Torsion of genus 2 Jacobians over ℚ

1. Inputs

For a number field \(K\subset\overline{\Q}\) we write \(G_K=\Gal(\overline{\Q}/K)\), and \(\chi_p:G_{\Q}\to\F_p^\times\) denotes the mod-\(p\) cyclotomic character. We use the following results.

1. Elliptic torsion. Derickx–Najman, Theorem 1.1, list the torsion groups of elliptic curves over quartic number fields. The exponent of each group in the list is at most \(24\), and none is divisible by \(30\). Hence no elliptic curve over a number field of degree \(1\), \(2\) or \(4\) has a point of order \(30\); for degrees \(1\) and \(2\), extend the ground field to a field of degree \(4\). For degree \(1\) this is also Mazur's Theorem (8). For degree \(2\) it is also a special case of Kenku–Momose, Theorem (0.1), applied to \((m,N)=(1,30)\): the curve \(X_1(30)\) has genus \(9\), and \((1,30)\) is not one of the exceptions listed there.

2. Good reduction. Let \(K\) be a finite extension of \(\Q_2\), let \(A/K\) be an abelian variety, and let \(\ell\) be an odd prime. We use the following results of Serre–Tate.

   1. (Theorem 1 and its Corollary 2, p. 493.) \(A\) has good reduction if and only if the inertia group of \(K\) acts trivially on \(T_\ell(A)\). Good reduction is invariant under \(K\)-isogeny.

   2. (Lemma 2, p. 495.) Suppose that \(A\) has good reduction, and let \(B\) be the special fibre of its Néron model, an abelian variety over the residue field \(k\). For every odd \(m\), reduction gives an isomorphism \(A(\overline K)[m]\to B(\overline k)[m]\), compatible with the action of \(\Gal(\overline K/K)\), which acts on the target through \(\Gal(\overline k/k)\). In particular, \(A(K)[m]\) injects into \(B(k)\), and \(T_\ell(A)\simeq T_\ell(B)\) equivariantly.

   3. (Theorem 6(a), p. 503.) An abelian variety over \(K\) with complex multiplication defined over \(K\), in the sense of Serre–Tate, §4, has potential good reduction. This applies to an elliptic curve \(E/K\) such that \(\Q\otimes\End_K(E)\) is an imaginary quadratic field.

   4. (§2, “The case of a finite residue field”, p. 498.) If \(A\) has potential good reduction, there is a finite totally ramified extension \(K'/K\) such that \(A_{K'}\) has good reduction. This consequence is stated explicitly in Clark–Xarles, §2.4, and in Clark, §2.1.

3. Endomorphisms over finite fields. Let \(B\) be an abelian variety over a finite field \(k\), with Frobenius endomorphism \(\pi\), and let \(f\) be the characteristic polynomial of \(\pi\). By Tate, Theorem 2(c), the algebra \(\Q\otimes\End_k(B)\) equals \(\Q[\pi]\) if and only if \(f\) has no multiple root.

4. Weil's theorems. Let \(B\) be an abelian surface over \(\F_2\), with Frobenius endomorphism \(\pi\), and let \(\ell\) be an odd prime. The characteristic polynomial of \(\pi\) on \(V_\ell(B)\) has the form

   \[ P(T)=T^4+aT^3+bT^2+2aT+4,\qquad a,b\in\Z, \]

   its complex roots have absolute value \(\sqrt2\), and \(\#B(\F_2)=P(1)\). The arithmetic Frobenius \(x\mapsto x^2\) of \(\Gal(\overline{\F}_2/\F_2)\) acts on \(B(\overline{\F}_2)\), and hence on \(T_\ell(B)\), as \(\pi\). An endomorphism of \(B_{\overline{\F}_2}\) is defined over \(\F_2\) if and only if it commutes with \(\pi\).

5. Lattices. Let \(G\) be a group and \(V\) a finite-dimensional \(\Q_p\)-representation of \(G\). If \(T_1,T_2\subset V\) are \(G\)-stable \(\Z_p\)-lattices, then \((T_1/pT_1)^{\mathrm{ss}}\simeq(T_2/pT_2)^{\mathrm{ss}}\). Indeed, the characteristic polynomial of each \(g\in G\) on \(T_i/pT_i\) is the reduction of its characteristic polynomial on \(V\), and the Brauer–Nesbitt theorem shows that a semisimple representation over a field is determined by these polynomials. We apply this to isomorphic representations by transporting a lattice through the isomorphism. Composition factors are always counted with multiplicity. Rational torsion is never treated as an isogeny invariant: only composition factors are compared.

Clark's Theorem 5 lists the possible orders of the rational torsion subgroup of an abelian surface over \(\Q\) with everywhere potentially good reduction; Theorem 1.4 of Clark–Xarles gives the same list for the larger class of surfaces with everywhere anisotropic reduction. The list contains \(30\) and \(60\). Thus, unlike the case of rational \(35\)-torsion, it does not settle the CM square case here. Section 4 replaces it with a local argument at \(2\).

