An obstruction to rational \(63\)-torsion on geometrically split abelian surfaces
21 October 2026 · Geometrically split, simple over ℚ
Theorem. Let \(A/\Q\) be an abelian surface that is simple over \(\Q\) and geometrically split. Then \(A(\Q)\) has no point of order \(63\).
Here “geometrically split” means that \(A_{\overline{\Q}}\) is isogenous to a product of elliptic curves. The theorem places no restriction on the degree of this isogeny, on the field of definition of the geometric endomorphisms, on the integral endomorphism ring, or on the polarization; in particular, it does not require a principal polarization. Consequently, the group [63] cannot occur as the rational torsion subgroup of a geometrically split genus-\(2\) Jacobian that is simple over \(\Q\).
The hypothesis of simplicity over \(\Q\) is essential. A product of elliptic curves over \(\Q\) with rational points of orders \(7\) and \(9\) has a rational point of order \(63\), and the census contains a genus-\(2\) Jacobian that is split over \(\Q\) and has torsion subgroup \([63]\).
- Complete proof (LaTeX source)
- Magma script checking the arithmetic facts used in the proof and its verification log
- Magma script for the consistency check with the census curve 21.b and its verification log
1. Inputs
We use the following results on torsion points of elliptic curves.
-
Kenku–Momose, Theorem (0.1): let \(N\) be a product of powers of \(2,3,5,7,11,13\), and let \(m\mid N\). If \(X_1(m,N)\) has positive genus and
\[ (m,N)\notin\{(1,11),(1,13),(1,14),(1,15),(1,16),(1,18),(2,10),(2,12)\}, \]then \(Y_1(m,N)(k)=\emptyset\) for every quadratic field \(k\). The curve \(X_1(21)=X_1(1,21)\) has genus \(5\). Hence no elliptic curve over a quadratic field has a point of order \(21\).
-
Mazur, Theorem (7′): if an elliptic curve over \(\Q\) has a rational point of order \(m\), then \(m\leq10\) or \(m=12\). In particular \(Y_1(21)(\Q)=\emptyset\).
-
Derickx–Etropolski–van Hoeij–Morrow–Zureick-Brown, Theorem A, classify the torsion of elliptic curves over cubic fields. Theorem A states in addition that the elliptic curve with Cremona label 162b1,
\[ y^2+xy+y=x^3-x^2-5x+5, \]over \(\Q(\zeta_9)^+=\Q[x]/(x^3-3x+1)\) is the unique elliptic curve over a cubic field with torsion \(\Z/21\Z\). We use the precise statement proved in their Section 6.2: the only closed points of degree \(3\) on \(X_1(21)\) are two points \(D_0\) and \(D_0'\).
For the reduction of abelian varieties we use the following results of Serre–Tate. Let \(K\) be a finite extension of \(\Q_p\) with residue field \(k\), let \(A\) be an abelian variety over \(K\), and let \(\ell\neq p\) be a prime.
(Néron–Ogg–Shafarevich, [ST, Theorem 1].) \(A\) has good reduction if and only if the inertia group of \(K\) acts trivially on \(T_\ell(A)\). By [ST, Corollary 2 of Theorem 1], two \(K\)-isogenous abelian varieties both have, or both do not have, good reduction.
([ST, Lemma 2].) For \(m\) prime to \(p\), reduction is an isomorphism from the inertia invariants of \(A[m]\) onto the \(m\)-torsion of the special fibre of the Néron model, compatible with the Galois action. Hence, if \(A\) has good reduction \(\widetilde A\), then \(A(K)[m]\) injects into \(\widetilde A(k)\), and \(T_\ell(A)\cong T_\ell(\widetilde A)\) compatibly with Frobenius.
([ST, §2, p. 498].) If \(A\) has potentially good reduction, then there is a finite extension \(K'/K\) with residue field \(k\) over which \(A\) has good reduction.
For an abelian surface \(B\) over \(\F_q\), \(\#B(\F_q)=P(1)\), where \(P\) is the characteristic polynomial of Frobenius, and the complex roots of \(P\) have absolute value \(\sqrt q\) (Weil). Hence \(\#B(\F_q)\leq(1+\sqrt q)^4\).
We also use the following standard facts: the independence of the semisimplified residual representation from the choice of a stable lattice (Brauer–Nesbitt), which applies to isomorphic \(p\)-adic representations after transporting a lattice through the isomorphism; the Skolem–Noether theorem; Poincaré's complete reducibility theorem; Galois descent for twists of projective varieties by a cocycle with values in the automorphism group; the Weil pairing; and the existence of a polarization defined over \(\Q\) on every abelian variety over \(\Q\).
Notation. For a number field \(K\subset\overline{\Q}\), put \(G_K=\Gal(\overline{\Q}/K)\). For a prime \(\ell\), \(T_\ell\) and \(V_\ell=T_\ell\otimes\Q_\ell\) denote the Tate module and the rational Tate module. For a stable lattice \(T'\) in a \(p\)-adic representation, \(T'/pT'\) is its residual module and \((\cdot)^{\mathrm{ss}}\) denotes semisimplification. We write \(\overline\chi_p\) for the mod-\(p\) cyclotomic character. A rational point of order \(p\) spans a trivial submodule of \(A[p]=T_p(A)/pT_p(A)\); by lattice independence, the residual module of every \(G_{\Q}\)-stable lattice in \(V_p(A)\) then has a trivial composition factor.
2. Setup and two lemmas on elliptic curves
Throughout the proof, \(A/\Q\) is an abelian surface that is simple over \(\Q\) and geometrically split, and we suppose that \(P\in A(\Q)\) has order \(63\). Put
of orders \(7\) and \(3\). Exactly one of the following holds:
- (N) \(A_{\overline{\Q}}\) is isogenous to \(E_1\times E_2\), where \(E_1\) and \(E_2\) are not isogenous;
- (CM) \(A_{\overline{\Q}}\) is isogenous to \(E^2\), where \(E\) has complex multiplication;
- (S) \(A_{\overline{\Q}}\) is isogenous to \(E^2\), where \(E\) has no complex multiplication.
Lemma 1 (combining coprime torsion). Let \(m,n\) be coprime positive integers and \(K\) a number field. Suppose \(E_m\) and \(E_n\) are \(K\)-isogenous elliptic curves over \(K\) with points of orders \(m\) and \(n\) in \(E_m(K)\) and \(E_n(K)\), respectively. Then their \(K\)-isogeny class contains an elliptic curve with a \(K\)-rational point of order \(mn\).
