is a real number then there is precisely one $\tau\in\mathfrak{T}$ such that

j(\mathcal{E}_{\tau})=j(\tau)=s.

In particular, the continuous function

F:\mathfrak{T}\to{\mathbb{R}},\ \tau\mapsto j(\tau)=j(\mathcal{E}_{\tau})

is a bijective map [12, Ch. V, Sect. 2, p. 417]

•

The real-valued function

f:[1/2,\infty]\to(-\infty,\infty),\quad y\mapsto\tau=1/2+iy\mapsto j(\tau)=j(%\mathcal{E}_{\tau})

is continuous injective, and

f(1/2)=1728,\ f(\sqrt{3}/2)=0,\ \lim_{y\to\infty}f(y)=-\infty

(see [12, Ch. V, Sect. 2, p. 414]).It follows that $f$ is decreasing and the image of the map $f$ lies in the closed semi-infinite interval $(-\infty,1728]$ .So, one may view as a decreasing function the map

f:[1/2,\infty)\to(-\infty,1728],\quad y\mapsto\tau=1/2+iy\to j(\tau)=j(%\mathcal{E}_{\tau}),

which is obviously bijective. In addition, the image of the open semi-infinite interval $(1/2,\infty)$ under $f$ is the open semi-infinite interval $(-\infty,1728)$ .

•

The bijectiveness of the continuous map $F:\mathfrak{T}\to{\mathbb{R}}$ implies that the set

\{j(\mathcal{E}_{\tau})\mid\tau\in\mathfrak{T}_{1}\}

coincides with the closed semi-infinite interval $[1728,\infty)$ .

Proof of Theorem 2.4.

Withous loss of generality we may and will assume that $E=\mathcal{E}_{\tau}$ with $\tau\in\mathfrak{T}$ .

Suppose that $\mathcal{E}_{\tau}$ is odd. In light of Lemma 3.9(i), $\mathrm{Re}(\tau)\neq 0$ .Since $\tau\in\mathfrak{T}$ , we get $\mathrm{Re}(\tau)=1/2$ , i.e., $\tau=\frac{1+\mathbf{i}y}{2}$ for some real positive $y$ . In other words,

\tau\in\mathfrak{T}_{2}.

Let $E^{\prime}$ be any odd CM elliptic curve with real $j(E^{\prime})$ such that $\mathcal{E}_{\tau}$ and $E^{\prime}$ are isogenous over ${\mathbb{C}}$ .Without loss of generality we may assume that

E^{\prime}=\mathcal{E}_{\tau^{\prime}},\ j(E^{\prime})=j(\mathcal{E}_{\tau^{%\prime}})=j(\tau^{\prime})

for a certain $\tau^{\prime}\in\mathfrak{T}$ . Applying Lemma 3.9(i) to $\mathcal{E}_{\tau^{\prime}}$ (instead of $\mathcal{E}_{\tau}$ ),we conclude that $\mathrm{Re}(\tau^{\prime})\neq 0$ and therefore

\tau^{\prime}\in\mathfrak{T}_{2}.

It follows that $\mathcal{J}({\mathbb{R}},E)$ lies in the open interval $(-\infty,1728)$ and thereforethe closure of $\mathcal{J}({\mathbb{R}},E)$ in ${\mathbb{R}}$ lies in the closed interval $(-\infty,1728]$ .

If $n,m$ are any positive odd integers such that

\frac{m}{n}y>1

then

\frac{m}{n}y/2>1/2,

and the set of allsuch $\frac{m}{n}y/2$ is dense in $(1/2,\infty)$ . If we put

\tau_{m,n}:=\frac{1+\mathbf{i}\frac{m}{n}y}{2}\in\mathfrak{T}

then the set of all $\tau_{m,n}$ is a dense subset of $\mathfrak{T}_{2}$ .By Example 3.7,all CM elliptic curves $\mathcal{E}_{\tau_{m,n}}$ are odd and isogenous to $\mathcal{E}_{\tau}$ . In addition,

f\left(\frac{m}{n}y/2\right)=j\left(\mathcal{E}_{\tau_{m,n}}\right)\in\mathcal%{J}({\mathbb{R}},\mathcal{E}_{\tau})\subset(-\infty,1728].

It follows from the continuity of $f$ that the set of all $j(\mathcal{E}_{\tau_{m,n}})$ is dense in $f((1/2,\infty])=(-\infty,1728)$ . This implies that the closure of $\mathcal{J}({\mathbb{R}},\mathcal{E}_{\tau})$ coincides with $(-\infty,1728]$ when $\mathcal{E}_{\tau}$ is odd.

Suppose that $\mathcal{E}_{\tau}$ is even. Then ${\mathbb{Q}}(\tau)={\mathbb{Q}}(\sqrt{d})$ where $d$ is a square-free negative integer.If $n,m$ are any positive integers such that

\frac{m}{n}\sqrt{|d|}>1

then the CM elliptic curves $E_{\frac{m}{n}\sqrt{d}}$ are isogenous to $\mathcal{E}_{\tau}$ ;in addition, they all are even, thanks to Lemma 3.9(i). On the other hand, obviously,the setof all such $\frac{m}{n}\sqrt{d}$ lies in $\mathfrak{T}_{1}$ (in particular, all $j\left(E_{\frac{m}{n}\sqrt{d}}\right)\in{\mathbb{R}}$ ) and dense there. This implies thatthe closure of $\mathcal{J}({\mathbb{R}},\mathcal{E}_{\tau})$ contains $F(\mathfrak{T}_{1})=[1728,\infty)$ .

Suppose that $d\not\equiv 1(\bmod 4)$ . Then the quadratic field ${\mathbb{Q}}(\sqrt{d})$ has discriminant $4d$ , which is even. It follows from Lemma 1.2that every elliptic curve $\mathcal{E}_{\tau^{\prime}}$ with

{\mathbb{Q}}(\tau^{\prime})={\mathbb{Q}}(\sqrt{d})={\mathbb{Q}}(\tau)

is even and isogenous to $\mathcal{E}_{\tau}$ .In particular, if $m,n$ are any positive integers such that $\frac{m}{n}\sqrt{|d|}>1$ then

\tau(m,n):=1/2+\frac{m}{n}\sqrt{d}\in\mathfrak{T}_{2},\quad{\mathbb{Q}}(\tau(m%,n))={\mathbb{Q}}(\sqrt{d}),

which implies that the elliptic curve $\mathcal{E}_{\tau(m,n)}$ is isogenous to $\mathcal{E}_{\tau}$ and even. Since $\tau(m,n)\in\mathfrak{T}_{2}$ , we get

j\left(\mathcal{E}_{\tau(m,n)}\right)\in{\mathbb{R}}\ \text{ and therefore}\ j%\left(\mathcal{E}_{\tau(m,n)}\right)\in\mathcal{J}({\mathbb{R}},\mathcal{E}_{%\tau}).

On the other hand, the set of all such $\tau(m,n)$ is obviously dense in $\mathfrak{T}_{2}$ . It followsthat the closure of $\mathcal{J}({\mathbb{R}},\mathcal{E}_{\tau})$ contains $f\left(\mathfrak{T}_{1}\right)=(-\infty,1728)$ .Recall that we have already checked that this closure contains $[1728,\infty)$ . It follows thatthe closure of $\mathcal{J}({\mathbb{R}},\mathcal{E}_{\tau})$ coincides with the whole ${\mathbb{R}}$ if $d\not\equiv 1(\bmod 4)$ .

Now suppose that $d\equiv 1(\bmod 4)$ . If $m,n$ are any positive odd integers such that $\frac{m}{n}\sqrt{|d|}>1/2$ then

\tau(m,n):=\frac{1}{2}+\frac{m}{n}\sqrt{d}\in\mathfrak{T}_{2},\quad j\left(%\mathcal{E}_{\tau(m,n)}\right)\in{\mathbb{R}},

and the elliptic curve $\mathcal{E}_{\tau(m,n)}$ is isogenous to $\mathcal{E}_{\tau}$ .By Corollary 3.11, $\mathcal{E}_{\tau(m,n)}$ is evenand therefore $j\left(\mathcal{E}_{\tau(m,n)}\right)\in\mathcal{J}({\mathbb{R}},\mathcal{E}_{%\tau})$ .Clearly, the set of all such $\tau(m,n)$ is dense in $\mathfrak{T}_{2}$ . It follows thatthe closure of $\mathcal{J}({\mathbb{R}},\mathcal{E}_{\tau})$ contains $F(\mathfrak{T}_{2})=(-\infty,1728)$ .Recall that we have already checked that this closure contains $[1728,\infty)$ . It follows thatthe closure of $\mathcal{J}({\mathbb{R}},\mathcal{E}_{\tau})$ coincides with the whole ${\mathbb{R}}$ if $d\equiv 1(\bmod 4)$ .This ends the proof of Theorem 2.4.∎

Remark 4.1.

Let $E_{{\mathbb{R}}}$ be an elliptic curve over ${\mathbb{R}}$ and let $E=E_{{\mathbb{R}}}\times_{{\mathbb{R}}}{\mathbb{C}}$ be its complexification.Then there is $\tau\in\mathfrak{H}$ that enjoys the following properties (see [12, Ch. V, Sect. 5.2, Th. 2.3 and its proof]).

(i)
$\mathrm{Re}(\tau)=0$ or $1/2$ .

(ii)

$E$ is isomorphic to $\mathcal{E}_{\tau}$ . In particular,

j(E_{{\mathbb{R}}})=j(E)=j(\tau)\in{\mathbb{R}}.

(iii)
$q:=\exp(2\pi\mathbf{i}\tau)$ is a nonzero real number such that $|q|<1$ and the real Lie group $E_{{\mathbb{R}}}({\mathbb{R}})$ of ${\mathbb{R}}$ -points on $E$ is isomorphic to the quotient ${\mathbb{R}}^{*}/q^{{\mathbb{Z}}}$ .(Here $q^{{\mathbb{Z}}}$ is the cyclic multiplicative subgroup of ${\mathbb{R}}^{*}$ generated by $q$ .)In particular, $E_{{\mathbb{R}}}({\mathbb{R}})$ is connected if and only if $q<0$ , i.e., $\mathrm{Re}(\tau)=1/2$ .
(iv)
Suppose that $E$ is an odd elliptic curve with CM. It follows from Lemma 3.9(i) that $\mathrm{Re}(\tau)=1/2$ . In light of (iii), $E_{{\mathbb{R}}}({\mathbb{R}})$ is connected.

5. Odd and even orders

Let $O$ be an order in a quadratic field $K$ with conductor $\mathfrak{f}$ .Lemmas 1.1 and 1.2 follow from the next two assertions.

Lemma 5.1.

The following conditions are equivalent.

(i)
$\mathrm{tr}(O)={\mathbb{Z}}$ .
(ii)
$\mathrm{tr}(O[1/n])$ contains $1$ for all odd positive integers $n$ .
(iii)
$\mathrm{discr}(O)$ is an odd integer.
(iv)
Both $\mathrm{discr}(K)$ and the conductor $\mathfrak{f}_{O}$ are odd integers.
(v)
There exists $w\in O$ such that $\mathrm{tr}(w)$ is an odd integer.
(vi)
There exists an integer $D\equiv 1(\bmod 4)$ that is not a square such that
$\frac{1+\sqrt{D}}{2}\in O.$

(vii)

There exists an integer $D\equiv 1(\bmod 4)$ that is not a square such that

O={\mathbb{Z}}+{\mathbb{Z}}\ \frac{1+\sqrt{D}}{2}={\mathbb{Z}}\left[\frac{1+%\sqrt{D}}{2}\right].

Lemma 5.2.

The following conditions are equivalent.

(i)
$\mathrm{tr}(O)=2{\mathbb{Z}}$ .
(ii)
$\mathrm{tr}(O[1/n])$ does not contain $1$ for all odd positive integers $n$ .
(iii)
$\mathrm{discr}(O)$ is an even integer.
(iv)
Either $\mathrm{discr}(K)$ or the conductor $\mathfrak{f}_{O}$ is an even integer.
(v)
There exists an integer $D$ that is not a square and such that $O={\mathbb{Z}}+{\mathbb{Z}}\ \sqrt{D}$ .

Proof of Lemmas 5.1 and 5.2.

