Elliptic Curves with Complex Multiplication

Drew Moore Discussed with Professor Emerton

1 Definition and Examples

Definition 1. We say that an E over a field L has complex multiplication by the imaginary quadratic field K if there is an R of K and an inclusion

[ · ]: R,→ End(E) α 7→ [α]

[ · ] is normalized if there is a map R → L such that [α]∗ω = αω, where ω is an invariant differential for E (and α ∈ R has been identified with its image in L). If L ⊃ K, this can always be done.

Remark (1). We also say that E has CM by the order R and that E ∈ Ell/L(R). For the rest of this document, I will for simplicity assume R = OK is the maximal order of K.

Remark (2). If L ⊂ C and End(E) )Z, then E has CM by some quadratic imaginary field. Indeed, the degree function on End(E) ensures that End(E) ⊗ Q is a quadratic imaginary field or Q. 2 3 Example 1. E/Q with Weirstrass equation y = x + ax has complex multiplication by 2 Z[i]. i corresponds to the map (x, y) 7→ (−x, iy). This works, because [i] (x, y) = (x, −y) = ∗ dx −dx dx [−1](x, y), and [i] y = iy = i y . √ Example 2. If j ∈ Q( 5) is one of two solutions to the set of equations

2 3 5 2 (j5 + 250j5 + 3125) − j5 j = 0 j5 = 125 √ and if j(E) = j, then E has CM by Z[ −5]. Indeed, the first equation gives the function p3 field of X0(5), where j5(τ) = ∆(5√τ)/∆(τ) is a uniformizer for X0(5). The second equation expresses the fact that√ the dual of [ −5] is itself, up to ±1. Specifically, the point on X0(5) corresponding to (E, [ −5]) is fixed by the Atkin-Lehner involution.

Example 3. If Λ = f is a fractional ideal of R, then C/Λ is (the complex analytic space of C-points of) an elliptic curve over C with complex multiplication by R. (This follows since αf ⊂ f for all α ∈ R). Later we will see that this elliptic curve can be defined over a number field, and that every elliptic curve in Ell/C(R) is of this form.

1 2 The action of the class group ClK on Ell/L(R)

2.1 Serre’s Construction

Example 3 leads us to define the action

 ¯ −1 ClK Ell/C(R) f ∗ (C/Λ) = C/(f · Λ) for f ∈ IK If two fractional ideals differ by a principal ideal, then as lattices they are homothetic; this shows that the action is well-defined. We would like to replace C by any field L. To do so, −1 note that as R-modules, we have that C/(f · Λ) is isomorphic to −1 ∼ f ⊗R (C/Λ) = HomR(f, C/Λ)

So we aim to define an elliptic curve ¯f ∗ E with CM by R such that ¯f ∗ E = HomR(f,E). By this, we mean that for A any L-algebra, we have a natural isomorphism of R-modules ∼ (¯f ∗ E)(A) = HomR(f,E(A)) (1)

Proposition 1 ([1]). If E ∈ Ell/L(R) is normalized respect to some map L → R, the functor L-Alg → R-Mod A 7→ HomR(f,E(A)) ¯ is representable by an elliptic curve f ∗ E ∈ Ell/L(R), so that equation (1) holds. This gives a well-defined action of ClK on Ell/L(R). Proposition 2. If L ⊂ C is algebraically closed (for example L = C), then the action of ClK on Ell/L(R) is simply transitive. Proposition 3. For m an integral ideal of R, there exists an isogeny (unique up to auto- morphisms of E) E → m¯ ∗ E such that if E[m] is the kernel, the points of E[m] are the m-torsion points of E. Further, if char L = 0, the L¯-points of E[m] form a free rank 1 (R/m)-.

Since m¯ ∗ E = HomR(m,E), we may define the map ϕ on A-points P as  ϕ: P 7→ α 7→ [α]P ∈ HomR(m,E(A)) Remarks.

1. While m¯ ∗ E only depends on the class of m in ClK , the map E → m¯ ∗ E does depend on the representative ideal m. For example, for π ∈ R not a unit, (π) = (1), but the corresponding isogenies are [π] 6= id. ∼ −1 2. When L = C and E(C) = C/Λ, this is just the natural map C/Λ → C/m Λ. 3. If n is another integral ideal, then the map E → m¯ n¯ ∗ E can be factored as E → n¯ ∗ E → m¯ ∗ n¯ ∗ E =∼ m¯ n¯ ∗ E.

