Visualizing Mordell-Weil Groups of Elliptic Curves Using Shafarevich-Tate Groups

William A. Stein

February 8, 2002 at University of Arizona

1 Introduction

Today I will tell you about a construction of elements of Shafarevich-Tate groups of abelian varieties A over ℚ.

A Construction of Elements of � (A)

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Birch and Swinnerton-Dyer Conjecture

• If L(A, 1) ≠ 0, then

#� (A) =^{?} \frac{L(A, 1)}{Ω_{A}} · \frac{#A(ℚ)_{tor} · #A(ℚ)_{tor}^{∨}}{∏_{p|N} c_{A,p}}

Find A in nature with conjecturally nontrivial � (A), and prove that � (A) is as big as expected.

• Construct A such that � (A) is nontrivial, then check that the BSD conjecture is not obviously false for A.

• Find a method for connecting the rank conjecture about elliptic curves to the rank 0 formula for abelian varieties.

What are the possibilities for #� (A)?

Question (Poonen, 1999 at AWS). Stoll and Poonen proved that if A is a Jacobian, then #� (A) is a square or twice a square. If A is not a Jacobian, is #� (A) always a square or twice a square?

Conjecture (Me, today). Let G be any finite abelian group (of odd order). Then there is an abelian variety A such that � (A) ≈ G × H, where gcd(#G, #H) = 1.

2 A Construction of Elements of � (A)

Theorem 2.1. Let E be an elliptic curve over ℚ, and suppose χ : (ℤ/ℓℤ)^{∗} → ℂ^{∗} is a Dirichlet character of prime modulus ℓ ∤ N_{E} and order n such that

• L(E, χ^{a}, 1) ≠ 0 for a = 1, …, n − 1,

• gcd (n, 2N_{E} ∏_{p|N_{E}} #Φ_{E}(\overline{𝔽}_{p})) = 1, and

• a_{ℓ} ≢ ℓ + 1 (mod p) for all p | n.

Let K be the degree n abelian extension of ℚ corresponding to χ. Then there exists a K-twist A of E^{⊕(n−1)} of rank 0 such that L(A, s) = ∏_{a=1}^{n−1} L(E, χ^{a}, s) and

E(ℚ)/nE(ℚ) ⊂ � (A/ℚ).

Remark 2.2. Note that K is contained in the totally real subfield ℚ(μ_{ℓ})^{+} of ℚ(μ_{ℓ}) because the order of χ(−1) divides the odd number n.

Sketch of Proof. Let R = Res_{K/ℚ}(E_{K}) be the Weil restriction of scalars of E_{K} down to ℚ. For any ℚ-scheme S, we have R(S) = E_{K}(S ×_{ℚ}K), and as Gal(\overline{ℚ}/ℚ)-modules

R(\overline{ℚ}) = E(\overline{ℚ} ⊗ K) ≅ E(\overline{ℚ}) ⊗_{ℤ} ℤ[Gal(K/ℚ)],

where τ ∈ Gal(\overline{ℚ}/ℚ) acts on ∑ P_{σ} ⊗ σ ∈ E(\overline{ℚ}) ⊗_{ℤ} ℤ[Gal(K/ℚ)] by

τ (∑ P_{σ} ⊗ σ) = ∑ τ(P_{σ}) ⊗ στ_{|K}.

The L-series of R is ∏_{a=1}^{n} L(E, χ^{a}, s), and R has good reduction at all p ∤ ℓ · N.

Let Δ : E ↪ R be the diagonal embedding, which sends P to ∑_{σ∈Gal(K/ℚ)} P ⊗σ, and let Σ : R → E be the summation map, which sends ∑ P_{σ} ⊗ σ to ∑ P_{σ}. Note that both Δ and Σ are defined over ℚ and that Σ ∘ Δ = [n]. If A = ker(Σ) then

A\overline{ℚ} = ker (+ : E⊕n ; \overline{ℚ} → E\overline{ℚ}) ≅ E^{⊕(n−1)},

the isomorphism being the one that sends (P_{1}, …, P_{n−1}) to (P_{1}, …, P_{n−1}, −(∑ P_{i})). In particular, A is a twist of E^{⊕(n−1)}. We summarize this information in the following diagram:

[figure: E[n]; E; Δ; [n]; E; A; R; Σ; E.]  (1)

Now pass to ℚ-rational points in diagram (1) and rearrange things to obtain the following diagram:

[figure: 0; E(ℚ); [n]; E(ℚ); E(ℚ)/nE(ℚ); ι; 0; 0; R(ℚ)/A(ℚ); E(ℚ); ker(H^{1}(ℚ, A) → H^{1}(ℚ, R)); 0.]

Here we have used that E(ℚ)[n] = 0, since E[p] is irreducible for p | n, and we’ve included the beginning of the long exact sequence of Galois cohomology associated to 0 → A → R → E → 0. Using the snake lemma, we see that ι is surjective and has kernel a subgroup of R(ℚ)/(A(ℚ)+E(ℚ)). One can use that a_{ℓ} ≢ ℓ+1 (mod p) for any p | n and that A(ℚ) is finite (which follows from Kato’s Euler system work!) to show that R(ℚ)/(A(ℚ) + E(ℚ)) contains no p-torsion for p | n, hence ker(ι) = 0.

