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 , an order in an imaginary quadratic extension of , or an order in a quaternion algebra over . 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 as an example, in an attempt to get a clue as to the general proof. The main source of our exposition is [1].

## Elliptic 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 over a field is
1. an irreducible smooth projective curve over of genus one with a specified point , or
2. a plane projective curve defined by a Weierstrass equation with nonzero discriminant. For , the Weierstrass equation can be simplified via a change of variables to or
3. () a one-dimensional complex torus, i.e. for some lattice in , or
4. a one-dimensional abelian variety over , i.e., a connected projective curve over with a group structure.

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

(a)(b) Suppose satisfies (a). By the Riemann-Roch theorem, for any , we have . Letting and , we can find two functions such that has exactly 2 poles at and has exactly 3 poles at . Now let , we get a linear relation between which gives us the Weierstrass equation (1). The map sends onto a plane projective curve defined by a Weierstrass equation. The extension degrees and implies that , hence has degree 1 and gives an isomorphism between and . Since is smooth, we know that the discriminant of the Weierstrass equation is nonzero. Moreover, is the point at infinity.

Conversely, suppose is a plane projective curve defined by a Weierstrass equation. The fact that the discriminant is nonzero means that is smooth. To see that has genus 1, consider the invariant differential It is holomorphic and non-vanishing, hence . By the Riemann-Roch theorem, , which implies that . Finally, the specified point is taken to be the point at infinity.

(b)(c) Suppose satisfies (c), . Define where the Weierstrass -function is given by Then is the plane projective curve cut out by (2) with and where are Eisenstein series of weights 4 and 6. Let be a basis of the lattice and . Then the equation (2) has three distinct roots , so the discriminant is nonzero. One can show that is injective using the properties of elliptic functions ([1, VI 3.6]). Since we know that induces an isomorphism on the cotangent spaces. Thus is an isomorphism between and .

Conversely, for any such that the discriminant is nonzero, there exists a unique lattice satisfying and by the Uniformization theorem ([2, I 4.3]). Thus the plane projective curve is isomorphic to the complex torus .

(a)(c) Suppose satisfies (a). Then is a genus one Riemann surface. Let be the fundamental loops of generating and let be the canonical differential on , then are -linearly independent. Let be the lattice spanned by them. The Abel-Jacobi map is biholomorphic by the Jacobi inversion theorem ([3, Theorem 5.1]).

Conversely, the one-dimensional complex torus is of genus 1. The specified point is taken to be .

(a)(d) Suppose satisfies (a). For any two points , if and only if , since by the Riemann-Roch theorem and contains constant functions. Therefore we have an injective map from to its degree-0 Picard group Again, by the Riemann-Roch theorem, for any degree-0 divisor . Let be a basis vector of this one-dimensional space. Then for some , so we get and thus is also surjective. So has a group structure inherited from and serves as the identity element. Moreover, is irreducible, so connected.

Conversely, suppose has a group structure. Then by generic smoothness we know is smooth. is connected and smooth, hence irreducible. Moreover, the canonical bundle is free of rank 1 and has degree , therefore the genus .

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

(c)(d) The group structure on is given by . The point serves as the identity element. The properties of elliptic functions also ensure a bijection between and ([1, VI 2.4]). These two group structures are compatible with each other.

## Isogeny and dual isogeny

Definition 2 Let be two elliptic curves. An isogeny from to is a morphism satisfying . The group of isogenies from to is denoted by . The ring of endomorphisms of an elliptic curve , i.e., the isogenies from to itself, is denoted by . For those isogenies defined over , we denote them by and respectively.

An isogeny is automatically a homomorphism. An isogeny is the same as the composition where is the pushforward map between Picard groups induced by the isogeny . All these three maps are group homomorphisms, hence the composition itself is a group homomorphism.

One can also consider the pullback map of Picard groups induced by , the composition turns out to be an isogeny from to as well and the composition has the effect .

Definition 3 Let be an isogeny of elliptic curves of degree , then there exists a unique isogeny such that , where denotes the multiplication-by- isogeny. is called the dual isogeny to .

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

Theorem 1 Let , be isogenies of elliptic curves.
1. Let , then on and on . In particular, .
2. .
3. .
4. and for any .
5. .
Example 1 Consider . In this case we can write down the isogenies, dual isogenies and endomorphism rings in very concrete terms.

Isogeny Suppose and are two elliptic curves and is an isogeny. Then, by the lifting property of the covering space, the holomorphic map can be lifted to a holomorphic map , since the source is simply-connected. For any , maps into a discrete subset of , hence must be constant. Therefore, ; in other words, is an elliptic function with respect to the lattice . But is also holomorphic, hence must be a constant and is a linear map. Now implies that for some . In summary, must be of the form There is a bijection

Dual isogeny Suppose corresponds to in the manner mentioned above. Then is a sublattice of . Let be a basis of , then we can find two integers such that forms a basis of . Moreover, is exactly the degree of . Now is a well-defined isogeny since . Also, it is easy to see that namely is just the dual isogeny of .

Endomorphism ring From the above discussion, we know that Suppose has a basis . Without loss of generality, we may multiply by and assume that is a basis of where . Let , then we can find integers such that and . So satisfies a quadratic equation Hence is an integral extension of . Now suppose , then we can find some with . Eliminating yields another quadratic equation in , So is an order in an imaginary quadratic extension of . In summary, can be either consisting of multiplication-by- maps, or an order in an imaginary quadratic field .