2 2.2 Rationality of j(E) and of End(E)

Though so far the only large family of elliptic curves with CM we have seen have been defined over C, the action of ClK on Ell/C(R) readily gives arithmetic/rationality information about such elliptic curves. Indeed, using this action, we get the following results:

Proposition 4. Suppose E ∈ Ell/C(R). Then

[Q(j(E)) : Q] ≤ hK the class number of K, so that j(E) ∈ Q¯ . Corollary. The map Ell (R) → Ell (R) is an equivalence. That is, every E/ with CM /Q¯ /C C by R can be defined over a number field L. Remark. So from now on in the characteristic 0 case, we will consider elliptic curves over a number field L containing K.

Proposition 5. Suppose that E ∈ End/L(R), with K ⊂ L ⊂ C. Then every endomorphism of E is defined over L.

3 The Tate Module and Reduction of CM Elliptic Curves

3.1 Preliminaries on the Tate Module

So that we may consider properties of good reduction, we consider E ∈ Ell/L(R), where char L may be nonzero. Let ` be a prime number different from char L. Set T = lim E[`n] and V = ⊗ T . ` ←−n ` Q` Z` ` We view T` as a Z`-sublattice of the 2-dimensional Q` vector space V`. Additionally, define R` = Z`⊗R and K` = Q`⊗QK. Then V` (resp. T`) is a K`-module (resp. R`-module).

Proposition 6. T` is free and rank one over R`. Additionally, for α ∈ K`, αT` ⊂ T` if and only if α ∈ R`.

Corollary 1. Let ρ` be the representation

sep ρ` : Gal(L /L) → Aut(T`)

× Then the image of ρ` is contained in R` ⊂ Aut(T`E). 3.2 Reduction of CM Elliptic Curves

For this section, let E ∈ Ell/L(R), with L ⊃ K a number field. Let µ be the roots of unity in K. (In particular, unless j = 0 or 1728, µ = {±1}.) The following proposition shows

3 that the reduction Ee of a CM elliptic curve E has CM, compatible with that on E. Fix a prime P of L.

Proposition 7. Suppose E has good reduction at P, and let Ee denote the E mod P. a) Reduction of endomorphisms End(E) ,→ End(Ee), ψ 7→ ψe, is well defined, preserves degrees, and gives Ee complex multiplication by R.

b) The image End(^E) of this map is its own commutant inside End(Ee). c) The map Ee → m ∗ Ee, for m ⊂ R integral, is the reduction of the map E → m ∗ E. d) Let q be a prime of K. Then Ee → q ∗ Ee is inseparable if and only if P | q. Theorem 1 ([2, Theorem 6]). a) E has potential good reduction at P. In particular j(E) is integral.

b) If nP is the exponent conductor at P of ρ` as an abelian character, then the exponent conductor at P of E is 2nP. Example 4. If we consider the example y2 = x3 − x, we can determine the conductor of 6 E/Q(i) explicitly using the Grossencharacter (which we define next section) to be (1 + i) . Thus, there exists a subextension of the field generated by the 3-torsion of E over which E has good reduction at every place. Specifically, we look at the µ4-fixed part of Q(i)(E[3]). Accordingly,√ some calculation shows that E has everywhere good reduction over the field Q(i, 3). 4 The Grössencharacter

4.1 In terms of ideals

Continue with the notation of subsection 3.2. Suppose P lies above the prime p of K.

Let φP = φq ∈ End(Ee) denote the q-th power endomorphism of Ee (where q is a power of ab p), and let σP ∈ GL be Frobenius. (These act on the Fq-points of Ee identically.)

Proposition 8. There is a unique α = αP ∈ R such that

[α] E E L (α) = NK (P) and commutes.

φq Ee Ee

Proof. First note: for every β ∈ R, [β] is defined over L by Proposition 5. In particular,

4 [fβ] is defined over Fq. In terms of Frobenius, this means that

σP σP  [fβ](P ) = [fβ](P ) ∀P ∈ Ee Fq [fβ] ◦ φP = φP ◦ [fβ]

Thus, φP commutes with the image of End(E) under the reduction map. Thus by part b of Proposition 7, we have that φq = [fα] for some unique α ∈ R. Because reduction is compatible with degrees, we see that

L L P |NQ (α)| = q = NQ ( ) This norm computation shows that the prime factorization of α in K only contains primes above p - either p or its conjugate p0. If we show that only p divides α, we can conclude L NK (P) = (α). For the sake of contradiction, suppose that p0 6= p and that p0 | α. But then the map [fα] can be factored as Ee → p¯0 ∗ Ee → Ee 0 0 and since P - p , the map Ee → p¯ ∗ E is separable. This contradicts the fact that [fα] = φP is purely inseparable.

