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{Torsion of elliptic curves over number fields\\ of degree $6$, $7$, $8$ and $9$} \author{Filip Najman} \address{University of Zagreb Faculty of Science, Department of Mathematics, Bijeni\v cka cesta 30, 10000 Zagreb, Croatia} \email{fnajman@math.hr} \thanks{The author was supported by the Croatian Science Foundation under the project no.\ IP-2022-10-5008, by the project ``Implementation of cutting-edge research and its application as part of the Scientific Center of Excellence for Quantum and Complex Systems, and Representations of Lie Algebras'', PK.1.1.10.0004, co-financed by the European Union through the European Regional Development Fund -- Competitiveness and Cohesion Programme 2021--2027, and by the European Union -- NextGenerationEU through the National Recovery and Resilience Plan 2021--2026, via an institutional grant of the University of Zagreb Faculty of Science, IK IA 1.1.3, Impact4Math.} \subjclass[2020]{11G05, 11G18, 14G05} \keywords{elliptic curves, torsion, number fields, modular curves, sporadic points} \setlength{\emergencystretch}{3em} \begin{document} \begin{abstract} We determine all the groups that appear as the torsion group of an elliptic curve over a number field of degree $d$ for $d=6$, $7$, $8$ and $9$. Apart from the groups that occur infinitely often, exactly $4$ groups occur in degree $6$, $3$ in degree $7$, $8$ in degree $8$ and $9$ in degree $9$, and among them the groups $\tg{2}{24}$ over sextic fields, $\tg{3}{15}$ over octic fields and $\tg{2}{26}$ over nonic fields were not known to occur. For $13$ of these $24$ pairs of a degree and a group we also determine all the elliptic curves that realize them. The proofs combine the global method of Derickx, Hecke sieves, formal immersion criteria at the prime $2$ with Hecke operators, and Riemann-Roch computations on modular curves and their quotients over finite fields. \end{abstract} \date{\today} \maketitle %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Introduction}\label{sec:intro} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Let $E$ be an elliptic curve over a number field $K$. By the Mordell-Weil theorem $E(K)\simeq E(K)_\tors\times \Z^r$, and by Merel's theorem \cite{merel} the torsion subgroup $E(K)_\tors$ is bounded in terms of $d=[K:\Q]$ alone. Hence, for every $d\geq 1$ the set \[ \Phi(d):=\{\,E(K)_\tors \;:\; [K:\Q]=d,\ E/K \text{ an elliptic curve}\,\} \] is finite. The set $\Phi(1)$ was determined by Mazur \cite{mazur}, building on work of Kubert \cite{kubert}, and $\Phi(2)$ by Kamienny \cite{kamienny}, building on work of Kenku and Momose \cite{KM}. Derickx, Etropolski, van Hoeij, Morrow and Zureick-Brown \cite{DEHMZ} determined $\Phi(3)$, using earlier work of Jeon, Kim and Schweizer \cite{JKS}, Parent \cite{parent} and Najman \cite{najman}. Derickx and Najman \cite{DN} determined $\Phi(4)$, using earlier work of Jeon, Kim and Park \cite{JKP} and Bruin and Najman \cite{bruin_najman2016}. Finally, $\Phi(5)$ was determined by the author \cite{N5}. In this paper we determine $\Phi(6)$, $\Phi(7)$, $\Phi(8)$ and $\Phi(9)$. Write $\Phi^\infty(d)\subseteq\Phi(d)$ for the subset of groups that occur as $E(K)_\tors$ for infinitely many $\overline\Q$-isomorphism classes of elliptic curves $E$ over number fields $K$ of degree $d$. A group in $\Phi(d)\setminus\Phi^\infty(d)$ will be called a \textit{sporadic torsion group in degree $d$}. This should not be confused with a \textit{sporadic point} on a curve $X$, which is a closed point of some degree $e$ such that $X$ has only finitely many closed points of degree at most $e$. A list of known sporadic and isolated points on the curves $X_1(m,n)$ is maintained on the webpage \cite{x1mn}. We have $\Phi^\infty(d)=\Phi(d)$ for $d=1,2,4$. The only sporadic torsion group in degree $3$ is $\torz{21}$ \cite{najman,DEHMZ}, and the sporadic torsion groups in degree $5$ are $\torz{28}$, $\torz{30}$ and $\tg{2}{18}$ \cite{N5}. The set $\Phi^\infty(6)$ was determined by Derickx and Sutherland \cite[Theorem 1.1]{DS}, the cyclic part being due to Derickx and van Hoeij \cite[Theorem 3]{DvH}: \begin{equation}\label{d6:eq:phiinf6} \begin{aligned} \Phi^\infty(6)=\ &\{\torz{n}\;:\;1\leq n\leq 30,\ n\neq 23,25,29\}\\ &\cup\ \{\tg{2}{2n}\;:\;1\leq n\leq 10\}\ \cup\ \{\tg{3}{3n}\;:\;1\leq n\leq 4\}\\ &\cup\ \{\tg{4}{4},\ \tg{4}{8},\ \tg{6}{6}\}. \end{aligned} \end{equation} The sets $\Phi^\infty(7)$, $\Phi^\infty(8)$ and $\Phi^\infty(9)$ were determined by Najman and Varivoda \cite[Theorem 1.1]{NV}: \begin{equation}\label{d7:eq:phiinf7} \Phi^\infty(7)=\{\torz{n}\;:\;1\leq n\leq 30,\ n\neq 25,29\}\ \cup\ \{\tg{2}{2n}\;:\;1\leq n\leq 10\}, \end{equation} \begin{equation}\label{d8:eq:phiinf8} \begin{aligned} \Phi^\infty(8)=\ &\{\torz{n}\;:\;1\leq n\leq 28\ \text{ or }\ n\in\{30,32,36\}\}\ \cup\ \{\tg{2}{2n}\;:\;1\leq n\leq 12\}\\ &\cup\ \{\tg{3}{3n}\;:\;1\leq n\leq 4\}\ \cup\ \{\tg{4}{4n}\;:\;1\leq n\leq 3\}\\ &\cup\ \{\tg{5}{5},\ \tg{6}{6}\}, \end{aligned} \end{equation} \begin{equation}\label{d9:eq:phiinf9} \Phi^\infty(9)=\{\torz{n}\;:\;1\leq n\leq 28\ \text{ or }\ n\in\{30,36\}\}\ \cup\ \{\tg{2}{2n}\;:\;1\leq n\leq 12\}. \end{equation} These sets have $44$, $38$, $52$ and $42$ elements. Some sporadic torsion groups in these degrees were already known. Van Hoeij \cite{vanHoeij} found points of degree $6$ on $X_1(25)$ and $X_1(37)$, of degree $7$ on $X_1(25)$, $X_1(33)$ and $X_1(36)$, of degree $8$ on $X_1(N)$ for $N=33,34,35,39,40$ and $42$, and of degree $9$ on $X_1(N)$ for $N=29,31,32,33,34,39,40$ and $42$. Derickx, Kamienny, Stein and Stoll proved that every elliptic curve over a sextic field with a point of order $37$ has $j$-invariant $-9317$ \cite[Proposition 1.4]{DKSS}. Gonz\'alez-Jim\'enez and Najman \cite{GN} found that the base change of the rational elliptic curve \lmfdbec{162}{d}{2} to a sextic field has torsion group $\tg{4}{12}$ (see also \cite[Remark 1]{DGJ}). Clark, Corn, Rice and Stankewicz \cite{CCRS14} determined the torsion groups of elliptic curves with complex multiplication over number fields of small degree. In particular, the groups $\torz{34}$, $\torz{39}$ and $\tg{5}{10}$ are torsion groups of elliptic curves with complex multiplication over octic fields, while all such groups over nonic fields lie in $\Phi^\infty(9)$. Closely related is the set $S(d)$ of primes $p$ for which there exists an elliptic curve over a number field of degree $d$ with a point of order $p$. Merel's theorem shows that $S(d)$ is finite. The sets $S(1)$, $S(2)$ and $S(3)$ follow from the work of Mazur, Kamienny, Kenku and Momose and Parent \cite{mazur,kamienny,KM,parent}, the sets $S(4),\dots,S(7)$ were determined by Derickx, Kamienny, Stein and Stoll \cite{DKSS}, and $S(8)$ by Khawaja \cite{khawaja}. Explicitly, $S(6)=\{2,3,5,7,11,13,17,19,37\}$ and $S(7)$ is the set of primes up to $23$ \cite[Theorem 1.2]{DKSS}, and so is $S(8)$ \cite[Theorem 1.2]{khawaja}. Derickx and Stoll \cite{DSt} studied the asymptotic behavior of $S(d)$ and developed tools that rule out most primes for any given $d$, and they have also determined $S(9)$, up to not being able to eliminate $89$ from $S(9)$ \cite{DSt9}. The author \cite{Sd} showed that $S(9)$ is the set of primes up to $31$. These results are the starting point of our proofs. Note that in degree $9$ our proof depends on \cite{Sd}. Our main result is the following theorem. \begin{theorem}\label{thm:main}\label{d6:thm:main}\label{d7:thm:main}\label{d8:thm:main}\label{d9:thm:main} Let $d\in\{6,7,8,9\}$, let $K$ be a number field of degree $d$ and let $E$ be an elliptic curve over $K$. Then $E(K)_\tors$ lies in $\Phi^\infty(d)$ or is one of the sporadic groups in degree $d$ given in the following table, and each of these sporadic groups occurs. \begin{center} \begin{tabular}{@{}cll@{}} \toprule $d$ & sporadic torsion groups in degree $d$ & $\#\Phi(d)$\\ \midrule $6$ & $\torz{25}$, $\torz{37}$, $\tg{4}{12}$, $\tg{2}{24}$ & $48$\\ $7$ & $\torz{25}$, $\torz{33}$, $\torz{36}$ & $41$\\ $8$ & $\torz{33}$, $\torz{34}$, $\torz{35}$, $\torz{39}$, $\torz{40}$, $\torz{42}$, $\tg{5}{10}$, $\tg{3}{15}$ & $60$\\ $9$ & $\torz{29}$, $\torz{31}$, $\torz{32}$, $\torz{33}$, $\torz{34}$, $\torz{39}$, $\torz{40}$, $\torz{42}$, $\tg{2}{26}$ & $51$\\ \bottomrule \end{tabular} \end{center} \end{theorem} Explicitly, $\Phi(6)$ consists of the groups \begin{alignat*}{2} &\torz{n}, \qquad && n=1,\dots,22,\ 24,\dots,28,\ 30,\ 37,\\ &\tg{2}{2n}, \qquad && n=1,\dots, 10,\ 12,\\ &\tg{3}{3n}, \qquad && n=1,\dots, 4,\\ &\tg{4}{4n}, \qquad && n=1,2,3,\\ &\tg{6}{6}, && \end{alignat*} $\Phi(7)$ consists of the groups \begin{alignat*}{2} &\torz{n}, \qquad && n=1,\dots,28,\ 30,\ 33,\ 36,\\ &\tg{2}{2n}, \qquad && n=1,\dots, 10, \end{alignat*} $\Phi(8)$ consists of the groups \begin{alignat*}{2} &\torz{n}, \qquad && n=1,\dots,28,\ 30,\ 32,\ 33,\ 34,\ 35,\ 36,\ 39,\ 40,\ 42,\\ &\tg{2}{2n}, \qquad && n=1,\dots, 12,\\ &\tg{3}{3n}, \qquad && n=1,\dots, 5,\\ &\tg{4}{4n}, \qquad && n=1,2,3,\\ &\tg{5}{5n}, \qquad && n=1,2,\\ &\tg{6}{6}, && \end{alignat*} and $\Phi(9)$ consists of the groups \begin{alignat*}{2} &\torz{n}, \qquad && n=1,\dots,34,\ 36,\ 39,\ 40,\ 42,\\ &\tg{2}{2n}, \qquad && n=1,\dots, 13. \end{alignat*} For many of the sporadic groups we also determine all the elliptic curves in question. Here two pairs $(K,E)$ and $(K',E')$, consisting of a number field and an elliptic curve over it, are isomorphic if there is a field isomorphism $\sigma:K\to K'$ such that $E^\sigma\simeq E'$ over $K'$. \begin{theorem}\label{thm:sporadic}\label{d6:thm:sporadic}\label{d7:thm:sporadic}\label{d8:thm:sporadic}\label{d9:thm:sporadic} Let $(d,G)$ be one of the pairs in the following table, let $K$ be a number field of degree $d$, and let $E/K$ be an elliptic curve such that $E(K)$ has a subgroup isomorphic to $G$. Then $E(K)_\tors\simeq G$, and up to isomorphism the pair $(K,E)$ is one of the pairs listed for $(d,G)$ in \Cref{table:sporadic}. The number of these pairs and their fields $K$ are as follows. \begin{center} \begin{tabular}{@{}cccl@{}} \toprule $d$ & $G$ & pairs $(K,E)$ & fields $K$\\ \midrule $6$ & $\torz{37}$ & $1$ & \lmfdbnf{6.6.300125.1}\\ & $\tg{2}{24}$ & $1$ & \lmfdbnf{6.2.231852.1}\\ $7$ & $\torz{25}$ & $1$ & \lmfdbnf{7.1.184607.1}\\ & $\torz{33}$ & $1$ & \lmfdbnf{7.3.792873.1}\\ & $\torz{36}$ & $2$ & \lmfdbnf{7.1.1609956.1}, \lmfdbnf{7.3.3192732.1}\\ $8$ & $\torz{33}$ & $1$ & \lmfdbnf{8.2.7486875.1}\\ & $\torz{34}$ & $3$ & \lmfdbnf{8.0.1257728.1}, \lmfdbnf{8.0.1265625.1}, \lmfdbnf{8.4.19360000.1}\\ & $\torz{35}$ & $1$ & \lmfdbnf{8.4.31878125.1}\\ & $\torz{39}$ & $1$ & \lmfdbnf{8.0.1601613.1}\\ & $\torz{40}$ & $1$ & \lmfdbnf{8.2.31500500.1}\\ & $\tg{3}{15}$ & $1$ & \lmfdbnf{8.0.2313441.1}\\ $9$ & $\torz{33}$ & $1$ & \lmfdbnf{9.3.393718863.1}\\ & $\tg{2}{26}$ & $1$ & \lmfdbnf{9.3.157505216.1}\\ \bottomrule \end{tabular} \end{center} \end{theorem} The fields and the curves are given explicitly in \Cref{sec:sporadic}, which also contains curves realizing the other sporadic groups. The elliptic curves with torsion groups $\tg{2}{24}$ over a sextic field, $\tg{3}{15}$ over an octic field and $\tg{2}{26}$ over a nonic field are new, and they are listed in \cite{x1mn} as the points \xpoint{2.24.6.a}, \xpoint{3.15.8.a} and \xpoint{2.26.9.a}. We found them in the course of this work, so these three groups were not known to occur in these degrees. We do not determine all the elliptic curves over sextic fields with torsion $\torz{25}$ or $\tg{4}{12}$, nor those over octic fields with torsion $\torz{42}$ or $\tg{5}{10}$, nor those over nonic fields with the sporadic cyclic torsion groups other than $\torz{33}$. This is not needed for \Cref{thm:main}, since all the groups properly containing these groups can be ruled out directly. The curve with torsion $\tg{3}{15}$ has $j$-invariant $1331/8$, so it is a twist of the rational elliptic curve \lmfdbec{338}{d}{2}, but it is not the base change of an elliptic curve over $\Q$ or $\Q(\zeta_3)$. Note also that $\torz{35}$, which is a sporadic torsion group in degree $8$, does not occur in degree $9$, and that $\torz{37}$, which occurs in degree $6$, occurs in none of the degrees $7$, $8$ and $9$. \subsection*{Outline of the proof and of the paper} As in \cite{DN} and \cite{N5}, the classification reduces to showing that certain modular curves $X_1(m,n)$ have no non-cuspidal points over number fields of degree $d$. We introduce the notation and the general tools in \Cref{sec:prelim}, and in \Cref{sec:groups} we show that $66$, $58$, $80$ and $81$ groups have to be ruled out in degrees $6$, $7$, $8$ and $9$. The remaining sections treat these groups by method, most curves once for all the degrees in which they are needed, and often the computation for the largest degree also covers the smaller ones (see \Cref{rem:smaller}). The global method of Derickx \cite[Proposition 4.3]{DN} and arguments with several diamond operators at the prime $2$ eliminate most of the groups $\torz n$ with large $n$ (\Cref{sec:global}). In degree $9$ we also use that a field of odd degree contains no imaginary quadratic field (\Cref{subsec:odd}). When the Jacobian has rank $0$ we use the criterion of \cite[Proposition 5.2]{DN}, which shows that the elliptic curve must have good reduction at some prime above a suitable $p$, together with computations with cuspidal divisor classes over finite fields (\Cref{sec:rankzero}). We exclude the possible good reductions with the Hecke sieve of \cite[Proposition 5.5]{DN}, often after pushing the divisors forward to a curve of smaller level whose Mordell-Weil group is known (\Cref{sec:sieve}). When every elliptic curve with a point of order $n$ has bad reduction at all the primes above $2$, but the Jacobian has positive rank or the gonality bound is too weak, we use a formal immersion criterion at $2$ (\Cref{sec:formal}). There \Cref{d6:lem:dyadic} shows that $T_2-\dm 2$ kills the $2$-torsion in the kernel of reduction, and the criterion becomes a condition on the Cartier operator on a quotient of $X_1(n)$ over $\F_2$, combined with elements of the Hecke algebra whose image on the Mordell-Weil group is finite, so that it also applies when the Jacobian has positive rank. When the Hecke sieve has to be applied on curves of large genus, as for $\torz{57}$ and $\torz{65}$ in degree $9$, we replace Riemann-Roch computations over finite fields by linear algebra in the canonical ring (\Cref{sec:positive}). The groups $\torz{57}$ and $\torz{65}$ in degrees $6$, $7$ and $8$, and the group $\torz{69}$, are treated in \Cref{sec:57_65}, partly on unramified covers of $X_0(57)$ and $X_0(69)$ that we construct over $\F_2$ by class field theory, and $\torz{54}$ in \Cref{sec:54}. Most of the groups $\tg{m}{n}$ with $m\geq 2$ are treated in \Cref{sec:noncyclic,sec:mxn}. In \Cref{sec:sporadic} we construct the sporadic points and determine all the points of the relevant degrees on several of the curves $X_1(m,n)$, and in this way we also exclude $\tg{2}{26}$ in degrees $6$, $7$ and $8$, $\torz{36}$ in degree $6$ and $\torz{39}$ in degrees $6$ and $7$. We prove \Cref{thm:main,thm:sporadic} in \Cref{sec:proof}, and the computations are described in \Cref{sec:computations}. \subsection*{Data availability and reproducibility} The code, the stored logs and the certificates of the computations described in this paper are available in four repositories, one for each degree: \begin{center} \url{https://github.com/F-Najman/sextic_torsion}\\ \url{https://github.com/F-Najman/septic_torsion}\\ \url{https://github.com/F-Najman/octic_torsion}\\ \url{https://github.com/F-Najman/nonic-torsion} \end{center} This paper refers to the revisions \texttt{dbcd360}, \texttt{52af365}, \texttt{5cac315} and \texttt{14d4464} of these repositories, and the links in the paper point to these revisions. The \texttt{README} files there list, statement by statement, which script establishes each claim and which log it produced, and also the checks that the proofs do not need (see \Cref{sec:computations}). Those of the septic, octic and nonic repositories use the numbering of this paper, and that of the sextic repository uses the numbering of a separate version of the proofs for degree $6$. The code in the nonic repository for the prime orders is not used in this paper. In the digital form of this article, the words reporting a computation, such as ``we \gitlinkVI{verify_37.m}{compute}'', are hyperlinks to the script that carries it out. \subsection*{AI disclosure and comments} AI models were extensively used in the work leading to this paper. After completing \cite{DN} (with Derickx) and \cite{N5} and giving talks about this work, I was asked several times whether we could do such a classification for number fields of degree $d>5$. My answer was that I thought it was not possible as it would require explicit computations that did not seem feasible to me. To check whether this was correct, in mid-October 2026 I asked AI whether this was possible, and whether it could do it. To my surprise, it was able to complete the classifications of possible torsion groups of elliptic curves over number fields, not only in degree 6, but later also in degrees 7, 8, 9 and finally 10. After some initial hesitation about releasing primarily AI-generated mathematics, I decided that the results were simply too interesting and the mathematics too beautiful to just stay on my computer. OpenAI's GPT-6.1 Sol produced almost all of the mathematical content of the first versions of the proofs and carried out most of the computations. Anthropic's Claude Opus 5.5 reviewed these versions, independently recomputed some of the computations (see \Cref{sec:computations}) and the explicit examples, merged the proofs for the four degrees, wrote the draft of the present manuscript and organized the code repositories. %I take full %responsibility for the content of the paper. The degree $6$ part of this paper, which I have checked carefully, will appear separately on arXiv. %\subsection*{Acknowledgements} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Notation and auxiliary results}\label{sec:prelim}\label{d6:sec:prelim}\label{d7:sec:prelim}\label{d8:sec:prelim}\label{d9:sec:prelim} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \subsection{Modular curves}\label{subsec:modular} Throughout, $m\mid n$ are positive integers with $mn>4$. We denote by $X:=X_1(m,n)$ the modular curve parametrizing triples $(E,P,Q)$ consisting of a generalized elliptic curve $E$ and points $P,Q$ of orders $m$ and $n$ generating a subgroup of $E[n]$ isomorphic to $\tg{m}{n}$, such that the Weil pairing of $P$ and $(n/m)Q$ is $\zeta_m$. It is a smooth projective geometrically irreducible curve over $\Z[\zeta_m,\tfrac1n]$, and it is a fine moduli space when $n>4$ (see \cite[\S 2]{DN} and \cite[\S 2]{DS}). Every curve $X_1(m,n)$ that we use has $n\geq 5$, so we use the fine moduli interpretation throughout. We write $Y_1(m,n)$ for the open subscheme obtained by removing the cusps, and $J:=J_1(m,n)$ for the Jacobian of $X$. We abbreviate $X_1(n):=X_1(1,n)$ and $J_1(n):=J_1(1,n)$. For $m\leq 2$ the curve $X$ is defined over $\Q$. For $d\geq1$ we write $X^{(d)}$ for the $d$-th symmetric power of $X$, so that $X^{(d)}(\Q(\zeta_m))$ is the set of effective $\Q(\zeta_m)$-rational divisors of degree $d$ on $X$. For a curve $X$ over a field $F$ with a set of cusps, such as $X_1(m,n)$ over $\Q(\zeta_m)$ or a quotient of it, we write $\ClCusp_F X$ for the subgroup of $J_X(F)$ generated by the classes of the $F$-rational divisors of degree $0$ supported on the cusps. For a curve $X$ we write $\gon_k X$ for its gonality over a field $k$, and $J_X$ or $J(X)$ for its Jacobian. For a divisor $D$ on a curve $X$ over a field $k$ we write $L(D):=\{f\in k(X)^\times:\ddiv(f)+D\geq0\}\cup\{0\}$ for its Riemann-Roch space and $\ell(D)=\dim L(D)$. Similarly, we write $\Omega(D)$ for the space consisting of $0$ and the non-zero differentials $\omega$ on $X$ with $\ddiv(\omega)\geq D$. Then $\Omega(0)=H^0(X,\Omega^1)$ is the space of holomorphic differentials on $X$, and if $D=\sum_P m_PP$ is effective, then $\Omega(D)$ is its subspace of the differentials that vanish to order at least $m_P$ at every $P$. The Riemann-Roch theorem states that $\dim L(D)=\deg D+1-g+\dim\Omega(D)$, where $g$ is the genus of $X$. For $i\geq1$ we put $R_i:=H^0(X,iK_X)$, where $K_X$ is a canonical divisor. The cusps of $X_1(m,n)$ correspond to N\'eron $k$-gons with $m\mid k\mid n$ together with a level structure (see \cite[\S 2]{DN}). All cusps are defined over $\Q(\zeta_n)$, and $\Gal(\Q(\zeta_n)/\Q(\zeta_m))$ acts on them through its action on $\mathbb G_m$. The cuspidal subscheme is finite \'etale over $\Z[\zeta_m,\tfrac1n]$. We denote by $c_0$ the cusp given by the N\'eron $n$-gon together with the level structure $(a,b)\mapsto (\zeta_m^a,b)$ (see \cite[Lemma 2.8]{DN}), which is $\Q(\zeta_m)$-rational. From the description of the cusps of $X_1(n)$ in \cite[\S 2]{DN} it follows that a cusp of $X_1(n)$ lying on a N\'eron $k$-gon is fixed by the diamond operator $\dm a$ (defined below) only if $a\equiv\varepsilon\pmod k$ and $a\equiv\varepsilon\pmod{n/k}$ for a common sign $\varepsilon=\pm1$. In particular $a$ is then congruent to $\varepsilon$ modulo every prime dividing $n$. For a cusp $c$ we write $\sigma_q(c)$ for its image under the automorphism $\zeta_n\mapsto\zeta_n^q$ of $\Q(\zeta_n)$. When we list the degrees of closed cusps or of Galois orbits of cusps, $a^r$ stands for $r$ of them of degree $a$. For a congruence subgroup $\Gamma\subseteq\SL_2(\Z)$ we write $\overline\Gamma$ for its image in $\PSL_2(\Z)$. The curve $X_1(m,n)_\C$ is the modular curve of the group $\Gamma_1(n)\cap\Gamma(m)$, and for $n>2$ we have $[\PSL_2(\Z):\overline{\Gamma_1(n)\cap\Gamma(m)}]=\tfrac m2\,n^2\prod_{\ell\mid n}(1-\ell^{-2})$, the product being over the primes dividing $n$. We write $i(m,n)$ for this index, and $i(n):=i(1,n)$, so that $i(2,n)=2i(n)$ for even $n$. For $a\in(\torz n)^\times$ we write $\dm a$ for the diamond automorphism $(E,P,Q)\mapsto (E,a^{-1}P,aQ)$ of $X_1(m,n)$, which preserves the Weil pairing. For $m\leq 2$ it is the usual diamond operator $(E,P,Q)\mapsto (E,aP,aQ)$, which is $(E,P)\mapsto(E,aP)$ on $X_1(n)$ and $(E,P,Q)\mapsto(E,P,aQ)$ on $X_1(2,n)$. It depends only on $a$ modulo $\pm1$, and we call $\Delta(n):=(\torz n)^\times/\{\pm1\}$ the group of diamond operators. We write $\Z[\Delta(n)]$ for its group ring. It acts on divisors, on divisor classes and on $J$. For $a_1,\dots,a_r\in(\torz n)^\times$ we write $X/\dm{a_1,\dots,a_r}$ for the quotient of $X$ by the subgroup of $\Delta(n)$ generated by $\dm{a_1},\dots,\dm{a_r}$. For a prime $q\nmid n$ with $q\equiv 1\pmod m$ we write $T_q$ for the Hecke correspondence $T_q(E,P,Q)=\sum_{G}(E/G,P\bmod G,Q\bmod G)$, the sum being over the $q+1$ subgroups $G\subseteq E[q]$ of order $q$, and we set \[ A_q:=T_q-q\dm q-1 . \] The congruence $q\equiv1\pmod m$ ensures that $T_q$ preserves the Weil pairing, and that $q$ splits completely in $\Q(\zeta_m)$. The operator $A_q$ commutes with $\Z[\Delta(n)]$. For a prime $q\mid n$ we write $U_q$ for the Hecke operator at $q$. In characteristic $q$, on the ordinary locus, the quotient by the connected subgroup of order $q$ gives the Frobenius, and the other $q$ subgroups all reduce to the kernel of the Verschiebung and send $(E,P)$ to $\dm q\Frob_q^{-1}(E,P)$. Hence, as correspondences on $X_{\F_q}$ and therefore on its Jacobian, \begin{equation}\label{d9:eq:ES} T_{q,\F_q}=\Frob_q+q\dm q\Frob_q^{-1}=\Frob_q+\dm q\Ver_q. \end{equation} This is the Eichler-Shimura relation (see \cite[(10.2.3)]{DiamondIm}). We will use reduction modulo primes of good reduction also for quotients of $X_1(m,n)$, by the following result of Liu and Lorenzini. \begin{lemma}[{\cite[Corollary 4.10]{LL}}]\label{lem:goodred} Let $f:X\to Y$ be a finite morphism of curves over a number field $F$, and let $\mathfrak p$ be a prime of $F$. Assume that $X$ has good reduction at $\mathfrak p$ and that $Y$ has positive genus. Then $Y$ has good reduction at $\mathfrak p$, and $f$ extends to a finite morphism of the smooth models over the localization of the ring of integers of $F$ at $\mathfrak p$. \end{lemma} In particular, every quotient of $X_1(m,n)$ of positive genus by a group of automorphisms defined over $\Q(\zeta_m)$ has good reduction at the primes above every prime $p\nmid n$, and we use these smooth models for reduction modulo the primes above $p$. \subsection{Trace divisors}\label{subsec:trace} Let $K$ be a number field of degree $d\in\{6,7,8,9\}$ and $E/K$ an elliptic curve with $\tg{m}{n}\subseteq E(K)$. By the Weil pairing $\Q(\zeta_m)\subseteq K$, so $\varphi(m)\mid d$, and we put $d_m:=d/\varphi(m)=[K:\Q(\zeta_m)]$. In particular $m\leq 2$ if $d$ is odd, as $\varphi(m)$ is even for $m>2$. Choosing generators $P,Q$ with the right Weil pairing we obtain a point $x\in Y_1(m,n)(K)$. We write $x^{(d_m)}$ for the sum of the $\Q(\zeta_m)$-conjugates of $x$, each counted $[K:\Q(\zeta_m)(x)]$ times. It is an effective $\Q(\zeta_m)$-rational divisor of degree $d_m$ on $X$ whose support contains no cusp, and we call it the \textit{trace divisor} of $x$, usually denoted by $D$. All the arguments below are applied to trace divisors, so they also cover the case in which the residue field $\Q(\zeta_m)(x)$ is smaller than $K$. If $d=7$, then the degree of $x$ divides $7$, and if $\tg mn\notin\Phi(1)$ then $x$ is not rational by Mazur's theorem \cite{mazur}, so $D$ is a closed point of degree $7$. This will be the case for all the groups that we have to rule out in degree $7$. If $d=9$, then $\Q(x)=K$ for all the groups that we need to exclude (\Cref{d9:rem:exactdegree}), so $D$ is a closed point of degree $9$. Isomorphism of pairs $(K,E)$ is as in \Cref{sec:intro}. \subsection{Hecke operators and reduction}\label{subsec:hecke} \begin{proposition}\label{d6:prop:tools}\label{d7:prop:tools}\label{d8:prop:tools}\label{d9:prop:tools} Let $X$ be $X_1(m,n)$ or its quotient by a subgroup of $\Delta(n)$, and let $q\nmid 2n$ be a prime with $q\equiv1\pmod m$. \begin{enumerate} \item \textup{(Eichler-Shimura)} $T_{q,\F_q}=\Frob_q+\dm q_{\F_q,*}\Ver_q$ on $J_{X,\F_q}$, where $\F_q$ is the residue field of a prime of $\Q(\zeta_m)$ above $q$. \item If $L\supseteq \Q(\zeta_m)$ is a number field in which $q$ splits completely, then $A_q\bigl(J_X(L)_\tors\bigr)=0$. \item $A_q(C)=0$ as a divisor for every $\Q(\zeta_m)$-rational cuspidal divisor $C$. \item If $J_X(\Q(\zeta_m))$ is finite, then $[A_q(D)]=0$ for every $\Q(\zeta_m)$-rational divisor $D$ on $X$. In particular $A_q(D)$ is principal for every $D\in X^{(d)}(\Q(\zeta_m))$ and every $d$. \end{enumerate} \end{proposition} \begin{proof} For $X=X_1(m,n)$ with $m\leq 2$ these are, respectively, \cite[(2.2)]{DN}, \cite[Proposition 2.6]{DN}, \cite[Lemma 2.8]{DN} and \cite[Lemma 5.4]{DN}. For $m\geq 3$ the curve $X_1(m,n)$ is, over $\Q(\zeta_m)$, isomorphic to the quotient of $X_1(mn)$ by the diamond operators congruent to $1$ modulo $n$ \cite[\S 2]{DS}, and this isomorphism is compatible with $T_q$ and $\dm q$ for $q\equiv1\pmod m$. Hence (1) follows from the Eichler-Shimura relation on $X_1(mn)$ \cite[(10.2.3)]{DiamondIm}, and the proofs of \cite[Proposition 2.6, Lemma 2.8 and Lemma 5.4]{DN} carry over verbatim. The same proofs work on the quotients by subgroups of $\Delta(n)$, as $T_q$ and $\dm q$ commute with the diamond operators. Part (3) is also the case $\sigma_q(C)=C$ of \Cref{d8:lem:cuspHecke} below. \end{proof} We also need the action of $T_q$ on all the cusps, not only on the rational ones, and also for $q=2$. \begin{lemma}\label{d8:lem:cuspHecke} Let $X$ be $X_1(m,n)$ or its quotient by a subgroup of $\Delta(n)$, let $q\nmid n$ be a prime with $q\equiv1\pmod m$, and let $\sigma_q\in\Gal(\Q(\zeta_n)/\Q(\zeta_m))$ be the automorphism with $\sigma_q(\zeta_n)=\zeta_n^q$. For every cusp $c$ of $X$ we have \[ T_qc=\sigma_q(c)+q\dm q\sigma_q^{-1}(c) \] as divisors. Hence $A_qC=(\sigma_qC-C)+q\dm q(\sigma_q^{-1}C-C)$ for every cuspidal divisor $C$, and $A_qC=0$ if $\sigma_q(C)=C$. The same holds for the reductions modulo a prime $p\nmid qn$, where $\sigma_q$ acts on the closed cusps through the reduction of the cusps, and in particular $A_qC=0$ for every cuspidal divisor $C$ on $X_{\F_{p^f}}$ if $q\equiv p^{fk}\pmod n$ for some $k$. \end{lemma} \begin{proof} This is the computation in the proof of \cite[Lemma 2.8]{DN}, carried out for an arbitrary cusp. Near a cusp the universal generalized elliptic curve is the Tate curve $\mathrm{Tate}(t)=\mathbb G_m/t^\Z$, and the points of the level structure are of the form $\zeta_n^{a}t^{b/n}$. The subgroups of order $q$ of $\mathrm{Tate}(t)$ are $\mu_q$, with quotient map $x\mapsto x^q$ to $\mathrm{Tate}(t^q)$, and the $q$ subgroups generated by $\zeta_q^it^{1/q}$, with quotient maps induced by the identity of $\mathbb G_m$ to $\mathrm{Tate}(\zeta_q^it^{1/q})$. The first sends $\zeta_n^at^{b/n}$ to $\zeta_n^{qa}(t^q)^{b/n}$, which gives the cusp $\sigma_q(c)$. The others send it to $\zeta_n^{a-ib}u^{qb/n}$ with $u=\zeta_q^it^{1/q}$, and as the label $a$ is only defined modulo multiples of $b$ and $q$ is invertible modulo $n$, all $q$ of them give the same cusp, with labels $(a,qb)$ for every point of the level structure. This is $\dm q\sigma_q^{-1}(c)$, as $\dm q$ multiplies both labels by $q$ (for $m\geq3$ note that $q\equiv1\pmod m$, so $\dm q$ fixes the point of order $m$). Letting $t\to0$ gives the formula on $X_1(m,n)$. As $T_q$, $\dm q$ and $\sigma_q$ commute with the diamond operators, the formula also holds on the quotients. The cuspidal subscheme is finite \'etale over $\Z[\zeta_m,\tfrac1n]$ and $T_q$ extends over it, so the formula reduces modulo $p$, and if $q\equiv p^{fk}\pmod n$ then $\sigma_q$ acts on the geometric cusps as a power of the Frobenius of $\F_{p^f}$, which fixes every closed cusp. \end{proof} Some of our computations use models of $X_1(n)$ and its quotients given by $q$-expansions, which parametrize the dual objects $(E/\dm P,\ \text{the induced }\mu_n\text{-structure})$. The isomorphism between the two models conjugates $\dm a$ to $\dm{a}^{-1}$, and on the dual model the Eichler-Shimura relation reads $T_q=\dm q\Frob_q+q\Frob_q^{-1}$, with annihilator $T_q-\dm q-q$, and the formula of \Cref{d8:lem:cuspHecke} reads $T_qc=\dm q\sigma_q(c)+q\sigma_q^{-1}(c)$. When we use such a model, we test both forms of the relevant divisors, so the results do not depend on the identification of the two conventions. We will also use the following results. \begin{itemize} \item (\textit{Manin-Drinfeld \cite{manin,drinfeld}}) the class of a degree $0$ cuspidal divisor on $X$ is torsion in $J(\Qbar)$; \item (\textit{Katz \cite[Appendix]{katz}}) if $p>2$ is a prime and $A$ is an abelian variety over a number field $L$ with good reduction at a prime $\wp$ of $L$ above $p$ which is unramified in $L$, then reduction modulo $\wp$ is injective on $A(L)_\tors$; \item (\textit{Parent \cite[Proposition 2.3]{parent}}) if $A$ is an abelian variety over a number field $L$ with good reduction at a prime $\lambda$ above $2$ which is unramified in $L$, then the kernel of reduction modulo $\lambda$ on $A(L)_\tors$ is trivial or of exponent $2$. Indeed, this kernel is contained in the group of points of the formal group of the N\'eron model of $A$ at $\lambda$ with coordinates in the maximal ideal, and multiplication by $2$ maps it into the subgroup of points with coordinates in the square of the maximal ideal, which is torsion-free as $\lambda$ is unramified over $2$. We apply this to $J$ for $2\nmid n$ and to the Jacobians of the quotients of $X_1(n)$ of positive genus (see \Cref{lem:goodred}); \item (\textit{Abramovich \cite[proof of Theorem 0.1]{abramovich}, with the bound $\lambda_1\geq 975/4096$ of Kim-Sarnak \cite[Appendix 2]{kim}}) for a congruence subgroup $\Gamma\subseteq\PSL_2(\Z)$, \[ \gon_\C X_\Gamma\geq\tfrac{325}{2^{15}}\,[\PSL_2(\Z):\Gamma]; \] \item (\textit{Kato \cite[Corollary 14.3]{kato}}) if $A$ is a quotient of $J_1(n)$ over $\Q$ with $L(A,1)\neq0$, then $A(\Q)$ is finite; \item (\textit{Clark-Genao-Pollack-Saia \cite{CGPS}}) the exact values of the least degree $d_{\mathrm{CM}}X_1(n)$ of a CM point on $X_1(n)$ that we quote are taken from the list \texttt{dcm\_list\_all\_\allowbreak min\_orders\_\allowbreak X1\_10k.m} in the \href{https://github.com/fsaia/least-cm-degree}{repository} accompanying \cite{CGPS}. \end{itemize} In degree $6$ we \gitlinkVI{global_table.py}{compute} all the values of Abramovich's bound and all the orders of diamond operators that we quote, and in degrees $7$, $8$ and $9$ they are computed by the scripts linked in the corresponding proofs and tables. We use Kato's theorem in the form of the \textit{winding quotient}: we compute, for each Galois orbit of even Dirichlet characters $\chi$ modulo $n$, the winding submodule of the space of modular symbols with character $\chi$, and if it is the whole space, then all the factors with character $\chi$ have finitely many rational points. Finally, we state the standard reduction results. \begin{lemma}\label{d6:lem:additive} Let $K$ be a number field, let $E/K$ be an elliptic curve with $\tg{m}{n}\subseteq E(K)$, and let $\wp$ be a prime of $K$ above a rational prime $p\nmid n$. Then $E$ does not have additive reduction at $\wp$. \end{lemma} \begin{proof} This is contained in the proof of \cite[Proposition 5.2]{DN}. If $E$ had additive reduction at $\wp$, the special fibre of the N\'eron model would have $a\cdot\#\F_\wp$ points over $\F_\wp$, where $a\leq4$ is the order of the component group and $\#\F_\wp$ is a power of $p$. Reduction is injective on the prime-to-$p$ torsion of $E(K)$, so $mn\mid a\leq 4$, contradicting $mn>4$. \end{proof} The following proposition is \cite[Proposition 5.2]{DN}, stated also for quotients of $X_1(m,n)$ and for one prime of $\Q(\zeta_m)$ above $p$ at a time. We include the short proof. \begin{proposition}\label{d6:prop:main}\label{d7:prop:main}\label{d8:prop:main}\label{d9:lem:cuspred} Let $F=\Q(\zeta_m)$, let $p\nmid n$ be a prime, let $X$ be $X_1(m,n)$ or its quotient of positive genus by a subgroup of $\Delta(n)$, and let $d\geq1$. Assume that $J_X(F)$ is finite and that one of the following holds: \begin{enumerate} \item[\textup{(a)}] $p>2$ and $\gon_FX>d$; \item[\textup{(b)}] $p=2$ and $\gon_FX>2d$. \end{enumerate} Then no effective $F$-rational divisor of degree $d$ on $X$ whose support contains no cusp reduces to a cuspidal divisor modulo a prime of $F$ above $p$. In particular, if $X=X_1(m,n)$, then every elliptic curve $E$ over a degree $d$ extension $K$ of $F$ with $\tg{m}{n}\subseteq E(K)$ has, for every prime $\mathfrak p$ of $F$ above $p$, good reduction at at least one prime of $K$ above $\mathfrak p$. \end{proposition} Here $d$ plays the role of $d_m$ in \Cref{subsec:trace}. \begin{proof} Let $D$ be such a divisor, and suppose that its reduction modulo a prime $\mathfrak p$ of $F$ above $p$ is cuspidal. The cusps are defined over $\Q(\zeta_n)$, which is unramified above $p$, so the reduction of $D$ lifts to a cuspidal divisor $C$ over a number field $L\supseteq F$ which is unramified above $p$. The classes $[D-dc_0]$ and $[C-dc_0]$ are torsion, by assumption and by the Manin-Drinfeld theorem, so $[D-C]$ is a torsion point of $J_X(L)$ which reduces to $0$ modulo a prime above $\mathfrak p$. If $p$ is odd, then $[D-C]=0$ by Katz's result, so $D\sim C$ over $L$, and $\ell(D)\geq2$ as $D\neq C$. If $p=2$, then $2[D-C]=0$ by Parent's result, so $2D\sim2C$ over $L$, and $\ell(2D)\geq2$. As $D$ is $F$-rational, $\ell(D)$ and $\ell(2D)$ do not change under extension of scalars from $F$ to $L$, which contradicts $\gon_FX>d$, respectively $\gon_FX>2d$. For the last statement, let $x\in Y_1(m,n)(K)$ be the point given by $E$, and let $D$ be its trace divisor (\Cref{subsec:trace}), which has degree $d$. If $E$ had bad reduction at all the primes of $K$ above $\mathfrak p$, then by \Cref{d6:lem:additive} it would have multiplicative reduction at all of them, so all the geometric points of $D$ would reduce to cusps modulo the primes above $\mathfrak p$, and the reduction of $D$ modulo $\mathfrak p$ would be cuspidal. \end{proof} \begin{lemma}[{\cite[Lemma 5.3]{DN}}]\label{d6:lem:badred} Let $p\nmid n$ be a prime and let $K\supseteq \Q(\zeta_m)$ be a number field. Suppose that for every prime $\wp$ of $K$ above $p$ there is no elliptic curve $E'$ over the residue field $\F_\wp$ with $\tg{m}{n}\subseteq E'(\F_\wp)$. Then no elliptic curve $E/K$ with $\tg{m}{n}\subseteq E(K)$ has good reduction at a prime of $K$ above $p$. \end{lemma} Whenever we invoke \Cref{d6:lem:badred}, the hypothesis follows from the Hasse bound, from Waterhouse's description \cite[Theorem 4.1]{waterhouse} of the possible traces of Frobenius, or from an explicit enumeration of all elliptic curves over the relevant finite fields, which we carry out separately in each degree (the scripts for the degrees \gitlinkVI{torsion_over_Fq.m}{$6$}, \gitlinkVII{finite_fields_2_degree7.m}{$7$}, \gitlinkVIII{large_levels/finite_fields_2_octic.m}{$8$} and \gitlinkIX{common/finite_field_prime_orders.m}{$9$} are linked). For example, over $\F_{64}$ the possible traces of Frobenius are the odd integers $t$ with $|t|\leq 16$ together with $t\in\{0,\pm8,\pm16\}$, so no elliptic curve over $\F_{64}$ has exactly $51$, $55$, $63$, $75$ or $77$ points. If $\#E(\F_{64})=81$, then the Frobenius $\pi$ satisfies $(\pi+8)^2=0$, so $\pi=-8$ as $\End(E)$ has no non-zero nilpotents, and $E(\F_{64})=\ker(1-\pi)=E[9]$. Similarly, if $\#E(\F_{64})=49$, then $\pi=8$ and $E(\F_{64})=E[7]$. In general, an elliptic curve over $\F_{2^e}$ with even trace of Frobenius is supersingular, so its trace is $0$ or $\pm2^{(e+1)/2}$ for odd $e$, and $0$, $\pm2^{e/2}$ or $\pm2^{e/2+1}$ for even $e$ \cite[Theorem 4.1]{waterhouse}. \subsection{Divisors and level automorphisms}\label{subsec:divisors} \begin{lemma}\label{d8:lem:closed}\label{d9:lem:closed}\label{d7:lem:closed} Let $X$ be a curve over a field $F$, and let $D$ and $A$ be effective $F$-rational divisors of degree $d$ on $X$ with $A\sim D$. \begin{enumerate} \item If $\gon_F X>d$, then $A=D$. \item If $D$ is a closed point of degree $d$ and $A\neq D$, then there is a function on $X$ over $F$ of degree exactly $d$. \end{enumerate} \end{lemma} \begin{proof} If $A\neq D$, a function $f$ with $\ddiv(f)=A-D$ is non-constant, and its degree is at most $d$. This proves (1). For (2), if $D$ were contained in the support of $A$, then $A=D$ as both have degree $d$. So the supports of $A$ and $D$ are disjoint, and a function with divisor $D-A$ has degree exactly $d$. \end{proof} An automorphism of $X_1(m,n)$ of the form $(E,P,Q)\mapsto(E,P',Q')$, where $P',Q'$ are integral linear combinations of $P$ and $Q$, will be called a \textit{level automorphism}. The diamond operators are level automorphisms. We will often use the following simple lemma. For subgroups of $\Delta(n)$, its first part is the argument at the end of the proof of \cite[Proposition 7.2]{N5}. \begin{lemma}\label{lem:orbits} Let $G$ be a finite group of level automorphisms of $X_1(m,n)$, and let $D$ be a divisor on $X_1(m,n)$ whose support consists of non-cuspidal points without complex multiplication. Then no element $g\neq1$ of $G$ fixes a geometric point in the support of $D$. If moreover $gD=D$ for all $g\in G$, then the geometric points in the support of $D$ form a union of $G$-orbits of length $\#G$, and $D=\sum_O m_O\sum_{x\in O}x$, where $O$ runs over these orbits. In particular $\#G\mid\deg D$. \end{lemma} \begin{proof} If $g$ fixes a geometric point $(E,P,Q)$ in the support of $D$, then there is an automorphism $u$ of $E$ with $u(P,Q)=g(P,Q)$. As $E$ has no complex multiplication, $u=\varepsilon$ with $\varepsilon=\pm1$, so $g(P,Q)=(\varepsilon P,\varepsilon Q)$. As $g$ is given by integral linear combinations which do not depend on the point, and $P,Q$ generate a group isomorphic to $\tg mn$, it follows that $g$ maps every triple $(E',P_1,Q_1)$ to $(E',\varepsilon P_1,\varepsilon Q_1)$, which is isomorphic to $(E',P_1,Q_1)$. So $g$ is the identity of $X_1(m,n)$. If $gD=D$ for all $g\in G$, then $D$ has the same coefficient at $x$ and at $gx$, so the geometric points in the support of $D$ form a union of $G$-orbits and the coefficients of $D$ are constant on each of them. By the first part, each orbit has length $\#G$. \end{proof} We will also often use the following lemma, whose proof is the same as that of \Cref{d6:prop:main}. \begin{lemma}\label{d9:lem:stab} Let $F=\Q(\zeta_m)$, let $p\nmid n$ be a prime, and let $X$ be $X_1(m,n)$ or its quotient of positive genus by a subgroup of $\Delta(n)$. Let $g$ be a level automorphism of $X$ defined over $F$, which maps cusps to cusps and extends to the smooth model of $X$ over the primes above $p$, such that $(g-1)J_X(F)$ is finite. Let $D$ be an effective $F$-rational divisor of degree $d$ on $X$, and let $C$ be an effective cuspidal divisor of degree $e$ defined over a number field $L\supseteq F$ which is unramified above $p$, such that $g$ fixes the reduction of $D-C$ modulo a prime above $p$. Assume that one of the following holds: \begin{enumerate} \item[\textup{(a)}] $p$ is odd, $C$ is $F$-rational and $\gon_F X>d+e$, \item[\textup{(b)}] $p$ is odd and $\gon_\C X>d+e$, \item[\textup{(c)}] $p=2$ and $\gon_\C X>2(d+e)$, \item[\textup{(d)}] $p=2$, $C$ is $F$-rational, $g$ has odd order and acts trivially on $J_X(F)[2]$, and $\gon_F X>d+e$. \end{enumerate} Then $gD+C=D+gC$. In particular, if the support of $D$ contains no cusp, then $gD=D$ and $gC=C$. \end{lemma} \begin{proof} The class $y:=[gD-D-gC+C]\in J_X(L)$ equals $(g-1)[D-dc_0]-(g-1)[C-ec_0]+(d-e)[gc_0-c_0]$, so it is torsion, by the assumption on $g$ and by the Manin-Drinfeld theorem. It reduces to $0$, so $y=0$ if $p$ is odd and $2y=0$ if $p=2$. In case (d) the class $y$ lies in $J_X(F)[2]$, so $gy=y$, and if $r$ is the order of $g$, then $ry=(1+g+\dots+g^{r-1})y$ is the class of the divisor $(g^r-1)(D-C)=0$. As $r$ is odd and $2y=0$, it follows that $y=0$. Hence $gD+C\sim D+gC$ in cases (a), (b) and (d), and $2gD+2C\sim2D+2gC$ in case (c), over $L$, and over $F$ in cases (a) and (d). \Cref{d9:lem:closed} gives the claim. \end{proof} Every level automorphism of $X$ defined over $F$ satisfies the assumption on $g$ if $J_X(F)$ is finite. For $m=1$ and $g=\dm a$ the assumption holds if all the factors of $J_X$ on which $\dm a$ acts non-trivially have finitely many rational points, in particular if every character $\chi$ for which the winding submodule is not the whole space satisfies $\chi(a)=1$. We also use the following form of the Mordell-Weil sieve with quadratic characters. \begin{lemma}\label{d9:lem:quadchar} Let $C$ be a smooth projective curve over $\F_q$ with $q$ odd, and let $f$ be a function on $C$ with $\ddiv(f)=2A$ for a divisor $A$. For a divisor $Z=\sum n_PP$ whose support is disjoint from that of $\ddiv(f)$ put $\chi_f(Z)=\prod_P N_{\F_q(P)/\F_q}(f(P))^{n_P}$ modulo $\F_q^{\times2}$. Then $\chi_f$ induces a homomorphism $\chi_f:\Pic^0(C)\to\F_q^\times/\F_q^{\times2}$, which is trivial on every subgroup of odd order. \end{lemma} \begin{proof} By Weil reciprocity, for a function $g$ whose divisor is disjoint from that of $f$ we have $\chi_f(\ddiv g)=g(\ddiv f)=g(A)^2$, which is a square. So $\chi_f$ is trivial on principal divisors, and by the moving lemma it induces a homomorphism on $\Pic^0(C)$, whose values have order dividing $2$. \end{proof} \subsection{Complex multiplication and fields of odd degree}\label{subsec:odd} \begin{lemma}\label{d7:lem:CM}\label{d8:lem:CM}\label{d9:lem:CM} Let $E$ be an elliptic curve with complex multiplication over a number field $K$ of degree $d$. \begin{enumerate} \item If $d=7$, then $E(K)_\tors$ is isomorphic to one of $\{0\}$, $\torz2$, $\torz3$, $\torz4$, $\torz6$ and $\tg22$. \item If $d=8$, then $E(K)_\tors$ belongs to $\Phi^\infty(8)\cup\{\torz{34},\torz{39},\tg{5}{10}\}$. \item If $d=9$, then $E(K)_\tors$ is isomorphic to $\torz n$ with $n\in\{1,2,3,4,6,9,14,18,19,27\}$ or to $\tg22$. In particular it lies in $\Phi^\infty(9)$. \end{enumerate} \end{lemma} \begin{proof} Part (1) follows from the Strong Odd Degree Theorem of Bourdon and Pollack \cite[Theorem 1.2]{BP}. Apart from the four groups $\{0\}$, $\torz2$, $\torz4$ and $\tg22$, the possible groups are $\torz{\ell^k}$ and $\torz{2\ell^k}$ for primes $\ell\equiv3\pmod4$, and these occur in degree $7$ only if $7$ is divisible by $\tfrac{\ell-1}2$ times the class number of $\Q(\sqrt{-\ell})$. This leaves $\ell=3$, and for $\ell=3$ and $k\geq2$ the degree has to be divisible by $3$. For (2), by \cite[Section 4.8]{CCRS14}, $E(K)_\tors$ is isomorphic to $\torz n$ with \[ n\in\{1,\dots,8,10,12,13,15,16,20,21,28,30,34,39\}, \] to $\tg{2}{n}$ with $n\in\{2,4,6,8,10,12,16,20\}$, to $\tg{4}{n}$ with $n\in\{4,8,12\}$, or to one of $\tg33$, $\tg36$, $\tg55$, $\tg{5}{10}$ and $\tg66$. Comparing with \eqref{d8:eq:phiinf8} proves (2). Part (3) is \cite[Section 4.9]{CCRS14}. \end{proof} Let $d\in\{7,8,9\}$, and let $\mathcal G_d$ be the union of $\Phi^\infty(d)$ and the sporadic groups of \Cref{thm:main} in degree $d$, which is closed under passing to subgroups (see \Cref{sec:groups}). By \Cref{d7:lem:CM}, if $E(K)$ contains a subgroup which is not in $\mathcal G_d$, then $E$ does not have complex multiplication, so in degrees $7$, $8$ and $9$ \Cref{lem:orbits} applies to all the trace divisors that we consider. For fields of odd degree we have the following stronger statement, which we use in degree $9$. \begin{lemma}\label{d9:lem:odd} Let $K$ be a number field of odd degree and let $E/K$ be an elliptic curve. Then $\End_K(E)=\Z$ and $\Aut_K(E)=\{\pm1\}$. Let $x$ be a point of odd degree on $Y_1(m,n)$ and let $g$ be a level automorphism of $X_1(m,n)$ which is not the identity. Then $g$ fixes no geometric point above $x$. In particular, if $D$ is an effective divisor whose support consists of non-cuspidal points of odd degree and $gD=D$, then every orbit of the group generated by $g$ on the geometric points in the support of $D$ has length equal to the order of $g$, so the order of $g$ divides $\deg D$. \end{lemma} \begin{proof} An endomorphism of $E$ acts on the invariant differential by multiplication by an element of $K$, which gives an injection $\End_K(E)\to K$. The image of an endomorphism outside $\Z$ would generate an imaginary quadratic subfield of $K$, which is impossible as $[K:\Q]$ is odd. If $g$ fixes a geometric point $(E,P,Q)$ above $x$, then there is an automorphism $\alpha$ of $E$ with $\alpha(P,Q)=g(P,Q)$. It is unique, as $X_1(m,n)$ is a fine moduli space, so it is defined over the residue field $\Q(x)$, and it is not $\pm1$, as $g$ is not the identity. This contradicts the first part applied to $E/\Q(x)$. The same applies to every non-trivial power of $g$, which proves the last assertion. \end{proof} We will also need the following special case of a result of Derickx and Stoll. \begin{lemma}[{\cite[Proposition 2.2]{DSt}}]\label{d9:lem:tree} Let $X$ be $X_1(n)$ or its quotient by a subgroup of $\Delta(n)$, and let $q\nmid n$ be a prime. Let $L$ be a divisor on $X_\C$ whose support consists of non-cuspidal points at which the elliptic curve has no complex multiplication. If $A_qL=0$, then $L=0$. \end{lemma} Here $A_q=h_1(T_q)-h_2(T_q)$ for the polynomials $h_1(x)=x$ and $h_2(x)=1+q\dm q$, in the notation of \cite{DSt}. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{The groups that have to be considered}\label{sec:groups}\label{d6:sec:groups}\label{d7:sec:groups}\label{d8:sec:groups}\label{d9:sec:groups} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Every finite subgroup of $E(K)$ is isomorphic to $\tg{m}{n}$ with $m\mid n$. For $d\in\{6,7,8,9\}$ let $\mathcal G_d$ be the union of $\Phi^\infty(d)$ and the sporadic groups in degree $d$ listed in \Cref{thm:main}. Each $\mathcal G_d$ is closed under passing to subgroups, and by the realizations in \Cref{sec:sporadic} it is contained in $\Phi(d)$. Recall from \Cref{sec:intro} that $S(6)=\{2,3,5,7,11,13,17,19,37\}$, that $S(7)$ and $S(8)$ are the sets of primes up to $23$, and that $S(9)$ is the set of primes up to $31$ \cite{Sd}. \begin{lemma}\label{lem:minimal}\label{d6:lem:66}\label{d7:lem:58}\label{d8:lem:80}\label{d9:lem:94} For $d\in\{6,7,8,9\}$ let $\mathcal M_d$ be the set of groups $\tg mn$ with $m\mid n$ such that $\varphi(m)\mid d$, every prime dividing $n$ lies in $S(d)$, $\tg mn\notin\mathcal G_d$, and every proper subgroup of $\tg mn$ lies in $\mathcal G_d$. Then $\mathcal M_6$ consists of the following $66$ groups: \[ \begin{array}{l} \torz{n}\ \text{ for }\ n\in\{32,33,34,35,36,38,39,40,42,44,45,48,49,50,\\ \qquad\quad 51,52,54,55,56,57,60,63,65,75,77,81,85,91,95,\\ \qquad\quad 119,121,125,133,143,169,187,209,221,247,289,323,361\},\\[2pt] \torz{37n}\ \text{ for }\ n\in\{2,3,5,7,11,13,17,19,37\},\\[2pt] \tg{2}{2n}\ \text{ for }\ n\in\{11,13,14,15\},\\[2pt] \tg{3}{3n}\ \text{ for }\ n\in\{5,6,7,8,9\},\\[2pt] \tg{4}{4n}\ \text{ for }\ n\in\{4,5,6\},\\[2pt] \tg{6}{12},\quad \tg{7}{7},\quad \tg{9}{9}, \end{array} \] $\mathcal M_7$ consists of the following $58$ groups: \[ \begin{array}{l} \torz{n}\ \text{ for }\ n\in\{32,34,35,38,39,40,42,44,45,46,48,49,50,51,52,54,55,56,57,60,\\ \qquad\quad 63,65,66,69,72,75,77,81,85,91,95,99,115,119,121,125,133,143,161,\\ \qquad\quad 169,187,209,221,247,253,289,299,323,361,391,437,529\},\\[2pt] \tg{2}{2n}\ \text{ for }\ n\in\{11,12,13,14,15,18\}, \end{array} \] $\mathcal M_8$ consists of the following $80$ groups: \[ \begin{array}{l} \torz{n}\ \text{ for }\ n\in\{38,44,45,46,48,49,50,51,52,54,55,56,57,60,63,\\ \qquad\quad 64,65,66,68,69,70,72,75,77,78,80,81,84,85,91,95,99,105,\\ \qquad\quad 115,117,119,121,125,133,143,161,169,175,187,209,221,247,\\ \qquad\quad 253,289,299,323,361,391,437,529\},\\[2pt] \tg{2}{2n}\ \text{ for }\ n\in\{13,14,15,16,17,18,20,21\},\\[2pt] \tg{3}{3n}\ \text{ for }\ n\in\{6,7,8,9,10,11,13\},\\[2pt] \tg{4}{4n}\ \text{ for }\ n\in\{4,5,6\},\\[2pt] \tg{5}{5n}\ \text{ for }\ n\in\{3,4,5,7\},\\[2pt] \tg{6}{12},\quad \tg{8}{8},\quad \tg{10}{10}, \end{array} \] and $\mathcal M_9$ consists of the following $81$ groups: \[ \begin{array}{l} \torz{n}\ \text{ for }\ n\in\{35,38,44,45,46,48,49,50,51,52,54,55,56,57,58,60,62,63,64,65,\\ \qquad\quad 66,68,69,72,75,77,78,80,81,84,85,87,91,93,95,99,115,117,119,121,125,\\ \qquad\quad 133,143,145,155,161,169,187,203,209,217,221,247,253,289,299,319,323,\\ \qquad\quad 341,361,377,391,403,437,493,527,529,551,589,667,713,841,899,961\},\\[2pt] \tg{2}{2n}\ \text{ for }\ n\in\{14,15,16,17,18,20,21\}. \end{array} \] Every group in $\Phi(d)\setminus\mathcal G_d$ contains a member of $\mathcal M_d$. \end{lemma} \begin{proof} By the Weil pairing and by the description of $S(d)$ above, every group in $\Phi(d)$ is of the form $\tg mn$ with $\varphi(m)\mid d$ and with all the primes dividing $n$ in $S(d)$. As $\mathcal G_d$ is closed under passing to subgroups, a group in $\Phi(d)\setminus\mathcal G_d$ contains a subgroup of minimal order which is not in $\mathcal G_d$, and this subgroup lies in $\mathcal M_d$. If $\torz n\in\mathcal M_d$ then $\Z/(n/p)\Z\in\mathcal G_d$ for every prime $p\mid n$, so $n$ is at most the largest element of $S(d)$ times the largest $n'$ with $\torz{n'}\in\mathcal G_d$, and if $\tg mn\in\mathcal M_d$ with $m>1$ then $\torz n\in\mathcal G_d$. It remains to run through these finitely many groups, which we do by computer in each degree (the scripts for \gitlinkVI{group_list.py}{$d=6$}, \gitlinkVII{group_list.py}{$d=7$}, \gitlinkVIII{common/structural_completeness_complete_20261026.py}{$d=8$} and \gitlinkIX{common/structural_completeness_complete_20261027.py}{$d=9$} are linked). The computation for $d=9$ uses the primes up to $89$ instead of $S(9)$, and gives in addition the $13$ groups $\torz p$ with $37\leq p\leq 89$, which are not in $\mathcal M_9$ as $p\notin S(9)$. \end{proof} \begin{remark}\label{d9:rem:exactdegree} No group in $\Phi(1)\cup\Phi(3)$ has a subgroup isomorphic to a group in $\mathcal M_9$ \cite{mazur,DEHMZ}, so if $E/K$ has a subgroup isomorphic to a group in $\mathcal M_9$ and $[K:\Q]=9$, the corresponding point $x$ has $\Q(x)=K$, and the trace divisor is a closed point of degree $9$. The same holds for the points on $X_1(33)$, $X_1(39)$ and $X_1(2,26)$ of degree $9$ that we consider in \Cref{sec:sporadic,sec:noncyclic}. \end{remark} To rule out a group $\tg mn\in\mathcal M_d$ we show that no effective $\Q(\zeta_m)$-rational divisor of degree $d_m$ on $X_1(m,n)$ with non-cuspidal support is the trace divisor (\Cref{subsec:trace}) of a point $x\in Y_1(m,n)(K)$ with $[K:\Q]=d$. Several members of $\mathcal M_d$ are reduced to others by a $2$-isogeny. \begin{lemma}\label{d6:lem:4n}\label{d7:lem:4n}\label{d8:lem:4n}\label{d9:lem:4n} Let $k\geq 2$. There is a $\Q$-rational morphism $X_1(4k)\to X_1(2,2k)$, $(E,P)\mapsto \bigl(E/\dm{2kP},\,Q,\,P\bmod \dm{2kP}\bigr)$, where $Q$ generates $E[2]/\dm{2kP}$. Hence, if $\tg{2}{2k}$ is not a subgroup of the torsion of an elliptic curve over a number field $K$, then neither is $\torz{4k}$. \end{lemma} \begin{proof} The morphism is the one described in \cite[\S 2]{DN} (see also \cite[\S 2]{DS}). It maps non-cuspidal points to non-cuspidal points, and the image of a point defined over $K$ is defined over $K$, as it only uses the rational $2$-isogeny $E\to E/\dm{2kP}$. \end{proof} In each degree \Cref{d6:lem:4n} reduces $\torz{4k}$ to $\tg{2}{2k}$ whenever the latter is in $\mathcal M_d$. It does not apply to $\torz{48}$ for $d=6,8,9$ and to $\torz{52}$ for $d=9$, as $\tg{2}{24}$, respectively $\tg{2}{26}$, lies in $\Phi(d)$, and we treat these groups in \Cref{d6:prop:48,d9:prop:48,d9:prop:52}. \Cref{sec:proof} lists where each of the groups in $\mathcal M_d$ is excluded. \section{The global method and diamond operators}\label{sec:global}\label{d6:sec:global}\label{d7:sec:global}\label{d8:sec:global}\label{d9:sec:global} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Most of the groups $\torz n$ with large $n$ are excluded by the global method of Derickx \cite[Proposition 4.3]{DN} (\Cref{subsec:global}) and by its version at the prime $2$, in which we use several diamond operators at once (\Cref{subsec:several}). The group $\torz{81}$ is excluded in all four degrees in \Cref{subsec:81}, and the groups $\torz{133}$ and $\torz{247}$ in degree $9$ need further arguments, which we give in \Cref{subsec:133_247}. Throughout the section we use the following consequence of Kato's theorem, as in \cite[\S 7 and the proof of Proposition 4.1]{N5}. The isogeny decomposition \[ J_1(n)\sim\bigoplus_{n'\mid n}\bigoplus_{[f]}A_{[f]}^{\sigma_0(n/n')}, \] where the inner sum is over the Galois orbits of newforms of weight $2$ and level $n'$, is equivariant for the diamond operators, $\dm a$ acting on $A_{[f]}$ through $\chi_f(a)$, where $\chi_f$ is the nebentypus of $f$, and $A_{[f]}(\Q)$ is finite if $L(f,1)\neq0$ by Kato's theorem. Hence, for $\psi=\sum_a c_a\dm a\in\Z[\Delta(n)]$, the group $\psi\bigl(J_1(n)(\Q)\bigr)$ is finite if $\sum_a c_a\chi_f(a)=0$ for every newform $f$ of level dividing $n$ with $L(f,1)=0$. All the analytic ranks that we take from the LMFDB \cite{lmfdb} are recorded there as proved. We \gitlinkVI{lmfdb_ranks.py}{check} them and the character values for the levels that occur in degree $6$, and the scripts that check them in the other degrees are linked in the proofs below. \subsection{The global method}\label{subsec:global} The main tool is the global method of Derickx \cite[Proposition 4.3]{DN}, which we recall for a number field of degree $d$, as in \cite[Section 4]{N5}. Let $n$ be an integer, let $q\nmid n$ be a prime and let $a\in(\torz n)^\times$ be such that $(\dm a-1)J_1(n)(\Q)$ is finite. If $q=2$ assume moreover that $(\dm a-1)J_1(n)(\Q)$ has odd order. Put $k_q=2q+1$ if $a\equiv\pm q^{\pm1}\pmod n$ and $k_q=2(q+1)$ otherwise. If $dk_q<\gon_\Q X_1(n)$, then every effective rational divisor of degree $d$ on $X_1(n)$ whose support contains no cusp is a sum of orbits under $\dm a$. Using Abramovich's bound, the inequality holds as soon as \[ b(n):=\frac{325}{2^{15}}\cdot\frac{i(n)}{k_q}>d,\qquad\text{where }i(n)=[\PSL_2(\Z):\overline{\Gamma_1(n)}]. \] For $q=2$ the proof of \cite[Proposition 4.3]{DN} applies without change, by \Cref{d8:lem:cuspHecke}, which allows $q=2$ as $n$ is odd, and as reduction modulo $2$ is injective on the torsion of odd order by Parent's result. We use $q=3$, except for $n=115$ in degree $9$, where we use $q=2$. For the levels $115$, $161$, $253$, $299$, $391$, $437$ and $529$ we obtain the finiteness of $(\dm a-1)J_1(n)(\Q)$ from the following lemma, which we also use for $n=99$ in \Cref{prop:77_99}. \begin{lemma}\label{d7:lem:winding} Let $n\in\{99,115,161,253,299,391,437,529\}$. Then $L(f,1)\neq0$ for every newform $f$ of weight $2$ and level dividing $n$ with non-trivial character. In particular, $(\dm a-1)J_1(n)(\Q)$ is finite for every $a\in(\torz n)^\times$. \end{lemma} \begin{proof} For each $n$ and each Galois orbit of non-trivial even Dirichlet characters $\chi$ modulo $n$, we \gitlinkVII{winding_probe.m}{compute}, using modular symbols, the plus quotient $S_\chi$ of the space of cusp forms of weight $2$ and character $\chi$, and its winding submodule $W_\chi$, the Hecke module generated by the projection of the winding element. In every case $W_\chi=S_\chi$. For $n=529$ and $\chi$ of order $253$ we \gitlinkVII{winding_probe_krylov_modp_v2.m}{verify} this instead by computing the projection $v$ of the winding element and the operator $T_2$, and checking that $v,vT_2,\dots,vT_2^{44}$ have rank $45=\dim S_\chi$ modulo the prime $2531$, which splits completely in the field of values of $\chi$ and at which all the denominators are units. As a check that no factor is missed, the sum of $\varphi(\ord\chi)\dim S_\chi$ over all orbits, including the trivial character, equals the genus of $X_1(n)$, which is $281$, $441$, $925$, $2421$, $3433$, $5985$, $7525$ and $11265$, respectively. It follows that $L(f,1)\neq0$ for every newform $f$ of level dividing $n$ with non-trivial character. As $\dm a-1$ kills the factors of $J_1(n)$ with trivial character, Kato's theorem shows that $(\dm a-1)J_1(n)(\Q)$ is finite. \end{proof} \begin{proposition}\label{prop:global}\label{d6:prop:global}\label{d7:prop:global}\label{d8:prop:global}\label{d9:prop:global} Let $n$ and $d$ be as in one of the rows of \Cref{table:global}. Then the group $\torz n$ is not a subgroup of the torsion of an elliptic curve over a number field of degree $d$. \end{proposition} \begin{proof} For each row we take $q$ and $a$ as in \Cref{table:global}, and we first show that $(\dm a-1)J_1(n)(\Q)$ is finite, and that it has odd order in the row with $q=2$. For the $11$ levels of the rows with $d=6,\dots,9$ we use the elements $a$ from \cite[Table 2]{N5}. For $n\in\{119,121,125,169,221,289,361\}$ every newform of weight $2$ and level dividing $n$ with positive analytic rank has trivial character, and for the remaining $n$ the newforms of positive analytic rank and non-trivial character have character orbits \lmfdcharorbit{143}{h}, \lmfdcharorbit{187}{g}, \lmfdcharorbit{209}{f} and \lmfdcharorbit{323}{k}, and $a$ lies in the kernel of each of them. So $(\dm a-1)J_1(n)(\Q)$ is finite, as is also shown in the proofs of \cite[Propositions 4.1 and 4.2]{N5}. For $n=133$ and $n=247$ the newforms of positive analytic rank have trivial character or a character in one of the orbits listed in \Cref{d6:table:large}, and these characters are trivial on $a$: for $n=133$ as $64=2^6$ and the characters have order $3$ or $6$, and for $n=247$ as $144=12^2$ and the characters are trivial on $12$. We \gitlinkVI{global_table.py}{check} the entries of \Cref{table:global} in the rows with $d=6,\dots,9$. An \gitlinkVII{certify_inherited_large.py}{LMFDB search} confirms the data above for these $13$ levels, and that all the other newforms of these levels have analytic rank $0$. For $n\in\{115,161,253,299,391,437,529\}$ and $a=3$ the finiteness follows from \Cref{d7:lem:winding}. For the $17$ levels that occur only in degree $9$ we compute (with \githubbareIX{common/winding_probe.m}, and with \githubbareIX{large_levels/winding_probe_krylov.m} for the levels $551$, $667$, $713$, $841$, $899$ and $961$) that the winding submodule is the whole space of modular symbols with character $\chi$ for every Galois orbit of even characters $\chi$ modulo $n$ with $\chi(a)\neq1$, and we \gitlinkIX{large_levels/check_nonic_characters.m}{check} the character values. For the characters of large order at the levels $841$, $899$ and $961$ (and again at the level $529$) we check this modulo a suitable prime, as in the proof of \Cref{d7:lem:winding}. For $n=115$ and $d=9$ we use $q=2$ and $a=16$. Then $(\dm{16}-1)J_1(115)$ is isogenous to the part of $J_1(115)$ with characters of order $11$, $22$ and $44$, which has dimension $400$, and it has finitely many rational points by \Cref{d7:lem:winding}. For each of these three character orbits we compute the norm of $\det\bigl((1+3\chi(3))-T_3\bigr)$ on the corresponding space of cusp forms, which is a power of the number of $\F_3$-points of the corresponding factor, and we \gitlinkIX{large_levels/C115_image_even_orders_certified.m}{find} that all three are odd. Hence $(\dm{16}-1)J_1(115)(\Q)$ has odd order, as it injects into the group of $\F_3$-points. By \Cref{table:global} we have $b(n)>d$ in every row, so the trace divisor $D$ of a point of order $n$ over a number field of degree $d$ is a sum of $\dm a$-orbits. Its support consists of non-CM points, for $d=6$ as $d_{\mathrm{CM}}X_1(n)>6$, and for $d\geq7$ by \Cref{d7:lem:CM}. So by \Cref{lem:orbits} each of these orbits has length $\ord\dm a$, the order of $\dm a$ in $\Delta(n)$, and $\ord\dm a$ divides $d$, which contradicts \Cref{table:global}. \end{proof} \begin{table}[H] \centering \small \begin{tabular}{rcccrcrrr} \toprule $n$ & $d$ & $q$ & $a$ & $\ord\dm a$ & $k_q$ & $i(n)$ & $b(n)$ & $d_{\mathrm{CM}}X_1(n)$\\ \midrule $115$ & $7,8$ & $3$ & $3$ & $44$ & $7$ & $6336$ & $8.97$ & \\ $115$ & $9$ & $2$ & $16$ & $11$ & $6$ & $6336$ & $10.47$ & \\ $119$ & $6$--$9$ & $3$ & $3$ & $48$ & $7$ & $6912$ & $9.79$ & $96$\\ $121$ & $6$--$9$ & $3$ & $2$ & $55$ & $8$ & $7260$ & $9.00$ & $110$\\ $125$ & $6$--$9$ & $3$ & $3$ & $50$ & $7$ & $7500$ & $10.62$ & $50$\\ $133$ & $7,8$ & $3$ & $64$ & $3$ & $8$ & $8640$ & $10.71$ & \\ $143$ & $6$--$9$ & $3$ & $67$ & $12$ & $8$ & $10080$ & $12.49$ & $120$\\ $145$ & $9$ & $3$ & $3$ & $14$ & $7$ & $10080$ & $14.28$ & \\ $155$ & $9$ & $3$ & $16$ & $5$ & $8$ & $11520$ & $14.28$ & \\ $161$ & $7$--$9$ & $3$ & $3$ & $66$ & $7$ & $12672$ & $17.95$ & \\ $169$ & $6$--$9$ & $3$ & $3$ & $39$ & $7$ & $14196$ & $20.11$ & $52$\\ $187$ & $6$--$9$ & $3$ & $122$ & $16$ & $8$ & $17280$ & $21.42$ & $160$\\ $203$ & $9$ & $3$ & $6$ & $7$ & $8$ & $20160$ & $24.99$ & \\ $209$ & $6$--$9$ & $3$ & $34$ & $18$ & $8$ & $21600$ & $26.77$ & $180$\\ $217$ & $9$ & $3$ & $3$ & $15$ & $7$ & $23040$ & $32.64$ & \\ $221$ & $6$--$9$ & $3$ & $3$ & $48$ & $7$ & $24192$ & $34.27$ & $96$\\ $247$ & $7,8$ & $3$ & $144$ & $3$ & $8$ & $30240$ & $37.49$ & \\ $253$ & $7$--$9$ & $3$ & $3$ & $55$ & $7$ & $31680$ & $44.88$ & \\ $289$ & $6$--$9$ & $3$ & $3$ & $136$ & $7$ & $41616$ & $58.96$ & $136$\\ $299$ & $7$--$9$ & $3$ & $3$ & $33$ & $7$ & $44352$ & $62.84$ & \\ $319$ & $9$ & $3$ & $3$ & $140$ & $7$ & $50400$ & $71.41$ & \\ $323$ & $6$--$9$ & $3$ & $101$ & $18$ & $8$ & $51840$ & $64.27$ & $288$\\ $341$ & $9$ & $3$ & $32$ & $2$ & $8$ & $57600$ & $71.41$ & \\ $361$ & $6$--$9$ & $3$ & $3$ & $171$ & $7$ & $64980$ & $92.06$ & $114$\\ $377$ & $9$ & $3$ & $5$ & $28$ & $8$ & $70560$ & $87.47$ & \\ $391$ & $7$--$9$ & $3$ & $3$ & $176$ & $7$ & $76032$ & $107.72$ & \\ $403$ & $9$ & $3$ & $92$ & $2$ & $8$ & $80640$ & $99.97$ & \\ $437$ & $7$--$9$ & $3$ & $3$ & $198$ & $7$ & $95040$ & $134.66$ & \\ $493$ & $9$ & $3$ & $3$ & $112$ & $7$ & $120960$ & $171.38$ & \\ $527$ & $9$ & $3$ & $3$ & $240$ & $7$ & $138240$ & $195.87$ & \\ $529$ & $7$--$9$ & $3$ & $3$ & $253$ & $7$ & $139656$ & $197.87$ & \\ $551$ & $9$ & $3$ & $12$ & $12$ & $8$ & $151200$ & $187.45$ & \\ $589$ & $9$ & $3$ & $3$ & $45$ & $7$ & $172800$ & $244.83$ & \\ $667$ & $9$ & $3$ & $3$ & $308$ & $7$ & $221760$ & $314.20$ & \\ $713$ & $9$ & $3$ & $2$ & $55$ & $8$ & $253440$ & $314.20$ & \\ $841$ & $9$ & $3$ & $12$ & $58$ & $8$ & $353220$ & $437.91$ & \\ $899$ & $9$ & $3$ & $3$ & $420$ & $7$ & $403200$ & $571.28$ & \\ $961$ & $9$ & $3$ & $3$ & $465$ & $7$ & $461280$ & $653.58$ & \\ \bottomrule \end{tabular} \medskip \caption{The pairs $(n,d)$ of \Cref{prop:global}, the group $\torz n$ being excluded in all the degrees $d$ in the second column. The order of $\dm a$ is its order in $\Delta(n)$, and $b(n)$ is rounded down to two decimal places (for $n=121$ we have $b(n)=9.0007\ldots$). The least degree $d_{\mathrm{CM}}X_1(n)$ of a CM point on $X_1(n)$ is given only in the rows with $d=6,\dots,9$, as we use it only for $d=6$.} \label{table:global}\label{d6:table:global}\label{d7:table:global}\label{d8:table:global}\label{d9:table:global} \end{table} \subsection{Several diamond operators}\label{subsec:several} For the levels $n=77,85,91,95,133,247$ the global method does not apply in degree $6$, either because $\ord\dm a\mid 6$ for every $a$ with $(\dm a-1)J_1(n)(\Q)$ finite, or because the gonality bound is too weak. In \cite{N5} the levels $91$, $95$, $133$ and $247$ were handled by the global method, with $a=3,11,64,144$, but the orders of these diamond operators, $6$, $3$, $3$ and $3$, divide $6$. For these levels we work at the prime $2$, and we use more general elements of the group ring of the diamond operators, such as $1+\dm{31}$ and $(\dm6-1)(\dm3-1)$, together with an argument with $X_0(91)$. We use the same method in degree $7$ at the levels $77$, $85$, $91$, $95$ and $99$, in degree $8$ at the levels $85$, $91$ and $95$, and in degree $9$ at the levels $95$ and $99$ (\Cref{prop:77_99,d8:prop:91}). \begin{proposition}\label{d6:prop:several}\label{d7:prop:several}\label{d8:prop:several}\label{d9:prop:several} Let $n$ be odd, let $\psi\in\Z[\Delta(n)]$ be such that $\psi\bigl(J_1(n)(\Q)\bigr)$ is finite, and write $\|\psi\|$ for the sum of the absolute values of its coefficients. Let $d\in\{6,7,8,9\}$ and assume that $\gon_\C X_1(n)>2d\|\psi\|$. Let $K$ be a number field of degree $d$, let $E/K$ be an elliptic curve with a point of order $n$, and suppose that $E$ has bad reduction at every prime of $K$ above $2$. Then the trace divisor $D$ satisfies $\psi(D)=0$. \end{proposition} \begin{proof} By \Cref{d6:lem:additive} the reduction of $E$ at every prime of $K$ above $2$ is multiplicative, so the reduction of $D$ modulo a prime $\wp$ of $L:=\Q(\zeta_n)$ above $2$ is cuspidal. As the cuspidal subscheme is finite \'etale over $\Z[\tfrac1n]$ and all cusps are defined over $L$, it lifts to an effective cuspidal divisor $C$ of degree $d$ on $X_1(n)_L$. The class $\psi[D-C]=\psi[D-dc_0]+\psi[dc_0-C]$ is torsion, as $\psi\bigl(J_1(n)(\Q)\bigr)$ is finite and by Manin-Drinfeld, and it reduces to $0$ modulo $\wp$, so $2\psi[D-C]=0$ by Parent's result, as $2$ is unramified in $L$. Writing $\psi=\psi^+-\psi^-$, where $\psi^+$ and $\psi^-$ are the positive and negative parts of $\psi$, the effective divisors $2\psi^+(D)+2\psi^-(C)$ and $2\psi^-(D)+2\psi^+(C)$, of degree $2d\|\psi\|$, are linearly equivalent. As $\gon_LX_1(n)\geq\gon_\C X_1(n)>2d\|\psi\|$, they are equal by \Cref{d8:lem:closed}, so $\psi(D)=\psi(C)$. The diamond operators preserve the set of cusps and its complement, and $D$ has no cusps in its support, so $\psi(D)=0$. \end{proof} We will apply \Cref{d6:prop:several} with $\psi=\dm a-1$, in which case its conclusion is that $D$ is invariant under $\dm a$, and with a few other elements. Note that the case $\psi=\dm a-1$ is the degree $d$ version of \cite[Proposition 7.2]{N5}, and that it is also the case of \Cref{d9:lem:stab}(c) in which $C$ is the lift of the reduction of $D$. \begin{table}[H] \centering \begin{tabular}{rlrl} \toprule $n$ & characters of the newforms of positive analytic rank & $d_{\mathrm{CM}}X_1(n)$ & $\tfrac{325}{2^{15}}i(n)$\\ \midrule $77$ & trivial & $60$ & $28.56$\\ $85$ & trivial, \lmfdcharorbit{85}{j} (order $4$) & $32$ & $34.27$\\ $91$ & trivial, \lmfdcharorbit{91}{g} (order $3$), \lmfdcharorbit{91}{u} (order $6$) & $24$ & $39.99$\\ $95$ & \lmfdcharorbit{95}{l} (order $12$) & $72$ & $42.84$\\ $133$ & trivial, \lmfdcharorbit{133}{f} (order $3$), \lmfdcharorbit{133}{s} (order $6$) & $36$ & $85.69$\\ $247$ & trivial, \lmfdcharorbit{247}{be} (order $12$) & $72$ & $299.92$\\ \bottomrule \end{tabular} \medskip \caption{The newforms of weight $2$ and level dividing $n$ with positive analytic rank, listed by their character orbits, and the data used in \Cref{prop:77_99,d6:prop:large}, with $i(n)$ as in \Cref{table:global}. Their Conrey labels are $85.4$, $91.9$, $91.30$, $95.8$, $133.39$, $133.31$ and $247.8$.} \label{d6:table:large} \end{table} \begin{proposition}\label{prop:77_99}\label{d7:prop:77_95}\label{d8:prop:85_95}\label{d9:prop:95_99} The group $\torz n$ is not a subgroup of the torsion of an elliptic curve over a number field of degree $d$ for $n\in\{77,91\}$ and $d\in\{6,7\}$, for $n=85$ and $d\in\{6,7,8\}$, for $n=95$ and $d\in\{6,7,8,9\}$, and for $n=99$ and $d\in\{7,9\}$. \end{proposition} \begin{proof} Let $E$ be such an elliptic curve over a number field $K$ of degree $d$, and let $D$ be the trace divisor of the corresponding point of $X_1(n)$. Its support consists of non-CM points, for $d=6$ as $d_{\mathrm{CM}}X_1(n)>6$ (\Cref{d6:table:large}) and for $d\geq7$ by \Cref{d7:lem:CM}. No elliptic curve over $\F_{2^e}$ with $e\leq d$ has a point of order $n$. For $e\leq6$ this follows from the Hasse bound, as such a curve has at most $81$ points, and for $n=77$ and $e=6$ from the discussion after \Cref{d6:lem:badred}. For $e=7$ the Hasse interval $[107,151]$ contains no multiple of $n$, for $e=8$ and $n=85$ we \gitlinkVIII{large_levels/finite_fields_2_octic.m}{compute} this, and for $e\in\{8,9\}$ and $n\in\{95,99\}$ the Hasse bound leaves only the group order $285$ for $e=8$ and the group orders $475$ and $495$ for $e=9$, with traces $-28$, $38$ and $18$, which are even and not traces of supersingular curves (see the discussion after \Cref{d6:lem:badred}). So by \Cref{d6:lem:badred} the curve $E$ has bad reduction at every prime of $K$ above $2$. For $n=77,85,91,99$ take $\psi=\dm a-1$ with $a=3,2,3,2$, so that $\dm a$ has order $30,8,6,15$ in $\Delta(n)$, and for $n=95$ take $\psi=1+\dm{31}$. By \Cref{d6:table:large} and an \gitlinkVII{certify_inherited_large.py}{LMFDB search}, for $n=77,85,91$ every newform of weight $2$ and level dividing $n$ with positive analytic rank has trivial character or a character $\chi$ with $\chi(a)=1$, and for $n=95$ every such newform has a character $\chi$ with $\chi(31)=-1$, as we \gitlinkVI{check_diamonds.m}{compute}. For $n=99$ we use \Cref{d7:lem:winding}. So $\psi\bigl(J_1(n)(\Q)\bigr)$ is finite. As $\|\psi\|=2$ and $\gon_\C X_1(n)>28.56,34.27,39.99,42.84,42.84$ for $n=77,85,91,95,99$ (\Cref{d6:table:large}, and $i(99)=4320$), which is larger than $4d$ in every case, \Cref{d6:prop:several} gives $\psi(D)=0$. For $n=95$ this says that $D+\dm{31}D=0$, which is impossible for a non-zero effective divisor. Otherwise $\dm aD=D$, and by \Cref{lem:orbits} the order of $\dm a$ divides $d$, a contradiction unless $(n,d)$ is $(91,6)$ or $(85,8)$. In these two cases $D$ is a single $\dm a$-orbit of length $d$ with multiplicity $1$, so its image on $X_1(n)/\dm a$ is a single point, which is rational as $D$ is rational. Hence $X_0(n)$ has a rational non-cuspidal point, which is impossible, as $X_0(91)$ has no rational non-cuspidal points \cite{kenku91} and there are no cyclic $85$-isogenies over $\Q$ \cite{mazur78,kenku82}. \end{proof} \begin{proposition}\label{d6:prop:large} The groups $\torz{133}$ and $\torz{247}$ are not subgroups of the torsion of an elliptic curve over a sextic field. \end{proposition} \begin{proof} Let $n\in\{133,247\}$, and let $D$ be the trace divisor of a point of order $n$ over a sextic field $K$. By the Hasse bound no elliptic curve over $\F_{2^f}$ with $f\leq6$ has a point of order $n$, so by \Cref{d6:lem:badred} the elliptic curve has bad reduction at every prime of $K$ above $2$, and \Cref{d6:prop:several} applies to $D$. As $d_{\mathrm{CM}}X_1(n)>6$ (\Cref{d6:table:large}), the support of $D$ consists of non-CM points. For $n=133$, let $\chi_f$ and $\chi_s$ be characters in the orbits \lmfdcharorbit{133}{f} and \lmfdcharorbit{133}{s}. We \gitlinkVI{check_diamonds.m}{compute} that $\chi_f(6)=1$ and $\chi_s(3)=1$, so \[ \psi=(\dm6-1)(\dm3-1),\qquad \|\psi\|=4,\qquad 12\|\psi\|=48<85.69, \] annihilates every factor of positive analytic rank, and \Cref{d6:prop:several} gives $\psi(D)=0$. Hence the divisor $D_1:=\dm3D-D$ satisfies $\dm6D_1=D_1$. Its support consists of non-CM points, so by \Cref{lem:orbits} its geometric points form a union of $\dm6$-orbits of length $\ord\dm6=18$. But these are among the at most $12$ geometric points in the supports of $D$ and $\dm3D$, and hence $D_1=0$. So $\dm3D=D$, and since $\dm3$ has order $9$, we get $9\mid\deg D=6$, which is a contradiction. For $n=247$, the characters in \lmfdcharorbit{247}{be} are trivial on $12$ and on $56$, so \Cref{d6:prop:several} gives $\dm{12}D=\dm{56}D=D$. We \gitlinkVI{check_diamonds.m}{compute} that $\dm{12}$ and $\dm{56}$ generate a subgroup of order $9$ of $\Delta(247)$, so by \Cref{lem:orbits}, $D$ is a sum of orbits of length $9$ of this subgroup, and $9\mid\deg D=6$, which is a contradiction. \end{proof} \begin{lemma}\label{d8:lem:3x6} Let $s$ and $t$ be commuting permutations of a set $S$, of orders $3$ and $6$, which generate a group isomorphic to $\torz3\times\torz6$ acting freely on $S$. If a finitely supported divisor $Z$ on $S$ satisfies $(s-1)(t-1)Z=0$, then $3\mid\deg Z$. \end{lemma} \begin{proof} Fix a point $z\in S$ and write the coefficient of $s^it^jz$ in $Z$ as $z_{ij}$, for $i\in\torz3$ and $j\in\torz6$. The condition says that $z_{ij}-z_{i+1,j}-z_{i,j+1}+z_{i+1,j+1}=0$ for all $i,j$, so $z_{ij}=r_i+b_j$ for some integers $r_i$ and $b_j$, and the sum of the coefficients on the orbit of $z$ is $6\sum_ir_i+3\sum_jb_j$. So $3$ divides the sum of the coefficients on each orbit, and hence $3\mid\deg Z$. \end{proof} \begin{proposition}\label{d8:prop:91} The group $\torz{91}$ is not a subgroup of the torsion of an elliptic curve over an octic field. \end{proposition} \begin{proof} Put $t=\dm3$ and $s=\dm{16}$, which have orders $6$ and $3$ in $\Delta(91)$ and generate a subgroup isomorphic to $\torz3\times\torz6$. As in the proof of \Cref{prop:77_99}, the group $(t-1)J_1(91)(\Q)$ is finite. Let $D$ be the trace divisor of a point of order $91$ over an octic field $K$. Its support consists of non-CM points by \Cref{d8:lem:CM}, so by \Cref{lem:orbits} the group generated by $s$ and $t$ acts freely on it. We \gitlinkVIII{large_levels/finite_fields_2_octic.m}{compute} that the only $e\leq 8$ such that an elliptic curve over $\F_{2^e}$ has a point of order $91$ is $e=8$. Suppose first that $E$ has bad reduction at every prime of $K$ above $2$. Then \Cref{d8:prop:several} with $\psi=t-1$ applies, as $\gon_\C X_1(91)>39.99>32$, and gives $tD=D$, which is impossible by \Cref{lem:orbits}, as $t$ has order $6$. So $E$ has good reduction at a prime of $K$ above $2$ of residue degree $8$, which is the unique prime of $K$ above $2$, and the reduction $\overline D$ of $D$ is a closed point of degree $8$ of $X_1(91)_{\F_2}$. We \gitlinkVIII{large_levels/check_good8_91_certified.m}{compute} that there are exactly two elliptic curves over $\F_{256}$ with a point of order $91$, up to isomorphism. Both have $j=0$ and cyclic group of order $273$, and for a point $P$ of order $91$ on either of them the points $uP$, for $u$ running over the automorphisms of the curve, are the points $aP$ with $a\in\{\pm1,\pm16,\pm74\}$. Hence $s\overline D=\overline D$. Consider the divisor $Z:=(s-1)(t-1)D=stD+D-sD-tD$ of degree $0$. Its class $[Z]=(s-1)(t-1)[D-8c_0]+8[(s-1)(t-1)c_0]$ is rational and torsion, as $(t-1)J_1(91)(\Q)$ is finite and by Manin-Drinfeld, and it reduces to $[(t-1)(s-1)\overline D]=0$. By Parent's result $2[Z]=0$, so $2Z$ is principal. Its positive part $2stD+2D$ has degree $32<\gon_\C X_1(91)$, so $2Z=0$ and $(s-1)(t-1)D=0$. As the group generated by $s$ and $t$ acts freely on the support of $D$, \Cref{d8:lem:3x6} gives $3\mid 8$, which is a contradiction. \end{proof} \subsection{The group \texorpdfstring{$\torz{81}$}{Z/81Z}}\label{subsec:81} \begin{proposition}\label{prop:81}\label{d8:prop:81}\label{d9:prop:81} The group $\torz{81}$ is not a subgroup of the torsion of an elliptic curve over a number field of degree $d\in\{6,7,8,9\}$. \end{proposition} \begin{proof} By \cite[Theorem 3.1(2)]{DEHMZ} the group $J_1(81)(\Q)$ is finite. We first show that $J_1(81)(\Q)[2]=0$. For $q=5,7$ the operator $A_q$ annihilates $J_1(81)(\Q)_\tors$ by \Cref{d8:prop:tools}(2), so $J_1(81)(\Q)[2]$ is contained in the intersection of the kernels of $A_5$ and $A_7$ on $J_1(81)[2]\simeq H_1(X_1(81)(\C),\Z)\otimes\F_2$. We \gitlinkVIII{large_levels/rational_two_torsion_8gb.m}{compute} the action of the Hecke and diamond operators on the integral homology $H_1(X_1(81)(\C),\Z)$, a lattice of rank $380$, using modular symbols, and find that the kernel of $A_5$ on $H_1(X_1(81)(\C),\Z)\otimes\F_2$ has dimension $8$, while the intersection of the kernels of $A_5$ and $A_7$ is $0$. Hence $J_1(81)(\Q)[2]=0$, so reduction modulo $2$ is injective on $J_1(81)(\Q)$ by Parent's result, and by \eqref{d9:eq:ES} the operator $A_2=T_2-1-2\dm2$ kills $J_1(81)(\Q)$. Let $D$ be the trace divisor of a point of order $81$ over a number field of degree $d\leq9$. As $A_2c_0=0$ by \Cref{d8:lem:cuspHecke}, applying $A_2$ to $[D-dc_0]$ we get $T_2D\sim D+2\dm2D$. Both sides are effective of degree $3d\leq27$, while $\gon_\C X_1(81)>28.92$ by Abramovich's bound, as $i(81)=2916$. So $T_2D=D+2\dm2D$ by \Cref{d9:lem:closed}, and the finite non-empty set of $j$-invariants of the points in the support of $D$ is closed under cyclic $2$-isogenies. But for $E_\tau=\C/(\Z+\tau\Z)$ the map $E_\tau\to E_{2\tau}$ is a cyclic $2$-isogeny, and $j(2^k\tau)=q^{-2^k}+744+O(q^{2^k})$, with $q=e^{2\pi i\tau}$, is unbounded as $k\to\infty$. \end{proof} \subsection{The groups \texorpdfstring{$\torz{133}$ and $\torz{247}$}{Z/133Z and Z/247Z}}\label{subsec:133_247} In degree $9$ we treat the levels $133$ and $247$ with \Cref{d9:lem:tree}. \begin{proposition}\label{d9:prop:133} The group $\torz{133}$ is not a subgroup of the torsion of an elliptic curve over a nonic field. \end{proposition} \begin{proof} Let $D$ be the trace divisor of a point of order $133$ over a nonic field, and put $h=\dm{64}$, of order $3$ in $\Delta(133)$. We \gitlinkIX{C133/winding133.m}{compute} that the winding submodule is the whole space for every character $\chi$ with $\chi(64)\neq1$, so $(h-1)J_1(133)(\Q)$ is finite. Hence $A_3(h-1)[D-9c_0]=0$, so $A_3(h-1)D$ is principal. Its positive part $T_3hD+D+3\dm3D$ has degree $72$, while $\gon_\C X_1(133)\geq\tfrac{325}{2^{15}}\cdot8640>85$, so $A_3(h-1)D=0$. By \Cref{d9:lem:CM} the points in the support of $D$ have no complex multiplication, so $hD=D$ by \Cref{d9:lem:tree}, and $D$ is the pull-back of an effective rational divisor $D_Y$ of degree $3$ on $Y=X_1(133)/\dm h$, by \Cref{d9:lem:odd}. Write $H_{19}=\{1,7,11\}\subseteq(\torz{19})^\times$. As $64\equiv1\pmod7$ and $64\equiv7\pmod{19}$, a non-cuspidal point of $Y$ over $\F_{2^e}$ gives an elliptic curve $E$ over $\F_{2^e}$ with a point of order $7$, up to sign (after replacing $E$ by a suitable twist), and a Frobenius stable $H_{19}$-orbit of points of order $19$, on which the Frobenius acts as an element $\lambda\in H_{19}$. We \gitlinkIX{C133/allbad_cubic.m}{enumerate} the elliptic curves over $\F_{2^e}$, $e\leq3$, with a point of order $7$: they have trace $a=-2$ for $e=2$ and $a=-5$ for $e=3$, and $\lambda^2-a\lambda+2^e\not\equiv0\pmod{19}$ for $\lambda\in H_{19}$. Hence $Y_{\F_2}$ has no non-cuspidal closed points of degree at most $3$, and the reduction of $D_Y$ modulo $2$ is cuspidal. Let $Z=X_1(133)/\dm{78}$, where $\dm{78}$ has order $18$ and contains $h$. As $78\equiv1\pmod 7$ and $78$ generates $(\torz{19})^\times$, the curve $Z$ parametrizes triples $(E,P_7,C_{19})$ consisting of a point of order $7$ and a cyclic subgroup of order $19$. The group $\dm{78}$ acts freely on $X_1(133)$, also on the cusps and in characteristic $2$, as an automorphism of an elliptic curve fixing a point of order $7$ is trivial, so $Z$ has genus $35$ and good reduction at $2$. The push-forward $D_Z$ of $D_Y$ to $Z$ is an effective rational divisor of degree $3$ with non-cuspidal support and cuspidal reduction. Let $\delta=\dm2$, of order $3$ on $Z$, and put \[ \psi_0=(1+\delta+\delta^2)(T_2^4+4T_2^3+T_2^2-8T_2-3),\qquad \psi_3=(2-\delta-\delta^2)(T_2-2\delta^2). \] We \gitlinkIX{C133/split_finite_operators.m}{compute} that the factors of $J_Z$ outside the winding quotient have characters of order $1$ and $3$, of total dimension $4$ and $2$ respectively, that the polynomial $X^4+4X^3+X^2-8X-3$ kills $T_2$ on the former, and that $T_2=2\delta^2$ on the latter. Hence $\psi_0(J_Z(\Q))$ and $\psi_3(J_Z(\Q))$ are finite. By \eqref{d9:eq:ES}, modulo $2$ we have $\psi_0^*=(1+\delta^*+\delta^{*2})(M^4+M^2+1)$ and $\psi_3^*=(\delta^*+\delta^{*2})M$, where $M=\Cart\delta^*$. The space $U=(\Cart-1)\bigl(\psi_0^*H^0(Z_{\F_2},\Omega^1)+\psi_3^*H^0(Z_{\F_2},\Omega^1)\bigr)$ has dimension $22$. The closed cusps of $Z_{\F_2}$ have degrees $1^6,3^2$, and we \gitlinkIX{C133/reproduce_FI3.m}{compute} that the restriction of $U$ to each of the $\binom83+2=58$ effective cuspidal divisors of degree $3$ is surjective. By \Cref{d8:prop:cartier} this is a contradiction. \end{proof} Note that $\psi_0\psi_3=0$, as $(1+\delta+\delta^2)(2-\delta-\delta^2)=0$, so it is essential that \Cref{d8:prop:cartier} allows several operators $\psi_i$. \begin{proposition}\label{d9:prop:247} The group $\torz{247}$ is not a subgroup of the torsion of an elliptic curve over a nonic field. \end{proposition} \begin{proof} Let $D$ be the trace divisor of a point of order $247$ over a nonic field and put $h=\dm{144}$, of order $3$. We \gitlinkIX{C247/winding247.m}{compute} that every character modulo $247$ whose winding submodule is not the whole space is trivial on $144$, so $(h-1)J_1(247)(\Q)$ is finite. As in the proof of \Cref{d9:prop:133}, $A_3(h-1)D$ is principal and its positive part has degree $72$, while $\gon_\C X_1(247)\geq\tfrac{325}{2^{15}}\cdot30240>299$, so $A_3(h-1)D=0$ and $hD=D$ by \Cref{d9:lem:CM,d9:lem:tree}. Let $H\subseteq\Delta(247)$ be the subgroup of order $18$ generated by $40$, which contains $144$. As $40\equiv1\pmod{13}$ and $40$ generates $(\torz{19})^\times$, the curve $Z=X_1(247)/H$ parametrizes triples $(E,P_{13},C_{19})$, and as in \Cref{d9:prop:133} the action of $H$ is free, so $Z$ has genus $129$, $i(n)/18=1680$ as index, and good reduction at $2$, $3$ and $17$. As above, $D$ gives an effective rational divisor $D_Z$ of degree $3$ on $Z$ whose support consists of points without complex multiplication. Let $t=\dm2$, of order $6$ on $Z$, and put $s=t^2$. We \gitlinkIX{C247/winding247.m}{compute} the winding submodules for the characters of $\Delta(247)/H$: for the characters of order $1$, $2$, $3$ and $6$ the dimensions of the spaces are $21$, $20$, $22$ and $22$, and those of the winding submodules $14$, $20$, $22$ and $22$. Hence the abelian variety $A=(t-1)J_Z$, of dimension $108$, has finitely many rational points. Put $A'=(s-1)J_Z$ and $A''=(1+s+s^2)A$, of dimensions $88$ and $20$. We \gitlinkIX{C247/export_norm_certificates.m}{compute} from the characteristic polynomials of $T_3$ and $T_{17}$ that $\#A'(\F_3)$ and $\#A''(\F_{17})$ are odd. If $T\in A(\Q)[2]$, then $(s-1)T\in A'(\Q)[2]=0$, so $sT=T$ and $T=3T=(1+s+s^2)T\in A''(\Q)[2]=0$. Hence $A(\Q)$ is finite of odd order. Let $c$ be the image on $Z$ of a rational cusp of $X_1(247)$, so that $T_2c=c+2tc$ by \Cref{d8:lem:cuspHecke}. The class $y=(t-1)[D_Z-3c]\in A(\Q)$ has odd order, and $A_2y=(T_2-1-2t)y$ reduces to $0$ modulo $2$ by \eqref{d9:eq:ES}, so $A_2y=0$. As $A_2(t-1)c=0$, we get \[ T_2(tD_Z)+tD_Z+D_Z\sim T_2D_Z+2t^2D_Z. \] Both sides are effective of degree $15$, while $\gon_\C Z\geq\tfrac{325}{2^{15}}\cdot1680>16$, so they are equal, and $A_2(t-1)D_Z=0$. By \Cref{d9:lem:tree} with $q=2$ we get $tD_Z=D_Z$. But an automorphism of an elliptic curve without complex multiplication which preserves a cyclic subgroup of order $19$ and maps a point $P_{13}$ of order $13$ to $2^kP_{13}$ is $\pm1$, so $2^k\equiv\pm1\pmod{13}$ and $6\mid k$. Hence the orbits of the group generated by $t$ on the support of $D_Z$ have length $6$, which does not divide $3$. \end{proof} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{The rank zero criterion and cuspidal divisor classes}\label{sec:rankzero}\label{d6:sec:rank0} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% When the Jacobian has rank $0$ we use \Cref{d6:prop:main}, which shows that the elliptic curve must have good reduction at some prime above a suitable $p$. At $p=2$ we replace its gonality condition by a computation (\Cref{subsec:doubled}), and we also use computations in the group generated by the cuspidal divisor classes over a finite field (\Cref{d6:sec:classes,subsec:X1_32_36}). Most of the propositions of this section are about sextic fields, while \Cref{d6:lem:fullrank,d6:lem:doubled,d6:lem:classtest} are also used in the other degrees. \subsection{The rank zero criterion}\label{subsec:rank0} To apply \Cref{d6:prop:main} to $\tg{7}{7}$ and $\tg{9}{9}$ we need the following lemma, as these cases are not covered by \cite[Theorem 4.1]{DS}. \begin{lemma}\label{d6:lem:fullrank} For $m=7$ and $m=9$ we have $\rk J_1(m,m)(\Q(\zeta_m))=0$. \end{lemma} \begin{proof} Put $F=\Q(\zeta_m)$. We use the argument of the proof of \cite[Theorem 4.1]{DS}, where $F$ is quadratic. Over $F$ the curve $X_1(m,m)$ is isomorphic to the quotient $X_H$ of $X_1(m^2)$ by the diamond operators congruent to $1$ modulo $m$ \cite[\S 2]{DS}, so it suffices to show that $B(F)$ is finite for the Jacobian $B$ of $X_H$, which is an abelian variety over $\Q$. The space of cusp forms of weight $2$ on $X_H$ is \[ V_m:=\bigoplus_{\varepsilon}S_2(\Gamma_0(m^2),\varepsilon), \] the sum being over the even Dirichlet characters $\varepsilon$ modulo $m$, so after extending scalars to $\overline\Q_\ell$ the $\ell$-adic representation of $B$ is the sum of the representations $\rho_f$ attached to the newforms $f$ occurring in $V_m$, with multiplicities. We have $B(F)=R(\Q)$ for the restriction of scalars $R:=\operatorname{Res}_{F/\Q}B_F$, whose $\ell$-adic representation is induced from the restriction of that of $B$ to $\Gal(\overline\Q/F)$. Hence, over $\overline\Q_\ell$, it is the sum of the twists $\rho_f\otimes\chi$, where $\chi$ runs over the characters of $\Gal(F/\Q)\simeq(\torz m)^\times$. The twist of a newform of level dividing $m^2$ and character $\varepsilon$ modulo $m$ by a character $\chi$ modulo $m$ is an eigenform of level dividing $m^2$ and character $\varepsilon\chi^2$, which is again even and defined modulo $m$, so $\rho_f\otimes\chi\simeq\rho_g$ for a newform $g$ occurring in $V_m$. The multiset of these newforms $g$ is stable under the action of $\Aut(\C)$ on the coefficients, so by Faltings' isogeny theorem \cite{faltings} the abelian variety $R$ is isogenous over $\Q$ to a product of the abelian varieties $A_{[g]}$ attached to the Galois orbits of the newforms $g$ occurring in $V_m$. By Kato's theorem it therefore suffices to show that $L(g,1)\neq0$ for every newform $g$ occurring in $V_m$. Using modular symbols we \gitlinkVI{full_level_rank.m}{compute} the dimension of the winding submodule of each Galois orbit of newforms occurring in $V_m$, and it is equal to the dimension of the orbit in every case. The newforms in question have level $49$ for $m=7$, and levels $27$ and $81$ for $m=9$. \end{proof} \begin{proposition}\label{d6:prop:rank0_cases} The groups $\torz{55}$, $\torz{75}$, $\tg{4}{24}$, $\tg{7}{7}$ and $\tg{9}{9}$ are not subgroups of the torsion of an elliptic curve over a sextic field. \end{proposition} \begin{proof} In each case suppose that $E$ is an elliptic curve over a sextic field $K$ containing the group in question, and let $d=d_m=[K:\Q(\zeta_m)]$. We show that the hypotheses of \Cref{d6:prop:main} hold for a suitable prime $p$, and that $E$ cannot have good reduction at any prime of $K$ above $p$ by \Cref{d6:lem:badred}, which is a contradiction. For $n=55,75$ we have $\rk J_1(n)(\Q)=0$ by \cite[Theorem 3.1(2)]{DEHMZ}, and Abramovich's bound gives $\gon_\Q X_1(n)>14.28,23.80$, respectively, both larger than $12$, so \Cref{d6:prop:main}(b) applies with $p=2$ and $d=6$. No elliptic curve over $\F_{2^f}$ with $f\leq 5$ has a point of order $n$ by the Hasse bound. Over $\F_{64}$ a point of order $55$ or $75$ would force $\#E(\F_{64})=55$ or $75$, which is impossible (see the discussion after \Cref{d6:lem:badred}). For $\tg{4}{24}$ we have $d=3$. By \cite[Theorem 4.1]{DS}, $\rk J_1(4,24)(\Q(i))=0$, and $i(4,24)=768$ gives $\gon_\C X_1(4,24)>7.61>3$, so \Cref{d6:prop:main}(a) applies with $p=5$. The prime $5$ splits in $\Q(i)$, so the residue fields of $K$ at the primes above $5$ are $\F_5$, $\F_{25}$ or $\F_{125}$. The Hasse intervals of these fields contain no multiple of $96$, since $(\sqrt{25}+1)^2<96$ and $96<(\sqrt{125}-1)^2<(\sqrt{125}+1)^2<192$. For $\tg{7}{7}$ and $\tg{9}{9}$ we have $K=\Q(\zeta_m)$ and $d=1$. The genera of $X_1(7,7)$ and $X_1(9,9)$ are $3$ and $10$, so $\gon_{\Q(\zeta_m)}X_1(m,m)>1$, and \Cref{d6:lem:fullrank} gives $\rk J_1(m,m)(\Q(\zeta_m))=0$. Take $p=29$ for $m=7$ and $p=19$ for $m=9$. Then $p\equiv1\pmod m$, so the residue fields of $K$ at the primes above $p$ are $\F_p$, and since $(\sqrt p+1)^22d$, because the kernel of reduction on the torsion of the Jacobian can contain elements of order $2$. For $X_1(3,15)$ and $X_1(3,21)$ the gonality bounds that we have are too weak for this, and we replace the gonality condition by a computation. We use it for $\tg{3}{15}$ in \Cref{d6:prop:doubled_cases} and for $\tg{3}{21}$ in \Cref{d8:prop:3_21}. \begin{lemma}\label{d6:lem:doubled} Let $m\mid n$ with $n$ odd and $mn>4$, let $\mathfrak p$ be a prime of $\Q(\zeta_m)$ above $2$ with residue field $k$, and let $d\geq1$. Put $X=X_1(m,n)$ and assume that $\rk J(\Q(\zeta_m))=0$ and that \[ \dim L(2\overline C)=1 \] for every effective cuspidal divisor $\overline C$ of degree $d$ on $X_k$. Then every elliptic curve $E$ over a degree $d$ extension $K$ of $\Q(\zeta_m)$ with $\tg{m}{n}\subseteq E(K)$ has good reduction at at least one prime of $K$ above $\mathfrak p$. \end{lemma} \begin{proof} Suppose the opposite, and let $D$ be the trace divisor of the corresponding point. By \Cref{d6:lem:additive} the reduction of $E$ at every prime of $K$ above $\mathfrak p$ is multiplicative, so $D$ reduces to an effective cuspidal divisor $\overline C$ of degree $d$ on $X_k$. Let $M=\Q(\zeta_n)$, choose a prime $\mathfrak P$ of $M$ above $\mathfrak p$, and let $F\subseteq M$ be the fixed field of its decomposition group in $\Gal(M/\Q(\zeta_m))$. The prime $\mathfrak P\cap F$ of $F$ has residue field $k$, and $F$ is unramified at $2$ as $n$ is odd. All cusps are defined over $M$ and the cuspidal subscheme is finite \'etale at $2$, so the closed points in the support of $\overline C$ lift to orbits of $\Gal(M/F)$ of the same size, and $\overline C$ lifts to an effective $F$-rational cuspidal divisor $C$ of degree $d$. As in the proof of \Cref{d6:prop:main}, the class $[D-C]=[D-dc_0]+[dc_0-C]\in J(F)$ is torsion, by the rank assumption on $J(\Q(\zeta_m))$ and by Manin-Drinfeld, and it reduces to $0$ modulo $\mathfrak P\cap F$, so $2[D-C]=0$ by Parent's result, and $2D\sim2C$. By semicontinuity \cite[Theorem III.12.8]{hartshorne}, the dimension of the Riemann-Roch space of $2C$ on $X_F$ satisfies \[ \dim L(2C)\leq\dim L(2\overline C)=1, \] so $2C$ is the only effective divisor in its class and $2D=2C$, which is impossible as $D$ is non-cuspidal and $C$ is cuspidal. \end{proof} We check the condition of \Cref{d6:lem:doubled} using differentials. Write $\overline C=\sum_Pm_PP$. The space $\Omega(2\overline C)$ is cut out in $H^0(X_k,\Omega^1)$ by the vanishing of the first $2m_P$ coefficients of the expansions in a local parameter at each $P$. These coefficients lie in the residue field of $P$, so these are $\sum_P2m_P\deg P=2d$ linear conditions over $k$, and by the Riemann-Roch theorem $\dim L(2\overline C)=1+2d-\operatorname{codim}\Omega(2\overline C)$. So $\dim L(2\overline C)=1$ if and only if these conditions are linearly independent. As $\Omega(2\overline C)$ is the intersection of the subspaces $\Omega(2m_PP)$, and $\Omega(2jP)$ depends only on the closed cusp $P$ and on $j\leq d/\deg P$, each of these subspaces is computed only once, by expanding a basis of $H^0(X_k,\Omega^1)$ at $P$, and for each $\overline C$ it remains to intersect at most $d$ of them. For $\tg{3}{15}$ we also need the following lemma, which we use again over octic fields in \Cref{d8:prop:3_15}. \begin{lemma}\label{d8:lem:3_15_C}\label{lem:3_15inj} Let $F=\Q(\zeta_3)$ and $X=X_1(3,15)$ over $F$, a curve of genus $9$ with $\rk J_X(F)=0$ \cite[Theorem 4.1]{DS}. Then $\ClCusp_F X\simeq\torz3\times\torz3\times\torz{24}\times\torz{96}\times\torz{96}$ and $3J_X(F)\subseteq\ClCusp_F X$. Reduction modulo the prime of $F$ above $2$ is injective on $J_X(F)$. \end{lemma} \begin{proof} All the cusps are defined over $\Q(\zeta_{15})$, and the Frobenius at a prime above $7$ generates $\Gal(\Q(\zeta_{15})/F)$, so the $20$ Galois orbits of cusps over $F$, of degrees $1^{16},4^4$, remain irreducible modulo $7$. We identify them with their reductions at the primes above $2$, $7$, $13$ and $19$ using the valuations of integral modular units, in the normal form $y^2+a_1xy+a_3y=x^3$ with $P=(0,0)$ of order $3$. We \gitlinkVIII{C3x15/diagonal_3_15_D4.m}{compute} that the reductions of $\ClCusp_F X$ at these four primes all have invariants $[3,3,24,96,96]$, so $\ClCusp_F X$ has these invariants, and that $\#J_X(\F_7)=149299200$, $\#J_X(\F_{13})=12230590464$ and $\#J_X(\F_{19})=247669456896$, whose greatest common divisor is $5971968=3\#\ClCusp_F X$. As reduction at these primes is injective on $J_X(F)$, the order of $J_X(F)$ divides $3\#\ClCusp_F X$, so $J_X(F)/\ClCusp_F X$ has order dividing $3$. In particular $J_X(F)[2^\infty]=(\ClCusp_F X)[2^\infty]$, on which reduction above $2$ is injective, and reduction above $2$ is injective on the odd part of $J_X(F)$. \end{proof} \begin{proposition}\label{d6:prop:doubled_cases} The group $\tg{3}{15}$ is not a subgroup of the torsion of an elliptic curve over a sextic field. \end{proposition} \begin{proof} Let $E$ be an elliptic curve over a sextic field $K$ with $\tg{3}{15}\subseteq E(K)$, and let $F$ and $X$ be as in \Cref{d8:lem:3_15_C}. Then $K\supseteq F$ and $d=[K:F]=3$. Let $\mathfrak p$ be the prime of $F$ above $2$, with residue field $\F_4$. As $2$ is inert in $F$, the residue fields of $K$ at the primes above $2$ are $\F_4$, $\F_{16}$ or $\F_{64}$, and the Hasse intervals of these fields contain no multiple of $45$. So $E$ has bad reduction at every prime of $K$ above $2$ by \Cref{d6:lem:badred}, and it suffices to show that $E$ has good reduction at some prime above $2$. We argue as in the proof of \Cref{d6:lem:doubled}, with one extra step. The curve $X_{\F_4}$ has $16$ cusps of degree $1$ and $8$ of degree $2$, so there are $\binom{18}{3}+16\cdot8=944$ effective cuspidal divisors of degree $3$. We \gitlinkVI{cusp_double_rr_3n_localref.m}{compute} that $\dim L(2\overline C)=1$ for $912$ of them, so the proof of \Cref{d6:lem:doubled} excludes these as reductions of the trace divisor $D$. The remaining $32$ divisors are supported on the cusps of degree $1$ and satisfy $\dim L(\overline C)=1$. The $16$ $F$-rational cusps of $X$ (see the proof of \Cref{d8:lem:3_15_C}) reduce to $16$ distinct cusps of degree $1$, as the cuspidal subscheme is finite \'etale at $2$, so they are all the cusps of $X_{\F_4}$ of degree $1$. It follows that each of the $32$ remaining divisors $\overline C$ lifts to an $F$-rational cuspidal divisor $C$, and if $D$ reduces to $\overline C$, then $[D-C]\in J_X(F)$ reduces to $0$. By \Cref{d8:lem:3_15_C}, reduction modulo $\mathfrak p$ is injective on $J_X(F)$, so $D\sim C$. By semicontinuity \cite[Theorem III.12.8]{hartshorne} we have $\dim L(C)\leq\dim L(\overline C)=1$, so $D=C$, which is a contradiction. \end{proof} \subsection{Arguments with cuspidal divisor classes}\label{d6:sec:classes} The Hecke sieve (\Cref{d6:sec:sieve}) uses only the fact that $A_q$ kills $J(\Q(\zeta_m))$. When $J(\Q(\zeta_m))$ is known explicitly, and in particular when it is generated by cuspidal divisor classes, we can instead compare the classes of the possible reductions with the reduction of $J(\Q(\zeta_m))$ itself. This is the method of \cite[\S 5.2]{DEHMZ} and \cite[\S 8]{N5}, which we use in the following form. \begin{lemma}\label{d6:lem:classtest} Let $X$ be a curve over a number field $F$ with an $F$-rational cusp $c_\ast$, let $\mathfrak p$ be a prime of good reduction of $X$ with residue field $k$, and let $G\subseteq J(k)$ be a subgroup containing the reduction of $J(F)$. Let $D\in X^{(d)}(F)$, and write its reduction modulo $\mathfrak p$ as $D'+C'$, where $D'$ is effective of degree $e$ with non-cuspidal support and $C'$ is effective and cuspidal. Then \[ [D'+C'-d\,\overline{c_\ast}]\in G. \] \end{lemma} \begin{proof} This is the reduction of the class $[D-dc_\ast]\in J(F)$. \end{proof} So an effective divisor $D'$ of degree $e$ with non-cuspidal support can be the non-cuspidal part of the reduction of $D$ only if $[D'+C'-d\,\overline{c_\ast}]\in G$ for some effective cuspidal divisor $C'$ of degree $d-e$ on $X_k$. To check this for many $D'$ at once, we compute for each $j\geq0$ the set $\mathcal S_j\subseteq J(k)/G$ of the classes of $[j\overline{c_\ast}-C']$, where $C'$ runs over the effective cuspidal divisors of degree $j$ on $X_k$. The condition then says that the image of $[D'-e\overline{c_\ast}]$ in $J(k)/G$ lies in $\mathcal S_{d-e}$. Note that $\mathcal S_j\subseteq\mathcal S_{j+1}$, as $[j\overline{c_\ast}-C']=[(j+1)\overline{c_\ast}-(C'+\overline{c_\ast})]$, so a divisor $D'$ which fails the test in degree $d+1$ also fails it in degree $d$. In practice $G$ is either the image of the reduction map on $J(F)$, when $J(F)$ is known and reduction is injective on it, or the subgroup generated by the classes $[\overline c-\deg(\overline c)\overline{c_\ast}]$ of all closed cusps $\overline c$ of $X_k$, when $J(F)$ is generated by cuspidal divisor classes. In the latter case $G$ contains the reduction of $J(F)$ without any assumption on which cusps are rational, and it contains the classes of all cuspidal divisors of degree $0$ on $X_k$. So every $\mathcal S_j$ is trivial, and the condition is simply $[D'-e\overline{c_\ast}]\in G$. Usually we test only one representative $D'$ of each orbit of a group of automorphisms of $X$ over $F$, such as the diamond operators or the level automorphisms of \Cref{subsec:divisors}, and the numbers of divisors $D'$ that we give are then numbers of orbit representatives. This is justified by the identity \[ [gD'-e\,\overline{c_\ast}]=g[D'-e\,\overline{c_\ast}]+e[g\overline{c_\ast}-\overline{c_\ast}] \] for such an automorphism $g$, which maps cusps to cusps. In both cases above $G$ is stable under $g$ and contains $[g\overline{c_\ast}-\overline{c_\ast}]$, as $gc_\ast$ is again an $F$-rational cusp. Moreover, $g$ permutes the effective cuspidal divisors of each degree, and the sets of cuspidal divisors that we allow in our computations are stable under $g$ as well, so the condition holds for $D'$ if and only if it holds for $gD'$. \subsection{The curve \texorpdfstring{$X_1(32)$}{X\_1(32)}}\label{subsec:X1_32_36} The proof below uses the Hecke sieve and the types of decompositions of \Cref{subsec:sieve_method}. \begin{proposition}\label{d6:prop:32}\label{d7:prop:32} The group $\torz{32}$ is not a subgroup of the torsion of an elliptic curve over a number field of degree $d\in\{6,7\}$. \end{proposition} \begin{proof} We have $\rk J_1(32)(\Q)=0$ and $J_1(32)(\Q)=\ClCusp_\Q X_1(32)\simeq (\torz2)^3\times\torz4\times\torz{120}\times\torz{11640}$ \cite[Theorem 3.1(2), Corollary 4.14 and Table 2]{DEHMZ}, and $\gon_\Q X_1(32)=8$ \cite[Theorem 4 and Table 1]{DvH}. So for $d=6,7$, \Cref{d6:prop:main}(a) with $p=3$ excludes the entirely cuspidal reduction of the trace divisor $D$ modulo $3$. Let $D'+C'$ be the reduction of $D$, as in \Cref{d6:lem:classtest}, let $c_\ast$ be a rational cusp, and let $G$ be the subgroup of $J_1(32)(\F_3)$ generated by the reductions of the rational cuspidal divisors of degree $0$. We \gitlinkVI{analyze_survivors.m}{compute} that $G\simeq(\torz2)^3\times\torz4\times\torz{120}\times\torz{11640}$, so $G$ is the reduction of $J_1(32)(\Q)$, and by \Cref{d6:lem:classtest} we have $[D'+C'-d\,\overline{c_\ast}]\in G$. The non-cuspidal closed points of $X_1(32)_{\F_3}$ of degrees $3,\dots,7$ form $1,1,2,19$ and $39$ diamond orbits. The primes $q=5,7$ are admissible for cuspidal parts of degree at most $1$, and the \gitlinkVII{hecke_sieve.m}{Hecke sieve} at $3$ with $q=5,7$ leaves, for $d=7$, $8$ possible non-cuspidal parts $D'$ up to the diamond operators, of degrees $3,4,5,5,6,6,6,6$. For these $8$ divisors we \gitlinkVII{cuspidal_class_filter.m}{compute} the classes $[D'+C'-7\overline{c_\ast}]$ for all $1688,412,82,82,12,12,12,12$ effective cuspidal divisors $C'$ of the complementary degree, and none of these $2312$ classes lies in $G$. As $G$ is stable under the diamond operators, this excludes all the possible reductions for $d=7$, and by the monotonicity of these tests in the degree (\Cref{d6:sec:classes,subsec:sieve_method}) also for $d=6$. \end{proof} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{The Hecke sieve and push-forwards to smaller levels}\label{sec:sieve}\label{d6:sec:sieve}\label{d7:sec:sieve}\label{d8:sec:sieve}\label{d9:sec:sieve} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% The groups of this section are excluded by reducing the trace divisor of a point modulo a small prime $p$ and applying the Hecke sieve of \cite[Proposition 5.5]{DN} to the possible reductions. For some of them we also compare the class of the reduction with the reduction of a known Mordell-Weil group, on the curve itself or after pushing the trace divisor forward to a modular curve of smaller level. In this way we exclude $\torz n$ for $n\in\{33,34,35,38,40,42,49,50\}$ over sextic fields, for $n\in\{34,35,38,40,42,46,49\}$ over septic fields, for $n\in\{38,44,46,50\}$ over octic fields and for $n\in\{35,38,44,46,50,58,62\}$ over nonic fields. The sieve computation on $X_1(32)$ is completed with cuspidal divisor classes in \Cref{d6:prop:32}, and the same sieve over $\Q(\zeta_m)$ is used for the groups $\tg mn$ with $m\geq3$ in \Cref{sec:mxn}. \subsection{The sieve and the push-forwards}\label{subsec:sieve_method} Let $X=X_1(m,n)$ with $\rk J(\Q(\zeta_m))=0$, let $\mathfrak p$ be a prime of $\Q(\zeta_m)$ above a prime $p\nmid n$, with residue field $k=\F_{p^f}$, let $K\supseteq\Q(\zeta_m)$ be a number field of degree $d$, and let $D$ be the trace divisor of a point $x\in Y_1(m,n)(K)$, which has degree $d_m$. Reducing modulo $\mathfrak p$ we write \[ D_k=D'+C', \] where $D'$ is effective of degree $e$ with non-cuspidal support and $C'$ is effective and cuspidal of degree $d_m-e$. Note that $e\geq 1$ if and only if the elliptic curve corresponding to $x$ has good reduction at some prime of $K$ above $\mathfrak p$, so the reductions with $e=0$ are excluded by \Cref{d6:prop:main} whenever its condition on the gonality holds. As in \cite[\S 6]{N5} we call the multiset of degrees of the closed points occurring in $D'$, a partition of $e$, the \textit{type} of the decomposition, and write it in brackets, as in $[3,3]$. The Hecke sieve \cite[Proposition 5.5]{DN} says that if $[A_q(D')]\neq 0$ and $A_q(C')=0$ for some prime $q\nmid 2pn$ with $q\equiv1\pmod m$, then this decomposition cannot occur, since $[A_q(D)]=0$ by \Cref{d6:prop:tools}(4) and its reduction is $[A_q(D')]+[A_q(C')]$. By \Cref{d8:lem:cuspHecke}, $A_q(C')=0$ if $\sigma_q$ fixes every closed cusp occurring in $C'$, and we then call $q$ \textit{admissible} for $C'$. Every prime $q\equiv p^{fj}\pmod n$, for some $j\geq0$, is admissible for every $C'$, and every $q$ is admissible for $C'$ if each closed cusp occurring in $C'$ is the reduction of a $\Q(\zeta_m)$-rational cuspidal divisor of the same degree, that is, if it comes from a Galois orbit of cusps over $\Q(\zeta_m)$ which stays irreducible modulo $\mathfrak p$. As $A_q$ commutes with the diamond operators, it suffices to test one $D'$ in each orbit of the diamond operators, and when $D'$ is a sum of several closed points, it suffices to take one of them up to the diamond operators. We also use \Cref{d6:lem:classtest}, on $X$ itself or after pushing the trace divisor forward to another curve. Let $m\leq2$, let $\pi:X\to Y$ be a morphism of modular curves over $\Q$ which maps cusps to cusps and non-cusps to non-cusps, let $p$ be a prime of good reduction of both curves such that $\pi$ extends over $\Z_{(p)}$, as all the degeneracy and forgetful maps that we use do, and let $y_0$ be a rational cusp of $Y$. Assume that $J_Y(\Q)$ is finite, and let $G\subseteq J_Y(\F_p)$ be a subgroup containing the reduction of $J_Y(\Q)$, as in \Cref{d6:sec:classes}. If the reduction of $D$ is $D'+C'$ as above, with $\deg D'=e$, then \Cref{d6:lem:classtest}, applied to $\pi_*D$, gives \makeatletter \begin{equation}\label{d8:eq:pushclass}\ltx@label{{d9:eq:pushclass}}% second label, d9: \label{d9:eq:pushclass} [\pi_*D'-e\,\overline{y_0}]\in G+\mathcal S_{d-e}, \end{equation} \makeatother where $\mathcal S_j$ is the set of the classes $[j\overline{y_0}-B]$ for the effective cuspidal divisors $B$ of degree $j$ on $Y_{\F_p}$. Note that we allow all the cuspidal divisors $B$ over $\F_p$, also those that are not reductions of rational cuspidal divisors. If every closed cusp of $Y_{\F_p}$ is the reduction of a rational cuspidal divisor of the same degree, then $\mathcal S_j\subseteq G$, as every class $[j\overline{y_0}-B]$ is then the reduction of a class in $J_Y(\Q)$, and \eqref{d8:eq:pushclass} reduces to $[\pi_*D'-e\,\overline{y_0}]\in G$. These tests are monotone in the degree. A prime $q$ which is admissible for all the cuspidal divisors of degree $j$ is also admissible for those of smaller degree, as each of them is contained in one of degree $j$ obtained by adding a multiple of the reduction of $c_0$, and $\mathcal S_j\subseteq\mathcal S_{j+1}$ (see \Cref{d6:sec:classes}). Hence, if a computation excludes all the possible non-cuspidal parts $D'$ for the fields of degree $d$, then it also excludes them for the fields of every smaller degree. In particular, a computation of all the possible reductions in degree $9$ also covers the degrees $6$, $7$ and $8$. The numbers of non-cuspidal closed points of $X_{\F_p}$ that we use are listed in \Cref{d7:table:sieve}, and for $X_1(2,26)$ and $X_1(2,28)$ in \Cref{d9:table:2n}. They are computed from the models (in degree $9$ also from the zeta functions), and checked independently by counting the elliptic curves over $\F_{p^f}$ with the required level structure (see \Cref{sec:computations}). \begin{table}[H] \centering \small \begin{tabular}{lrl} \toprule $X$ & $p$ & non-cuspidal closed points of degree $1,2,\dots$\\ \midrule $X_1(29)$ & $3$ & $0,0,14,28,14,133,280,770,2282$\\ $X_1(31)$ & $3$ & $0,0,0,15,91,120,330,930,1980$\\ $X_1(32)$ & $3$ & $0,0,8,8,16,140,312$\\ $X_1(33)$ & $5$ & $0,10,60,230,602,2660$\\ $X_1(34)$ & $3$ & $0,0,0,26,32,80,256$\\ $X_1(35)$ & $3$ & $0,0,12,0,36,126,348,813,2240$\\ $X_1(38)$ & $3$ & $0,0,3,0,72,87,216,801,2127$\\ $X_1(39)$ & $2$ & $0,0,0,0,0,12,0$\\ $X_1(40)$ & $3$ & $0,0,0,32,72,80,408$\\ $X_1(42)$ & $5$ & $0,0,4,183,648,2286,11160$\\ $X_1(44)$ & $3$ & $0,0,0,0,60,160,240,920,2260$\\ $X_1(46)$ & $3$ & $0,0,0,55,44,55,264,781,1958$\\ $X_1(49)$ & $3$ & $0,0,0,21,63,154,379$\\ $X_1(50)$ & $3$ & $0,0,0,0,0,60,200,1045,2040$\\ $X_1(55)$ & $2$ & $0,0,0,0,0,0,20,0,40$\\ $X_1(58)$ & $3$ & $0,0,0,0,0,140,224,756,2380$\\ $X_1(62)$ & $3$ & $0,0,0,0,180,150,330,930,1600$\\ $X_1(2,22)$ & $3$ & $0,0,0,0,60,100,240$\\ \bottomrule \end{tabular} \medskip \caption{Non-cuspidal closed points of $X_{\F_p}$ of small degree. For each curve the list ends at the largest degree in which we use it. For $X_1(2,26)$ and $X_1(2,28)$ see \Cref{d9:table:2n}.} \label{d7:table:sieve}\label{d8:table:sieve}\label{d9:table:sieve} \end{table} \subsection{The curves over \texorpdfstring{$\Q$}{Q} in degrees \texorpdfstring{$6$}{6} and \texorpdfstring{$7$}{7}}\label{subsec:sieve_deg6}\label{subsec:sieve_deg7} \begin{proposition}\label{d6:prop:sieve} The group $\torz{33}$ is not a subgroup of the torsion of an elliptic curve over a sextic field. \end{proposition} \begin{proof} We have $\rk J_1(33)(\Q)=0$ \cite[Theorem 3.1(2)]{DEHMZ} and $\gon_\Q X_1(33)=10$ \cite[Theorem 4 and Table 1]{DvH}, so \Cref{d6:prop:main}(a) with $p=5$ excludes the entirely cuspidal reduction. As $Y_1(33)(\F_5)=\emptyset$ (\Cref{d7:table:sieve}), the cuspidal part has degree at most $4$. The only Galois orbit of cusps over $\Q$ that splits modulo $5$ has degree $10$ and splits into $2$ places of degree $5$, which cannot occur in such a cuspidal part, so every prime $q\nmid330$ is admissible. The non-cuspidal closed points of $X_1(33)_{\F_5}$ of degrees $2,\dots,6$ form $1,6,24,61,269$ diamond orbits, and we \gitlinkVI{hecke_sieve_cached.m}{compute} that $[A_7(D')]\neq0$ for every possible non-cuspidal part $D'$ of every type of degree at most $6$, up to the diamond operators (the computation is split over several scripts, which are listed in the \texttt{README} of the sextic repository). \end{proof} \begin{proposition}\label{d7:prop:sieve} The group $\torz{34}$ is not a subgroup of the torsion of an elliptic curve over a number field of degree $6$ or $7$. \end{proposition} \begin{proof} We have $\rk J_1(34)(\Q)=0$ \cite[Theorem 3.1(2)]{DEHMZ} and $\gon_\Q X_1(34)=10$ \cite[Theorem 4 and Table 1]{DvH}, so \Cref{d6:prop:main}(a) with $p=3$ excludes the entirely cuspidal reduction. Every closed cusp of $X_1(34)_{\F_3}$ of degree at most $7$ is the reduction of a rational cusp, as the other closed cusps have degree $8$, so every prime $q\nmid102$ is admissible. The non-cuspidal closed points of $X_1(34)_{\F_3}$ of degrees $4$, $5$, $6$ and $7$ form $5$, $4$, $11$ and $32$ diamond orbits, and we \gitlinkVII{hecke_sieve.m}{compute} with the Hecke sieve at $p=3$ and $q\in\{5,7,11\}$ that every possible non-cuspidal part $D'$, of every type of degree at most $7$, satisfies $[A_q(D')]\neq0$ for one of these $q$. This excludes all the possible reductions for $d=6$ and for $d=7$. \end{proof} For $X_1(40)$ one possible reduction survives the Hecke sieve with the admissible primes, and it requires a correction by cuspidal divisors which are not reductions of rational ones. \begin{lemma}\label{d7:lem:40} Let $P$ be the closed point of degree $5$ of $X_1(40)_{\F_3}$ which survives the Hecke sieve at $3$ with $q=11$. Then $A_7(P+C)$ is not principal for any effective cuspidal divisor $C$ of degree $2$ on $X_1(40)_{\F_3}$. \end{lemma} \begin{proof} The closed cusps of $X_1(40)_{\F_3}$ which are not reductions of rational cuspidal divisors are $4$ cusps of degree $2$, of type $k=5$ in the description of the cusps of $X_1(40)$ as $(\mu_{40/k}\times\torz k)'/\{\pm1\}$ \cite[Section 2]{DN}. They come from $2$ Galois orbits of degree $4$ over $\Q$, each of which splits into $2$ closed points of degree $2$ over $\F_3$. Write a geometric cusp of this type as $(\zeta_8^a,b)$ with $a\in(\torz8)^\times$ and $b\in(\torz5)^\times$, modulo $(a,b)\sim(-a,-b)$. The Frobenius at $3$ sends $(a,b)$ to $(3a,b)$, the Galois element $\sigma_7$ sends it to $(7a,b)$, and the diamond operator $\delta=\dm7$ sends it to $(7a,7b)$. Hence $\sigma=\sigma_7$ interchanges the $2$ closed points in each orbit, and $\delta^2=\sigma$ on these $4$ points. As $\sigma^{-1}=\sigma$ on them, \Cref{d8:lem:cuspHecke} gives \[ A_7c=(\sigma-1)c+7\delta(\sigma-1)c \] for such a cusp $c$, while $A_7C=0$ for the other $80$ effective cuspidal divisors $C$ of degree $2$, which are reductions of rational cuspidal divisors. We do not identify the cusps of the model with their labels $(a,b)$, so we test both permutations $\delta$ of order $4$ of these $4$ cusps with $\delta^2=\sigma$. They are the actions of $\dm7$ and of $\dm7^{-1}=\dm{23}$, as $23^2\equiv7^2\pmod{40}$, so the two cases also correspond to the two conventions for the diamond operators of \Cref{subsec:hecke}. We \gitlinkVII{hecke_cusp_correction.m}{compute} that $A_7P$ and the $8$ divisors $A_7(P+c)$, for the $4$ cusps $c$ and both permutations $\delta$, are non-principal. \end{proof} \begin{proposition}\label{d7:prop:40} The group $\torz{40}$ is not a subgroup of the torsion of an elliptic curve over a number field of degree $6$ or $7$. \end{proposition} \begin{proof} Let $d\in\{6,7\}$. We have $\rk J_1(40)(\Q)=0$ \cite[Theorem 3.1(2)]{DEHMZ} and $\gon_\Q X_1(40)=12$ \cite[Theorem 4 and Table 1]{DvH}, so the entirely cuspidal reduction at $3$ is excluded by \Cref{d6:prop:main}(a). By \Cref{d7:table:sieve} the non-cuspidal part $D'$ of the reduction is a single closed point of degree $4,\dots,d$, and there are $4,9,10$ and $51$ diamond orbits of points of degree $4,5,6$ and $7$. The prime $q=11$ is admissible for cuspidal parts of degree at most $3$, and $q=7$ for those of degree $1$. For $d=7$ we \gitlinkVII{hecke_sieve_commonzero.m}{compute} that the Hecke sieve excludes all the representatives of degree $4$ and $6$, all but one of degree $5$, and \gitlinkVII{hecke_sieve_smart.m}{all} those of degree $7$, for which the cuspidal part is empty and every $q\nmid 120$ is admissible, and by the monotonicity in \Cref{subsec:sieve_method} this also holds for $d=6$. The remaining point $P$ of degree $5$ is excluded by \Cref{d7:lem:40}: for $d=7$ its cuspidal part $C'$ has degree $2$, and for $d=6$ it has degree $1$ and $A_7(P+C')=A_7(P+C'+\overline{c_0})$, as $A_7\overline{c_0}=0$, where $C'+\overline{c_0}$ has degree $2$. \end{proof} \begin{proposition}\label{d7:prop:42} The group $\torz{42}$ is not a subgroup of the torsion of an elliptic curve over a number field of degree $6$ or $7$. \end{proposition} \begin{proof} We have $\rk J_1(42)(\Q)=0$ \cite[Theorem 3.1(2)]{DEHMZ}, and Derickx and van Hoeij proved $\gon_\Q X_1(42)>8$ \cite[proof of Proposition 6]{DvH}, so \Cref{d6:prop:main}(a) with $p=5$ excludes the entirely cuspidal reduction. The closed cusps of $X_1(42)_{\F_5}$ have degrees $1^{12},2^6,3^4,6^2$, while the Galois orbits over $\Q$ have degrees $1^{12},2^6,6^4$. A cusp with stabilizer $H\subseteq(\torz{42})^\times$ gives a closed cusp which is fixed by $\sigma_q$ for every $q$ in the subgroup generated by $H$ and $5$, so every prime $q$ is admissible for cuspidal parts of degree at most $2$, and the primes $q\equiv 5^i\pmod{42}$ for cuspidal parts of degree $3$ or $4$. For septic fields the possible types of the non-cuspidal part and the tests are the following: \[ \begin{array}{c|r|c|l} \text{type}&\text{diamond representatives}&\deg C'&q\\\hline {[3]}&2&4&17\\ {[4]}&33&3&17\\ {[5]}&108&2&11,13,17\\ {[6]}&392&1&11\\ {[7]}&1860&0&11,13,17\\ {[3,3]}&8&1&11,13,17\\ {[3,4]}&366&0&11,13,17 \end{array} \] We compute that every one of these $2769$ divisors $D'$ satisfies $[A_q(D')]\neq0$ for one of the listed primes. The types $[3]$, $[5]$, $[6]$ and $[3,3]$ were tested in the \gitlinkVI{hecke_sieve_42_roots.m}{sextic computation}, which is split over several scripts listed in the \texttt{README} of the sextic repository, with primes that are admissible here, and the types \gitlinkVII{N42_lower7.m}{$[4]$}, \gitlinkVII{N42_mixed34_shard.m}{$[3,4]$} and \gitlinkVII{N42_test7_shard.m}{$[7]$} anew. We \gitlinkVII{audit_42_degree7.py}{check} that the tests cover all $2769$ divisors. By the monotonicity in \Cref{subsec:sieve_method} this also covers sextic fields. \end{proof} \subsection{The curve \texorpdfstring{$X_1(35)$}{X1(35)}}\label{subsec:sieve_35} The octic points of $X_1(35)$ are determined in \Cref{d8:prop:35}. \begin{proposition}\label{prop:35}\label{d9:prop:35} The group $\torz{35}$ is not a subgroup of the torsion of an elliptic curve over a number field of degree $6$, $7$ or $9$. \end{proposition} \begin{proof} Let $d\in\{6,7,9\}$. We have $\gon_\Q X_1(35)=12$ \cite[Theorem 4 and Table 1]{DvH}, and $J_1(35)(\Q)\simeq\torz{13}\times\torz{109148520}$ is finite and generated by rational cuspidal divisors \cite[Theorem 3.1(2), Corollary 4.14 and Table 2]{DEHMZ}. So by \Cref{d6:prop:main}(a) with $p=3$ the reduction of $D$ modulo $3$ is $D'+C'$ with $e=\deg D'\geq1$. The closed cusps of $X_1(35)_{\F_3}$ have degrees $1^{12},4^3,6^2,12$, the same as the Galois orbits of cusps over $\Q$, so every closed cusp of $X_1(35)_{\F_3}$ is the reduction of a rational cuspidal divisor of the same degree. Let $H$ be the reduction of $J_1(35)(\Q)$. By \eqref{d8:eq:pushclass} with $\pi=\mathrm{id}$, $y_0=c_0$ and $G=H$, as then $\mathcal S_{d-e}\subseteq H$, we have $[D'-e\,\overline{c_0}]\in H$. We \gitlinkIX{C35/presentation_v2/build.m}{compute} \[ J_1(35)(\F_3)\simeq(\torz4)^2\times\torz8\times\torz{520}\times\torz{218297040}, \] in which $H$ has index $2^{11}\cdot5$. By \Cref{d7:table:sieve} the possible types of $D'$ of degree at most $9$ are $[3]$, $[5]$, $[6]$, $[7]$, $[8]$, $[9]$, $[3,3]$, $[5,3]$, $[6,3]$ and $[3,3,3]$. This gives $5961$ cases, and we \gitlinkIX{C35/presentation_v2/classify_FB_tail.m}{find} that exactly $6$ of them have class in $H$, all of type $[8]$. This excludes the degrees $6$ and $7$. So $d=9$, and $D$ reduces to $P+\overline c$ for one of these $6$ points $P$ of degree $8$ and a rational cusp $c$. The $6$ points are the reductions of the closed points of degree $8$ of $X_1(35)$ in the diamond orbit of the octic point of level $35$ of \Cref{d8:prop:realizations}, as we \gitlinkIX{C35/known_octic35.m}{check}. So $D$ reduces to $\overline B+\overline c$ for such an octic point $B$, and as reduction modulo $3$ is injective on $J_1(35)(\Q)$, we get $D\sim B+c$. As $\gon_\Q X_1(35)>9$, \Cref{d9:lem:closed} gives $D=B+c$, which is a contradiction, as the support of $D$ contains no cusp. \end{proof} \subsection{The curves \texorpdfstring{$X_1(38)$, $X_1(46)$ and $X_1(50)$}{X1(38), X1(46) and X1(50)}}\label{subsec:sieve_38}\label{subsec:sieve_38_44}\label{subsec:sieve_46_50} Over septic fields, $\torz{50}$ is excluded in \Cref{d7:prop:25}. \begin{proposition}\label{prop:38}\label{d9:prop:38}\label{d8:prop:46_50}\label{d9:prop:46_50} The group $\torz{38}$ is not a subgroup of the torsion of an elliptic curve over a number field of degree $d\in\{6,7,8,9\}$, nor is $\torz{46}$ for $d\in\{7,8,9\}$, nor $\torz{50}$ for $d\in\{6,8,9\}$. \end{proposition} \begin{proof} Let $n\in\{38,46,50\}$. We have $\rk J_1(n)(\Q)=0$ \cite[Theorem 3.1(2)]{DEHMZ}, $\gon_\Q X_1(38)=12$ \cite[Theorem 4 and Table 1]{DvH} and $\gon_\Q X_1(n)\geq10$ for $n=46,50$ \cite[Table 1]{NV}, so \Cref{d6:prop:main}(a) with $p=3$ excludes the entirely cuspidal reduction. The closed cusps of $X_1(n)_{\F_3}$ have degrees $1^{18},9^2$ for $n=38$, $1^{22},11^2$ for $n=46$ and $1^{20},4^4,10^2$ for $n=50$, the same as the Galois orbits of cusps over $\Q$, so every prime $q\nmid3n$ is admissible for every cuspidal part. Let $\pi:X_1(n)\to Y=X_1(n/2)$ be the map $(E,P)\mapsto(E,2P)$. The groups $J_1(19)(\Q)$, $J_1(23)(\Q)$ and $J_1(25)(\Q)$ are cyclic of orders $4383$, $408991$ and $227555$ and generated by rational cuspidal divisors \cite[Theorem 3.1(2), Corollary 4.14 and Table 2]{DEHMZ}. Their reductions $G$ have index $2$, $23$ and $11$ in $J_1(19)(\F_3)\simeq\torz{8766}$, $J_1(23)(\F_3)\simeq\torz{9406793}$ and $J_1(25)(\F_3)\simeq\torz{2503105}$, and every effective cuspidal divisor of degree at most $9$ on $Y_{\F_3}$ has zero image in $J_Y(\F_3)/G$, so \eqref{d8:eq:pushclass} says that $[\pi_*D'-e\,\overline{y_0}]\in G$. In particular neither of the two tests depends on $d$. By \Cref{d7:table:sieve} the possible types of $D'$ of degree at most $9$ are $[3]$, $[5]$, $[6]$, $[3,3]$, $[7]$, $[8]$, $[5,3]$, $[9]$, $[6,3]$ and $[3,3,3]$ for $n=38$, with $1$, $8$, $11$, $3$, $24$, $89$, $24$, $239$, $33$ and $6$ cases up to the diamond operators, $[4]$, $[5]$, $[6]$, $[7]$, $[8]$, $[4,4]$, $[9]$ and $[5,4]$ for $n=46$, with $5$, $4$, $5$, $24$, $71$, $275$, $178$ and $220$ cases, and $[6]$, $[7]$, $[8]$ and $[9]$ for $n=50$, with $6$, $20$, $107$ and $204$ cases. We \gitlinkIX{common/hecke_sieve_quotient_D9.m}{compute} that each of these cases fails \eqref{d8:eq:pushclass} or satisfies $[A_q(D')]\neq0$ for some prime $q$: for $n=38$ with $q\in\{5,7,11,13,17\}$ for all $438$ cases (see the \gitlinkIX{C38/logs/N38_p3_quotient_D9_rerun.log}{log}), for $n=46$ with $q\in\{5,7,11,13,17,19\}$ for all $782$ cases (see the \gitlinkIX{C46_C50/logs/N46_p3_quotient_D9_rerun.log}{log}), and for $n=50$ with $q\in\{7,11,13,17,19\}$, for the $133$ cases of degree at most $8$ in the \gitlinkVIII{common/hecke_sieve_quotient.m}{octic computation} and for the $204$ cases of type $[9]$ in the \gitlinkIX{C46_C50/logs/N50_p3_good9.log}{nonic computation}. This excludes all $d\in\{6,7,8,9\}$. \end{proof} \subsection{The curve \texorpdfstring{$X_1(44)$}{X1(44)}}\label{subsec:sieve_44} \begin{proposition}\label{d8:prop:44}\label{d8:prop:38_44}\label{d9:prop:44} The group $\torz{44}$ is not a subgroup of the torsion of an elliptic curve over a number field of degree $8$ or $9$. \end{proposition} \begin{proof} Let $d\in\{8,9\}$. We have $\rk J_1(44)(\Q)=0$ \cite[Theorem 3.1(2)]{DEHMZ} and $\gon_\Q X_1(44)\geq10$ \cite[Table 1]{NV}, so \Cref{d6:prop:main}(a) with $p=3$ excludes the entirely cuspidal reduction. By \Cref{d7:table:sieve} the non-cuspidal part $D'$ is a single closed point of degree $5,\dots,d$, so $C'$ has degree at most $4$. The closed cusps of $X_1(44)_{\F_3}$ have degrees $1^{15},2^5,5^3,10$, and those of degree at most $4$ are reductions of rational cuspidal divisors of the same degree, so every prime $q\nmid132$ is admissible. There are $6,16,24$ and $92$ diamond orbits of points of degree $5,6,7$ and $8$, and we \gitlinkVIII{common/hecke_sieve_smart.m}{compute} that for a representative $D'$ of each of them one of $[A_q(D')]$, $q\in\{5,7,13,17,19\}$, is non-zero. For $d=9$ it remains to treat $D'=P$ of degree $9$, with $C'=0$. Then $[A_5P]=0$ by \Cref{d6:prop:tools}(4), that is, $B=T-U$ is principal, where $T=T_5P$ and $U=P+5\dm5P$, both of degree $54$ and computed directly on the moduli model, and $P$ may be taken among the $226$ representatives of the diamond orbits of the $2260$ points of degree $9$. The genus of $X_1(44)$ is $36$, and $\dim R_2=105$, $\dim R_4=245$. We \gitlinkIX{C44/build_canonical_v2.m}{compute} the canonical model over $\F_3$, of degree $70$ in $\PP^{35}$, as the canonical embedding of a plane model of degree $19$, from which we also obtain the local parameters, and the multiplication $R_2\times R_2\to R_4$, and we \gitlinkIX{C44/matrix22_all.m}{check} with \Cref{d9:lem:mult} (\Cref{sec:positive}), with $\mathcal L=\mathcal M=2K_X$, that $\ell(B)=0$ in all $226$ cases, which is a contradiction. \end{proof} \subsection{The curve \texorpdfstring{$X_1(49)$}{X1(49)}}\label{subsec:sieve_49} Over octic and nonic fields, $\torz{49}$ is excluded by formal immersion at $2$ in \Cref{d8:prop:49}. Over sextic and septic fields we use the Hecke operators $A_5$ and $A_2$. The prime $q=2$ is not covered by \Cref{d6:prop:tools}(2), so we first show the following. \begin{lemma}\label{lem:J49}\label{d7:lem:J49} The operator $A_2=T_2-2\dm2-1$ annihilates $J_1(49)(\Q)$. \end{lemma} \begin{proof} By \cite[Theorem 3.1(2), Corollary 4.14 and Table 2]{DEHMZ} the group $J_1(49)(\Q)$ is finite with invariant factors $[7,52367710906884085342]$, so its $2$-primary part is $\{0,z\}$ for an element $z$ of order $2$. Reduction modulo $2$ is injective on the odd part, and on the reduction of a rational point Frobenius acts trivially and Verschiebung acts as multiplication by $2$, so $A_2$ annihilates the odd part by the Eichler-Shimura relation. By \Cref{d6:prop:tools}(2) the element $z$ lies in the kernel of $A_5$ on $J_1(49)[2]$, and we \gitlinkVI{two_torsion_49_T2.m}{compute}, using the action of the Hecke algebra on $H_1(J_1(49)(\C),\Z)/2H_1(J_1(49)(\C),\Z)\simeq J_1(49)[2]$, that this kernel has dimension $2$ and meets the kernel of $T_2$ trivially. As $T_2$ and $\dm2$ preserve $\{0,z\}$, it follows that $T_2z=\dm2z=z$, so $A_2z=-2z=0$. \end{proof} \begin{proposition}\label{d6:prop:49}\label{d7:prop:49} The group $\torz{49}$ is not a subgroup of the torsion of an elliptic curve over a number field of degree $6$ or $7$. \end{proposition} \begin{proof} Let $d\in\{6,7\}$, and let $D$ be the trace divisor of a point of order $49$ over a number field of degree $d$. We have $\rk J_1(49)(\Q)=0$ by \cite[Theorem 3.1(2)]{DEHMZ}, and Abramovich's bound gives $\gon_\Q X_1(49)>11.66>7$, so \Cref{d6:prop:main}(a) with $p=3$ excludes the entirely cuspidal reduction. The Galois orbits of cusps over $\Q$ have degrees $1^{21},6^3,21$, and $3$ is a primitive root modulo $49$, so each of them stays irreducible modulo $3$. So every closed cusp of $X_1(49)_{\F_3}$ is the reduction of a rational cuspidal divisor of the same degree, and $A_5C'=A_2C'=0$ for every cuspidal part $C'$ by \Cref{d8:lem:cuspHecke}. By \Cref{d7:table:sieve} the non-cuspidal part $D'$ of the reduction is a single closed point of degree $4\leq\deg D'\leq d$. Moreover $[A_5(D)]=0$ by \Cref{d6:prop:tools}(4), and $[A_2(D)]=0$ by \Cref{lem:J49}, as $A_2c_0=0$, so $[A_5(D')]=[A_2(D')]=0$. We \gitlinkVI{hecke_sieve_cached.m}{compute} that $[A_5(D')]\neq0$ for the representatives of the $1$ and $3$ diamond orbits of degrees $4$ and $5$, and \gitlinkVI{hecke_sieve_cyclic_commonzero.m}{also} for the representatives of the $8$ diamond orbits of degree $6$, and we \gitlinkVII{N49_degree7.m}{compute} that $[A_2(D')]\neq0$ for the representatives of the $19$ diamond orbits of degree $7$. \end{proof} \subsection{The curves \texorpdfstring{$X_1(58)$ and $X_1(62)$}{X1(58) and X1(62)}}\label{subsec:sieve_58_62} \begin{proposition}\label{d9:prop:58_62} The groups $\torz{58}$ and $\torz{62}$ are not subgroups of the torsion of an elliptic curve over a nonic field. \end{proposition} \begin{proof} Let $n=58$ or $62$, put $N=n/2$, so that $N=29$ or $31$, and let $\pi:X_1(n)\to Y=X_1(N)$, $(E,P)\mapsto(E,2P)$, which over a non-cuspidal point $(E,P_N)$ has fibre $E[2]\setminus\{0\}$. We have $\gon_\Q X_1(29)=11$ and $\gon_\Q X_1(31)=12$ \cite[Theorem 4 and Table 1]{DvH}, and \[ J_1(29)(\Q)\simeq\torz4\times\torz4\times\torz{64427244},\qquad J_1(31)(\Q)\simeq\torz{10}\times\torz{1772833370} \] \cite[Theorem 3.1(2), Corollary 4.14 and Table 2]{DEHMZ}, and in both cases $J_1(N)(\Q)$ is generated by rational cuspidal divisors. Let $D_Y=\pi_*D$. The closed cusps of $Y_{\F_3}$ have degrees $1^{N'},N'$ with $N'=(N-1)/2$, as do the orbits of cusps over $\Q$, since $3$ has order $N-1$ modulo $N$. If $D_Y$ reduced to a cuspidal divisor, it would be supported on the rational cusps, and \Cref{d6:prop:main}(a) would give a contradiction. So $D$ reduces to $D'+C'$ with $D'\neq0$. By the Hasse bound a non-cuspidal point of $X_1(n)$ over $\F_{3^e}$ has $e\geq5$, so $D'$ is a single closed point $Q_e$ of degree $e\in\{5,\dots,9\}$, and $C'$ has degree at most $4$, so it is supported on the rational cusps. Write $\pi(Q_e)=P_k$ with $e=fk$ and $f\in\{1,2,3\}$. If $f=2$, then $(k,e)\in\{(3,6),(4,8)\}$, and an elliptic curve over $\F_{27}$ or $\F_{81}$ with a point of order $N$ has $N$ or $3N$ points, an odd number, so its $2$-division polynomial is irreducible and the fibre has no point of degree $2$ over $P_k$. If $f=3$ then $k=3$, as $Y$ has no non-cuspidal points over $\F_9$, and for $N=31$ also no points of degree $3$. So $\pi_*Q_e$ is $3P_3$, for $N=29$, or a point $P_k$ of degree $k\in\{5,\dots,9\}$ at which the elliptic curve has a rational point of order $2$, and the cuspidal part of $\pi_*D$ is supported on the rational cusps. Hence \[ [P_k-k\,\overline{c_0}]\in H\quad\text{or}\quad[3P_3-9\,\overline{c_0}]\in H, \] where $H\subseteq J_Y(\F_3)$ is the reduction of $J_Y(\Q)$. We \gitlinkIX{C58_C62/C58_X129/classify58.m}{compute} \[ \begin{gathered} J_1(29)(\F_3)\simeq(\torz4)^3\times(\torz8)^2\times\torz{386563464},\\ J_1(31)(\F_3)\simeq(\torz2)^2\times\torz{10}\times\torz{3352427902670}, \end{gathered} \] so that $H$ has index $2^9\cdot3$, respectively $2^2\cdot31\cdot61$. For $n=58$ there are $14$, $84$, $168$, $476$ and $1582$ tests in degrees $3$, $6$, $7$, $8$ and $9$, and none of the classes lies in $H$. For $n=62$ there are $90$, $90$, $180$, $630$ and $1200$ tests in degrees $5,\dots,9$, and \gitlinkIX{C58_C62/C62_X131/classify62.m}{exactly $5$} classes lie in $H$, all of degree $9$. These $5$ points are the reductions of the closed point $B$ of degree $9$ of $X_1(31)$, and of its diamond translates, given by the point of order $31$ of \Cref{d9:prop:realizations}. Indeed, we \gitlinkIX{C58_C62/C62_X131/known31/verify_known31.py}{check} that the field of definition of $B$ is inert at $3$ and that the reductions of the $15$ diamond translates of $B$ are exactly these $5$ points. So $D_Y$ and a diamond translate $B'$ of $B$ have the same reduction, and as $[D_Y-B']\in J_1(31)(\Q)$ we get $D_Y\sim B'$, so $D_Y=B'$ by \Cref{d9:lem:closed}. As $X_1(31)$ is a fine moduli space, the elliptic curve of $D$ is then a curve of \Cref{d9:prop:realizations} over its field of definition, and its torsion is $\torz{31}$, so it has no point of order $2$, which is a contradiction. \end{proof} \section{Formal immersion at the prime \texorpdfstring{$2$}{2}}\label{sec:formal}\label{d6:sec:formal}\label{d7:sec:formal}\label{d8:sec:formal}\label{d9:sec:formal} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% At the levels $n$ of this section the trace divisor can reduce modulo $2$ to an entirely cuspidal divisor, and in most cases neither \Cref{d6:prop:main}(b) nor \Cref{d6:prop:several} excludes this: the lower bounds that we have for the gonality of $X_1(n)$ are too weak (for $n=51$ Abramovich's bound gives only $\gon_\C X_1(51)>11.42$, while \Cref{d6:prop:main}(b) needs a bound larger than $12$ already for $d=6$), or the Jacobian has positive rank. We exclude the entirely cuspidal reduction by a formal immersion argument at $2$ (\Cref{d8:prop:cartier}). It allows elements of the Hecke algebra with finite image on the Mordell-Weil group, so it applies also when the Jacobian has positive rank. We treat each level once, for all the degrees in which this method is needed, and usually a single computation, in the largest degree, covers all of them (\Cref{rem:smaller}). The level $75$ is included here because its proof uses the same kind of quotient of $X_1(n)$. Reduction modulo $2$ need not be injective on the torsion of the Jacobian (the point $-1$ of $\mu_2$ over $\Z_2$ reduces to $1$). The following lemma shows that a suitable Hecke operator kills such points. \begin{lemma}\label{d6:lem:dyadic} Let $A$ be an abelian scheme over $\Z_2$, and let $T$ and $\delta$ be endomorphisms of $A$ such that $T=F+\delta V$ on the special fibre $A_{\F_2}$, where $F$ and $V$ are the Frobenius and the Verschiebung of $A_{\F_2}$. If $z\in A(\Q_2)$ satisfies $2z=0$ and reduces to $0$ modulo $2$, then $(T-\delta)z=0$. \end{lemma} \begin{proof} Choose coordinates $X=(X_1,\dots,X_g)$ of the formal group of $A$, that is, functions vanishing on the zero section whose differentials at the origin of $A_{\F_2}$ form a basis of the cotangent space. Then the points of $A(\Q_2)$ that reduce to $0$ are the points with coordinates in $2\Z_2$ (see the proof of Lemma$'$ in \cite[Appendix]{katz}). Write \[ [2](X)=2X+Q(X)+R(X), \] where the coordinates of $Q$ are homogeneous of degree $2$ and $R$ has only terms of degree at least $3$, all with coefficients in $\Z_2$. Let $M$ be a matrix over $\Z_2$ whose reduction $\overline M$ is the matrix of the linear part of $V$. As $[2]=V\circ F$ on $A_{\F_2}$ and $F$ squares the coordinates, the reduction of $Q$ is $\overline Q(X)=\overline M(X_1^2,\dots,X_g^2)^t$. Write the coordinates of $z$ as $2u$ with $u\in\Z_2^g$. As $2z=0$, we have \[ 0=[2](2u)\equiv4u+4Q(u)\pmod 8, \] so $\overline u+\overline M(\overline u_1^2,\dots,\overline u_g^2)^t=0$, and as $a^2=a$ for $a\in\F_2$, we get $(\overline M-1)\overline u=0$. Put $\alpha=T-\delta$. As the linear part of $F$ is $0$, the linear part of $\alpha$ on $A_{\F_2}$ is $\overline N(\overline M-1)$, where $\overline N$ is the linear part of $\delta$. Hence the linear term of $\alpha$ at $2u$ is divisible by $4$, and so are all the other terms, so the coordinates of $\alpha z$ lie in $4\Z_2$. Finally, a non-zero point $v$ with coordinates in $4\Z_2$ is not killed by $2$. Indeed, if $s\geq2$ is the least valuation of a coordinate of $v$, then in such a coordinate the linear term of $[2](v)$ has valuation $s+1$, and all the other terms have valuation at least $2s>s+1$. As $2\alpha z=\alpha(2z)=0$, we get $\alpha z=0$. \end{proof} We apply this to quotients of $X_1(n)$. Let $n$ be odd, let $H\subseteq\Delta(n)$ be a subgroup, and put $Y=X_1(n)/H$. We assume that $H$ has odd order, or that $H$ acts freely on the geometric points of $X_1(n)$ in characteristic $2$, cusps included. In both cases $Y$ is the generic fibre of a smooth proper curve over $\Z_{(2)}$, the quotient of the smooth model of $X_1(n)$, and its cuspidal subscheme is finite \'etale over $\Z_{(2)}$. We write $c_0$ for the image of the standard cusp, and $\Cart$ for the Cartier operator on the differentials of $Y_{\F_2}$: if $x$ is a separating variable, every differential can be written uniquely as $(f^2+g^2x)\,dx$ with $f,g$ rational functions on $Y_{\F_2}$, and $\Cart\bigl((f^2+g^2x)\,dx\bigr)=g\,dx$. Equivalently, in terms of a local parameter $t$ we have $\Cart\bigl(\sum_na_nt^n\,dt\bigr)=\sum_na_{2n+1}^{1/2}t^n\,dt$. The operator $\Cart$ preserves $H^0(Y_{\F_2},\Omega^1)$ and is $\F_2$-linear on it, and at a cusp at which $q$ is a local parameter it acts on $q$-expansions by $\sum_na_nq^n\,dq/q\mapsto\sum_na_{2n}^{1/2}q^n\,dq/q$. If all $a_n$ lie in $\F_2$, this is Atkin's operator $U_2\colon\sum_na_nq^n\mapsto\sum_na_{2n}q^n$ applied to $\sum_na_nq^n$. Note that here $U_2$ is only an operation on $q$-expansions, as $n$ is odd and the Hecke operators $U_q$ are defined only for $q\mid n$. We describe the fixed points of the diamond operators with the following lemma. \begin{lemma}\label{lem:fixed} Let $n$ be odd, and let $\dm a\neq1$ fix a geometric point $x$ of $X_1(n)$ in characteristic $0$ or $2$. If $x$ is a cusp, then $a$ is congruent to the same sign $\pm1$ modulo every prime dividing $n$, so $n$ is not squarefree. If $x$ is not a cusp, then $a^2\equiv-1$ or $a^2\pm a+1\equiv0\pmod n$, and $a\not\equiv\pm1\pmod m$ for every divisor $m\geq5$ of $n$. \end{lemma} \begin{proof} If $x=(E,P)$ is not a cusp, then $u(P)=aP$ for an automorphism $u$ of $E$, and $u\neq\pm1$ as $\dm a\neq1$. Up to sign, $u^2=-1$ or $u^2+u+1=0$, which gives the congruences. If $a\equiv\pm1\pmod m$ for a divisor $m\geq5$ of $n$, then $u$ or $-u$ fixes the point $(n/m)P$ of order $m$, which is impossible, as the kernel of $v-1$ has order at most $4$ for every automorphism $v\neq1$ of $E$. For a cusp in characteristic $0$ this is \Cref{subsec:modular}, and it also holds in characteristic $2$, as the cuspidal subscheme is finite \'etale over $\Z[\tfrac1n]$. \end{proof} Let $\mathbb T_Y$ be the ring of endomorphisms of $J_Y$ generated by the Hecke operators $T_\ell$ for primes $\ell\nmid n$, the operators $U_\ell$ for primes $\ell\mid n$, and the diamond operators, all of which descend to $Y$ as they commute with $H$. By the N\'eron mapping property every $\psi\in\mathbb T_Y$ extends to the N\'eron model of $J_Y$ over $\Z_2$, which is an abelian scheme, and we write $\psi^*$ for its action on the cotangent space $H^0(Y_{\F_2},\Omega^1)$ of the special fibre. As $\psi$ commutes with the Verschiebung of the special fibre, $\psi^*$ commutes with $\Cart$, which is the pull-back by the Verschiebung \cite{oda}. By the Eichler-Shimura relation $T_2=F+\dm2V$ and as the Frobenius $F$ pulls back differentials to $0$, we have \begin{equation}\label{d8:eq:T2star} T_2^*=\Cart\,\dm2^*\quad\text{on } H^0(Y_{\F_2},\Omega^1), \end{equation} and hence \begin{equation}\label{eq:T2minus} (T_2-\dm2)^*=(\Cart-1)\dm2^*\quad\text{on } H^0(Y_{\F_2},\Omega^1). \end{equation} In the following criterion, the elements $\psi_i$ turn the class of the difference of a rational divisor and a cuspidal divisor with the same reduction into a torsion class, \Cref{d6:lem:dyadic} shows that $T_2-\dm2$ kills this class, and by \eqref{eq:T2minus} the formal immersion condition becomes a condition on the space $U$ below. \begin{proposition}\label{d8:prop:cartier}\label{d6:prop:cartier}\label{d7:prop:cartier} Let $Y=X_1(n)/H$ be as above, let $\psi_1,\dots,\psi_r\in\mathbb T_Y$ be such that each $\psi_i\bigl(J_Y(\Q)\bigr)$ is finite, and put \[ U:=(\Cart-1)\Bigl(\sum_{i=1}^r\psi_i^*H^0(Y_{\F_2},\Omega^1)\Bigr). \] Let $\overline C$ be an effective cuspidal divisor of degree $d$ on $Y_{\F_2}$ such that the restriction map $U\to H^0(\overline C,\Omega^1|_{\overline C})$ is surjective, i.e.