Proof. Let \(f:E_m\to E_n\) be a \(K\)-isogeny, and let \(H\) be the subgroup of \(\ker f\) consisting of the elements whose order is divisible only by primes dividing \(n\). It is a characteristic subgroup of \(\ker f\), hence \(G_K\)-stable, and \(f\) factors over \(K\) as
The degree of \(a\) is prime to \(m\), so \(a\) is injective on \(E_m[m]\) and maps the given point of order \(m\) to a point of order \(m\) in \(E'(K)\). The degree of \(b\) is prime to \(n\); since \(b\circ\hat b=[\deg b]\) is injective on \(E_n[n]\), so is the dual isogeny \(\hat b\), which maps the given point of order \(n\) to a point of order \(n\) in \(E'(K)\). The sum of the two points has order \(mn\). ∎
Lemma 2. Let \(X\) be an abelian surface over a field \(k\), let \(E\subset X\) be an elliptic curve defined over \(k\), and put \(F=X/E\). If \(Q\in X(k)\) has prime order \(q\), then either \(Q\in E(k)\), or the image of \(Q\) in \(F(k)\) has order \(q\).
Proof. The image has order dividing \(q\). If it is zero, then \(Q\in E(\overline k)\cap X(k)=E(k)\). ∎
3. Nonisogenous factors
Suppose that \(A_{\overline{\Q}}\) is isogenous to \(E_1\times E_2\), where \(E_1\) and \(E_2\) are not isogenous. By Poincaré's theorem, every elliptic curve in \(A_{\overline{\Q}}\) is isogenous to \(E_1\) or to \(E_2\), and so is the image of a nonzero homomorphism \(E_i\to A_{\overline{\Q}}\). For \(i=1,2\), the space \(\Hom(E_i,A_{\overline{\Q}})\otimes\Q\) has dimension one over the division algebra \(\End^0(E_i)\). Hence all nonzero homomorphisms \(E_i\to A_{\overline{\Q}}\) have the same image, and \(A_{\overline{\Q}}\) contains exactly two elliptic curves.
The group \(G_{\Q}\) permutes these two elliptic curves. If it fixed them, they would be defined over \(\Q\), and \(A\) would not be simple over \(\Q\). Hence they are exchanged. They are defined over a quadratic field \(K\), and the nontrivial automorphism \(\sigma\) of \(K\) interchanges them; call them \(E\) and \(E^\sigma\).
Put \(F=A_K/E\). The composite \(E^\sigma\hookrightarrow A_K\to F\) has finite kernel \(E\cap E^\sigma\), so it is a \(K\)-isogeny. Applying \(\sigma\), we find that \(F^\sigma\) is \(K\)-isogenous to \(E\).
Let \(q\in\{3,7\}\). By Lemma 2 over \(K\), either \(P_q\in E(K)\), or the image of \(P_q\) in \(F(K)\) has order \(q\); in the second case, its conjugate under \(\sigma\) is a point of order \(q\) in \(F^\sigma(K)\). Thus the \(K\)-isogeny class of \(E\) contains curves with \(K\)-rational points of orders \(3\) and \(7\). By Lemma 1, it contains an elliptic curve over the quadratic field \(K\) with a point of order \(21\). This contradicts the theorem of Kenku and Momose.
4. The CM case: a local argument at \(2\)
Suppose that \(A_{\overline{\Q}}\) is isogenous to \(E^2\), where \(E\) has complex multiplication. The curve \(E\) is defined over a number field and its \(j\)-invariant is an algebraic integer, so \(E\) has potentially good reduction at every prime ([ST, §2]). Over a finite extension of \(\Q_2\), the surface \(A\) becomes isogenous to \(E^2\) and \(E\) acquires good reduction. By (a), \(A\) acquires good reduction there as well, so \(A_{\Q_2}\) has potentially good reduction.
By (c), there is a finite extension \(K'/\Q_2\) with residue field \(\F_2\) over which \(A\) has good reduction \(\widetilde A\). Since \(63\) is odd, (b) embeds the subgroup of \(A(\Q)\subseteq A(K')\) generated by \(P\) into \(\widetilde A(\F_2)\). Therefore
a contradiction. In fact, \(\#\widetilde A(\F_2)\in\{1,\dots,16,19,20,25\}\) for every abelian surface over \(\F_2\) (see Section 11). Alternatively, Clark's Theorem 5 lists the possible orders of the rational torsion subgroup of an abelian surface over \(\Q\) with everywhere potentially good reduction, and \(63\) is not among them. This case does not use simplicity over \(\Q\).
5. Non-CM squares: the endomorphism field
From now on, suppose that \(A_{\overline{\Q}}\) is isogenous to \(E^2\), where \(E\) has no complex multiplication. Then
Let \(L\) be the smallest field over which all geometric endomorphisms of \(A\) are defined. It is a finite Galois extension of \(\Q\). Put \(\Gamma=\Gal(L/\Q)\). The group \(G_{\Q}\) acts on \(D\) by \(g(\varphi)=g\circ\varphi\circ g^{-1}\), and this action factors through a faithful action of \(\Gamma\). By Skolem–Noether, \(g(\varphi)=R_g\varphi R_g^{-1}\) for some \(R_g\in D^\times=\GL_2(\Q)\), unique up to \(\Q^\times\), and \(g\mapsto[R_g]\) embeds \(\Gamma\) in \(\operatorname{Aut}(M_2(\Q))=\PGL_2(\Q)\). On Tate modules, \(g(\varphi x)=g(\varphi)\,g(x)\) for \(\varphi\in D\) and \(x\in V_\ell(A)\).
Finite subgroups of \(\PGL_2(\Q)\). A finite subgroup of \(\PGL_2(\mathbb R)\) is cyclic or dihedral: it fixes a point of the hyperbolic plane, and its action on the tangent plane embeds it in \(O(2)\). If an element \([M]\in\PGL_2(\Q)\) has finite order \(n>2\), then \(M\) has distinct eigenvalues whose ratio is a primitive \(n\)-th root of unity \(\zeta\), and
Hence \(n\in\{3,4,6\}\), and
where \(D_{2n}\) is the dihedral group of order \(2n\) and \(V_4=C_2\times C_2\).
Trace-one representatives. If \([M]\) has order \(n\in\{3,4,6\}\), then \(\operatorname{tr}M\neq0\), and we may scale \(M\) to a representative \(R\) with \(\operatorname{tr}R=1\). Then \(1/\det R=2+\zeta+\zeta^{-1}=c\), where \(c=1,2,3\) for \(n=3,4,6\), so that
Since \(R(1-R)=1/c\) is a scalar, the trace-one representative of \([M]^{-1}\) is \(1-R\). An element \([M]\) of order \(2\) has a representative \(R\) with \(\operatorname{tr}R=0\), and then \(R^2=-\det R\) is a nonzero scalar.