S Letting IL denote the fractional ideals of L coprime to S, We define a map

S × αE/L : IL → R P 7→ αP

Proposition 9. The map S × αE/L : IL → C is a classical . I.e., it has a conductor c, and has an infinity type. Its infinity L −1 type is equal to (NK ) . By this I mean that if m = (γ) is a principal ideal in L and γ ≡ 1 mod c, then

L αE/L(m) = NK (γ)

Example 5. The Grossencharacter of E : y2 = x3 − x is

ψ(p) = π where π satisfies π ≡ 1 mod (1 + i)3, p = (π)

6 Thus, the conductor E/Q(i) is (1 + i) .

5 4.2 In terms of idéles

Since αE/L is a classical Grossencharacter, there exists a corresponding continuous idele class group Grossencharacter × × × ψE/L : AL /L → C × such that for primes of good reduction of E, ψE/L(P) = αP, and if t∞ ∈ (R ⊗ L) , L ψE/L(1,..., 1, t∞) = NK (t∞).

We would like to compare ψE/L and ρE,`. We will consider ρE,` as an idele character via class field theory: × × ab × ρE,` : AL /L  GL → K`

× Proposition 10. For a ∈ AL an idele, let a` ∈ Q` ⊗ L be its `-part. Then we have an equality L −1 L −1 ψE/H (a) · NK (a∞) = ρE,`(a) · NK (a`)

where the left (resp. right) side of the equation a priori belongs to R ⊗ K (resp. Q` ⊗ K) but in fact belongs to K, and equality holds in K.

This proposition follows because for primes of good reduction P, reduction E → Ee gives an isomorphism on Tate modules. Thus, we can study ρE,` mod P, where the connection with ψE/H becomes clear since ψ was defined by lifting Frobenius.

Corollary. The conductor exponent at P of αE/H (or ψE/H ) is equal to half the conductor exponent at P of E.

5 Class Theory for K

Theorem 2. The of K is K(j(E)), for any E ∈ Ell/L(R), L ⊃ K. The main ingredient of the proof is the map

σ ∼ F : GK → ClK E = F (σ) ∗ E (over Q)

The main difficulty is showing that this definition does not depend on the choice of E, so that it is a group homomorphism. This is easy for us, since we define the action of ClK arithmetically via Serre’s construction in section 1.

From this point on, fix an elliptic curve E ∈ Ell/H (R) where H is the Hilbert Class Field of K.

Proposition 11. For m an integral ideal of K, set Hm = H(E[m]). × (a) The extension Hm/H has a subgroup of (R/m) .

6 µ (b) If we set Km = (Hm) to be the µ-invariant field (where we consider µ acting via × × (R/m) ), then Km/H has Galois group isomorphic to (R/m) /µ.

(c) The extension Km/K is abelian. We have an exact sequence

1 → G(Hm/H) → G(Hm/K) → G(H/K) → 1 with the first and third groups abelian. Thus, we get an action of G(H/K) on G(Hm/H). If α = ψ(P) mod m, then the action of σ ∈ G(H/K) is given by ασ = ψ(Pσ) mod m. Since Pσ and P have equal ideal norms in K, we have

ψ(Pσ) ∈ µ ψ(P)

Hence, if we quotient by µ, the middle term becomes abelian. Example 6. Consider the elliptic curve with j(E) = j as in Example 2. In this case, we can compute √ √ √ K = Q( −5) H = Q( −5, 5) 0 89OK = p · p pOH = P1P2 √ ψ(P1) = −ψ(P2) = 3 + 4 −5 which shows that H(E[89]) is a non- of K.

Theorem 3. If Km is the field defined in proposition 11, then Km is the ray class field of modulus m for K.

The first step is to show that the conductor of Km/K divides m, and here we can use the formula given by Proposition 10. Then, we show that the induced map from the ray class group ClK,m → Gal(Km/K) is an isomorphism using Proposition 11(b).

References

[1] Jean-Pierre Serre. “Complex multiplication”. In: Theory (Proc. In- structional Conf., Brighton, 1965). Thompson, Washington, D.C., 1967, pp. 292–296. [2] Jean-Pierre Serre and . “Good reduction of abelian varieties”. In: Ann. of Math. (2) 88 (1968), pp. 492–517. [3] Joseph H. Silverman. Advanced Topics in the Arithmetic of Elliptic Curves (Graduate Texts in Mathematics). Springer, Sept. 1999. [4] Joseph H. Silverman. The Arithmetic of Elliptic Curves (Graduate Texts in Mathemat- ics). 2nd ed. 2009. Springer, Mar. 2016.

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