Verifying the Full Birch and Swinnerton-Dyer Conjecture for Specific Elliptic Curves of Analytic Rank 0 or 1

Stephen Donnelly and Stefan Patrikis and William A. Stein and Michael Stoll^{∗}

July 26, 2004

Abstract

Suppose E is an elliptic curve over Q with conductor at most 87 and rank one or conductor at most 19 and rank zero. Then the full Birch and Swinnerton-Dyer conjecture is true for E. We prove this by ...

Algorithm 0.1. Let E be an elliptic curve over Q of analytic rank at most 1. The following algorithm computes � (E/Q)[p] for all primes p.

1. [Choose K] Choose the first quadratic imaginary field K that satisfies the Heegner hypothesis, is such that E/K has analytic rank 1, and whose discriminant is divisible by at least two primes. Let D be the discriminant of K. Note that the condition that D be divisible by two primes and that (D, N_{E}) = 1, implies that Q(E[p]) is linear disjoint from K for all primes p. In fact, this is a necessary and sufficient condition, since if D is divisible by only one prime p, then because of the Weil pairing Q(E[p]) will automatically contain K. [[Note: we’ll need a density argument here to know that such a K exists.]]

Proof. Having chosen the discriminant D this way, for any prime p, there is a prime factor of D coprime with N_{E}p. Therefore K is not a subfield of Q(E[p]) and since [K : Q] = 2 we actually have linear disjointness. This in turn implies that G(Q(E[p])/Q) ≡ G(K(E[p])/K).

2. [Find p-torsion] Decide for which primes p, there is a curve E′ that is Q-isogenous to E such that E′(K)[p] ≠ 0. (We enumerate the Q-isogeny class of E using [standard method]. Then for each E′ in the Q-isogeny class, compute the torsion subgroup of E′(K) using [standard method], and see whether or not p divides its order.) Let B be the product of 2 and these primes.

∗Some of these authors might not even know they are authors, so don’t blame them for any mistakes below. Blame William Stein.

Proof. Let p be a prime so that for any elliptic curve E′ which is Q-rational p-isogenous to E we have E′(K)[p] = 0.

Proposition 0.2. H^{i}(K(E[p])/K, E[p]) = 0 for all i > 0.

Since G(Q(E[p])/Q) ≡ G(K(E[p])/K) there is an isomorphism H^{i}(G(Q(E[p])/Q), E[p]) ≡ H^{i}(G(K(E[p])/K), E[p]). Therefore it is enough to show this on the level of Q.

First observe that G(Q(E[p])/Q) = H is precisely the image G of the Galois representation ρ_{p} : G(Q̄/Q) → Aut(E[p]) ≅ GL_{2}(F_{p}).

There are two cases to consider: if p ∤ #G, the cohomology clearly vanishes because multiplication by #G kills the cohomology but is an isomorphism on the p-group E[p]. Otherwise, let p|#G. By Proposition 15 of [?], G either contains SL_{2}(F_{p}) or is contained in a Borel subgroup of GL_{2}(F_{p}).

If G contains SL_{2}(F_{p}) then G contains −I_{2} ∈ SL_{2}(F_{p}) and so a nontrivial scalar. Lemma 4.1 [state it here or something] implies that det : G → F_{p}^{∗} is surjective, so G = GL_{2}(F_{p}) and contains all of the nontrivial scalars, which we denote by Z. Applying the inflation-restriction exact sequence to the subgroup Z of G we get that H^{1}(G/Z, E[p]^{Z}) → H^{1}(G, E[p]) → H^{1}(Z, E[p])^{G/Z} → . . .. Since #Z − p − 1 ∤ p and Z leaves no subgroup of E[p] invariant the central group is trivial. By induction this holds in all dimensions > 0.

Therefore G is contained in a Borel subgroup. This implies that the representation ρ_{p} is reducible since E[p] is reducible under the action of a Borel subgroup. This implies the existence of a Q-rational p-isogeny.