2. Combining coprime torsion in an elliptic isogeny class

Proof. Choose a \(K\)-isogeny \(f:E_m\to E_n\). Let \(C\subseteq\ker f\) be the subgroup of points whose order divides a power of \(n\). It is stable under \(\Gal(\overline K/K)\), because \(\ker f\) is. Hence \(E'=E_m/C\) is an elliptic curve over \(K\), and \(f\) factors as

with \(a\) and \(b\) defined over \(K\). The order of \(\ker a=C\) is prime to \(m\), so \(a\) maps the given point \(P\) of order \(m\) to a point of order \(m\). The order of \(\ker b\simeq\ker f/C\) is prime to \(n\). Let \(Q\in E_n(K)\) be the given point of order \(n\), and let \(\hat b\) be the dual isogeny, so that \(b\circ\hat b=[\deg b]\). If \(\hat b(kQ)=0\), then \((\deg b)kQ=0\), hence \(kQ=0\). Thus \(\hat b(Q)\in E'(K)\) has order \(n\), and \(a(P)+\hat b(Q)\in E'(K)\) has order \(mn\). Since \(E'\) is \(K\)-isogenous to \(E_m\), this proves the lemma. ∎

Applying the lemma first with \((m,n)=(2,3)\) and then with \((m,n)=(6,5)\), we see: if a \(K\)-isogeny class of elliptic curves contains curves with \(K\)-rational points of each of the orders \(2\), \(3\) and \(5\), then it contains a curve with a \(K\)-rational point of order \(30\).

For the rest of the proof, suppose that \(A(\Q)\) contains a point of order \(30\), and choose nonzero points \(P_p\in A(\Q)[p]\) for \(p\in\{2,3,5\}\). Exactly one of the following holds:

• \(A_{\overline{\Q}}\) is isogenous to \(E_1\times E_2\), where \(E_1\) and \(E_2\) are not geometrically isogenous (Section 3);
• \(A_{\overline{\Q}}\) is isogenous to \(E^2\), where \(E\) has complex multiplication (Section 4);
• \(A_{\overline{\Q}}\) is isogenous to \(E^2\), where \(E\) has no complex multiplication (Sections 5–9).

3. Nonisogenous geometric factors

Suppose that \(A_{\overline{\Q}}\) is isogenous to \(E_1\times E_2\), where \(E_1\) and \(E_2\) are not geometrically isogenous. Fix an isogeny \(\psi:A_{\overline{\Q}}\to E_1\times E_2\). If \(C\subset A_{\overline{\Q}}\) is an elliptic curve, then \(\psi(C)\) is an elliptic curve isogenous to \(C\). Its two projections cannot both be nonzero, since \(\Hom(E_1,E_2)=0\). Hence \(\psi(C)\) equals \(E_1\times0\) or \(0\times E_2\), and \(C\) is the identity component of \(\psi^{-1}(E_1\times0)\) or of \(\psi^{-1}(0\times E_2)\). Thus \(A_{\overline{\Q}}\) has exactly two elliptic subvarieties \(C_1\neq C_2\).

The group \(G_{\Q}\) permutes \(\{C_1,C_2\}\). If it fixed both, each would be defined over \(\Q\), and \(A\) would contain an elliptic curve over \(\Q\), contradicting simplicity over \(\Q\). This is the only place where simplicity over \(\Q\) is used. Hence the stabilizer of \(C_1\) is \(G_K\) for a quadratic field \(K\), and the nontrivial automorphism \(\sigma\) of \(K\) interchanges the two subvarieties. Write \(E=C_1\) and \(E^\sigma=C_2\); both are elliptic curves over \(K\).

Put \(F=A_K/E\), with quotient map \(q:A_K\to F\). The restriction of \(q\) to \(E^\sigma\) has finite kernel \(E\cap E^\sigma\) and nonzero image, so it is a \(K\)-isogeny \(E^\sigma\to F\). Conjugating by \(\sigma\), we obtain a \(K\)-isogeny \(E\to F^\sigma\).

Let \(p\in\{2,3,5\}\). If \(q(P_p)=0\), then \(P_p\in E(K)\) has order \(p\). Otherwise \(q(P_p)\in F(K)\) has order \(p\), and its conjugate \(\sigma(q(P_p))\in F^\sigma(K)\) also has order \(p\). In both cases, a curve in the \(K\)-isogeny class of \(E\) has a \(K\)-rational point of order \(p\). By Lemma 1, this class contains an elliptic curve over the quadratic field \(K\) with a point of order \(30\). This contradicts input 1.

4. The CM square case: a local obstruction at \(2\)

We prove a stronger local statement. It does not use simplicity over \(\Q\) or the point of order \(2\).

If \(A/\Q\) is geometrically isogenous to the square of a CM elliptic curve, then so is \(A_{\Q_2}\), and \(A(\Q)\subseteq A(\Q_2)\). Hence the proposition excludes the CM square case of the theorem.

Proof. Let \(M\) be the imaginary quadratic field of complex multiplication of \(E\). Then \(\End^0(A_{\overline{\Q}_2})\simeq M_2(M)\), with center \(M\). Suppose that \(A(\Q_2)\) contains a point of order \(15\).

Step 1: good reduction over a totally ramified extension. Choose a finite extension \(K_0/\Q_2\) such that \(E\) and all its endomorphisms are defined over \(K_0\), and \(A_{K_0}\) is \(K_0\)-isogenous to \(E^2\). By Serre–Tate, Theorem 6(a), \(E\) has good reduction over a finite extension \(K_1/K_0\). Then \(E^2\) has good reduction over \(K_1\), and so does the isogenous surface \(A_{K_1}\), by Serre–Tate, Corollary 2 of Theorem 1. Thus \(A\) has potential good reduction. By Serre–Tate, p. 498, there is a finite totally ramified extension \(K'/\Q_2\) over which \(A\) has good reduction. Since \(K'/\Q_2\) is totally ramified, the residue field of \(K'\) is \(\F_2\). Let \(B/\F_2\) be the special fibre of the Néron model of \(A_{K'}\).