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 , we are able to conclude that over an arbitrary field of characteristic 0, the endomorphism ring 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.

## The Tate module

As we saw, the endomorphisms of elliptic curves over can be described as endomorphisms of the lattice , which is a rank 2 free -module. But in the general case, there is no possible way to assign a rank 2 free -module functorially, since a rational quaternion algebra has no two-dimensional -representation and therefore cannot act on for supersingular elliptic curves. However, we can assign a rank 2 free -module (Corollary 1) functorially for , namely the -adic Tate module .

Definition 4 Let be an elliptic curve and be the -torsion part of (over ). Let be a prime. The -adic Tate module is defined to be the inverse limit with respect to the multiplication-by- maps .
Remark 2 In the complex case, we can regard the lattice as the first cohomology group and the -torsion part as the first cohomology group . Analogously, in the positive characteristic case, we may interpret as the first étale cohomology group and as the first étale cohomology group .
Remark 3 Since acts naturally on and the action commutes with , it produces a continuous -adic Galois representation .

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

Theorem 2 Let be an elliptic curve. Then
1. , where is a positive integer with prime factorization .
2. If , then .
3. If , then either for any or for any .
Remark 4 See the connection between and in Theorem 2.

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

Corollary 1 Let be an elliptic curve. Then
1. If , then .
2. If , then or .

## Classification 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 is a torsion-free -module. is a (not necessarily commutative) ring of characteristic 0 and has no zero divisors.
Proof Let be an integer. Suppose satisfies . Taking degrees gives . If , then is a non-constant isogeny ([1, III 4.2 (a)]), . Thus and itself is .

In particular, is a ring of characteristic 0. If satisfy , similarly taking degrees will give , which implies or . So has no zero divisors. ¡õ

Theorem 4 Let be an elliptic curve with . Then is commutative.
Proof Let be an invariant differential on . For any , there exists a rational function such that since is a one-dimensional -vector space. But since . So is a constant in . Now, we have a map Then is a ring homomorphism because and ([1, III 5.2]). By assumption , so any nonconstant isogeny is a separable morphism. Thus and then . We conclude that is an injection. Because is commutative, it follows that is commutative. % ¡õ

For an isogeny , induces a homomorphism on the -torsion parts, hence a -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 be two elliptic curves, and be a prime. Then the natural map is injective.
Proof Step 1 For any finitely generated subgroup of , we claim that the group is also finitely generated. We can extend the degree map continuously to . Then is discrete in , since every isogeny has degree at least one. So is a discrete subgroup of a finite-dimensional vector space and thus must be finitely generated. This proves the claim.

Step 2 Because is torsion-free, we have Since and , it suffices to show that for each finitely generated subgroup such that , the natural map is injective. Since is finitely generated and torsion-free, it is free. Let be a basis for and . We can write Now suppose , then for each , there exist integers such that annihilates . Hence factors through since is separable, namely there exists such that . Therefore . We can write Now implies that But is arbitrary, so then and itself is zero. This completes the proof of the injectivity of . ¡õ

Now the following consequence is relatively easy and less surprising.

Theorem 6 is a free -module of rank at most 4.
Proof Let be a prime. Since and are both isomorphic to , hence is a rank 4 free -module. Using the injectivity in Theorem 5 we know that is a -vector space of dimension at most 4, hence , as a discrete subgroup of , is a free -module and has rank at most 4. (Added 2021/02/03: thank Carlo Pagano for correction of this proof.) ¡õ

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 , an order in an imaginary quadratic extension of , or an order in a quaternion algebra over . Let us make precise the meaning of a quaternion algebra.

Definition 5 A (definite) quaternion algebra over is an four-dimensional -algebra generated by whose multiplication satisfies

Keep in mind that we will apply the following little tricky theorem to and as the dual isogeny.

Theorem 7 Let be a (not necessarily commutative) ring of characteristic zero having no zero divisors. Assume has the following properties:
1. .
2. has an involution such that for any and ,
• For any , . Moreover, if and only if .

Then is of one of the following types:

1. .
2. is an order in an imaginary quadratic extension of .
3. is an order in a quaternion algebra over .
Proof Let , it suffices to show that is either , an imaginary quadratic extension of or a quaternion algebra over . Define the norm map and trace map from to using the involution, Then since . Suppose such that , then since . So if and then is a negative rational number.

If , we are done.

Otherwise, we can find such that . Replacing by we may assume . Hence is a negative rational number. If , then we are done.

Otherwise, we can find such that . Replacing by , we may assume . Hence is also a negative rational number and . Now it suffices to show that is linearly independent over to complete the proof.

Suppose there is a linear relation Taking the trace we get . Multiplying by on the left and on the right and using , we obtain But are linearly independent over by construction, which implies that . This completes the proof. ¡õ

Theorem 8 The endomorphism ring of an elliptic curve is either , an order in an imaginary quadratic extension of , or an order in a quaternion algebra over . Moreover, only the first two cases are possible when .
Proof The result follows from Theorem 6 about the rank, Theorem 1 about the properties of the dual isogeny and Theorem 7. When , is commutative by Theorem 4, hence the third case is impossible. ¡õ
Remark 5 We will see if is defined over a finite field, then (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 , then we say that an elliptic curve has complex multiplication or (historically) that is singular, if . If , we say that is supersingular if is an order in a rational quaternion algebra, otherwise we say that 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.