\ such that $\dim U-\dim\bigl(U\cap\Omega(\overline C)\bigr)=d$. Then no effective rational divisor of degree $d$ on $Y$ with non-cuspidal support reduces to $\overline C$ modulo $2$. \end{proposition} \begin{proof} Suppose that $D$ reduces to $\overline C$. The cuspidal subscheme of $Y$ is finite \'etale at $2$, so $\overline C$ lifts uniquely to a cuspidal divisor $C$, which is fixed by a Frobenius element at $2$, so it is defined over the decomposition field $L\subseteq\Q(\zeta_n)$ of a prime $\lambda$ above $2$, which satisfies $L_\lambda=\Q_2$. For each $i$ the class $\psi_i[D-C]=\psi_i[D-dc_0]-\psi_i[C-dc_0]$ is torsion, by the assumption on $\psi_i$ and by Manin-Drinfeld, and it reduces to $0$ modulo $\lambda$, as $\psi_i$ commutes with reduction. By Parent's result it is killed by $2$, so by \Cref{d6:lem:dyadic}, applied to the N\'eron model of $J(Y)$ over $\Z_2$ with $T=T_2$ and $\delta=\dm2$, we get $(T_2-\dm2)\psi_i[D-C]=0$. Consider the morphism $f:Y^{(d)}\to J(Y)^r$ over $\Z_2$ given by $f(B)=\bigl((T_2-\dm2)\psi_i[B-dc_0]\bigr)_i$. Then $f(D)=f(C)$, and $D$ and $C$ have the same reduction $\overline C$. By \eqref{eq:T2minus}, and since $\dm2^*$ is invertible and commutes with $\Cart$ and with the $\psi_i^*$, the image of the cotangent map of $f$ at $\overline C$ is the restriction of $U$ to $\overline C$, as the cotangent map of $Y^{(d)}\to J(Y)$, $B\mapsto[B-dc_0]$, at $\overline C$ is the restriction map $H^0(Y_{\F_2},\Omega^1)\to H^0(\overline C,\Omega^1|_{\overline C})$. By assumption it is surjective, so $f$ is a formal immersion at $\overline C$, and by \cite[Lemma 4.2]{DKSS} (which is due to Parent \cite[Lemme 4.13]{parentformal}) $f$ is injective on the residue class of $\overline C$. Hence $D=C$, which is a contradiction since $D$ is non-cuspidal. \end{proof} As in \Cref{subsec:doubled}, we compute the subspaces $\Omega(jP)$ of $H^0(Y_{\F_2},\Omega^1)$ once for each closed cusp $P$ of $Y_{\F_2}$ and each $j$, as spaces of differentials or from local expansions at $P$, and $\Omega(\overline C)$ as their intersection. When a group of automorphisms of $Y$ which commute with $\Cart$ and preserve $U$ acts on the cusps, we compute $\Omega(jP)$ for one cusp in each orbit and transport it to the others. When $J(Y)(\Q)$ is finite we take $r=1$ and $\psi_1=1$, and then $U=(\Cart-1)H^0(Y_{\F_2},\Omega^1)$. \begin{remark}\label{rem:smaller} If the restriction of $U$ to $\overline C$ is surjective, then so is its restriction to every effective divisor $\overline C'\leq\overline C$, as the restriction map $H^0(\overline C,\Omega^1|_{\overline C})\to H^0(\overline C',\Omega^1|_{\overline C'})$ is surjective. Every effective cuspidal divisor of degree $d'9$, \Cref{d9:lem:closed} gives $D=C$, which is a contradiction. It remains to show that reduction modulo $2$ is injective on $J_1(45)(\Q)$. It is injective on the odd part. The $2$-primary part of $J_1(45)(\Q)$ is $\torz4\times\torz{32}$ \cite[Theorem 4.13 and Table 2]{DEHMZ}, whose automorphism group is a $2$-group, so the subgroup $H=\{1,16,31\}$ of $\Delta(45)$ acts trivially on it, and $\pi^*\pi_*=3$ on it, where $\pi:X\to Y=X/H$. Hence it suffices to show that reduction modulo $2$ is injective on $J_Y(\Q)[2^\infty]$. Over $F=\Q(\zeta_3)$ the curve $Y$ is isomorphic to $X_1(3,15)$: the quotient of $(E,P)$ by the subgroup generated by $15P$ gives a point of order $15$ together with the kernel of the dual isogeny, with a generator determined by $15P$ and the Weil pairing, and this data is invariant exactly under $H$. This isomorphism comes from the moduli interpretation, so it is compatible with reduction modulo the prime of $F$ above $2$. By \Cref{d8:lem:3_15_C}, reduction modulo this prime is injective on $J_1(3,15)(F)$, and hence on $J_Y(\Q)\subseteq J_Y(F)$. \end{proof} \subsection{The curve \texorpdfstring{$X_1(49)$}{X1(49)}}\label{subsec:formal_49} \begin{proposition}\label{prop:49}\label{d8:prop:49}\label{d9:prop:49} The group $\torz{49}$ is not a subgroup of the torsion of an elliptic curve over a number field of degree $8$ or $9$. \end{proposition} \begin{proof} Let $X=X_1(49)$ and $J=J_1(49)$, let $d\in\{8,9\}$, and let $D$ be the trace divisor of a point of order $49$ over a number field of degree $d$. We have \[ J(\Q)\simeq\torz7\times\torz{52367710906884085342} \] by \cite[Theorem 3.1(2), Corollary 4.14 and Table 2]{DEHMZ}, so $J(\Q)[2^\infty]\simeq\torz2$. We first show that reduction modulo $2$ is injective on $J(\Q)$. It is injective on the odd part. The diamond group $\Delta(49)$ has order $21$, so it acts trivially on $J(\Q)[2^\infty]$, and for the forgetful map $\pi:X\to X_0(49)$ we have $\pi^*\pi_*=\sum_{\delta\in\Delta(49)}\delta=21$ on this group. Hence $\pi_*$ is injective on $J(\Q)[2^\infty]$. The curve $X_0(49)$ is the elliptic curve \lmfdbec{49}{a}{4}, $y^2+xy=x^3-x^2-2x-1$, whose rational point $(2,-1)$ of order $2$ reduces to the non-zero point $(0,1)$ modulo $2$. As reduction commutes with $\pi_*$, it is injective on $J(\Q)[2^\infty]$. No elliptic curve over $\F_{2^e}$ with $e\leq 8$ has a point of order $49$. For $e\leq5$ this follows from the Hasse bound, for $e=6$ a point of order $49$ forces $E(\F_{64})=E[7]$ (see the discussion after \Cref{d6:lem:badred}), and for $e=7,8$ the only possible group orders are $147$ and $245$, with traces of Frobenius $-18$ and $12$, which are impossible by \cite[Theorem 4.1]{waterhouse}. So $X_{\F_2}$ has no non-cuspidal closed points of degree at most $8$, and we \gitlinkIX{C49/prime_and_49_good_branches.m}{compute} that it has $42$ of degree $9$, in $2$ diamond orbits. For $d=9$ a non-cuspidal reduction of $D$ modulo $2$ is one of these points $P$, and $[A_3P]=0$ as $J(\Q)$ is finite, but $A_3P$ is \gitlinkIX{C49/prime_and_49_good_branches.m}{not principal} for a representative $P$ of each orbit. So the reduction $\overline D$ of $D$ modulo $2$ is entirely cuspidal. The Galois orbits of cusps of $X$ have degrees $1^{21},6^3,21$, and the closed cusps of $X_{\F_2}$ have degrees $1^{21},3^6,21$. If $\overline D$ is supported on the cusps of degree $1$, it is the reduction of an effective rational cuspidal divisor $C$, and $[D-C]\in J(\Q)$ reduces to $0$, so $D\sim C$. As $\gon_\Q X>11.66>9$, this contradicts \Cref{d8:lem:closed}. So $\overline D$ contains a cusp of degree $3$, and at most $d-3\leq6$ cusps of degree $1$, counted with multiplicity. The $21$ cusps of degree $1$ lie on the N\'eron $49$-gon, and $\Delta(49)$ acts on them simply transitively, as a diamond operator $\dm a$ multiplies the label of such a cusp in $(\torz{49})^\times/\{\pm1\}$ by $a$. Fix a set $\mathcal A$ of $18$ of these cusps. For each of the at most $6$ cusps of degree $1$ in the support of $\overline D$ and each of the $3$ cusps not in $\mathcal A$, exactly one diamond operator maps the former to the latter. As $6\cdot3<21$, there is a $\delta\in\Delta(49)$ such that $\delta\overline D$ is supported on $\mathcal A$ and the $6$ cusps of degree $3$, and as $\delta D$ is again the trace divisor of a point of order $49$, we may assume that $\delta=1$. The space $U=(\Cart-1)H^0(X_{\F_2},\Omega^1)$ has dimension $68$, and we \gitlinkIX{C49/cartier_mixed9_export.m}{compute} that its restriction to each of the \[ 6\binom{23}6+\binom72\binom{20}3+\binom83=629678 \] effective divisors of degree $9$ supported on these cusps which contain a cusp of degree $3$ is surjective. By \Cref{rem:smaller}, adding a cusp in $\mathcal A$, the same holds for the divisors of degree $8$ of this form, and \Cref{d8:prop:cartier} with $H=\{1\}$ gives a contradiction. \end{proof} \subsection{The curve \texorpdfstring{$X_1(51)$}{X1(51)}}\label{subsec:formal_51} The following lemma treats the quotients $Y_{51}$ (used only in \Cref{sec:computations}) and $Y_{63}$ (used in \Cref{subsec:formal_63}). \begin{lemma}\label{d6:lem:quotients_51_63} Let $Y_{51}:=X_1(51)/\dm{16}$ and $Y_{63}:=X_1(63)/\dm8$. The diamond operators $\dm{16}$ and $\dm8$ have order $2$ and act freely on $X_1(51)$ and $X_1(63)$, also in characteristic $2$, so $Y_{51}$ and $Y_{63}$ have good reduction at $2$ and have genus $33$ and $49$. The function field of $Y_{51}$ is obtained by adjoining to the function field of $X_1(17)$ a root of the $3$-division polynomial of the universal elliptic curve, and that of $Y_{63}$ by adjoining to the function field of $X_1(9)$ a root of the $7$-division polynomial of the universal elliptic curve. The groups $J(Y_{51})(\Q)$ and $J(Y_{63})(\Q)$ are finite. \end{lemma} \begin{proof} Write $(r,s,a)=(17,3,16)$ and $(9,7,8)$ in the two cases, so that $a\equiv-1\pmod r$ and $a\equiv1\pmod s$. A point of order $rs$ corresponds to a pair of points $(P_r,P_s)$ of orders $r$ and $s$, and modulo the common sign, $\dm a$ identifies $(P_r,P_s)$ with $(P_r,-P_s)$. So $Y$ parametrizes pairs $(E,P_r)$ together with the $x$-coordinate of $P_s$, which gives the description of the function field. As $a\equiv-1\pmod r$ with $r\geq9$ and $a\equiv1\pmod s$, \Cref{lem:fixed} shows that $\dm a$ fixes no geometric point of $X_1(rs)$, in characteristic $0$ or $2$. So the quotient map is \'etale, also at $2$, and the genera follow from Riemann-Hurwitz, as $X_1(51)$ and $X_1(63)$ have genus $65$ and $97$. The Jacobian of $X_1(rs)/\dm a$ is isogenous to the part of $J_1(rs)$ on which $\dm a$ acts trivially, whose isogeny factors correspond to the newforms of level dividing $rs$ whose characters $\chi$ satisfy $\chi(a)=1$. For $rs=51$ we have $\rk J_1(51)(\Q)=0$ by \cite[Theorem 3.1(2)]{DEHMZ}. For $rs=63$ the only newform of level dividing $63$ with positive analytic rank is \lmfdbnewform{63}{2}{s}{a}, whose character $\chi$ satisfies $\chi(8)=-1$, as we \gitlinkVI{rank_quotient_63_character.m}{check}, so it does not contribute to $J(Y_{63})$, and $J(Y_{63})(\Q)$ is finite by Kato's theorem. \end{proof} \begin{proposition}\label{prop:51}\label{d6:prop:51_63}\label{d7:prop:51}\label{d8:prop:51}\label{d9:prop:51} The group $\torz{51}$ is not a subgroup of the torsion of an elliptic curve over a number field of degree $d\in\{6,7,8,9\}$. \end{proposition} \begin{proof} Let $X=X_1(51)$, which has genus $65$, and let $D$ be the trace divisor of a point of order $51$ over a number field of degree $d$. We have $\rk J_1(51)(\Q)=0$ \cite[Theorem 3.1(2)]{DEHMZ}. No elliptic curve over $\F_{2^e}$ with $e\leq8$ has a point of order $51$. For $e\leq5$ this follows from the Hasse bound, for $e=6$ and $e=8$ the only multiples of $51$ in the Hasse intervals are $51$ and $255$, with traces of Frobenius $14$ and $2$, which are impossible by \cite[Theorem 4.1]{waterhouse}, and for $e=7$ the Hasse interval $[107,151]$ contains no multiple of $51$. So $X_{\F_2}$ has no non-cuspidal closed points of degree at most $8$, and we \gitlinkIX{common/independent_counts_46_50_51.m}{compute} that it has $80$ of degree $9$, in $5$ diamond orbits. For $d=9$ a non-cuspidal reduction of $D$ modulo $2$ is one of these points $P$, and $[A_5P]=0$ as $J_1(51)(\Q)$ is finite, but we \gitlinkIX{C51/N51_good9_and_pure9_control.m}{compute} that $A_5P$ is not principal for a representative $P$ of each orbit. So the reduction of $D$ is entirely cuspidal. The closed cusps of $X_{\F_2}$ have degrees $1^{16},2^8,8^4$, and $U=(\Cart-1)H^0(X_{\F_2},\Omega^1)$ has dimension $62$. We \gitlinkVIII{common/generic_cartier_completion_export.m}{compute} the subspaces $\Omega(m\overline c)$ with $m\deg\overline c\leq8$, and we \gitlinkIX{C51/N51_good9_and_pure9_control.m}{compute} the subspace $\Omega(9\overline c)$ at a rational cusp $\overline c$, which we transport to the other rational cusps with the diamond operators. From these we \gitlinkIX{C51/N51_replay_degree9.py}{check} that the restriction of $U$ to each of the \[ [t^9](1-t)^{-16}(1-t^2)^{-8}(1-t^8)^{-4}=3333264 \] effective cuspidal divisors of degree $9$ is surjective. By \Cref{rem:smaller} the same holds in the degrees $6$, $7$ and $8$, and \Cref{d8:prop:cartier} with $H=\{1\}$ gives a contradiction. \end{proof} \subsection{The curve \texorpdfstring{$X_1(55)$}{X1(55)}}\label{subsec:formal_55} For the levels $55$ and $77$ the curve $Y$ is a cyclic unramified cover of $X_0(n)$, which we construct over $\F_2$ by class field theory, as we also do for $X_0(57)$ and $X_0(69)$ in \Cref{subsec:covers_57_69}. By class field theory, a geometrically connected unramified cyclic cover of degree $k$ of $Z_{\F_2}$, for a curve $Z$, in which a given rational point splits, corresponds to a surjective homomorphism $\Pic^0(Z_{\F_2})\to\torz k$, and it is determined by its values at enough closed points. For $k=2$ we construct it by an Artin-Schreier equation, and for $k=4$ by Artin-Schreier-Witt equations of length $2$. \begin{lemma}\label{d8:lem:Y55} Let $H\subseteq\Delta(55)$ be the subgroup of order $5$ generated by $\dm{16}$, and $Y=X_1(55)/H$. Then $Y\to X_0(55)$ is an unramified cyclic cover of degree $4$, also over $\F_2$, and $Y$ has genus $17$. The closed cusps of $Y_{\F_2}$ have degrees $1^4,2^2,4^2$, and the push-forward to $Y_{\F_2}$ of an effective cuspidal divisor of degree at most $9$ on $X_1(55)_{\F_2}$ is supported on the four rational cusps above the cusp of width $55$ of $X_0(55)$ and the cusp of degree $4$ above the cusp of width $11$. \end{lemma} \begin{proof} By \Cref{lem:fixed}, $\Delta(n)$ acts freely on $X_1(n)$ for $n\in\{55,77\}$, in characteristic $0$ and in characteristic $2$, as $n$ is squarefree, and for $n=55$ the congruence $a^2\equiv-1$ is impossible modulo $11$ and $a^2\pm a+1\equiv0$ modulo $5$, and for $n=77$ the first is impossible modulo $7$ and the second modulo $11$. So $X_1(55)\to Y\to X_0(55)$ are unramified, also over $\F_2$, and $Y$ has genus $1+(81-1)/5=17$. The group $\Delta(55)/H$ is cyclic of order $4$, generated by $\dm2$. The Frobenius elements at the cusps of $X_0(55)_{\F_2}$ of widths $55,11,5,1$ are $\dm2^0,\dm2^1,\dm2^2,\dm2^3$, which gives the degrees of the cusps of $Y_{\F_2}$. The source cusps of $X_1(55)_{\F_2}$ above the remaining cusps of $Y_{\F_2}$ have degrees $10$ and $20$, which proves the last assertion. \end{proof} We \gitlinkVIII{C55/C55_builtin_model.m}{construct} $Y_{\F_2}$ as follows. We have $\Pic(X_0(55)_{\F_2})\simeq\torz5\times\torz{20}\times\Z$, and there is exactly one homomorphism to $\torz4$ with the Frobenius values at the cusps given in the proof of \Cref{d8:lem:Y55}. We construct the corresponding Artin-Schreier-Witt cover, and check that it has genus $17$, that a rational cusp of width $55$ splits in it, and that its Frobenius elements at a set of generators of $\Pic(X_0(55)_{\F_2})/4$ are the right ones. By class field theory this cover is $Y_{\F_2}$. \begin{proposition}\label{prop:55}\label{d7:prop:55}\label{d8:prop:55}\label{d9:prop:55} The group $\torz{55}$ is not a subgroup of the torsion of an elliptic curve over a number field of degree $d\in\{7,8,9\}$. \end{proposition} \begin{proof} Let $D$ be the trace divisor of a point of order $55$ over a number field of degree $d$. We have $\rk J_1(55)(\Q)=0$ \cite[Theorem 3.1(2)]{DEHMZ}. We \gitlinkIX{C55/independent_counts55.m}{compute} that the non-cuspidal closed points of $X_1(55)_{\F_2}$ of degree at most $9$ are $20$ points of degree $7$, which form a single diamond orbit, and $40$ points of degree $9$, which form $2$ diamond orbits. The closed cusps of $X_1(55)_{\F_2}$ have degrees $1^{20},4^5,10^2,20$, and those of degree $1$ are reductions of rational cusps. So if the reduction of $D$ modulo $2$ is not entirely cuspidal, it is $P+C'$ with $P$ of degree $7$ and $C'$ supported on the reductions of the rational cusps, or $P$ of degree $9$. As $J_1(55)(\Q)$ is finite and $A_3$ annihilates the rational cusps, we get $[A_3P]=0$. We \gitlinkVII{N55_degree7.m}{compute} that $[A_3P]\neq0$ for a representative $P$ of the orbit of points of degree $7$, and \gitlinkIX{C55/N55_good9.m}{that} $A_3P$ is not principal for a representative $P$ of each orbit of points of degree $9$. So the reduction of $D$ is entirely cuspidal. For $d=7$ this contradicts \Cref{d6:prop:main}(b), as Abramovich's bound gives $\gon_\Q X_1(55)>14.28>14$. For $d\in\{8,9\}$ the reduction of the push-forward $\pi_*D$ of $D$ to the curve $Y$ of \Cref{d8:lem:Y55} is entirely cuspidal as well, and by \Cref{d8:lem:Y55} it is supported on the four rational cusps of width $55$ and the closed cusp of degree $4$ above the cusp of width $11$, which gives $[z^d](1-z)^{-4}(1-z^4)^{-1}$ possible divisors, that is, $201$ for $d=8$ and $280$ for $d=9$. As $J_Y(\Q)$ is finite, we apply \Cref{d8:prop:cartier} with $r=1$ and $\psi_1=1$. On the model of $Y_{\F_2}$ described above, $U=(\Cart-1)H^0(Y_{\F_2},\Omega^1)$ has dimension $15$, and we \gitlinkVIII{C55/C55_cartier_reproduce.m}{compute} that its restriction to each of the $201$ divisors of degree $8$ has rank $8$, which proves the claim for $d=8$, and we \gitlinkIX{C55/N55_coupled_FI_RR_v2.m}{compute} that its restriction to each of the $280$ divisors of degree $9$ is surjective, except for $8$ of them, which are supported on the rational cusps and satisfy $\ell(\overline C)=1$. Such a divisor $\overline C$ is the reduction of a rational cuspidal divisor $C$. If reduction modulo $2$ is injective on $J_Y(\Q)$, then $\pi_*D\sim C$, and by semicontinuity $\ell(C)\leq\ell(\overline C)=1$, so $\pi_*D=C$, which is a contradiction. It remains to show that reduction modulo $2$ is injective on $J_Y(\Q)$. By \cite[Theorem 4.13 and Table 2]{DEHMZ}, $J_1(55)(\Q)[2^\infty]\simeq\torz2\times\torz{16}$, and as $\pi_*\pi^*=5$, the group $J_Y(\Q)[2^\infty]$ has order at most $32$. Every closed cusp of $Y_{\F_2}$ is the reduction of a rational cuspidal divisor of the same degree, and we \gitlinkIX{C55/N55_quotient_injectivity_D9_v2.m}{compute} that $J_Y(\F_2)\simeq(\torz5)^2\times\torz{10}\times\torz{1200}$ and that the reductions of the rational cuspidal divisors generate a subgroup isomorphic to $\torz5\times\torz{10}\times\torz{1200}$, whose $2$-primary part has order $32$. Hence reduction is injective on $J_Y(\Q)[2^\infty]$, and on the odd torsion it always is. \end{proof} \subsection{The curve \texorpdfstring{$X_1(63)$}{X1(63)}}\label{subsec:formal_63} \begin{proposition}\label{prop:63}\label{d6:prop:63}\label{d7:prop:63}\label{d8:prop:63}\label{d9:prop:63} The group $\torz{63}$ is not a subgroup of the torsion of an elliptic curve over a number field of degree $d\in\{6,7,8,9\}$. \end{proposition} \begin{proof} Let $Y=Y_{63}$ be the quotient of \Cref{d6:lem:quotients_51_63}, and let $D$ be the push-forward to $Y$ of the trace divisor of a point of order $63$ over a number field of degree $d$. As $J(Y)(\Q)$ is finite by \Cref{d6:lem:quotients_51_63}, the operator $A_5=T_5-1-5\dm5$ annihilates it (\Cref{d6:prop:tools}), and it annihilates the rational cuspidal divisors on $Y$ (\Cref{d8:lem:cuspHecke}). We \gitlinkIX{C63/source_counts.m}{compute} that the non-cuspidal closed points of $X_1(63)_{\F_2}$ of degree at most $9$ have degrees $7$, $8$ and $9$, and that there are $54$, $54$ and $72$ of them. They give $27$, $27$ and $36$ closed points of $Y_{\F_2}$ of the same degrees, which form $3$, $3$ and $4$ orbits under the residual diamond operators $\dm2$ and $\dm5$. The closed cusps of $Y_{\F_2}$ have degrees $1^9,2^3,3^9,6$, and we \gitlinkIX{C63/cusp63.py}{check} that those of degree at most $2$ are reductions of rational cuspidal divisors of the same degree. So if the reduction of $D$ modulo $2$ is not entirely cuspidal, then $d\geq7$, and the reduction is $P+C'$, where $P$ is one of these closed points and $C'$ is the reduction of an effective rational cuspidal divisor of degree $d-\deg P\leq2$. As $A_5$ annihilates $J(Y)(\Q)$ and the rational cuspidal divisors, this gives $[A_5P]=0$. For a representative $P$ of each of the $10$ orbits we compute that $A_5P=T_5P-P-5\dm5P$ is not principal, for the points of degree \gitlinkVII{sieve63_quotient_T5.m}{$7$}, \gitlinkVIII{C63/sieve63_quotient_T5.m}{$8$} and \gitlinkIX{C63/good9_T5.m}{$9$}. So the reduction of $D$ is entirely cuspidal. The space $U=(\Cart-1)H^0(Y_{\F_2},\Omega^1)$ has dimension $45$, and we \gitlinkIX{C63/cartier_rr_D9.m}{compute} that its restriction to each of the $105040$ effective cuspidal divisors of degree $9$ on $Y_{\F_2}$ is surjective. By \Cref{rem:smaller} the same holds in the degrees $6$, $7$ and $8$, and \Cref{d8:prop:cartier} with $r=1$ and $\psi_1=1$ gives a contradiction. \end{proof} \subsection{The curve \texorpdfstring{$X_1(75)$}{X1(75)}}\label{subsec:formal_75} We use the following quotient of $X_1(75)$. Let $H\subseteq\Delta(75)$ be the subgroup of order $2$ generated by $\dm{26}$, and $Y=X_1(75)/H$. As $26\equiv1\pmod{25}$ and $26\equiv-1\pmod3$, \Cref{lem:fixed} shows that $H$ acts freely, also in characteristic $2$, so $Y$ has genus $73$ and good reduction at $2$. The curve $Y$ parametrizes elliptic curves with a point of order $25$ and a cyclic subgroup of order $3$. In characteristic $2$ an ordinary elliptic curve has a model $y^2+xy=x^3+a_2x^2+1/j$, and the non-zero points of a subgroup of order $3$ have a common $x$-coordinate $x_3$ with $x_3^4+x_3^3+1/j=0$. So we \gitlinkVIII{C75/C75_generic_canonical3_root.m}{obtain} a model of $Y_{\F_2}$ by adjoining $x_3$ to the function field of $X_1(25)_{\F_2}$, and we \gitlinkVIII{C75/C75_good7_certificate_root_v2.m}{check} that this quartic extension is irreducible and has genus $73$. \begin{proposition}\label{prop:75}\label{d8:prop:75}\label{d9:prop:75} The group $\torz{75}$ is not a subgroup of the torsion of an elliptic curve over a number field of degree $8$ or $9$. \end{proposition} \begin{proof} Let $d\in\{8,9\}$ and let $D$ be the trace divisor of a point of order $75$ over a number field of degree $d$. We have $\rk J_1(75)(\Q)=0$ \cite[Theorem 3.1(2)]{DEHMZ}, and Abramovich's bound gives $\gon_\Q X_1(75)>23.80>2d$. So by \Cref{d6:prop:main}(b) the elliptic curve has good reduction at some prime above $2$, and the reduction of $D$ modulo $2$ is not entirely cuspidal. We \gitlinkIX{C75/independent_counts75.m}{compute} that the only non-cuspidal closed points of $X_1(75)_{\F_2}$ of degree at most $9$ are $20$ points of degree $7$, which form a single diamond orbit. So the reduction of $D$ is $P+C'$, where $P$ is one of these points and $C'$ is cuspidal of degree $d-7\leq2$, and we \gitlinkIX{C75/N75_cusp_complements_D9.py}{check} that $C'$ is the reduction of a rational cuspidal divisor which is fixed by $\sigma_7$. The Jacobian of the curve $Y$ above has finitely many rational points, as it is a quotient of $J_1(75)$, so the operator $A_7=T_7-1-7\dm7$ annihilates $J_Y(\Q)$, and by \Cref{d8:lem:cuspHecke} it annihilates the image of $C'$ on $Y$. Hence the image $P$ on $Y_{\F_2}$ of the closed point of degree $7$ satisfies $[A_7P]=0$. We \gitlinkVIII{C75/C75_good7_certificate_root_v2.m}{compute} that $T_7P$ consists of four closed points of degree $14$, each with multiplicity $1$, and that $\ell(A_7P)=0$ for a representative $P$ of the orbit, which is a contradiction. \end{proof} \subsection{The curve \texorpdfstring{$X_1(77)$}{X1(77)}}\label{subsec:formal_77} \begin{lemma}\label{d8:lem:77cusps} The groups $(\dm a-1)J_1(77)(\Q)$, for every $a$, and $T_2J_1(77)(\Q)$ are finite. If $D$ is the trace divisor of a point of order $77$ over a number field of degree at most $9$, then the reduction of $D$ modulo $2$ is entirely cuspidal. \end{lemma} \begin{proof} We \gitlinkVIII{C77/winding77_hecke2_certificate.m}{compute} the winding submodules for all even characters modulo $77$ (see also \githubbareIX{C77/winding77.m}). Their dimensions add up to the genus $181$ of $X_1(77)$, every non-trivial character has full winding submodule, and the winding submodule of the trivial character, of dimension $7$, has codimension $1$, and $T_2$ acts as $0$ on the quotient. This proves the first claim. No elliptic curve over $\F_{2^e}$ with $e\leq9$ has a point of order $77$. For $e\leq5$ this follows from the Hasse bound, and for $6\leq e\leq9$ the only multiples of $77$ in the Hasse intervals are $77$, $231$ and $539$, for $e=6,8,9$, with traces of Frobenius $-12$, $26$ and $-26$, which are even and not traces of supersingular curves (see the discussion after \Cref{d6:lem:badred}). So the elliptic curve has bad reduction at all primes above $2$, which is multiplicative by \Cref{d6:lem:additive}, and this proves the second claim. \end{proof} \begin{lemma}\label{d8:lem:Y77} Let $H\subseteq\Delta(77)$ be the subgroup of order $5$ generated by $\dm{36}$, and $Y=X_1(77)/H$. Then $Y\to X_0(77)$ is an unramified cyclic cover of degree $6$, also over $\F_2$, and $Y$ has genus $37$. The closed cusps of $X_1(77)_{\F_2}$ have degrees $1^{30},3^{10},10^3,30$. Those of degree $1$ are the reductions of the $30$ rational cusps, whose images are the $6$ rational cusps of $Y$ above the cusp of width $77$, and those of degree $3$ map to the $2$ closed cusps of degree $3$ of $Y_{\F_2}$ above the cusp of width $11$. \end{lemma} \begin{proof} By the proof of \Cref{d8:lem:Y55}, $\Delta(77)$ acts freely on $X_1(77)$, also in characteristic $2$. So $Y\to X_0(77)$ is unramified, also over $\F_2$, and $Y$ has genus $1+(181-1)/5=37$. The last assertions follow from the description of the cusps, as we \gitlinkIX{common/cusps_order5_quotients.py}{check}. \end{proof} We construct $Y_{\F_2}$ as follows. The curve $X_0(77)_{\F_2}$ is obtained by adjoining to the function field of $X_0(11)_{\F_2}$ a root $v$ of $v^8+jv^7+v^4+1=0$, which gives $X_0(7)\times_{X(1)}X_0(11)$, and we check that its normalization has genus $7$ and four rational cusps of widths $1,7,11,77$. The cover $Y\to X_0(77)$ is the composite of a cyclic cubic and a quadratic cover. The cubic one is $X_1(7)\times_{X(1)}X_0(11)$, which we construct using the Tate normal form of $X_1(7)$, with parameter $t$, and whose automorphisms over $X_0(7)$ are generated by $t\mapsto1/(t+1)$. For the quadratic one we compute $\Pic(X_0(77)_{\F_2})\simeq\torz{30}\times\torz{60}\times\Z$, and there are six homomorphisms to $\torz6$ with the Frobenius values $\dm2^0,\dm2^2,\dm2^3,\dm2^5$ at the cusps of widths $77,11,7,1$. Exactly one of them is compatible with the splitting in $Y$ of the images on $X_0(77)$ of the points of order $77$ over $\F_{1024}$, and we \gitlinkVIII{C77/C77_validate_AS.m}{construct} its quadratic part by an Artin-Schreier equation, which we check against the Frobenius elements at a set of generators of $\Pic(X_0(77)_{\F_2})/2$. The composite has genus $37$ and a split rational cusp, and by class field theory it is $Y_{\F_2}$. \begin{proposition}\label{prop:77}\label{d8:prop:77}\label{d9:prop:77} The group $\torz{77}$ is not a subgroup of the torsion of an elliptic curve over a number field of degree $8$ or $9$. \end{proposition} \begin{proof} Let $d\in\{8,9\}$, let $D$ be the trace divisor of a point of order $77$ over a number field of degree $d$, and let $Y$ be as in \Cref{d8:lem:Y77}. By \Cref{d8:lem:77cusps} the reduction of $D$ modulo $2$ is entirely cuspidal, so it is supported on the closed cusps of degree $1$ and $3$, and by \Cref{d8:lem:Y77} the reduction of the push-forward of $D$ to $Y$ is supported on the $6$ rational cusps of width $77$ and the $2$ closed cusps of degree $3$ of $Y_{\F_2}$. The group $T_2J_Y(\Q)$ is finite by \Cref{d8:lem:77cusps}, as $J_Y$ is a quotient of $J_1(77)$. We apply \Cref{d8:prop:cartier} with $r=1$ and $\psi_1=T_2$, for which \[ U=(\Cart-1)T_2^*H^0(Y_{\F_2},\Omega^1)=\dm2^*\Cart(\Cart-1)H^0(Y_{\F_2},\Omega^1) \] by \eqref{d8:eq:T2star}. We \gitlinkVIII{C77/C77_fast_reproduce.m}{compute} that $\Cart(\Cart-1)H^0(Y_{\F_2},\Omega^1)$ has dimension $32$, and we \gitlinkVIII{C77/C77_deck_reproduce.m}{compute} the spaces $\Omega(jP)$ for $j\leq8$ at one rational cusp $P$, which we transport to the other five with the automorphisms of $Y$ over $X_0(77)$, as these commute with $\Cart$. We \gitlinkIX{C77/build_model_deck.m}{compute} the spaces for $j=9$ at a rational cusp and for $j\leq3$ at a cusp of degree $3$, which we transport in the same way, and we \gitlinkIX{C77/replay_fi9.py}{check} that the restriction of $U$ to each of the \[ \binom{14}5+2\binom{11}5+3\binom83+4=3098 \] effective divisors of degree $9$ supported on these $8$ cusps is surjective. By \Cref{rem:smaller} the same holds in degree $8$, and \Cref{d8:prop:cartier} gives a contradiction. \end{proof} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Riemann-Roch computations in the canonical ring}\label{sec:positive}\label{d9:sec:positive} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% In this section we treat levels $n$ at which the Hecke sieve has to be applied over nonic fields on curves of large genus. For most of them $J_1(n)$ has positive rank. The groups $\torz{57}$ and $\torz{65}$ over fields of degree $6$, $7$ and $8$ are treated in \Cref{sec:57_65}. On such curves our main tool for showing that a divisor class over a finite field is not principal is the following criterion, which replaces Riemann-Roch computations by linear algebra in the canonical ring. \begin{lemma}\label{d9:lem:mult} Let $X$ be a smooth projective geometrically integral curve over a field $k$, and let $B=T-U$ with $T,U$ effective divisors of the same degree. Let $\mathcal L,\mathcal M$ be line bundles on $X$, and choose $0\neq F\in H^0(X,\mathcal L(-T))$ and $H\in H^0(X,\mathcal M(-T))$. If the linear map \[ H^0(X,\mathcal L)\longrightarrow H^0(U,\mathcal L|_U)\oplus \frac{H^0(X,\mathcal L\otimes\mathcal M)}{F\cdot H^0(X,\mathcal M)},\qquad G\longmapsto\bigl(G|_U,\ GH \bmod F\cdot H^0(X,\mathcal M)\bigr) \] is injective, then $\ell(B)=0$. The same holds if several sections $H$ are used. \end{lemma} \begin{proof} Let $f\in L(T-U)$. As $F$ and $H$ vanish on $T$, we have $G:=fF\in H^0(X,\mathcal L(-U))$ and $fH\in H^0(X,\mathcal M)$. Hence $GH=F\cdot fH$, so $G$ lies in the kernel of the map, and $G=0$. As $F\neq0$ we get $f=0$. \end{proof} We apply the lemma with $\mathcal L=iK_X$ and $\mathcal M=jK_X$, so that the second component is the multiplication $R_i\times R_j\to R_{i+j}$. For the curves of this section we compute an integral basis of $S_2(\Gamma_1(n))$, or of the cusp forms on the quotient that we use, including the old forms, and its $q$-expansions beyond the Sturm bound. This gives the canonical model and the products in $R_2$ (or $R_3$, $R_4$), as products of $q$-expansions are determined by their coefficients up to the Sturm bound, and we check by comparing dimensions that the products of the $q$-expansions of the basis of $R_1=H^0(X,\Omega^1)$ span the spaces $R_i$ that we use. To evaluate sections at the support of $U$ we need local parameters there. At non-cuspidal points we obtain them from the deformation $a_6\mapsto a_6+t$ of an ordinary curve $y^2+xy=x^3+a_2x^2+a_6$, or $a_1\mapsto t$ of a supersingular curve $y^2+a_3y=x^3+a_4x+a_6$ (these are the normal forms of ordinary and supersingular elliptic curves over a field of characteristic $2$), by lifting the point of order $n$ by Hensel's lemma, and then compute the canonical coordinates from the standard modular units of \cite[Sections 2 and 2.1]{DvH}, whose linear relations with the canonical forms we verify to an order beyond the degree bound. For $n=87$ we obtain the local parameters from the canonical quadrics instead. When $\alpha=\dm a-1$ has finite image on $J_1(n)(\Q)$, the class $[\dm aD-D]$ is torsion, and for a non-cuspidal closed point $P$ the divisor $A_q(\dm aP-P)$ has positive and negative parts of the same degree. The divisors $T_qP$ are computed from the $q+1$ isogenies of degree $q$ of an elliptic curve representing $P$. As the identification of the diamond operators on the canonical model with those on the moduli problem is a matter of convention (see \Cref{subsec:hecke}), we test both forms of the annihilator and all the simultaneous diamond translates of the target points, which covers all conventions, unless the divisors are computed directly on the moduli model, as in Step 3 of \Cref{d9:prop:57}, or the two forms coincide, as for $n=87$. \begin{proposition}\label{d9:prop:87} The group $\torz{87}$ is not a subgroup of the torsion of an elliptic curve over a nonic field. \end{proposition} \begin{proof} The group $J_1(87)(\Q)$ is finite \cite[Theorem 3.1(2)]{DEHMZ}, and $i(87)=3360$, so $\gon_\C X_1(87)\geq\tfrac{325}{2^{15}}\cdot3360>33>18$, and \Cref{d6:prop:main}(b) excludes the entirely cuspidal reduction modulo $2$. We \gitlinkIX{common/finite_field_prime_orders.m}{compute} that no elliptic curve over $\F_{2^e}$ with $e\leq8$ has a point of order $87$, and \gitlinkIX{C87/direct5/isogeny5_targets.m}{that} over $\F_{2^9}$ exactly $36$ curves do, all ordinary with $522$ points and cyclic $87$-torsion, with $j$-invariants in $4$ Frobenius orbits of length $9$. So $D$ reduces to a single closed point of degree $9$. Let $H$ be the subgroup of $\Delta(87)$ generated by $\dm{16}$, which has order $7$, and let $Y=X_1(87)/H$, of genus $33$. As $16^2\equiv-5\pmod{87}$, the operator $\dm5$ acts trivially on $Y$, so $A_5=T_5-6$ kills $J_Y(\Q)$, and $A_5c_0=6c_0-6c_0=0$. The map $X_1(87)\to Y$ is Galois of degree $7$, so the image $P$ of the reduction of $D$ is a closed point of degree $9$ of $Y_{\F_2}$, one of $16$, and $Q-6P$ is principal, where $Q=T_5P$. For each of the $4$ elliptic curves above, the $6$ isogenies of degree $5$ have distinct target $j$-invariants which form one Frobenius orbit of length $54$, so $Q$ is a single closed point of degree $54$. We compute a canonical model of $Y_{\F_2}$, of degree $64$ in $\PP^{32}$, and the complete fibres of the $j$-map above the $4$ target $j$-invariants, which contain $16$ closed points of degree $54$ each. We \gitlinkIX{C87/canonical_replay/replay64.py}{check} with \Cref{d9:lem:mult}, with $\mathcal L=\mathcal M=2K$, that $\ell(Q-6P)=0$ for one $P$ in each diamond orbit and all $16$ points $Q$ in the corresponding fibre, i.e.\ in $64$ cases. \end{proof} \begin{proposition}\label{d9:prop:57} The group $\torz{57}$ is not a subgroup of the torsion of an elliptic curve over a nonic field. \end{proposition} \begin{proof} By \Cref{lem:57_65} and its proof, every diamond difference has finite image on $J_1(57)(\Q)$, and so do the operators $T_\ell-a_\ell$ for $\ell\in\{2,5,7,11,13\}$, where $a_2,a_5,a_7,a_{11},a_{13}=-2,-3,-5,1,2$ are the coefficients of \lmfdbnewform{57}{2}{a}{a}, as they act as $0$ on the corresponding factor. By the Hasse bound $X_1(57)_{\F_2}$ has no non-cuspidal points of degree at most $5$. \textit{Step 1: a quotient of genus $9$.} Let $Y_{57}=X_1(57)/H$ be the curve of genus $9$ of \Cref{d7:lem:double}, where $H$, the subgroup of $\Delta(57)$ of odd order, is generated by $\dm4$ and has order $9$, and let $\sigma=\dm5$, which generates $\Delta(57)/H$. The Jacobian $J_{Y_{57}}$ is isogenous to the product of $J_0(57)$ and an abelian fourfold $A$ with $L(A,1)\neq0$, on which $\sigma$ acts as $-1$, so $\alpha J_{Y_{57}}(\Q)$ is finite for $\alpha=\sigma-1$. We compute that $J_{Y_{57}}(\F_2)\simeq\torz{105}\times\torz{210}$ and that the $\alpha$-images of the classes of the cuspidal divisors generate a subgroup $V\simeq(\torz7)^2$, which is the whole $\alpha$-image as $\#A(\F_2)=49$. On $V$ the annihilator $A_5$, computed from the action of $T_5$ on the cusps, has kernel $V_0\simeq\torz7$, so the reduction of $\alpha J_{Y_{57}}(\Q)$ lies in $V_0$. The closed cusps of $X_1(57)$ of degree at most $3$ are reductions of rational cuspidal divisors, so if $D$ reduces to $P+C'$ with $P$ non-cuspidal, of degree $e\geq6$, then the class of the image of $P$ in $J_{Y_{57}}(\F_2)$, normalized at a rational cusp, satisfies $\alpha[\pi_*P-e\,\overline{c_0}]\in V_0$. We \gitlinkIX{C57/quotient9/finite57_v8_export.m}{enumerate} the closed points of $(Y_{57})_{\F_2}$ of degree at most $9$ above the $j$-invariants of the elliptic curves over $\F_{2^e}$ with a point of order $57$, and this condition leaves only $3$ possibilities for $\pi_*P$: $3P_2$ with $P_2$ of degree $2$ above $j=0$, and $2$ points of degree $8$. Similarly, the only closed cusps of $X_1(57)_{\F_2}$ of degree at most $9$ which are not reductions of rational cuspidal divisors of the same degree are $2$ cusps of degree $9$ splitting an orbit of degree $18$, and for both the condition fails. So either the reduction of $D$ is $P+C'$ with $P$ of degree $6$ or $8$ above one of the $3$ points above and $C'$ the reduction of a rational cuspidal divisor $C$, or it is the reduction of a rational cuspidal divisor $C$. \textit{Step 2: the cuspidal case.} For $r\in\{2,5,7,11,13\}$ and $\ell\in\{5,7,11,13\}$ put $\Phi_{r,\ell}=(T_r-a_r)(T_\ell-\dm\ell-\ell)$ on the $q$-expansion model. Each of them kills $J_1(57)(\Q)$, so $\Phi_{r,\ell}[D-C]=0$, and we use the map $X_1(57)^{(9)}\to J_1(57)^{20}$ given by all of them, whose cotangent space modulo $2$ we compute from the integral canonical lattice. The pull-backs of the differentials span a space of dimension $82$. The cuspidal divisors of degree $9$ in question are supported on the $18$ rational cusps and the $9$ closed cusps of degree $2$, and we \gitlinkIX{C57/source85/replay_all_cusp_FI_sage.py}{check} that the map is a formal immersion at all of them, which after moving a rational cusp of maximal multiplicity to $c_0$ by a diamond operator gives $796111$ configurations, using local parameters at the cusps of degree $2$ obtained from the Tate curve $E(q^{19})$ with the point $\zeta_3q$. As $D$ and $C$ have the same reduction and the same image, $D=C$ by \cite[Lemma 4.2]{DKSS}, which is a contradiction. \textit{Step 3: the non-cuspidal case.} The elliptic curves over $\F_{2^6}$ with a point of order $57$ are $2$ supersingular curves with $57$ points, and those over $\F_{2^8}$ are ordinary with $228$ points, with $j$-invariants in one Frobenius orbit. Put $\sigma=\dm5$ and $A_5=T_5-1-5\sigma$ on the moduli model. As $[\sigma D-D]$ and $[\sigma C-C]$ are torsion and killed by $A_5$, the divisor \[ B=A_5(\sigma P-P)=\sigma T-T+P+4\sigma P-5\sigma^2P,\qquad T=T_5P, \] is principal. We compute $T$ from the isogenies of degree $5$: it consists of closed points of degrees $18$ and $9$, with multiplicities $1$ and $2$, for the $2$ points of degree $6$, and of $2$ closed points of degree $24$ for the point of degree $8$. We \gitlinkIX{C57/source85/rr24_v3.m}{check} with \Cref{d9:lem:mult}, with $\mathcal L=\mathcal M=2K$ on $X_1(57)$, of genus $85$, that $\ell(B)=0$ in these $3$ cases, and the other convention as well. \end{proof} \begin{proposition}\label{d9:prop:65} The group $\torz{65}$ is not a subgroup of the torsion of an elliptic curve over a nonic field. \end{proposition} \begin{proof} By \Cref{lem:57_65} and its proof, every diamond difference has finite image on $J_1(65)(\Q)$, and so do the operators $T_\ell-a_\ell$ for $\ell\in\{2,3,7,11,17\}$, where $a_2,a_3,a_7,a_{11},a_{17}=-1,-2,-4,2,2$ are the coefficients of \lmfdbnewform{65}{2}{a}{a}, as they act as $0$ on the corresponding factor. By the Hasse bound $X_1(65)_{\F_2}$ has no non-cuspidal points of degree at most $5$. The elliptic curves over $\F_{2^6}$ with a point of order $65$ are supersingular with $65$ points, and those over $\F_{2^7}$, $\F_{2^8}$, $\F_{2^9}$ are ordinary, with $130$, $260$ and $520$ points, and give $48$, $48$ and $72$ closed points of degree $7$, $8$ and $9$ of $X_1(65)_{\F_2}$, in $2$, $2$ and $3$ diamond orbits. The degrees of the Galois orbits of cusps and of the closed cusps of $X_1(65)_{\F_2}$ are given in \Cref{subsec:57_65_red2}. \textit{Non-cuspidal reduction.} Suppose that $D$ reduces to $P+C'$ with $P$ a non-cuspidal closed point. Then $C'$ has degree at most $3$, so it is the reduction of a rational cuspidal divisor. Put $\sigma=\dm3^{-1}$, $\tau=\dm3$ and $A_3=T_3-1-3\tau$. As in \Cref{d9:prop:57}, with $T=T_3P$ the divisor $A_3(\sigma P-P)=\sigma T-T-\sigma P-2P+3\tau P$ is principal. We compute $X_1(65)$, of genus $121$, through its canonical forms and the units $x=f_7/f_8$, $y=f_8/f_9$ of \cite[Sections 2 and 2.1]{DvH}, and we \gitlinkIX{C65/composite65/canonical_R1_sieve.py}{check} with \Cref{d9:lem:mult}, with $\mathcal L=\mathcal M=K$, that this divisor and its analogue for the other convention are not principal, for every diamond translate of $T$: these are $336$ tests for the points of degree $7$, $8$, $9$. For the supersingular point of degree $6$, $T$ consists of $2$ closed points of degree $12$ and is computed on the moduli model, and we \gitlinkIX{C65/CM6/rr12.m}{check} that $A_3(\tau P-P)$, which equals $-\tau A_3(\sigma P-P)$ as $\tau=\sigma^{-1}$, and its analogue for the other form of the annihilator are not principal. \textit{Cuspidal reduction at a non-rational cusp.} The $4$ closed cusps of degree $6$ split an orbit of degree $24$. Suppose that $D$ reduces to such a cusp $c$ plus the reduction of a rational cuspidal divisor $C$ of degree $3$. Let $\pi:X_1(65)\to Y$ be the quotient by the kernel of the character $u\mapsto\left(\frac{65}{u}\right)$ of $\Delta(65)$, a curve of genus $9$. By \Cref{lem:goodred}, $Y$ has good reduction at $2$ and $\pi$ extends to the smooth models. We \gitlinkIX{C65/quotient9/finite65.m}{compute} $J_Y(\F_2)\simeq\torz4\times\torz{16}\times\torz{336}$, and that the images under $\alpha=\dm3-1$ of the classes of the cuspidal divisors generate a subgroup $V\simeq\torz4\times\torz8$, on which $T_3-\dm3-3$, the annihilator of the $q$-expansion model, has kernel isomorphic to $\torz8$. Each cusp $c$ maps to a rational cusp $c_Y$ of $Y$ of width $65$ with multiplicity $6$, and we check that $6\alpha[c_Y-\overline{c_0}]\notin\ker\bigl((T_3-\dm3-3)|_V\bigr)$. But this class is the reduction of $\alpha[\pi_*D-9c_0]-\alpha[\pi_*C-3c_0]$, and both terms lie in the finite group $\alpha J_Y(\Q)$ by \Cref{lem:57_65}, so they are killed by $T_3-\dm3-3$, which is a contradiction. \textit{Other cuspidal reductions.} They are reductions of rational cuspidal divisors $C$ supported on the $24$ rational cusps and the $6$ cusps of degree $4$. As in \Cref{d9:prop:57}, we use $\Phi_{r,\ell}=(T_r-a_r)(T_\ell-\dm\ell-\ell)$ for $r\in\{2,3,7,11,17\}$ and $\ell\in\{3,7,11,17\}$, whose pull-backs span a space of dimension $106$ modulo $2$, and we \gitlinkIX{C65/allbad/replay_all_cusp_FI_sage.py}{check} that the corresponding map is a formal immersion at all such divisors, which gives $2250325$ configurations after moving a rational cusp of maximal multiplicity to $c_0$. As $D$ and $C$ have the same reduction and the same image, $D=C$ by \cite[Lemma 4.2]{DKSS}, which is a contradiction. \end{proof} \begin{proposition}\label{d9:prop:85_91_93} The groups $\torz{85}$, $\torz{91}$ and $\torz{93}$ are not subgroups of the torsion of an elliptic curve over a nonic field. \end{proposition} \begin{proof} Let $n\in\{85,91,93\}$. By \Cref{d9:rem:exactdegree}, the trace divisor $D$ is a closed point of degree $9$. Put $h=\dm{16}$, $\dm{27}$ and $\dm4$ respectively, of order $2$, $2$ and $5$ in $\Delta(n)$. We \gitlinkIX{common/probe_winding.m}{compute} that the characters of the factors outside the winding quotient are trivial on $\dm2$ and $\dm{16}$ for $n=85$ ($2$ factors of dimension $2$), on $\dm3$ and hence on $\dm{27}$ for $n=91$, and that for $n=93$ there is a single such factor, of dimension $2$ and trivial character. So $(h-1)J_1(n)(\Q)$ is finite, and so are $(\dm2-1)J_1(85)(\Q)$ and $(\dm3-1)J_1(91)(\Q)$. \textit{Cuspidal reduction at $2$.