Step 2: the number of points. By input 2(b), \(A(K')[15]\) injects into \(B(\F_2)\). Since \(A(\Q_2)\subseteq A(K')\), the group \(B(\F_2)\) contains a point of order \(15\), so \(15\mid\#B(\F_2)\). Let \(\pi\) be the Frobenius endomorphism of \(B\), and \(P\) its characteristic polynomial, as in input 4. If \(\alpha_1,\ldots,\alpha_4\) are the roots of \(P\), then

so \(\#B(\F_2)\in\{15,30\}\).

Step 3: the only possible Weil polynomial. Write \(P(T)=T^4+aT^3+bT^2+2aT+4\) as in input 4. Then \(P(T)=T^2\,Q(T+2/T)\) with \(Q(x)=x^2+ax+b-4\). If \(\beta_1,\beta_2\) are the roots of \(Q\), then

Each \(\beta_i\) equals \(\alpha+2/\alpha\) for a root \(\alpha\) of \(P\). Since \(|\alpha|=\sqrt2\), we have \(2/\alpha=\overline\alpha\), so \(\beta_i=2\operatorname{Re}\alpha\) is real and \(|\beta_i|\le2\sqrt2\). Consequently \(|a|\le4\sqrt2\), so \(|a|\le5\). Next, \(\beta_1\beta_2\le(\beta_1+\beta_2)^2/4\). Also \((2\sqrt2-\beta_1)(2\sqrt2-\beta_2)\ge0\) and \((2\sqrt2+\beta_1)(2\sqrt2+\beta_2)\ge0\), which give \(\beta_1\beta_2\ge2\sqrt2|\beta_1+\beta_2|-8\). Hence

Moreover \(P(1)=5+3a+b\).

If \(P(1)=15\), then \(b=10-3a\). The upper bound in (1) gives \(a^2+12a-24\ge0\). Since \(a\) is an integer with \(|a|\le5\), this forces \(a\ge2\). The lower bound then gives \((3+2\sqrt2)a\le14\), so \(a\le14/(3+2\sqrt2)<2.41\). Hence \(a=2\) and \(b=4\).

If \(P(1)=30\), then \(b=25-3a\). The upper bound gives \(a^2+12a-84\ge0\), which forces \(a\ge5\). The lower bound gives \((3+2\sqrt2)a\le29\), so \(a\le29/(3+2\sqrt2)<4.98\). This is a contradiction. Therefore

Step 4: the endomorphism algebra of \(B\). The two factors in (2) are distinct irreducible quadratic polynomials, so \(P\) has no multiple root. By input 3, \(\End^0_{\F_2}(B)=\Q[\pi]\). The minimal polynomial of \(\pi\) divides \(P\) and vanishes at every root of \(P\), so it equals \(P\). Therefore

Step 5: specialization of the CM centre. Fix an odd prime \(\ell\). Since \(A_{K'}\) has good reduction, the inertia group of \(K'\) acts trivially on \(T_\ell(A)\). An element \(\tau\) of \(\Gal(\overline{\Q}_2/K')\) sends an endomorphism \(\varphi\) of \(A_{\overline{\Q}_2}\) to \(\tau\varphi\tau^{-1}\), and on \(T_\ell(A)\) this is conjugation by the action of \(\tau\). Endomorphisms act faithfully on \(T_\ell(A)\). Hence inertia fixes every geometric endomorphism, and all geometric endomorphisms of \(A\) are defined over a finite unramified extension \(K''/K'\). The Néron model of \(A_{K''}\) is the base change of that of \(A_{K'}\), so its special fibre is the base change of \(B\) to the residue field of \(K''\). By the Néron mapping property, each endomorphism of \(A_{K''}\) extends to the Néron model and restricts to an endomorphism of the special fibre. This gives a unital ring homomorphism

Under the isomorphism \(T_\ell(A)\simeq T_\ell(B)\) of input 2(b), \(\spe(\varphi)\) acts on \(T_\ell(B)\) as \(\varphi\) acts on \(T_\ell(A)\). Hence \(\spe\) is injective. Since that isomorphism is Galois-equivariant, \(\spe\) is equivariant for the action of \(\Gal(\overline{\Q}_2/K')\), which acts on \(\End^0(B_{\overline{\F}_2})\) through \(\Gal(\overline{\F}_2/\F_2)\). By input 4, the arithmetic Frobenius acts there as \(\psi\mapsto\pi\psi\pi^{-1}\).

The centre \(M\) of \(\End^0(A_{\overline{\Q}_2})\) is stable under \(\Gal(\overline{\Q}_2/K')\). This group acts on \(M\) through \(\Gal(\overline{\F}_2/\F_2)\), since inertia acts trivially, and by field automorphisms. Hence a Frobenius element acts on \(M\) either trivially or by complex conjugation \(m\mapsto\overline m\). Accordingly, either \(\pi\spe(m)\pi^{-1}=\spe(m)\) for all \(m\in M\), or \(\pi\spe(m)\pi^{-1}=\spe(\overline m)\) for all \(m\in M\). We need not know which case occurs: the CM field need not be contained in \(K'\).