The trivial group. If \(\Gamma=C_1\), then \(\End^0_{\Q}(A)=D\) contains a nontrivial idempotent, and \(A\) is not simple over \(\Q\). Hence \(\Gamma\neq C_1\).
Lemma 3. Every elliptic curve \(C\subset A_{\overline{\Q}}\) is defined over \(L\), and any two such curves are \(L\)-isogenous. If \(E\subset A_L\) is one of them, then \(A_L/E\) is \(L\)-isogenous to \(E\).
Proof. By Poincaré's theorem, \(C\) has a complement, and \(C\) is the image of \(\varphi=ne\in\End(A_{\overline{\Q}})\) for a rank-one idempotent \(e\in D\) and an integer \(n\geq1\). Since \(\varphi\) is defined over \(L\), so is its image. Let \(C'\) be another elliptic curve, the image of \(\varphi'=n'e'\). Since \(D\simeq M_2(\Q)\), there is \(X\in D\) with \(e'Xe\neq0\); choose \(k\geq1\) with \(kX\in\End(A_{\overline{\Q}})\), and put \(\psi=\varphi'\circ kX\circ\varphi\). Then \(\psi\) is defined over \(L\), its image lies in \(C'\), and \(\psi\circ\varphi=n\psi\neq0\), so \(\psi\) does not vanish on \(C\). Thus \(\psi|_C:C\to C'\) is an \(L\)-isogeny. Finally, if \(E'\subset A_L\) is an elliptic curve with \(E\cap E'\) finite (for instance the image of an integral multiple of \(1-e\)), then \(E'\to A_L/E\) is an \(L\)-isogeny, and \(E'\) is \(L\)-isogenous to \(E\). ∎
Corollary 4. There is an elliptic curve \(E''\) over \(L\), \(L\)-isogenous to every elliptic curve in \(A_L\), such that \(E''(L)\) contains a point of order \(21\).
Proof. Choose an elliptic curve \(E\subset A_L\) and put \(F=A_L/E\). By Lemma 2 over \(L\), for \(q\in\{3,7\}\), either \(P_q\in E(L)\), or its image in \(F(L)\) has order \(q\). By Lemma 3, \(F\) is \(L\)-isogenous to \(E\). So the \(L\)-isogeny class of \(E\) contains curves with \(L\)-rational points of orders \(3\) and \(7\), and Lemma 1 gives \(E''\). By Lemma 3, \(E''\) is \(L\)-isogenous to every elliptic curve in \(A_L\). ∎
6. Three lemmas on residual representations
In this section \(p\) is an odd prime. Put \(V=V_p(A)\), \(T=T_p(A)\), and \(D_p=D\otimes\Q_p\simeq M_2(\Q_p)\). The algebra \(D_p\) acts on \(V\), and \(g(\varphi x)=g(\varphi)\,g(x)\) for \(g\in G_{\Q}\), \(\varphi\in D_p\) and \(x\in V\). As a \(D_p\)-module, \(V\) is the sum of two copies of the standard module \(\Q_p^2\).
Lemma 5 (polarization). Let \(T'\subset V\) be a \(G_{\Q}\)-stable lattice such that every composition factor of \(T'/pT'\) is trivial as a \(G_L\)-module. Then \(\Q(\zeta_p)\subseteq L\).
Proof. A polarization of \(A\) defined over \(\Q\) and the Weil pairing give a nondegenerate alternating \(G_{\Q}\)-equivariant pairing on \(V\) with values in \(\Q_p(1)\), hence an isomorphism \(V\simeq V^*\otimes\Q_p(1)\). Transport the lattice \(\Hom_{\Z_p}(T',\Z_p)\otimes\Z_p(1)\) into \(V\) by this isomorphism. Its residual module is \((T'/pT')^*\otimes\mu_p\), whose \(G_L\)-composition factors are four copies of \(\overline\chi_p|_{G_L}\). By lattice independence over \(G_L\), these agree with the composition factors of \(T'/pT'\), which are trivial. Hence \(\overline\chi_p|_{G_L}=1\), that is, \(\Q(\zeta_p)\subseteq L\). The degree of the polarization plays no role. ∎
Lemma 6 (unramified order). Suppose \(\mathcal B\subset D_p\) is a \(\Z_p\)-order isomorphic to the ring of integers of the unramified quadratic extension of \(\Q_p\), such that \(R_g\mathcal BR_g^{-1}=\mathcal B\) for all \(g\in\Gamma\). Suppose also that some \(g_0\in\Gamma\) has a representative \(R_{g_0}\in\mathcal B^\times\) whose reduction in \(\mathcal B/p\mathcal B\simeq\F_{p^2}\) does not lie in \(\F_p\). If \(A(\Q)[p]\neq0\), then \(\Q(\zeta_p)\subseteq L\).
Proof. A stable maximal order. The field \(\mathcal B\otimes\Q_p\) makes \(\Q_p^2\) a one-dimensional vector space over it, so the \(\mathcal B\)-stable lattices in \(\Q_p^2\) are the fractional ideals of the discrete valuation ring \(\mathcal B\) times a fixed vector. Since \(p\) is a uniformizer of \(\mathcal B\), they form a single homothety class \(\{p^j\Lambda\}\). For \(g\in\Gamma\), the lattice \(R_g\Lambda\) is stable under \(R_g\mathcal BR_g^{-1}=\mathcal B\), so \(R_g\Lambda=p^j\Lambda\) for some \(j\). Rescaling, we choose for each \(g\in\Gamma\) a representative \(M_g\in\GL(\Lambda)\), unique up to \(\Z_p^\times\); for \(g_0\) we take \(M_{g_0}=R_{g_0}\), which lies in \(\GL(\Lambda)\) because \(\mathcal B^\times\) preserves \(\Lambda\). Thus \(\Gamma\) preserves the maximal order \(\mathcal R=\End_{\Z_p}(\Lambda)\simeq M_2(\Z_p)\).
