Our main goal in this chapter is to characterize the endomorphism rings of elliptic curves. Surprisingly, it turns out that there are only three possibilities: the ring of integers $\mathbb{Z}$, an order in an imaginary quadratic extension of $\mathbb{Q}$, or an order in a quaternion algebra over $\mathbb{Q}$. Among others, the dual isogeny and the Tate module play key roles in the proof. We will recall the basic notions of elliptic curves in the first section by looking at their various equivalent definitions. Before proving the general case, we will also illustrate the special case of $K=\mathbb{C}$ as an example, in an attempt to get a clue as to the general proof. The main source of our exposition is [1].

TopElliptic curves

Let us get started by recalling some basic notions about elliptic curves. The rich structure of an elliptic curve allows us to define it in various flavors. We may summarize them as follows.

Definition 1 An elliptic curve $E$ over a field $K$ is
  1. an irreducible smooth projective curve over $K$ of genus one with a specified point $O$, or
  2. a plane projective curve defined by a Weierstrass equation 
 \begin{equation*}
   y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6,\quad a_i\in K \tag{1}
 \end{equation*}
 with nonzero discriminant. For $\Char(K)\ne 2,3$, the Weierstrass equation can be simplified via a change of variables to 
 \begin{equation*}
   y^2=4x^3-g_2x-g_3, \quad g_2,g_3\in K, \tag{2}
 \end{equation*}
 or
  3. ($K=\mathbb{C}$) a one-dimensional complex torus, i.e. $E=\mathbb{C}/\Lambda$ for some lattice $\Lambda$ in $\mathbb{C}$, or
  4. a one-dimensional abelian variety over $K$, i.e., a connected projective curve over $K$ with a group structure.

Let us now elaborate on some of the connections between these definitions.

(a)$\Leftrightarrow$(b) Suppose $E$ satisfies (a). By the Riemann-Roch theorem, for any $n>2g-2=0$, we have $\dim \mathcal{L}(nO)=n-g+1=n$. Letting $n=2$ and $n=3$, we can find two functions $x,y\in K(E)$ such that $x$ has exactly 2 poles at $O$ and $y$ has exactly 3 poles at $O$. Now let $n=6$, we get a linear relation between $\{1,x,y,x^2,xy,y^2,x^3\}\subseteq\mathcal{L}(6O)$ which gives us the Weierstrass equation (1). The map $$\phi: E\rightarrow \mathbb{P}^2,\qquad \phi=[x,y,1]$$ sends $E$ onto a plane projective curve defined by a Weierstrass equation. The extension degrees $[K(E):K(x)]=2$ and $[K(E):K(y)]=3$ implies that $[K(E):K(x,y)]=1$, hence $\phi$ has degree 1 and gives an isomorphism between $E$ and $\phi(E)$. Since $E$ is smooth, we know that the discriminant of the Weierstrass equation is nonzero. Moreover, $\phi(O)=[0,1,0]$ is the point at infinity.

Conversely, suppose $E$ is a plane projective curve defined by a Weierstrass equation. The fact that the discriminant is nonzero means that $E$ is smooth. To see that $E$ has genus 1, consider the invariant differential $$\omega=\frac{dx}{2y+a_1x+a_3}\in\Omega_E,$$ It is holomorphic and non-vanishing, hence $\div(\omega)=0$. By the Riemann-Roch theorem, $2g-2=\deg\div(\omega)=0$, which implies that $g=1$. Finally, the specified point is taken to be the point at infinity.