} We have $i(n)=3456$, $4032$, $3840$, and $\gon_\C X_1(n)\geq\tfrac{325}{2^{15}}i(n)>34$, $>39$, $>38$. Suppose that $D$ reduces to a cuspidal divisor, and let $C$ be a lift over an unramified extension of $\Q_2$. For $n=91$ and $n=93$, \Cref{d9:lem:stab}(c) gives $hD=D$, as $2(9+9)=36$, which contradicts \Cref{d9:lem:odd}, as $h$ has order $2$, respectively $5$. For $n=85$ the closed cusps of $X_1(85)_{\F_2}$ have degrees $1^{32},4^8,8^8$, and $h$ acts freely on the rational cusps and fixes every non-rational closed cusp. If the reduction contains a non-rational cusp, write it as $Z+\overline R$ with $Z$ supported on the non-rational cusps and $R$ a rational cuspidal divisor of odd degree $r\in\{1,5\}$. Then \Cref{d9:lem:stab}(c), with $C=R$ and $2(9+r)\leq28$, gives $hR=R$, which is impossible as $h$ acts freely on the rational cusps and $r$ is odd. If the reduction is supported on the rational cusps, it is the reduction of a rational cuspidal divisor $C$. We use the polynomial $k(X)=(X^2+4X+2)(X^2+2X+2)$, the product of the characteristic polynomials of $T_3$ on the $2$ factors outside the winding quotient, so that $k(T_3)J_1(85)(\Q)$ is finite, and the operator $t=(T_3-\dm3-3)k(T_3)$ on the $q$-expansion model, which kills $J_1(85)(\Q)$. So $t[D-9c_0]=t[C-9c_0]$, and by \cite[Lemma 4.2]{DKSS} it suffices to show that the map $X_1(85)^{(9)}\to J_1(85)$, $x\mapsto t[x-9c_0]$, is a formal immersion modulo $2$ at every rational cuspidal divisor of degree $9$. By the criterion of \cite[Lemma 5.9 and Proposition 5.10]{DEHMZ}, applied as in \cite[Corollary 4.4 and Proposition 5.3]{DKSS}, whose proof works for any level, this holds at $\sum_cm_cc$ if the rows $t\dm{g_c}T_j$ with $1\leq j\leq m_c$ are linearly independent modulo $2$, where $\dm{g_c}$ is the diamond operator which maps $c_0$ to $c$. At most one cusp has $m_c\geq5$, and we may move it to $c_0$ by a diamond operator. We \gitlinkIX{C85_C91_C93/C85/rational_cusp_FI_positive.m}{compute} the $133\times225$ matrix of the rows $tT_j$, $j\leq9$, at $c_0$ and $t\dm{g_c}T_j$, $j\leq4$, at the other $31$ rational cusps $c$, and we \gitlinkIX{C85_C91_C93/C85/verify_FI_positive_readonly.py}{check} that its rank is $122$ and that for every linear relation between its rows, the sum over the cusps $c$ of the largest $j$ for which a row at $c$ occurs in the relation is at least $32$. As $\sum_cm_c=9<32$, no relation involves only rows with $j\leq m_c$, so the criterion holds, and $D=C$, which is a contradiction. \textit{Non-cuspidal reduction.} By the Hasse bound and an \gitlinkIX{common/odd_census.m}{enumeration} of the elliptic curves over $\F_{2^e}$, $e\leq9$, the non-cuspidal closed points of $X_1(n)_{\F_2}$ of degree at most $9$ are as follows. For $n=85$, there are $160$ points of degree $9$ from curves with $510$ points, in $5$ diamond orbits. For $n=91$ there are $3$ points of degree $8$, with $j=0$, from $2$ twists with $273$ points, all fixed by $h$, and $144$ points of degree $9$ from curves with $546$ points, in $4$ orbits. For $n=93$ there are $10$ points of degree $9$ from curves with $558$ points, in one orbit. For $n=91$, if $D$ reduces to $P_8+c$ with $P_8$ of degree $8$ and $c$ a rational cusp, \Cref{d9:lem:stab}(c) gives $hc=c$, which is impossible as $h$ acts freely on the rational cusps. In the remaining cases $D$ reduces to a single point $P$ of degree $9$, and with $\sigma=\dm2^{-1}$, $\dm3^{-1}$, $\dm2^{-1}$ respectively, the divisor $A_q(\sigma P-P)$, with $q=3$, $3$, $5$, is principal. We \gitlinkIX{C85_C91_C93/C91/canonical_R1_sieve.py}{check} with \Cref{d9:lem:mult}, with $\mathcal L=\mathcal M=K$, on $X_1(n)$ of genus $225$, $265$ and $261$, that these divisors are not principal, in $320$, $288$ and $120$ tests respectively, which include both conventions and all the simultaneous diamond translates of the targets. \end{proof} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{The groups \texorpdfstring{$\torz{57}$, $\torz{65}$ and $\torz{69}$}{Z/57Z, Z/65Z and Z/69Z}}\label{sec:57_65}\label{d6:sec:57_65}\label{d7:sec:57_69}\label{d8:sec:57_65} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% The groups $\torz{57}$ and $\torz{65}$ over fields of degree $6$, $7$ and $8$ require several arguments each. For both levels the Jacobian $J_1(n)$ has positive rank, and over finite fields of characteristic $2$ there are elliptic curves with a point of order $n$. We first restrict the possible reductions of the trace divisor modulo $2$ (\Cref{subsec:57_65_red2}). The remaining reductions are excluded on $X_1(n)$ and on its quotients by subgroups of $\Delta(n)$, using elements of the Hecke algebra whose image on the Mordell-Weil group is finite, and for $n=57$ some of these quotients are unramified covers of $X_0(57)$ which we construct over $\F_2$ by class field theory (\Cref{subsec:covers_57_69}). Most entirely cuspidal reductions are excluded with the Cartier operator (\Cref{d6:prop:cartier}). The level $69$ uses the same kind of double cover as $57$, so we treat it here as well, in degrees $7$, $8$ and $9$ (\Cref{subsec:63_69}). The group $\torz{63}$ is treated in \Cref{prop:63}, and the groups $\torz{57}$ and $\torz{65}$ over nonic fields in \Cref{sec:positive}. \subsection{Reduction modulo \texorpdfstring{$2$}{2}}\label{subsec:57_65_red2} We first collect the facts about the Mordell-Weil groups that we use. \begin{lemma}\label{lem:57_65} Let $n\in\{57,65\}$. The only newform $f$ of level dividing $n$ with $L(f,1)=0$ is \lmfdbnewform{57}{2}{a}{a} for $n=57$ and \lmfdbnewform{65}{2}{a}{a} for $n=65$. It has trivial character, and its Fourier coefficients satisfy $a_2=-2$ and $a_3=-1$ for $n=57$, and $a_2=-1$ for $n=65$. Hence $(\dm a-1)J_1(n)(\Q)$ is finite for every $a$, and so are $(T_2+2)J_1(57)(\Q)$, $(U_3+1)J_1(57)(\Q)$, $(T_2+1)J_1(65)(\Q)$ and $(T_2^6-1)J_1(65)(\Q)$. The same holds on all the quotients of $X_1(n)$ by subgroups of $\Delta(n)$. \end{lemma} \begin{proof} By the LMFDB, these are the only newforms of level dividing $n$ with positive analytic rank. We \gitlinkVIII{C57/winding57_hecke2_certificate.m}{verify} this by computing the winding submodules of all the even characters modulo $n$, whose dimensions add up to the genus of $X_1(n)$: every non-trivial character has full winding submodule, and the winding submodule of the trivial character has codimension $1$, with $T_2=-2$ and $U_3=-1$ on the quotient for $n=57$, and $T_2=-1$ for $n=65$ (see also \githubbareVIII{C57/winding57_hecke3_certificate.m} and \githubbareVIII{C65/winding65_hecke2_certificate.m}). By Kato's theorem, all the factors of $J_1(n)$ except the one corresponding to $f$ have finitely many rational points, and the operators above act as $0$ on this factor. The Jacobians of the quotients of $X_1(n)$ by subgroups of $\Delta(n)$ are quotients of $J_1(n)$. \end{proof} Over sextic fields we first show that for $n=57$ the reduction to a supersingular curve over $\F_{64}$ cannot occur. \begin{proposition}\label{d6:prop:good2} Let $E$ be an elliptic curve over a sextic field $K$ with a point of order $57$. Then $E$ has bad reduction at every prime of $K$ above $2$. \end{proposition} \begin{proof} There are no points of order $57$ over number fields of degree at most $3$ \cite{DEHMZ}, so the point $x\in Y_1(57)(K)$ has degree $6$ and its trace divisor $D$ is a closed point of degree $6$. Suppose that $E$ has good reduction at a prime $\wp$ of $K$ above $2$. By the Hasse bound $\F_\wp=\F_{64}$, so $\wp$ is the only prime of $K$ above $2$, and it is unramified. The reduction $\widetilde E$ has a point of order $57$ over $\F_{64}$, so $\#\widetilde E(\F_{64})=57$, and the trace of Frobenius is $8$. Thus $\widetilde E$ is supersingular and $j(\widetilde E)=0$. As $\widetilde E$ is supersingular, $[2]$ is purely inseparable \cite[Theorem V.3.1]{silverman}, so $[8]$ and the Frobenius endomorphism $\pi$ of $\widetilde E$ are purely inseparable isogenies of degree $64$. By \cite[Corollary II.2.12]{silverman} both are the $64$-th power Frobenius morphism followed by an isomorphism, so $\pi=\zeta\circ[8]$ for an automorphism $\zeta$ of $\widetilde E$. As $\pi$ acts trivially on $\widetilde E(\F_{64})$, which has odd order $57$, the automorphism $\zeta$ acts on it as multiplication by $8^{-1}$. In particular $\zeta$ maps $8\widetilde P$ to $\widetilde P$, where $\widetilde P$ is the reduction of the point of order $57$, so $(\widetilde E,8\widetilde P)\simeq (\widetilde E,\widetilde P)$ and $\dm8$ fixes the reduction of $x$. As $\dm8$ is defined over $\Q$, it also fixes the conjugates of this point, so the reduction of $D$ is fixed by $\dm8$. By \Cref{lem:57_65} the group $(\dm8-1)J_1(57)(\Q)$ is finite, and Abramovich's bound gives $\gon_\C X_1(57)>14.28>2\cdot6$. So \Cref{d9:lem:stab}(c), applied with $g=\dm8$ and $C=0$, gives $\dm8D=D$. The least degree of a CM point on $X_1(57)$ is $12$, so the support of $D$ is non-CM and, by \Cref{lem:orbits}, $\dm8$ acts freely on its geometric points. The operator $\dm8$ has order $3$ in $\Delta(57)$, so the image of $x$ on $X_1(57)/\dm8$ has degree $2$, and the image of $x$ on $X_0(57)$ has degree $1$ or $2$. The curve $X_0(57)$ has no rational non-cuspidal points, and its quadratic points were determined by Box \cite[\S 4.4]{Box}. The non-CM ones are defined over $\Q(\sqrt{-23})$ or $\Q(\sqrt{13})$, and the corresponding quadratic field is contained in $K$. The prime $2$ splits in $\Q(\sqrt{-23})$, which contradicts the uniqueness of the prime of $K$ above $2$. The two $j$-invariants over $\Q(\sqrt{13})$ are \[ \frac{3387888351672962316333\sqrt{13}-12215205167504087643323}{2}\quad\text{and}\quad \frac{-17787\sqrt{13}-61763}{2}, \] and writing them in the basis $1,t$ of the ring of integers, with $t=(1+\sqrt{13})/2$, they are congruent to $t$ modulo the prime above $2$, which is inert. As $t^2+t+1\equiv0$ modulo this prime, these $j$-invariants are non-zero modulo $2$, contradicting $j(\widetilde E)=0$. \end{proof} \begin{lemma}\label{d6:lem:57two} The diamond operator $\dm2$ acts trivially on $J_1(57)(\Q)[2]$. \end{lemma} \begin{proof} Let $J=J_1(57)$. By \Cref{d6:prop:tools}(2), $J(\Q)[2]$ is contained in the kernel of $A_5$ on $J[2]$. We \gitlinkVI{rational_two_torsion.m}{compute} the action of the Hecke operators on $H_1(J(\C),\Z)/2H_1(J(\C),\Z)\simeq J[2]$, a space of dimension $170$ over $\F_2$, and find that the kernel of $A_5$ has dimension $2$ and that $\dm2-1$ vanishes on it. Here we obtain the diamond operators over $\Z$ from the relation $q\dm q=T_q^2-T_{q^2}$, with $q=2$ and $q=5$. \end{proof} We \gitlinkVIII{common/independent_counts.m}{compute} that the numbers of non-cuspidal closed points of $X_1(n)_{\F_2}$ of degrees $1,\dots,8$ are $0,0,0,0,0,2,36,18$ for $n=57$ and $0,0,0,0,0,2,48,48$ for $n=65$. The Galois orbits of cusps of $X_1(57)$ and $X_1(65)$ have degrees $1^{18},2^9,18^2$ and $1^{24},4^6,12^2,24$, and the closed cusps of $X_1(57)_{\F_2}$ and $X_1(65)_{\F_2}$ have degrees $1^{18},2^9,9^2,18$ and $1^{24},4^6,6^4,12^2$, so the closed cusps of degree at most $2$ are reductions of Galois orbits of cusps of the same degree. So over a number field of degree $d\leq8$ the reduction of the trace divisor modulo $2$ is entirely cuspidal, or it is $G+\overline C$, where $G$ is a single non-cuspidal closed point of degree $e\in\{6,7,8\}$ and $\overline C$ is the reduction of an effective rational cuspidal divisor $C$ of degree $d-e$. \subsection{Unramified covers of \texorpdfstring{$X_0(57)$}{X0(57)} and \texorpdfstring{$X_0(69)$}{X0(69)}}\label{subsec:covers_57_69} For $n=57$ and $n=69$ we use the following quotient of $X_1(n)$. \begin{lemma}\label{d7:lem:double} Let $n=3\ell$ with $\ell\in\{19,23\}$, let $H$ be the subgroup of odd order of $\Delta(n)$, which has index $2$, and put $Y_n=X_1(n)/H$ and $Z_n=X_0(n)$. The natural map $Y_n\to Z_n$ is an unramified double cover, also over $\F_2$. The curves $Z_n$ and $Y_n$ have genus $5$ and $9$ for $n=57$, and $7$ and $13$ for $n=69$. \end{lemma} \begin{proof} As $H$ has odd order, $Y_n$ has good reduction at $2$. The elements $\dm a$ of $\Delta(n)$ outside $H$ have even order, so they are not solutions of $a^2\pm a+1\equiv0\pmod n$, whose order in $\Delta(n)$ divides $3$, and $a^2\equiv-1$ is impossible modulo $3$. As $n$ is squarefree, \Cref{lem:fixed} shows that they fix no geometric point of $X_1(n)$, in characteristic $0$ or $2$, so the non-trivial element of $\Delta(n)/H$ fixes no geometric point of $Y_n$. Hence $Y_n\to Z_n$ is unramified, and the genera follow from $g(X_0(57))=5$, $g(X_0(69))=7$ and the Riemann-Hurwitz formula. \end{proof} As in \Cref{subsec:formal_55}, a geometrically connected unramified double cover of $Z_{n,\F_2}$ is given by a character of $\Pic(Z_{n,\F_2})$ that does not factor through the degree, whose values are the Frobenius elements in $\Delta(n)/H\simeq\F_2$, and it is determined by these at enough closed points. The four cusps of $Z_{n,\F_2}$ are rational and are distinguished by their widths $1,3,\ell,n$. From the description of the cusps of $X_1(n)$ and the action of $2$ on their cyclotomic coordinates, their Frobenius elements in $\Delta(n)/H$ are $(0,1,1,0)$ for $n=57$ and $(1,0,1,0)$ for $n=69$, in the order of the widths. We obtain a model of $Z_{n,\F_2}$ by adjoining to the function field of $X_0(\ell)_{\F_2}$ a root $t$ of $t^4+t^3+1/j=0$, which gives the function field of $X_0(3)$ over the $j$-line, and we compute, for \gitlinkVII{C57_good7_final.m}{$n=57$} and \gitlinkVII{C69_good7_final.m}{$n=69$}, \[ \Pic(Z_{57,\F_2})\simeq\torz{15}\times\torz{30}\times\Z,\qquad \Pic(Z_{69,\F_2})\simeq\torz{22}\times\torz{44}\times\Z. \] For $n=57$ the values at the cusps determine a unique character. For $n=69$ two characters have these values, and we use the condition that the images on $Z_{69}$ of the points of order $69$ over $\F_{128}$ (which come from points of $Y_{69}$) must split in $Y_{69}$. This determines the character uniquely, and we construct $Y_{n,\F_2}$ as the corresponding class field, which has the required genus, a rational split cusp of width $n$, and the required Frobenius element at every generator of $\Pic(Z_{n,\F_2})$. To exclude the entirely cuspidal reduction for $\torz{57}$ over octic fields we use the quotient $W_{57}=X_1(57)/\dm7$, where $\dm7$ has order $3$. It has good reduction at $2$. By \Cref{lem:fixed}, as $57$ is squarefree and $a^2\equiv-1$ is impossible modulo $3$, the only non-trivial elements of $\Delta(57)$ which can fix a geometric point of $X_1(57)$, in characteristic $0$ or $2$, are $\dm7$ and $\dm{49}=\dm7^2$, the solutions of $a^2\pm a+1\equiv0\pmod{57}$. So $W_{57}\to X_0(57)$ is an unramified cyclic cover of degree $6$, also over $\F_2$, and $W_{57}$ has genus $6\cdot4+1=25$. We construct $W_{57,\F_2}$ as follows. The curve $X_0(19)_{\F_2}$ is an elliptic curve $E_{19}$ with $\#E_{19}(\F_2)=3$. By class field theory, as $\Pic^0(E_{19})\simeq E_{19}(\F_2)\simeq\torz3$, there is a unique geometrically connected unramified cyclic cubic cover of $E_{19}$ in which the cusp $c_{19}$ of width $19$ splits, namely $P\mapsto\Frob_2(P)-P+c_{19}$. The quotient of $X_1(19)_{\F_2}$ by the subgroup of order $3$ of $\Delta(19)$ is such a cover, as this subgroup contains the stabilizers of the elliptic points of order $3$, so it is this cover. Pulling back this cubic cover to $Z_{57,\F_2}$ gives a curve of genus $13$, and its composite with the double cover $Y_{57,\F_2}$ above, which we \gitlinkVIII{C57/data/C57_fixed_AS.m}{construct} again by an Artin-Schreier equation checked at a set of generators of $\Pic(Z_{57,\F_2})/2$, has genus $25$ and a split rational cusp, so by class field theory it is $W_{57,\F_2}$. Its closed cusps have degrees $1^6,2^3,3^2,6$. In the case $d=8$ of \Cref{prop:57} we also use the operator $U_3$ on $W_{57}$. To compute $U_3$ on the differentials of $W_{57,\F_2}$ we use $q$-expansions. Let $P$ be a closed cusp of degree $3$ of $W_{57,\F_2}$ above the cusp of width $1$ of $X_0(57)$, and let $q$ be the corresponding parameter of the Tate curve. We identify the $q$-expansion of a differential $\omega=f(q)\,dq/q$ at a geometric point above $P$ with the Fourier expansion of $f$ at $\infty$. \begin{lemma}\label{d8:lem:U3} In this identification, $U_3^*\bigl(\sum_n a_nq^n\,dq/q\bigr)=\sum_na_{3n}q^n\,dq/q$. An $\F_2$-linear endomorphism of $H^0(W_{57,\F_2},\Omega^1)$ is determined by the first $17$ coefficients of the $q$-expansions at $P$ of the images of a basis. \end{lemma} \begin{proof} The first assertion is the action of $U_3$ on Fourier expansions at $\infty$. For the second, if a holomorphic differential defined over $\F_2$ vanishes to order at least $17$ at a geometric point above $P$, it vanishes to order at least $17$ at all three of them, and as $3\cdot17>48=2g(W_{57})-2$ it is $0$. \end{proof} In practice, at such a cusp we use the parameter $r=t_3+1$, where $t_3$ is the parameter of $X_0(3)$ with $j^{-1}=t_3^3(t_3+1)$, and then $j^{-1}=r(1+r)^3$. As $E_4\equiv1\pmod2$, we have $j^{-1}\equiv\Delta=q\prod_{n\geq1}(1-q^n)^{24}\pmod 2$, and $r$ is obtained as a power series in $q$ by reversion. We \gitlinkVIII{C57/C57_U3_explicit_series_v2.m}{compute} the matrix of $U_3^*$ in this way, and we check that it has coefficients in $\F_2$ and commutes with $\Cart$. Depending on the normalization of the action of the Hecke algebra on differentials, the operator computed in this way is $U_3$ or its conjugate $w_{57}U_3w_{57}^{-1}$ by the Atkin-Lehner involution $w_{57}$. This does not matter, as both are endomorphisms of $J_1(57)$ which act on the newform \lmfdbnewform{57}{2}{a}{a} by $a_3=-1$, so for both of them the image of $J(W_{57})(\Q)$ under the operator plus $1$ is finite by \Cref{lem:57_65}. \subsection{The group \texorpdfstring{$\torz{57}$}{Z/57Z}}\label{subsec:57} \begin{proposition}\label{prop:57}\label{d6:prop:57}\label{d7:prop:57}\label{d8:prop:57}\label{d8:prop:57good} The group $\torz{57}$ is not a subgroup of the torsion of an elliptic curve over a number field of degree $d\in\{6,7,8\}$. \end{proposition} \begin{proof} Let $D$ be the trace divisor of a point of order $57$ over a number field of degree $d$. Put $g=\dm2$, which has order $9$ in $\Delta(57)$, and $h=g^3=\dm8$. By \Cref{lem:57_65} the groups $(g-1)J_1(57)(\Q)$ and $(h-1)J_1(57)(\Q)$ are finite, by \Cref{d6:lem:57two} $g$ and $h$ act trivially on $J_1(57)(\Q)[2]$, and the support of $D$ is non-CM, as the least degree of a CM point on $X_1(57)$ is $12$. By \Cref{subsec:57_65_red2}, the reduction of $D$ modulo $2$ is entirely cuspidal, or it is $G+\overline C$, where $G$ is a non-cuspidal closed point of degree $e\in\{6,7,8\}$ and $\overline C$ is the reduction of an effective rational cuspidal divisor $C$ of degree $d-e$. For $d=6$ only the first case occurs, by \Cref{d6:prop:good2} and \Cref{d6:lem:additive}. Suppose first that $e=6$. Then $d\in\{7,8\}$, and, as in \Cref{d6:prop:good2}, every elliptic curve over $\F_{64}$ with a point of order $57$ is supersingular with $j=0$ (as we \gitlinkVII{check_good2_57_65.m}{check}), and $h$ acts on its points of order $57$ through an automorphism of the curve, so $hG=G$, and $h$ fixes the reduction of $D-C$. As $d+(d-6)\leq10<14.28<\gon_\C X_1(57)$, \Cref{d9:lem:stab}(d) gives $hD=D$, so $3\mid d$ by \Cref{lem:orbits}, which is a contradiction. Suppose next that $e=7$, and let $s$ be the non-trivial automorphism of $Y_{57}$ over $Z_{57}$ (\Cref{d7:lem:double}), which is the non-trivial element of $\Delta(57)/H$. As $\dm5\notin H$ we have $\dm5=s$ on $Y_{57}$, and $(s-1)J(Y_{57})(\Q)$ is finite by \Cref{lem:57_65}. The operator $A_5=T_5-1-5s$ annihilates this group by the Eichler-Shimura relation and the injectivity of reduction at $5$, and it annihilates the images on $Y_{57}$ of the rational cuspidal divisors of $X_1(57)$, such as $C$, as divisors (\Cref{d8:lem:cuspHecke}). So the image $P$ of $G$ on $Y_{57,\F_2}$ satisfies \[ [(s-1)(T_5P-P-5sP)]=[sT_5P-T_5P+4(sP-P)]=0. \] We \gitlinkVII{C57_good7_final.m}{compute} that the elliptic curves over $\F_{128}$ with a point of order $57$ form $14$ isomorphism classes, with cyclic group of order $114$, and give exactly two closed points of degree $7$ on $Z_{57,\F_2}$, both of which split in $Y_{57}$. For each of them, the image under $T_5$ consists of three closed points of $Z_{57,\F_2}$ of degree $14$ and multiplicity $1$, and each of them splits in $Y_{57}$. So $T_5P$ is one of $2^3=8$ divisors on $Y_{57}$, and replacing $P$ by $sP$ applies $s$ to all of them. We compute that $\Pic(Y_{57,\F_2})\simeq\torz{105}\times\torz{210}\times\Z$ and that for both points $P$ and all $8$ choices the class $[sT_5P-T_5P+4(sP-P)]$ is non-zero, which is a contradiction. Suppose next that $e=8$. Then $d=8$ and the reduction of $D$ is a single closed point $P$ of degree $8$. Put $\delta=\dm5$, which generates $\Delta(57)$. As $(\delta-1)J_1(57)(\Q)$ is finite by \Cref{lem:57_65}, $A_5(\delta-1)$ annihilates it, and $A_5c_0=0$, so $[(\delta-1)A_5P]=0$ in $J_1(57)(\F_2)$. The $18$ points of degree $8$ form a single diamond orbit, so we may take $P$ to be a fixed one of them. We compute $(\delta-1)A_5P$ on a model of $X_1(57)_{\F_2}$ of genus $85$. Let $F_{19}$ be the function field of $X_1(19)_{\F_2}$, write the universal elliptic curve over it as $y^2+xy=x^3+a_2x^2+1/j$, and put $w=y/x+x$ for a point $(x,y)$ of order $3$. Then $x^4+x^3+1/j=0$ and $w^2+w=x+a_2$, so the function field of $X_1(57)_{\F_2}$ is \[ F_{19}(w),\qquad (w^2+w+a_2)^4+(w^2+w+a_2)^3+1/j=0, \] where the involution $w\mapsto w+1$ corresponds to negating the point of order $3$. We \gitlinkVIII{C57/relative_divisors_resume_20261026.m}{compute} that $T_5P=Q_1+Q_2$ with $Q_1,Q_2$ closed points of degree $24$, by computing the six $5$-isogenies of the corresponding elliptic curve and carrying both level structures through them, and that the divisor \[ (\delta-1)A_5P=\delta Q_1+\delta Q_2-Q_1-Q_2+4\delta P+P-5\delta^2P \] satisfies $\ell\bigl((\delta-1)A_5P\bigr)=0$, so it is not principal, which is a contradiction. Suppose finally that the reduction of $D$ is entirely cuspidal. As the closed cusps of $X_1(57)_{\F_2}$ of degree at most $8$ have degree $1$ or $2$ (\Cref{subsec:57_65_red2}), it is the reduction of a rational cuspidal divisor $C$. If $d\leq7$, then \Cref{d9:lem:stab}(d), applied with $g$ and $C$, gives $gD=D$, as $2d\leq14<14.28<\gon_\C X_1(57)$, which contradicts \Cref{lem:orbits}, as $9\nmid d$. If $d=8$, the reduction of the push-forward of $D$ to $W_{57}$ is supported on the $6$ rational and the $3$ quadratic cusps of $W_{57,\F_2}$, which gives $[x^8](1-x)^{-6}(1-x^2)^{-3}=3654$ possible divisors. We apply \Cref{d8:prop:cartier} with $\psi_1=T_2+2$ and $\psi_2=U_3+1$ (\Cref{lem:57_65}), for which, by \eqref{d8:eq:T2star} and with $U_3^*$ as in \Cref{d8:lem:U3}, \[ U=\Cart(\Cart-1)H^0(W_{57,\F_2},\Omega^1)+(U_3^*+1)(\Cart-1)H^0(W_{57,\F_2},\Omega^1). \] We \gitlinkVIII{C57/C57_joint_reproduce.m}{compute} that $U$ has dimension $23$ and that its restriction to each of the $3654$ divisors has rank $8$, which is a contradiction. \end{proof} \subsection{The group \texorpdfstring{$\torz{65}$}{Z/65Z}}\label{subsec:65} For $n=65$ we use the quotients $Y_{65}'=X_1(65)/\dm{4,51}$ and $W_{65}=X_1(65)/\dm{51}$. We use the coordinate $t$ on $X_1(5)$ for which the universal curve in Tate normal form has $b=c=t$, and a coordinate $z$ on the genus $0$ curve $X_1(13)/\dm3$. Over $\F_2$ the $j$-maps are \[ j_5(t)=\frac{t^{12}+t^8+t^4+1}{t^7+t^6+t^5},\qquad j_{13}(z)=\frac{z^{16}+z^4+1}{z^{28}+z^{25}+z^{24}+z^{21}+z^{20}+z^{17}+z^{16}+z^{13}}, \] and we \gitlinkVI{setup13_twist_parameter_certified.m}{compute} the coordinate $z$ and the map $j_{13}$ from a model of $X_1(13)$ over $\F_2$. \begin{lemma}\label{d6:lem:Y65} The subgroup of $\Delta(65)$ generated by $\dm4$ and $\dm{51}$ has order $6$ and acts freely on $X_1(65)$, also in characteristic $2$, so $Y_{65}'$ and $W_{65}$ have good reduction at $2$ and genus $21$ and $61$. The function field of $Y_{65,\F_2}'$ is generated by $t$ and $z$ with $j_5(t)=j_{13}(z)$, and the diamond operators $\dm3$ and $\dm6$ act on it by $\tau_T(t,z)=(1/t,z)$ and $\tau_U(t,z)=(t,z+1)$. The curve $W_{65}$ is the normalization of $X_1(5)\times_{X(1)}X_1(13)$. \end{lemma} \begin{proof} Every element $\dm a$ of this subgroup has $a\equiv\pm1\pmod5$ (the residues are $\pm1,\pm4,\pm9,\pm16,\pm36,\pm51$ modulo $65$), so, as $65$ is squarefree, \Cref{lem:fixed} shows that the subgroup acts freely, in characteristic $0$ and $2$, and the genera follow from Riemann-Hurwitz, as $X_1(65)$ has genus $121$. As $51\equiv1\pmod5$ and $51\equiv-1\pmod{13}$, the curve $Y_{65}'$ parametrizes elliptic curves with a point of order $5$ up to sign and a $\dm3$-orbit of points of order $13$ up to sign, so it is the normalization of the fibre product of $X_1(5)$ and $X_1(13)/\dm3$ over the $j$-line, which we \gitlinkVI{cusp65_genus41_rr_transport_full.m}{verify} to be irreducible, and in the same way $W_{65}$ parametrizes elliptic curves with points of order $5$ and $13$, each up to sign. The maps from $Y_{65}'$ to $X_1(5)$ and to $X_1(13)/\dm3$ commute with the diamond operators, so $\dm a$ acts on $t$ through $a\bmod5$ and on $z$ through $a\bmod13$. On $X_1(5)$ the operators $\dm2=\dm3$ act by $t\mapsto-1/t$, hence by $t\mapsto1/t$ over $\F_2$, as we \gitlinkVI{check_65_diamonds.m}{compute}, and on $X_1(13)/\dm3$ over $\F_2$ the operator $\dm2$ acts by $z\mapsto z+1$, as we \gitlinkVI{setup13_twist_parameter_certified.m}{compute} from the moduli interpretation. As $3\equiv-2\pmod5$ and $\dm3$ acts trivially on $X_1(13)/\dm3$, while $6\equiv1\pmod5$ and $\dm6=\dm2\dm3$ on $X_1(13)$, the operators $\dm3$ and $\dm6$ act on $Y_{65}'$ by $(t,z)\mapsto(1/t,z)$ and $(t,z)\mapsto(t,z+1)$. \end{proof} \begin{proposition}\label{prop:65}\label{d6:prop:65}\label{d7:prop:65}\label{d8:prop:65} The group $\torz{65}$ is not a subgroup of the torsion of an elliptic curve over a number field of degree $d\in\{6,7,8\}$. \end{proposition} \begin{proof} Let $D$ be the trace divisor of a point of order $65$ over a number field of degree $d$. By \Cref{subsec:57_65_red2}, its reduction modulo $2$ is entirely cuspidal, or it is $G+\overline C$, where $G$ is a non-cuspidal closed point of degree $e\in\{6,7,8\}$ and $\overline C$ is the reduction of an effective rational cuspidal divisor of degree $d-e$. Suppose first that the reduction is $G+\overline C$. On $Y_{65}'$ we have $\dm3=\tau_T$, so $A_3=T_3-1-3\tau_T$. For $\psi\in\{\tau_T-1,\tau_U-1,T_2+1\}$ the group $\psi J(Y_{65}')(\Q)$ is finite by \Cref{lem:57_65}, so $A_3\psi$ annihilates $J(Y_{65}')(\Q)$, and $A_3$ annihilates the images of the rational cuspidal divisors as divisors (\Cref{d8:lem:cuspHecke}). Hence the push-forward $P$ of $G$ to $Y_{65}'$ satisfies $[\psi A_3P]=0$ for these three $\psi$, and none of the following tests depends on $\overline C$. For $e=7$ we \gitlinkVII{sieve65_quotient_good7_certified.m}{compute} that all the elliptic curves over $\F_{128}$ with a point of order $65$ have cyclic group of order $130$, and that their points of order $65$ give $56$ points of $Y_{65}'(\F_{128})$, that is, $8$ closed points of degree $7$. For $e\in\{7,8\}$ the push-forwards of the points of degree $e$ form two orbits under the residual diamond operators, and for a representative $P$ of each of them we compute that the divisor \[ (\tau_T-1)A_3P=\tau_TT_3P-T_3P+2(\tau_TP-P) \] is not principal, for \gitlinkVII{sieve65_quotient_good7_certified.m}{$e=7$}, where $T_3P$ is a single closed point of degree $28$, and for \gitlinkVIII{C65/C65_good8_D8.m}{$e=8$}. For $e=6$, the only elliptic curve over $\F_{64}$ with a point of order $65$, up to isomorphism, is supersingular with cyclic group of order $65$. We \gitlinkVIII{C65/C65_good6_D8.m}{compute} the images on $Y_{65}'$ of its $48$ points of order $65$ and of their images under the four $3$-isogenies. This gives $3$ possibilities for the divisor $P$ and $10$ possibilities for $T_3P$ for each of them, and for each of the $30$ possibilities one of $[(\tau_T-1)A_3P]$, $[(\tau_U-1)A_3P]$ and $[(T_2+1)A_3P]$ is non-zero. Suppose finally that the reduction of $D$ is entirely cuspidal. Then so is that of its push-forward to $W_{65}$. The closed cusps of $W_{65,\F_2}$, with their degrees and widths, are $12$ of degree $1$ and width $65$, $6$ of degree $2$ and width $13$, and $2$ each of degree $6$ and widths $1$ and $5$. Those of width $5$ lie below cusps of $X_1(65)_{\F_2}$ of degree $12$, so they cannot occur, and those of degree $2$ lie below cusps of degree $4$, so they occur with even multiplicity. For $d=8$ this leaves $[t^8](1-t)^{-12}(1-t^4)^{-6}(1-t^6)^{-2}=83949$ possible divisors, which we \gitlinkVIII{C65/cusp65_genus61.py}{check} using the description of the cusps. The residual diamond group $\Delta(65)/\dm{51}$ is isomorphic to $\torz2\times\torz6$, generated by $\dm{27}$ and $\dm{41}$, and $\dm2^6=1$ on $W_{65}$. We apply \Cref{d8:prop:cartier} with $\psi_1=\dm{27}-1$, $\psi_2=\dm{41}-1$ and $\psi_3=T_2^6-1$ (\Cref{lem:57_65}), for which $\psi_3^*=\Cart^6-1$ by \eqref{d8:eq:T2star}, so \[ U=(\Cart-1)\bigl((\dm{27}^*-1)\Omega+(\dm{41}^*-1)\Omega+(\Cart^6-1)\Omega\bigr),\qquad \Omega=H^0(W_{65,\F_2},\Omega^1). \] We \gitlinkVIII{C65/C65_allbad_reproduce.m}{compute} that $U$ has dimension $48$, and, computing the spaces $\Omega(jP)$ at one cusp in each orbit of the residual diamond operators and transporting them, that the restriction of $U$ to each of the $83949$ divisors has rank $8$. Adding $8-d$ times a cusp of degree $1$ to a possible divisor of degree $d\in\{6,7\}$ gives one of these, so by \Cref{rem:smaller} the restriction of $U$ is also surjective at the possible divisors of degree $6$ and $7$, and \Cref{d8:prop:cartier} gives a contradiction. \end{proof} \subsection{The group \texorpdfstring{$\torz{69}$}{Z/69Z}}\label{subsec:63_69}\label{subsec:sieve_69}\label{subsec:formal_69} \begin{proposition}\label{prop:69}\label{d7:prop:69}\label{d8:prop:69}\label{d9:prop:69} The group $\torz{69}$ is not a subgroup of the torsion of an elliptic curve over a number field of degree $d\in\{7,8,9\}$. \end{proposition} \begin{proof} Let $D$ be the trace divisor of a point of order $69$ over a number field of degree $d$. We have $\rk J_1(69)(\Q)=0$ \cite[Theorem 3.1(2)]{DEHMZ}, and Abramovich's bound gives $\gon_\Q X_1(69)>20.94>2d$, so by \Cref{d6:prop:main}(b) the reduction of $D$ modulo $2$ is not entirely cuspidal. We \gitlinkIX{C69/independent_counts69.m}{compute} that the non-cuspidal closed points of $X_1(69)_{\F_2}$ of degree at most $9$ have degrees $7$, $8$ and $9$, and that there are $66$, $44$ and $44$ of them. The Galois orbits of cusps of $X_1(69)$ over $\Q$ \gitlinkVII{cusp69_lift_checks.py}{have} degrees $1^{22},2^{11},22^2$, and the closed cusps of $X_1(69)_{\F_2}$ have degrees $1^{22},2^{11},11^2,22$, so the closed cusps of degree $1$ and $2$ are reductions of Galois orbits of cusps of the same degree. So the reduction of $D$ is $G+\overline C$, where $G$ is one of these non-cuspidal points and $\overline C$ is the reduction of an effective rational cuspidal divisor of degree $d-\deg G\leq2$. Let $Y_{69}\to Z_{69}=X_0(69)$ be the unramified double cover of \Cref{d7:lem:double}, where $H$ is the subgroup of order $11$ of $\Delta(69)$, with the model of $Y_{69,\F_2}$ of \Cref{subsec:covers_57_69}. The Jacobians of $Y_{69}$ and $Z_{69}$ are quotients of $J_1(69)$, so their groups of rational points are finite. As $\dm5\in H$, the operator $A_5=T_5-6$ annihilates these groups (\Cref{d6:prop:tools}(2)), and it annihilates the images of the rational cuspidal divisors as divisors (\Cref{d8:lem:cuspHecke}). So the image $P$ of $G$ on $Z_{69,\F_2}$ satisfies $[T_5P-6P]=0$, and so does its image $P_Y$ on $Y_{69,\F_2}$. The images of the points $G$ on $Z_{69,\F_2}$ are $3$ closed points of degree $7$, $2$ of degree $8$ and $2$ of degree $9$, and we compute that $T_5P-6P$ is not principal on $Z_{69,\F_2}$ for two of those of degree \gitlinkIX{C69/N69_cover_good_D9.m}{$7$}, for both of degree \gitlinkVIII{C69/N69_cover_good_D8.m}{$8$} and for one of degree \gitlinkIX{C69/N69_cover_good_D9.m}{$9$}. For the remaining point $P$ of degree $7$ and the remaining one of degree $9$, the non-trivial automorphism of $Y_{69}$ over $Z_{69}$, which commutes with $T_5$, exchanges the two points of $Y_{69,\F_2}$ above $P$, so we may take $P_Y$ to be a fixed one of them, and for all the possible lifts $T$ of $T_5P$ to $Y_{69}$, eight for degree $7$ and two for degree $9$, we compute that $T-6P_Y$ is not principal. \end{proof} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{The group \texorpdfstring{$\torz{54}$}{Z/54Z}}\label{sec:54}\label{d7:sec:54}\label{d8:sec:54}\label{d9:sec:54} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Let $X=X_1(54)$ and $Y=X_1(27)$, and let $\pi:X\to Y$ be the map $(E,P)\mapsto(E,2P)$. In degree $6$ a class test on $Y$ at the prime $5$ leaves only elliptic curves whose reductions modulo $5$ have $j$-invariant $2$, and a $2$-isogeny excludes them. In degrees $7$, $8$ and $9$ we combine the possible reductions at $2$ and at $5$ of the push-forward to $Y$ of the trace divisor, which is a form of the Mordell-Weil sieve of Bruin and Stoll \cite{BS}. As $2\mid54$, at the prime $2$ we use a semistable model of $X$ over $\Z_2$. We first collect the facts at $5$ which the three proofs share. Let $E$ be an elliptic curve over a number field $K$ of degree $d\in\{6,7,8,9\}$ with a point $P$ of order $54$, and let $D$ be the trace divisor of $(E,2P)$ on $Y$. It is the push-forward under $\pi$ of the trace divisor of $(E,P)$ on $X$, and an effective rational divisor of degree $d$ on $Y$ with non-cuspidal support. We have $\rk J_1(54)(\Q)=0$ \cite[Theorem 3.1(2)]{DEHMZ}, and as $i(54)=972$, we have $\gon_\Q X\geq\gon_\C X>9.64>d$. So \Cref{d6:prop:main}(a) with $p=5$ shows that the reduction modulo $5$ of the trace divisor of $(E,P)$ is not entirely cuspidal, and neither is that of $D$. By \cite[Theorem 3.1(2), Theorem 4.13, Corollary 4.14 and Table 2]{DEHMZ}, $J_Y(\Q)=\ClCusp_\Q Y\simeq\torz3\times\torz3\times\torz{52497}$. As $5$ is a primitive root modulo $27$, every Galois orbit of cusps of $Y$ over $\Q$ reduces to a single closed cusp of $Y_{\F_5}$ of the same degree. We fix a rational cusp $c_\ast$ of $Y$ and take all the classes relative to it. We \gitlinkVI{certify_54_mod5.m}{compute} \[ J_Y(\F_5)\simeq\torz3\times\torz3\times\torz{342175446}, \] and that the reduction $H_5$ of $J_Y(\Q)$, which is generated by the classes of the closed cusps of $Y_{\F_5}$, has invariants $[3,3,52497]$ and index $6518$. So $H_5$ contains the classes of all cuspidal divisors of degree $0$ on $Y_{\F_5}$. If we write the reduction of $D$ modulo $5$ as $D'+C'$, as in \Cref{d6:lem:classtest}, with $D'$ of degree $e>0$, then \Cref{d6:lem:classtest} gives $[D'+C'-d\,\overline{c_\ast}]\in H_5$, and hence $[D'-e\,\overline{c_\ast}]\in H_5$. The map $\pi$, of degree $3$, is the scheme of non-zero $2$-torsion points of the universal elliptic curve. Above a non-cuspidal closed point $y$ of $Y_{\F_5}$ of degree $k$ there is one closed point of $X_{\F_5}$ of degree $3k$, a point of degree $k$ and a point of degree $2k$, or three points of degree $k$, according to whether the $2$-division cubic has $0$, $1$ or $3$ roots over the residue field of $y$, and their push-forwards are $3y$, $y$ and $2y$, and $y$, respectively. We also \gitlinkVI{certify_54_mod5.m}{compute} that the numbers of non-cuspidal closed points of $Y_{\F_5}$ of degrees $1,\dots,6$ are $0,9,57,171,612,2700$, and that the elliptic curves of $0,0,30,171,396,1890$ of them have a point of order $2$ over the residue field. Finally, let $\phi:E\to\widetilde E=E/\dm{27P}$ be the quotient isogeny. The point $\phi(P)$ has order $27$, and $\ker\hat\phi=\phi(E[2])$ is generated by a $K$-rational point $R$ of order $2$, so $\widetilde P=\phi(P)+R$ is a point of order $54$ of $\widetilde E$ over $K$, with $2\widetilde P=\phi(2P)$. So all of the above also holds for the trace divisor $\widetilde D$ of $(\widetilde E,\phi(2P))$ on $Y$. As $E$ and $\widetilde E$ are isogenous, they have good reduction at the same primes. \begin{proposition}\label{d6:prop:54} The group $\torz{54}$ is not a subgroup of the torsion of an elliptic curve over a sextic field. \end{proposition} \begin{proof} Let $E$, $P$ and $D$ be as above, with $d=6$, and write the reduction of $D$ modulo $5$ as $D'+C'$, with $D'$ of degree $e>0$. As $e\leq6$, it follows from the description of the fibres of $\pi$ and the numbers of points above that $D'$ is $3y$ for a closed point $y$ of degree $2$, a single closed point of degree $3$, $4$, $5$ or $6$, or a sum of two closed points of degree $3$ (possibly equal), where in the last two cases the elliptic curve of each point has a point of order $2$ over its residue field. We \gitlinkVI{certify_54_mod5.m}{compute} that $[D'-e\,\overline{c_\ast}]\in H_5$ for only six of the $9+30+171+396+1890+465$ divisors of these types, namely $z_i+z_j$ with $1\leq i\leq j\leq3$, where $z_1,z_2,z_3$ are the closed points of degree $3$ of $Y_{\F_5}$ whose elliptic curves have $j$-invariant $2$. For a prime $\wp$ of $K$ above $5$, the reduction of $(E,2P)$ modulo $\wp$ is a point of $Y$ with values in the residue field of $\wp$, so it lies on a closed point of $Y_{\F_5}$, which is non-cuspidal if and only if $E$ has good reduction at $\wp$. The reduction of $D$ modulo $5$ is a sum of positive multiples of these closed points, so the support of $D'$ consists of those for the primes $\wp$ at which $E$ has good reduction, and as $e>0$ there is at least one such prime. As each of them is one of $z_1,z_2,z_3$, the reduction of $E$ modulo each such $\wp$ has $j$-invariant $2$. Let $\wp$ be such a prime. The above also applies to $\widetilde E$, which has good reduction at $\wp$, so the reduction of $\widetilde E$ modulo $\wp$ also has $j$-invariant $2$. But this reduction is the quotient of the reduction of $E$ by a subgroup of order $2$. The curve $y^2=x^3+x+1$ has $j$-invariant $1728\cdot4/31=2$, so over $\overline\F_5$ the reduction of $E$ is isomorphic to it. Its points of order $2$ are $(r,0)$ with $r^3+r+1=0$, and as $x^3+x+1$ has no root in $\F_5$, $r$ has degree $3$ over $\F_5$. Moving $(r,0)$ to $(0,0)$ gives the model $y^2=x(x^2+3rx+3r^2+1)$, and the quotient of a curve $y^2=x(x^2+ax+b)$ by $\dm{(0,0)}$ is $y^2=x(x^2-2ax+a^2-4b)$. So the quotient by $\dm{(r,0)}$ is $y^2=x(x^2+4rx+2r^2+1)$, whose $j$-invariant is \[ \frac{1}{r^2+r+1}=2r^2+r+4\notin\F_5, \] which is a contradiction. \end{proof} In degrees $7$, $8$ and $9$ we also use the prime $2$. \begin{lemma}\label{d7:lem:54good2}\label{d8:lem:54good2}\label{d9:lem:54good2} Let $E$ be an elliptic curve over a number field $K$ of degree $d\in\{7,8,9\}$ with a point $P$ of order $54$. Then $E$ has good reduction at some prime of $K$ above $2$. Moreover, every prime of $K$ above $2$ at which $E$ has good reduction is unramified and has residue degree at least $6$, and the reduction of $E$ at it is ordinary. \end{lemma} \begin{proof} We have seen that $J_1(54)(\Q)$ is finite and that $\gon_\Q X>d$. The rational cuspidal subgroup $(\torz3)^3\times(\torz9)^2\times\torz{1102437}\times\torz{1529080119}$ of $J_1(54)(\Q)$ has index dividing $3$ in it \cite[Theorem 4.13 and Table 2]{DEHMZ}, so $J_1(54)(\Q)$ has odd order. As $\Gamma_1(54)=\Gamma_1(27)\cap\Gamma_0(2)$ and the $\Gamma_1(27)$-structure is rigid, Deligne and Rapoport \cite[Chapter VI, Theorem 6.9]{DR} give a regular semistable model of $X$ over $\Z_2$. Its special fibre consists of two copies of $Y_{\F_2}$ glued at the supersingular points, and the cusps reduce to smooth points. All the cusps are defined over subfields of $\Q(\zeta_{27})$, and $2$ is a primitive root modulo $27$, so every Galois orbit of cusps over $\Q$ reduces to a single closed point of the same degree. Suppose that $E$ has bad reduction at every prime of $K$ above $2$, and let $D_X$ be the trace divisor of $(E,P)$ on $X$. By \Cref{d6:lem:additive}, applied to the point $2P$ of order $27$, the elliptic curve $E$ has multiplicative reduction at every prime of $K$ above $2$, so $D_X$ reduces to a sum of cusps. By the above, this is also the reduction of a $\Q$-rational cuspidal divisor $C$. As $D_X$ and $C$ have the same reduction on a regular model, the class $[D_X-C]\in J_1(54)(\Q)$ maps to $0$ in the special fibre of the N\'eron model of $J_1(54)$. Every torsion point in the kernel of reduction modulo $2$ has order a power of $2$, and $J_1(54)(\Q)$ has odd order, so $[D_X-C]=0$. As $\gon_\Q X>d$, this forces $D_X=C$, which is impossible as $D_X$ is not cuspidal. Now let $\mathfrak p$ be a prime of $K$ above $2$ at which $E$ has good reduction. Reduction modulo $\mathfrak p$ is injective on the points of odd order, so the reduction of $(E,2P)$ is a non-cuspidal point of $Y$ over the residue field of $\mathfrak p$. As we \gitlinkVII{N54_via27_final.m}{compute} that $Y_{\F_2}$ has no non-cuspidal closed points of degree at most $5$, the residue degree $f$ of $\mathfrak p$ is at least $6$, and as its ramification index $e$ satisfies $ef\leq d\leq9$, we have $e=1$. If the reduction were supersingular, its $j$-invariant would be $0$, so $E$ would have a Weierstrass equation over the completion $K_{\mathfrak p}$ with integral coefficients and $a_1\in2\mathcal O_{K_{\mathfrak p}}$, and as the reduction has no point of order $2$, the point $27P$ of order $2$ would lie in the formal group. But then $[2](T)=2T-a_1T^2+O(T^3)$, where the terms of degree at least $3$ have integral coefficients, and for $v_2(T)\geq1$ the valuation of $2T$ is smaller than that of all the other terms. So, as $K_{\mathfrak p}$ is unramified over $\Q_2$, the formal group has no non-zero point of order $2$ over $K_{\mathfrak p}$, which is a contradiction. \end{proof} As $J_Y(\Q)$ has odd order, reduction at $2$ is injective on it, and we \gitlinkIX{C54/certify54.py}{compute} that its reduction is the whole group $J_Y(\F_2)\simeq(\torz3)^2\times\torz{52497}$ (see also \githubbareVII{N54_via27_final.m} and \githubbareVIII{C54/C54_lower_reproduce.m}). As $2$ is a primitive root modulo $27$, the Galois orbits of cusps of $Y$ over $\Q$ also stay irreducible modulo $2$. We match the closed cusps modulo $2$ and modulo $5$ using the orders of vanishing of the functions $b^4/\Delta$ and $b$, where $y^2+(1-c)xy-by=x^3-bx^2$ is the universal elliptic curve in Tate normal form and $\Delta$ is its discriminant. These functions are units on $Y_1(27)_{\Z[1/27]}$, so their orders of vanishing along the cusps do not change under reduction. This gives a map $\kappa:H_5\to J_Y(\F_2)$, sending the reduction modulo $5$ of a class in $J_Y(\Q)$ to its reduction modulo $2$, and $\kappa$ is an isomorphism. \begin{proposition}\label{d8:prop:54}\label{d7:prop:54} The group $\torz{54}$ is not a subgroup of the torsion of an elliptic curve over a number field of degree $7$ or $8$. \end{proposition} \begin{proof} Let $D$ be as above, and suppose first that $d=8$. At $2$, the reduction is not entirely cuspidal by \Cref{d8:lem:54good2}. We \gitlinkVIII{C54/C54_lower_reproduce.m}{compute} that the non-cuspidal closed points of $Y_{\F_2}$ of degrees $1,\dots,8$ number $0,0,0,0,0,9,9,18$, so the reduction $\overline D$ is a closed point of degree $6$ plus a cuspidal divisor of degree $2$, a closed point of degree $7$ plus one of the $9$ rational cusps, or a closed point of degree $8$. The effective cuspidal divisors of degree $2$ give $48$ distinct classes, so we obtain a set $S_2\subseteq J_Y(\F_2)$ of $9\cdot48+9\cdot9+18=531$ classes $[\overline D-8\overline{c_\ast}]$, and the class $a=[\overline D-8\overline{c_\ast}]$ of the reduction at $5$ must satisfy \begin{equation}\label{d8:eq:54} a\in H_5\quad\text{and}\quad\kappa(a)\in S_2. \end{equation} By the description of the fibres of $\pi$ and the numbers of points above, the non-cuspidal part of $\overline D$ is $P_k$ for a closed point $P_k$ of degree $k\in\{3,\dots,8\}$ with a rational $2$-torsion point, $3P_2$ for a closed point $P_2$ of degree $2$, or $P_3+P_3'$, $P_3+P_4$, $P_3+P_5$ or $P_4+P_4'$, where the points may coincide. Reducing the first point by the diamond operators, which preserve condition \eqref{d8:eq:54}, and allowing all the classes of cuspidal divisors of the complementary degree, we obtain the following tests: \[ \begin{array}{c|r|r|r} \text{non-cuspidal part}&\text{supports}&\text{cuspidal classes}&\text{tests}\\ \hline P_3&4&1836&7344\\ P_4&19&636&12084\\ P_5&44&192&8448\\ P_6&212&48&10176\\ P_7&828&9&7452\\ P_8&3579&1&3579\\ 3P_2&1&48&48\\ P_3+P_3'&120&48&5760\\ P_3+P_4&684&9&6156\\ P_3+P_5&1584&1&1584\\ P_4+P_4'&3249&1&3249 \end{array} \] For the $828$ points of degree $7$ we \gitlinkVII{N54_good7_bsgs.m}{compute} the elliptic curves over $\F_{5^7}$ with a point of order $27$ and a rational $2$-torsion point directly, and check the completeness of this list against an \gitlinkVII{additional_counts.m}{independent count} of the points of $X_1(54)$ over $\F_{5^7}$. For the points of degree $8$ we \gitlinkVIII{C54/C54_good8_enumeration_v2.m}{compute} the elliptic curves over $\F_{5^8}$ with a point of order $27$ and a rational point of order $2$, by running through all the $j$-invariants and their twists. We do not assume that the $3$-primary part of the group is cyclic: if it is $\torz3\times\torz{3^k}$, we take all the cyclic subgroups of order $27$. This gives $3579$ diamond orbits of closed points of $Y_{\F_5}$ of degree $8$, $2704$ with one and $875$ with three rational points of order $2$, which we \gitlinkVIII{C54/C54_good8_positive_verify_v2.m}{verify} separately. They account for $9(2704+3\cdot875)=47961$ closed points of degree $8$ of $X_1(54)_{\F_5}$, and the remaining $126$ lie above points of degree $4$ with one rational $2$-torsion point, whose push-forward is $2P_4$. Their sum $48087$ is the number of closed points of degree $8$ of $X_1(54)_{\F_5}$, which we compute independently. Of the $65880$ tests, we compute (\githubbareVIII{C54/C54_lower_reproduce.m} for the $62301$ with a non-cuspidal part of degree at most $7$, \githubbareVIII{C54/C54_good8_class_shard.m} for the others) that $153$ give a class in $H_5$, and none of their images under $\kappa$ lies in $S_2$. This contradicts \eqref{d8:eq:54}. Suppose now that $d=7$. By \Cref{d7:lem:54good2} the reduction of $D$ at $2$ is not entirely cuspidal, so, by the numbers of non-cuspidal closed points of $Y_{\F_2}$ above, it is a closed point of degree $6$ plus one of the $9$ rational cusps, or a closed point of degree $7$. The reduction of $D$ at $5$ is $D'+C'$, where $D'$ is of one of the types of degree at most $7$ in the table above, which lists all the possible non-cuspidal parts of degree at most $8$, and $C'$ is cuspidal of the complementary degree. Adding $\overline{c_\ast}$ to the reductions of $D$ at $2$ and at $5$ does not change their classes relative to $c_\ast$, and gives possibilities considered above for $d=8$ (see the discussion after \Cref{d6:lem:classtest}). So the class $[D'+C'-7\overline{c_\ast}]$ satisfies \eqref{d8:eq:54}, which the computation above excludes. \end{proof} \begin{proposition}\label{d9:prop:54} The group $\torz{54}$ is not a subgroup of the torsion of an elliptic curve over a nonic field. \end{proposition} \begin{proof} Let $E$, $P$, $D$, $\widetilde E$ and $\widetilde D$ be as above, with $d=9$. Neither $D$ nor $\widetilde D$ reduces modulo $5$ to a cuspidal divisor. Modulo $2$, by \Cref{d9:lem:54good2}, $D$ reduces to an ordinary non-cuspidal closed point of degree $6$, $7$, $8$ or $9$ of $Y_{\F_2}$ plus a cuspidal divisor of the complementary degree, and as the closed cusps of $Y_{\F_2}$ are reductions of rational cuspidal divisors, the classes of these reductions, taken relative to $c_\ast$, form a set $S_2$ of $2412$ classes. Hence the class $a=[\overline D-9\overline{c_\ast}]$ of the reduction of $D$ modulo $5$ satisfies $a\in H_5$ and $\kappa(a)\in S_2$, and the same holds for $\widetilde D$. Using the description of the fibres of $\pi$ and the numbers of points above, we run through all the possible reductions of $D$ modulo $5$, which are single points of degree $3,\dots,9$, the multiples $3P_2$, $3P_3$, $2P_3$, $2P_4$, the sums $P_3+P_3'$, $P_3+P_4$, $P_3+P_5$, $P_3+P_6$, $P_4+P_4'$, $P_4+P_5$, $P_3+P_3'+P_3''$ and $3P_2+P_3$, where the points in the sums may coincide, each completed by every cuspidal divisor of the remaining degree, and we \gitlinkIX{C54/lower9.m}{find} (for $3P_2+P_3$ in a \gitlinkIX{C54/mixed_inert23/run.m}{separate computation}) that apart from single points of degree $8$ and $9$ only $3$ possibilities satisfy both conditions, of the forms $P_3+P_6$, $P_4+P_5$ and $P_4+P_5$. For the points of degree $8$, none of the $3579$ representatives up to the diamond operators has class in $H_5$. For the points of degree $9$ there are $16135$ representatives, and we first use \Cref{d9:lem:quadchar}: a function $f$ on $Y_{\F_5}$ with $\ddiv f=2A$ defines a homomorphism $\chi_f:J_Y(\F_5)\to\F_5^\times/\F_5^{\times2}$, which is trivial on $H_5$, as $H_5$ has odd order. It \gitlinkIX{C54/quadratic_prepass9.m}{excludes} $8087$ of them, and of the remaining $8048$ exactly $4$ satisfy both conditions. Finally, for these $7$ possibilities we compute the reductions of $\widetilde D$ for all the possible choices of the reduction of $\dm{27P}$, which are $1$, $3$, $3$ and $3$, $1$, $1$, $1$ choices respectively, using the isogenies of degree $2$ of the elliptic curves over finite fields, and we \gitlinkIX{C54/projection2_v2/run.m}{find} that none of the $13$ resulting classes lies in $H_5$, which is a contradiction. \end{proof} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{The groups \texorpdfstring{$\tg{2}{2n}$}{Z/2Z x Z/2nZ}}\label{sec:noncyclic}\label{d7:sec:noncyclic}% \label{d8:sec:noncyclic}\label{d9:sec:noncyclic} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% The groups $\tg{2}{2n}$ treated here are those with $2n\in\{22,28,30\}$ for $d=6$, $2n\in\{22,24,28,30\}$ for $d=7$, $2n\in\{28,30,32,36,42\}$ for $d=8$ and $2n\in\{28,30,32,34,36,40,42\}$ for $d=9$, and we show that none of them is a subgroup of the torsion of an elliptic curve over a number field of degree $d$. In the last two subsections we treat the groups $\torz{48}$ for $d\in\{6,8,9\}$ and $\torz{52}$ for $d=9$, using the morphisms $X_1(4k)\to X_1(2,2k)$ of \Cref{d6:lem:4n} with $k=12$ and $k=13$. The group $\tg{2}{26}$ for $d\in\{6,7,8\}$, the points of degree $9$ on $X_1(2,26)$ and the points of degree $6$ on $X_1(2,24)$ are treated in \Cref{sec:sporadic}. The method is the same in all degrees. In all these cases the Jacobian of $X_1(2,2n)$ has rank $0$ over $\Q$, and we exclude the entirely cuspidal reduction of the trace divisor at an odd prime $p$. We test only one representative of each orbit of a group of level automorphisms of $X_1(2,2n)$ (see \Cref{d6:sec:classes}), and we push the possible reductions of the trace divisor forward to $X_1(2n)$, to curves $X_1(2,m)$ and $X_1(m)$ with $m$ a proper divisor of $2n$, and to quotients of $X_1(2,2n)$ by groups of level automorphisms. On each of these curves the class of the push-forward is constrained by the reduction of the Mordell-Weil group of its Jacobian (\Cref{d6:lem:classtest} and \eqref{d8:eq:pushclass}). The few cases that survive these tests are excluded by the Hecke sieve, by Riemann-Roch computations, or by showing that a level automorphism which fixes the reduction of the trace divisor would fix the trace divisor itself. Let $X=X_1(2,N)$ with $N=2n$. It parametrizes triples $(E,P,Q)$ with $P$ of order $2$ and $Q$ of order $N$ such that $P\notin\dm Q$, i.e.