Step 6: trivial action. Suppose \(\pi\) commutes with \(\spe(M)\). By input 4, \(\spe(M)\) is contained in \(\End^0_{\F_2}(B)\), which is the algebra (3). Composing the unital embedding \(M\to\Q(\sqrt{-2})\times\Q(i)\) with the two projections gives unital ring homomorphisms from the field \(M\) to \(\Q(\sqrt{-2})\) and to \(\Q(i)\). They are injective, and all three fields have degree \(2\), so \(M\simeq\Q(\sqrt{-2})\) and \(M\simeq\Q(i)\). This is impossible.

Step 7: nontrivial action. Suppose \(\pi\spe(m)\pi^{-1}=\spe(\overline m)\) for all \(m\in M\). Put \(V=V_\ell(B)\otimes_{\Q_\ell}\overline{\Q}_\ell\), and let \(\iota_1,\iota_2\) be the two embeddings of \(M\) into \(\overline{\Q}_\ell\), with \(\iota_2(m)=\iota_1(\overline m)\). Since \(M\otimes\overline{\Q}_\ell\simeq\overline{\Q}_\ell\times\overline{\Q}_\ell\), we have \(V=V_1\oplus V_2\), where \(\spe(m)\) acts on \(V_j\) as \(\iota_j(m)\). If \(v\in V_1\), then

so \(\pi V_1\subseteq V_2\), and similarly \(\pi V_2\subseteq V_1\). With respect to the decomposition \(V=V_1\oplus V_2\), the matrix of \(\pi\) has zero diagonal blocks, so \(\tr(\pi\mid V)=0\). This contradicts (2).

Both cases are impossible, which proves the proposition. ∎

The reduction \(B\) may have more endomorphisms than \(A\); only the injective specialization of the centre \(M\) is used. We do not enlarge \(K'\) to make \(M\) defined over it, because such an enlargement need not remain totally ramified.

5. The non-CM square case: the endomorphism field

In this section and the next four, \(A_{\overline{\Q}}\) is isogenous to \(E_0^2\) for an elliptic curve \(E_0\) without complex multiplication. Then

The group \(G_{\Q}\) acts on the finitely generated \(\Z\)-module \(\End(A_{\overline{\Q}})\) by \(g\cdot\varphi=g\varphi g^{-1}\), and each endomorphism is defined over a number field. Let \(L\) be the fixed field of the kernel of this action, that is, the smallest field over which all geometric endomorphisms of \(A\) are defined. It is a finite Galois extension of \(\Q\). Put \(\Gamma=\Gal(L/\Q)\). It acts faithfully on \(D\) by \(\Q\)-algebra automorphisms. By Skolem–Noether, \(\Aut_{\Q}(M_2(\Q))=\PGL_2(\Q)\), so \(\Gamma\) is a finite subgroup of \(\PGL_2(\Q)\). For \(\gamma\in\Gamma\) we choose a representative \(S_\gamma\in\GL_2(\Q)\), so that \(\gamma\) acts on \(D\) as \(x\mapsto S_\gamma xS_\gamma^{-1}\).

A finite subgroup of \(\PGL_2(\mathbb R)\) is cyclic or dihedral. Indeed, it acts by isometries on the hyperbolic plane, it fixes a point there, and its action on the tangent plane at that point embeds it in \(O(2)\). Let an element of \(\PGL_2(\Q)\) have order \(n>2\), and let \(S\) be a representative. Then the ratio of the two eigenvalues of \(S\) is a primitive \(n\)-th root of unity \(\zeta\), and

Hence \(n\in\{3,4,6\}\). It follows that

where \(C_n\) is cyclic of order \(n\) and the subscript on a dihedral group denotes its order; thus \(D_4=C_2\times C_2\). When \(\Gamma\) is \(C_n\) or \(D_{2n}\) with \(n\ge3\), its rotation subgroup is the unique cyclic subgroup \(\langle r\rangle\) of order \(n\). It is normal, and \(\gamma r\gamma^{-1}\in\{r,r^{-1}\}\) for every \(\gamma\in\Gamma\).

Trace-one representatives. Let \(r\in\Gamma\) have order \(n\ge3\), and let \(S\) be a representative. Then \(\tr S\neq0\): otherwise \(S^2=-\det S\) is scalar by Cayley–Hamilton, and \(r\) would have order at most \(2\). Put \(R=S/\tr S\). Its characteristic polynomial is \(X^2-X+d\), where \(1/d=(\tr R)^2/\det R=2+\zeta+\zeta^{-1}\). Thus

with discriminants \(-3\), \(-1\) and \(-1/3\), respectively. Since \(R(1-R)=d\), the trace-one representative of \(r^{-1}\) is \(1-R\).

Elliptic subvarieties. Let \(C\subset A_{\overline{\Q}}\) be an elliptic curve. By Poincaré's reducibility theorem there is an elliptic curve \(C'\subset A_{\overline{\Q}}\), a complement of \(C\), such that the addition map \(\mu:C\times C'\to A_{\overline{\Q}}\) is an isogeny. Choose an isogeny \(\psi:A_{\overline{\Q}}\to C\times C'\) with \(\psi\circ\mu=[N]\) for some integer \(N\ge1\). Composing \(\psi\) with the first projection and the inclusion of \(C\) gives an endomorphism \(\varepsilon_C\) of \(A_{\overline{\Q}}\) with image \(C\), whose restriction to \(C\) is multiplication by \(N\). Since \(\varepsilon_C\) is defined over \(L\), so is its image \(C\). Thus every elliptic curve in \(A_{\overline{\Q}}\) is defined over \(L\). Each of them is isogenous to \(E_0\): composing its inclusion with an isogeny \(A_{\overline{\Q}}\to E_0^2\) and a suitable projection gives a nonzero homomorphism to \(E_0\).