(b)$\Leftrightarrow$(c) Suppose $E$ satisfies (c), $E\cong \mathbb{C}/\Lambda$. Define $$\phi:\mathbb{C}/\Lambda\rightarrow \mathbb{P}^2,\qquad z\mapsto [\wp_\Lambda(z),\wp_\Lambda'(z),1]$$ where the Weierstrass $\wp$-function is given by $$\wp_\Lambda(z)=\frac{1}{z^2}+\sum_{w\in\Lambda-\{0\}}\left(\frac{1}{(z-w)^2}-\frac{1}{w^2}\right).$$ Then $\phi(E)$ is the plane projective curve cut out by (2) with $g_2=g_2(\Lambda)$ and $g_3=g_3(\Lambda)$ where $$g_2(\Lambda)=60\sum_{w\in\Lambda-\{0\}}\frac{1}{w^{4}},\quad g_3(\Lambda)=140\sum_{w\in\Lambda-\{0\}}\frac{1}{w^{6}},$$ are Eisenstein series of weights 4 and 6. Let $\omega_1,\omega_2$ be a basis of the lattice $\Lambda$ and $\omega_3=\omega_1+\omega_2$. Then the equation (2) has three distinct roots $\wp'_\Lambda(\omega_i/2)$, so the discriminant is nonzero. One can show that $\phi$ is injective using the properties of elliptic functions ([1, VI 3.6]). Since $$\phi^*(dx/y)=d\wp_\Lambda(z)/\wp_\Lambda'(z)=dz,$$ we know that $\phi$ induces an isomorphism on the cotangent spaces. Thus $\phi$ is an isomorphism between $E$ and $\phi(E)$.

Conversely, for any $g_2, g_3\in\mathbb{C}$ such that the discriminant is nonzero, there exists a unique lattice $\Lambda\subseteq\mathbb{C}$ satisfying $g_2=g_2(\Lambda)$ and $g_3=g_3(\Lambda)$ by the Uniformization theorem ([2, I 4.3]). Thus the plane projective curve $E$ is isomorphic to the complex torus $\mathbb{C}/\Lambda$.

(a)$\Leftrightarrow$(c) Suppose $E$ satisfies (a). Then $E$ is a genus one Riemann surface. Let $\alpha, \beta$ be the fundamental loops of $E$ generating $H_1(E,\mathbb{Z})$ and let $\omega$ be the canonical differential on $E$, then $\int_\alpha\omega, \int_\beta\omega$ are $\mathbb{R}$-linearly independent. Let $\Lambda$ be the lattice spanned by them. The Abel-Jacobi map $$E\rightarrow \Jac(E)=\mathbb{C}/\Lambda,\qquad P\mapsto \int_{O}^P\omega+\Lambda$$ is biholomorphic by the Jacobi inversion theorem ([3, Theorem 5.1]).

Conversely, the one-dimensional complex torus $\mathbb{C}/\Lambda$ is of genus 1. The specified point is taken to be $0+\Lambda$.

(a)$\Leftrightarrow$(d) Suppose $E$ satisfies (a). For any two points $P,Q \in E$, $(P)\sim (Q)$ if and only if $P=Q$, since $\dim\mathcal{L}(Q)=1$ by the Riemann-Roch theorem and $\mathcal{L}(Q)$ contains constant functions. Therefore we have an injective map from $E$ to its degree-0 Picard group $$\phi: E\rightarrow \Pic^0(E),\qquad P\mapsto (P)-(O).$$ Again, by the Riemann-Roch theorem, $\dim\mathcal{L}(D+(O))=1$ for any degree-0 divisor $D$. Let $f$ be a basis vector of this one-dimensional space. Then $\div(f)=-D-(O)+(P)$ for some $P$, so we get $D\sim (P)-(O)$ and thus $\phi$ is also surjective. So $E$ has a group structure inherited from $\Pic^0(E)$ and $O$ serves as the identity element. Moreover, $E$ is irreducible, so connected.

Conversely, suppose $E$ has a group structure. Then by generic smoothness we know $E$ is smooth. $E$ is connected and smooth, hence irreducible. Moreover, the canonical bundle $\Omega_E$ is free of rank 1 and has degree $2g-2=0$, therefore the genus $g=1$.

(b)$\Rightarrow$(d) The group structure on $E$ can be written down in terms of secants and tangents given by regular functions of the coordinates $x,y$ with coefficients $a_i$. The point at infinity serves as the identity element.

(c)$\Rightarrow$(d) The group structure on $E$ is given by $(z_1+\Lambda)+(z_2+\Lambda)=(z_1+z_2)+\Lambda$. The point $0+\Lambda$ serves as the identity element. The properties of elliptic functions also ensure a bijection between $\mathbb{C}/\Lambda$ and $\Pic^0(\mathbb{C}/\Lambda)$ ([1, VI 2.4]). These two group structures are compatible with each other.

