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 .
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.
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.
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 .
Here we list some easy properties of the dual isogeny as follows ([1, III 6.2]).
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.
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 .
For fields of positive characteristic, the -torsion part can ``collapse'' somewhat since the isogeny is not separable ([1, III 6.4]).
From Theorem 2, the next corollary about the structure of the Tate modules follows immediately.
In this section, we will classify the endomorphism rings of elliptic curves over an arbitrary field. Let us start by making several simple observations.
In particular, is a ring of characteristic 0. If satisfy , similarly taking degrees will give , which implies or . So has no zero divisors. ¡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.
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.
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.
Keep in mind that we will apply the following little tricky theorem to and as the dual isogeny.
Then is of one of the following types:
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
Elliptic curves with different endomorphism rings behave fundamentally differently in many aspects, so various terminologies were invented.
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.
The Arithmetic of Elliptic Curves, Springer, 2010.
Advanced Topics in the Arithmetic of Elliptic Curves, Springer, 1994.
Introduction to Algebraic Curves, American Mathematical Society, 1989.