6. Polarization duality and an irreducible projective factor

Let \(p\) be a prime, \(T=T_p(A)\) and \(V=V_p(A)\). The algebra \(D\otimes\Q_p\simeq M_2(\Q_p)\) acts faithfully on \(V\), and \(G_{\Q}\) acts compatibly: \(g(xv)=(g\cdot x)(gv)\) for \(x\in D\otimes\Q_p\) and \(v\in V\), where \(G_{\Q}\) acts on \(D\) through \(\Gamma\).

Proof. Every abelian variety over \(\Q\) has a polarization over \(\Q\). Combined with the Weil pairing, it gives a nondegenerate alternating \(G_{\Q}\)-equivariant pairing \(V\times V\to\Q_p(1)\), hence an isomorphism \(V\simeq V^*\otimes\Q_p(1)\). The lattice \(\Hom_{\Z_p}(T'',\Z_p)\otimes\Z_p(1)\) is \(G_{\Q}\)-stable, with reduction \((T''/pT'')^*\otimes\chi_p\). Transport it into \(V\) through this isomorphism. By input 5 over \(H\), its residual semisimplification is \(Y\). On the other hand, dualizing and twisting are exact, so it is also \(Y^*\otimes\chi_p|_H\). ∎

The polarization is used only on the rational Tate module, so its degree may be divisible by \(p\).

To make the hypothesis precise, write \(\mathcal R=\End_{\Z_p}(\Lambda)\) for a lattice \(\Lambda\subset\Q_p^2\). An element \(\gamma\in\Gamma\) preserves \(\mathcal R\) exactly when \(S_\gamma\Lambda\) is homothetic to \(\Lambda\). We may then rescale \(S_\gamma\) so that \(S_\gamma\in\GL(\Lambda)\). It is unique up to \(\Z_p^\times\), and its reduction gives a homomorphism \(\Gamma\to\PGL(\Lambda/p\Lambda)\). The hypothesis is that no line in \(U=\Lambda/p\Lambda\) is stable under this projective action.

Proof. Put \(T'=\mathcal R T\). It is a full lattice in \(V\), because \(\mathcal R\) is finite over \(\Z_p\). It is \(G_{\Q}\)-stable, because \(T\) and \(\mathcal R\) are. Choosing a basis of \(\Lambda\) identifies \(\mathcal R\) with \(M_2(\Z_p)\). The matrix unit \(e_{21}\) maps \(e_{11}T'\) isomorphically onto \(e_{22}T'\), with inverse \(e_{12}\). This gives

where \(W_0\) is free of rank \(2\) and \(\mathcal R\) acts on the first factor.

Put \(\overline V=T'/pT'=U\otimes_{\F_p}\overline W_0\). By input 5, \(\overline V\) and \(T/pT=A[p]\) have the same composition factors. Since \(A(\Q)[p]\neq0\), the module \(\overline V\) has a trivial composition factor.

For \(g\in G_{\Q}\) with image \(\gamma\in\Gamma\), and \(x\in\mathcal R\), we have \(g\circ\overline x\circ g^{-1}=\overline{S_\gamma xS_\gamma^{-1}}\) on \(\overline V\). Pull back the central extension

along the projective action of \(G_{\Q}\) on \(U\). The result is the group \(\widetilde G\) of pairs \((g,\theta)\), with \(g\in G_{\Q}\) and \(\theta\in\GL(U)\) a lift of the projective class of \(\overline{S_\gamma}\). For such a pair, \(g(\overline x\otimes1)g^{-1}=\theta\overline x\theta^{-1}\otimes1\) for \(x\in\mathcal R\), because scalars cancel. Hence the operator \((\theta^{-1}\otimes1)\circ g\) on \(\overline V\) commutes with \(M_2(\F_p)\otimes1\). The commutant of \(M_2(\F_p)\otimes1\) in \(\End(U\otimes\overline W_0)\) is \(1\otimes\End(\overline W_0)\). Hence \((\theta^{-1}\otimes1)\circ g=1\otimes w(g,\theta)\) for a unique \(w(g,\theta)\in\GL(\overline W_0)\). By uniqueness, \(w\) is a homomorphism. Let \(W\) denote \(\overline W_0\) with this action of \(\widetilde G\), and let \(\widetilde G\) act on \(U\) through \(\theta\). Then

as \(\widetilde G\)-modules, where \(\widetilde G\) acts on \(\overline V\) through its projection to \(G_{\Q}\). Since this projection is surjective, a composition series of \(\overline V\) as a \(G_{\Q}\)-module is one as a \(\widetilde G\)-module, so \(\overline V\) has a trivial \(\widetilde G\)-composition factor.

The module \(U\) is irreducible by hypothesis. Since \(G_L\) acts trivially on \(D\), the pairs \((h,1)\) with \(h\in G_L\) lie in \(\widetilde G\). They form a normal subgroup

which acts trivially on \(U\).