TopIsogeny and dual isogeny

Definition 2 Let $E_1, E_2$ be two elliptic curves. An isogeny from $E_1$ to $E_2$ is a morphism $\phi: E_1\rightarrow E_2$ satisfying $\phi(O)=O$. The group of isogenies from $E_1$ to $E_2$ is denoted by $\Hom(E_1, E_2)$. The ring of endomorphisms of an elliptic curve $E$, i.e., the isogenies from $E$ to itself, is denoted by $\End(E)$. For those isogenies defined over $K$, we denote them by $\Hom_K(E_1, E_2)$ and $\End_K(E)$ respectively.

An isogeny is automatically a homomorphism. An isogeny $\phi:E_1\rightarrow E_2$ is the same as the composition $$\phi: E_1\cong \Pic^0(E_1)\xrightarrow{\phi_*}\Pic^0(E_2)\cong E_2,$$ where $\phi_*$ is the pushforward map between Picard groups induced by the isogeny $\phi$. All these three maps are group homomorphisms, hence the composition $\phi$ itself is a group homomorphism.

One can also consider the pullback map $\phi^*$ of Picard groups induced by $\phi$, the composition$$\hat\phi: E_2\cong\Pic^0(E_2)\xrightarrow{\phi^*}\Pic^0(E_1)\cong E_1.$$ turns out to be an isogeny from $E_2$ to $E_1$ as well and the composition $\phi\circ\hat\phi$ has the effect $\phi_*\circ\phi^*=[\deg\phi]$.

Definition 3 Let $\phi:E_1\rightarrow E_2$ be an isogeny of elliptic curves of degree $m\ne0$, then there exists a unique isogeny $\hat\phi:E_2\rightarrow E_1$ such that $\hat\phi\circ\phi=[m]$, where $[m]$ denotes the multiplication-by-$m$ isogeny. $\hat\phi$ is called the dual isogeny to $\phi$.

Here we list some easy properties of the dual isogeny as follows ([1, III 6.2]).

Theorem 1 Let $\phi,\psi: E_1\rightarrow E_2$, $\lambda:E_2\rightarrow E_3$ be isogenies of elliptic curves.
  1. Let $m=\deg\phi$, then $\hat\phi\circ\phi=[m]$ on $E_1$ and $\phi\circ\hat\phi=[m]$ on $E_2$. In particular, $\phi=\hat{\hat{\phi}}$.
  2. $\widehat{\lambda\circ\phi}=\hat\phi\circ\hat\lambda$.
  3. $\widehat{\phi+\psi}=\hat\phi+\hat\psi$.
  4. $\widehat{[m]}=[m]$ and $\deg[m]=m^2$ for any $m\in\mathbb{Z}$.
  5. $\deg\hat\phi=\deg\phi$.
Example 1 Consider $K=\mathbb{C}$. In this case we can write down the isogenies, dual isogenies and endomorphism rings in very concrete terms.

Isogeny Suppose $E_1=\mathbb{C}/\Lambda_1$ and $E_2=\mathbb{C}/\Lambda_2$ are two elliptic curves and $\phi: E_1\rightarrow E_2$ is an isogeny. Then, by the lifting property of the covering space, the holomorphic map $\mathbb{C}\rightarrow\mathbb{C}/\Lambda_1\xrightarrow{\phi}\mathbb{C}/\Lambda_2$ can be lifted to a holomorphic map $f: \mathbb{C}\rightarrow\mathbb{C}$, since the source $\mathbb{C}$ is simply-connected. For any $\lambda\in\Lambda_1$, $f(z+\lambda)-f(z)$ maps into a discrete subset of $\mathbb{C}$, hence must be constant. Therefore, $f'(z+\lambda)=f'(z)$; in other words, $f'(z)$ is an elliptic function with respect to the lattice $\Lambda_1$. But $f'(z)$ is also holomorphic, hence $f'(z)$ must be a constant and $f(z)$ is a linear map. Now $f(0)=0$ implies that $f(z)=\alpha z$ for some $z\in\mathbb{C}$. In summary, $\phi$ must be of the form $$\phi:\mathbb{C}/\Lambda_1\rightarrow \mathbb{C}/\Lambda_2,\qquad z+\Lambda_1\mapsto \alpha z+\Lambda_2, \quad (\alpha\Lambda_1\subseteq\Lambda_2).$$ There is a bijection $$\Hom(E_1,E_2) \cong \{\alpha\in\mathbb{C}\mid \alpha\Lambda_1\subseteq\Lambda_2\}.$$