Recall ((1) and (2)) that

\mathrm{tr}(O)={\mathbb{Z}}\ \text{ or }2{\mathbb{Z}},\quad\mathrm{discr}(O)=%\mathrm{discr}(K)\cdot\mathfrak{f}^{2}.

(22)

Step 1 Let $n$ be an odd positive integer. Then

\mathrm{tr}(O[1/n])={\mathbb{Z}}[1/n]\cdot\mathrm{tr}(O)\subset{\mathbb{Q}}.

It follows that

\mathrm{tr}(O[1/n])={\mathbb{Z}}[1/n]\ni 1\ \text{ if }\mathrm{tr}(O)={\mathbb%{Z}};

\mathrm{tr}(O[1/n])=2\cdot{\mathbb{Z}}[1/n]\not\ni 1\ \text{ if }\mathrm{tr}(O%)=2{\mathbb{Z}}.

Now the equivalence of (i) and (ii) follows from first equality of (22) (for both Lemmas).The equivalence of (iii) and (iv) follows from second equality of (22) (for both Lemmas).

Step 2 Let us concentrate for a while on the proof of Lemma5.1. The equivalence of (i) and (v) follows from first equality of (22). In addition,(vii) obviously implies (vi) while (vi) implies (v), because if we put $w=\frac{1+\sqrt{D}}{2}$ then

\mathrm{tr}(w)=\mathrm{tr}\left(\frac{1+\sqrt{D}}{2}\right)=\frac{1+\sqrt{D}}{%2}+\frac{1-\sqrt{D}}{2}=1

is an odd integer. On the other hand, (vii) implies that $\mathrm{disc}(O)=D$ , which is odd. It follows that(vii) implies (iii). So, in order to finish the proof of Lemma 5.1, it suffices to check that

(1o)
(i) implies (vii);
(2o)
(iii) implies (i).

Now let us switch to Lemma 5.2.Clearly, (v) implies both(i). On the other hand, (v) implies that $\mathrm{disc}(O)=4D$ , which is odd. It follows that (v) implies (iii).So, in order to finish the proof of Lemma 5.2, it suffices to check that

(1e)
(i) implies (v);
(2e)
(iii) implies (i).

Step 3 We have $K={\mathbb{Q}}(\sqrt{d})$ where $d\neq 1$ is a square-free integer. Clearly,

\mathrm{tr}(\sqrt{d})=0.

(23)

Suppose that $d\not\equiv 1(\bmod 4)$ .It is known [2, Ch. 2, Sect. 7, Th.1] that

\mathrm{disc}(K)=4d,\ \ O={\mathbb{Z}}+\mathfrak{f}\sqrt{d}\ {\mathbb{Z}}={%\mathbb{Z}}+\sqrt{\mathfrak{f}^{2}d}\ {\mathbb{Z}},\quad\mathrm{disc}(O)=%\mathrm{disc}(K)\mathfrak{f}^{2}=4d\mathfrak{f}^{2}.

In particular, $\mathrm{disc}(O)$ is even. On the other hand, if we put $D:=\mathfrak{f}^{2}d$ then $D$ is not a square (because $d$ is square-free) and

K={\mathbb{Q}}(\sqrt{d})={\mathbb{Q}}(\sqrt{D}),\quad O={\mathbb{Z}}+\sqrt{D}%\ {\mathbb{Z}},\quad\mathrm{discr}(O)=4d\mathfrak{f}^{2}=4D.

It follows from (23) that $\mathrm{tr}(O)=2{\mathbb{Z}}$ . This implies that $O$ enjoys properties (i), (iii), (v) of Lemma 5.2.On the other hand, $O$ does not enjoy any of properties (i) and (iii) of Lemma 5.1. In light of Step 2, this proves both Lemmas in the (last remaining) case when $d\not\equiv 1(\bmod 4)$ .

Suppose that $d\equiv 1(\bmod 4)$ .It is known [2, Ch. 2, Sect. 7, Th.1] that

\mathrm{disc}(K)=d,\quad O={\mathbb{Z}}+\mathfrak{f}\frac{1+\sqrt{d}}{2}{%\mathbb{Z}},\ \mathrm{disc}(O)=\mathrm{disc}(K)\mathfrak{f}^{2}=d\mathfrak{f}^%{2}.

It follows from (23) that

\mathrm{tr}(O)={\mathbb{Z}}\ \text{ if }\mathfrak{f}\ \text{ is odd};\quad%\mathrm{tr}(O)=2{\mathbb{Z}}\ \text{ if }\mathfrak{f}\ \text{ is even}.

(24)

If $\mathfrak{f}$ is even then $\mathrm{tr}(O)=2{\mathbb{Z}}$ , $\mathrm{disc}(O)$ is even, $m=\mathfrak{f}/2$ is a positive integer, and

O={\mathbb{Z}}+\mathfrak{f}\frac{1+\sqrt{d}}{2}{\mathbb{Z}}={\mathbb{Z}}+m%\left(1+\sqrt{d}\right){\mathbb{Z}}={\mathbb{Z}}+m\sqrt{d}\ {\mathbb{Z}}=

{\mathbb{Z}}+\sqrt{m^{2}d}\ {\mathbb{Z}}={\mathbb{Z}}+\sqrt{D}\ {\mathbb{Z}}

where $D:=m^{2}d$ , which is not a square, because $d$ is square-free. We get

K={\mathbb{Q}}(\sqrt{d})={\mathbb{Q}}(\sqrt{D}),\quad O={\mathbb{Z}}+\sqrt{D}%\ {\mathbb{Z}},\quad\mathrm{discr}(O)=\mathfrak{f}^{2}d=(2m)^{2}d=4D.

This implies that $O$ enjoys all the properties (i), (iii), (v) of Lemma 5.2.On the other hand, $O$ does not enjoy any of properties (i) and (iii) of Lemma 5.1.In light of Step 2, this proves both Lemmas in the case when $d\not\equiv 1(\bmod 4)$ and $\mathfrak{f}$ is even.

Now suppose that $\mathfrak{f}$ is odd. It follows from (24) that $\mathrm{tr}(O)={\mathbb{Z}}$ and $\mathrm{discr}(O)$ is odd. In addition, if we put $m:=(\mathfrak{f}-1)/2\in{\mathbb{Z}}$ then $\mathfrak{f}=2m+1$ and

O={\mathbb{Z}}+\mathfrak{f}\frac{1+\sqrt{d}}{2}{\mathbb{Z}}={\mathbb{Z}}+\frac%{\mathfrak{f}+\mathfrak{f}\sqrt{d}}{2}\ {\mathbb{Z}}={\mathbb{Z}}+\left(m+%\frac{1+\mathfrak{f}\sqrt{d}}{2}\right)\ {\mathbb{Z}}=

{\mathbb{Z}}+\frac{1+\mathfrak{f}\sqrt{d}}{2}{\mathbb{Z}}={\mathbb{Z}}+\frac{1%+\sqrt{\mathfrak{f}^{2}d}}{2}\ {\mathbb{Z}}={\mathbb{Z}}+\frac{1+\sqrt{D}}{2}{%\mathbb{Z}}

where $D:=\mathfrak{f}^{2}d$ is not a square, because $d$ is square-free. Since $\mathfrak{f}$ is odd, $\mathfrak{f}^{2}\equiv 1(\bmod 4)$ .Since $d\equiv 1(\bmod 4)$ , the product $D:=\mathfrak{f}^{2}d$ is also congruent to $1$ modulo $4$ . We get

K={\mathbb{Q}}(\sqrt{d})={\mathbb{Q}}(\sqrt{D}),\quad O={\mathbb{Z}}+\frac{1+%\sqrt{D}}{2}\ {\mathbb{Z}},\quad\mathrm{discr}(O)=\mathfrak{f}^{2}d=D.

This implies that $O$ enjoys properties (i), (iii), (vii) of Lemma 5.1.On the other hand, $O$ does not enjoy any of properties (i) and (iii) of Lemma 5.2. In light of Step 2, this proves both Lemmas in the case when $d\equiv 1(\bmod 4)$ and $\mathfrak{f}$ is odd. This ends the proof of both Lemmas.∎

Remark 5.3.

Let $n$ be a positive odd integer, $\phi:K_{1}\to K_{2}$ an isomorphism of quadratic fields,and $O_{1}$ and $O_{2}$ are orders in $K_{1}$ and $K_{2}$ respectively.Suppose that

\phi(O_{1}[1/n])=O_{2}[1/n].

Then

\mathrm{tr}_{K_{1}/{\mathbb{Q}}}(O_{1}[1/n])=\mathrm{tr}_{K_{2}/{\mathbb{Q}}}(%O_{2}[1/n]).

It follows from Lemmas 5.1(ii) and 5.2(ii) that $O_{1}$ and $O_{2}$ have the same parity,i.e., either they both are odd or both even.

Remark 5.4.

Recall that if $\mathfrak{f}$ is an odd integer then $\mathfrak{f}^{2}\equiv 1\ (\bmod 4)$ .On the other hand, if $\mathfrak{f}$ is an even integer then $\mathfrak{f}^{2}\equiv 0\ (\bmod 4).$

Let $O$ be an order in a quadratic field $K$ . Now the explicit formulas for its discriminant (see Step 3 above) imply the following.

(e)

$O$ is even if and only if $\mathrm{disc}(O)\equiv 0\ (\bmod 4)$ . If this is the case then there is an integer $D$ that is not a square and such that

K={\mathbb{Q}}(\sqrt{D}),\quad O={\mathbb{Z}}+{\mathbb{Z}}\ \sqrt{D},\quad%\mathrm{discr}(O)=4D.

(o)

$O$ is odd if and only if $\mathrm{disc}(O)\equiv 1\ (\bmod 4)$ . If this is the case then there is an integer $D\equiv 1\ (\bmod 4)$ that is not a square and such that

K={\mathbb{Q}}(\sqrt{D}),\quad O={\mathbb{Z}}+{\mathbb{Z}}\ \frac{1+\sqrt{D}}{%2},\quad\mathrm{discr}(O)=D.

6. Isogenies of Elliptic Curves

Proof of Proposition 1.4.

There is a (dual) isogeny $\psi:E_{2}\to E_{1}$ such that $\psi\circ\phi:E_{1}\to E_{1}$ is multiplication by $n$ in $E_{1}$ and $\phi\circ\psi:E_{2}\to E_{2}$ is multiplication by $n$ in $E_{2}$ [6, p. 7]. Then we get an isomorphism of ${\mathbb{Q}}$ -algebras

\Phi:\mathrm{End}^{0}(E_{1})=\mathrm{End}(E_{1})\otimes{\mathbb{Q}}\to\mathrm{%End}(E_{2})\otimes{\mathbb{Q}}=\mathrm{End}^{0}(E_{2}),\ u_{1}\mapsto\frac{1}{%n}\phi u_{1}\psi,

whose inverse is an isomorphism of ${\mathbb{Q}}$ -algebras

\Psi:\mathrm{End}^{0}(E_{2})=\mathrm{End}(E_{2})\otimes{\mathbb{Q}}\to\mathrm{%End}(E_{1})\otimes{\mathbb{Q}}=\mathrm{End}^{0}(E_{1}),\ u_{2}\mapsto\frac{1}{%n}\psi u_{2}\phi.

Clearly, both $\Phi$ and $\Psi$ sends $1$ to $1$ , sums to sums and therefore are hom*omorphisms of ${\mathbb{Q}}$ -vector spaces. On the other handif $u_{1}\in\mathrm{End}^{0}(E_{1})$ then

\Psi(\Phi(u_{1}))=\frac{1}{n}\psi\Phi(u_{1})\phi=\frac{1}{n}\psi\left(\frac{1}%{n}\phi u_{1}\psi\right)\phi=\frac{1}{n}\left(\frac{1}{n}\psi\phi\right)u_{1}%\left(\frac{1}{n}\psi\phi\right)=u_{1}.

Since both $\mathrm{End}^{0}(E_{1})$ and $\mathrm{End}^{0}(E_{2})$ are ${\mathbb{Q}}$ -vector spaces of the same dimension 2, $\Phi$ is an isomorphism of ${\mathbb{Q}}$ -vector spaces and $\Psi$ is itsinverse, hence, also an isomorphism of ${\mathbb{Q}}$ -vector spaces.The only thing that remains to check is that both $\Phi$ and $\Psi$ are compatible with multiplication. Let us check it.If $u_{1},v_{1}\in\mathrm{End}^{0}(E_{1})$ then

\Phi(u_{1})\Phi(v_{2})=\frac{1}{n}\phi u_{1}\psi\frac{1}{n}\phi v_{1}\psi=%\frac{1}{n}\phi u_{1}\frac{1}{n}(\psi\phi)v_{1}\psi=\frac{1}{n}\phi u_{1}v_{1}%\psi=\Phi(u_{1}v_{1}).