If \(W\) had a \(\widetilde G\)-stable line \(W_1\), then \(\overline V\) would have the filtration \(0\subset U\otimes W_1\subset U\otimes W\). Both graded pieces are twists of \(U\) by characters, hence irreducible of dimension \(2\). By the Jordan–Hölder theorem, \(\overline V\) would have no trivial composition factor. Thus \(W\) is irreducible.

The restriction \(W|_{\widetilde H}\) is semisimple. Indeed, choose a simple \(\widetilde H\)-submodule of \(W\). Its \(\widetilde G\)-translates are simple \(\widetilde H\)-submodules, because \(\widetilde H\) is normal. Their sum is a nonzero \(\widetilde G\)-submodule of \(W\), hence all of \(W\).

Therefore \(\overline V|_{\widetilde H}\simeq W|_{\widetilde H}\oplus W|_{\widetilde H}\) is semisimple. It has a trivial composition factor, so \(\overline V^{\widetilde H}\neq0\). Since \(\overline V^{\widetilde H}=U\otimes W^{\widetilde H}\), also \(W^{\widetilde H}\neq0\). The subspace \(W^{\widetilde H}\) is \(\widetilde G\)-stable, because \(\widetilde H\) is normal, and \(W\) is irreducible, so \(W^{\widetilde H}=W\). Hence \(G_L\) acts trivially on \(\overline V\).

Apply Lemma 3 to \(T''=T'\) and \(H=G_L\). Then \(Y\) is four copies of the trivial character, while \(Y^*\otimes\chi_p\) is four copies of \(\chi_p|_{G_L}\). Hence \(\chi_p|_{G_L}=1\), which is equivalent to \(\Q(\zeta_p)\subseteq L\). ∎

7. Rotations of order \(3\) or \(6\)

Suppose that

and let \(r\) generate the rotation subgroup, of order \(n\in\{3,6\}\). Let \(R\) be its trace-one representative. Its characteristic polynomial is \(X^2-X+1\) or \(X^2-X+\frac13\), with discriminant \(-3\) or \(-\frac13\), respectively. Modulo \(5\) these discriminants are \(2\) and \(3\), and neither is a square. Hence the reduction of the characteristic polynomial modulo \(5\) is irreducible over \(\F_5\). Thus \(\Q_5[R]\) is the unramified quadratic extension of \(\Q_5\), and \(\Z_5[R]\) is its ring of integers.

An \(R\)-stable \(\Z_5\)-lattice \(\Lambda\subset\Q_5^2\) is a finitely generated torsion-free module over the discrete valuation ring \(\Z_5[R]\), and \(\Q_5^2\) has dimension \(1\) over \(\Q_5[R]\). Hence \(\Lambda\) is free of rank \(1\) over \(\Z_5[R]\). Since the uniformizer of \(\Z_5[R]\) is \(5\), any two such lattices are homothetic under \(\Q_5^\times\). Now suppose \(r\) fixes the homothety class of a lattice \(\Lambda\), so \(R\Lambda=c\Lambda\) for some \(c\in\Q_5^\times\). Comparing indices gives \(v_5(\det R)=2v_5(c)\). Since \(\det R\in\{1,\frac13\}\) is a \(5\)-adic unit, \(c\) is a unit and \(R\Lambda=\Lambda\). Hence \(r\) fixes exactly one homothety class \([\Lambda_0]\).

Let \(\gamma\in\Gamma\). Since \(\gamma^{-1}r\gamma=r^{\pm1}\), we have \(S_\gamma^{-1}RS_\gamma=c'R^{\pm1}\) for a scalar \(c'\). As \(R^{\pm1}\Lambda_0=\Lambda_0\), we get \(R(S_\gamma\Lambda_0)=c'S_\gamma\Lambda_0\). Thus the class of \(S_\gamma\Lambda_0\) is fixed by \(r\), hence equals \([\Lambda_0]\). It follows that

is a \(\Gamma\)-stable maximal order. The projective action of \(r\) on \(\Lambda_0/5\Lambda_0\) is given by the reduction of \(R\in\GL(\Lambda_0)\). Its characteristic polynomial is irreducible over \(\F_5\), so it has no eigenvector. Hence no line in \(\Lambda_0/5\Lambda_0\) is stable, and the residual projective action is irreducible.

Since \(A(\Q)[5]\neq0\), Lemma 4 gives \(\Q(\zeta_5)\subseteq L\). Then \(\Gamma\) has a quotient \(\Gal(\Q(\zeta_5)/\Q)\simeq C_4\). But \(C_3\) and \(C_6\) have order prime to \(4\), and the abelianizations of \(D_6\) and \(D_{12}\) are \(C_2\) and \(C_2\times C_2\). None of these groups has a cyclic quotient of order \(4\), a contradiction.