A tensor decomposition. Put \(T'=\mathcal RT\). It is a lattice in \(V\), and it is \(G_{\Q}\)-stable because \(g(\varphi t)=g(\varphi)\,g(t)\) and \(g(\mathcal R)=\mathcal R\). By Morita equivalence,
with \(W_0\) free of rank \(2\). Let \(\widetilde G\) be the group of pairs \((g,M)\) with \(g\in G_{\Q}\), \(M\in\GL(\Lambda)\) and \([M]=[M_{g|_L}]\) in \(\PGL(\Lambda)\). For \((g,M)\in\widetilde G\), the map \(t\mapsto M^{-1}g(t)\) is an \(\mathcal R\)-linear automorphism of \(T'\), hence equals \(1\otimes w\) for a unique \(w=w(g,M)\in\GL(W_0)\). Thus \(\widetilde G\) acts linearly on \(\Lambda\) through \(M\) and on \(W_0\) through \(w\), and \((g,M)\) acts on \(T'\) as \(g=M\otimes w\). Reducing modulo \(p\),
as \(\widetilde G\)-modules, and these actions factor through a finite quotient of \(\widetilde G\). The subgroup \(H=\{(h,1):h\in G_L\}\) is normal in \(\widetilde G\), acts trivially on \(U\), and acts on \(\overline V\) as \(G_L\) does.
\(U\) is irreducible. The element \((g,R_{g_0})\), for any lift \(g\) of \(g_0\), acts on \(U\simeq\mathcal B/p\mathcal B=\F_{p^2}\) by multiplication by an element of \(\F_{p^2}\setminus\F_p\). Its characteristic polynomial over \(\F_p\) is irreducible, so it fixes no line of \(U\).
\(W\) is irreducible. Since \(A(\Q)[p]\neq0\), the module \(\overline V\) has a trivial composition factor, as a \(G_{\Q}\)-module and hence as a \(\widetilde G\)-module. If \(W\) had a stable line \(\ell\), then \(\overline V\) would have the composition factors \(U\otimes\ell\) and \(U\otimes(W/\ell)\), both irreducible of dimension \(2\), which is impossible.
\(G_L\) acts trivially on \(\overline V\). The restriction \(W|_H\) is semisimple: if \(S\subseteq W\) is a simple \(H\)-submodule, then its \(\widetilde G\)-translates are simple \(H\)-submodules because \(H\) is normal, and their sum is a nonzero \(\widetilde G\)-submodule, hence equal to \(W\). The module \(\overline V|_H\simeq W|_H\oplus W|_H\) has a trivial composition factor, so \(W^H\neq0\). Since \(H\) is normal, \(W^H\) is \(\widetilde G\)-stable, so \(W^H=W\). Hence \(G_L\) acts trivially on \(\overline V=U\otimes W\), and Lemma 5 gives \(\Q(\zeta_p)\subseteq L\). ∎
Lemma 7 (split blocks). Suppose \(\mathcal B=\Z_pe_1\oplus\Z_pe_2\subset D_p\), where \(e_1,e_2\) are orthogonal idempotents of rank one with \(e_1+e_2=1\), is preserved by \(\Gamma\), and some element of \(\Gamma\) exchanges \(e_1\) and \(e_2\). Let \(\Gamma_0\) be the stabilizer of \(e_1\), of index \(2\), and \(K=L^{\Gamma_0}\). Then \(G_K\) acts on the line \(e_1D_pe_2\) by a character \(\eta\) that factors through \(\Gal(L/K)\); its values are roots of unity, hence lie in \(\Z_p^\times\). If the reduction \(\overline\eta\) is nontrivial and \(A(\Q)[p]\neq0\), then \(\Q(\zeta_p)\subseteq L\).
Proof. Put \(T'=\mathcal BT\), a \(G_{\Q}\)-stable lattice, and \(T_i=e_iT'\), so that \(T'=T_1\oplus T_2\). The group \(G_K\) preserves each \(T_i\), since \(g(e_it)=e_ig(t)\), while every element of \(G_{\Q}\setminus G_K\) maps \(T_1\) onto \(T_2\) and \(T_2\) onto \(T_1\). Put \(V_i=T_i\otimes\Q_p\) and \(M_i=T_i/pT_i\), both of dimension \(2\).
Choose \(0\neq X\in e_1D_pe_2\). It restricts to an isomorphism \(V_2\to V_1\), and for \(g\in G_K\) and \(v\in V_2\) we have \(g(Xv)=\eta(g)X\,g(v)\). Hence \(V_1\simeq\eta\otimes V_2\) as \(G_K\)-representations, and lattice independence gives
Since \(A(\Q)[p]\neq0\), the \(G_K\)-module \(M_1\oplus M_2\) has a trivial composition factor, so one of the \(M_i\) has one. An element \(\tau\in G_{\Q}\setminus G_K\) identifies \(M_{3-i}\) with the conjugate of \(M_i\) by \(\tau\), and the conjugate of a trivial composition factor is trivial. So both \(M_1\) and \(M_2\) have a trivial composition factor. Write \(M_2^{\mathrm{ss}}=\mathbf 1\oplus\alpha\). Then \(M_1^{\mathrm{ss}}=\overline\eta\oplus\overline\eta\alpha\) contains \(\mathbf 1\), and \(\overline\eta\neq\mathbf 1\), so \(\alpha=\overline\eta^{-1}\). All four composition factors are powers of \(\overline\eta\), hence trivial on \(G_L\), and Lemma 5 gives \(\Q(\zeta_p)\subseteq L\). ∎
7. The prime \(7\): the endomorphism field is cyclic of order \(2\), \(3\) or \(6\)
We apply the lemmas of Section 6 with \(p=7\), using \(A(\Q)[7]\ni P_7\neq0\). The field \(\Q(\zeta_7)\) is cyclic of degree \(6\) over \(\Q\). So if \(\Q(\zeta_7)\subseteq L\), then \(\Gamma\) has a quotient isomorphic to \(C_6\). Modulo \(7\), the discriminants of the trace-one rotations are
and \(6\) is not a square modulo \(7\).
\(\Gamma=D_6\) or \(D_{12}\). Let \(r\) be a rotation of order \(n=3\) or \(6\) generating the rotation subgroup, and \(R\) its trace-one representative. Every element of \(\Gamma\) conjugates \(r\) to \(r^{\pm1}\), so it maps \(R\) to \(R\) or to \(1-R\). Hence \(\mathcal B=\Z_7[R]\), a \(\Z_7\)-order since \(R^2-R+1/c=0\) with \(c\in\{1,3\}\), is preserved by \(\Gamma\). Its discriminant is a nonzero square modulo \(7\), so by Hensel's lemma the roots \(\lambda_1,\lambda_2\) of \(X^2-X+1/c\) lie in \(\Z_7^\times\) and \(\lambda_1-\lambda_2\) is a unit. Thus \(\mathcal B\simeq\Z_7\times\Z_7\), with idempotents \(e_1=(R-\lambda_2)/(\lambda_1-\lambda_2)\) and \(e_2=(R-\lambda_1)/(\lambda_2-\lambda_1)\). The rotations fix \(R\), hence \(e_1\) and \(e_2\). Since \(\lambda_1+\lambda_2=1\), the reflections, which map \(R\) to \(1-R\), exchange \(e_1\) and \(e_2\). The stabilizer of \(e_1\) is the rotation subgroup. For \(X\in e_1D_7e_2\) we have \(RXR^{-1}=(\lambda_1/\lambda_2)X\), so \(\eta\) maps a lift of \(r\) to \(\lambda_1/\lambda_2\), a primitive \(n\)-th root of unity. As \(7\nmid n\), reduction modulo \(7\) is injective on \(n\)-th roots of unity, so \(\overline\eta\) has order \(n\geq3\). Lemma 7 gives \(\Q(\zeta_7)\subseteq L\). But the abelianizations of \(D_6\) and \(D_{12}\) are \(C_2\) and \(C_2\times C_2\), which have no quotient \(C_6\). This is a contradiction.