Dual isogeny Suppose $\phi$ corresponds to $\alpha$ in the manner mentioned above. Then $\alpha\Lambda_1$ is a sublattice of $\Lambda_2$. Let $\{\omega_1,\omega_2\}$ be a basis of $\Lambda_2$, then we can find two integers $n_1, n_2$ such that $\{n_1\omega_1, n_2\omega_2\}$ forms a basis of $\alpha\Lambda_1$. Moreover, $n_1n_2$ is exactly the degree of $\phi$. Now $$\hat\phi: \mathbb{C}/\Lambda_2\rightarrow\mathbb{C}/\Lambda_1,\qquad z+\Lambda_2\mapsto n_1n_2\alpha^{-1}z+\Lambda_1$$ is a well-defined isogeny since $n_1n_2\alpha^{-1}\Lambda_2\subseteq\Lambda_1$. Also, it is easy to see that $$\hat\phi\circ\phi=[n_1n_2]=[\deg\phi],$$ namely $\hat\phi$ is just the dual isogeny of $\phi$.

Endomorphism ring From the above discussion, we know that $$\End(\mathbb{C}/\Lambda)\cong\{\alpha\in\mathbb{C}\mid \alpha\Lambda\subseteq\Lambda\}.$$ Suppose $\Lambda$ has a basis $\{\omega_1,\omega_2\}$. Without loss of generality, we may multiply $\Lambda$ by $\omega_1^{-1}$ and assume that $\{1, \tau\}$ is a basis of $\Lambda$ where $\tau=\omega_2/\omega_1\in\mathbb{C}\backslash\mathbb{R}$. Let $\alpha\in\mathcal{R}=\{\alpha\in\mathbb{C}\mid \alpha\Lambda\subseteq\Lambda\}$, then we can find integers $a, b, c, d$ such that $\alpha=a+b\tau$ and $\alpha\tau=c+d\tau$. So $\alpha$ satisfies a quadratic equation $$\alpha^2-(a+d)\alpha+(ad-bc)=0.$$ Hence $\mathcal{R}$ is an integral extension of $\mathbb{Z}$. Now suppose $\mathcal{R}\ne\mathbb{Z}$, then we can find some $\alpha$ with $b\ne0$. Eliminating $\alpha$ yields another quadratic equation in $\tau$, $$b\tau^2-(a-d)\tau-c=0.$$ So $\mathcal{R}\subseteq\mathbb{Q}(\tau)$ is an order in an imaginary quadratic extension of $\mathbb{Q}$. In summary, $\End(\mathbb{C}/\Lambda)$ can be either $\mathbb{Z}$ consisting of multiplication-by-$m$ maps, or an order in an imaginary quadratic field $\mathbb{Q}(\omega_2/\omega_1)$.

Elliptic curves in the latter case, which are relatively rare compared to the first case, are said to have complex multiplication, or CM for short. We will see more interesting examples and properties of elliptic curves with complex multiplication later in the third chapter.

Remark 1 Using the Lefschetz principle ([1, VI.6]), which says that algebraic geometry over an algebraic closed field of characteristic 0 is the ``same'' as over $\mathbb{C}$, we are able to conclude that over an arbitrary field of characteristic 0, the endomorphism ring $\End(E)$ only has the above two classes as well. However, some interesting phenomena happen in positive characteristic. An extra class, called supersingular elliptic curves, with endomorphism ring an order in a rational quaternion algebra, can appear.

TopThe Tate module

As we saw, the endomorphisms of elliptic curves over $\mathbb{C}$ can be described as endomorphisms of the lattice $\Lambda$, which is a rank 2 free $\mathbb{Z}$-module. But in the general case, there is no possible way to assign a rank 2 free $\mathbb{Z}$-module functorially, since a rational quaternion algebra has no two-dimensional $\mathbb{Q}$-representation and therefore $\End(E)\otimes\mathbb{Q}$ cannot act on $H^1(E, \mathbb{Q})$ for supersingular elliptic curves. However, we can assign a rank 2 free $\mathbb{Z}_{\ell}$-module (Corollary 1) functorially for $\ell\ne\Char(K)$, namely the $\ell$-adic Tate module $T_\ell(E)$ .