This proves that $\Phi$ is a hom*omorphism of ${\mathbb{Q}}$ -algebras and therefore is an isomorphism of quadratic fields $\mathrm{End}^{0}(E_{1})$ and $\mathrm{End}^{0}(E_{2})$ .Hence, its inverse $\Psi$ is also a field isomorphism.

Now I claim that

\Phi(\mathrm{End}(E_{1})[1/n])=\mathrm{End}(E_{2})[1/n],\quad\Psi(\mathrm{End}%(E_{2})[1/n])=\mathrm{End}(E_{1})[1/n].

(25)

Indeed, it follows from the definition of $\Phi$ and $\Psi$ that

\Phi(\mathrm{End}(E_{1}))\subset\frac{1}{n}\mathrm{End}(E_{2})\subset\mathrm{%End}(E_{2})[1/n],

\Psi(\mathrm{End}(E_{2}))\subset\frac{1}{n}\mathrm{End}(E_{1})\subset\mathrm{%End}(E_{1})[1/n].

It follows that

\Phi(\mathrm{End}(E_{1})[1/n])\subset\mathrm{End}(E_{2})[1/n],\quad\Psi(%\mathrm{End}(E_{2})[1/n])\subset\mathrm{End}(E_{1})[1/n]

and therefore

\mathrm{End}(E_{1})[1/n])=\Psi(\Phi(\mathrm{End}(E_{1})[1/n]))\subset\Psi(%\mathrm{End}(E_{2})[1/n]),

\mathrm{End}(E_{2})[1/n]=\Phi(\Psi(\mathrm{End}(E_{2})[1/n]\subset\Phi(\mathrm%{End}(E_{1})[1/n]).

We get

\mathrm{End}(E_{1})[1/n])\subset\Psi(\mathrm{End}(E_{2})[1/n])]\subset\mathrm{%End}(E_{1})[1/n],

\mathrm{End}(E_{2})[1/n]\subset\Phi(\mathrm{End}(E_{1})[1/n])\subset\mathrm{%End}(E_{2})[1/n],

which proves (25).

So, $\Phi$ is an isomorphism of quadratic fields $\mathrm{End}^{0}(E_{1})$ and $\mathrm{End}^{0}(E_{2})$ such that

\Phi(\mathrm{End}(E_{1})[1/n])=\mathrm{End}(E_{2})[1/n].

Now the desired result follows from Remark 5.3.∎

Proof of Proposition 1.5.

We may assume that

E_{1}=E_{1,{\mathbb{R}}}\times_{{\mathbb{R}}}{\mathbb{C}},\quad E_{2}=E_{2,{%\mathbb{R}}}\times_{{\mathbb{R}}}{\mathbb{C}}.

Let $\phi:E_{1}\to E_{2}$ be an isogeny and $\bar{\phi}:E_{1}\to E_{2}$ its “complex-conjugate”.If $\bar{\phi}=\phi$ then $\phi$ could be descended to an isogeny $E_{1,{\mathbb{R}}}\to E_{2,{\mathbb{R}}}$ over ${\mathbb{R}}$ .

Suppose that $\bar{\phi}\neq\phi$ . Then its difference $\psi:=\bar{\phi}-\phi:E_{1}\to E_{2}$ is a nonzero hom*omorphism of elliptic curves and therefore is an isogeny.Its complex-conjugate $\bar{\psi}=\overline{\bar{\phi}-\phi}$ coincides with $\phi-\bar{\phi}=-\psi$ . On the other hand, it is well known that not all endomorphismsof the CM elliptic curve $E_{1}$ are defined over ${\mathbb{R}}$ , i.e., there is an endomorphism $\alpha$ of $E_{1}$ such that its complex-conjugate $\bar{\alpha}$ does not coincide with $\alpha$ (see Lemma 6.1 below).This means that $\beta=\bar{\alpha}-\alpha:E_{1}\to E_{1}$ is a nonzero endomorphism of $E_{1}$ ,i.e., is an isogeny. Clearly, the complex conjugate $\bar{\beta}$ of $\beta$ coincides with

\overline{\bar{\alpha}-\alpha}=\alpha-\bar{\alpha}=-\beta.

So,

\bar{\psi}=-\psi,\quad\bar{\beta}=-\beta.

This implies that the composition $\lambda:=\psi\circ\beta:E_{1}\to E_{2}$ is an isogeny, whose complex conjugate $\bar{\lambda}=\overline{\psi\circ\beta}$ equals

\bar{\psi}\circ\bar{\beta}=(-\psi)\circ(-\beta)=\psi\circ\beta=\lambda.

So, the nonzero hom*omorphism $\lambda$ coincides with its “complex-conjugate”,i.e., is defined over ${\mathbb{R}}$ . This means thatthere is a ${\mathbb{R}}$ -hom*omorphism $\lambda_{{\mathbb{R}}}:E_{1,{\mathbb{R}}}\to E_{2,{\mathbb{R}}}$ of elliptic curves, whose “comlexification” $E_{1}\to E_{2}$ coincides with $\lambda$ . Since $\lambda\neq 0$ , we obtain that $\lambda_{{\mathbb{R}}}\neq 0$ and therefore is an isogeny.∎

The following assertion (and its proof that I am going to reproduce) is pretty well known but I was unable to find a reference.

Lemma 6.1.

Let $E_{{\mathbb{R}}}$ be an elliptic curve over ${\mathbb{R}}$ . Then the ring $\mathrm{End}_{{\mathbb{R}}}(E_{{\mathbb{R}}})$ of ${\mathbb{R}}$ -endomorphisms of $E_{{\mathbb{R}}}$ is ${\mathbb{Z}}$ and the corresponding endomorphism algebra

\mathrm{End}_{{\mathbb{R}}}^{0}(E_{{\mathbb{R}}}):=\mathrm{End}_{{\mathbb{R}}}%(E_{{\mathbb{R}}})\otimes{\mathbb{Q}}

is ${\mathbb{Q}}$ .

Proof.

Let us consider the complexification $E=E_{{\mathbb{R}}}\times_{{\mathbb{R}}}{\mathbb{C}}$ of $E_{{\mathbb{R}}}$ . We have

{\mathbb{Z}}\subset\mathrm{End}_{{\mathbb{R}}}(E_{{\mathbb{R}}})\subset\mathrm%{End}(E),\quad{\mathbb{Q}}\subset\mathrm{End}_{{\mathbb{R}}}^{0}(E_{{\mathbb{R%}}})\subset\mathrm{End}^{0}(E).

If $\mathrm{End}_{{\mathbb{R}}}(E_{{\mathbb{R}}})\neq{\mathbb{Z}}$ then

\mathrm{End}(E)\neq{\mathbb{Z}},\quad\mathrm{End}^{0}(E)\neq{\mathbb{Q}},

i.e., $E$ is an elliptic curve with CMand $K:=\mathrm{End}^{0}(E)$ is an imaginary quadratic field. This implies that

K=\mathrm{End}^{0}(E)=\mathrm{End}_{{\mathbb{R}}}^{0}(E_{{\mathbb{R}}})

and $\mathrm{End}_{{\mathbb{R}}}(E_{{\mathbb{R}}})$ is an order in the quadratic field $K$ . Let $\Omega^{1}_{{\mathbb{R}}}(E_{{\mathbb{R}}})$ be the space of differentials of the first kind on $E_{{\mathbb{R}}}$ ,which is a ${\mathbb{R}}$ -vector space of dimension $1$ . By functoriality, $\mathrm{End}_{{\mathbb{R}}}(E_{{\mathbb{R}}})$ acts on $\Omega^{1}_{{\mathbb{R}}}(E_{{\mathbb{R}}})$ ,which gives us the injective ring hom*omorphism

\delta:\mathrm{End}_{{\mathbb{R}}}(E_{{\mathbb{R}}})\hookrightarrow\mathrm{End%}_{{\mathbb{R}}}(\Omega^{1}_{{\mathbb{R}}}(E_{{\mathbb{R}}}))={\mathbb{R}}

[10, Ch. I, Sect. 2.8]. By ${\mathbb{Q}}$ -linearity, $\delta$ extends to the hom*omorphism of ${\mathbb{Q}}$ -algebras

\delta_{{\mathbb{Q}}}:K=\mathrm{End}_{{\mathbb{R}}}^{0}(E_{{\mathbb{R}}})=%\mathrm{End}_{{\mathbb{R}}}(E_{{\mathbb{R}}})\otimes{\mathbb{Q}}\to\mathrm{End%}_{{\mathbb{R}}}(\Omega^{1}_{{\mathbb{R}}}(E_{{\mathbb{R}}}))={\mathbb{R}},

which is also injective (since $K$ is a field). So, $\delta_{{\mathbb{Q}}}:K\to{\mathbb{R}}$ is a field embedding. But such an embedding does not exist,since $K$ is an imaginary quadratic field. The obtained contradiction proves the desired result.∎

7. Quadratic Exercises

In light of Theorem 7.4(ii) (see below), it is natural to “classify” (count the number of) divisors $\beta$ of the discriminant $D$ that are relatively prime to $D/\beta$ . Let us start with the following definition (notation).

Definition 7.1 (Definition-Notation).

Let $n$ be a nonzero integer. We write $P_{n}$ for the (finite) set of prime divisors of $n$ . By Main Theorem of Arithmetic,

n=\pm\prod_{p\in P_{n}}p^{e_{n,p}}

where $e_{n,p}$ are certain positive integers uniquely determined by $n$ .

Let us call a divisor $r$ of $n$ a saturated divisor if the integers $r$ and $n/r$ are relatively prime.

Clearly, $r$ is a saturated divisor of $n$ if and only if $|n|/r$ is one. It is also clear that $r$ is a saturated divisor of $n$ if and only if $-r$ is also one.

The following assertion is an easy exercise in elementary number theory.

Proposition 7.2.

(i)

Let $S$ be a subset of $P_{n}$ . Then both integers

\mathbf{r}_{n,S}=\prod_{p\in S}p^{e_{n,p}}

and $-\mathbf{r}_{n,S}$ are saturated divisors of $n$ .(As usual, if $S$ is the empty set then $r_{n,\emptyset}=1$ .)Conversely, if $c$ is a saturated divisor of $n$ then $-c$ is also a saturated divisor of $n$ and

c=\pm\prod_{p\in P_{c}}p^{e_{n,p}}=\pm\mathbf{r}_{n,P_{c}}.

(ii)
The number of positive saturated divisors of $n$ is the number of subsets of $P_{n}$ , i.e., $2^{\#(P_{n})}$ where $\#(P_{n})$ is the cardinality of $P_{n}$ , i.e., the number of prime divisors of $n$ .

(iii)

Let $c_{1}$ and $c_{2}$ be positive saturated divisors of $n$ . Then their greatest common divisor $c_{0}$ and least common multiple $c_{12}$ are also saturated divisors of $n$ . In addition, the integers $c_{10}=c_{1}/c_{0}$ and $c_{20}=c_{2}/c_{0}$ are also positive saturated divisors of $n$ .We also have

c_{0}=\mathbf{r}_{n,S_{0}},\ c_{12}=\mathbf{r}_{n,S_{12}},\ c_{10}=\mathbf{r}_%{n,S_{10}},\ c_{20}=\mathbf{r}_{n,S_{10}},\ \frac{c_{1}c_{2}}{c_{0}^{2}}=%\mathbf{r}_{n,S_{1}\Delta S_{2}}

where

S_{1}=P_{c_{1}},\ S_{2}=P_{c_{2}},\ S_{0}=S_{1}\cap S_{2},\ S_{12}=S_{1}\cup S%_{2},\ S_{10}=S_{1}\setminus S_{0},\ S_{20}=S_{2}\setminus S_{0}

and

S_{1}\Delta S_{2}=(S_{1}\setminus S_{2})\cup(S_{2}\setminus S_{1})

is the symmetric difference of $S_{1}$ and $S_{2}$ .