\(\Gamma=D_8\) or \(C_4\). Let \(r\) be a rotation of order \(4\) and \(R\) its trace-one representative, with \(R^2-R+\tfrac12=0\). As before, every element of \(\Gamma\) maps \(R\) to \(R\) or \(1-R\), so \(\mathcal B=\Z_7[R]\) is preserved by \(\Gamma\). Since \(X^2-X+4\), the reduction of \(X^2-X+\tfrac12\), has discriminant \(-15\equiv6\), a nonsquare modulo \(7\), it is irreducible over \(\F_7\). Hence \(\mathcal B\) is the ring of integers of the unramified quadratic extension of \(\Q_7\). Moreover \(R\in\mathcal B^\times\), since \(\det R=\tfrac12\) is a unit, and its reduction is a root of an irreducible quadratic, so it does not lie in \(\F_7\). Lemma 6 gives \(\Q(\zeta_7)\subseteq L\). But \(C_4\) has no quotient \(C_6\), and the abelianization \(C_2\times C_2\) of \(D_8\) has none either. This is a contradiction.
\(\Gamma=V_4\). The map \([M]\mapsto v_7(\det M)\bmod2\) is a well-defined homomorphism \(\Gamma\to\Z/2\Z\). Choose \(r\neq1\) in its kernel, and a representative \(R\) of trace \(0\), scaled by a power of \(7\) so that \(\det R\in\Z_7^\times\). Then \(R^2=a\) with \(a=-\det R\in\Q^\times\cap\Z_7^\times\). Since \(\Gamma\) is abelian, every \(R_g\) conjugates \(R\) to a trace-zero representative of \(r\) with the same square, that is, to \(\pm R\). Hence \(\mathcal B=\Z_7[R]\) is preserved by \(\Gamma\). Its discriminant \(4a\) is a unit.
- If \(a\) is not a square modulo \(7\), then \(X^2-a\) is irreducible modulo \(7\), so \(\mathcal B\) is the ring of integers of the unramified quadratic extension of \(\Q_7\). Also \(R\in\mathcal B^\times\) and its reduction does not lie in \(\F_7\). Lemma 6 applies with \(g_0=r\).
- If \(a\) is a square modulo \(7\), then \(a=b^2\) with \(b\in\Z_7^\times\), and \(\mathcal B\simeq\Z_7\times\Z_7\), with idempotents \(e_\pm=(1\pm R/b)/2\). Some element of \(\Gamma\) conjugates \(R\) to \(-R\): otherwise every \(R_g\) would lie in the centralizer \((\Q_7[R])^\times\simeq\Q_7^\times\times\Q_7^\times\) of \(R\), and \(\Gamma\) would embed in its image \(\Q_7^\times\) in \(\PGL_2(\Q_7)\), whose finite subgroups are cyclic. That element exchanges \(e_+\) and \(e_-\). The stabilizer of \(e_+\) is \(\{1,r\}\), and \(r\) acts on \(e_+D_7e_-\) by \(RXR^{-1}=(b/(-b))X=-X\). Thus \(\overline\eta\) is the nontrivial quadratic character, and Lemma 7 applies.
In both cases \(\Q(\zeta_7)\subseteq L\), which is impossible since \([L:\Q]=4\).
Together with \(\Gamma\neq C_1\) (Section 5), this shows:
Proposition 8. In case (S), \(\Gamma\) is cyclic of order \(2\), \(3\) or \(6\).
8. The quadratic case
If \(\Gamma=C_2\), then \(L\) is a quadratic field, and Corollary 4 gives an elliptic curve over \(L\) with an \(L\)-rational point of order \(21\). This contradicts the theorem of Kenku and Momose.
9. The cyclic sextic case: no rational \(3\)-torsion
In this section \(\Gamma=\langle r\rangle\simeq C_6\). Let \(R\) be the trace-one representative of \(r\), so \(R^2-R+\tfrac13=0\), and let \(K=L^{\langle r^2\rangle}\) be the quadratic subfield of \(L\). We show that \(A[3]^{\mathrm{ss}}\) has no trivial composition factor. Since \(P_3\) is a rational point of order \(3\), this excludes \(\Gamma=C_6\).
9.1. \(L\) is totally real
Fix an embedding \(\overline{\Q}\subset\mathbb C\), and let \(c\in G_{\Q}\) be complex conjugation. Put \(H=H_1(A(\mathbb C),\mathbb R)\), a real vector space of dimension \(4\) with the complex structure \(J_A\) of the complex torus \(A(\mathbb C)\). The algebra \(D_{\mathbb R}=D\otimes\mathbb R\simeq M_2(\mathbb R)\) acts faithfully on \(H\), commuting with \(J_A\). Since the simple \(M_2(\mathbb R)\)-module is \(\mathbb R^2\), we have \(H\simeq\mathbb R^2\otimes W\) with \(\dim W=2\), the algebra \(D_{\mathbb R}\) acting on the first factor. The commutant of \(M_2(\mathbb R)\otimes1\) is \(1\otimes\End(W)\), so \(J_A=1\otimes J\) with \(J^2=-1\).