Definition 4 Let $E/K$ be an elliptic curve and $E[m]=\ker[m]$ be the $m$-torsion part of $E$ (over $\overline{K}$). Let $\ell$ be a prime. The $\ell$-adic Tate module $T_\ell(E)$ is defined to be the inverse limit $\varprojlim E[\ell^n]$ with respect to the multiplication-by-$\ell$ maps $[\ell]$.
Remark 2 In the complex case, we can regard the lattice $\Lambda$ as the first cohomology group $H^1(E,\mathbb{Z})$ and the $m$-torsion part $E[m]\cong m^{-1}\Lambda/\Lambda\cong \mathbb{Z}/m\mathbb{Z}\times\mathbb{Z}/m\mathbb{Z}$ as the first cohomology group $H^1(E,\mathbb{Z}/m\mathbb{Z})$. Analogously, in the positive characteristic case, we may interpret $T_\ell(E)$ as the first étale cohomology group $H^1_\mathrm{et}(E,\mathbb{Z}_{\ell})$ and $E[m]$ as the first étale cohomology group $H^1_\mathrm{et}(E,\mathbb{Z}/m\mathbb{Z})$.
Remark 3 Since $\Gal(\overline{K}/K)$ acts naturally on $E[\ell^n]$ and the action commutes with $[\ell]$, it produces a continuous $\ell$-adic Galois representation $\Gal(\overline{K}/K)\rightarrow \Aut(T_\ell(E))$.

For fields of positive characteristic, the $p^r$-torsion part can ``collapse'' somewhat since the isogeny $[p^r]$ is not separable ([1, III 6.4]).

Theorem 2 Let $E/K$ be an elliptic curve. Then
  1. $E[m]\cong\prod_q E[q^{r_q}]$, where $m$ is a positive integer with prime factorization $m=\prod_q q^{r_q}$.
  2. If $\ell\ne\Char(K)$, then $E[\ell^r]\cong\mathbb{Z}/\ell^r\mathbb{Z}\times\mathbb{Z}/\ell^r\mathbb{Z}$.
  3. If $p=\Char(K)$, then either $E[p^r]\cong 0$ for any $r\ge1$ or $E[p^e]\cong \mathbb{Z}/p^r\mathbb{Z}$ for any $r\ge1$.
Remark 4 See the connection between $E[p^r]$ and $\End(E)$ in Theorem 2.

From Theorem 2, the next corollary about the structure of the Tate modules follows immediately.

Corollary 1 Let $E/K$ be an elliptic curve. Then
  1. If $\ell\ne\Char(K)$, then $T_\ell(E)\cong \mathbb{Z}_{\ell}\times \mathbb{Z}_{\ell}$.
  2. If $p=\Char(K)$, then $T_p(E)\cong 0$ or $\mathbb{Z}_p$.

TopClassification of endomorphism rings

In this section, we will classify the endomorphism rings of elliptic curves over an arbitrary field. Let us start by making several simple observations.

Theorem 3 $\Hom(E_1, E_2)$ is a torsion-free $\mathbb{Z}$-module. $\End(E)$ is a (not necessarily commutative) ring of characteristic 0 and has no zero divisors.
Proof Let $m$ be an integer. Suppose $\phi\in\Hom(E_1,E_2)$ satisfies $[m]\circ\phi=0$. Taking degrees gives $(\deg[m])(\deg\phi)=0$. If $m\ne0$, then $[m]$ is a non-constant isogeny ([1, III 4.2 (a)]), $\deg[m]\ne0$. Thus $\deg\phi=0$ and $\phi$ itself is $[0]$.

In particular, $\End(E)$ is a ring of characteristic 0. If $\phi,\psi\in \End(E)$ satisfy $\phi\circ\psi=0$, similarly taking degrees will give $(\deg\phi)(\deg\psi)=0$, which implies $\phi=[0]$ or $\psi=[0]$. So $\End(E)$ has no zero divisors.