(iv)

If $S_{1}$ and $S_{2}$ are disjoint subsets of $P_{n}$ then $\mathbf{r}_{n,S_{1}}$ and $\mathbf{r}_{n,S_{2}}$ are relatively prime and

\mathbf{r}_{n,S_{1}\cup S_{2}}=\mathbf{r}_{n,S_{1}}\mathbf{r}_{n,S_{2}}.

(v)
Suppose that $n$ is a negative integer that is congruent to $1$ modulo $4$ . Then the numberof positive saturated divisors $r$ of $n$ with $r<\sqrt{|n|}$ equals $2^{\#(P_{n})-1}$ .

Proof.

(i) Let $S^{\prime}=P_{n}\setminus S$ . Then

n=\pm\mathbf{r}_{n,S}\mathbf{r}_{n,S^{\prime}},\quad\frac{n}{\mathbf{r}_{n,S}}%=\pm\mathbf{r}_{n,S^{\prime}}.

If $p$ is a prime divisor of both $\mathbf{r}_{n,S}$ and $\mathbf{r}_{n,S^{\prime}}$ then it must belong to both sets $S$ and $S^{\prime}$ . However, $S$ and $S^{\prime}$ do not meet each other. Hence, $\mathbf{r}_{n,S}$ and $\mathbf{r}_{n,S^{\prime}}=\pm n/\mathbf{r}_{n,S}$ have no common prime factors and therefore are relatively prime. This implies that the divisor $\mathbf{r}_{n,S}$ (and therefore $-\mathbf{r}_{n,S}$ ) of $n$ is saturated.

Conversely, suppose that $c$ is a saturated divisor of $n$ . Then $P_{c}\subset P_{n}$ and

c=\pm\prod_{p\in P_{r}}p^{e_{c,p}},\quad 1\leq e_{c,p}\leq e_{n,p}\ \forall p%\in P_{c}.

I claim that

e_{c,p}=e_{n,p}\ \forall p\in P_{c},\ \text{i.e.},\ d=\pm\mathbf{r}_{P_{c}}.

Indeed, suppose that $e_{c,p}<e_{n,p}$ for some $p\in P_{d}$ . Then $p$ divides both $c$ and $n/c$ ,which contradicts the saturatedness of $c$ . This ends the proof.

(ii) It follows from (i) that the set of positive saturated divisors of $n$ coincides with

\{\mathbf{r}_{n,S}\mid S\subset P_{n}\}.

It is also clear that if $S_{1}$ and $S_{2}$ are distinct subsets of $P_{n}$ then $\mathbf{r}_{n,S_{1}}\neq\mathbf{r}_{n,S_{2}}$ ,because their sets of prime divisors (i.e., $S_{1}$ and $S_{2}$ ) do not coincide.So, the number of positive saturated divisors of $n$ equals the number of subsets of $P_{n}$ , which is $2^{\#(P_{n})}$ .

(iii) Since

c_{1}=\mathbf{r}_{n,S_{1}}=\prod_{p\in S_{1}}p^{e_{n,p}},\ c_{2}=\mathbf{r}_{n%,S_{2}}=\prod_{p\in S_{1}}p^{e_{n,p}},

the GCD $c_{0}$ of $c_{1}$ and $c_{2}$ equals $\prod_{p\in S_{1}\cap S_{2}}p^{e_{n,p}}$ and the LCM $c_{12}$ of $c_{1}$ and $c_{2}$ equals $\prod_{p\in S_{1}\cup S_{2}}p^{e_{n,p}}$ .Now the saturatedness of both $c_{0}$ and $c_{12}$ follows from the already proven (i). In order to handle $c_{10}$ and $c_{20}$ , notice that

c_{10}=\frac{\mathbf{r}_{n,S_{1}}}{\mathbf{r}_{n,S_{0}}}=\mathbf{r}_{n,S_{1}%\setminus S_{0}},\quad c_{20}=\frac{\mathbf{r}_{n,S_{2}}}{\mathbf{r}_{n,S_{0}}%}=\mathbf{r}_{n,S_{2}\setminus S_{0}}.

Again, it follows from (i) that both $c_{10}$ and $c_{20}$ are saturated divisors of $n$ .Since $S_{1}\Delta S_{2}$ is the disjoint union of $S_{1}\setminus S_{0}$ and $S_{2}\setminus S_{0}$ ,

\frac{c_{1}c_{2}}{c_{0}^{2}}=c_{10}c_{20}=\mathbf{r}_{n,S_{1}\setminus S_{0}}%\mathbf{r}_{n,S_{2}\setminus S_{0}}=\mathbf{r}_{n,S_{1}\Delta S_{2}},

which is also saturated, in light of (i).

(iv) follows readily from (iii).

(v) Since $n\equiv 1(\bmod 4)$ , the positive odd integer $|n|=(-n)$ is congruent to $3$ modulo $4$ and therefore is not a square. Hence, if $\beta$ is a positive saturated divisor of $n$ then $|n|/\beta$ is also a positive saturated divisor of $n$ while $\beta\neq|n|/\beta$ . Clearly,in the pair $\{\beta,|n|/\beta\}$ there is precisely one element that is strictly less than $\sqrt{|n|}$ .In light of (ii), the number of positive saturated divisors of $n$ that satisfies this inequality is $2^{\#(P_{n})}/2=2^{\#(P_{n})-1}$ .∎

Let $O=O_{D}$ be an odd order with discriminant $D$ in a quadratic field $K$ . It follows from Lemma 5.1 combined with Remark 5.4that the odd integer $D$ is not a square, $D\equiv 1\ (\bmod 4)$ and

K={\mathbb{Q}}(\sqrt{D}),\quad O={\mathbb{Z}}+{\mathbb{Z}}\ \frac{1+\sqrt{D}}{%2}\subset{\mathbb{Q}}+\sqrt{D}\ {\mathbb{Q}}=K.

If $\alpha_{1},\dots,\alpha_{n}$ are elements of $K$ then we write $[\alpha_{1},\dots,\alpha_{n}]$ for the additive subgroup $\sum_{i=1}^{n}{\mathbb{Z}}\ \alpha_{i}$ of $K$ generated by the set $\{\alpha_{1},\dots,\alpha_{n}\}$ . As usual, if $A$ and $B$ are additive subgroups of $K$ then we write $AB$ for the additive subgroup of $K$ generated by all the products $ab$ ( $a\in A,b\in B$ ).For example, if $A=[\alpha_{1},\dots,\alpha_{n}],\ B=[\mu_{1},\dots,\mu_{m}]$ then

AB=[\alpha_{1}\mu_{1},\dots,\alpha_{i}\mu_{j},\dots,\alpha_{n}\mu_{m}]=\sum_{i%=1}^{n}\sum_{j=1}^{m}{\mathbb{Z}}\ \alpha_{i}\mu_{j}.

We write $\sigma$ for the only nontrivial automorphism of the field $K$ defined by the formula

\sigma(a+b\sqrt{D})=a-b\sqrt{D}\quad\forall a,b\in{\mathbb{Q}}.

Clearly,

{\mathbb{Q}}=K^{\sigma}:=\{u\in K\mid\sigma(u)=u\}.

(26)

It is known [2, Ch. II, Sect. 7] that

\sigma(O)=O.

In what follows, $\beta$ is always an odd integer. Let us put

m_{\beta}:=\frac{\beta-1}{2}\in{\mathbb{Z}},\quad w_{\beta}=w_{\beta,D}:=\frac%{\beta+\sqrt{D}}{2}=\frac{1+\sqrt{D}}{2}+m_{\beta}\in O.

(27)

We have

w_{1}=\frac{1+\sqrt{D}}{2},\quad m_{1}=0;\quad w_{1}=w_{\beta}-m_{\beta}.

(28)

This implies that

O=[w_{1},1]=[w_{\beta},1]\quad\forall\ \beta.

(29)

We also have

O=O_{D}={\mathbb{Z}}\ \frac{1+\sqrt{D}}{2}+{\mathbb{Z}}\ \sqrt{D}=\left[w_{1},%\sqrt{D}\right],

(30)

in light of

\sqrt{D}=2w_{1}-1\in O,\quad 1=2w_{1}-\sqrt{D}\in\left[w_{1},\sqrt{D}\right],

combined with first equality of (29).

Let us consider the ${\mathbb{Z}}$ -lattice

M(\beta)=M(\beta,D):=[w_{\beta},\beta]={\mathbb{Z}}\ \frac{\beta+\sqrt{D}}{2}+%{\mathbb{Z}}\ \beta\subset K

(31)

in $K$ .It follows from (30) that $M(\beta)$ is a subgroup of $O=[w_{\beta},1]$ and the corresponding index

[O:M(\beta)]=|\beta|.

(32)

Taking into account that $\{w_{\beta},\beta\}$ is a basis of the ${\mathbb{Q}}$ -vector space $K$ , we get

M(\beta)\cap{\mathbb{Q}}={\mathbb{Z}}\ \beta=M(\beta)\cap{\mathbb{Z}}\subset{%\mathbb{Z}}.

(33)

Examples 7.3.

(i)

If $\beta=1$ then

w_{1}=\frac{1+\sqrt{D}}{2},M(1)=[w_{1},1]={\mathbb{Z}}\ \frac{1+\sqrt{D}}{2}+{%\mathbb{Z}}\ 1=O.

(ii)

If $\beta=D$ then, in light of (30),

M(D)=\left[\frac{D+\sqrt{D}}{2},D\right]=\sqrt{D}\ \left(\frac{1+\sqrt{D}}{2}{%\mathbb{Z}}+\sqrt{D}{\mathbb{Z}}\right)=\sqrt{D}\ \left[w_{1},1\right]=\sqrt{D%}\ O.

(iii)

M(-\beta)=[\beta,w_{-\beta}]=[\beta,\beta+w_{-\beta}]=[\beta,w_{\beta}]=[-%\beta,w_{\beta}]=M(\beta),

because

\beta+w_{-\beta}=\beta+\frac{-\beta+\sqrt{D}}{2}=\frac{\beta+\sqrt{D}}{2}=w_{%\beta}.

So,

M(-\beta)=M(\beta).

(34)

(iv)

Since

\sigma(w_{\beta})=\sigma\left(\frac{\beta+\sqrt{D}}{2}\right)=\frac{\beta-%\sqrt{D}}{2}=\beta-w_{\beta},

we get

\sigma(M(\beta))=[\beta,\sigma(w_{\beta})]=[\beta,\beta-w_{\beta}]=M(\beta).

So,

\sigma(M(\beta))=M(\beta).

(35)

Actually, we have checked that

M(\beta,D)=M(\beta)=\frac{\beta+\sqrt{D}}{2}\ {\mathbb{Z}}+\beta\ Z=\frac{%\beta+(-\sqrt{D})}{2}\ {\mathbb{Z}}+\beta\ Z.

This implies that $M(\beta,D)$ does not depend on the choice of a square root of $D$ .

(v)

We have

M(\beta)=[w_{\beta},\beta]=[w_{\beta},2w_{\beta}-\beta]=[w_{\beta},\sqrt{D}],

because

2w_{\beta}-\beta=2\cdot\frac{\beta+\sqrt{D}}{2}-\beta=\sqrt{D}.

So,

M(\beta)=[w_{\beta},\sqrt{D}]\ni\sqrt{D}.

(36)

(iv)

Suppose that $\beta$ divides $D$ . Then

\tilde{\beta}=-\frac{D}{\beta}

is an odd integer and

\frac{w_{\tilde{\beta}}}{\tilde{\beta}}=\frac{w_{-\beta}}{\sqrt{D}}.

(37)

Indeed,

\frac{w_{-\beta}}{\sqrt{D}}=\frac{\frac{-\beta+\sqrt{D}}{2}}{\sqrt{D}}=\frac{%\sqrt{D}}{-2D/\beta}+\frac{1}{2}=

\frac{1}{2}+\frac{\sqrt{D}}{2\tilde{\beta}}=\frac{1}{\tilde{\beta}}\cdot\frac{%\tilde{\beta}+\sqrt{D}}{2}=\frac{w_{\tilde{\beta}}}{\tilde{\beta}},

which proves (37).It follows that

M(\beta)=M(-\beta)=[w_{-\beta},\sqrt{D}]=\sqrt{D}\left[\frac{w_{-\beta}}{\sqrt%{D}},1\right]=

\sqrt{D}\left[\frac{w_{\tilde{\beta}}}{\tilde{\beta}},1\right]=\frac{\sqrt{D}}%{\tilde{\beta}}\left[w_{\tilde{\beta}},\tilde{\beta}\right]=\frac{\sqrt{D}}{%\tilde{\beta}}M(\tilde{\beta}).