Complex conjugation on \(A(\mathbb C)\) induces an involution \(c_*\) of \(H\) that anticommutes with \(J_A\), and the action of \(c\) on \(D\) is \(\varphi\mapsto c_*\varphi c_*^{-1}\). By Skolem–Noether, there is \(M\in\GL_2(\mathbb R)\) with \(c_*(\varphi\otimes1)c_*^{-1}=M\varphi M^{-1}\otimes1\); then \((M^{-1}\otimes1)c_*\) commutes with \(M_2(\mathbb R)\otimes1\), so \(c_*=M\otimes N\) with \(N\in\GL(W)\). The relation \(c_*J_A=-J_Ac_*\) gives \(NJ=-JN\). In a basis of \(W\) with \(J=\left(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right)\), this forces \(N=\left(\begin{smallmatrix}a&b\\b&-a\end{smallmatrix}\right)\), so \(N^2=(a^2+b^2)\cdot1\) with \(a^2+b^2>0\). From \(c_*^2=1\) we get \(M^2\otimes N^2=1\), so \(M^2=\mu\cdot1\) with \(\mu=(a^2+b^2)^{-1}>0\). If \(M\) is not a scalar, its minimal polynomial is \(X^2-\mu\), its eigenvalues are \(\pm\sqrt\mu\), and \(\det M=-\mu<0\).
On the other hand, the action of \(c\) on \(D\) is \(\Ad(R_c)\) with \([R_c]\in\Gamma\), so \(M\) is a real multiple of \(R_c\), and \(\det M\) has the sign of \(\det R_c\); this sign is well defined on \(\PGL_2(\mathbb R)\). Every element of \(\Gamma\) is \([R^k]\) with \(\det R^k=3^{-k}>0\). Hence \(M\) is a scalar, \(c\) acts trivially on \(D\), and \(c\in G_L\). Thus \(L\subset\mathbb R\), and since \(L/\Q\) is Galois, \(L\) is totally real. In particular \(K\) is a real quadratic field.
9.2. An isogenous surface with an action of \(\Z[\zeta_6]\)
We have \(\End^0_{\Q}(A)=D^\Gamma=\Q[R]\simeq\Q(\sqrt{-3})\), the centralizer of \(R\). Put \(u=3R-1\). Then
so \(u\) is a primitive sixth root of unity. Put \(\mathcal O=\Z[u]\). Choose \(m\geq1\) with \(mu\in\End_{\Q}(A)\), let \(\iota:A\to A\times A\) be \(a\mapsto(ma,(mu)(a))\), and put \(A'=\iota(A)\). It is \(\Q\)-isogenous to \(A\), since \(\ker\iota\subseteq A[m]\). The endomorphism \(U(x,y)=(y,y-x)\) of \(A\times A\) is defined over \(\Q\) and satisfies \(U^2-U+1=0\). It preserves \(A'\): given \(a\), choose \(a''\) with \(ma''=a\); since \((mu)^2=m(mu)-m^2\) in \(\End(A)\),
So \(\mathcal O\subseteq\End_{\Q}(A')\), with \(u\) acting as \(U\). The isogeny \(A\to A'\) identifies \(D\) with \(\End^0(A'_{\overline{\Q}})\), compatibly with the Galois action and with \(u\mapsto U\); we use the same letters on \(A'\). By lattice independence, \(A'[3]^{\mathrm{ss}}\simeq A[3]^{\mathrm{ss}}\).
9.3. The twist
For \(g\in G_{\Q}\), write \(g|_L=r^{k(g)}\) with \(k(g)\in\Z/6\Z\), and put \(\xi(g)=u^{-2k(g)}\). Then \(\xi:G_{\Q}\to\mu_3=\langle u^2\rangle\subset\Aut_{\Q}(A')\) is a homomorphism. Since its values are defined over \(\Q\), it is a cocycle, and Galois descent gives an abelian surface \(B/\Q\) and an isomorphism \(f:A'_{\overline{\Q}}\to B_{\overline{\Q}}\) with \({}^gf=f\circ\xi(g)\) for all \(g\in G_{\Q}\). Transporting endomorphisms by \(f\), the Galois action on \(D\) becomes
Put \(T=3S=u^{-2}(1+u)\in\mathcal O\), so that \(T^2=-3\). We deduce:
- \(\mathcal O\) commutes with \(\xi\), so \(f\mathcal Of^{-1}\subseteq\End_{\Q}(B)\); we write \(u\) and \(T\) for the corresponding endomorphisms of \(B\);
- \(B_{\overline{\Q}}\simeq A'_{\overline{\Q}}\) is isogenous to the square of an elliptic curve without CM;
- since \(\Ad(S^2)=1\), all geometric endomorphisms of \(B\) are defined over \(K\), and \({}^\tau\varphi=T\varphi T^{-1}\) for \(\tau\in G_{\Q}\setminus G_K\);
- for \(x\in A'(\overline{\Q})\), \(g(f(x))=f\big(\xi(g)\,g(x)\big)\).
9.4. A cyclic \(3\)-isogeny
Choose a rank-one idempotent \(e\in D\) and \(n\geq1\) with \(ne\in\End(B_K)\), and let \(E_0=ne(B)\), an elliptic curve over \(K\). Let \(\tau\in G_{\Q}\setminus G_K\). Since \(3\,{}^\tau(ne)=-T(ne)T\), we get \(\tau(E_0)=T(E_0)\). Moreover \(T(E_0)\neq E_0\): otherwise \(T|_{E_0}\) would be an endomorphism of \(E_0\) with square \(-3\), and \(E_0\) would have CM. As \(T\) is defined over \(\Q\), the restrictions \(T|_{E_0}:E_0\to\tau E_0\) and \(T|_{\tau E_0}:\tau E_0\to E_0\) are \(K\)-isogenies conjugate under \(\tau\), hence of the same degree \(d\), and their composite is \([-3]\) on \(E_0\). So \(d^2=9\), and \(T|_{E_0}\) is a cyclic \(K\)-isogeny of degree \(3\).
9.5. Conclusion
Lemma 9. If \(\Gamma\simeq C_6\), then \(A[3]^{\mathrm{ss}}\) has no trivial composition factor. In particular \(A(\Q)[3]=0\).
Proof. Suppose \(A[3]^{\mathrm{ss}}\) has a trivial composition factor.
- Reduction to \(A'[T]\). Since \(T^2=-3\), multiplication by \(T\) maps \(A'[3]\) into \(A'[T]=\ker T\), with kernel \(A'[T]\). As \(\#A'[3]=81\) and \(\#A'[T]=\deg T=9\), the map is onto, and it is \(G_{\Q}\)-equivariant because \(T\) is defined over \(\Q\). Hence \(A'[3]^{\mathrm{ss}}\simeq(A'[T]^{\mathrm{ss}})^{2}\), and \(A'[T]^{\mathrm{ss}}\) has a trivial composition factor.
- Passage to \(B[T]\). In \(\mathcal O\), the elements \(T\) and \(\zeta_3-1=u^2-1\) generate the same prime ideal above \(3\). So \(\mu_3\) acts trivially on \(A'[T]\), and by the last item of 9.3, \(f:A'[T]\to B[T]\) is \(G_{\Q}\)-equivariant. Hence \(B[T]^{\mathrm{ss}}\) has a trivial composition factor.