\ $P\neq nQ$. If $n$ is even, the maps \makeatletter \begin{equation}\label{d9:eq:aut2N}\ltx@label{{d8:eq:aut2N}}% second label, d8: \label{d8:eq:aut2N} (E,P,Q)\longmapsto\bigl(E,P+n\beta Q,\ \gamma P+aQ\bigr),\qquad \beta,\gamma\in\{0,1\},\ a\in(\torz{2n})^\times, \end{equation} \makeatother are level automorphisms, and taken modulo $-1$ they form a group of $2\varphi(N)$ automorphisms of $X$: as $n(\gamma P+aQ)=nQ$, the new point of order $2$ is not in the subgroup generated by the new point of order $N$. If $n$ is odd, the level automorphisms permute the three points of order $2$ and act by $\Delta(n)$ on the point of order $n$, and they form a group isomorphic to $S_3\times\Delta(n)$. These automorphisms, and the other ones that we use, commute with the Hecke operators $T_q$ for odd primes $q\nmid N$, as these are defined by isogenies of odd degree. We identify the Galois orbits of cusps of $X$ over $\Q$ and their reductions using the valuations of integral modular units. In the normalization $y^2=x^3+cx^2+b(1-b-c)x$, $P=(0,0)$, $Q=(b,b)$, of \eqref{d6:eq:tate2}, the functions $b$, $b(1-b-c)$, $c^2-4b(1-b-c)$ and $x(kQ)$, $1\leq k\leq N/2$, are units on $Y_1(2,N)$ over $\Z[\tfrac1{2N}]$, so their valuations at the cusps do not change under reduction modulo an odd prime $p\nmid N$. We also use the four maps $X\to X_1(2n)$ which keep $Q$ or $P+Q$, or divide $E$ by $\dm P$ or $\dm{P+nQ}$ and keep the image of $Q$, which has order $2n$ in all cases. By \cite[Theorem 3.1(3)]{DEHMZ} (see also \cite[Theorem 4.1]{DS}), $\rk J_1(2,N)(\Q)=0$ for all the $N=2n$ above. By \cite[Table 1]{NV} we have $\gon_\Q X\geq 8,8,12,12,10,10,10$ for $N=22,24,26,28,30,32,36$ and $\gon_\Q X_1(2,34)>9$, and for $N=40,42$ we have $i(2,N)=1152$, so $\gon_\C X\geq\tfrac{325}{2^{15}}\cdot1152>11.42$, hence $\gon_\C X\geq12$. So $\gon_\Q X>d$ in every degree $d$ in which we use $X$, and by \Cref{d6:prop:main}(a) the reduction of the trace divisor modulo an odd prime $p\nmid N$ is not entirely cuspidal. By \Cref{d6:lem:4n}, excluding $\tg{2}{2n}$ also excludes $\torz{4n}$. The numbers of non-cuspidal closed points of $X_{\F_p}$ that we use are listed in \Cref{d9:table:2n}, except those of $X_1(2,22)$, which are in \Cref{d7:table:sieve}, and those of $X_1(2,24)$, which are in the proof of \Cref{d7:prop:2_24}. \begin{table}[H] \centering \small \begin{tabular}{lrl} \toprule $X$ & $p$ & non-cuspidal closed points of degree $1,\dots,9$\\ \midrule $X_1(2,26)$ & $3$ & $0,0,0,0,36,122,288,936,2328$\\ $X_1(2,28)$ & $3$ & $0,0,0,0,48,64,312,606,2112$\\ $X_1(2,32)$ & $3$ & $0,0,0,0,0,192,320,960,1824$\\ $X_1(2,34)$ & $3$ & $0,0,0,36,48,48,144,738,2160$\\ $X_1(2,36)$ & $5$ & $0,0,48,120,720,2904,10416,50334,218080$\\ $X_1(2,40)$ & $3$ & $0,0,0,32,96,64,480,504,2048$\\ $X_1(2,42)$ & $5$ & $0,0,0,126,648,2080,11376,48816,211776$\\ \bottomrule \end{tabular} \medskip \caption{Non-cuspidal closed points of $X_1(2,2n)_{\F_p}$ of small degree, checked against the zeta functions.} \label{d9:table:2n} \end{table} \subsection{The curves \texorpdfstring{$X_1(2,22)$}{X1(2,22)} and \texorpdfstring{$X_1(2,24)$}{X1(2,24)}}\label{subsec:X1_2_22_24} The curve $X_1(2,24)$ is also used for $\torz{48}$ in \Cref{subsec:48}. \begin{proposition}\label{d6:prop:2_22}\label{d7:prop:2_22} The group $\tg{2}{22}$ is not a subgroup of the torsion of an elliptic curve over a number field of degree $6$ or $7$. \end{proposition} \begin{proof} Let $X=X_1(2,22)$, a curve of genus $16$, and let $d\in\{6,7\}$. By \Cref{d7:table:sieve} the reduction of the trace divisor modulo $3$ is $x+C$, with $x$ a single non-cuspidal closed point of degree $e\in\{5,\dots,d\}$ and $C$ cuspidal of degree $d-e\leq2$. The closed cusps have degrees $1^{15},5^3$, and the cusps of degree $1$ are reductions of rational cusps, so $A_5C=0$ and $[A_5x]=0$. We \gitlinkVII{sieve_2_22_degree7.m}{compute} that $A_5x$ is not principal for the representatives of all $12,20,48$ diamond orbits of points of degree $5,6,7$, which covers both degrees. \end{proof} \begin{proposition}\label{d7:prop:2_24} The group $\tg{2}{24}$ is not a subgroup of the torsion of an elliptic curve over a septic field. \end{proposition} \begin{proof} Let $X=X_1(2,24)$, a curve of genus $13$, and work at $p=5$. Using the valuations of integral modular units, we \gitlinkVII{fine_2_24_degree7.m}{find} $19$ classes of closed cusps of $X_{\F_5}$ which are reductions of rational cuspidal divisors, of degrees $1^8,2^6,4^5$, and we fix one of the $8$ rational cusps as $c_\ast$. We compute \[ J(\F_5)\simeq(\torz2)^2\times(\torz4)^2\times\torz{12}\times\torz{24}\times(\torz{480})^2, \] and that the classes of these divisors generate a subgroup $G\simeq\torz2\times(\torz4)^2\times\torz{120}\times\torz{240}$ of order $921600$, which is the order of $J(\Q)$ \cite[Proposition 4.16 and Table 3]{DEHMZ}, so $G$ is the reduction of $J(\Q)$ and reduction is injective on $J(\Q)$. By \Cref{d6:prop:2_24_points}(2), $X$ has $8$ non-cuspidal closed points of degree $6$. Adding any of the $8$ rational cusps to them gives a set $\mathcal R$ of $64$ reducible effective rational divisors of degree $7$, with $64$ distinct reduction classes, and $\mathcal R$ is stable under the diamond operators. The non-cuspidal closed points of $X_{\F_5}$ of degrees $1,\dots,7$ number $0,0,16,96,672,2712,11184$, so the non-cuspidal part of the reduction of the trace divisor $D$ is a single closed point of degree $3,\dots,7$, a sum of two cubic points, or a sum of a cubic and a quartic point. Taking the first point up to the diamond operators, there are $4,32,168,684,2796$ cases of the first kind and $64$ and $384$ of the other two. For each of these $4132$ cases we \gitlinkVII{fine_2_24_degree7.m}{compute} the classes of all the possible reductions $\overline D$, with all the cuspidal parts of the complementary degree, and find that only $16$ of them lie in $G$, all coming from $2$ closed points of degree $6$. These $16$ classes are exactly the reduction classes of $16$ divisors in $\mathcal R$, as we \gitlinkVII{verify_n24_reducible_witnesses.m}{verify} using the two sextic points. Hence $[D-7c_\ast]=[R-7c_\ast]$ for some $R\in\mathcal R$, so $D\sim R$, and $D=R$ by \Cref{d7:lem:closed}, which is impossible since $D$ is irreducible. \end{proof} \subsection{The curve \texorpdfstring{$X_1(2,28)$}{X1(2,28)}}\label{subsec:X1_2_28} Let $X=X_1(2,28)$, a curve of genus $25$, with $\gon_\Q X\geq12$ \cite[Table 1]{NV}, and work at $p=3$. The closed cusps of $X_{\F_3}$ have degrees $1^{12},2^6,3^4,6^2$, and those of degree $1$ and $2$ are reductions of rational cuspidal divisors of the same degree, so $q=5$ and $q=11$ are admissible for every cuspidal divisor supported on them. All the cusps are defined over $\Q(\zeta_{28})$, and the Frobenius at $3$ has order $6$ in $\Gal(\Q(\zeta_{28})/\Q)$, so $\sigma_5$ either fixes a cusp $c$ of degree $3$ or maps it to the other cusp of degree $3$ with the same valuations of the unit functions above. Hence $A_5c=(\sigma_5c-c)+5\dm5(\sigma_5c-c)$ is $0$ or of the form $v+5w$, where $v$ and $w$ are differences of two such cusps, which gives $17$ possible values of $A_5c$. Let $Y=X_1(28)$. By \cite[Theorem 3.1(2), Corollary 4.14 and Table 2]{DEHMZ}, $J_Y(\Q)\simeq\torz2\times\torz4\times\torz{12}\times\torz{936}$ is generated by rational cuspidal divisor classes, and we \gitlinkVIII{C2x28/N28_D8_forgetful_class_probe.m}{compute} that its reduction $G$ has index $4$ in $J_Y(\F_3)\simeq(\torz4)^2\times\torz{24}\times\torz{936}$. \begin{proposition}\label{d6:prop:2_28}\label{d7:prop:2_28}\label{d8:prop:2_28}\label{d9:prop:2_28} The group $\tg{2}{28}$ is not a subgroup of the torsion of an elliptic curve over a number field of degree $d\in\{6,7,8,9\}$. \end{proposition} \begin{proof} By \Cref{d9:table:2n} the reduction of the trace divisor $D$ is $x+C$, with $x$ a single non-cuspidal closed point of degree $e\in\{5,\dots,d\}$ and $C$ cuspidal of degree $d-e$. Suppose first that $d\leq8$. If $d-e\leq2$, then $[A_qx]=0$ for $q=5,11$. If $d=8$ and $e=5$, then $C$ may contain a cusp of degree $3$, and $A_5x+A_5C$ is principal, where $A_5C$ is one of the $17$ values above, which include $0$. Up to the group \eqref{d8:eq:aut2N} there are $2,5,13,29$ possible points $x$ of degree $5,6,7,8$, which we \gitlinkVIII{C2x28/export_N28_D8_fullaut_Hecke.m}{enumerate}. For the $2$ points of degree $5$ we \gitlinkVIII{C2x28/N28_D8_quintic_cusp_corrections.m}{compute} that $A_5x+A_5C$ is not principal for all $17$ values, and for the points of degree $6$, $7$ and $8$ we \gitlinkVIII{C2x28/N28_D8_fullaut_high_Hecke_shard.m}{compute} that $A_5x$ or $A_{11}x$ is not principal, except for $2$ points of degree $8$, which we call $x_1$ and $x_2$. As $d-e\leq2$ for $d\leq7$, this proves the claim for $d\in\{6,7\}$. To exclude $x_1$ and $x_2$ for $d=8$ we use the forgetful map $\pi:X\to Y$, $(E,P,Q)\mapsto(E,Q)$. For a rational cusp $c_\ast$ of $Y$, the class $[\pi_*x_2-8\overline{c_\ast}]$ is not in $G$, which excludes $x_2$. The class $[\pi_*x_1-8\overline{c_\ast}]$ is the reduction of $[L-8c_\ast]$ for an explicit rational cuspidal divisor $L$ of degree $8$ on $Y$, so if $D$ reduces to $x_1$ (after applying an automorphism of $X$, which maps trace divisors to trace divisors), then $\pi_*D\sim L$ by injectivity of reduction. We \gitlinkVIII{C2x28/N28_D8_quotient_charzero_certify_v2.m}{compute} that $\ell(L)=1$ on $Y$ over $\Q$ and that the unique effective divisor $y$ in its class is a closed point of degree $8$, with residue field $K=\Q(t)$, where \[ 16t^8-78t^7+149t^6-118t^5-22t^4+132t^3-120t^2+48t-8=0. \] Hence $\pi_*D=y$, and $D$ is a closed point of degree $8$ with the same residue field $K$. But the $2$-division polynomial of the elliptic curve corresponding to $y$ factors over $K$ into factors of degrees $1$ and $2$, so this curve has only one point of order $2$ over $K$, and $y$ does not lift to a point of $X$ over $K$. This is a contradiction. Suppose now that $d=9$. Up to the $24$ automorphisms \eqref{d9:eq:aut2N} there are $2$, $5$, $13$, $29$ and $88$ cases. We use the four maps to $Y$, and we \gitlinkIX{C2x28/four_quotients.m}{compute} that only $10$ cases satisfy \eqref{d9:eq:pushclass} for all four maps. For these we use $A_5$, with the $17$ possible values of $A_5c$ if the complement of a sextic point is a cusp $c$ of degree $3$. We compute that $A_5$ of the non-cuspidal part, with any of these corrections, is not principal, except for one case, a point $x$ of degree $8$ with a cuspidal complement $c$ of degree $1$, which is the reduction of a rational cusp. The level automorphism $h(P,Q)=(P+14Q,13Q)$, of order $2$, fixes $x$ and none of the rational cusps. If $D$ reduced to $x+c$, then \Cref{d9:lem:stab}(a), with $d=9$, $e=1$ and $\gon_\Q X\geq12>10$, would give $hc=c$, which is a contradiction. \end{proof} \subsection{The curve \texorpdfstring{$X_1(2,30)$}{X1(2,30)}}\label{subsec:X1_2_30}\label{d6:subsec:2_30} Let $X=X_1(2,30)$, a curve of genus $25$. In all four degrees we push the divisors forward to quotient curves. Let $Y=X_1(30)$ and consider the three morphisms \[ \pi_1(E,P,Q)=(E,Q),\qquad \pi_2(E,P,Q)=(E,P+16Q),\qquad \pi_3(E,P,Q)=(E,P+Q) \] from $X$ to $Y$. They are defined over $\Q$, they map cusps to cusps and they commute with the diamond operators. This is because $15Q$, $P$ and $P+15Q$ are the non-zero points of $E[2]$, and adding $16Q$, which has order $15$, to them gives points of order $30$. These three maps do not suffice, and we also use the quotient $W$ of $X$ by the automorphism $(E,P,Q)\mapsto(E,15Q,P+Q)$ of order $3$, which permutes the three non-zero points of $E[2]$ cyclically. We denote the quotient map by $\alpha:X\to W$. The curve $W$ sees a part of $J_X$ that is invisible on $Y$ (see \Cref{rem:2_30}). Recall that $X_1(15)$ is an elliptic curve. The curve $W$ is the double cover $f:W\to X_1(15)$ given by a square root of the discriminant of the universal elliptic curve over $X_1(15)$ in Tate normal form, every cusp of $X_1(15)$ is totally ramified in it, and it has genus $9$, as we \gitlinkVI{a3_m2_30_model.m}{compute}. \begin{remark}\label{rem:2_30} The covering $X\to X_1(15)$, $(E,P,Q)\mapsto(E,2Q)$, is Galois with group $\SL_2(\F_2)\simeq S_3$, acting on the ordered bases $(P,15Q)$ of $E[2]$, and $W$ is its quotient by the subgroup $A_3$ of order $3$. The space $H^0(X,\Omega^1)$ decomposes into $1$ copy of the trivial representation of $S_3$, $8$ copies of the sign representation and $8$ copies of the $2$-dimensional one, as the genera $1$, $9$, $9$ and $25$ of $X_1(15)$, $W$, $Y$ and $X$ show. Each $\pi_i$ is invariant under a subgroup of order $2$ of $S_3$, so the differentials pulled back from $Y$ involve only the trivial and the $2$-dimensional representations, while those on $W$ are the invariants of $A_3$, which are the trivial and the sign representations. So the test on $W$ detects the part of $J_X$, of dimension $8$, on which $S_3$ acts by the sign character and which is invisible on $Y$. \end{remark} The map $\alpha$ is unramified, also over $\F_7$: a non-trivial element of $A_3$ fixing a non-cuspidal point $(E,P,Q)$ would be induced by an automorphism of $E$ fixing the point $2Q$ of order $15$, which is the identity, and the identity does not permute the points of order $2$ non-trivially. The stabilizers of the cusps in $S_3$ have order $2$, so $A_3$ acts freely on the cusps as well. Above a non-cuspidal closed point $y$ of $W_{\F_7}$ of degree $k$, the $2$-division cubic of the corresponding elliptic curve has square discriminant, so it either splits, and then $\alpha$ has three points of degree $k$ above $y$, or it is irreducible, and then there is one point of degree $3k$ above $y$, whose push-forward is $3y$. The Galois orbits of cusps of $W$ have degrees $1,1,4,2,4,1,2,1$, and we denote them by $C_1,\dots,C_8$ in this order. Let $c_W=\alpha(c_0)$, a rational cusp of $W$. By \cite[Proposition 4.16 and Table 3]{DEHMZ}, $J_X(\Q)\simeq(\torz2)^5\times\torz{24}\times(\torz{8160})^2$ is generated by rational cuspidal divisor classes, and push-forward maps rational cuspidal divisors of degree $0$ on $X$ to rational cuspidal divisors of degree $0$ on $W$. So for every effective rational divisor $D$ of degree $d$ on $X$, the class $[\alpha_*D-d\,c_W]=\alpha_*[D-d\,c_0]$ lies in the subgroup $H$ of $J_W(\Q)$ generated by the classes $[C_i-(\deg C_i)c_W]$. The group $J_W(\Q)$ is finite, as $\alpha^*$ maps it to $J_X(\Q)$ with finite kernel, so reduction modulo $7$ is injective on it, and we identify $H$ with its reduction. We \gitlinkVI{filter_a3_m2_30.m}{compute} \[ J_W(\F_7)\simeq(\torz2)^6\times\torz{102}\times\torz{10608},\qquad H\simeq(\torz2)^5\times\torz8. \] Similarly, by \cite[Theorem 3.1(2), Corollary 4.14 and Table 2]{DEHMZ}, $J_Y(\Q)=\ClCusp_\Q Y\simeq\torz4\times\torz{8160}$, and we \gitlinkVI{m2_30_quotient_prefilter.m}{compute} that $J_Y(\F_7)\simeq(\torz4)^3\times\torz{136}\times\torz{16320}$, in which the reduction $G$ of $J_Y(\Q)$ has index $4352$. We apply \eqref{d8:eq:pushclass} to the three maps $\pi_i$ with $p=7$ and this $G$, and the images in $J_Y(\F_7)/G$ of the sets $\mathcal S_0,\dots,\mathcal S_5$ of \eqref{d8:eq:pushclass} have $1,9,41,67,68,68$ elements. \begin{proposition}\label{d6:prop:2_30}\label{d7:prop:2_30}\label{d8:prop:2_30}\label{d9:prop:2_30} The group $\tg{2}{30}$ is not a subgroup of the torsion of an elliptic curve over a number field of degree $d\in\{6,7,8,9\}$. \end{proposition} \begin{proof} Let $D$ be the trace divisor and work at $p=7$. The entirely cuspidal reduction is excluded as explained at the beginning of the section. Suppose first that $d\leq7$. We determine all the possible reductions on $W$, and then test their lifts with the map $\pi_2$. If $d=6$ we replace $D$ by $D+c_0$, whose reduction has the same non-cuspidal part, as the argument below uses only that $D$ is an effective rational divisor of degree $7$ whose reduction is not entirely cuspidal (the condition on $H$ holds for every effective rational divisor, and the test with $\pi_2$ is \Cref{d6:lem:classtest} applied to $\pi_{2*}D$). The divisor $T:=\alpha_*\overline D$ is effective of degree $7$, and its class $[T-7\overline{c_W}]$ lies in $H$. For each of the $256$ classes $h\in H$ we \gitlinkVII{filter_2_30_degree7_a3_first.m}{compute} the Riemann-Roch space of a divisor of degree $7$ in the class $h+[7\overline{c_W}]$, which gives all the effective divisors in that class. This gives $1404$ effective divisors $T$, of which $816$ are not entirely cuspidal. For these $816$ divisors $T$ we run through all the divisors $\overline D$ with $\alpha_*\overline D=T$, using the description of the fibres of $\alpha$ above. For $448$ of them no such $\overline D$ exists, as the multiplicities are not divisible by $3$. For the remaining $368$ we obtain $1440$ possibilities for the non-cuspidal part $D'$ of $\overline D$, of degree $e$ say. By \eqref{d8:eq:pushclass} for $\pi_2$, the class $[\pi_{2*}D'-e\,\overline{y_0}]$ lies in $G+\mathcal S_{7-e}$, and we compute that this fails for all $1440$ possibilities. Suppose now that $d=8$, and let $x$ be the corresponding point of $X$. No elliptic curve over a number field of degree $1$, $2$ or $4$ has a point of order $30$ \cite{mazur,KM,kamienny}, \cite[Theorem 1.1]{DN}, so $x$ has degree $8$ and $D$ is a closed point of degree $8$. As $\alpha$ is a cyclic cover of degree $3$, the push-forward $D_W:=\alpha_*D$ is also a closed point of degree $8$, and $[D_W-8c_W]\in H$. For each of the $256$ classes $h\in H$ we choose a rational cuspidal divisor $L=\sum_iv_iC_i$ of degree $8$ with $[L-8c_W]=h$, and we \gitlinkVIII{C2x30/src/classes256.m}{compute} $\ell(\overline L)$ on $W_{\F_7}$. As $\ell(L)\leq\ell(\overline L)$, and as the members of $|L|$ reduce to members of $|\overline L|$, this determines the possible divisors $D_W$. Write $B=\sum_i\lfloor v_i/2\rfloor f(C_i)$ and $C=\sum_i(v_i\bmod 2)C_i$. Then $L=f^*B+C$, as $f$ is totally ramified at the cusps, and the pull-backs of the sections of $L(B)$ on $X_1(15)$ give lower bounds for $\ell(L)$. We find the following: \begin{enumerate} \item[a)] for $35$ classes, $\ell(\overline L)=0$, \item[b)] for $116$ classes, $\ell(\overline L)=1$ and $L$ is effective and cuspidal, \item[c)] for $96$ classes, $C\neq0$ and $\ell(L)=\ell(\overline L)=\ell(B)$, so every divisor in $|L|$ is of the form $f^*B'+C$, \item[d)] for $4$ classes, $L=f^*B$ with $\deg B=4$ and $\ell(L)=\ell(\overline L)=4$, so every divisor in $|L|$ is the pull-back of a divisor in $|B|$, \item[e)] for the remaining $5$ classes, $\ell(\overline L)=2$. \end{enumerate} By injectivity of reduction, $[D_W-8c_W]$ is one of these $256$ classes, and $D_W\in|L|$ for the corresponding $L$. This is impossible in cases a), b) and c), as $D_W$ is non-cuspidal. In case d), $D_W=f^*B'$ for an effective divisor $B'$ of degree $4$, which is a closed point as $D_W$ is closed. Let $F$ be its residue field, of degree $4$. The universal elliptic curve and the point of order $15$ are defined over $F$, and the residue field $K$ of $D$, of degree $2$ over $F$, contains all the $2$-torsion. So the Galois group of the $2$-division polynomial over $F$ has order at most $2$, and the curve has a point of order $2$ over $F$, hence a point of order $30$ over a field of degree $4$, which is impossible. In case e), $D_W$ reduces to one of the $8$ divisors in the pencil $|\overline L|$ over $\F_7$. For each of the $5$ pencils we \gitlinkVIII{C2x30/src/exceptional_fibres.m}{compute} these $8$ divisors and their lifts to $X_{\F_7}$, using the description of the fibres of $\alpha$ above. Only one pencil has members that lift to a non-cuspidal divisor on $X_{\F_7}$, namely $6$ closed points of degree $8$, each with three lifts, so $\overline D$, which is not entirely cuspidal, is one of $18$ divisors. Finally, \eqref{d8:eq:pushclass} for the three maps $\pi_i$, with $e=8$, says that $[\pi_{i*}\overline D-8\overline{y_0}]\in G$ for $i=1,2,3$. We \gitlinkVIII{C2x30/src/three_quotients.m}{compute} that for each of the $18$ divisors at least one of the three classes does not lie in $G$. This is a contradiction. Suppose finally that $d=9$. The class $[\alpha_*D-9c_W]$ lies in $H$, and we \gitlinkIX{C2x30/classes256_D9.m}{compute} that the $956$ effective cuspidal divisors of degree $9$ on $W$ represent all $256$ classes of $H$. For each class we choose such a divisor $L=\sum_iv_iC_i$, and write $L=f^*B+C$ with $B=\sum_i\lfloor v_i/2\rfloor f(C_i)$ and $C=\sum_i(v_i\bmod 2)C_i$. Note that $C\neq0$, as $\deg L=9$ is odd and $\deg f^*B$ is even. Riemann-Roch computations over $\F_7$ give $\ell(\overline L)=1$ for $84$ of these classes, $\ell(\overline L)=\ell_{X_1(15)}(\overline B)$ for $144$ others, and $\ell(\overline L)=2$ for the remaining $28$. As $\ell(L)\geq1$, and $\ell(L)\geq\ell_{X_1(15)}(B)$ as $f^*L(B)\subseteq L(L)$, semicontinuity shows that the first two equalities also hold over $\Q$. Hence for these $228$ classes every effective divisor in $|L|$ over $\Q$ contains a cusp: it is $L$ itself in the first case, and $f^*B'+C$ in the second. As $\alpha_*D$ is effective with non-cuspidal support, its class is one of the $28$ pencil classes, and its reduction is one of the $8\cdot28=224$ members of these pencils over $\F_7$. We lift them to $X_{\F_7}$, using the description of the fibres of $\alpha$ above, and we find $42$ possible reductions of $D$ that are not entirely cuspidal, $24$ closed points of degree $9$ and $18$ of degree $8$ together with a cusp. Finally we \gitlinkIX{C2x30/three_quotients_D9.m}{compute} that each of the $42$ cases fails \eqref{d9:eq:pushclass} for one of the three maps $\pi_i$. \end{proof} \subsection{The curves \texorpdfstring{$X_1(2,32)$}{X1(2,32)}, \texorpdfstring{$X_1(2,34)$}{X1(2,34)} and \texorpdfstring{$X_1(2,40)$}{X1(2,40)}}\label{subsec:X1_2_32_34_40} Over octic fields the groups $\tg{2}{34}$ and $\tg{2}{40}$ are excluded in \Cref{d8:prop:34,d8:prop:40}, where we determine the octic points on $X_1(34)$ and $X_1(40)$. \begin{proposition}\label{d8:prop:2_32}\label{d9:prop:2_32_2_34_2_40} The group $\tg{2}{32}$ is not a subgroup of the torsion of an elliptic curve over a number field of degree $8$ or $9$, and the groups $\tg{2}{34}$ and $\tg{2}{40}$ are not subgroups of the torsion of an elliptic curve over a nonic field. \end{proposition} \begin{proof} We work modulo $3$. For $X_1(2,32)$, by \Cref{d9:table:2n} the reduction of $D$ is a closed point of degree $6,\dots,d$ plus a cuspidal divisor, and there are $8$, $10$, $32$ and $57$ cases of degree $6,\dots,9$ up to the $32$ automorphisms \eqref{d9:eq:aut2N}. We use the four maps to $X_1(32)$, whose Mordell-Weil group $(\torz2)^3\times\torz4\times\torz{120}\times\torz{11640}$ is generated by rational cuspidal divisors \cite[Theorem 3.1(2), Corollary 4.14 and Table 2]{DEHMZ} and has index $68$ in $J_1(32)(\F_3)$, allowing all the cuspidal complements over $\F_3$, and we \gitlinkIX{C2x32_C2x34_C2x40/C2x32/four_quotients.m}{compute} that each of the $107$ cases fails \eqref{d9:eq:pushclass} with $d=9$ for one of them. As $\mathcal S_{8-e}\subseteq\mathcal S_{9-e}$, this also excludes $\tg{2}{32}$ over octic fields. For $X_1(2,34)$ the possible types of the non-cuspidal part are $[4],\dots,[9]$, $[4,4]$ and $[4,5]$, which gives $239$ cases up to the $48$ level automorphisms, where in the types with two points we normalize only the first one. We use the two maps to $X_1(34)$ which keep $Q$ and $P+Q$. The group $J_1(34)(\Q)\simeq\torz{8760}\times\torz{595680}$ is generated by rational cuspidal divisors \cite[Theorem 3.1(2), Corollary 4.14 and Table 2]{DEHMZ}, its reduction has index $128$ in $J_1(34)(\F_3)$ and contains the classes of all the cuspidal divisors, and we \gitlinkIX{C2x32_C2x34_C2x40/C2x34/four_quotients_cached.m}{compute} that each of the $239$ cases fails \eqref{d9:eq:pushclass} for one of the two maps. For $X_1(2,40)$ the possible types are $[4],\dots,[9]$, $[4,4]$ and $[4,5]$, which gives $234$ cases up to the $32$ automorphisms \eqref{d9:eq:aut2N}. We use the map $X_1(2,40)\to X_1(2,20)$, $(E,P,Q)\mapsto(E,P,2Q)$, and the map which divides $E$ by $\dm{20Q}$ and keeps the images of $P$ and $Q$, which are independent of orders $2$ and $20$. The group $J_1(2,20)(\Q)\simeq\torz4\times\torz{60}\times\torz{120}$ is generated by rational cuspidal divisors \cite[Proposition 4.16 and Table 3]{DEHMZ}, its reduction has index $4$ in $J_1(2,20)(\F_3)$, and it contains the classes of all the cuspidal divisors. These two maps leave $27$ cases, all with non-cuspidal part of degree $8$ or $9$. For them we use the map to $X_1(40)$, with $J_1(40)(\Q)\simeq(\torz2)^3\times\torz8\times\torz{120}\times\torz{895440}$ of index $4592$ in $J_1(40)(\F_3)$, and we \gitlinkIX{C2x32_C2x34_C2x40/C2x40/filter_cyclic40.m}{compute} that all $27$ cases fail \eqref{d9:eq:pushclass}. \end{proof} \subsection{The curve \texorpdfstring{$X_1(2,36)$}{X1(2,36)}}\label{subsec:X1_2_36} Over septic fields the group $\tg{2}{36}$ is excluded in \Cref{d7:prop:36}, where we determine the septic points on $X_1(36)$. Let $X=X_1(2,36)$, a curve of genus $41$, and work at $p=5$. The closed cusps of $X_{\F_5}$ have degrees $1^{24},2^8,6^4$, and we use the group of $24$ automorphisms of the form \eqref{d8:eq:aut2N}. We use the map $X\to X_1(2,18)$, $(E,P,Q)\mapsto(E,P,2Q)$, and the four maps to $X_1(36)$ described at the beginning of the section. By \cite[Theorem 3.1(3), Proposition 4.16 and Table 3]{DEHMZ}, $J_1(2,18)(\Q)\simeq\torz2\times\torz{42}\times\torz{126}$, and by \cite[Theorem 3.1(2), Corollary 4.14 and Table 2]{DEHMZ}, $J_1(36)(\Q)\simeq\torz{12}\times\torz{252}\times\torz{7812}$, both generated by rational cuspidal divisor classes. As $5$ generates $(\torz{18})^\times$, every closed cusp of $X_1(2,18)_{\F_5}$ is the reduction of a Galois orbit of cusps of the same degree, so the reduction of $J_1(2,18)(\Q)$ contains the classes of all the cuspidal divisors, and \eqref{d8:eq:pushclass} says that the push-forward to $X_1(2,18)$ of the non-cuspidal part of the reduction of the trace divisor has class in it. Let $g(E,P,Q)=(E,P+18Q,P+Q)$, an automorphism of $X$ of order $4$, with $g^2(E,P,Q)=(E,P,19Q)$. Every power of $g$ fixes the point $4Q$ of order $9$, and a non-trivial automorphism of an elliptic curve does not fix a point of order $9$, so the group generated by $g$ acts freely on the non-cuspidal points of $X$, and we check that it acts on the cusps with orbits of length $4$. So the quotient $W=X/\dm g$ is a curve of genus $11$, and $X\to W$ is unramified. Let $t$ be the parameter of $X_1(9)$ in Tate normal form, \[ r=t^2-t+1,\quad c=t(r-1),\quad b=rc,\quad E_9:\ y^2+(1-c)xy-by=x^3-bx^2, \] and $j_9(t)=j(E_9)$. Writing $R=18Q$, $S=9Q$, $U=P$ and $V=P+R$, the functions $t$ of $(E,4Q)$ and $w=4(x(S)-x(R))/(x(U)-x(V))$ are invariant under $g$, and we check that \[ \Q(W)=\Q(t,w),\qquad 16(w^2+16)^3=j_9(t)\,w^4. \] To see this, after moving $R$ to $(0,0)$ and writing $E:y^2=x(x^2+\alpha x+\beta)$, we have $(x(S)-x(R))^2=\beta$ and $(x(U)-x(V))^2=\alpha^2-4\beta$, so $w^2=16\beta/(\alpha^2-4\beta)$, which gives the equation, and the degrees match. Over nonic fields we also need the following two lemmas, and a quotient $Z$ of $X$ of genus $9$, defined in \Cref{d9:lem:Z9}. \begin{lemma}\label{d9:lem:Zmult} Let $X$ be $X_1(m,n)$, with good reduction at an odd prime $p$ and $J_X(\Q)$ finite, let $h$ be a level automorphism of $X$, and let $D$ be an effective rational divisor whose support contains no cusp. Suppose that $D$ reduces to $x+\overline c$, where $hx=x$ and $c$ is a cusp defined over a number field unramified at $p$, with the same residue field as $\overline c$, and suppose that the $h$-orbit of $c$ consists of $r$ cusps $c_i=h^{i-1}c$, $1\leq i\leq r$, with distinct reductions. Then $\ell\bigl(x+\sum_i\overline{c_i}\bigr)\geq r$. \end{lemma} \begin{proof} As in \Cref{d9:lem:stab}, the class $[hD-D-hc+c]$ is torsion and reduces to $0$, so it is $0$, and the class of $L=D-c$ is $h$-invariant. Put $M=L+\sum_ic_i$. For each $i$ we have $M\sim h^{i-1}D+\sum_{j\neq i}c_j$, which is effective, and the corresponding sections vanish at all $c_j$ with $j\neq i$ but not at $c_i$, as $h^{i-1}D$ contains no cusp. So these $r$ sections are linearly independent, and the claim follows by semicontinuity, as $M$ reduces to $x+\sum_i\overline{c_i}$. \end{proof} \begin{lemma}\label{d9:lem:Z9} Let $X=X_1(2,36)$, $h(P,Q)=(P,13Q)$, which has order $3$, and $Z=X/\dm h$, of genus $9$. The reduction of $J_Z(\Q)$ modulo $5$ is contained in a subgroup $H\simeq(\torz3)^3\times(\torz{24})^3$ of index $8$ of $J_Z(\F_5)\simeq(\torz3)^3\times(\torz{24})^2\times\torz{192}$. \end{lemma} \begin{proof} The curve $Z$ is the cover of $X_1(2,12)$ obtained by adjoining a root of the cubic factor of the $3$-division polynomial of $E/\dm{4Q_{12}}$ which corresponds to the three cyclic subgroups of order $3$ other than the kernel of the dual isogeny. Its Jacobian has rank $0$, as a quotient of $J_X$, so $A_7J_Z(\Q)=0$. We \gitlinkIX{C2x36/diagonal_quotient/proof_generator.m}{compute} that the classes of the cuspidal divisors of $Z_{\F_5}$ generate the subgroup $H$, that $A_7H=0$ by \Cref{d9:prop:tools}, and that for a cubic point $z$ the class $a=[z-3\overline{c_0}]$ generates $J_Z(\F_5)/H$ and $A_7a$ has order $8$, so $\ker(A_7)=H$. \end{proof} \begin{proposition}\label{d8:prop:2_36}\label{d9:prop:2_36} The group $\tg{2}{36}$ is not a subgroup of the torsion of an elliptic curve over a number field of degree $8$ or $9$. \end{proposition} \begin{proof} Let $D$ be the trace divisor. Suppose first that $d=8$. By \Cref{d9:table:2n} the non-cuspidal part of the reduction of $D$ is a closed point of degree $3,\dots,8$, or a sum of two closed points, of type $[3,3]$, $[3,4]$, $[3,5]$ or $[4,4]$. We \gitlinkVIII{C2x36/enumerate_2_36_D8_direct_all_degrees.m}{find} $2,5,30,126,434,2115$ representatives of the closed points of degrees $3,\dots,8$ under the group of automorphisms, and for the sums of two points we reduce only the first point. We \gitlinkVIII{C2x36/filter_2_36_D8_quotient.m}{compute} that only $461$ of the $2712$ single points and $2376$ pairs pass the test with the map to $X_1(2,18)$. For these we use three of the maps to $X_1(36)$, given by $(E,P,Q)\mapsto(E',Q')$, where $E'$ is $E$, $E/\dm P$ or $E/\dm{P+18Q}$ and $Q'$ is the image of $Q$, and we apply \eqref{d8:eq:pushclass} with all the cuspidal divisors over $\F_5$. We \gitlinkVIII{C2x36/filter_2_36_D8_cyclic36_fast.m}{compute} that only one divisor passes all three tests, a single closed point $\overline x$ of degree $8$ with multiplicity $1$. We check that $g\overline x=\overline x$. If $D$ reduces to $\overline x$, then \Cref{d9:lem:stab}(a), with $C=0$ and $\gon_\Q X\geq10>8$, gives $gD=D$, so $D$ is the pull-back of a rational divisor $D_W$ of degree $2$ on $W$. The Jacobian of $W$ has rank $0$, as it is a quotient of $J_X$, so for an odd prime $q\nmid 36$ the operator $A_q=T_q-1-q\dm q$, which descends to $W$, annihilates $J_W(\Q)$ and the image of the rational cusp $c_0$. So $[A_q\overline x_W]=0$ for the image $\overline x_W$ of $\overline x$ on $W_{\F_5}$, which is a closed point of degree $2$. We \gitlinkVIII{C2x36/certify_2_36_D8_final.m}{compute} that $\ell(A_7\overline x_W)=0$, which is a contradiction. Suppose now that $d=9$. By \Cref{d9:table:2n} the possible types of the non-cuspidal part of the reduction of $D$ are $[3],\dots,[9]$, $[3,3]$, $[3,4]$, $[3,5]$, $[3,6]$, $[4,4]$, $[4,5]$ and $[3,3,3]$. Normalizing the first point by the $24$ automorphisms gives $25936$ cases. The map to $X_1(2,18)$, where the reduction of $J_1(2,18)(\Q)$ has index $12$ in $J_1(2,18)(\F_5)$, leaves $2064$ cases. \Cref{d9:lem:Z9}, applied to the image of $D$ on $Z$, leaves $267$ cases, and the four maps to $X_1(36)$, where the reduction of $J_1(36)(\Q)$ has index $214272$ in $J_1(36)(\F_5)$, leave a single case, which is a closed point $x$ of degree $8$ plus a rational cusp $c$. We \gitlinkIX{C2x36/cyclic_tail_after_genus9/tail_all267.m}{compute} all of this. The automorphism $g$ fixes $x$, and the $24$ rational cusps of $X_{\F_5}$ form $6$ orbits of length $4$ under $g$, so by \Cref{d9:lem:Zmult} it suffices to show that $\ell(x+\sum_{j=0}^3g^jc)<4$ for every rational cusp $c$. We construct $W_{\F_5}$ explicitly from the model above, as the extension $\F_5(t,w)$ of the function field of $X_1(9)$, and $X_{\F_5}\to W_{\F_5}$ as the unramified cyclic cover of degree $4$ given by $\zeta^4=f$, where $\ddiv(f)=4D_0$ with $D_0$ of order $4$, which we \gitlinkIX{C2x36/orbit_section_criterion/check_kummer_certified.m}{check}. Then $\ell_X(\pi^*A)=\sum_{j=0}^3\ell_W(A+jD_0)$ for every divisor $A$ on $W_{\F_5}$, where $\pi:X\to W$ is the quotient map. The point $x$ is the pull-back of a closed point $x_W$ of degree $2$ of $W_{\F_5}$, and each orbit $\{g^jc\}$ is the pull-back of a rational cusp $c_W$. We compute $\ell_W(x_W+c_W+jD_0)=1,0,0,0$ for $j=0,1,2,3$, so $\ell_X(x+\sum_jg^jc)=1<4$, which is a contradiction. \end{proof} \subsection{The curve \texorpdfstring{$X_1(2,42)$}{X1(2,42)}}\label{subsec:X1_2_42} Let $X=X_1(2,42)$, a curve of genus $61$, and work at $p=5$. The cusps of $X$ have degrees $1^{18},2^9,6^6$ over $\Q$, and the closed cusps of $X_{\F_5}$ have degrees $1^{18},2^9,3^6,6^3$. Put $S_1=P$, $S_2=21Q$ and $S_3=P+21Q$. The $36$ automorphisms $(E,P,Q)\mapsto(E,S_i,S_j+2uQ)$, with $i\neq j$ and $u\in\{1,2,4,5,8,10\}$, form a group isomorphic to $(\SL_2(\F_2)\times(\torz{21})^\times)/\{\pm1\}$. We push the trace divisor forward to $X_1(2,14)$ and to $X_1(21)$, whose Jacobians have finite Mordell-Weil groups generated by rational cuspidal divisor classes, namely $J_1(2,14)(\Q)\simeq(\torz2)^2\times\torz6\times\torz{18}$ \cite[Theorem 3.1(3), Proposition 4.16 and Table 3]{DEHMZ} and $J_1(21)(\Q)\simeq\torz{364}$ \cite[Theorem 3.1(2), Corollary 4.14 and Table 2]{DEHMZ}. For $d=8$ we also use a quotient of $X$ of genus $31$. \begin{lemma}\label{d8:lem:2_42_quotient} Let $X=X_1(2,42)$ and let $\iota$ be the automorphism $(E,P,Q)\mapsto(E,P,29Q)$, which negates the $3$-primary part of $Q$ and fixes $P$ and $3Q$. Then $Z=X/\dm\iota$ has genus $31$ and good reduction at every prime $p\nmid 42$, and $J_Z(\Q)$ is finite. Over the function field of $X_1(2,14)$, with the universal elliptic curve $y^2=x^3+cx^2+b(1-b-c)x$, $P=(0,0)$, $3Q=(b,b)$, the function field of $Z$ is obtained by adjoining a root $x_3$ of the $3$-division polynomial $3x_3^4+4cx_3^3+6b(1-b-c)x_3^2-b^2(1-b-c)^2$. For a rational effective divisor of degree $8$ on $X$ whose reduction modulo $5$ is $N+C$, with $N$ non-cuspidal and $C$ cuspidal of degree at most $1$ (resp.\ $4$), the divisor $A_{11}\pi_*N$ (resp.