8. The dihedral group of order \(8\)

Suppose \(\Gamma=D_8\), with rotation subgroup \(\langle r\rangle\) of order \(4\). Let \(R\) be the trace-one representative of \(r\). Its characteristic polynomial \(X^2-X+\frac12\) has discriminant \(-1\), which is congruent to \(2^2\) modulo \(5\). By Hensel's lemma, its roots \(\lambda_1,\lambda_2\) lie in \(\Z_5\). They are units, since \(\lambda_1\lambda_2=\frac12\), and \(\lambda_1-\lambda_2=\pm\sqrt{-1}\) is a unit. Their ratio \(\zeta=\lambda_1/\lambda_2\) satisfies \(\zeta+\zeta^{-1}=1/d-2=0\), so \(\zeta^2=-1\), and \(\zeta\) is a primitive fourth root of unity. The ring

is a split quadratic order with orthogonal idempotents \(e_1,e_2\), and \(R=\lambda_1e_1+\lambda_2e_2\). Since \(R\) is not scalar, each \(e_i\) is a rank-one idempotent of \(M_2(\Q_5)\).

The rotations act by conjugation by powers of \(R\), so they fix \(R\), \(e_1\) and \(e_2\). A reflection \(s\) satisfies \(srs^{-1}=r^{-1}\). Hence \(S_sRS_s^{-1}\) represents \(r^{-1}\) and has trace \(1\), so it is the trace-one representative \(1-R\) of \(r^{-1}\). Since \(\lambda_1+\lambda_2=1\), it sends \(e_1\) to \((\lambda_1-R)/(\lambda_1-\lambda_2)=e_2\). Hence \(\mathcal B\) is \(\Gamma\)-stable: rotations fix \(e_1\) and \(e_2\), and reflections interchange them. This holds even though \(\Gamma\) need not preserve any maximal order.

Let \(K=L^{\langle r\rangle}\), a quadratic field. Put \(T'=\mathcal BT_5(A)\). As in Section 6, it is a \(G_{\Q}\)-stable full lattice in \(V=V_5(A)\). We have

Each \(T_i\) is a direct summand of \(T'\), hence free. The four-dimensional \(M_2(\Q_5)\)-module \(V\) is the sum of two copies of the standard module, so \(V_i=e_iV\) has dimension \(2\) and \(T_i\) has rank \(2\). The group \(G_K\) preserves each \(T_i\). Every element \(\tau\in G_{\Q}\setminus G_K\) satisfies \(\tau T_1=T_2\) and \(\tau T_2=T_1\).

The space \(e_1(D\otimes\Q_5)e_2\) is one-dimensional. Let \(X\) span it. The group \(G_K\) acts on \(D\otimes\Q_5\) through \(\langle r\rangle\), fixing \(e_1\) and \(e_2\), so

for a character \(\eta:G_K\to\Z_5^\times\), trivial on \(G_L\). Since \(RXR^{-1}=(\lambda_1/\lambda_2)X\), its value on a lift of \(r\) is \(\zeta\). Thus \(\eta\) has exact order \(4\). Its reduction \(\overline\eta\) also has exact order \(4\), because \(\zeta^2=-1\) and \(-1\not\equiv1\) modulo \(5\).

As a nonzero element of \(e_1M_2(\Q_5)e_2\), the matrix \(X\) maps the line \(e_2\Q_5^2\) isomorphically onto \(e_1\Q_5^2\). Since \(V\) is a sum of two copies of \(\Q_5^2\), the map \(X:V_2\to V_1\) is an isomorphism. For \(g\in G_K\) and \(v\in V_2\) we have \(g(Xv)=\eta(g)X(gv)\). Hence \(X\) is a \(G_K\)-isomorphism \(\eta\otimes V_2\to V_1\). Put \(\overline T_i=T_i/5T_i\). Input 5, applied to the lattices \(T_1\) and \(X(T_2)\) of \(V_1\), gives

as \(G_K\)-modules. This step does not require \(X\) to carry \(T_2\) onto \(T_1\).

By input 5, \(T'/5T'=\overline T_1\oplus\overline T_2\) has the same composition factors as \(A[5]\), so it has a trivial \(G_K\)-composition factor. Hence one of \(\overline T_1,\overline T_2\) has one. An element \(\tau\in G_{\Q}\setminus G_K\) maps \(\overline T_1\) onto \(\overline T_2\) and intertwines the action of \(g\in G_K\) on \(\overline T_1\) with that of \(\tau g\tau^{-1}\) on \(\overline T_2\). Conjugation preserves trivial composition factors, so both \(\overline T_1\) and \(\overline T_2\) have one. By (4), \(\overline T_1^{\,\mathrm{ss}}\) contains both \(\mathbf 1\) and \(\overline\eta\otimes\mathbf 1=\overline\eta\). Since \(\overline\eta\neq\mathbf 1\) and \(\dim\overline T_1=2\),

Since \(\overline\eta\) is trivial on \(G_L\), the restriction of \(T'/5T'\) to \(G_L\) has four trivial composition factors. Lemma 3, with \(T''=T'\) and \(H=G_L\), gives \(\chi_5|_{G_L}=1\), that is, \(\Q(\zeta_5)\subseteq L\). Then \(D_8\) would have a cyclic quotient of order \(4\). This is impossible, since its abelianization is \(C_2\times C_2\).

Alternatively, apply Lemma 3 with \(H=G_K\). The \(G_K\)-composition factors of \(T'/5T'\) are \(\mathbf 1,\mathbf 1,\overline\eta,\overline\eta^{-1}\), and the trivial character is the only one of multiplicity at least \(2\). The dual twist consists of \(\chi_5,\chi_5,\chi_5\overline\eta^{-1},\chi_5\overline\eta\), in which \(\chi_5|_{G_K}\) has multiplicity at least \(2\). Hence \(\chi_5|_{G_K}=1\), which would put the quartic field \(\Q(\zeta_5)\) inside the quadratic field \(K\).