- Two lines in \(B[T]\). The group \(B[T]\) is a two-dimensional \(\F_3\)-vector space. Put
\(G=\ker(T|_{E_0})\subset B[T]\), a \(G_K\)-stable line with character \(\alpha\). Then
\(\tau(G)=\ker(T|_{\tau E_0})\), on which \(G_K\) acts by \(\alpha^\tau(g)=\alpha(\tau^{-1}g\tau)\).
Also \(\ker(T|_{\tau E_0})=T(E_0[3])\), since \(T(E_0[3])\) is killed by \(T^2=-3\) and both groups
have order \(3\). As \(G_K\)-modules, \(T(E_0[3])\simeq E_0[3]/G\), and the Weil pairing gives
\(E_0[3]/G\simeq\Hom(G,\mu_3)\). Hence
\[ \alpha^\tau=\alpha^{-1}\,\overline\chi_3|_{G_K}. \]
- If \(\tau G=G\), then \(\alpha^\tau=\alpha\), so \(\overline\chi_3|_{G_K}=\alpha^2=1\), because \(\alpha\) takes values in \(\F_3^\times=\{\pm1\}\).
- If \(\tau G\neq G\), then \(B[T]=G\oplus\tau G\) as \(G_K\)-modules, with characters \(\alpha\) and \(\alpha^\tau\). The trivial composition factor forces one of them to be trivial, and then both are. So \(\overline\chi_3|_{G_K}=1\).
In both cases \(\zeta_3\in K\), that is, \(K=\Q(\sqrt{-3})\). This contradicts 9.1. ∎
10. The cyclic cubic case: the curve 162b1
By Proposition 8 and Sections 8 and 9, it remains to treat \(\Gamma=\langle r\rangle\simeq C_3\), with \(L\) a cyclic cubic field.
10.1. The field \(L\) and the isogeny class
By Corollary 4, there is an elliptic curve \(E''\) over \(L\) and a point \(P''\in E''(L)\) of order \(21\), where \(E''\) is \(L\)-isogenous to every elliptic curve in \(A_L\). The pair \((E'',P'')\) defines a point \(x\in Y_1(21)(L)\), and \(Y_1(21)\) is a fine moduli space because \(21\geq4\). By Mazur's theorem, \(x\) is not rational, so its field of definition is \(L\) and \(x\) is a geometric point of a closed point of degree \(3\) on \(X_1(21)\).
Let \(E_{162}\) denote the curve 162b1. Over \(\Q(\zeta_9)^+\) its torsion subgroup is \(\Z/21\Z\) (Section 11), so it has exactly \(12\) points of order \(21\), forming six classes \(\{\pm Q\}\). Since \(\Aut(E_{162})=\{\pm1\}\), these give six distinct points of \(Y_1(21)\), each defined over \(\Q(\zeta_9)^+\) and not over \(\Q\). They form two closed points of degree \(3\), which by [DEvHMZB, Section 6.2] are \(D_0\) and \(D_0'\), and they are all the geometric points of \(D_0\) and \(D_0'\). Hence \(x\) is one of them. Therefore \(L=\Q(\zeta_9)^+\), and \((E'',P'')\simeq(E_{162},\pm Q)\) over \(L\) for some \(Q\). In particular, every elliptic curve in \(A_L\) is \(L\)-isogenous to \(E_{162}\).
10.2. Untwisting
Now \(R^2-R+1=0\), so \(u:=R\) is a primitive sixth root of unity in \(\End^0_{\Q}(A)=D^\Gamma=\Q[R]\), and we put \(\mathcal O=\Z[u]\) and \(F=\Q(u)\simeq\Q(\sqrt{-3})\). As in 9.2, we replace \(A\) by a \(\Q\)-isogenous surface \(A'\) with \(\mathcal O\subseteq\End_{\Q}(A')\). For \(g\in G_{\Q}\), write \(g|_L=r^{k(g)}\) with \(k(g)\in\Z/3\Z\), and put \(\xi(g)=u^{2k(g)}\). This is a homomorphism \(G_{\Q}\to\mu_3\subset\Aut_{\Q}(A')\) with kernel \(G_L\). As in 9.3, it gives a twist \(B/\Q\) and \(f:A'_{\overline{\Q}}\to B_{\overline{\Q}}\) with \({}^gf=f\circ\xi(g)\). The Galois action on \(D\) becomes
So every geometric endomorphism of \(B\) is defined over \(\Q\), and \(\End^0_{\Q}(B)\simeq M_2(\Q)\). Let \(e\in\End^0_{\Q}(B)\) be a rank-one idempotent and \(E_1\) the image of an integral multiple of \(e\), an elliptic curve over \(\Q\). By Lemma 3 applied to \(B\), whose endomorphism field is \(\Q\), the surface \(B\) is \(\Q\)-isogenous to \(E_1^2\).
We claim that \(E_1\) is \(\Q\)-isogenous to \(E_{162}\). Since \(\xi|_{G_L}=1\), the isomorphism \(f\) is defined over \(L\), so \(B_L\simeq A'_L\), and the \(\Q\)-isogeny \(A'\to A\) maps \(f^{-1}(E_1)\) onto an elliptic curve in \(A_L\). Hence \(E_1\) is \(L\)-isogenous to an elliptic curve in \(A_L\), and by 10.1 to \(E_{162}\). The group \(\Hom_{\overline{\Q}}(E_1,E_{162})\) is free of rank one over \(\Z\), because \(E_1\) has no CM. The group \(G_{\Q}\) acts on it through a character \(\varepsilon:G_{\Q}\to\Aut(\Z)=\{\pm1\}\). This character is trivial on \(G_L\), because there is a nonzero homomorphism defined over \(L\). So \(\varepsilon\) factors through \(\Gal(L/\Q)\simeq C_3\), and is therefore trivial. Hence every geometric homomorphism \(E_1\to E_{162}\) is defined over \(\Q\), which proves the claim. In particular \(a_p(E_1)=a_p(E_{162})\) for every prime \(p\nmid6\).