Choosing a basis, suppose that the action of G on E[p] = F_{p}^{2} is given by

( \begin{matrix}χ & ∗ \\ 0 & ψ \end{matrix} ) for characters χ and ψ. If χ is trivial, all matrices of the above

form fix ( \begin{matrix}1 \\ 0 \end{matrix} ). In particular, there is a point of E[p] fixed by the action of G,

a contradiction since we have assumed that E(Q)[p] = 0. Now suppose that ψ

is trivial. Matrices of the above form preserve the line generated by ( \begin{matrix}1 \\ 0 \end{matrix} ), so

this line forms a G(Q̄/Q)-stable subspace of E[p]. In particular, there exists an isogeny over Q to a curve E′ having this line as kernel. The image of the complementary line generated by ( \begin{matrix}0 \\ 1 \end{matrix} ) is a 1-dimensional subspace of

E′[p], and if ψ = 1, G(Q̄/Q) clearly acts trivially on this subspace (we have

an isomorphism of Galois modules E/ < ( \begin{matrix}1 \\ 0 \end{matrix} ) >≅ E′). Thus, E′(Q)[p] is

nontrivial, contradicting our assumption.

Let W be the (unique) p-Sylow subgroup of ( \begin{matrix}∗ & ∗ \\ 0 & ∗ \end{matrix} ) consisting of matrices

of the form ( \begin{matrix}1 & ∗ \\ 0 & 1 \end{matrix} ). We may assume W ⊂ H, for otherwise H has order

prime to p, and the cohomology clearly vanishes.

We begin by explicitly computing the third cohomology group using the fact

that W is cyclic (generated by w = ( \begin{matrix}1 & 1 \\ 0 & 1 \end{matrix} ), for instance). Recal that

for cyclic groups we can compute cohomology using the particularly simple projective resolution

... → Z[W] → Z[W] → Z → 0

where the boundary maps alternate between w − 1 and Norm = ∑_{i=0}^{p−1} w^{i} (i.e., the maps are given by multiplication in the group ring Z[W]). Then we immediately see that

H^{i}(W, E[p]) = \{ \begin{matrix}ker(1 − w)/im(Norm(w)) =< ( \begin{matrix}1 \\ 0 \end{matrix} ) > & if i is even; \\ ker(Norm(w))/im(1 − w) = F_{p}^{2}/ < ( \begin{matrix}1 \\ 0 \end{matrix} ) > & if i is odd \end{matrix} \} .

Since χ and ψ are nontrivial by assumption, the H/W -invariants for both of these groups are trivial. Thus, H^{i}(W, E[p])^{H/W} = 0 for i > 0. Let us then consider the Hochschild-Serre spectral sequence

H^{i}(H/W, H^{j}(W, F_{p}^{2})) ⇒ H^{i+j}(H, F_{p}^{2}).

For i > 0, since |H/W | is prime to p, and H^{j}(W, F_{p}^{2}) is a p-group (∀j), the group H^{i}(H/W, H^{j}(W, F_{p}^{2})) is trivial. But when i = 0 we have just computed that H^{i}(H/W, H^{j}(W, F_{p}^{2})) = H^{j}(W, F_{p}^{2})^{H/W} = 0, so the entire spectral sequence is trivial, and we conclude that H^{n}(H, E[p]) = 0 for all n ≥ 0.

The condition that E(K)[p] = 0 implies that another cohomology group is trivial.

Proposition 0.3. Let E be an elliptic curve with E(K)[p] = 0, where p > 3 or, if p = 3, K ≠ Q(μ_{3}). Then H^{i}(F/K, E(F)[p]) = 0 for all i ≥ 1 and F/K abelian extension.

Proof. We may write the abelian group G(F/K) as a direct sum P ⊕P′, where P is its Sylow p-subgroup and (p, #P′) = 1. We claim that the subgroup of E(F)[p] invariant under P′ is trivial. Let G = G(F/K)/H, where H is the subgroup of G(F/K) that acts trivially on E(F)[p]. If (#G, p) = 1, P ⊆ H, so P′ surjects onto G. As there is no nontrivial element of E(F)[p] invariant under all of G(F/K) (by the assumption on E(K)[p], the same then holds for P′.

If p|#G, we cannot have E(F)[p] = Z/pZ: the latter group has automorphism group isomorphic to (Z/pZ)^{∗}, of order p − 1, but if p|#G, G would give rise to at least p distinct automorphisms. Thus, E(F)[p] is the full p-torsion subgroup of E, and we can identify G with a subgroup of GL_{2}(Z/pZ) acting on E(F)[p] = (Z/pZ)^{2}.