To show that the image of ι lies in the subgroup � (A/ℚ) of H^{1}(ℚ, A), uses that gcd(n, 2 · N_{E} · c) = 1, where c is the product of all Tamagawa numbers of E and A. These last steps are fairly technical and use some nontrivial machinery. (That n is odd is only used to show that ι maps into � (A/ℚ).)

3 Data Collection

Next we collect some data that both gives evidence for the Birch and Swinnerton-Dyer conjecture and for my conjecture that if G is an abelian group then there is an abelian variety A such that � (A) ≈ G × H with gcd(#H, #G) = 1. We will always choose E below so that N_{E} is prime, E is isolated in its isogeny class (hence ρ_{E,p} is surjective for all p), and c_{E,p} = 1 for all p | N.

Let #� _{an}(A)^{∗} denote the prime-to-2ℓ part of

\frac{L(A, 1)}{Ω_{A}} · \frac{#A(ℚ)_{tor} · #A^{∨}(ℚ)_{tor}}{∏_{p|ℓN_{E}} c_{A,p}} .

Remark 5.4 of Edixhoven’s Néron models and tame ramification can be used to show that

Φ_{A,ℓ}(\overline{𝔽}_{ℓ}) = E(\overline{𝔽}_{ℓ})[n] ≈ (ℤ/nℤ)^{2},

so c_{A,ℓ} = 1, since E(𝔽_{ℓ})[p] = 0 for all p | n. Since K is only ramified at ℓ and the formation of Néron models commutes with unramified base change, c_{A,p} = c_{E,p}^{n−1} = 1 for p | N_{E}. Since A(ℚ) ⊂ A(K) ≈ E(K)^{⊕(n−1)}, and E(K)_{tor} = 0 (since all ρ_{E,p} are surjective), we have #A(ℚ)_{tor} = #A^{∨}(ℚ)_{tor} = 1. I think (but have not proven, yet!) that

Ω_{A/ℚ} = (\frac{1}{\sqrt{ℓ}} · Ω_{E/ℚ})^{n−1} .

To prove this, it would (mostly) suffice to show that Ω_{A/K} = Ω_{A/ℚ}^{n} · ℓ^{\binom{n}{2}}, where \binom{n}{2} = n(n − 1)/2. Assume this formula for Ω_{A/ℚ}, we can very quickly compute � _{an}(A)^{∗} using modular symbols.

The elliptic curves 61A of rank 1, 389A of rank 2, and 5077A of rank 3 each have prime conductor, trivial torsion subgroup, and Tamagawa number c_{p} = 1. In the table below, p_{d} denotes a d-digit prime number (where d is written in Roman numerals), and a − means that some hypothesis of Theorem 2.1 is not satisfied. (This table took under ten minutes to compute on a Pentium III 933.)

n ℓ  #� ∗_{an} for 61A  #� ∗_{an} for 389A  #� ∗_{an} for 5077A
3 487  3  3^{4}  3^{3}
9 487  3^{2}·19^{2}  3^{8}  3^{6}·17^{2}
27 487  3^{3}·19^{2}·p_{vi}^{2}  3^{12}·163^{2}  3^{9}·17^{2}·433^{2}·p_{vi}^{2}
81 487 3^{4}·19^{2}·p_{iv}^{2}·p_{vi}^{2}·p_{vii}^{2}  3^{16}·163^{2}·p_{xix}^{2}  3^{12}·17^{2}·433^{2}·p_{iv}^{2}·p_{v}^{2}·p_{vi}^{2}·p_{vii}^{2}·p_{ix}^{2}
5 251  5  5^{2}  −
25 251  5^{2}·151^{2}·p_{v}^{2}  5^{4}·149^{2}·p_{iv}^{2}  −
125 251 5^{3}·151^{2}·p_{v}^{2}·p_{xviii}^{2} 5^{6}·149^{2}·p_{iv}^{2}·p_{v}^{2}·p_{x}^{2}·p_{xi}^{2}  −
7 197  7·29^{2}  7^{2}·13^{4}  7^{3}
49 197  7^{2}·29^{2}·p_{x}^{2}  7^{4}·13^{4}·p_{ix}^{2}  7^{6}·p_{iv}^{2}·p_{iv}^{2}·p_{v}^{2}
11 89  11·67^{2}  11^{2}  11^{3}·67^{2}
13 53  13  13^{2}  −
17 103  17·613^{2}  17^{2}·101^{2}  17^{3}·67^{2}
19 191  19·37^{2}  19^{2}  19^{5}·37^{2}

The BSD conjecture and this table (and my “conjecture” about Ω_{A}) imply that for the integers n in the first column of the table, there is an A such that

� (A) ≈ (ℤ/nℤ) × H

with gcd(n, #H) = 1. This is evidence for Conjecture 1, and also gives lots of examples to show that #� (A) is neither a square or twice a square in general.

Challenge: Let E be one of the curves considered in the table, let r be its rank, and notice that in the table n^{r} | #� ∗_{an}. The BSD conjecture predicts that this divisibility should always hold. Prove that it does for infinitely many ℓ.