This implies that $M(\beta)=\left(\sqrt{D}/\tilde{\beta}\right)M(\tilde{\beta})$ and therefore

M(\tilde{\beta})=\frac{\tilde{\beta}}{\sqrt{D}}M(\beta)=\left(-\frac{\sqrt{D}}%{\beta}\right)M(\beta)=\frac{\sqrt{D}}{\beta}M(\beta).

(38)

Theorem 7.4.

(i)
The lattice $M(\beta)$ is an ideal in the ring $O$ if and only if $\beta$ divides $D$ .

(ii)

Suppose that $\beta$ divides $D$ . Then the following conditions are equivalent.

(1)

$M(\beta)$ is a proper $O$ -ideal, i.e.,

O=\{\mu\in K\mid\mu M(\beta)\subset M(\beta)\}.

(39)

(2)
$\beta$ is a saturated divisor of $D$ , i.e.,the integers $\beta$ and $D/\beta$ are relatively prime.

If these conditions hold then $M(\beta)$ is a projective (invertible) $O$ -module of rank $1$ and

M(\beta)M(\beta)=\beta O.

(40)

(iii)

Suppose that integers $\beta_{1}$ and $\beta_{2}$ are relatively prime saturated divisors of $D$ . Then their product $\beta_{1}\beta_{2}$ is also a saturated divisor of $D$ and

M(\beta_{1})M(\beta_{2})=M(\beta_{1}\beta_{2}).

(41)

In the course of the proof we will needthe following multiplication table that will be proven later in this section.

Lemma 7.5 (Multiplication Table).

Let $\beta_{1},\beta_{2}$ , and (of course) $\beta$ be odd integers. Then the following equalities and inclusions hold.

(i)
$w_{\beta}w_{-\beta}=\frac{D-\beta^{2}}{4}\in{\mathbb{Z}}.$
(ii)
$w_{\beta}^{2}=\beta w_{\beta}+\frac{D-\beta^{2}}{4}.$
(iii)
$w_{1}w_{\beta}=(\beta-m_{\beta})w_{\beta}+\frac{D-\beta^{2}}{4}.$
In particular, $w_{1}w_{\beta}\in M(\beta)$ if and only if $\beta\mid D$ .

(iv)

M(\beta_{1})M(\beta_{2})\ni\beta_{1}w_{\beta_{2}}=w_{\beta_{1}\beta_{2}}+m_{%\beta_{1}}\sqrt{D}\in M(\beta_{1}\beta_{2});

(42)

M(\beta_{1})M(\beta_{2})\ni w_{\beta_{1}}\beta_{2}=w_{\beta_{1}\beta_{2}}+m_{%\beta_{2}}\sqrt{D}\in M(\beta_{1}\beta_{2}).

(v)

w_{\beta_{1}}w_{\beta_{2}}=\frac{\beta_{1}\beta_{2}\frac{\gamma+1}{2}+\frac{%\beta_{1}+\beta_{2}}{2}\sqrt{D}}{2}\quad\text{where}\ \gamma:=\frac{D}{\beta_{%1}\beta_{2}}.

(43)

Proof of Theorem 7.4 (modulo Lemma 7.5.

(i) We need to find out for which $\beta$

\frac{1+\sqrt{D}}{2}\cdot\beta\in M(\beta),\quad\frac{1+\sqrt{D}}{2}\cdot\frac%{\beta+\sqrt{D}}{2}\in M(\beta).

The first inclusion holds without any additional conditions on $\beta$ . Indeed,

\frac{1+\sqrt{D}}{2}\cdot\beta=\frac{\beta+\beta\sqrt{D}}{2}=\beta\frac{\beta+%\sqrt{D}}{2}+\frac{\beta-\beta^{2}}{2}=

\beta\frac{\beta+\sqrt{D}}{2}+\frac{\beta-1}{2}\beta=\beta w_{\beta}+m(\beta)%\beta\in[w_{\beta},\beta]=M(\beta).

Concerning the second inclusion, recall ((28) and Lemma 7.5(i,ii)) that

w_{1}=w_{\beta}-m,\quad(D-\beta^{2})/4\in{\mathbb{Z}},\quad w_{\beta}^{2}=%\beta w_{\beta}+(D-\beta^{2})/4.

This implies that

w_{1}w_{\beta}=w_{\beta}-m)w_{\beta}=w_{\beta}^{2}-mw_{\beta}=

\left(\beta w+(D-\beta^{2})/4\right)-mw_{\beta}=(\beta-m)w_{\beta}+\frac{D-%\beta^{2}}{4\beta}\cdot\beta.

Since $(\beta-m)w_{\beta}$ obviously lies in $M(\beta)$ , the product $w_{1}w_{\beta}$ liesin $M(\beta)$ if and only if the integer $(D-\beta^{2})/4$ lies in $M(\beta)$ . In light of (33), this means that $(D-\beta^{2})/4$ is divisible by $\beta$ . Since $4$ and odd $\beta$ are relatively prime, the divisibility condition means that $\beta|(D-\beta^{2})$ , i.e., $\beta$ divides $D$ . This ends the proof of (i).

(ii) Since $\beta$ divides $D$ , the ratio $\tilde{\beta}=(D-\beta^{2})/(4\beta)$ is an integer and

\beta\tilde{\beta}=\frac{D-\beta^{2}}{4}.

Case 1 Suppose that $D/\beta$ is relatively prime to $\beta$ . Then $\beta$ and $\tilde{\beta}$ are also relatively prime.

By Lemma 7.5(i) and Example 7.3(iii),

w_{\beta}w_{-\beta}=(D-\beta^{2})/4=\beta\tilde{\beta},\quad M(\beta)=M(-\beta%)=[\beta,w_{-\beta}].

This implies that

M(\beta)M(\beta)=M(\beta)M(-\beta)=[\beta,w_{\beta}][\beta,w_{-\beta}]=

[\beta^{2},\beta w_{-\beta},w_{\beta}\beta,w_{\beta}w_{-\beta}]=[\beta^{2},%\beta w_{-\beta},\beta w_{\beta},\beta\tilde{\beta}]=

\beta\ [\beta,\tilde{\beta},w_{-\beta},w_{\beta}].

Since $\beta$ and $\tilde{\beta}$ are relatively prime integers,

[\beta,\tilde{\beta}]={\mathbb{Z}}.

This implies that

M(\beta)M(\beta)=\beta\left({\mathbb{Z}}+{\mathbb{Z}}w_{\beta}+{\mathbb{Z}}w_{%-\beta}\right)=\beta\left(\left({\mathbb{Z}}+w_{\beta}{\mathbb{Z}}\right)+%\left({\mathbb{Z}}+w_{-\beta}{\mathbb{Z}}\right)\right)=

\beta(O+O)=\beta O

(here we use (29)). So, we have proven that

M(\beta)M(\beta)=\beta O.

(44)

According to the already proven (i), $M$ is an ideal in the ring $O$ and therefore is an $O$ -submodule in the field $K$ . The fractional $O$ -ideal

\tilde{M}(\beta):=\frac{1}{\beta}M(\beta)\subset K

is also an $O$ -submodule in $K$ . It follows from (44) that

M(\beta)\tilde{M}(\beta)=M(\beta)\left(\frac{1}{\beta}M(\beta))\right)=\frac{1%}{\beta}\left(M(\beta)M(\beta)\right)=\frac{1}{\beta}(\beta O)=O.

Hence, $M(\beta)$ is an invertible $O$ -submodule of the field $K$ . By [3, Ch. 2, Sect. 5, n 6, Th. 4], $M(\beta)$ is a projective $O$ -module of rank $1$ .

Now let us check that $M(\beta)$ is a proper $O$ -ideal. Indeed,let $u$ be an element of $K$ such that $uM(\beta)\subset M(\beta)$ . We need to check that $u\in O$ . Indeed, we have

O=M(\beta)\tilde{M}(\beta)\supset\left(uM(\beta)\right)\tilde{M}=u(M(\beta)%\tilde{M}(\beta))=uO.

So, $uO\supset O$ . Since $1\in O$ , we get $u=u\cdot 1\in O$ , i.e., $u\in O$ .(See also [3, Ch. 2, Sect. 5. Ex. 9b].)

Case 2 Suppose that the integer $D/\beta$ is not relatively prime to $\beta$ . We are going to prove that the $O$ -ideal $M(\beta)=M(\beta,D)$ is not proper.

Let $f>1$ be the greatest common divisor of $D/\beta$ and $\beta$ . Let us consider the integers

\beta^{\prime}=\frac{\beta}{f},\quad D^{\prime}=\frac{D}{f^{2}}=\beta^{\prime}%\frac{D/\beta}{f}.

Then $\beta^{\prime}$ is obviously an odd integer that divides $D^{\prime}$ . Let us stress that it follows from the very definition of $f$ that $\beta^{\prime}$ and $D^{\prime}/\beta^{\prime}$ are relatively prime.In addition, $\sqrt{D}/f\in K$ is obviously a square root of $D^{\prime}$ . On the other hand, $f$ is obviously odd, hence its square $f^{2}$ is congruent to $1$ modulo $4$ . Since $D\equiv 1\ (\bmod 4)$ , we conclude that $D^{\prime}$ is also congruent to $1$ modulo $4$ . Now let us consider the order

O^{\prime}:=O_{D^{\prime}}={\mathbb{Z}}+{\mathbb{Z}}\ \frac{1+\sqrt{D^{\prime}%}}{2}

of discriminant $D^{\prime}$ in $K$ ,its element

w^{\prime}_{\beta^{\prime}}:=w_{\beta^{\prime},D^{\prime}}=\frac{\beta^{\prime%}+\sqrt{D^{\prime}}}{2}\in K,

and the subgroup

M^{\prime}(\beta^{\prime}):=M(\beta^{\prime},D^{\prime})=[w^{\prime}_{\beta^{%\prime}},1]={\mathbb{Z}}\ \frac{\beta^{\prime}+\sqrt{D^{\prime}}}{2}+{\mathbb{%Z}}\subset K.

We have

D^{\prime}=\frac{D}{f^{2}}<D,

It follows that

O\neq O^{\prime},

because the orders $O$ and $O^{\prime}$ have distinct discriminants. On the other hand,

w^{\prime}_{\beta^{\prime}}=\frac{\beta/f+\sqrt{D/f^{2}}}{2}=\frac{1}{f}\frac{%\beta+\sqrt{D}}{2}=\frac{1}{f}w_{\beta,D}=\frac{1}{f}w_{\beta}

and therefore

M^{\prime}(\beta^{\prime})=\left[\frac{1}{f}w_{\beta,D},\frac{1}{f}\beta\right%]=\frac{1}{f}[w_{\beta,D},\beta]=\frac{1}{f}M(\beta,D)=\frac{1}{f}M(\beta).

(45)

Now we are ready to apply to $D^{\prime}$ , $O^{\prime}$ , and $M^{\prime}(\beta^{\prime})$ (instead of $D$ , $O$ , and $M(\beta)$ )all the already proven results of this section (including Case 1 of the proof of (ii)), taking into account that the odd integers $\beta^{\prime}$ and $D^{\prime}/\beta^{\prime}$ are relatively prime.We get that:

(a)
$M^{\prime}(\beta^{\prime})$ is an ideal in the ring $O^{\prime}$ ;
(b)
$M^{\prime}(\beta^{\prime})M^{\prime}(\beta^{\prime})=\beta^{\prime}O^{\prime}$ ;

(c)

$M^{\prime}(\beta^{\prime})$ is a projective $O^{\prime}$ -module of rank 1that is a proper ideal of $O^{\prime}$ - the latter means that

O^{\prime}=\{u\in K\mid uM^{\prime}(\beta^{\prime})\subset M^{\prime}(\beta)\}%\subset K.

(46)

On the other hand, in light of (45), $M^{\prime}(\beta^{\prime})=\frac{1}{f}M(\beta)$ .Combining it with (b) and (46), we conclude that $M(\beta)$ is an ideal in $O^{\prime}$ , and

M(\beta)M(\beta)=f^{2}\beta^{\prime}O^{\prime}=f\beta O^{\prime},

(47)

O^{\prime}=\{u\in K\mid uM(\beta)\subset M(\beta)\}.