We can choose a basis of (Z/pZ)^{2} so that G contains the subgroup ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ),

where x ∈ Z/pZ. Being abelian, G must be contained in the normalizer

of this subgroup, so G ⊆ \{( \begin{matrix}a & b \\ 0 & a \end{matrix} ) |a, b ∈ Z/pZ\}, and we claim that G

contains an element with a ≠ 1. Since E[p] ⊂ E(F)[p], the representation G(Q̄/K) → Aut(E[p]) factors through G(F/K) (recall that the image of the representation is G(K(E[p]/K)). We argued before that this image contained the scalars corresponding to squares in F_{p}^{∗}, so P′ contains at least \frac{p−1}{2} (which is > 1 for p > 3) elements that leave no subgroup of E(F)[p] invariant. Now, the result will follow from an application of the inflation-restriction exact sequence:

0 → H^{1}(P, E(F)[p]^{P′}) → H^{1}(F/K, E(F)[p]) → H^{1}(P′, E(F)[p])

The first group vanishes since E(F)[p]^{P′} = 0, and the third group vanishes since the order of P′ is prime to p, and thus to the order of E(F)[p]. We conclude that the middle group is trivial, as desired. We can therefore extend the sequence to the second cohomology groups, deduce the same triviality result, and by induction conclude that H^{m}(F/K, E(F)[p]) = 0 for all m ≥ 1.

If p = 3, E(F)[3] = E[3] ⇒ μ_{3} ⊂ F ⇒ K = Q(μ_{3}). The last implication holds because G(F/Q) is abelian since G(F/K) and G(K/Q) are, so it has a unique index 2 subgroup; both K and Q(μ_{3}) correspond to index 2 subgroups by elementary Galois theory). This contradicts our assumption on K, so we must have E(F)[3] = 0, in which case the cohomology is clearly trivial.

Gross’s account of Kolyvagin’s proof of triviality of p torsion for the Shafarevich-Tate group relies on two cohomological statements. One of them is precisely the proposition that H^{i}(K(E[p])/K, E[p]) = 0 (by the Hochshild-Serre spectral sequence he obtains the isomorphism H^{1}(K, E[p]) ≅ H^{1}(K(E[p]), E[p])^{G(K(E[p])/K)}). The other one is that H^{i}(K_{n}/K, E(K_{n})[p]) = 0 which follows from the previous proposition since K_{n}/K is abelian (he uses this to construct Kolyvagin’s cohomology classes and to show that restriction gives an isomorphism H^{1}(K, E[p]) ≅ H^{1}(K_{n}, E[p])^{G_{n}}, where K_{n} denotes the ring class field of conductor n over K and G_{n} = G(K_{n}/K)). S. Donnelly has suggested replacing Kolyvagin’s original hypothesis with these weaker and more easily computable ones.

Kolyvagin’s work implies that � (E/K)[p] = 0 for all p ∤ BI_{K}, where I_{K} = [E(K) : Zy_{K}]. The remainder of the algorithm will deal with these bad primes.

3. [Root number] Compute the root number of E using the algorithm in [section blah of Cohen’s Algorithms for ...].

4. [Compute Mordell-Weil]

• If the root number is −1, compute E(Q) (using ..., [possible because of [] implies that the algebraic rank equals the analytic rank since the analytic rank is at most 1]), and let z be a generator modulo torsion.

• If the root number is +1, compute E^{D}(Q), and let z be a generator modulo torsion.

Proof. The fact that this yields an effective computation of the Mordell-Weil group comes from the following proposition:

Proposition 0.4. For a quadratic imaginary field K = Q(\sqrt{D}), E(K) decomposes, up to 2-torsion, as a direct sum E(Q) ⊕ E_{D}(Q), where E_{D} is the quadratic twist of E. (Recall that from a Weierstrass equation of E/Q, we obtain a Weierstrass equation of E_{D}/Q as follows:

\begin{matrix}E : y^{2} = x^{3} + ax + b \\ E_{D} : y^{2} = x^{3} + D^{2}ax + D^{3}b \end{matrix} )

Proof. Denote complex conjugation by τ. We will decompose E(K) into its eigenspaces under the action of τ. First, we kill E(K)[2] (by tensoring with Z[\frac{1}{2}], for instance), so that, using the fact that τ^{2} = 1, we can define the projections from E(K) to its +1 and −1 eigenspaces under τ: for P ∈ E(K),

P = \frac{1 + τ}{2} P + \frac{1 − τ}{2} P.