Theorem 4 Let $E/K$ be an elliptic curve with $\Char(K)=0$. Then $\End(E)$ is commutative.
Proof Let $\omega$ be an invariant differential on $E$. For any $\phi\in\End(E)$, there exists a rational function $a_\phi\in\overline{K}(E)$ such that $\phi^*(\omega)=a_\phi\omega$ since $\Omega_E$ is a one-dimensional $\overline{K}(E)$-vector space. But $$\div(a_\phi)=\div(\phi^*\omega)-\div(\omega)=\phi^*\div(\omega)-\div(\omega)=0$$ since $\div(\omega)=0$. So $a_\phi$ is a constant in $\overline{K}$. Now, we have a map $$v: \End(E)\rightarrow\overline{K},\qquad \phi\mapsto a_\phi.$$ Then $v$ is a ring homomorphism because $(\phi\circ\psi)^*\omega=\psi^*(\phi^*(\omega))$ and $(\phi+\psi)^*=\phi^*\omega+\psi^*\omega$ ([1, III 5.2]). By assumption $\Char(K)=0$, so any nonconstant isogeny $\phi\in\End(E)$ is a separable morphism. Thus $\phi^*(\omega)\ne0$ and then $a_\phi\ne0 $. We conclude that $v$ is an injection. Because $\overline{K}$ is commutative, it follows that $\End(E)$ is commutative. %

For an isogeny $\phi:E_1\rightarrow E_2$, $\phi$ induces a homomorphism on the $\ell^n$-torsion parts, hence a $\mathbb{Z}_{\ell}$-module homomorphism of the Tate modules. The next theorem allows us to extract information from the Tate modules to get information about the isogenies. We will see more of this idea in the next chapter.

Theorem 5 Let $E_1, E_2$ be two elliptic curves, and $\ell\ne \Char(K)$ be a prime. Then the natural map $$T_\ell: \Hom(E_1,E_2)\otimes \mathbb{Z}_\ell\rightarrow \Hom(T_\ell(E_1), T_\ell(E_2))$$ is injective.
Proof Step 1 For any finitely generated subgroup $M$ of $\Hom(E_1,E_2)$, we claim that the group $$M^{\mathrm{div}}=\mathbb{Q}M\cap\Hom(E_1,E_2)$$ is also finitely generated. We can extend the degree map continuously to $\mathbb{Q}M\rightarrow \mathbb{Q}$. Then $M^{\mathrm{div}}$ is discrete in $\mathbb{Q}M$, since every isogeny has degree at least one. So $M^{\mathrm{div}}$ is a discrete subgroup of a finite-dimensional vector space $\mathbb{Q}M$ and thus must be finitely generated. This proves the claim.

Step 2 Because $\Hom(E_1,E_2)$ is torsion-free, we have $$M^{\mathrm{div}}=\{\phi\in\Hom(E_1,E_2): [m]\circ\phi\in M \text{ for some integer } m\ge1\}.$$ Since $M\subseteq M^{\mathrm{div}}$ and $(M^{\mathrm{div}})^{\mathrm{div}}=M^{\mathrm{div}}$, it suffices to show that for each finitely generated subgroup $M$ such that $M=M^{\mathrm{div}}$, the natural map $$T_\ell: M\otimes\mathbb{Z}_{\ell}\rightarrow\Hom(T_{\ell}(E_1),T_{\ell}(E_2))$$ is injective. Since $M$ is finitely generated and torsion-free, it is free. Let $\psi_1,\ldots,\psi_t$ be a basis for $M$ and $\phi\in M\otimes \mathbb{Z}_{\ell} $. We can write $$\phi=c_1\psi_1+\cdots+c_t\psi_t,\qquad c_i\in \mathbb{Z}_{\ell}.$$ Now suppose $T_\ell(\phi)=0$, then for each $n\ge1$, there exist integers $a_i\equiv c_i \pmod{\ell^n}$ such that $\psi= a_1\psi_1+\cdots+ a_t\psi_t\in M$ annihilates $E_1[\ell^n]$. Hence $\psi$ factors through $[\ell^n]$ since $[\ell^n]$ is separable, namely there exists $\lambda\in \Hom(E_1,E_2)$ such that $\psi=[\ell^n]\circ \lambda$. Therefore $\lambda\in M^{\mathrm{div}}=M$. We can write $$\lambda=[b_1]\circ \psi_1+\cdots+[b_t]\circ \psi_t,\qquad b_i\in\mathbb{Z}.$$ Now $\psi=[\ell^n]\circ\psi$ implies that $$c_i\equiv a_i\equiv \ell^n b_i\equiv 0\pmod{\ell^n}.$$ But $n\ge 1$ is arbitrary, so then $c_i=0$ and $\phi$ itself is zero. This completes the proof of the injectivity of $T_\ell$.

