Torsion of genus 2 Jacobians over ℚ

groups›[2,2,14]›[2,2,2,10]›impossibility proof

The 2-rank of abelian surfaces split over ℚ

21 October 2026 · A consequence of Mazur's theorem

We show that the groups [2,2,14] and [2,2,2,10] do not occur in the rational torsion of an abelian surface split over ℚ. Here “split over ℚ” means isogenous over ℚ to a product of elliptic curves over ℚ. The degree of the isogeny is arbitrary, and no polarization is used. For a finite abelian group M and a prime p, write rp(M) = dim𝔽p M[p].

Proposition

Let A/ℚ be an abelian surface split over ℚ, and put T = A(ℚ)tors.

  1. If T has a point of order 7, then r2(T) ≤ 2.
  2. If r2(T) = 4, then T has no point of order 5 or 7.

In particular, T contains no subgroup isomorphic to [2,2,14] ≅ (ℤ/2)3 × ℤ/7 or to [2,2,2,10] ≅ (ℤ/2)4 × ℤ/5.

Proof. Choose an isogeny E × E′ → A defined over ℚ, with E and E′ elliptic curves over ℚ. The image of E × {0} is an elliptic subvariety E₁ ⊂ A defined over ℚ, and the quotient E₂ = A/E₁ is an elliptic curve over ℚ. The exact sequence 0 → E₁ → A → E₂ → 0 of abelian varieties over ℚ gives, for every integer m ≥ 1, an exact sequence

0 → E₁(ℚ)[m] → A(ℚ)[m] → E₂(ℚ)[m].

Taking m = p prime, we obtain two consequences:

  1. rp(T) ≤ rp(E₁(ℚ)tors) + rp(E₂(ℚ)tors);
  2. if T has a point of order p, then E₁(ℚ) or E₂(ℚ) has one. Indeed, if E₁(ℚ)[p] = 0, the sequence embeds A(ℚ)[p] into E₂(ℚ)[p].

By Mazur's theorem, the torsion subgroup of an elliptic curve over ℚ is ℤ/n with 1 ≤ n ≤ 10 or n = 12, or ℤ/2 × ℤ/2m with 1 ≤ m ≤ 4. We use three consequences. First, its 2-rank is at most 2. Second, the only group in the list with a point of order 7 is ℤ/7, which has 2-rank 0. Third, the groups of 2-rank 2 are ℤ/2 × ℤ/2m with m ≤ 4, of orders 4, 8, 12, 16, so none of them has a point of order 5 or 7.

(1) By (b), one of the curves, say Ei, has a rational point of order 7. Its torsion is then ℤ/7, so r2(Ei(ℚ)tors) = 0. The other curve has 2-rank at most 2. By (a), r2(T) ≤ 0 + 2 = 2.

(2) If r2(T) = 4, then (a) forces r2(E₁(ℚ)tors) = r2(E₂(ℚ)tors) = 2. Neither curve then has a rational point of order 5 or 7, so by (b) neither does T. ∎

The same argument bounds the odd part of T when r2(T) = 4. Its order divides 9, because both curves then have odd torsion of order 1 or 3. Among the census groups this excludes nothing beyond [2,2,2,10].

Consequences for the census

The group [2,2,14] has 2-rank 3 and a point of order 7. The group [2,2,2,10] has 2-rank 4 and a point of order 5. Neither can be the rational torsion subgroup of a genus-2 Jacobian split over ℚ, for any splitting degree. The corresponding cells in the census are labelled impossible. No other census group meets the hypotheses of the proposition.

The hypothesis that the splitting is defined over ℚ is essential. A Jacobian that is geometrically split but simple over ℚ need not contain an elliptic curve over ℚ, and the argument says nothing about that class or about geometrically simple Jacobians. Both groups have geometrically simple examples in the census.

Reference

B. Mazur, Modular curves and the Eisenstein ideal, Publications Mathématiques de l'IHÉS 47 (1977), 33–186, Theorem (8). Mazur's classification gives E(ℚ)tors ≅ ℤ/nℤ for 1 ≤ n ≤ 10 or n = 12, or ℤ/2ℤ × ℤ/2mℤ for 1 ≤ m ≤ 4.