But if P = (x, y) ∈ E(K) satisfies τP = P, then P ∈ E(Q); if τP = −P = (x, −y), then x ∈ Q and y ∈ \sqrt{D}Q. In particular, (Dx, D\sqrt{D}y) ∈ E_{D}(Q), and conversely we can obtain any such point in the −1 eigenspace of E(K) from a point of E_{D}(Q). Thus, having eliminated the 2-torsion, the decomposition of E(K) into its τ-eigenspaces just reads

E(K) = E(Q) ⊕ E_{D}(Q).

This part of the algorithm uses the root number to determine a generator of E(K)/E(K)_{tors}. Since we are dealing with elliptic curves of rank 0 or 1 over K, the generator of the free group over Q comes from the generator of either E(Q) or E^{D}(Q). The root number will determine which one.

5. [Height of Heegner point] Compute the height h_{K}(y_{K}) relative to K of the Heegner point associated to K using the Gross-Zagier formula:

h_{K}(y_{K}) = α · L′(E/K, 1) = \begin{cases}α · L′(E_{D}/Q, 1) · L(E/Q, 1) & if E has rank 0 \\ α · L(E_{D}/Q, 1) · L′(E/Q, 1) & if E has rank 1, \end{cases}

where

α = \frac{u^{2}\sqrt{|D|}}{| ∫_{E(ℂ)} ω ∧ \overline{iω}|}.

Here u is half the number of units in the ring of integers of K. We may compute

∫_{E(ℂ)} ω ∧ \overline{iω} = ∫_{E(ℂ)} \frac{dx}{y} ∧ \overline{i\frac{dx}{y}} = −i ∫_{ℂ/Λ} ℘′(z)\frac{dz}{℘′(z)} ∧ \overline{℘′(z)\frac{dz}{℘′(z)}}
= −i ∫_{ℂ} dz ∧ dz̄ = ∫ (dx + idy) ∧ (−idx − dy) = −2 ∫ dx ∧ dy = 2a,

where a is the volume of the lattice ℂ/Λ.

[[ and ∥ω_{E}∥^{2} is the volume of E(C), which is twice the volume of the period lattice Zω_{1} + Zω_{2} associated to E and ω_{E}. The differential ω_{E} is the c · ω, where c is the Manin constant for E and ω is a Néron differential on E (we are assuming E is modular here, which is OK.) [[say something about computing cω... cite Section 2.14 of Cremona’s book.]] If π : X_{0}(N) → E is the modular parametrization, then π^{∗}(ω) = c·ω_{E}, where ω_{E} = f(q)\frac{dq}{q} ∈ H^{0}(X_{0}(N), Ω_{X_{0}(N)}) is the normalized cuspidal eigenform corresponding to E.]]

6. [Index of Heegner point] Compute

I_{K} = \sqrt{h_{K}(y_{K})/h_{K}(z)} = [E(K) : Zy_{K}],

up to primes that divide the B from step 2. Note that h_{K}(z) = 2 · h_{Q}(z), and that we do not compute E(K) or Zy_{K} directly, but instead compute the index using properties of heights.

7. [Annihilate � ] Then � (E/Q)[p] = 0 for all primes p ∤ B · I_{K}.

8. [p-descent] For each prime p | B · I_{K}, do a p-descent and compute � (E/Q)[p]. (Note that this is likely not too difficult because there is a p-torsion point over K on a curve F that is Q-isogenous to E. Ideas: If an isogeny from E to F has degree divisible by p, then E has a rational p-isogeny, which makes p-descent eiser. If an isogeny from E to F has degree coprime to p, then � (F/Q)[p] ≅ � (E/Q)[p], and F has a K-rational p-torsion point, so p-descent on F should be relatively easy.) To reduce the number of p for which one must do a p-descent, use several K.

[K exists by Murty-Murty or Bump-Friedberg-Hoffstein.... ]