Now the following consequence is relatively easy and less surprising.

Theorem 6 $\Hom(E_1,E_2)$ is a free $\mathbb{Z}$-module of rank at most 4.
Proof Let $\ell\ne\Char(K)$ be a prime. Since $T_\ell(E_1)$ and $T_\ell(E_2)$ are both isomorphic to $\mathbb{Z}_\ell\times \mathbb{Z}_{\ell}$, hence $\Hom(T_\ell(E_1),T_\ell(E_2))\cong M_2(\mathbb{Z}_{\ell})$ is a rank 4 free $\mathbb{Z}_{\ell}$-module. Using the injectivity in Theorem 5 we know that $\Hom(E_1,E_2)\otimes \mathbb{Z}_{\ell}$ is a free $\mathbb{Z}_{\ell}$ module of rank at most $4$, since $\mathbb{Z}_{\ell}$ is a principal ideal domain. So $\Hom(E_1,E_2)$ is a finitely generated $\mathbb{Z}$-module. But $\Hom(E_1,E_2)$ is a torsion-free $\mathbb{Z}$-module by Theorem 3, hence it is a free $\mathbb{Z}$-module and has rank at most 4.

We are now in a position to prove the three possibilities for the endomorphism ring of an elliptic curve mentioned in the beginning of this chapter, i.e., the integers $\mathbb{Z}$, an order in an imaginary quadratic extension of $\mathbb{Q}$, or an order in a quaternion algebra over $\mathbb{Q}$. Let us make precise the meaning of a quaternion algebra.

Definition 5 A (definite) quaternion algebra over $\mathbb{Q}$ is an four-dimensional $\mathbb{Q}$-algebra generated by $\{1,\alpha,\beta,\alpha\beta\}$ whose multiplication satisfies $$\alpha^2,\beta^2\in\mathbb{Q},\quad \alpha^2,\beta^2<0,\quad \beta\alpha=-\alpha\beta.$$

Keep in mind that we will apply the following little tricky theorem to $\mathcal{R}=\End(E)$ and $\hat\alpha$ as the dual isogeny.

Theorem 7 Let $\mathcal{R}$ be a (not necessarily commutative) ring of characteristic zero having no zero divisors. Assume $\mathcal{R}$ has the following properties:
  1. $\rank \mathcal{R}\le 4$.
  2. $\mathcal{R}$ has an involution $\alpha\mapsto\hat \alpha$ such that for any $\alpha,\beta\in \mathcal{R}$ and $a\in\mathbb{Z}\subseteq \mathcal{R}$, $$\widehat{\alpha+\beta}=\hat{\alpha}+\hat{\beta},\quad \widehat{\alpha\beta}=\hat{\beta}\hat{\alpha},\quad \hat{\hat{\alpha}}=\alpha,\quad \hat a=a.$$
  • For any $\alpha\in \mathcal{R}$, $\alpha\hat{\alpha}\in\mathbb{Z}_{\ge0}$. Moreover, $\alpha\hat{\alpha}=0$ if and only if $\alpha=0$.

Then $\mathcal{R}$ is of one of the following types:

  1. $\mathcal{R}\cong\mathbb{Z}$.
  2. $\mathcal{R}$ is an order in an imaginary quadratic extension of $\mathbb{Q}$.
  3. $\mathcal{R}$ is an order in a quaternion algebra over $\mathbb{Q}$.
Proof Let ${\cal K}={\cal R}\otimes\mathbb{Q}$, it suffices to show that ${\cal K}$ is either $\mathbb{Q}$, an imaginary quadratic extension of $\mathbb{Q}$ or a quaternion algebra over $\mathbb{Q}$. Define the norm map and trace map from ${\cal K}$ to $\mathbb{Q}$ using the involution, $$\mathrm{N}\alpha=\alpha\hat{\alpha},\quad \mathrm{T}\alpha=\alpha+\hat{\alpha}.$$ Then $\mathrm{T}\alpha\in\mathbb{Q}$ since $\mathrm{T}\alpha=1+\mathrm{N}\alpha-\mathrm{N}(\alpha-1)$. Suppose $\alpha\in{\cal K}$ such that $\mathrm{T}\alpha=0$, then $\alpha^2=-\mathrm{N}\alpha$ since $\alpha^2-(\mathrm{T}\alpha)\alpha+\mathrm{N}\alpha=(\alpha-\alpha)(\alpha-\hat{\alpha})=0$. So if $\mathrm{T}\alpha=0$ and $\alpha\ne0$ then $\alpha^2$ is a negative rational number.