Since $O^{\prime}\neq O$ , the ideal $M(\beta)$ of $O$ is not proper.(Notice that (47) implies that $M(\beta)$ is a proper ideal in $O^{\prime}$ .)

This ends the proof of (ii).

(iii) In light of Proposition 7.2(iii), $\beta_{1}\beta_{2}$ is a saturated divisor of $D$ . Let us put

\gamma=\frac{D}{\beta_{1}\beta_{2}}\in{\mathbb{Z}}.

Then $\gamma$ and $\beta_{1}\beta_{2}$ are relatively prime and

D=\beta_{1}\beta_{2}\gamma.

We have

M(\beta_{1})M(\beta_{2})=[\beta_{1},w_{\beta_{1}}][\beta_{2},w_{\beta_{2}}]=[%\beta_{1}\beta_{2},b_{1}w_{\beta_{2}},b_{2}w_{\beta_{1}},w_{\beta_{1}}w_{\beta%_{2}}];

in particular,

\beta_{1}\beta_{2}\in M(\beta_{1})M(\beta_{2}).

In addition, in light of Lemma 7.5,

M(\beta_{1})M\beta_{2})\ni\beta_{1}w_{\beta_{2}}=\frac{\beta_{1}\beta_{2}+%\beta_{1}\sqrt{D}}{2},\ \beta_{2}w_{\beta_{1}}=\frac{\beta_{1}\beta_{2}+\beta_%{2}\sqrt{D}}{2}.

Since

M(\beta_{1})=M(-\beta_{1}),\quad M(\beta_{2})=M(-\beta_{2}),

$M(\beta_{1})M\beta_{2})=M(-\beta_{1})M(-\beta_{2})$ contains both

\frac{(-\beta_{1})(-\beta_{2})-\beta_{1}\sqrt{D}}{2}=\frac{\beta_{1}\beta_{2}-%\beta_{1}\sqrt{D}}{2},\quad\frac{(-\beta_{1})(-\beta_{2})-\beta_{2}\sqrt{D}}{2%}=\frac{\beta_{1}\beta_{2}-\beta_{2}\sqrt{D}}{2}.

Hence, $M(\beta_{1})M\beta_{2})$ contains both

\frac{\beta_{1}\beta_{2}+\beta_{1}\sqrt{D}}{2}-\frac{\beta_{1}\beta_{2}-\beta_%{1}\sqrt{D}}{2}=\beta_{1}\sqrt{D}

and

\frac{\beta_{1}\beta_{2}+\beta_{2}\sqrt{D}}{2}-\frac{\beta_{1}\beta_{2}-\beta_%{2}\sqrt{D}}{2}=\beta_{2}\sqrt{D}.

Since the integers $\beta_{1}$ and $\beta_{2}$ are relatively prime,

\sqrt{D}\in M(\beta_{1})M\beta_{2}).

Taking into account that $1=\beta_{1}-2m_{\beta_{1}}$ with $m_{\beta_{1}}\in{\mathbb{Z}}$ , we obtain that $M(\beta_{1})M\beta_{2})$ contains

\frac{\beta_{1}\beta_{2}+\beta_{1}\sqrt{D}}{2}-m_{\beta_{1}}\sqrt{D}=\frac{%\beta_{1}\beta_{2}+(\beta_{1}-2m_{\beta_{1}})\sqrt{D}}{2}=\frac{\beta_{1}\beta%_{2}+\sqrt{D}}{2}=w_{\beta_{1}\beta_{2}}.

This implies that

M(\beta_{1})M\beta_{2})\supset{\mathbb{Z}}w_{\beta_{1}\beta_{2}}+{\mathbb{Z}}(%\beta_{1}\beta_{2})=M(\beta_{1}\beta_{2}).

On the other hand, recall (Example 7.3(v)) that $\sqrt{D}\in M(\beta_{1}\beta_{2})$ .In light of Example 7.3(v), in order to prove the desired equality, it suffices to check that $w_{\beta_{1}}w_{\beta_{2}}\in M(\beta_{1}\beta_{2})$ . Let us do it.

Recall that $D\equiv 1\ (\bmod 4)$ .By Lemma 7.5,

w_{\beta_{1}}w_{\beta_{2}}=\frac{\beta_{1}\beta_{2}\frac{\gamma+1}{2}+\frac{%\beta_{1}+\beta_{2}}{2}\sqrt{D}}{2}.

Case 1 Suppose that $\gamma\equiv 1\ (\bmod 4)$ . Then $\beta_{1}\beta_{2}\equiv 1\ (\bmod 4)$ .This means that $\beta_{1}\equiv\beta_{2}\bmod 4$ . It follows that

\beta_{1}+\beta_{2}\equiv 2\beta_{1}\ (\bmod 4)4=2(2m_{\beta_{1}}+1)\ (\bmod 4%)\equiv 2\ (\bmod 4).

This implies that the integer $(\beta_{1}+\beta_{2})/2$ is congruent to $1$ modulo $2$ , i.e., is odd.On the other hand, $\gamma+1\equiv 2\ (\bmod 4)$ , which implies that $(\gamma+1)/2$ is an odd integer. It follows that there are integers $n_{1}$ and $n_{2}$ such that

\frac{\gamma+1}{2}=2n_{1}+1,\quad\frac{\beta_{1}+\beta_{2}}{2}=2n_{2}+1

and therefore

w_{\beta_{1}}w_{\beta_{2}}=\frac{(2n_{1}+1)\beta_{1}\beta_{2}+(2n_{2}+1)\sqrt{%D}}{2}=

(n_{1}n_{2})\beta_{1}\beta_{2}+n_{2}\sqrt{D}+w_{\beta_{1}\beta_{2}}\in M(\beta%_{1}\beta_{2}).

Since, $\beta_{1}\beta_{2},\sqrt{D}\in M(\beta_{1}\beta_{2})$ , we conclude that $w_{\beta_{1}}w_{\beta_{2}}\in M(\beta_{1}\beta_{2})$ . So, we are done.

Case 2 Suppose that $\gamma\equiv 3\ (\bmod 4)$ . Then $\beta_{1}\beta_{2}\equiv 3\ (\bmod 4)$ . This means that $\beta_{1}\not\equiv\beta_{2}\bmod 4$ . It follows that $\beta_{1}+\beta_{2}$ is divisible by $4$ (recall that both $\beta_{1}$ and $\beta_{2}$ are odd) and therefore the integer $(\beta_{1}+\beta_{2})/2$ is even. It follows that there are integers $n_{1}$ and $n_{2}$ such that

\frac{\gamma+1}{2}=2n_{1},\quad\frac{\beta_{1}+\beta_{2}}{2}=2n_{2}.

Hence,

w_{\beta_{1}}w_{\beta_{2}}=\frac{(2n_{1})\beta_{1}\beta_{2}+(2n_{2})\sqrt{D}}{%2}=

n_{1}\beta_{1}\beta_{2}+n_{2}\sqrt{D}\in M(\beta_{1})M(\beta_{2}),

because $\beta_{1}\beta_{2},\sqrt{D}\in M(\beta_{1}\beta_{2})$ . This ends the proof.

(v) Applying (iv) to

\beta_{1}=\beta,\ \beta_{2}=D/\beta,\ \gamma=1,

we get

M(\beta)M(D/\beta)=M(D).

ByExample 7.3, $M(D)=\sqrt{D}\ O$ . This implies that

\sqrt{D}\ O=M(\beta)M(D/\beta).

Multiplying both sides by $M(\beta)$ , we get (in light of the already proven (40))

M(\beta)\sqrt{D}\ O=M(\beta)M(\beta)M(D/\beta)=\beta\ M(D/\beta),

which means that

\sqrt{D}M(\beta)=\beta\ M(D/\beta),

i.e.,

M(D/\beta)=\frac{\sqrt{D}}{\beta}M(\beta).

∎

Corollary 7.6.

(i)
Let $\beta$ be a positive divisor of $D$ . Then $M(\beta)$ is a prime ideal in $O$ if and only if $\beta$ is a prime.If this is the case then $M(\beta)$ is a maximal ideal in $O$ and the residue field $O/M(\beta)$ is the finite prime field $\mathbb{F}_{\beta}$ that consists of $\beta$ elements.

(ii)

Let $P_{D}$ be the (finite nonempty) set of positive prime divisors of $D$ . Let $S_{1}$ and $S_{2}$ be two subsets of $P_{D}$ and $S_{0}:=S_{1}\cap S_{2}$ . Then

M(\mathbf{r}_{D,S_{1}})M(\mathbf{r}_{D,S_{2}})=\mathbf{r}_{D,S_{0}}M(\mathbf{r%}_{D,S_{1}\Delta S_{2}}).

Proof.

(i) Recall (33) that $M(\beta)\cap{\mathbb{Z}}=\beta{\mathbb{Z}}.$ If $\beta=\beta_{1}\beta_{2}$ where $\beta_{1},\beta_{2}>1$ are integersthen none of $\beta_{1},\beta_{2}$ lies in $M(\beta)$ while their product $\beta$ lies in ${\mathbb{Z}}$ . So, $M(\beta)$ is not a prime idealif $\beta$ is not a prime number.

Suppose that $\beta$ is a prime number. In light of (32), the quotient ring $O/M(\beta)$ consists of $\beta$ elements.Since $\beta$ is prime, $O/M(\beta)$ is the prime finite field $\mathbb{F}_{\beta}$ . Hence, $M(\beta)$ is a maximal idealin $O$ ; hence it is a prime ideal in $O$ .

(ii) In light of Proposition 7.2 applied to $n=D$ , the positive integer $\mathbf{r}_{D,S_{0}}$ is the GCD of $\mathbf{r}_{D,S_{1}}$ and $\mathbf{r}_{D,S_{2}}$ . In addition,if we put $S_{10}=S_{1}\setminus S_{0},\ S_{20}=S_{2}\setminus S_{0}$ then $S_{1}\Delta S_{2}$ coincides with the disjoint union of $S_{10}$ and $S_{20}$ ,every pair of integers from $\mathbf{r}_{D,S_{0}},\mathbf{r}_{D,S_{10}},\mathbf{r}_{D,S_{20}}$ is mutually prime and

\mathbf{r}_{D,S_{1}}=\mathbf{r}_{D,S_{0}},\mathbf{r}_{D,S_{10}},\quad\mathbf{r%}_{D,S_{2}}=\mathbf{r}_{D,S_{0}},\mathbf{r}_{D,S_{20}}.

In light of Theorem 7.4(ii, iii),

M(\mathbf{r}_{D,S_{1}})M(\mathbf{r}_{D,S_{2}})=\left(M(\mathbf{r}_{D,S_{10}})M%(\mathbf{r}_{D,S_{0}})\right)\left(M(\mathbf{r}_{D,S_{0}})M(\mathbf{r}_{D,S_{2%0}})\right)=

M(\mathbf{r}_{D,S_{10}})\ \left(M(\mathbf{r}_{D,S_{0}})M(\mathbf{r}_{D,S_{0}})%\right)\ M(\mathbf{r}_{D,S_{20}})=

M(\mathbf{r}_{D,S_{10}})\ \left(\mathbf{r}_{D,S_{0}}O\right)\ M(\mathbf{r}_{D,%S_{20}})=

\mathbf{r}_{D,S_{0}}M(\mathbf{r}_{D,S_{10}})M(\mathbf{r}_{D,S_{20}})=\mathbf{r%}_{D,S_{0}}M(\mathbf{r}_{D,S_{1}\Delta S_{2}}).

This ends the proof.∎

Proof of Lemma 7.5.

(i) We have

w_{\beta}w_{-\beta}=\frac{\beta+\sqrt{D}}{2}\frac{-\beta+\sqrt{D}}{2}=-\frac{%\beta+\sqrt{D}}{2}\frac{\beta-\sqrt{D}}{2}=

-\frac{\beta^{2}-D}{4}=\frac{D-\beta^{2}}{4}.

Since $\beta$ is odd, $\beta^{2}\equiv 1\ (\bmod 4)$ . Since $D\equiv 1\ (\bmod 4)$ , the difference $D-\beta^{2}$ is divisible by $4$ , i.e., $(D-\beta^{2})/4$ is an integer.

(ii)

w_{\beta}+\sigma(w_{\beta})=\frac{\beta+\sqrt{D}}{2}+\frac{\beta-\sqrt{D}}{2}=%\beta,\quad w_{\beta}\sigma(w_{\beta})=\frac{\beta+\sqrt{D}}{2}\frac{\beta-%\sqrt{D}}{2}=\frac{\beta^{2}-D}{4}.