9. Completion of the proof

Sections 7 and 8 leave

so \([L:\Q]\in\{1,2,4\}\). By Section 5, every elliptic curve in \(A_{\overline{\Q}}\) is defined over \(L\). Choose one, \(E\), and put \(F=A_L/E\), with quotient map \(q:A_L\to F\).

The elliptic curves \(E\) and \(F\) are \(L\)-isogenous. Indeed, let \(E'\) be a complement of \(E\) as in Section 5, so that \(E\cap E'\) is finite; it is defined over \(L\). Both \(E\) and \(E'\) are isogenous to \(E_0\), so there is a nonzero homomorphism \(h:E\to E'\) over \(\overline{\Q}\). Let \(\varepsilon_E\) be the endomorphism of Section 5, with image \(E\) and restriction \([N]\) to \(E\). Then \(\varphi=h\circ\varepsilon_E\), viewed as an endomorphism of \(A_{\overline{\Q}}\), is defined over \(L\), and \(\varphi(E)=E'\). The composite

is defined over \(L\). Both maps are isogenies: the first is \(Nh\), and the second has finite kernel \(E\cap E'\).

For each \(p\in\{2,3,5\}\), the exact sequence

shows that \(P_p\) lies in \(E(L)\) or maps to a point of order \(p\) in \(F(L)\). Thus the common \(L\)-isogeny class of \(E\) and \(F\) contains curves with \(L\)-rational points of orders \(2\), \(3\) and \(5\). By Lemma 1, it contains an elliptic curve over \(L\) with a point of order \(30\). Since \([L:\Q]\in\{1,2,4\}\), this contradicts input 1.

All three geometric types of Section 2 lead to a contradiction, which proves the theorem. ∎

10. Where the hypotheses are used

Simplicity over \(\Q\) is used only in Section 3, where it forces Galois to interchange the two geometric elliptic subvarieties. The other sections prove two stronger statements:

• If \(A/\Q_2\) is an abelian surface that is geometrically isogenous to the square of a CM elliptic curve, then \(A(\Q_2)\) has no point of order \(15\) (Proposition 2). In particular, such an \(A/\Q\) has no rational point of order \(15\).
• If \(A/\Q\) is an abelian surface that is geometrically isogenous to the square of an elliptic curve without complex multiplication, then \(A(\Q)\) has no point of order \(30\) (Sections 5–9), whether or not \(A\) is simple over \(\Q\).

Hence an abelian surface over \(\Q\) with a rational point of order \(30\) that is geometrically split is geometrically isogenous to a product of two elliptic curves that are not geometrically isogenous, and it is not simple over \(\Q\). In Section 3 the points of orders \(2\), \(3\) and \(5\) enter only through Lemma 1. In Section 4 the points of orders \(3\) and \(5\) together give the contradiction, and the point of order \(2\) is not used. In the non-CM square case, the point of order \(5\) drives Sections 7 and 8, and the points of orders \(2\) and \(3\) enter only through Lemma 1 in Section 9.

No step uses the degree of a geometric splitting isogeny, the maximality of the integral endomorphism ring, or a principal polarization. No numerical computation is required: the only finite check, the determination of the Weil polynomial in Section 4, is carried out by hand.

11. Consequences for the census

The groups [30], [60] and [2,30] are impossible in the class “geometrically split, simple over \(\Q\)”. They are the only census groups containing a point of order \(30\).

Simplicity over \(\Q\) is essential. All three groups have examples in the class “split over \(\Q\)”: 30.b, 60.a and 2.30.a. By Section 10, the two elliptic factors of each of these Jacobians are not geometrically isogenous. Thus these exclusions do not change the combined geometrically split ∞? grade of any of the three groups, which remains certified exact, through families split over \(\Q\). The group [30] also keeps its geometrically simple example 30.a.

References

1. M. Derickx and F. Najman, Classification of torsion of elliptic curves over quartic fields, arXiv:2412.16016v2 (2025), Theorem 1.1.

2. B. Mazur, Modular curves and the Eisenstein ideal, Publications Mathématiques de l'IHÉS 47 (1977), 33–186, Theorem (8).

3. M. A. Kenku and F. Momose, Torsion points on elliptic curves defined over quadratic fields, Nagoya Mathematical Journal 109 (1988), 125–149, Theorem (0.1).

4. J.-P. Serre and J. Tate, Good reduction of abelian varieties, Annals of Mathematics 88 (1968), 492–517: Theorem 1 and its Corollary 2 (p. 493), Lemma 2 (p. 495), the case of a finite residue field in §2 (p. 498), and Theorem 6(a) (p. 503).

5. J. Tate, Endomorphisms of abelian varieties over finite fields, Inventiones Mathematicae 2 (1966), 134–144, Theorem 2(c).

6. P. L. Clark and X. Xarles, Local bounds for torsion points on abelian varieties, Canadian Journal of Mathematics 60 (2008), 532–555, §2.4 and Theorem 1.4. Theorem 1.4 concerns abelian surfaces with everywhere anisotropic reduction, a class that includes potentially good reduction.

7. P. L. Clark, Bounds for torsion on abelian varieties with integral moduli, arXiv:math/0407264v2 (2004), §2.1 and Theorem 5.

Related proofs: order 35, and orders 33 and 34.