If ${\cal K}=\mathbb{Q}$, we are done.

Otherwise, we can find $\alpha\in{\cal K}$ such that $\alpha\not\in \mathbb{Q}$. Replacing $\alpha$ by $\alpha-\frac{1}{2}\mathrm{T}\alpha$ we may assume $\mathrm{T}\alpha=0$. Hence $\alpha^2$ is a negative rational number. If ${\cal K}=\mathbb{Q}(\alpha)$, then we are done.

Otherwise, we can find $\beta\in{\cal K}$ such that $\beta\not\in\mathbb{Q}(\alpha)$. Replacing $\beta$ by $\beta-\frac{1}{2}\mathrm{T}\beta-\frac{\mathrm{T}(\alpha\beta)}{2\alpha^2}\cdot\alpha$, we may assume $\mathrm{T}\alpha=\mathrm{T}(\alpha\beta)=0$. Hence $\beta$ is also a negative rational number and $\alpha\beta=-\hat{\beta}\hat{\alpha}=-\beta\alpha$. Now it suffices to show that $\{1,\alpha,\beta,\alpha\beta\}$ is linearly independent over $\mathbb{Q}$ to complete the proof.

Suppose there is a linear relation $$a+b\alpha+c\beta+d\alpha\beta=0,\quad a,b,c,d\in\mathbb{Q}.$$ Taking the trace we get $a=0$. Multiplying by $\alpha$ on the left and $\beta$ on the right and using $\alpha\beta^2=\beta^2\alpha$, we obtain $$b\alpha^2\cdot\beta+c\beta^2\cdot\alpha+d\alpha^2\beta^2\cdot1=0.$$ But $\{\alpha,\beta,1\}$ are linearly independent over $\mathbb{Q}$ by construction, which implies that $b=c=d=0$. This completes the proof.

Theorem 8 The endomorphism ring $\End(E)$ of an elliptic curve $E/K$ is either $\mathbb{Z}$, an order in an imaginary quadratic extension of $\mathbb{Q}$, or an order in a quaternion algebra over $\mathbb{Q}$. Moreover, only the first two cases are possible when $\Char(K)=0$.
Proof The result follows from Theorem 6 about the rank, Theorem 1 about the properties of the dual isogeny and Theorem 7. When $\Char(K)=0$, $\End(E)$ is commutative by Theorem 4, hence the third case is impossible.
Remark 5 We will see if $E$ is defined over a finite field, then $\End(E)\ne\mathbb{Z}$ (Theorem 4), so only the latter two cases can happen.

Elliptic curves with different endomorphism rings behave fundamentally differently in many aspects, so various terminologies were invented.

Definition 6 If $\Char(K)=0$, then we say that an elliptic curve $E/K$ has complex multiplication or (historically) that $E$ is singular, if $\End(E)\ne\mathbb{Z}$. If $\Char(K)>0$, we say that $E/K$ is supersingular if $\End(E)$ is an order in a rational quaternion algebra, otherwise we say that $E$ is ordinary.
Remark 6 We will see that supersingular types are far rarer than ordinary types (Corollary 1).

By now we have achieved our goal of giving a primary classification of the endomorphism rings of elliptic curves in this chapter. In the next chapter we will concentrate more on the positive characteristic case. In particular, we will give a refined characterization of the endomorphism rings of supersingular elliptic curves using more useful tools and deep results.

References

[1]Joseph H. Silverman, The Arithmetic of Elliptic Curves, Springer, 2010.

[2]Joseph H. Silverman, Advanced Topics in the Arithmetic of Elliptic Curves, Springer, 1994.

[3]Phillip A. Griffiths, Introduction to Algebraic Curves, American Mathematical Society, 1989.