By Vieta’s formulas,

w_{\beta}^{2}-\beta w_{\beta}+\frac{\beta^{2}-D}{4}=0,\ \text{i.e.},\ w_{\beta%}^{2}=\beta w_{\beta}+\frac{D-\beta^{2}}{4}.

(iii) Recall that $w_{1}=w_{\beta}-m_{\beta}$ . In light of already proven (i),

w_{1}w_{\beta}=(w_{\beta}-m(\beta))w_{\beta}=w_{\beta}^{2}-m(\beta)w_{\beta}=

\beta w_{\beta}+\frac{D-\beta^{2}}{4}-m(\beta)w_{\beta}=(\beta-m(\beta))w_{%\beta}+\frac{D-\beta^{2}}{4}.

This proves the first assertion, of (ii). On the other hand, since $w_{\beta}\in M(\beta)$ , the product $w_{1}w_{\beta}$ lies in $M(\beta)$ if and only if $(D-\beta^{2})/4\in M(\beta)$ .In light of (33), this is equivalent to the divisibility of $(D-\beta^{2})/4$ , which ends the proof.

(iv)

\beta_{1}w_{\beta_{2}}=\beta_{1}\frac{\beta_{2}+\sqrt{D}}{2}=\frac{\beta_{1}%\beta_{2}+\beta_{1}\sqrt{D}}{2}=

\frac{\beta_{1}\beta_{2}+(1+2m_{\beta_{1}})\sqrt{D}}{2}=\frac{\beta_{1}\beta_{%2}+\sqrt{D}}{2}=w_{\beta_{1}\beta_{2}}+m_{\beta_{1}}\sqrt{D}.

So,

\beta_{1}w_{\beta_{2}}=w_{\beta_{1}\beta_{2}}+m_{\beta_{1}}\sqrt{D}.

Since both $w_{\beta_{1}\beta_{2}}$ and $\sqrt{D}$ lie in $M(\beta_{1}\beta_{2})$ , the same is true for $\beta_{1}w_{\beta_{2}}$ . This proves the first formula from (iv).Interchanging $\beta_{1}$ and $\beta_{2}$ , we get the second formula from (iv).

(v)

w_{\beta_{1}}w_{\beta_{2}}=\frac{\beta_{1}+\sqrt{D}}{2}\cdot\frac{\beta_{2}+%\sqrt{D}}{2}=\frac{(\beta_{1}\beta_{2}+D)+(\beta_{1}+\beta_{2})\sqrt{D}}{4}=

\frac{(\beta_{1}\beta_{2}+\beta_{1}\beta_{2}\gamma)+(\beta_{1}+\beta_{2})\sqrt%{D}}{4}=

\frac{\beta_{1}\beta_{2}(\gamma+1)+(\beta_{1}+\beta_{2})\sqrt{D}}{4}=\frac{%\beta_{1}\beta_{2}\frac{\gamma+1}{2}+\frac{\beta_{1}+\beta_{2}}{2}\sqrt{D}}{2}.

∎

Now and till the end of this section we assume that $D<0$ , i.e., $={\mathbb{Q}}(\sqrt{D})K$ is an imaginary quadratic field.We will view $K$ as the subfield ${\mathbb{Q}}+{\mathbb{Q}}\ \sqrt{D}$ of ${\mathbb{C}}$ such that

\sqrt{D}\in\mathfrak{H}\subset{\mathbb{C}}.

Then the nontrivial automorphism $\sigma:K\to K$ coincides with the restriction of the complex conjugation to $K$ . This implies(in light of (26)) that

K\cap{\mathbb{R}}=K^{\sigma}={\mathbb{Q}}.

(48)

Recall [2, Ch. 2] that a full module in $K$ is a subgroup $M$ of $K$ that is a free abelian (sub)group of rank $2$ ;the multiplier ring $\mathbf{O}(M)$ of $M$ is defined as

\{u\in K\mid u(M)\subset M\}\subset K.

It is well known that $\mathbf{O}(M)$ is an order in $K$ .

On the other hand, if $O$ is an order in $K$ then it is said that a full module $M$ belongs to $O$ if $\mathbf{O}(M)=O$ . Such a module $M$ is also often called a (fractional) proper $O$ -ideal.For example, $uO$ is a proper $O$ -ideal for all $u\in K^{*}$ ; such an ideal is called principal.Obviously, if $M$ is a proper $O$ -ideal then $uM$ is also one for all $u\in K^{*}$ . The proper $O$ -ideals $M$ and $uM$ are called similar. Recall that $\sigma(O)=O$ . It follows that $\sigma(M)$ is also a proper $O$ -ideal. There is a nonzero rational number $\mathbf{N}(M)$ that is called the norm of $M$ such that

M\ \sigma(M)=\mathbf{N}(M)\ O

(49)

[2, Ch. 2, Sect. 7.4]. It follows that

M\ \left(\frac{1}{\mathbf{N}(M)}\sigma(M)\right)=O,

which implies that $M$ is an invertible $O$ -submodule of the field $K$ . By [3, Ch. 2, Sect. 5, n 6, Th. 4], $M$ is a projective $O$ -module of rank $1$ .

If $M_{1},M_{2}$ are two proper $O$ -ideals then $M_{1}M_{2}$ is also a proper $O$ -ideal and the binary operation

M_{1},M_{2}\mapsto M_{1}M_{2}

defines the structure of a commutative group on the set of all proper $O$ -ideals where $O$ corresponds to the zeroof the group law while the inverse of $M$ is $\frac{1}{\mathbf{N}(M)}\sigma(M)$ [2, Ch. 2, Sect. 7.4, Th. 2].(Notice thatthe hom*omorphism of $O$ -modules

M_{1}\otimes_{O}M_{2}\to M_{1}M_{2},\ m_{1}\otimes m_{2}\mapsto m_{1}m_{2}

is a surjective hom*omorphism of projective $O$ -modules of rank $1$ and therefore is an isomorphism.This implies that the $O$ -modules $M_{1}\otimes_{O}M_{2}$ and $M_{1}M_{2}$ are canonically isomorphic.)

The quotient of this group modulo the subgroup of principal ideals is called the class group of $O$ and denoted by $\mathrm{Cl}(O)$ ;its elements are similarity classes of proper $O$ -ideals. It is known [2, Ch. 2, Sect. 7.4, Th. 3] that $\mathrm{Cl}(O)$ is a finite commutative group. We write $\mathrm{Cl}(O)[2]$ for the kernel of multiplication by $2$ in $\mathrm{Cl}(O)$ . In light of (49),elements of $\mathrm{Cl}(O)[2]$ are precisely similarity classes of those proper $O$ -ideals $M$ such that $M$ is similar to $\sigma(M)$ .

When $O$ is a maximal order in $K$ , the structure of the elementary $2$ -group $\mathrm{Cl}(O)[2]$ is well known ([8, Sect. V.1, Th. 39], [13, Teil II, $ 12]. The next assertion describes explicitly thestructure of $\mathrm{Cl}(O)[2]$ when $O$ is any odd order (in an imaginary quadratic field).

Theorem 7.7.

Let $K$ be an imaginary quadratic field, $O$ an odd order in $K$ with discriminant $D$ .Let $P_{D}$ be the set of prime divisors of $D$ and $s_{D}$ the cardinality of $P_{D}$ .Let $\mathcal{B}(P_{D})$ be the Booleanalgebra of subsets of $P_{D}$ where the sum of two (sub)sets is their symmetric difference. Let us consider the map

F_{D}:\mathcal{B}(P_{D})\to\mathrm{Cl}(O)

that assigns to each $S\subset P_{n}$ the similarity class of the ideal $M(\mathbf{r}_{D,S})$ .

Then $F_{D}$ is a group hom*omorphism, whose image coincides with $\mathrm{Cl}(O)[2]$ while the kernel consists of two elements, namely, $\emptyset$ and $P_{D}$ .

The following assertion is a natural generalization of a special case of([8, Sect. V.1, Cor. 1 to Th. 39], [13, p. 112, Korollar]).

Corollary 7.8.

Let $O$ be an odd order with discriminant $D$ in an imaginary quadratic field $K$ .Then the order of $\mathrm{Cl}(O)[2]$ is $2^{s_{D}-1}$ where $s_{D}$ is the number of prime divisors of $D$ .

Proof of Corollary 7.8 (modulo Theorem 7.7).

One may view $F_{D}$ as the surjective group hom*omorphism $\mathcal{B}(P_{D})\to\mathrm{Cl}(O)[2]$ ,whose kernel has order $2$ . Recall that the order of $\mathcal{B}(P_{D})$ is $2^{s_{D}}$ .Then the order of $\mathrm{Cl}(O)[2]$ is the order of $\mathcal{B}(P_{D})$ divided by $2$ , i.e., $2^{s_{D}}/2=2^{s_{D}-1}$ . This ends the proof.∎

Proof of Theorem 7.7.

It follows from Corollary 7.6(ii) that $F_{D}$ is a group hom*omorphism. Since the order of every element of $\mathcal{B}(P_{D})$ is either $2$ or $1$ , the image $F_{D}(\mathcal{B}(P_{D}))$ lies in $\mathrm{Cl}(O)[2]$ .

If $S=P_{D}$ then

\mathbf{r}_{D,P_{D}}=|D|=-D.

Now it follows from Examples 7.3(ii,iii)that $M(-D)=\sqrt{D}O$ is a principal ideal and therefore $\ker(F_{D})$ contains $P_{D}$ .Suppose that

S\in\ker(F_{D}),\quad S\neq\emptyset,P_{D}.

We need to arrive to a contradiction.First,

S^{\prime}:=S\Delta P_{D}=P_{D}\setminus S\in\ker(F_{S}),\quad S^{\prime}\neq%\emptyset,P_{D}.

Then

|D|=\mathbf{r}_{D,P_{D}}=\mathbf{r}_{D,P_{S}}\mathbf{r}_{D,P_{S}^{\prime}};%\quad\mathbf{r}_{D,P_{S}},\mathbf{r}_{D,P_{S}^{\prime}}>1.

(50)

This implies that both $\mathbf{r}_{D,P_{S}}$ and $\mathbf{r}_{D,P_{S}^{\prime}}$ are divisors of odd $D$ ; hence, they are odd integers as well.In addition, it follows from (50) that either $\mathbf{r}_{D,P_{S}}<\sqrt{|D|}$ or $\mathbf{r}_{D,P_{S}^{\prime}}<\sqrt{|D|}$ .Replacing if necessary $S$ by $S^{\prime}$ , we may and will assume that

\beta:=\mathbf{r}_{D,P_{S}}<\sqrt{|D|}.

In addition, $\beta>1$ is an odd divisor of $D$ .

Second, the condition $S\in\ker(F_{D})$ means that the ideal $M(\beta)$ is principal, i.e., there is a nonzero $u\in O\subset K$ such that $M(\beta)=uO$ . The latter means that $[w_{\beta},\beta]=u[w_{1},1]$ , which implies that

\left[\frac{w_{\beta}}{\beta},1\right]=\frac{u}{\beta}[w_{1},1]

where

w_{1}=\frac{1+\sqrt{D}}{2},\quad\tau_{\beta}:=\frac{w_{\beta}}{\beta}=\frac{%\beta+\sqrt{D}}{2\beta}\mathfrak{H}\subset{\mathbb{C}}..

This implies that (in the notation of Section 3)

\Lambda_{\tau_{\beta}}=\frac{u}{\beta}\Lambda_{w_{1}}.

(51)

Notice that

\mathrm{Re}(w_{1})=\frac{1}{2},\ \mathrm{Im}(w_{1})=\frac{\sqrt{|D|}}{2}>\frac%{1}{2};

\mathrm{Re}(\tau_{\beta})=\frac{1}{2},\ \mathrm{Im}(\tau_{\beta})=\frac{\sqrt{%|D|}}{2\beta}>\frac{1}{2};

the latter inequality holds, because $\beta<\sqrt{|D|}$ . On the other hand, since $\beta>1$ ,

\mathrm{Im}(\tau_{\beta})<\frac{\sqrt{|D|}}{2}=\mathrm{Im}(w_{1}).

This implies that $w_{1}$ and $\tau_{\beta}$ are distinct points of $\mathfrak{T}_{2}$ . In particular,

j(\tau_{\beta}\neq j(w_{1}).