10.3. Frobenius at \(5\)
Since \(\End^0_{\Q}(B)\simeq M_2(\Q)\) commutes with \(G_{\Q}\), the module \(V_\ell(B)\) is \(\Q_\ell^2\otimes eV_\ell(B)\), with \(M_2(\Q)\) acting on the first factor, and \(eV_\ell(B)\simeq V_\ell(E_1)\) as \(G_{\Q}\)-modules. The subfield \(F\subset M_2(\Q)\) makes \(\Q^2\) a one-dimensional \(F\)-vector space. Hence
with \(F\) acting on the first factor and \(G_{\Q}\) on the second. By the last relation of 9.3, transporting through \(f\), an element \(g\in G_{\Q}\) acts on \(V_\ell(A')\) as \(\xi(g)^{-1}\otimes\rho_{E_1}(g)\). Also \(V_\ell(A)\simeq V_\ell(A')\).
The field \(L=\Q(\zeta_9)^+\) is unramified at \(5\), so \(\xi\) is unramified at \(5\), and so is \(V_\ell(E_1)\) for \(\ell\neq5\), since \(E_1\) is \(\Q\)-isogenous to \(E_{162}\), which has conductor \(162\). By (a), \(A\) has good reduction at \(5\). The prime \(5\) is inert in \(L\) (equivalently, \(5\not\equiv\pm1\pmod9\)), so \(\mathrm{Frob}_5\) generates \(\Gal(L/\Q)\), and \(\zeta=\xi(\mathrm{Frob}_5)^{-1}\) is a primitive cube root of unity in \(F\). Over \(\overline{\Q}_\ell\), the eigenvalues of \(\mathrm{Frob}_5\) on \(V_\ell(A)\) are \(\omega\beta_1,\omega\beta_2,\omega^{-1}\beta_1,\omega^{-1}\beta_2\), where \(\omega\) is the image of \(\zeta\) under an embedding \(F\to\overline{\Q}_\ell\) and \(\beta_1,\beta_2\) are the roots of \(X^2-a_5X+5\). With \(a_5=a_5(E_{162})=0\), the characteristic polynomial of \(\mathrm{Frob}_5\) is
By (b), the reduction \(\widetilde A\) of \(A\) at \(5\) satisfies
Since \(63\) is prime to \(5\), (b) embeds \(\langle P\rangle\simeq\Z/63\Z\) into \(\widetilde A(\F_5)\). But \(63\) does not divide \(21\). This contradiction completes the proof of the theorem. ∎
The value \(21\) depends only on the \(\Q\)-isogeny class, so no choice of lattice, polarization or isogenous model avoids it. The bound is attained: the census curve 21.b has torsion \(\Z/21\Z\), and its Frobenius polynomials agree with the formula above for all primes \(5\leq p<300\) (Section 11).
11. Computations and a consistency check
The script qsimple_63_arithmetic.m checks, with an assertion for each claim, every numerical fact used above:
- the residues modulo \(7\) of Section 7: \(-3\equiv4\) and \(-\tfrac13\equiv2\) are squares, \(-1\equiv6\) is not, and \(X^2-X+4\) and \(X^2-a\), for \(a\) a nonsquare, are irreducible over \(\F_7\);
- that \(\#B(\F_2)\in\{1,\dots,16,19,20,25\}\) for every abelian surface \(B\) over \(\F_2\), by exhaustive enumeration of the \(2\)-Weil polynomials \(X^4+aX^3+bX^2+2aX+4\) (Section 4);
- the identities in \(\Q(\sqrt{-3})\) of Sections 9 and 10, including \(S^2=-\tfrac13\), \(T^2=-3\), and the equality of the ideals \((T)\) and \((\zeta_3-1)\);
- that \(\Q[x]/(x^3-3x+1)\) is the cyclic cubic field \(\Q(\zeta_9)^+\), of discriminant \(81\), in which \(5\) is inert;
- that 162b1 has the model above, conductor \(162\), non-integral \(j\)-invariant \(-140625/8\) (hence no CM), and \(a_5=0\), and that its torsion subgroup over \(\Q(\zeta_9)^+\) is \(\Z/21\Z\), with \(12\) points of order \(21\), for instance \((1-2\alpha,2-2\alpha)\) where \(\alpha^3-3\alpha+1=0\);
- the polynomial \(P_5(X)=X^4-5X^2+25\) and the value \(P_5(1)=21\) of Section 10.
The script qsimple_63_census_21b.m is a consistency check and is not used in the proof. The census curve 21.b,
(LMFDB 324.a.648.1) is listed in the class “geometrically split, simple over \(\Q\)” with torsion \(\Z/21\Z\). The script checks that, for every prime \(5\leq p<300\), the characteristic polynomial of Frobenius of its Jacobian equals \((X^2-a_pX+p)^2\) if \(p\equiv\pm1\pmod9\) and \(N_{F/\Q}(X^2-a_p\zeta X+p\zeta^2)\) otherwise, with \(a_p=a_p(E_{162})\), as predicted by Section 10. It also checks that \(\#J(\F_5)=21\) and \(J(\Q)_{\mathrm{tors}}\simeq\Z/21\Z\). So the argument stops exactly at \([21]\).
Both scripts were run in Magma V2.30-1, and both logs end with ALL_CHECKS_PASSED. No step
depends on the generalized Riemann hypothesis.
12. Consequence for the census
The group [63] is impossible in the class “geometrically split, simple over \(\Q\)”. The theorem places no restriction on the geometric splitting degree, the endomorphism field, the integral geometric endomorphism ring, or the polarization.
Simplicity over \(\Q\) is essential: [63] has an example in the class “split over \(\Q\)” (the curve 63.a). Thus this exclusion does not imply ∅ on the combined geometrically split side, and the combined geometrically split ∞? grade of [63] is unchanged.
References
M. A. Kenku and F. Momose, Torsion points on elliptic curves defined over quadratic fields, Nagoya Math. J. 109 (1988), 125–149, Theorem (0.1).
B. Mazur, Modular curves and the Eisenstein ideal, Publ. Math. IHÉS 47 (1977), 33–186, Theorem (7′).
M. Derickx, A. Etropolski, M. van Hoeij, J. S. Morrow and D. Zureick-Brown, Sporadic cubic torsion, Algebra & Number Theory 15 (2021), 1837–1864, arXiv:2007.13929, Theorem A and Section 6.2 (numbering of arXiv version 2).
J.-P. Serre and J. Tate, Good reduction of abelian varieties, Ann. of Math. (2) 88 (1968), 492–517, Theorem 1 and its Corollary 2, Lemma 2, and §2 (p. 496 and p. 498).
P. L. Clark, Bounds for torsion on abelian varieties with integral moduli, arXiv:math/0407264v2 (2004), Theorem 5. The same list is Theorem 1.4 of P. L. Clark and X. Xarles, Local bounds for torsion points on abelian varieties, Canad. J. Math. 60 (2008), 532–555, for abelian surfaces with everywhere anisotropic reduction, which includes potentially good reduction.