On the other hand, in light ofRemark 3.2, the equality (51) implies that the complex elliptic curves $\mathcal{E}_{\tau_{\beta}}$ and $\mathcal{E}_{w_{1}}$ are isomorphic, i.e.,

j(\tau_{\beta})=j(\mathcal{E}_{\tau_{\beta}})=j(\mathcal{E}_{w_{1}})=j(w_{1}).

This implies that $j(\tau_{\beta})=j(w_{1})$ , which gives us the desired contradiction that proves our assertion about $\ker(F_{D})$ .

Now let us check the surjectiveness of

F_{D}:\mathcal{B}(P_{D})\to\mathrm{Cl}(O)[2].

Let $M$ be a proper $O$ -ideal,whose similarity class has order 1 or 2 in $\mathrm{Cl}(O)$ , i.e., $M^{2}=M\ M$ is a principal ideal $uO$ for some nonzero $u\in K$ . Notice that $M=[\omega_{1},\omega_{2}]$ where $\{\omega_{1},\omega_{2}\}$ is a basis of the ${\mathbb{Q}}$ -vector space $K$ . Then $M$ is similar to

\frac{1}{\omega_{2}}M=[\omega_{1}/\omega_{2},1]=[\tau,1]

where $\tau:=\omega_{1}/\omega_{2}\not\in{\mathbb{Q}}$ . Since $\tau\in K\subset{\mathbb{C}}$ , it follows from (48) that $\tau\in{\mathbb{C}}\setminus{\mathbb{R}}$ .Replacing if necessary $\tau$ by $-\tau$ , we may and will assume that

\tau\in\mathfrak{H}\subset{\mathbb{C}},\quad M=[\tau,1]\subset K\subset{%\mathbb{C}}.

So, $[\tau,1]$ is a proper $O$ -ideal, whose square is a principal $O$ -ideal.This means that $j(\tau)\in{\mathbb{R}}$ (see [9, Ch. 5, Sect. 5.4, Claim (5.4.4)].Hence, there is

\tau^{\prime}\in\mathfrak{T}\subset\mathfrak{H}\ \text{ such that }\ j(\tau^{%\prime})=j(\tau).

It follows from the basic properties of the modular function $j$ that there exists a unimodular matrix

B=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\mathrm{SL}(2,{\mathbb{Z}})

such that

\tau^{\prime}=\frac{a\tau+b}{c\tau+d}.

This implies that

\tau^{\prime}\in K,\ \ {\mathbb{Q}}(\tau^{\prime})=K\supset\left[\tau^{\prime}%,1\right]=\frac{1}{c\tau+d}\left[a\tau+b,c\tau+d\right]=\frac{1}{c\tau+d}\left%[\tau,1\right].

It follows that $\Lambda_{\tau^{\prime}}=[\tau^{\prime},1]$ is a proper $O$ -ideal that is similar to $M$ . In light of Key Lemma 3.9, there is a positive odd integer $\beta$ dividing $D$ such that

\tau^{\prime}=\frac{1}{2}+\frac{\sqrt{D}}{2\beta}\in\mathfrak{T}_{2}\subset%\mathfrak{T}\subset\mathfrak{H}.

This implies that

\Lambda_{\tau^{\prime}}=\left[\frac{1}{2}+\frac{\sqrt{D}}{2\beta},1\right]=%\frac{1}{\beta}\left[\frac{\beta+\sqrt{D}}{2},\beta\right]=\frac{1}{\beta}M(%\beta).

This implies that $M(\beta)$ is a proper $O$ -ideal that is similar to $\Lambda_{\tau^{\prime}}$ and therefore to $M$ .It follows from Theorem 7.4 that $\beta$ is a saturated divisor of $D$ . This implies that if $S$ is the set of prime divisors of $\beta$ then $\beta=\mathbf{r}_{D,S}$ . This means that $F_{D}(S)$ is the similarity class of $M(\beta)$ , which coincides with the similarity class of $M$ .It follows that the similarity class of $M$ lies in the image of $F_{D}$ . This ends the proof.∎

8. Odd elliptic curves revisited

Theorem 8.1.

Let $D$ be a negative integer that is congruent to $1$ modulo $4$ . Let $s_{D}$ be the number of prime divisors of $D$ .

(1)

Let $E$ be an elliptic curve over ${\mathbb{C}}$ with $j(E)\in{\mathbb{R}}$ . Then the following two conditions are equivalent.

(i)
$E$ has CM and its endomorphism ring $\mathrm{End}(E)$ is an order of discriminant $D$ . (In particular, $E$ is odd.)

(ii)

There exists a positive saturated divisor $\beta$ of $D$ such that $\beta<\sqrt{|D|}$ and

j(E)=j(\tau)\ \text{where}\ \tau=\frac{1}{2}+\frac{\sqrt{D}}{2\beta}=\frac{1}{%2}+\frac{\sqrt{|D|}}{2\beta}\mathbf{i}\in\mathfrak{T}_{2}\subset\mathfrak{H}.

(2)
Let $\mathcal{J}_{D}({\mathbb{R}})$ be the set of $j$ -invariants of all $E$ that enjoy the equivalent properties (i)-(ii).Then $\mathcal{J}_{D}({\mathbb{R}})$ consists of $2^{s_{D}-1}$ distinct real numbers and lies in the semi-open interval $\left[j\left(\frac{1+\sqrt{D}}{2}\right),1728\right)$ .

Proof.

(1) Suppose that $E$ enjoys the properties (i). Since $j(E)\in{\mathbb{R}}$ ,there is (precisely) one $\tau\in\mathfrak{T}$ such that $E$ is isomorphic to $\mathcal{E}_{\tau}$ and $j(E)=j(\tau)$ . So, we may assume that $E=\mathcal{E}_{\tau}$ .In particular, $\mathcal{E}_{\tau}$ is odd. It follows from Key Lemma 3.9(i) that $\mathrm{Re}(\tau)\neq 0$ ,i.e.,

\tau\in\mathfrak{T}_{2},\ \mathrm{Re}(\tau)=\frac{1}{2}.

Since $O=\mathrm{End}(\mathcal{E}_{\tau})\cong O_{\tau}\subset{\mathbb{C}}$ is an order with odd discriminant $D$ ,

O_{\tau}={\mathbb{Z}}\left[\frac{1+\sqrt{D}}{2}\right]={\mathbb{Z}}+{\mathbb{Z%}}\ \frac{1+\sqrt{D}}{2}\subset{\mathbb{C}}.

Since $\frac{1+\sqrt{D}}{2}\in O_{\tau}$ , it follows from Key Lemma 3.9(ii) that there is a positive odd divisor $\beta$ of $D$ such that

\tau=\frac{1}{2}+\frac{\sqrt{D}}{2\beta}=\frac{1}{2}+\frac{\sqrt{|D|}}{2\beta}%\mathbf{i}.

Hence,

K:={\mathbb{Q}}(\tau)={\mathbb{Q}}(\sqrt{D})\subset{\mathbb{C}}

is an imaginary quadratic field and

{\mathbb{Z}}+{\mathbb{Z}}\ \frac{1+\sqrt{D}}{2}=O_{\tau}\subset K.

Since $\tau\in\mathfrak{T}_{2}$ , its imaginary part is strictly greater than $1/2$ , i.e,

\beta<\sqrt{|D|}.

Recall that

O_{\tau}=\{\lambda\in K\mid\lambda[\tau,1]\subset[\tau,1]\}.

We have

[\tau,1]=\left[\frac{1}{2}+\frac{\sqrt{D}}{2\beta},1\right]=\frac{1}{2\beta}%\left[\beta,\frac{\beta+\sqrt{D}}{2}\right]=\frac{1}{2\beta}M(\beta,D)=\frac{1%}{2\beta}M(\beta).

It follows that $M(\beta)$ is a proper ideal in $O_{\tau}$ , i.e.,

O_{\tau}=\{\lambda\in K\mid\lambda M(\beta)\in M(\beta)\}.

By Theorem 7.4(ii), $\beta$ is relatively prime to $D/\beta$ , which means that $\beta$ is a saturated divisor of $D$ . This proves that (i) implies (ii).

Conversely, suppose that (ii) holds. We may assume that $E=\mathcal{E}_{\tau}$ where $\tau=\frac{1}{2}+\frac{\sqrt{D}}{2\beta}\in\mathfrak{T}_{2}$ ; here $\beta$ is a positive saturated divisor of $D$ , which is odd, since $D$ is odd.Then

j(\mathcal{E}_{\tau})=j(\tau)\in{\mathbb{R}}.

Clearly, $K:={\mathbb{Q}}(\tau)={\mathbb{Q}}(\sqrt{D})$ is an imaginary quadratic field. Hence, $\mathcal{E}_{\tau}$ has CM.By Key Lemma 3.9(ii),

O_{\tau}\supset{\mathbb{Z}}+{\mathbb{Z}}\ \frac{1+\sqrt{D}}{2}={\mathbb{Z}}%\left[\frac{1+\sqrt{D}}{2}\right]=:O_{D},

where $O_{D}$ is an odd order of discriminant $D$ . We need to prove the equality $O_{\tau}=O_{D}$ .

As above, $[\tau,1]=\frac{1}{2\beta}M(\beta)$ and therefore $O_{\tau}$ coincides with

\{\lambda\in K\mid\lambda M(\beta)\in M(\beta)\}.

In light of Theorem 7.4(ii), the latter coincides with $O_{D}$ , because $\beta$ is a saturated divisor of $D$ .

(2) In order to prove the last assertion of our theorem,it suffices to recall that the function

[1/2,\infty]\to{\mathbb{R}},\quad t\mapsto j(1/2+\mathbf{i}t)

is strictly decreasing and $j(1/2+\mathbf{i}/2)=1728$ .

∎

Acknowledgements. I am grateful to Jean-Louis Colliot-Thélène and Alena Pirutka for an interesting and stimulating questionabout the distribution of $j$ -invariants of certain elliptic curves of CM type and for their useful comments to a preliminary version of this paper,which may be viewed as an extension (or a variant) of my answer [4, Sect. 8]. I thank David Masser and Serge Vladuts for their interest in this paper and stimulatingdiscussions.

References

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[6] B. Gross, Arithmetic on elliptic curves with complex multiplication. Lecture Notes in Math. 776, Springer-Verlag,Berlin Heidelberg New York, 1980.
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Odd and Even Elliptic Curves with Complex Multiplication (2024)

Abstract.

1. Orders in quadratic fields

Definition 1.1 (Definition-Lemma).

Definition 1.2 (Definition-Lemma).

Definition 1.3.

Proposition 1.4.

Proposition 1.5.

Remark 1.6.

Examples 1.7.

2. j𝑗jitalic_j-invariants of CM elliptic curves

Proposition 2.1.

Remark 2.2.

Corollary 2.3.

Theorem 2.4.

Remark 2.5.

Theorem 2.6.

3. The upper half-plane and elliptic curves

Definition 3.1.

Remark 3.2.

Remark 3.3.

Remark 3.4.

Lemma 3.5.

Proof.

Proof of Proposition 2.1.

Example 3.6.

Example 3.7.

Example 3.8.

Lemma 3.9 (Key Lemma).

Proof.

Example 3.10.

Corollary 3.11.

Proof.

4. j𝑗jitalic_j-invariants: the real case

Proof of Theorem 2.4.

Remark 4.1.

5. Odd and even orders

Lemma 5.1.

Lemma 5.2.

Proof of Lemmas 5.1 and 5.2.

Remark 5.3.

Remark 5.4.

6. Isogenies of Elliptic Curves

Proof of Proposition 1.4.

Proof of Proposition 1.5.

Lemma 6.1.

Proof.

7. Quadratic Exercises

Definition 7.1 (Definition-Notation).

Proposition 7.2.

Proof.

Examples 7.3.

Theorem 7.4.

Lemma 7.5 (Multiplication Table).

Proof of Theorem 7.4 (modulo Lemma 7.5.

Corollary 7.6.

Proof.

Proof of Lemma 7.5.

Theorem 7.7.

Corollary 7.8.

Proof of Corollary 7.8 (modulo Theorem 7.7).

Proof of Theorem 7.7.

8. Odd elliptic curves revisited

Theorem 8.1.

Proof.

References

References

2. $j$ -invariants of CM elliptic curves

4. $j$ -invariants: the real case