\ $A_{17}\pi_*N$) is principal on $Z_{\F_5}$, where $\pi:X\to Z$ is the quotient map. \end{lemma} \begin{proof} A fixed point of $\iota$ would be induced by an automorphism of an elliptic curve fixing a point of order $14$, which is the identity, and the identity does not negate a point of order $3$. We \gitlinkVIII{C2x42/check_C242_forget3sign_Tate.py}{check} that $\iota$ acts freely on the $72$ cusps as well, so $\pi$ is unramified, also modulo every prime $p\nmid42$, and $g(Z)=31$ as $g(X)=61$. A point of order $3$ is determined up to sign by its $x$-coordinate, which gives the function field. The Jacobian of $Z$ is a quotient of $J_X$, which has rank $0$. For a prime $q\nmid 42$ the isogenies of degree $q$ commute with $\iota$, so $T_q$ descends to $Z$, and $A_q$ annihilates $J_Z(\Q)$ and the image of the rational cusp $c_0$. We check that $\sigma_{11}$ fixes every closed cusp of $X_{\F_5}$ of degree $1$, while $\sigma_{17}$ fixes every closed cusp as $17\equiv5^5\pmod{42}$, so $A_qC=0$ in the respective cases by \Cref{d8:lem:cuspHecke}. \end{proof} \begin{lemma}\label{d8:lem:2_42_h} Let $X=X_1(2,42)$ and $h(E,P,Q)=(E,P+21Q,Q)$. Let $x$ be a non-cuspidal closed point of $X$ of degree $d\mid 8$, and $D=(8/d)x$. Then the reduction of $D$ modulo $5$ is not fixed by $h$. \end{lemma} \begin{proof} Suppose $h\overline D=\overline D$. Then $[hD-D]\in J_X(\Q)$ reduces to $0$, so $hD\sim D$, and $hD=D$ by \Cref{d8:lem:closed}, as $\gon_\Q X\geq12$. So $hx=x$. The map $X\to X_1(42)$, $(E,P,Q)\mapsto(E,Q)$, has degree $2$, and $h$ is its non-trivial automorphism over $X_1(42)$. The map is unramified on the non-cuspidal locus, as an automorphism of an elliptic curve fixing a point of order $42$ is the identity. As $hx=x$, the image of $x$ is a non-cuspidal point of $X_1(42)$ of degree $d/2$, which divides $4$. This is impossible, as no elliptic curve over a number field of degree $1$, $2$ or $4$ has a point of order $42$ \cite{mazur,KM,kamienny}, \cite[Theorem 1.1]{DN}. \end{proof} For $d=9$ let $G_9$ be the group of level automorphisms generated by $a(P,Q)=(21Q,P+Q)$, which permutes the three points of order $2$ cyclically, and $b(P,Q)=(P,25Q)$, which multiplies the point of order $7$ by $4$, and let $Y=X/G_9$, which has genus $5$, with quotient map $\pi_Y:X\to Y$. \begin{lemma}\label{d9:lem:Y5} The reduction of $J_Y(\Q)$ modulo $5$, taken relative to a rational cusp $c_\ast$, is contained in a subgroup $K\simeq(\torz2)^3\times\torz{24}$ of index $18$ of $J_Y(\F_5)\simeq\torz2\times(\torz6)^2\times\torz{48}$. \end{lemma} \begin{proof} The function field of $Y$ is obtained from the Tate normal form of $X_1(7)$, with parameter $v$, by putting $t=(v^3-3v+1)/(v(v-1))-8$, adjoining $u$ with $u^2=t$, and the coordinates of a point of order $3$ on the elliptic curve \[ E_u:y^2=x^3-\frac{f_2}{48f_1}x+\frac{f_3}{864f_1^2}, \] where $f_1=t^2+13t+49$, $f_2=t^2+5t+1$ and $f_3=t^4+14t^3+63t^2+70t-7$. This field has degree $8$ over $\Q(u)$, and $j=f_1f_2^3/t$ has degree $16$ as a function of $u$, so it has degree $128=1152/9$ over $\Q(j)$ and is the full invariant field of $G_9$. We \gitlinkIX{C2x42/genus5_quotient/probe_finite.m}{compute} its reduction modulo $5$, which has genus $5$. The Jacobian $J_Y$ has rank $0$, as a quotient of $J_X$, so $A_{11}=T_{11}-1-11\dm{11}$ kills $J_Y(\Q)$. We compute $A_{11}$ on $J_Y(\F_5)$, on the cusps by \Cref{d9:prop:tools} and on three ordinary cubic places from the isogenies of degree $11$, and find that its kernel is $K$. \end{proof} \begin{proposition}\label{d8:prop:2_42}\label{d9:prop:2_42} The group $\tg{2}{42}$ is not a subgroup of the torsion of an elliptic curve over a number field of degree $8$ or $9$. \end{proposition} \begin{proof} Let $D$ be the trace divisor. Suppose first that $d=8$. By \Cref{d9:table:2n} the non-cuspidal part of the reduction of $D$ is a single closed point of degree $4,\dots,8$ or a sum of two closed points of degree $4$. We \gitlinkVIII{C2x42/enumerate_2_42_D8_direct_all_degrees.m}{find} $5,18,64,316,1372$ representatives of the closed points of degrees $4,\dots,8$ under the group of $36$ automorphisms. For the sums of two quartic points we reduce only the first point and take all $126$ choices of the second, which gives $1775$ single points and $630$ pairs to test. We first use the maps $X\to X_1(2,14)$, $(E,P,Q)\mapsto(E,P,3Q)$, and $X\to X_1(21)$, $(E,P,Q)\mapsto(E,2Q)$, and we apply \eqref{d8:eq:pushclass}, with all the cuspidal divisors over $\F_5$. We \gitlinkVIII{C2x42/filter_2_42_D8_small_quotients_shards.m}{compute} that only $288$ of the $2405$ cases pass. Next we use the three maps $X\to X_1(21)$ given by $(E,P,Q)\mapsto(E/G,2Q\bmod G)$ for $G=\dm P,\dm{21Q},\dm{P+21Q}$, and \gitlinkVIII{C2x42/filter_2_42_D8_three_2isogenies_v2.m}{find} that $41$ cases pass: $2$ quartic points with a cuspidal part of degree $4$, a septic point with a cusp of degree $1$, $4$ octic points and $34$ pairs of quartic points. Up to the full group of automorphisms these are $30$ cases. For these we use the curve $Z$ of \Cref{d8:lem:2_42_quotient}. We \gitlinkVIII{C2x42/sieve_C242_genus31_remaining.m}{compute} that $A_{17}\pi_*N$ is not principal for both quartic points, and that $A_{11}\pi_*N$ is not principal for the septic point, \gitlinkVIII{C2x42/probe_C242_genus31_Hecke_v3.m}{all $4$ octic points} and $21$ of the $23$ remaining pairs. For one of the two remaining pairs $A_{17}\pi_*N$ is not principal, as we \gitlinkVIII{C2x42/refine_C242_genus31_pair118_v2.m}{compute}, and the other pair is fixed by the automorphism $h$ of \Cref{d8:lem:2_42_h}, so it is excluded by that lemma. Suppose now that $d=9$. By \Cref{d9:table:2n} the possible types of the non-cuspidal part of the reduction of $D$ are $[4],\dots,[9]$, $[4,4]$ and $[4,5]$, and up to the $36$ level automorphisms there are $11533$ cases. We \gitlinkIX{C2x42/genus5_quotient/filter_919/filter919.m}{compute} the following sequence of necessary conditions. \begin{enumerate} \item[a)] The two maps to $X_1(2,14)$, keeping $(P,3Q)$ or dividing by $\dm{14Q}$, where the reduction of $J_1(2,14)(\Q)$ has index $4$ in $J_1(2,14)(\F_5)$, leave $919$ cases. \item[b)] \Cref{d9:lem:Y5}, applied to the image of $D$ on $Y$ with all the cuspidal complements allowed, leaves $119$ cases. \item[c)] The four maps to $X_1(21)$, keeping $2Q$ on $E$ and on its three quotients by points of order $2$, where the reduction of $J_1(21)(\Q)$ has index $6$ in $J_1(21)(\F_5)$, leave $44$ cases. \item[d)] If the non-cuspidal part has degree $8$ and is fixed by a level automorphism $g$ of even order, then \Cref{d9:lem:stab}(b), with a cuspidal complement of degree $1$ and $9+1<\gon_\C X$, gives $gD=D$, which contradicts \Cref{d9:lem:odd}. This leaves $18$ cases. \end{enumerate} Exactly $3$ of the remaining cases are single closed points $P_9$ of degree $9$, and each is fixed by an element $h\in G_9$ of order $3$. As $J_X(\Q)$ is finite, \Cref{d9:lem:stab}(a) gives $hD=D$, and by \Cref{d9:lem:odd} $D$ is the pull-back of a closed point of degree $3$ of $X/\dm h$, so $\pi_{Y*}D/3$ is a rational effective divisor of degree $3$ on $Y$, and the class of its reduction, taken relative to $c_\ast$, must lie in $K$. This excludes $2$ of these $3$ cases. For the third, $h=ab^2$, the image of $D$ on $V=X/\dm h$ is a rational effective divisor of degree $3$ whose reduction is a cubic point $P_3$ of $V_{\F_5}$. As $J_V(\Q)$ is finite, $B=T_{11}P_3-P_3-11\dm{11}P_3$ is principal. We compute $T_{11}P_3$ from the $12$ isogenies of degree $11$, a sum of $4$ closed points of degree $9$, and we \gitlinkIX{C2x42/diagonal_genus17_lattice/export.m}{check} by a Riemann-Roch computation on $V_{\F_5}$ that $\ell(B)=0$. Here $V$ has genus $17$. Indeed, $h$ fixes no cusp, as it permutes the points of order $2$ cyclically, while at a cusp a point of order $2$ is distinguished. A non-cuspidal fixed point of $h$ needs an automorphism of order $3$ of $E$, so $j=0$, and counting the compatible level structures for the two automorphisms of order $3$ gives $6$ fixed points for each. So $h$ has $12$ fixed points, and $120=3(2g(V)-2)+24$ by the Riemann-Hurwitz formula. Among the $15$ remaining cases, $8$ are closed points of degree $6$ with $j=0$, which are fixed by $b$. The image of such a point on $Y$ is three times a point of degree $2$, and a cuspidal complement which is not the reduction of a rational cuspidal divisor is a closed cusp $c$ of degree $3$, which is fixed by $b$ and maps to $3c_Y$ with $c_Y$ one of the two cusps of $Y$ defined over $\Q(\sqrt{21})$, which are $\F_5$-rational as $5$ splits in $\Q(\sqrt{21})$. Both of these contribute the same non-zero class of order $2$ modulo $K$, and for $4$ of these $8$ points the condition of \Cref{d9:lem:Y5} forces this contribution. For them the whole reduction is fixed by $b$. The cuspidal lift $C$ of $c$ satisfies $bC=C$, so $[bD-D]=[bD-D-bC+C]$ is torsion and reduces to $0$, hence it is $0$, and $bD=D$ by \Cref{d9:lem:closed}, as $\gon_\Q X>9$. So $\pi_{Y*}D/3$ is a rational divisor of degree $3$ on $Y$ reducing to $P_2+c_Y$, and we compute that $[P_2+c_Y-3\overline{c_\ast}]\notin K$. For the remaining $11$ cases, $4$ of degree $6$ with $j=0$ and $7$ others, the same argument shows that the cuspidal complement is the reduction of a rational cuspidal divisor $C$. Then $A_{11}N$ is principal, where $N$ is the non-cuspidal part of the image on $X_1(42)$ under $(E,P,Q)\mapsto(E,Q)$, as $A_{11}$ kills $J_1(42)(\Q)\simeq\torz{182}\times\torz{1092}\times\torz{131040}$ \cite[Theorem 3.1(2), Corollary 4.14 and Table 2]{DEHMZ} and the rational cuspidal divisors. We \gitlinkIX{C2x42/cyclic42_A11_seven/test.m}{compute} that $\ell(A_{11}N)=0$ in $10$ cases, and in the last one for the map $(E,P,Q)\mapsto(E,P+Q)$. \end{proof} \subsection{The group \texorpdfstring{$\torz{48}$}{Z/48Z}}\label{subsec:48} Over septic fields the group $\torz{48}$ is excluded by \Cref{d7:lem:4n} and \Cref{d7:prop:2_24}. In the other degrees we need a separate argument. \begin{proposition}\label{d6:prop:48} The group $\torz{48}$ is not a subgroup of the torsion of an elliptic curve over a sextic field. \end{proposition} \begin{proof} Suppose $E_1$ is an elliptic curve over a sextic field $K$ with a point $P_1$ of order $48$. By \Cref{d6:lem:4n}, the curve $E'=E_1/\dm{24P_1}$ has $\tg{2}{24}\subseteq E'(K)$, so by \Cref{d6:prop:2_24_points}(3) we may assume that $K=K_{2,24}$ and $E'=E_{2,24}$. Let $\phi:E_1\to E'$ be the quotient isogeny. Its kernel has order $2$, and $\ker\hat\phi=\phi(E_1[2])$ is generated by a $K$-rational point of order $2$, so $E_1$ is isomorphic to one of the three curves $2$-isogenous to $E_{2,24}$ over $K$. We \gitlinkVI{verify_sporadic_curves.m}{compute} that their torsion groups over $K$ are $\torz{24}$, $\tg{2}{12}$ and $\torz{24}$, which is a contradiction. \end{proof} For octic and nonic fields we use the Hecke sieve and the push-forwards of \Cref{d9:sec:sieve}. \begin{proposition}\label{d8:prop:48}\label{d9:prop:48} The group $\torz{48}$ is not a subgroup of the torsion of an elliptic curve over a number field of degree $8$ or $9$. \end{proposition} \begin{proof} We have $\rk J_1(48)(\Q)=0$ \cite[Theorem 3.1(2)]{DEHMZ} and $\gon_\Q X_1(48)\geq\gon_{\F_5}X_1(48)\geq10$ \cite[Table 1]{NV}, so the entirely cuspidal reduction modulo $5$ is excluded by \Cref{d9:lem:cuspred}. The numbers of non-cuspidal closed points of $X_1(48)_{\F_5}$ of degrees $1,\dots,9$ are $0$, $0$, $16$, $64$, $656$, $2712$, $10784$, $48992$ and $217520$, so for $d=9$ the possible types of $D'$ are $[k]$ for $3\leq k\leq9$, $[3,3]$, $[3,4]$, $[3,5]$, $[3,6]$, $[4,4]$, $[4,5]$ and $[3,3,3]$. We use the map $\pi:X_1(48)\to Y=X_1(2,24)$, $(E,P)\mapsto(E/\dm{24P},T,\overline P)$, where $T$ is the non-zero point of the kernel of the dual isogeny and $\overline P$ the image of $P$, which has order $24$. It has degree $2$, as $P$ and $25P$ have the same image, and it commutes with the diamond operators. The group $J_Y(\Q)\simeq\torz2\times(\torz4)^2\times\torz{120}\times\torz{240}$ is generated by rational cuspidal divisors \cite[Proposition 4.16 and Table 3]{DEHMZ}, and we \gitlinkIX{C48/replay_D9_integer.py}{compute} that its reduction $G$ has index $4608$ in $J_Y(\F_5)$ (the group $J_Y(\F_5)$ and the order $921600$ of $G$ are given in the proof of \Cref{d7:prop:2_24}). We apply \eqref{d9:eq:pushclass} with $d=9$, allowing all the effective cuspidal divisors of degree $9-e$ on $Y_{\F_5}$. For the types other than $[9]$ there are $20882$ cases, and $48$ of them satisfy \eqref{d9:eq:pushclass}. For the type $[9]$ we enumerate the $216208$ closed points of degree $9$ of $Y_{\F_5}$, in $13513$ orbits under the $16$ level automorphisms of $Y$ which preserve $G$ and the sets $\mathcal S_j$. Of these orbits, $1644$ do not lift to $X_1(48)$. Before computing the classes of the others, we evaluate four characters $\chi_f$ as in \Cref{d9:lem:quadchar}, for functions $f$ on $Y_{\F_5}$ with $\ddiv(f)=2A$, which we check to be trivial on $G$. They exclude $11173$ orbits, and of the remaining $696$ only $11$ satisfy \eqref{d9:eq:pushclass}, which gives $24$ cases on $X_1(48)$. We \gitlinkIX{C48/character_filter9/run.m}{compute} all of this. For the remaining $72$ cases we use the Hecke sieve on $X_1(48)_{\F_5}$. The cusps of $X_1(48)$ are defined over $\Q(\zeta_{48})$ and the orbits of closed cusps of $X_1(48)_{\F_5}$ have degrees $1^{16},2^{12},4^4$. As $29\equiv5^3\pmod{48}$, the automorphism $\sigma_{29}$ fixes every closed cusp, so $29$ is admissible for every $C'$. As the closed cusps of degree $1$ are defined over $\Q$ or $\Q(i)$, the prime $13$ is admissible if $C'$ is $0$ or a single closed cusp of degree $1$. So we use $A_{29}$ if $D'$ has degree $6$ or $7$, $A_{13}$ if $D'$ has degree $8$ and $A_7$ if $D'$ has degree $9$, and we \gitlinkIX{C48/certify48.py}{compute} that $A_qD'$ is not principal in all $72$ cases. This also covers $d=8$: as $\mathcal S_{8-e}\subseteq\mathcal S_{9-e}$, every $D'$ that satisfies \eqref{d8:eq:pushclass} for $d=8$ also satisfies it for $d=9$, so it is, up to the diamond operators, one of the $48$ cases above, and the prime that we use for it is admissible for its cuspidal part of degree $8-e$: $29$ is admissible for every $C'$, and $13$ is used only if $D'$ has degree $8$, when $C'=0$ (see \Cref{subsec:sieve_method}). \end{proof} \subsection{The group \texorpdfstring{$\torz{52}$}{Z/52Z}}\label{subsec:52} Over sextic, septic and octic fields the group $\torz{52}$ is excluded in \Cref{d6:prop:2_26}. Over nonic fields we use the uniqueness of the elliptic curve with a subgroup $\tg{2}{26}$, which follows from the same proposition. \begin{proposition}\label{d9:prop:52} The group $\torz{52}$ is not a subgroup of the torsion of an elliptic curve over a nonic field. \end{proposition} \begin{proof} Let $F/K$ have a point $U$ of order $52$, and let $\phi:F\to E=F/\dm{26U}$. The point $\phi(U)$ has order $26$ and is independent of the generator of the kernel of the dual isogeny, as the dual isogeny maps $13\phi(U)$ to $26U\neq0$. So $E$ has a subgroup isomorphic to $\tg{2}{26}$, and by \Cref{d9:prop:2_26} and \Cref{d9:rem:exactdegree} the pair $(K,E)$ is isomorphic to the pair $(K_{2,26},E_{2,26})$ of \Cref{d9:prop:realizations}. Hence $F$ is the quotient of $E_{2,26}$ by one of its three subgroups of order $2$. If $F$ has full $2$-torsion over $K$, then $\tg{2}{26}\subseteq F(K)$, and the same reductions as for $E_{2,26}$ show that $\#F(K)_\tors$ divides $52$, so $F$ has no point of order $4$. Otherwise the point of order $4$ would be a half of the generator of the kernel of the dual isogeny. Writing $E_{2,26}:y^2=x^3+cx^2+b(1-b-c)x$ with $2$-torsion points $(r_i,0)$, the quotient by $(r_i,0)$ is $y^2=x^3-2Ax^2+\delta^2x$ with $A=c+3r_i$ and $\delta=r_j-r_k$, and a half of $(0,0)$ on it has $x=\pm\delta$ and $y^2=2x(\delta^2-Ax)$. We \gitlinkIX{C2x26/explicit/replay_sage.py}{check}, by reducing modulo primes above $3$ and $7$, that all $6$ of these values are non-squares in $K_{2,26}$. \end{proof} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{The groups \texorpdfstring{$\tg mn$}{Z/mZ x Z/nZ} with \texorpdfstring{$m\geq3$}{m>=3}}\label{sec:mxn}\label{d8:sec:cyclotomic} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Let $E$ be an elliptic curve over a number field $K$ with $\tg mn\subseteq E(K)$ for some $m\geq3$. By the Weil pairing $\varphi(m)$ divides $[K:\Q]$, and $\varphi(m)$ is even, so $K$ is a sextic or an octic field (see \Cref{subsec:trace}). Over sextic fields we treat here the groups $\tg{3}{18}$, $\tg{3}{21}$, $\tg{3}{24}$, $\tg{3}{27}$, $\tg{4}{16}$, $\tg{4}{20}$ and $\tg{6}{12}$. The groups $\tg{4}{24}$, $\tg{7}{7}$ and $\tg{9}{9}$ are ruled out in \Cref{d6:prop:rank0_cases}, and $\tg{3}{15}$ in \Cref{d6:prop:doubled_cases}. Over octic fields we treat here the groups $\tg{3}{3n}$ for $n=6,7,8,9$, $\tg{4}{4n}$ for $n=4,5,6$, $\tg{5}{5n}$ for $n=3,4,5$, $\tg{6}{12}$, $\tg88$ and $\tg{10}{10}$. For most groups that we treat in both degrees, the computation for octic fields also covers sextic fields. The groups $\tg{5}{10}$ and $\tg{3}{15}$ do occur over octic fields (\Cref{d8:prop:5_10,d8:prop:3_15}), and the uniqueness in \Cref{d8:prop:3_15} also rules out $\tg{3}{30}$. The groups $\tg{3}{33}$, $\tg{3}{39}$ and $\tg{5}{35}$ are ruled out in \Cref{d8:prop:33,d8:prop:39,d8:prop:35}. In this section $F=\Q(\zeta_m)$ with $m\geq3$, and the trace divisor of a point over $K$ has degree $d_m=[K:F]$ over $F$. For $m\in\{3,4,6\}$ we have $d_m=3$ over sextic fields and $d_m=4$ over octic fields, and for $m\in\{5,10\}$ we have $d_m=2$. By \cite[Theorem 4.1]{DS}, $\rk J_1(m,n)(F)=0$ for \[ (m,n)\in\{(3,15),(3,18),(3,21),(3,24),(3,27),(4,16),(4,20),(4,24),(5,15),(5,20),(6,12)\}. \] In \cite{DS} the case $m=5$ is listed without a written proof, so we also checked it: by the argument recalled before \Cref{d8:lem:X10} below, it suffices that all the twists $f\otimes\chi$ of the newforms $f$ occurring in $J_1(5,15)$ and $J_1(5,20)$ by characters $\chi$ modulo $5$ have non-zero $L$-value at $1$. These twists have levels dividing $75$ and $100$, and there are no newforms of positive analytic rank at these levels \cite{lmfdb}. For a prime $q\nmid n$ with $q\equiv1\pmod m$, the operator $A_q$ annihilates $J_1(m,n)(F)$ when this group is finite, and it annihilates the $F$-rational cusps as divisors (\Cref{d8:prop:tools}). In each case over an octic field we also use an explicit group of automorphisms of $X_1(m,n)$ over $F$, given by replacing the level structure $(P,Q)$ by integral linear combinations of $P$ and $Q$ which preserve the orders and the Weil pairing, taken modulo $\pm1$. They commute with $T_q$ for $q\equiv1\pmod m$. By \Cref{d8:lem:CM}, an elliptic curve over an octic field with one of the groups that we exclude in this section has no complex multiplication, so \Cref{lem:orbits} applies to its trace divisor. As the cusps of $X_1(m,n)$ are defined over $\Q(\zeta_n)$, they are finite \'etale at every prime $p\nmid n$, and if $\gon_F X_1(m,n)>d_m$ and $p$ is odd, the reduction of the trace divisor modulo a prime of $F$ above $p$ is not entirely cuspidal by \Cref{d8:prop:main}(a). \subsection{The groups \texorpdfstring{$\tg{3}{27}$, $\tg{5}{15}$ and $\tg{5}{20}$}{Z/3Z x Z/27Z, Z/5Z x Z/15Z and Z/5Z x Z/20Z}}\label{subsec:3x27_5x15_5x20} \begin{proposition}\label{d8:prop:local}\label{d6:prop:3_27} There is no elliptic curve over a number field of degree $6$ or $8$ with a subgroup isomorphic to $\tg{3}{27}$, and there is no elliptic curve over an octic field with a subgroup isomorphic to $\tg{5}{15}$ or $\tg{5}{20}$. \end{proposition} \begin{proof} In each case we show that the assumptions of \Cref{d8:prop:main} hold for a suitable prime $p$, and that the elliptic curve cannot have good reduction at a prime above $p$ by \Cref{d6:lem:badred}. For $(m,n)=(5,15),(5,20)$ we have $d_m=2$, and $i(5,15)=480$ and $i(5,20)=720$ give $\gon_\C X_1(m,n)>4.76$ and $>7.14$. We take $p=11$, which splits completely in $F$, so the residue fields are $\F_{11}$ and $\F_{121}$. Their Hasse intervals $[6,18]$ and $[100,144]$ contain no multiple of $75$, and the only multiple of $100$ is $100$ itself, over $\F_{121}$. Then the trace of Frobenius is $22$, so the Frobenius is the scalar $11$ and the group is $\tg{10}{10}$, which has no point of order $20$. For $(m,n)=(3,27)$ we take $p=2$. We have $d_m=3$ or $4$, and $i(3,27)=972$ gives $\gon_\C X_1(3,27)>9.64>2d_m$. The prime $2$ is inert in $F$, so the residue fields of $K$ above $2$ are $\F_4$, $\F_{16}$ and $\F_{64}$, and also $\F_{256}$ when $K$ is an octic field. We \gitlinkVIII{C5x15_C5x20_C3x27/check_local_cyclotomic.m}{check} that no elliptic curve over these fields has a subgroup $\tg{3}{27}$. For the first two this follows from the Hasse bound. Over $\F_{64}$ it would force $E(\F_{64})=E[9]\simeq(\torz9)^2$ (see the discussion after \Cref{d6:lem:badred}), which has no point of order $27$. Over $\F_{256}$ the only multiple of $81$ in the Hasse interval is $243$, which would require the trace of Frobenius $14$, and this is impossible by \cite[Theorem 4.1]{waterhouse}. \end{proof} \subsection{The group \texorpdfstring{$\tg{3}{21}$}{Z/3Z x Z/21Z}}\label{subsec:3x21} \begin{proposition}\label{d8:prop:3_21}\label{d6:prop:3_21} There is no elliptic curve over a number field of degree $6$ or $8$ with a subgroup isomorphic to $\tg{3}{21}$. \end{proposition} \begin{proof} Let $E$ be an elliptic curve over a number field $K$ of degree $6$ or $8$ with $\tg{3}{21}\subseteq E(K)$, so that $d_m=[K:F]$ is $3$ or $4$. Let $X=X_1(3,21)$ over $F=\Q(\zeta_3)$, of genus $25$, and let $\mathfrak p$ be the prime of $F$ above $2$, with residue field $\F_4$. The closed cusps of $X_{\F_4}$ have degrees $1^{24},3^8$, so there are $\binom{27}{4}+24\cdot8=17742$ effective cuspidal divisors $\overline C$ of degree $4$ on $X_{\F_4}$, and we \gitlinkVIII{C3x21/double_cusp_3_21_D4_dt.m}{compute} that $\ell(2\overline C)=1$ for each of them (see also \githubbareVIII{C3x21/double_cusp_3_21_D4.m}), using the spaces $\Omega(2jP)$ as in \Cref{subsec:doubled}. Every effective cuspidal divisor of degree $3$ on $X_{\F_4}$ is contained in one of degree $4$, so by \Cref{rem:smaller} the same holds for the divisors of degree $3$. As $\rk J_X(F)=0$, \Cref{d6:lem:doubled} with $d=d_m$ shows that $E$ has good reduction at some prime of $K$ above $\mathfrak p$. If $K$ is a sextic field, then the residue fields of $K$ above $\mathfrak p$ are $\F_4$, $\F_{16}$ and $\F_{64}$. No elliptic curve over $\F_{64}$ has exactly $63$ points (see the discussion after \Cref{d6:lem:badred}), and the Hasse intervals of these fields contain no other multiple of $63$, so $E$ has bad reduction at every prime of $K$ above $2$ by \Cref{d6:lem:badred}, which is a contradiction. If $K$ is an octic field, we \gitlinkVIII{C3x21/preflight_3_21_Frob.m}{compute} that the non-cuspidal closed points of $X_{\F_4}$ of degrees $1,\dots,4$ number $0,0,0,144$, so the trace divisor reduces to one of the $144$ non-cuspidal closed points of degree $4$. Write $Q_3=7Q$ and $Q_7=3Q$. The $72$ automorphisms $(P,Q)\mapsto(aP+bQ_3,\ cP+dQ_3+5uQ_7)$, with $\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\SL_2(\F_3)$ and $u\in(\torz7)^\times$, taken modulo $\pm1$, have $2$ orbits on these points, and we \gitlinkVIII{C3x21/sieve_3_21_D4_p2.m}{compute} that $A_{13}x$ is not principal for a representative $x$ of each orbit. This is a contradiction. \end{proof} \subsection{The groups \texorpdfstring{$\tg{3}{24}$ and $\tg{4}{24}$}{Z/3Z x Z/24Z and Z/4Z x Z/24Z}}\label{subsec:3x24_4x24} Over sextic fields the group $\tg{4}{24}$ is ruled out in \Cref{d6:prop:rank0_cases}. \begin{proposition}\label{d8:prop:3_24}\label{d6:prop:3_24} There is no elliptic curve over a number field of degree $6$ or $8$ with a subgroup isomorphic to $\tg{3}{24}$, and there is no elliptic curve over an octic field with a subgroup isomorphic to $\tg{4}{24}$. \end{proposition} \begin{proof} For $(m,n)=(3,24)$ let $X=X_1(3,24)$ over $F=\Q(\zeta_3)$, of genus $25$, and work at a prime $\mathfrak p$ above $7$. The closed cusps of $X_{\F_7}$ have degrees $1^{24},2^{12}$, the $24$ cusps of degree $1$ are reductions of $F$-rational cusps, and the non-cuspidal closed points of degrees $1,\dots,4$ number $0,0,48,456$. As $i(3,24)=576$ gives $\gon_\C X>5.71>d_m$, the reduction of the trace divisor at $\mathfrak p$ is not entirely cuspidal by \Cref{d8:prop:main}(a), so it is a point of degree $3$, plus the reduction of an $F$-rational cusp if $d_m=4$, or a point of degree $4$. Writing $R_3=8Q$ and $R_8=3Q$, the $48$ automorphisms $(P,R_3,R_8)\mapsto(aP+bR_3,\ cP+dR_3,\ uR_8)$, with $\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\SL_2(\F_3)$ and $u\in(\torz8)^\times$, taken modulo $\pm1$, have $1$ orbit on the points of degree $3$ and $13$ on those of degree $4$, and we \gitlinkVIII{C3x24/sieve_3_24_D4_Aut48.m}{compute} that $A_{13}x$ is not principal for all $14$ representatives $x$, which excludes all these reductions. For $(m,n)=(4,24)$ let $X=X_1(4,24)$ over $F=\Q(i)$, of genus $33$, and work at a prime above $5$. As $i(4,24)=768$, we have $\gon_\C X>7.61>4$. The closed cusps of $X_{\F_5}$ have degrees $1^{16},2^{24}$, and the non-cuspidal closed points of degrees $1,\dots,4$ number $0,0,0,96$. So the reduction is a single point of degree $4$. The $64$ automorphisms $(P,Q)\mapsto(aP+6bQ,cP+dQ)$, with $b,c\in\torz4$, $d\in\{1,5,7,11\}$ and $a\equiv d^{-1}(1+6bc)\pmod4$, taken modulo $\pm1$, have $4$ orbits on these points, and we \gitlinkVIII{C4x24/sieve_4_24_D4_fullaut_root.m}{compute} that $A_{13}x$ is not principal for a representative $x$ of each of them. \end{proof} \subsection{The group \texorpdfstring{$\tg{3}{18}$}{Z/3Z x Z/18Z}}\label{subsec:3x18} For $\tg{3}{18}$ the Hecke sieve at a single prime does not suffice, and we do not know $J_1(3,18)(\Q(\zeta_3))$ explicitly. Instead we compare the reductions at two primes, using the following lemma, which does not require the Mordell-Weil group to be generated by cuspidal divisor classes. We use the same comparison for $\tg{4}{16}$ and $\tg{6}{12}$ over octic fields below. \begin{lemma}\label{d6:lem:matched} Let $X$ be a curve over a number field $F$ with $J(F)$ finite and with an $F$-rational cusp $c_\ast$. Let $\mathfrak p_1,\mathfrak p_2$ be primes of good reduction, with residue fields $k_1,k_2$, such that reduction modulo $\mathfrak p_1$ is injective on $J(F)$, let $H_i\subseteq J(k_i)$ be the image of $\ClCusp_F X$ under reduction modulo $\mathfrak p_i$, and let $\psi:H_1\to H_2$ be the composite of the inverse of the reduction map $\ClCusp_F X\to H_1$ with the reduction map $\ClCusp_F X\to H_2$. If $J(k_1)/H_1$ has exponent $e$, then for every $D\in X^{(d)}(F)$, with reductions $\overline D_i$ and $\overline{c_{\ast,i}}$ of $D$ and $c_\ast$ modulo $\mathfrak p_i$, we have \[ \psi\bigl(e[\overline D_1-d\overline{c_{\ast,1}}]\bigr)=e[\overline D_2-d\overline{c_{\ast,2}}]. \] \end{lemma} \begin{proof} Let $y=[D-dc_\ast]\in J(F)$. The reduction of $ey$ modulo $\mathfrak p_1$ lies in $H_1$, so it is the reduction of some $c\in\ClCusp_F X$. Then $ey-c$ has trivial reduction modulo $\mathfrak p_1$, so $ey=c$. Reducing modulo $\mathfrak p_2$ gives the claim. \end{proof} Let $F=\Q(\zeta_3)$ and let $X=X_1(3,18)$ over $F$, a curve of genus $10$ with $\rk J_X(F)=0$ \cite[Theorem 4.1]{DS}. We work at primes $\mathfrak p_1$ and $\mathfrak p_2$ of $F$ above $7$ and $13$, with residue fields $\F_7$ and $\F_{13}$. As $7$ and $13$ both generate $\Gal(\Q(\zeta_{18})/F)$, the Galois orbits of cusps over $F$ stay irreducible modulo $\mathfrak p_1$ and $\mathfrak p_2$, so at both primes the closed cusps have degrees $1^{18},3^6$, and every closed cusp is the reduction of an $F$-rational cuspidal divisor of the same degree. To compare the two primes we need to identify the reductions of the cuspidal divisor classes. We use the model of $X$ of \cite{DSmodels}, as implemented in \cite{mdmagma}, on which the universal curve is $y^2+a_1xy+a_3y=x^3$ with $P=(0,0)$ of order $3$, normalized by $y(Q)=x(Q)^2$ for the point $Q$ of order $18$. A point of this curve with $x=0$ is $\pm P$, and one with $y=0$ is $P$, so $x(Q)$ and $y(Q)$ have no zeros or poles on $Y_1(3,18)$, and the normalization makes $a_1$ and $a_3$ regular functions on $Y_1(3,18)$ over $\Z[\zeta_3,\tfrac16]$. The functions \[ a_3,\quad a_1^3-27a_3,\quad x(kQ+jP),\ y(kQ+jP)\ \ (1\leq k\leq 9,\ 0\leq j\leq 2), \] \[ x(kQ)-x(kQ+P)\ \ (1\leq k\leq 9),\qquad x(kQ+P)-x(kQ+2P)\ \ (1\leq k\leq 8) \] are units on $Y_1(3,18)$ over $\Z[\zeta_3,\tfrac16]$. Indeed, the discriminant of the curve is $a_3^3(a_1^3-27a_3)$, the points $kQ+jP$ are never equal to $O$ or $\pm P$, and in every fibre $kQ+P\neq\pm kQ$ and $kQ+2P\neq\pm(kQ+P)$ for these $k$. So grouping the cusps by their orders of vanishing at these functions gives the same $24$ blocks, of degrees $1^{18},3^6$, at both primes, each of them the reduction of a Galois orbit of cusps over $F$. Let $H_1$ and $H_2$ be the reductions of $\ClCusp_F X$ modulo $\mathfrak p_1$ and $\mathfrak p_2$. We \gitlinkVI{diagonal_3_18_cusp_classes_certificate.m}{compute} that $H_1$ and $H_2$ both have invariant factors $[3,3,3,3,63,63,63]$ and that $J_X(\F_7)/H_1\simeq\torz3\times\torz3\times\torz9$. As $7$ is odd, unramified in $F$ and a prime of good reduction for $X$, reduction modulo $\mathfrak p_1$ is injective on $J_X(F)=J_X(F)_\tors$, so the map $\psi$ of \Cref{d6:lem:matched} is defined, and the blocks determine it. Fix an $F$-rational cusp $c_\ast$. By \Cref{d6:lem:matched} with $e=9$, the reductions $\overline D_1$ and $\overline D_2$ of an effective $F$-rational divisor $D$ of degree $d$ satisfy \begin{equation}\label{d6:eq:matched318} \psi\bigl(9[\overline D_1-d\overline{c_{\ast,1}}]\bigr)=9[\overline D_2-d\overline{c_{\ast,2}}]. \end{equation} \begin{proposition}\label{d6:prop:3_18}\label{d8:prop:3_18} The group $\tg{3}{18}$ is not a subgroup of the torsion of an elliptic curve over a number field of degree $6$ or $8$. \end{proposition} \begin{proof} Let $D$ be the trace divisor, of degree $d_m\in\{3,4\}$. As $\gon_FX\geq6$ \cite[Table 1]{NV}, \Cref{d6:prop:main}(a) excludes the entirely cuspidal reduction at $\mathfrak p_1$ and at $\mathfrak p_2$. The non-cuspidal closed points of degrees $1,\dots,4$ number $0,27,144,513$ over $\F_7$ and $0,54,846,7506$ over $\F_{13}$. Suppose first that $d_m=3$. Then the non-cuspidal part of the reduction of $D$ modulo $\mathfrak p_i$ is either a single point of degree $3$ or a point of degree $2$, in which case the cuspidal complement is a cusp of degree $1$, which is the reduction of an $F$-rational cusp. The diamond operators act on these points with orbits of length $3$. At $\mathfrak p_1$ we \gitlinkVI{diagonal_3_18_cusp_classes_certificate.m}{compute} that $A_{13}$ excludes all $9$ orbits of points of degree $2$ and $36$ of the $48$ orbits of points of degree $3$, and at $\mathfrak p_2$ that $A_7$ excludes all $18$ orbits of points of degree $2$ and $270$ of the $282$ orbits of points of degree $3$. This leaves $36$ possible reductions at each prime, and we compute that none of the $36\cdot36$ pairs $(\overline D_1,\overline D_2)$ satisfies \eqref{d6:eq:matched318} with $d=3$. Suppose now that $d_m=4$, and write $Q'=6Q$, of order $3$. The $27$ automorphisms $(P,Q)\mapsto(aP+bQ',\ cP+dQ)$, with $b,c\in\torz3$, $d\in\{1,5,7\}$ and $a\equiv d^{-1}\pmod3$, preserve the Weil pairing. At $7$ they have $1$, $8$ and $19$ orbits on the points of degrees $2,3,4$. We \gitlinkVIII{C3x18/sieve_3_18_D4_fullaut_root_v4.m}{compute} that $A_{13}$ or $A_{19}$ excludes the quadratic orbit (with all the possible cuspidal parts, and also all the sums of two quadratic points), $4$ of the cubic orbits and $11$ of the quartic orbits. The remaining $4$ cubic orbits have length $9$, so a point $x$ in them is fixed by an automorphism $g$ of order $3$. Its cuspidal complement is the reduction of a single $F$-rational cusp, and as $\gon_FX\geq6>5$, \Cref{d9:lem:stab}(a) gives $gD=D$, and \Cref{lem:orbits} gives $3\mid\deg D=4$, which is impossible. So the reduction of $D$ at $7$ is one of the $216$ quartic points in the remaining $8$ orbits. These $216$ quartic points give $135$ distinct classes $9[x-4\overline{c_\ast}]$. At $13$ we \gitlinkVIII{C3x18/diagonal_3_18_D4_9multiple_aut27_root.m}{test} all the possible reductions against \eqref{d6:eq:matched318} with $d=4$: $9234$ quadratic points with a cuspidal part, $15228$ cubic points with a cusp, $1485$ pairs of quadratic points and $278$ representatives of the quartic points, $26225$ cases in all. Only $270$ cases of a cubic point with a cusp, involving $54$ cubic points in $6$ orbits of length $9$, and $9$ quartic representatives pass. The former are excluded by \Cref{d9:lem:stab}(a) and \Cref{lem:orbits} as at $7$, and for the latter we compute that $A_7x$ is not principal. \end{proof} \subsection{The group \texorpdfstring{$\tg{4}{16}$}{Z/4Z x Z/16Z}}\label{subsec:4x16} \begin{proposition}\label{d6:prop:sieve_m}\label{d8:prop:4_16} The group $\tg{4}{16}$ is not a subgroup of the torsion of an elliptic curve over a number field of degree $6$ or $8$. \end{proposition} \begin{proof} Let $X=X_1(4,16)$ over $F=\Q(i)$, of genus $13$, and let $D$ be the trace divisor, of degree $d_m\in\{3,4\}$. We have $\gon_FX\geq8$ by \cite[Table 1]{NV}. We use the primes of $F$ above $5$, $3$ and $13$, with residue fields $\F_5$, $\F_9$ and $\F_{13}$. The cusps are defined over $\Q(\zeta_{16})$, and both $5$ and $13$ generate $\Gal(\Q(\zeta_{16})/F)$, so the $24$ Galois orbits of cusps over $F$, of degrees $1^{16},2^4,4^4$, stay irreducible modulo the primes above $5$ and $13$. The prime $3$ is inert in $F$, and its Frobenius $\sigma_9$ has order $2$ in $\Gal(\Q(\zeta_{16})/F)$, so modulo $3$ the $4$ quadratic orbits split, and $X_{\F_9}$ has $24$ cusps of degree $1$. We identify the orbits at all these primes using the valuations of integral modular units, in the normal form $y^2+xy+By=x^3+Bx^2$ with $P=(0,0)$ of order $4$. The non-cuspidal closed points of degrees $1,\dots,4$ number $0,0,64,160$ over $\F_5$ and $0,0,320,1808$ over $\F_9$. Suppose first that $d_m=3$. By \Cref{d6:prop:main}(a), as $\gon_FX>3$, and by these numbers, the trace divisor reduces modulo the prime above $3$ to one of the $320$ non-cuspidal closed points $x$ of degree $3$ of $X_{\F_9}$, which form $80$ diamond orbits. The prime $q=5$ splits in $F$, so $A_5$ annihilates $J_X(F)$, and the standard cusp $c_0$ satisfies $A_5(c_0)=0$ by \Cref{d6:prop:tools}(3). Hence $[A_5(x)]=0$, while we \gitlinkVI{sieve_4_16_p3.m}{compute} that $[A_5(x)]\neq0$ for one representative $x$ of each diamond orbit, which is a contradiction. Suppose now that $d_m=4$. We \gitlinkVIII{C4x16/diagonal_4_16_D4_p5_p3.m}{compute} \[ \begin{gathered} J_X(\F_5)\simeq(\torz2)^6\times(\torz4)^2\times(\torz{40})^2\times(\torz{80})^2,\\ \ClCusp_F X\simeq(\torz2)^3\times(\torz4)^2\times\torz{20}\times(\torz{40})^2\times\torz{80}, \end{gathered} \] so $J_X(F)/\ClCusp_F X$ is killed by $2$, as reduction at $5$ is injective on $J_X(F)$. At $5$ the reduction of $D$ is a cubic point plus one of the $16$ rational cusps, or a quartic point, which gives $1184$ possibilities. At $3$ it is a cubic point plus one of the $24$ cusps of $X_{\F_9}$ of degree $1$, or a quartic point. The $64$ automorphisms $(P,Q)\mapsto(aP+4bQ,cP+dQ)$, with $b,c\in\torz4$, $d\in(\torz{16})^\times$ and $a\equiv d^{-1}\pmod4$, taken modulo $\pm1$, have $5$ orbits on the cubic points and $33$ on the quartic points at $3$, which gives $153$ cases. Fix an $F$-rational cusp $c_\ast$. Comparing the classes $2[\overline D-4\overline{c_\ast}]$ at the two primes by \Cref{d6:lem:matched} with $e=2$, only $17$ cases at $3$ pass, all of them a cubic point $x$ plus a cusp. The cusps of degree $1$ of $X_{\F_9}$ are the $16$ reductions of $F$-rational cusps and the $8$ points of the $4$ pairs $\{c,c'\}$ into which the quadratic orbits over $F$ split, and $\sigma_5$ interchanges the two points in each pair. By \Cref{d8:lem:cuspHecke}, $A_5c=0$ for the former, and $A_5c=(c'-c)+5\dm5(c'-c)$ for the latter. The operator $\dm5$ maps each pair to a pair, which we determine from the valuations of the unit functions at $Q$ and at $5Q$, but this leaves open the order of its two cusps. So there are $1+8\cdot2=17$ possible values of $A_5$ on the cuspidal part, and we \gitlinkVIII{C4x16/sieve_4_16_D4_F9_cusp_corrections_refined.m}{compute} that in $16$ of the $17$ cases above, $A_5$ of the reduction is not principal for any of these values. In the remaining case the class $2[D-4c_\ast]$ is the class of $L-8c_\ast$ for an explicit $F$-rational cuspidal divisor $L$ of degree $8$, so $2D\sim L$. At a prime above $13$ we \gitlinkVIII{C4x16/M4N16_D4_final_RR13.m}{compute} that $\ell(\overline L)=1$ and that the unique effective divisor in $|\overline L|$ is a sum of two distinct quartic points. This is a contradiction, as the reduction of $2D$ is an effective divisor in $|\overline L|$ with even coefficients. \end{proof} \subsection{The group \texorpdfstring{$\tg{6}{12}$}{Z/6Z x Z/12Z}}\label{subsec:6x12} \begin{proposition}\label{d8:prop:6_12}\label{d6:prop:6_12} There is no elliptic curve over a number field of degree $6$ or $8$ with a subgroup isomorphic to $\tg{6}{12}$. \end{proposition} \begin{proof} Let $X=X_1(6,12)$ over $F=\Q(\zeta_3)$, of genus $9$, with $\gon_FX\geq6$ \cite[Table 1]{NV}, and let $D$ be the trace divisor, of degree $3$ or $4$. At a prime above $7$ the closed cusps have degrees $1^{16},2^8$ and are reductions of $F$-rational cuspidal divisors of the same degree, and the non-cuspidal closed points of degrees $1,\dots,4$ number $0,0,96,468$. As $\gon_FX>4$, the entirely cuspidal reduction is excluded by \Cref{d8:prop:main}(a), so the reduction of $D$ is a cubic point, plus a cusp if $\deg D=4$, or a quartic point. As $\dm{13}=1$ on $X$, we have $A_{13}=T_{13}-14$, and it annihilates $J_X(F)$ and the $F$-rational cuspidal divisors. So $[A_{13}x]=0$ for the cubic point $x$, and we \gitlinkVIII{C6x12/control_6_12_D4_cubics_root.m}{compute} that $A_{13}x$ is not principal for the $48$ representatives under the diamond involution. This proves the claim for sextic fields. Over an octic field it follows that the reduction of $D$ at $7$ is a single quartic point, with multiplicity $1$, and in particular $D$ is a closed point of degree $4$. Fix an $F$-rational cusp $c_\ast$. The cuspidal divisors of degree $0$ over $F$ are combinations of the $24$ Galois orbits of cusps over $F$, and we \gitlinkVIII{C6x12/diagonal_6_12_D4_root.m}{compute} \[ J_X(\F_7)\simeq\torz6\times\torz{12}\times(\torz{24})^2\times(\torz{48})^2,\qquad \ClCusp_F X\simeq\torz6\times(\torz{12})^3\times(\torz{24})^2, \] so $J_X(\F_7)$ modulo the reduction of $\ClCusp_F X$ is isomorphic to $\torz4\times\torz4$. For a quartic point $x$ at $7$ let $r\in\{2,4\}$ be the order of $[x-4\overline{c_\ast}]$ modulo this reduction. Then $r[D-4c_\ast]\in\ClCusp_F X$, as in the proof of \Cref{d6:lem:matched}. We compare with the reductions at a prime above $13$, where all the $32$ cusps are rational and the non-cuspidal closed points of degrees $1,\dots,4$ number $0,144,576,7116$, testing all the $112020$ possible reductions with all the cuspidal parts. Only $12$ quartic points at $7$ remain, all with $r=2$. For each of them $2D\sim L$ for an $F$-rational cuspidal divisor $L$ of degree $8$, which we identify using the valuations of $141$ integral modular units, and after adding principal divisors of such units we obtain six distinct effective cuspidal divisors $L_1,\dots,L_6$ of degree $8$. We \gitlinkVIII{C6x12/RR_6_12_L8_char0_six_pencils_root.m}{compute} that $\ell(L_i)=2$, that $|L_i|$ is a base point free pencil, and that its ramification divisor is a sum of four closed points of degree $8$, for each $i$. If $2D\sim L_i$, then $2D$ is a fibre of the corresponding map $X\to\PP^1$, so every point of $D$ is a ramification point. This is impossible, as $D$ is a closed point of degree $4$ and the ramification divisor is a sum of closed points of degree $8$. \end{proof} \subsection{The group \texorpdfstring{$\tg{10}{10}$}{Z/10Z x Z/10Z}}\label{subsec:10x10} For $\tg{10}{10}$ and $\tg{5}{25}$ we need the rank over $\Q(\zeta_5)$ of Jacobians which are not covered by \cite[Theorem 4.1]{DS}. By the argument of the proof of \Cref{d6:lem:fullrank}, if $B$ is a quotient of $J_1(N)$ over $\Q$ and $F$ is an abelian field, then $B(F)$ is finite if, for every newform $f$ occurring in $B$ and every character $\chi$ of $\Gal(F/\Q)$, the twist $\rho_f\otimes\chi$ is the representation of a newform $g$ with $L(g,1)\neq0$. \begin{lemma}\label{d8:lem:X10} Let $F=\Q(\zeta_5)$. The Jacobian of the modular curve $X(10)$ over $F$, parametrizing elliptic curves with a basis of the $10$-torsion with fixed Weil pairing, has rank $0$ over $F$. \end{lemma} \begin{proof} Let $H=\dm{11}\subseteq(\torz{100})^\times$, of order $10$. By the description of \cite[Section 2]{DS}, $X(10)$ is isomorphic over $F$ to $X_1(100)/H$, which is defined over $\Q$. The non-trivial even characters modulo $100$ that are trivial on $H$ factor through $(\torz{10})^\times$, so the newforms occurring in the Jacobian of $X_1(100)/H$ have level dividing $100$ and character trivial or quadratic of conductor $5$. Their local representations at $5$ have conductor exponent at most $2$ and tame determinant, so they are tamely ramified, as a wildly ramified two-dimensional representation with conductor exponent at most $2$ has a wildly ramified determinant. So their twists by characters $\chi$ modulo $5$ again have level dividing $100$ and even character modulo $10$. We \gitlinkVIII{common/winding_probe.m}{compute} that for each of the six Galois orbits of even characters modulo $100$, of orders $1,20,10,5,4,2$, the winding submodule is the full space of cusp forms, of relative dimensions $7,13,12,12,9,6$, which add up to the genus $231$ of $X_1(100)$ when counted with the degrees $1,8,4,4,2,1$ of the fields of values. So every newform of level dividing $100$ has non-zero $L$-value at $1$, and the lemma follows from the argument above. \end{proof} \begin{proposition}\label{d8:prop:10_10} There is no elliptic curve over an octic field with a subgroup isomorphic to $\tg{10}{10}$. \end{proposition} \begin{proof} Let $F=\Q(\zeta_5)$ and $X=X(10)_F$, a curve of genus $13$ with $36$ cusps, all defined over $F$, and $\gon_\C X\geq\lceil\tfrac{325}{2^{15}}\cdot 360\rceil=4$. A trace divisor $D$ has degree $2$ over $F$. We use the following model. Let \begin{align*} a(t)&=t^{20}-228t^{15}+494t^{10}+228t^5+1,\\ b(t)&=t^{30}+522t^{25}-10005t^{20}-10005t^{10}-522t^5+1,\\ h(t)&=t^5(t^{10}+11t^5-1)^5, \end{align*} so that $b^2-a^3=1728h$. The elliptic curve $y^2=x^3-27a(t)x+54b(t)$ over $F(t)$ has $j$-invariant $-a^3/h$, of degree $60$, and its $5$-torsion representation is $\pm1$, so a quadratic twist of it has full $5$-torsion, and $F(t)$ is the function field of $X(5)_F$. Adjoining the $2$-torsion, we \gitlinkVIII{C10x10/check_full10_Klein5_root.m}{obtain} the function field $F(t)(r,s)$, $r^3-27a(t)r+54b(t)=0$, $s^2=-3r^2+108a(t)$, of $X$, and \gitlinkVIII{C10x10/diagonal_full10_D2_root.m}{check} that it has genus $13$. Let $C\subseteq J_X(F)$ be generated by the classes of the differences of the cusps, and fix a cusp $c_\ast$. We work at primes above $11$ and $31$, which split completely in $F$, at which reduction is injective on $J_X(F)$ by \Cref{d8:lem:X10}. We \gitlinkVIII{C10x10/diagonal_full10_D2_root.m}{compute} \[ J_X(\F_{11})\simeq\torz2\times\torz6\times(\torz{30})^4\times(\torz{90})^4, \] in which the reduction of $C$ is $\torz2\times\torz6\times(\torz{30})^8$, so $3J_X(F)\subseteq C$. We identify the cusps at the two primes using the valuations of the units $t$, $t-u$ for the ten roots $u$ of $u^{10}+11u^5-1$, and the differences of the roots of the cubic, which are units on the integral model away from $2$ and $5$. There are no non-cuspidal points of $X$ over $\F_{11}$ and $\F_{31}$. By \Cref{d8:prop:main}(a), as $\gon_FX\geq4>2$, the reduction of $D$ at $11$ is not entirely cuspidal. So $D$ reduces to one of the $150$ non-cuspidal closed points $P$ of degree $2$ of $X_{\F_{11}}$, and similarly to one of the $810$ such points $Q$ of $X_{\F_{31}}$, and $3[P-2\overline{c_\ast}]$ and $3[Q-2\overline{c_\ast}]$ are the reductions of the same element of $C$. We compute that this holds for none of the $150\cdot810$ pairs. \end{proof} \subsection{The group \texorpdfstring{$\tg{5}{25}$}{Z/5Z x Z/25Z}}\label{subsec:5x25} \begin{lemma}\label{d8:lem:J25} Let $F=\Q(\zeta_5)$. Then $J_1(25)(F)$ is finite of odd order. \end{lemma} \begin{proof} The non-zero spaces of cusp forms of weight $2$ on $\Gamma_1(25)$ have characters of order $5$ or $10$. Let $f$ be a newform in them with character $\varepsilon$ and let $\chi$ be a character modulo $5$. The local representation of $f$ at $5$ has conductor exponent $2$, which is the codimension of the inertia invariants plus the Swan conductor. If the Swan conductor were $2$, the inertia invariants would be the whole space, which is impossible, as an unramified representation has Swan conductor $0$. So the Swan conductor is at most $1$. Twisting by the tame character $\chi$ does not change the Swan conductor, which depends only on the action of the wild inertia group, so $f\otimes\chi$ has conductor exponent at most $3$ at $5$, and it is unramified elsewhere. So $f\otimes\chi$ has level dividing $125$, and its character $\varepsilon\chi^2$ is trivial on $26$, as $26\equiv1\pmod{25}$. We \gitlinkVIII{C5x25/winding125_H26_root.m}{compute} that the spaces of cusp forms of level $125$ with characters of orders $5$ and $10$ trivial on $26$ are generated by their winding elements. So $L(f\otimes\chi,1)\neq0$ for all $f$ and $\chi$, and $J_1(25)(F)$ is finite by the argument above. The prime $11$ splits completely in $F$, and we \gitlinkVIII{C5x25/rank_J25_Qzeta5_root.m}{compute} that $\#J_1(25)(\F_{11})=5014032203125$, which is odd. As reduction at $11$ is injective on the torsion, $J_1(25)(F)$ has odd order. \end{proof} \begin{proposition}\label{d8:prop:5_25} There is no elliptic curve over an octic field with a subgroup isomorphic to $\tg{5}{25}$. \end{proposition} \begin{proof} Let $F=\Q(\zeta_5)$, so $[K:F]=2$, and let $D$ be the trace divisor on $X=X_1(25)_F$ of the point of order $25$, obtained by forgetting the point of order $5$. It is an effective $F$-rational divisor of degree $2$. The prime above $2$ in $F$ has residue field $\F_{16}$, and we compute $\#X(\F_{16})=18$, all of these points being cusps. These $18$ cusps are reductions of $F$-rational cusps: the $10$ cusps on the N\'eron $25$-gon are rational, the $8$ cusps on the N\'eron $5$-gon with $5$-torsion in $\mu_5$ are defined over $F$, and the other $10$ cusps form two orbits of length $5$ under the Frobenius $16$, which generates $\Gal(\Q(\zeta_{25})/F)$. If $D$ reduces to a sum of cusps, let $C$ be its lift. Then $[D-C]\in J_1(25)(F)$ reduces to $0$, so $D\sim C$ by \Cref{d8:lem:J25}, as reduction is injective on the odd torsion. We \gitlinkVIII{C5x25/certificate_5_25_target25_cusp_spaces_root.m}{compute} that $\dim\Omega(\overline C)=10$ for all the $171$ effective divisors $\overline C$ of degree $2$ supported on the $18$ cusps, so $\ell(\overline C)=1$, and $\ell(C)=1$ by semicontinuity. So $D=C$, which is a contradiction. So the reduction of $D$ is a closed point $P$ of degree $2$, corresponding to an elliptic curve over $\F_{256}$ with a subgroup $\tg{5}{25}$. By the Hasse bound its order is $250$, and its group is $\tg{5}{50}$. We \gitlinkVIII{C5x25/certificate_5_25_target25_good_root.m}{find} $4$ such curves, giving $100$ closed points of degree $2$ of $X$ over $\F_{16}$, in $12$ diamond orbits. At the prime $11$, $A_{11}$ annihilates $J_1(25)(F)$ and the rational cusp $c_0$, so $[A_{11}P]=0$, and hence also $[(\dm6-1)A_{11}P]=0$. It suffices to test this consequence, and we compute that $(\dm6-1)A_{11}P$ is not principal for the representatives $P$ of all $12$ orbits, which is a contradiction. \end{proof} \subsection{The group \texorpdfstring{$\tg{4}{20}$}{Z/4Z x Z/20Z}}\label{subsec:4x20} For $\tg{4}{20}$ over octic fields we use the following elliptic curve. \begin{lemma}\label{d8:lem:Z420} Let $F=\Q(i)$ and let $Z$ be the normalization of the curve \[ \frac{16(w^2+16)^3}{w^4}=\frac{(t^2+10t+5)^3}{t}. \] Then $Z$ has genus $1$, and $Z(F)$ consists of $4$ points with $t\in\{0,\infty\}$ or $w\in\{0,\infty\}$ and $4$ points at which both sides are equal to $1728$. \end{lemma} \begin{proof} We \gitlinkVIII{C4x20/C420_compact_completion_final.m}{check} that the maps \[ t=\frac{(x-3)^2(x+2)}{(x+1)^2},\qquad w=\frac{y(x+1)^2}{4(x+2)} \] define a birational map from $E':y^2=(x-3)(x+1)(x+2)$ to $Z$. The curves $E'$ and its quadratic twist by $-1$ have rank $0$ over $\Q$, by $2$-descent, so $E'(F)$ is finite. Reduction modulo the prime $3$, which is inert in $F$, injects $E'(F)_\tors$ into $E'(\F_9)\simeq\torz4\times\torz4$, and of the three points of order $2$ only $(-1,0)$ is divisible by $2$ over $F$, so $E'(F)\simeq\torz2\times\torz4$, consisting of $O$, $(-2,0)$, $(-1,0)$, $(3,0)$ and $(-1\pm2i,\pm(4\mp2i))$. The first four map to the points where $t$ or $w$ is $0$ or $\infty$, and the last four to $t=-11\pm2i$, $w=\pm2i$, where both sides equal $1728$. \end{proof} \begin{proposition}\label{d8:prop:4_20}\label{d6:prop:4_20} There is no elliptic curve over a number field of degree $6$ or $8$ with a subgroup isomorphic to $\tg{4}{20}$. \end{proposition} \begin{proof} Let $F=\Q(i)$ and $X=X_1(4,20)$ over $F$, of genus $25$, with $\gon_\C X>\tfrac{325}{2^{15}}\cdot576>5$. Let $D$ be the trace divisor, of degree $3$ or $4$. We work at the prime $3$, which is inert in $F$, with residue field $\F_9$. The closed cusps of $X_{\F_9}$ have degrees $1^{24},2^{12}$, the $24$ cusps of degree $1$ are reductions of $F$-rational cusps, and the non-cuspidal closed points of degrees $1,\dots,4$ number $0,96,320,888$. The entirely cuspidal reduction is excluded by \Cref{d8:prop:main}(a), as $\gon_\C X>5$, so the non-cuspidal part of the reduction of $D$ is a point of degree $2$, $3$ or $4$, or a sum of two points of degree $2$. Let $G$ be the group of $96$ automorphisms $(P,Q)\mapsto(aP+bQ_4,\ cP+dQ_4-uQ_5)$, where $Q_4=5Q$, $Q_5=4Q$, $\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\SL_2(\torz4)$ and $u\in(\torz5)^\times$, taken modulo $(-I,-1)$. It has $2,4,12$ orbits on the points of degrees $2,3,4$. For $q\equiv1\pmod4$, $q\nmid 60$, the operator $A_q$ annihilates $J_X(F)$, and $q=13,17$ are admissible for cuspidal parts of degree at most $1$, and $q=29\equiv9\pmod{20}$ for those of degree at most $2$. We \gitlinkVIII{C4x20/C420_full_reproduce.m}{compute} that $A_qx$ is not principal for one of these $q$ for all the representatives of the points of degree $2$ and $3$ and for $10$ of the $12$ quartic representatives, and that for the sums $x+x'$ of two quadratic points, with $x$ one of the $2$ representatives and $x'$ any of the $96$ quadratic points, $A_q(x+x')$ is not principal except in $2$ cases. This proves the claim for sextic fields. In the remaining cases over an octic field the reduction $\overline D$ has no cuspidal part. If $g\in G$ fixes $\overline D$, then $gD=D$ by \Cref{d9:lem:stab}(a) with $C=0$, as $\gon_FX>4$. So $D$ is invariant under the stabilizer of $\overline D$ in $G$, and by \Cref{lem:orbits} the order of this stabilizer divides $\deg D=4$. The $2$ remaining quartic points lie in orbits of length $12$, so their stabilizers have order $8$, which is impossible. The $2$ remaining sums of two quadratic points have stabilizers of order $4$, which are conjugate in $G$ to the subgroup $H_1$ generated by the automorphisms with parameters $\left(\begin{pmatrix}3&2\\2&3\end{pmatrix},1\right)$ and $\left(\begin{pmatrix}0&-1\\1&0\end{pmatrix},2\right)$. Replacing $D$ by its image under an element of $G$, which is again a trace divisor, we may assume that $D$ is $H_1$-invariant. Then $D$ is a single $H_1$-orbit of a non-CM point, so its image on $X/H_1$ is an $F$-rational point. We \gitlinkVIII{C4x20/C420_compact_completion_final.m}{check} that there is a non-constant map $X/H_1\to Z$ to the curve of \Cref{d8:lem:Z420}, preserving the $j$-invariant. Here $t=b-11-1/b$, where $y^2+(1-b)xy-by=x^3-bx^2$ is the Tate normal form of $(E,Q_5)$. As $t$ does not change when $Q_5$ is replaced by a multiple, it is a parameter on $X_0(5)$, and $j=(t^2+10t+5)^3/t$. Further, $w=4(x(U)-x(T))/(x(P_2)-x(P_2+T))$, where $U=P-Q_4$, $T=2U$ and $P_2=-2Q_4$, with $j=16(w^2+16)^3/w^4$. So the image of $D$ on $Z$ is an $F$-rational point which is not a cusp, hence it has $j=1728$ by \Cref{d8:lem:Z420}, and the elliptic curve has complex multiplication, contradicting \Cref{d8:lem:CM}. \end{proof} \subsection{The group \texorpdfstring{$\tg88$}{Z/8Z x Z/8Z}}\label{subsec:8x8} \begin{proposition}\label{d8:prop:8_8} There is no elliptic curve over an octic field with a subgroup isomorphic to $\tg88$. \end{proposition} \begin{proof} By the Weil pairing $K\supseteq\Q(\zeta_8)$, and a basis of $E[8]$ with fixed Weil pairing gives a non-cuspidal point of $X(8)$ of degree at most $2$ over $\Q(\zeta_8)$. But Bars, Kontogeorgis and Xarles \cite[Remark 5.3]{BKX} showed that the only points of degree at most $2$ over $\Q(\zeta_8)$ on $X(8)$ are the $24$ cusps. \end{proof} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{The sporadic points}\label{sec:sporadic}\label{d6:sec:sporadic}\label{d7:sec:sporadic}% \label{d8:sec:sporadic}\label{d9:sec:sporadic}\label{d9:sec:33_39} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% We now describe the sporadic torsion groups in degree $d\in\{6,7,8,9\}$, that is, the groups in $\Phi(d)\setminus\Phi^\infty(d)$. \Cref{table:sporadic} lists the elliptic curves with these torsion groups that we give explicitly, and the results in which they are given. Each subsection after \Cref{subsec:sporadic_realizations} treats one modular curve in degrees in which the corresponding group is sporadic, and three of them also treat degrees in which it does not occur: $X_1(36)$ in degree $6$, $X_1(39)$ in degrees $6$ and $7$, and $X_1(2,26)$ in degrees $6$, $7$ and $8$. In most cases we determine all the points of degree $d$ on the curve, which shows that the curves we give are the only ones with the given torsion group and excludes the groups properly containing it. For the curves $X_1(N)$ we argue as follows in degree $d$. We work at a prime $p$ at which reduction is injective on $J_1(N)(\Q)$, which is finite and generated by rational cuspidal divisor classes \cite[Theorem 3.1(2), Corollary 4.14 and Table 2]{DEHMZ}, and, except for $X_1(25)$, the fact that $\gon_\Q X_1(N)>d$, so by \Cref{d8:lem:closed} distinct effective rational divisors of degree $d$ have distinct classes. As in \Cref{sec:54}, we identify the Galois orbits of cusps over $\Q$ with their reductions using the valuations of the modular units $b$ and $b^4/\Delta$ of the Tate normal form, which are units on $Y_1(N)$ over $\Z[\tfrac1N]$. We write $H\subseteq J_1(N)(\F_p)$ for the subgroup generated by the reductions of the rational cuspidal divisors of degree $0$, which is the reduction of $J_1(N)(\Q)$. Let $D$ be an effective rational divisor of degree $d$ on $X_1(N)$ whose support contains no cusp. Then $[\overline D-d\overline{c_0}]\in H$, and if this is also the class of the reduction of a known effective rational divisor $A$ of degree $d$, then $D=A$. In particular, $\overline D$ is not the reduction of an effective rational cuspidal divisor. Finally, the diamond operators change neither the elliptic curve nor the residue field of a closed point. So if every such $D$ is a diamond translate of one of the known closed points of degree $d$, then, as $X_1(N)$ is a fine moduli space, every elliptic curve over a number field of degree $d$ with a point of order $N$ is, together with its field, isomorphic to a pair given by one of these points, and its torsion group is the one given in \Cref{subsec:sporadic_realizations}. In the proofs for the curves $X_1(N)$ below, $D$ and $H$ always have this meaning. \subsection{Realizations}\label{subsec:sporadic_realizations} Recall that for $b,c$ in a field, the curve \begin{equation}\label{d6:eq:tate} E(b,c):\quad y^2+(1-c)xy-by=x^3-bx^2 \end{equation} in Tate normal form has the point $(0,0)$, which is a point of order $N$ exactly when $(b,c)$ comes from a point of $Y_1(N)$, and that the curve \begin{equation}\label{d6:eq:tate2} E'(b,c):\quad y^2=x^3+cx^2+(1-b-c)bx \end{equation} has the point $P=(0,0)$ of order $2$ and the point $Q=(b,b)$. The curve $E_{25}^{(6)}$ of \Cref{d6:prop:examples}, the curves of \Cref{d7:prop:realizations,d8:prop:realizations} and the first eight curves of \Cref{d9:prop:realizations} come from the lists of low degree points of van Hoeij \cite{vanHoeij}. For a point $(x,y)$ on the model of $X_1(N)$ of \cite{vanHoeij} we put \[ r=\frac{x^2y-xy+y-1}{x^2y-x},\qquad s=\frac{xy-y+1}{xy},\qquad b=rs(r-1),\qquad c=s(r-1), \] and denote by $E(b,c)$ the elliptic curve \eqref{d6:eq:tate}, on which the point $(0,0)$ has order $N$. In the proofs of \Cref{d7:prop:realizations,d8:prop:realizations,d9:prop:realizations} we bound $E(K)_\tors$ by reduction at primes $\wp$ of $K$ above odd primes $p$ unramified in $K$ at which $E$ has good reduction. \Cref{table:certificates} lists them as $(p,f_\wp,\#\widetilde E(\F_\wp))$, with $f_\wp$ the residue degree of $\wp$. As $E(K)_\tors$ injects into $\widetilde E(\F_\wp)$, its order divides the $\gcd$ of these numbers, which in each case is $N$. \begin{table}[htbp] \centering \small \renewcommand{\arraystretch}{1.1} \begin{tabular}{c|l|l|l|l} $d$ & $G$ & $K$ & $E$ & \\\hline $6$ & $\torz{25}$ & \lmfdbnf{6.2.28037.1} & $E_{25}^{(6)}$ & \Cref{d6:prop:examples}\\ & $\torz{37}$ & \lmfdbnf{6.6.300125.1} & $E_{37}$, $j=-9317$ & \Cref{d6:prop:37}\\ & $\tg{4}{12}$ & \lmfdbnf{6.0.419904.2} & $E_{4,12}$, base change of \lmfdbec{162}{d}{2} & \Cref{d6:prop:examples}\\ & $\tg{2}{24}$ & \lmfdbnf{6.2.231852.1} & $E_{2,24}$ & \Cref{d6:prop:2_24_points}\\\hline $7$ & $\torz{25}$ & \lmfdbnf{7.1.184607.1} & $E_{25}$ & \Cref{d7:prop:realizations}\\ & $\torz{33}$ & \lmfdbnf{7.3.792873.1} & $E_{33}$ & \\ & $\torz{36}$ & \lmfdbnf{7.1.1609956.1} & $E_{36}$ & \\ & $\torz{36}$ & \lmfdbnf{7.3.3192732.1} & $E_{36}'$ & \\\hline $8$ & $\torz{33}$ & \lmfdbnf{8.2.7486875.1} & & \Cref{d8:prop:realizations}\\ & $\torz{34}$ & \lmfdbnf{8.0.1257728.1} & $j=1728$ & \\ & $\torz{34}$ & \lmfdbnf{8.0.1265625.1} & $j$ of degree $2$ & \\ & $\torz{34}$ & \lmfdbnf{8.4.19360000.1} & $j$ of degree $4$ & \\ & $\torz{35}$ & \lmfdbnf{8.4.31878125.1} & & \\ & $\torz{39}$ & \lmfdbnf{8.0.1601613.1} & $j=0$ & \\ & $\torz{40}$ & \lmfdbnf{8.2.31500500.1} & & \\ & $\torz{42}$ & \lmfdbnf{8.4.23040000.1} & & \\ & $\torz{42}$ & \lmfdbnf{8.2.7993287.1} & two curves, $2$-isogenous & \\ & $\tg{5}{10}$ & $\Q(\zeta_{20})$ & $j=1728$ & \Cref{d8:prop:5_10}\\ & $\tg{3}{15}$ & \lmfdbnf{8.0.2313441.1} & $j=1331/8$ & \Cref{d8:prop:3_15}\\\hline $9$ & $\torz{29}$ & \lmfdbnf{9.5.507490937.1} & & \Cref{d9:prop:realizations}\\ & $\torz{31}$ & \lmfdbnf{9.3.114479303.1} & base change from a cubic subfield & \\ & $\torz{32}$ & \lmfdbnf{9.1.32344469.1} & & \\ & $\torz{33}$ & \lmfdbnf{9.3.393718863.1} & & \\ & $\torz{34}$ & \lmfdbnf{9.3.131123051.1} & & \\ & $\torz{39}$ & \lmfdbnf{9.3.195458751.1} & base change from a cubic subfield & \\ & $\torz{40}$ & \lmfdbnf{9.1.30626693.1} & & \\ & $\torz{42}$ & \lmfdbnf{9.3.272097792.1} & base change of \lmfdbec{162}{c}{3} & \\ & $\tg{2}{26}$ & \lmfdbnf{9.3.157505216.1} & $E_{2,26}$, base change from a cubic subfield & \\ \end{tabular} \caption{The elliptic curves $E$ over number fields $K$ of degree $d$ with $E(K)_\tors\simeq G$ given in this section, for the sporadic torsion groups $G$ in degree $d$. The curves with $j=0$ and $j=1728$ have complex multiplication, in accordance with \cite{CCRS14}.} \label{table:sporadic} \end{table} \begin{proposition}\label{d6:prop:examples} \begin{enumerate} \item Let $K_{25}^{(6)}=\Q(x,y)$ with $x^3-x^2+1=0$ and $y^2+(x^2-2x-1)y+x=0$, which is the field \lmfdbnf{6.2.28037.1}, and let $E_{25}^{(6)}=E(b,c)$, with $b$ and $c$ given by the formulas above. Then $(0,0)$ has order $25$ and $E_{25}^{(6)}(K_{25}^{(6)})_\tors\simeq\torz{25}$. \item Let $K_{4,12}=\Q(\alpha,i)$ with $\alpha^3-3\alpha-4=0$, which is the field \lmfdbnf{6.0.419904.2}, and let $E_{4,12}$ be the base change of \lmfdbec{162}{d}{2}: $y^2+xy+y=x^3-x^2+4x-1$ to $K_{4,12}$. Then $E_{4,12}(K_{4,12})_\tors\simeq\tg{4}{12}$. \end{enumerate} \end{proposition} \begin{proof} The curve in (2) was found by Gonz\'alez-Jim\'enez and Najman \cite{GN}. We \gitlinkVI{verify_known_examples.m}{verify} both claims directly in Magma \cite{magma}. \end{proof} \begin{proposition}\label{d7:prop:realizations} Let $N$, $f$ and $y$ be one of the following: \begin{alignat*}{2} &N=25:\quad &&f=x^7+x^6-x^5-x^4+x^2+x-1,\qquad y=x^2+x,\\ &N=33:\quad &&f=x^7-6x^6+9x^5-5x^3-x+1,\\ & &&y=\tfrac13\bigl(-3x^6+16x^5-17x^4-9x^3+11x^2+3x+4\bigr),\\ &N=36:\quad &&f=x^7-x^6+2x^5-4x^4+3x^3+2x^2-2x+1,\\ & &&y=\tfrac16\bigl(-2x^6+3x^5-5x^4+11x^3-9x^2+4x+7\bigr),\\ &N=36:\quad &&f=x^7-2x^6+4x^5-x^4-8x^3+2x^2+5x+1,\\ & &&y=\tfrac12\bigl(-13x^6+36x^5-79x^4+73x^3+51x^2-65x-16\bigr), \end{alignat*} where $x$ is a root of $f$, and put $K=\Q(x)$ and $E=E(b,c)$. Then $E(K)_\tors\simeq\torz N$. The fields $K$ are listed, in the same order, in \Cref{table:sporadic}, and we denote the four pairs $(K,E)$ by $(K_{25},E_{25})$, $(K_{33},E_{33})$, $(K_{36},E_{36})$ and $(K_{36}',E_{36}')$. \end{proposition} \begin{proof} We \gitlinkVII{verify_sporadic_examples.m}{compute} that $(0,0)$ has order $N$ and that $E$ has good reduction at the primes listed in \Cref{table:certificates}, with the numbers of points given there. \end{proof} \begin{proposition}\label{d8:prop:realizations} Let $N$, $f$ and $g$ be one of the following, let $x$ be a root of $f$ and $y$ the corresponding root of $g$, and put $K=\Q(x,y)$ and $E=E(b,c)$. \[ \renewcommand{\arraystretch}{1.15} \begin{array}{r@{\quad}l} N & f\text{ and }g\\\hline 33 & x^4-3x^3+3x^2-3x+1\\ & y^2+(x^3-3x^2+x-3)y+2x^2-x+2\\ 34 & x^8-6x^7+16x^6-28x^5+42x^4-44x^3+27x^2-8x+1\\ & 157y+3x^7+53x^6-320x^5+716x^4-1036x^3+1516x^2-1197x+64\\ 34 & x^8+x^7-x^5-x^4-x^3+x+1\\ & y+x^7-x^5-x^4-x^3+1\\ 34 & x^8-2x^6+8x^4-7x^2+1\\ & 3y+3x^7-5x^5+23x^3-16x\\ 35 & x^8-10x^7+36x^6-55x^5+19x^4+35x^3-25x^2-5x+5\\ & y+6x^7-65x^6+268x^5-531x^4+473x^3-49x^2-177x+74\\ 39 & x^8-4x^7+10x^6-25x^5+37x^4-25x^3+10x^2-4x+1\\ & 3y-2x^7+6x^6-13x^5+34x^4-33x^3-x^2-3x\\ 40 & 5x^8-15x^7+15x^6-10x^5+13x^4-8x^3-7x^2+10x-2\\ & 6653y-5455x^7+25010x^6-27370x^5+10105x^4-24648x^3\\ & \qquad{}+20135x^2+10773x-14311\\ 42 & x^8-4x^7+6x^5-3x^2-2x+1\\ & 7y+3x^7-11x^6-3x^5+16x^4-x^3+7x^2-12x-6 \end{array} \] Then $[K:\Q]=8$ and $E(K)_\tors\simeq\torz N$. The fields $K$ are listed, in the same order, in \Cref{table:sporadic}. \end{proposition} \begin{proof} For $N=33$ the element $x+y$ generates $K$, with minimal polynomial $t^8-15t^7+95t^6-331t^5+741t^4-1100t^3+1009t^2-688t+359$. We \gitlinkVIII{realizations/verify_octic_examples.m}{compute} that $(0,0)$ has order $N$ on $E$ and that $E$ has good reduction at the primes listed in \Cref{table:certificates}, with the numbers of points given there. \end{proof} Van Hoeij's list contains two more octic points on $X_1(42)$, for which the same holds. They are defined over the same field \lmfdbnf{8.2.7993287.1}, and the two elliptic curves are $2$-isogenous but not isomorphic. \begin{proposition}\label{d8:prop:5_10} Let $z=\zeta_{20}$, $i=z^5$ and $K=\Q(z)$. The elliptic curve $E:y^2+(1-i)xy-iy=x^3-ix^2$ over $K$ has $E(K)_\tors\simeq\tg{5}{10}$. \end{proposition} \begin{proof} The curve has $j$-invariant $1728$, and this example is contained in \cite[Section 4.8]{CCRS14}. The points $P=(-1+z^2+z^3-z^4,\,-1+z^4-z^7)$ and $Q=(1+z^5,\,-2)$ have orders $5$ and $10$, and the Weil pairing of $P$ and $2Q$ is the primitive fifth root of unity $-z^6$, as we \gitlinkVIII{realizations/verify_cm_5_10.m}{compute}, so they generate a subgroup isomorphic to $\tg{5}{10}$. The reductions of $E$ modulo the primes $(41,z-2)$ and $(61,z-8)$ of $K$ are good, with $50$ points each, so $\#E(K)_\tors$ divides $50$. \end{proof} \begin{proposition}\label{d9:prop:realizations} Let $N$, $f$ and $g$ be one of the following, let $x$ be a root of $f$, let $y$ be determined by $g(x,y)=0$, and put $K=\Q(x)$ and $E=E(b,c)$. \[ \renewcommand{\arraystretch}{1.15} \begin{array}{r@{\quad}l} N & f\text{ and }g\\\hline 29 & x^9-8x^8+23x^7-26x^6+2x^5+17x^4-11x^3+2x^2+1\\ & y-x^8+6x^7-11x^6+4x^5+6x^4-6x^3+3x^2\\ 31 & x^9-2x^8+x^6-x^5+14x^4-28x^3+19x^2-2x-1\\ & 119y-9x^8+28x^7-16x^6-46x^5+11x^4-125x^3+440x^2-386x-150\\ 32 & x^9-x^7-x^6+2x^5+2x^4-x^3-2x^2+1\\ & y-x^7-x^6+x^5+x^4-x^3-3x^2-x+1\\ 33 & x^9-5x^8-3x^7+25x^6+20x^5-19x^4-15x^3+4x^2+2x-1\\ & 543y-113x^8+638x^7+71x^6-3534x^5-544x^4+5545x^3+1342x^2-2083x-197\\ 34 & x^9-2x^8-x^7+3x^6+2x^5+x^3-2x-1\\ & y+40x^8-108x^7+36x^6+94x^5+14x^4-9x^3+47x^2-33x-57\\ 39 & x^9-5x^8+8x^7-10x^5+4x^4+4x^2-4x+1\\ & 3y+7x^8-33x^7+46x^6+15x^5-67x^4+7x^3+2x^2+29x-18\\ 40 & x^9-6x^8+14x^7-17x^6+12x^5-4x^4-x^3+3x^2-2x+1\\ & y+x^2-2x\\ 42 & x^9+7x^6-3x^4+4x^3-3x^2+3x-1\\ & 4y-x^8-x^7+x^6-10x^5-4x^4+5x^3-13x^2+2x-7 \end{array} \] Then $[K:\Q]=9$ and $E(K)_\tors\simeq\torz N$. The fields $K$ are listed, in the same order, in \Cref{table:sporadic}. We denote the points of $X_1(31)$, $X_1(33)$ and $X_1(39)$ given by the second, fourth and sixth entry by $x_{31}$, $x_{33}$ and $x_{39}$. Moreover, let $K_{2,26}=\Q(z)$, where $z^9+5z^8+15z^7+14z^6-14z^5-25z^4-7z^3+5z^2+4z+1=0$, which is the field \lmfdbnf{9.3.157505216.1}, and put \[ \begin{aligned} 907b&=62z^8+1386z^7+6639z^6+18950z^5+20738z^4-6786z^3\\ &\qquad{}-25017z^2-14822z-2411,\\ 907c&=-3864z^8-22538z^7-75946z^6-112918z^5-25630z^4+94470z^3\\ &\qquad{}+104296z^2+46530z+8797. \end{aligned} \] Then the elliptic curve $E_{2,26}=E'(b,c)$ over $K_{2,26}$, as in \eqref{d6:eq:tate2}, satisfies $E_{2,26}(K_{2,26})_\tors\simeq\tg{2}{26}$, generated by $(0,0)$ and $(b,b)$. We denote by $x_{2,26}$ the corresponding point of $X_1(2,26)$. \end{proposition} \begin{proof} The list of \cite{vanHoeij} contains $19$ points of degree $9$ on the curves $X_1(N)$ with $N\geq29$, all of which we \gitlinkIX{realizations/verify_nonic_examples.m}{check}. For the first eight entries above we compute that $f$ is irreducible, that $(0,0)$ has order $N$ on $E$, and that $E$ has good reduction at the primes listed in \Cref{table:certificates}, with the numbers of points given there. For $E_{2,26}$ we \gitlinkIX{C2x26/explicit/verify_native.m}{check} that $(0,0)$ and $(b,b)$ have orders $2$ and $26$ and that $(0,0)\neq13(b,b)$. The polynomial defining $K_{2,26}$ is irreducible modulo $3$, and $E_{2,26}$ has good reduction at the prime above $3$, with $19604$ points over $\F_{3^9}$. Modulo $7$ the polynomial factors into irreducible factors of degrees $2,2,2,3$, and $E_{2,26}$ has good reduction at the four primes above $7$, with $52$, $52$, $52$ and $364$ points. Hence $\#E_{2,26}(K_{2,26})_\tors$ divides $\gcd(19604,52,364)=52$. \end{proof} \begin{table}[htbp] \centering \small \renewcommand{\arraystretch}{1.1} \begin{tabular}{c|r|l} $d$ & $N$ & $(p,f_\wp,\#\widetilde E(\F_\wp))$\\\hline $7$ & $25$ & $(3,7,2175)$, $(5,2,25)$\\ & $33$ & $(5,3,132)$, $(13,7,62752635)$\\ & $36$ & $(29,1,36)$\\ & $36$ & $(5,3,108)$, $(19,2,360)$\\\hline $8$ & $33$ & $(7,8,5760579)$, $(13,8,815675619)$\\ & $34$ & $(5,2,34)$\\ & $34$ & $(7,4,2448)$, $(11,2,102)$, $(13,4,28900)$\\ & $34$ & $(7,4,2482)$, $(13,4,28424)$\\ & $35$ & $(7,8,5763520)$, $(11,2,105)$\\ & $39$ & $(7,2,39)$\\ & $40$ & $(3,4,80)$, $(31,3,29560)$\\ & $42$ & $(41,1,42)$\\\hline $9$ & $29$ & $(3,3,29)$\\ & $31$ & $(53,1,62)$, $(83,1,93)$\\ & $32$ & $(3,3,32)$\\ & $33$ & $(5,2,33)$\\ & $34$ & $(11,3,1292)$, $(13,4,28594)$\\ & $39$ & $(13,2,156)$, $(19,3,6981)$\\ & $40$ & $(3,4,80)$, $(5,3,120)$\\ & $42$ & $(37,1,42)$ \end{tabular} \caption{Primes $\wp$ of good reduction of the elliptic curves $E$ over the fields $K$ of degree $d$ of \Cref{d7:prop:realizations,d8:prop:realizations,d9:prop:realizations}, in the order of these propositions, above odd primes $p$ unramified in $K$, with the residue degree $f_\wp$ and $\#\widetilde E(\F_\wp)$.} \label{table:certificates} \end{table} We \gitlinkIX{independent/task2/check_entry.m}{also check} that the curves with $N=31$ and $N=39$, and $E_{2,26}$, are base changes of elliptic curves over the cubic subfields of their fields of definition of discriminants $-23$, $-31$ and $-44$, and that the curve with $N=42$ is the base change of \lmfdbec{162}{c}{3}, which has torsion $\torz{21}$ over the cubic field of discriminant $81$ \cite{najman}. A second elliptic curve with a point of order $42$ over the same field is the base change of a curve over its cubic subfield of discriminant $-648$. \subsection{The curve \texorpdfstring{$X_1(25)$}{X1(25)}}\label{subsec:sporadic_25} \begin{proposition}\label{d7:prop:25} Every closed point of degree $7$ on $X_1(25)$ is in the orbit under $\Delta(25)$ of the point $(x,y)$ of \Cref{d7:prop:realizations}. Hence, if $E$ is an elliptic curve over a septic field $K$ with a point of order $25$, then $(K,E)\simeq(K_{25},E_{25})$ and $E(K)_\tors\simeq\torz{25}$. In particular, $\torz{50}$, $\torz{75}$, $\torz{125}$ and $\tg{2}{50}$ are not subgroups of the torsion of an elliptic curve over a septic field. \end{proposition} \begin{proof} Let $X=X_1(25)$ and $J=J_1(25)$. As $\gon_\Q X=5<7$ \cite[Theorem 4 and Table 1]{DvH}, we use part (2) of \Cref{d7:lem:closed}. By \cite[Theorem 3]{DvH}, $X$ has only finitely many closed points of degree $7$, so by Hilbert's irreducibility theorem there is no rational function of degree $7$ on $X$. By \cite[Theorem 3.1(2), Corollary 4.14 and Table 2]{DEHMZ}, $J(\Q)$ is cyclic of order $227555$ and generated by rational cuspidal divisor classes. As this order is odd, reduction at $2$ is injective on $J(\Q)$, and we \gitlinkVII{certify_septimic_25.m}{compute} $J(\F_2)\simeq\torz{227555}$, so reduction is an isomorphism $J(\Q)\simeq J(\F_2)$. The cusps are defined over $\Q(\zeta_{25})$ and $2$ is a primitive root modulo $25$, so every closed cusp of $X_{\F_2}$ is the reduction of a Galois orbit of cusps of the same degree. The closed cusps of $X_{\F_2}$ have degrees $1^{10},4^2,10$. Let $D$ be a closed point of degree $7$ on $X$. If $\overline D$ is the reduction of an effective rational divisor $A$ of degree $7$, then $D\sim A$, so $D=A$ by \Cref{d7:lem:closed}. If $\overline D$ is cuspidal, it is the reduction of a rational cuspidal divisor $A$, which is impossible as $D$ is not cuspidal. The non-cuspidal closed points of $X_{\F_2}$ of degrees $1,\dots,7$ number $0,0,0,0,1,5,10$, so the non-cuspidal part of $\overline D$ is a single closed point of degree $5$, $6$ or $7$. If it has degree $5$, the cuspidal part has degree $2$, and we compute that for all $55$ effective cuspidal divisors $C$ of degree $2$ the class of $P_5+C-7\overline{c_0}$ is also the class of an effective cuspidal divisor of degree $7$, so this case is impossible as before. The sextic point of \Cref{d6:prop:examples}(1) reduces modulo $2$ to a closed point of degree $6$, and we compute that the reductions of its translates by $\Delta(25)$ are all the $5$ closed points of degree $6$ of $X_{\F_2}$. So if $\overline D$ is a closed point of degree $6$ plus a rational cusp, it is the reduction of a reducible effective rational divisor of degree $7$, which is impossible. Similarly, the reductions of the translates of the septic point of \Cref{d7:prop:realizations} are all the $10$ closed points of degree $7$ of $X_{\F_2}$, so if $\overline D$ is a closed point of degree $7$, then $D$ is one of these translates. The rest of the proof is as explained at the beginning of the section. \end{proof} \subsection{The curve \texorpdfstring{$X_1(33)$}{X1(33)}}\label{subsec:sporadic_33} \begin{proposition}\label{d7:prop:33}\label{d8:prop:33}\label{d9:prop:33}\label{d9:cor:66} Let $d\in\{7,8,9\}$. Every effective rational divisor of degree $d$ on $X_1(33)$ whose support contains no cusp is one of the $10$, $5$, respectively $10$ closed points of degree $d$ in the $\Delta(33)$-orbit of the point of level $33$ of \Cref{d7:prop:realizations}, \Cref{d8:prop:realizations}, respectively \Cref{d9:prop:realizations} (the point $x_{33}$). Hence, if $E$ is an elliptic curve over a number field $K$ of degree $d$ with a point of order $33$, then $(K,E)$ is isomorphic to the pair with $N=33$ of the same proposition, and $E(K)_\tors\simeq\torz{33}$. In particular, $\torz{66}$, $\torz{99}$ and $\tg{3}{33}$ are not subgroups of the torsion of an elliptic curve over a number field of degree $d$. \end{proposition} \begin{proof} Let $X=X_1(33)$. We have $\gon_\Q X=10$ \cite[Theorem 4 and Table 1]{DvH}, and $J_X(\Q)\simeq\torz5\times\torz{42373650}$ \cite[Theorem 3.1(2), Corollary 4.14 and Table 2]{DEHMZ}. We work at $p=2$. The Galois orbits of cusps over $\Q$ have degrees $1^{10},2^5,10^2$, and modulo $2$ one of the orbits of degree $10$ splits into two closed points of degree $5$, so the closed cusps of $X_{\F_2}$ have degrees $1^{10},2^5,5^2,10$, and a cuspidal divisor on $X_{\F_2}$ need not be the reduction of a rational one. We \gitlinkIX{C33/rebound/classify_degree9_33.m}{compute} that $J_X(\F_2)\simeq\torz{15}\times\torz{42373650}$ and that $H$ has invariant factors $[5,42373650]$, so reduction modulo $2$ is injective on $J_X(\Q)$, including its $2$-primary part. The non-cuspidal closed points of $X_{\F_2}$ of degrees $1,\dots,9$ number $0,0,0,0,2,20,30,25,50$, so the reduction of an effective divisor of degree $9$ contains at most one of them. We run through all the $155770$ effective divisors of degree $9$ on $X_{\F_2}$ with at most one non-cuspidal point, and compute which of them have class, normalized at a rational cusp, in $H$. These are the $144250$ reductions of effective rational cuspidal divisors, the $600$ divisors $\overline S+\overline{C'}$ with $S$ one of the ten closed points of degree $7$ of the statement and $C'$ an effective rational cuspidal divisor of degree $2$, the $50$ divisors $\overline O+\overline c$ with $O$ one of the five closed points of degree $8$ of the statement and $c$ a rational cusp, the reductions of the ten closed points of degree $9$ of the statement, and ten further closed points of degree $9$, whose classes are the reductions of $[C_i-9c_0]$ for explicit $\Z$-linear combinations $C_1,\dots,C_{10}$ of degree $9$ of rational cuspidal divisors. Let $D$ be as in the statement, let $c_\ast$ be a rational cusp, and put $D_9=D+(9-d)c_\ast$. The reduction of $D_9$ is one of these divisors, so by the injectivity $D_9$ is linearly equivalent to an effective rational cuspidal divisor, to a divisor $S+C'$ or $O+c$ as above, to one of the ten closed points of degree $9$ of the statement, or to one of the $C_i$. In the first four cases \Cref{d9:lem:closed} shows that $D_9$ is equal to this divisor. As the support of $D$ contains no cusp, the first case is impossible, and comparing the parts of the supports outside the cusps in the next three cases gives $D=S$ (so $d=7$), $D=O$ (so $d=8$), respectively that $D$ is one of the ten closed points of degree $9$ (so $d=9$). It remains to show that no effective rational divisor is linearly equivalent to one of the $C_i$, and for this it suffices that $\ell(\overline{C_i})=0$ on $X_{\F_5}$, by semicontinuity. Let $A=\sum_c\max_i\{0,v_c(C_i)\}c$, of degree $48$, so that $B_i=A-C_i$ is effective of degree $39$. By Riemann-Roch $\ell(A)=48+1-21=28$, and we \gitlinkIX{C33/F5_rr/verify33.m}{compute} a basis of $L(\overline A)$ over $\F_5$ and check that the $28\times39$ matrix of its restrictions to $\overline{B_i}$ has rank $28$ for every $i$, so $L(\overline{C_i})=L(\overline A-\overline{B_i})=0$. The rest of the proof is as explained at the beginning of the section. \end{proof} \subsection{The curve \texorpdfstring{$X_1(34)$}{X1(34)}}\label{subsec:sporadic_34} \begin{proposition}\label{d8:prop:34} Every effective rational divisor of degree $8$ on $X_1(34)$ with non-cuspidal support is one of the $7$ closed points in the $\Delta(34)$-orbits of the three octic points of level $34$ of \Cref{d8:prop:realizations}, which have lengths $1$, $2$ and $4$. Hence, if $E$ is an elliptic curve over an octic field $K$ with a point of order $34$, then $(K,E)$ is isomorphic to one of the three pairs of \Cref{d8:prop:realizations} with $N=34$, and $E(K)_\tors\simeq\torz{34}$. In particular, $\torz{68}$ and $\tg{2}{34}$ are not subgroups of the torsion of an elliptic curve over an octic field. \end{proposition} \begin{proof} We work at $p=3$. By \cite[Table 2]{DEHMZ}, $J_1(34)(\Q)\simeq\torz{8760}\times\torz{595680}$, and we \gitlinkVIII{C34/root_reproduce_octic_34.m}{compute} $J_1(34)(\F_3)\simeq(\torz4)^3\times\torz{8760}\times\torz{1191360}$, in which $H$ has invariants $[8760,595680]$, so reduction is injective on $J_1(34)(\Q)$. We have $\gon_\Q X_1(34)=10$ \cite[Theorem 4 and Table 1]{DvH}. The Galois orbits of cusps over $\Q$ and the closed cusps over $\F_3$ both have degrees $1^{16},8^2$, so the entirely cuspidal reduction is excluded. The non-cuspidal closed points of degrees $1,\dots,8$ number $0,0,0,26,32,80,256,734$, so the non-cuspidal part of $\overline D$ is a single closed point of degree $4,\dots,8$, or a sum of two quartic points. All the cuspidal parts have class in $H$. Of these single points and the $351$ sums of two quartic points, exactly one octic point and $6$ sums of two quartic points have class in $H$. The three octic points of level $34$ of \Cref{d8:prop:realizations} reduce to an octic point, to twice a quartic point and to a sum of two distinct quartic points, respectively, and their diamond translates give exactly these $7$ classes. The rest of the proof is as explained at the beginning of the section, and the three pairs are not isomorphic, as their $j$-invariants have degrees $1$, $2$ and $4$. \end{proof} \subsection{The curve \texorpdfstring{$X_1(35)$}{X1(35)}}\label{subsec:sporadic_35} \begin{proposition}\label{d8:prop:35} Every effective rational divisor of degree $8$ on $X_1(35)$ with non-cuspidal support is one of the $6$ translates under $\Delta(35)$ of the octic point of \Cref{d8:prop:realizations}. Hence, if $E$ is an elliptic curve over an octic field $K$ with a point of order $35$, then $(K,E)$ is isomorphic to the pair of \Cref{d8:prop:realizations}, and $E(K)_\tors\simeq\torz{35}$. In particular, $\torz{70}$, $\torz{105}$, $\torz{175}$ and $\tg{5}{35}$ are not subgroups of the torsion of an elliptic curve over an octic field. \end{proposition} \begin{proof} The argument in the proof of \Cref{prop:35} applies to $d=8$, as $\gon_\Q X_1(35)=12>8$. Let $D$ be an effective rational divisor of degree $8$ on $X_1(35)$ whose support contains no cusp. Its reduction modulo $3$ is not cuspidal, its non-cuspidal part is one of the $5961$ divisors of degree at most $9$ tested there, and the only ones of these with class in the reduction of $J_1(35)(\Q)$ are the reductions of the $6$ translates under $\Delta(35)$ of the octic point $B$ of \Cref{d8:prop:realizations}. So $D$ and a translate $\delta B$ have the same reduction, hence $D\sim\delta B$, as reduction modulo $3$ is injective on $J_1(35)(\Q)$, and $D=\delta B$ by \Cref{d8:lem:closed}. The rest of the proof is as explained at the beginning of the section. \end{proof} \subsection{The curve \texorpdfstring{$X_1(36)$}{X1(36)}}\label{subsec:sporadic_36} \begin{proposition}\label{d7:prop:36}\label{d6:prop:36} The group $\torz{36}$ is not a subgroup of the torsion of an elliptic curve over a sextic field. The non-cuspidal closed points of degree $7$ on $X_1(36)$ are the points in the two $\Delta(36)$-orbits of the septic points of \Cref{d7:prop:realizations}. Hence, if $E$ is an elliptic curve over a septic field $K$ with a point of order $36$, then $(K,E)$ is isomorphic to $(K_{36},E_{36})$ or $(K_{36}',E_{36}')$, and $E(K)_\tors\simeq\torz{36}$. In particular, $\torz{72}$ and $\tg{2}{36}$ are not subgroups of the torsion of an elliptic curve over a septic field. \end{proposition} \begin{proof} Let $X=X_1(36)$ and $J=J_1(36)$, and work at $p=5$. We have $\gon_\Q X=8$ \cite[Theorem 4 and Table 1]{DvH}, and $J(\Q)$ has invariant factors $[12,252,7812]$ and is generated by rational cuspidal divisor classes \cite[Theorem 3.1(2), Corollary 4.14 and Table 2]{DEHMZ}. The Galois orbits of cusps over $\Q$ have degrees $1^9,2^6,3,4,6^2$, and the closed cusps of $X_{\F_5}$ have degrees $1^{15},2^5,3,6^2$. The computations in this proof were done in one interactive Magma session, whose \gitlinkVII{logs/x1_36/quotient_filter.log}{log} of the class computations and exported \gitlinkVII{data/certificate_36_quotient_data.py}{certificate} are kept in the repository. We compute \[ J(\F_5)\simeq(\torz4)^2\times(\torz{12})^2\times\torz{36}\times(\torz{7812})^2, \] and that the reductions of the rational cuspidal divisors of degree $0$ generate a subgroup $H$ with invariant factors $[12,252,7812]$, so $H$ is the injective image of $J(\Q)$. Let $D$ be a non-cuspidal closed point of degree $7$ on $X$. As explained at the beginning of the section, it suffices to show that the class of $\overline D-7\overline{c_0}$, which lies in $H$, is the class of the reduction of an effective rational cuspidal divisor or of a translate of one of the two septic points. There are $19159$ effective rational cuspidal divisors of degree $7$, with distinct classes, and among the $188865$ effective cuspidal divisors of degree $7$ on $X_{\F_5}$, those with class in $H$ are exactly these. So $\overline D$ is not cuspidal. The non-cuspidal closed points of $X_{\F_5}$ of degrees $1,\dots,7$ number $0,0,48,144,660,2856,10728$, so the non-cuspidal part of $\overline D$ is a single closed point of degree $3,\dots,7$, a sum of two cubic points (possibly equal) or a sum of a cubic and a quartic point. We take the first point up to $\Delta(36)$ and allow all the cuspidal parts of the complementary degree. To speed up the cases of degree $6$ and $7$ we first use the map $\pi:X_1(36)\to X_1(2,18)$ given by the $2$-isogeny $E\to E/\dm{18P}$, as in the proof of \Cref{d7:lem:4n}. By \cite[Proposition 4.16 and Table 3]{DEHMZ}, $J_1(2,18)(\Q)$ has invariant factors $[2,42,126]$ and is generated by rational cuspidal divisor classes. We compute $J_1(2,18)(\F_5)\simeq\torz6\times\torz{84}\times\torz{252}$. As $5$ generates $(\torz{18})^\times$, the closed cusps of $X_1(2,18)_{\F_5}$ are reductions of Galois orbits of the same degree, and their classes generate the reduction $H'$ of $J_1(2,18)(\Q)$. Every cusp class lies in $H'$, so the class of the push-forward of the non-cuspidal part of $\overline D$ (minus the appropriate multiple of a rational cusp) must lie in $H'$. The cases that pass this test are then tested in $J(\F_5)$. The results are the following, and we \gitlinkVII{replay_36_quotient_certificate.py}{replay} the class computations of the certificate independently with integer arithmetic (see the \gitlinkVII{logs/x1_36/quotient_replay.log}{log}): \[ \begin{array}{c|r|r|r} \text{type of the non-cuspidal part} & \text{cases} & \text{passing the test on } X_1(2,18) & \text{remaining}\\\hline {[3]} & 8 & \text{--} & 0\\ {[4]} & 24 & \text{--} & 0\\ {[5]} & 110 & \text{--} & 0\\ {[3,3]} & 384 & \text{--} & 0\\ {[3,4]} & 1152 & \text{--} & 1\\ {[6]} & 480 & 54 & 0\\ {[7]} & 1788 & 150 & 1 \end{array} \] Here ``remaining'' counts the cases with a cuspidal part such that the class of $\overline D-7\overline{c_0}$ lies in $H$ and is not the class of the reduction of an effective rational cuspidal divisor. The two septic points of \Cref{d7:prop:realizations} reduce to a closed point of degree $7$ and to the sum of a cubic and a quartic point, respectively, and each of their $\Delta(36)$-orbits gives $6$ distinct classes. The two remaining cases are among these $12$ classes. So $D$ is a translate of one of the two septic points, and the rest of the proof for septic fields is as explained at the beginning of the section. Finally, let $D$ be the trace divisor of a point of order $36$ over a sextic field, and let $c_\ast$ be a rational cusp. The computations above apply to the effective rational divisor $D+c_\ast$ of degree $7$, as they allow all the cuspidal parts, and by the point counts the reduction of $D+c_\ast$ is cuspidal or its non-cuspidal part has one of the types $[3]$, $[4]$, $[5]$, $[3,3]$ and $[6]$. So by the computations above $[\overline D+\overline{c_\ast}-7\overline{c_0}]$ is the class of the reduction of an effective rational cuspidal divisor $A$ of degree $7$. Then $D+c_\ast\sim A$, and $D+c_\ast=A$ by \Cref{d7:lem:closed}, as $\gon_\Q X=8>7$, which is impossible, as the support of $D$ contains no cusp. \end{proof} \subsection{The curve \texorpdfstring{$X_1(37)$}{X1(37)}}\label{subsec:sporadic_37} \begin{proposition}\label{d6:prop:37} Let $E_{1225}:y^2=g(x)=x^3-10395x+444150$, which is a model of \lmfdbec{1225}{b}{2}, let $w$ be a root of \[ f_{37}=x^6-3150x^5+796635x^4-75770100x^3+3111596775x^2-44606598750x-85333003875, \] put $K_{37}:=\Q(w)$ and $\delta:=g(w)$, and let \[ E_{37}:\quad y^2=x^3-10395\delta^2x+444150\delta^3 . \] Then $K_{37}$ is the cyclic totally real sextic field \lmfdbnf{6.6.300125.1}, the point $(\delta w,\delta^2)$ of $E_{37}$ has order $37$ and $E_{37}(K_{37})_\tors\simeq\torz{37}$. If $K$ is a sextic field and $E/K$ is an elliptic curve with a point of order $37$, then the pair $(K,E)$ is isomorphic to $(K_{37},E_{37})$. In particular, no group properly containing $\torz{37}$ is a sextic torsion group. \end{proposition} \begin{proof} The map $(x,y)\mapsto(\delta x,\delta^2y)$ is an isomorphism from $E^*:\delta y^2=g(x)$ to $E_{37}$, which sends $P=(w,1)$ to $(\delta w,\delta^2)$. We \gitlinkVI{verify_sporadic_curves.m}{compute} that $K_{37}$ is the cyclic totally real field \lmfdbnf{6.6.300125.1}, and we \gitlinkVI{verify_37.m}{compute} that $P$ has order $37$ and that the irreducible factors over $\Q$ of the $37$-division polynomial of $E_{1225}$ have degrees $6,6,6,666$. As $E^*$ is a quadratic twist of $E_{1225}$, the $x$-coordinates $x_r$ of the points $rP$, for $1\leq r\leq18$, are $18$ distinct roots of this division polynomial. They lie in $K_{37}$, so they have degree at most $6$, and hence they are exactly the $18$ roots of the three sextic factors. In particular $\Q(x_r)=K_{37}$ for every $r$. Writing $rP=(x_r,y_r)$, we have $\delta y_r^2=g(x_r)$ with $y_r\neq0$, as $rP$ does not have order $2$. Now let $(K,E)$ be as in the statement. By \cite[Proposition 1.4]{DKSS} we have $j(E)=-9317=j(E_{1225})$, so $E$ is isomorphic to $E_d:dy^2=g(x)$ for some $d\in K^\times$. A point of order $37$ on $E_d$ has $x$-coordinate $x'\in K$, which is a root of the $37$-division polynomial of $E_{1225}$ of degree at most $6$, hence a root of one of the sextic factors. So $K=\Q(x')$, and there is an isomorphism $\tau:K\to K_{37}$ with $\tau(x')=x_r$ for some $r$. As $d$ and $g(x')$ differ by a non-zero square in $K$, the elements $\tau(d)$ and $g(x_r)=\delta y_r^2$ differ by a square in $K_{37}$, so $E_d^\tau\simeq E^*\simeq E_{37}$ over $K_{37}$. Thus $(K,E)\simeq(K_{37},E_{37})$. Finally, we \gitlinkVI{verify_37.m}{compute} that $f_{37}$ has roots modulo $29$ and modulo $41$, and that at the corresponding primes of degree $1$ of $K_{37}$, which are of good reduction, the reduction of $E_{37}$ has exactly $37$ points. As reduction is injective on the torsion of order prime to the residue characteristic, $\#E_{37}(K_{37})_\tors$ divides $37$. \end{proof} \subsection{The curve \texorpdfstring{$X_1(39)$}{X1(39)}}\label{subsec:sporadic_39} Let $X=X_1(39)$. By \cite[Theorem 3.1(2), Corollary 4.14 and Table 2]{DEHMZ}, the group $J_X(\Q)\simeq\torz7\times\torz{31122}\times\torz{3236688}$ is generated by rational cuspidal divisor classes, and $\gon_\Q X=14$ \cite[Theorem 4 and Table 1]{DvH}. The Galois orbits of cusps over $\Q$ and the closed cusps of $X_{\F_2}$ both have degrees $1^{12},2^6,12^2$, so every effective cuspidal divisor on $X_{\F_2}$ is the reduction of an effective rational cuspidal divisor of the same degree. We \gitlinkVIII{C39/hecke_good_39.m}{compute} that the non-cuspidal closed points of $X_{\F_2}$ of degrees $1,\dots,8$ number $0,0,0,0,0,12,0,49$, that the $12$ points of degree $6$ form one diamond orbit, and that the $49$ points of degree $8$ form $5$ diamond orbits. \begin{lemma}\label{lem:39inj}\label{d9:lem:39inj} Reduction modulo $2$ is injective on $J_1(39)(\Q)$. \end{lemma} \begin{proof} As $J_1(39)(\Q)$ is generated by rational cuspidal divisor classes, it suffices that the reductions of these classes generate a subgroup of $J_1(39)(\F_2)$ of order $\#J_1(39)(\Q)$, and we \gitlinkVIII{independent/classification/classify8.m}{compute} that this is the case. \end{proof} Hence no effective rational divisor $D$ of degree $d<14$ on $X$ whose support contains no cusp reduces to a cuspidal divisor modulo $2$: otherwise $D$ and an effective rational cuspidal divisor $C$ have the same reduction, so $D\sim C$ by \Cref{d9:lem:39inj}, and $D=C$ by \Cref{d8:lem:closed}, as $\gon_\Q X=14>d$. For $d\geq7$ we cannot use \Cref{d6:prop:main}(b) instead, as it would require $\gon_\Q X>2d$. \begin{proposition}\label{d7:prop:39} The group $\torz{39}$ is not a subgroup of the torsion of an elliptic curve over a number field of degree $6$ or $7$. \end{proposition} \begin{proof} Let $D$ be the trace divisor on $X$ of a point of order $39$ over a number field of degree $d\in\{6,7\}$. By the above and the point counts, the reduction of $D$ modulo $2$ is $D'+C'$, where $D'$ is one of the $12$ points of degree $6$, and $C'=0$ if $d=6$ and $C'$ is the reduction of a rational cusp if $d=7$. So every prime $q\nmid78$ is admissible for $C'$, and as the $12$ points form one diamond orbit, it suffices to take one of them as $D'$. We \gitlinkVII{hecke_sieve_commonzero.m}{compute} that $[A_q(D')]\neq0$ for $q=5$ or $q=7$, so the Hecke sieve at $2$ excludes this reduction. \end{proof} \begin{proposition}\label{d8:prop:39} The octic point of level $39$ of \Cref{d8:prop:realizations} is the only effective rational divisor of degree $8$ on $X_1(39)$ with non-cuspidal support. Hence, if $E$ is an elliptic curve over an octic field $K$ with a point of order $39$, then $(K,E)$ is isomorphic to the pair of \Cref{d8:prop:realizations}, and $E(K)_\tors\simeq\torz{39}$. In particular, $\torz{78}$, $\torz{117}$ and $\tg{3}{39}$ are not subgroups of the torsion of an elliptic curve over an octic field. \end{proposition} \begin{proof} We work at $p=2$, where reduction is injective on $J_X(\Q)$ by \Cref{d9:lem:39inj}. By the above, $\overline D$ is not cuspidal, so by the point counts it is a sextic point plus the reduction of a rational cuspidal divisor of degree $2$, or an octic point. As $A_5$ annihilates $J_X(\Q)$ and the rational cuspidal divisors, $[A_5P]=0$ for the non-cuspidal part $P$. We \gitlinkVIII{C39/hecke_good_39.m}{compute} that $A_5P$ is not principal for the representatives of the sextic orbit and of $4$ of the octic orbits. The fifth octic orbit consists of a single point, which is the reduction of the octic point of \Cref{d8:prop:realizations}. The rest of the proof is as explained at the beginning of the section. \end{proof} \begin{lemma}\label{d9:lem:39red} Let $E$ be an elliptic curve over a nonic field $K$ with a point $P$ of order $39$, and let $D$ be the trace divisor of $(E,P)$. Then $K$ has a unique prime above $2$, it is unramified of residue degree $9$, $E$ has good ordinary reduction there, and $D$ reduces to a closed point of degree $9$ of $X_1(39)_{\F_2}$. \end{lemma} \begin{proof} By the above, $D$ does not reduce to a cuspidal divisor, so by the point counts the non-cuspidal part of its reduction is a point of degree $6$, $8$ or $9$, and its cuspidal part is the reduction of a rational cuspidal divisor. The computations of \Cref{d8:prop:39}, with the operator $A_5$, which kills the rational cuspidal complements, exclude the orbit of degree $6$ and four of the five orbits of degree $8$. The fifth consists of the reduction of the octic point $O$ of \Cref{d8:prop:39}, and if $D$ reduced to $\overline O+\overline c$ with $c$ a rational cusp, then $D\sim O+c$ by \Cref{d9:lem:39inj}, so $D=O+c$ by \Cref{d9:lem:closed}, which is impossible. So $D$ reduces to a single closed point of degree $9$, which implies the other assertions, as the prime-to-$2$ part of the torsion reduces injectively. The supersingular curves over $\F_{2^9}$ have $481$, $513$ or $545$ points, none divisible by $39$, so the reduction is ordinary. \end{proof} \begin{proposition}\label{d9:prop:117} The group $\torz{117}$ is not a subgroup of the torsion of an elliptic curve over a nonic field. \end{proposition} \begin{proof} Let $E$ have a point of order $117$ over $K$, and apply \Cref{d9:lem:39red} to $P$ three times this point. Then $117$ divides $\#E(\F_{2^9})$, which lies in the Hasse interval $[468,558]$, so $\#E(\F_{2^9})=468$. We \gitlinkIX{C78_C117/C78_X139/census39_F512.py}{compute} that the elliptic curves over $\F_{2^9}$ with $468$ points are three curves with $j$-invariants in $\F_8$, whose points of order $39$ give $36$ points of $X_1(39)$ over $\F_{2^9}$, in four Frobenius orbits. These are exactly the reductions of the four closed points of degree $9$ in the diamond orbit of $x_{39}$. So the trace divisor $D$ of the point of order $39$ of \Cref{d9:lem:39red} and a diamond translate of $x_{39}$ have the same reduction, hence are linearly equivalent by \Cref{d9:lem:39inj}, and equal by \Cref{d9:lem:closed}. Thus $E$ is isomorphic over $K$ to the curve of \Cref{d9:prop:realizations}, whose torsion is $\torz{39}$, which is a contradiction. \end{proof} \begin{proposition}\label{d9:prop:78} The group $\torz{78}$ is not a subgroup of the torsion of an elliptic curve over a nonic field. \end{proposition} \begin{proof} Let $E/K$ have a point $Q$ of order $78$, put $P=2Q$ and $R=39Q$, and let $\phi:E\to E'=E/\dm R$, which is defined over $K$, and $P'=\phi(P)$, of order $39$. Let $D$ and $D'$ be the trace divisors of $(E,P)$ and $(E',P')$. As the curve of \Cref{d9:prop:realizations} with a point of order $39$ has torsion $\torz{39}$, neither $D$ nor $D'$ is a diamond translate of $x_{39}$, so by \Cref{d9:lem:39red} the prime $\mathfrak p$ of $K$ above $2$ is unique and unramified with residue field $\F_{2^9}$, $E$ and $E'$ have good ordinary reduction at $\mathfrak p$, and $D$ and $D'$ reduce to closed points of degree $9$. The reduction of $\phi$ is an isogeny of degree $2$ of an ordinary curve, so its kernel is either the connected subgroup of order $2$, in which case it is the Frobenius isogeny, or the unique \'etale subgroup of order $2$, in which case it is the Verschiebung up to isomorphism. Hence the reduction of $(E',P')$ is the image of the reduction of $(E,P)$ under the Frobenius, or under $\dm2$ composed with the inverse of the Frobenius. As closed points are stable under the Frobenius, the reduction of $D'$ is that of $D$ or of $\dm2D$. By \Cref{d9:lem:39inj} and \Cref{d9:lem:closed} we get $D'=D$ or $D'=\dm2D$. Hence the $K$-points $(E',P')$ and $(E,P)$, respectively $(E,2P)$, of $X_1(39)$ lie above the same closed point with residue field $K$, so they differ by an automorphism $\sigma$ of $K$. As $X_1(39)$ is a fine moduli space, $E'\simeq E^\sigma$ over $K$, and $\phi$ gives a $K$-rational isogeny $\psi:E\to E^\sigma$ of degree $2$. The order $r$ of $\sigma$ divides $\#\Aut(K)$, which divides $[K:\Q]=9$, so $r$ is odd. The composition $\psi^{\sigma^{r-1}}\circ\dots\circ\psi^\sigma\circ\psi$ is an endomorphism of $E$ defined over $K$ of degree $2^r$, so by \Cref{d9:lem:odd} it is multiplication by an integer, and $2^r$ is a square, which is impossible as $r$ is odd. \end{proof} \subsection{The curve \texorpdfstring{$X_1(40)$}{X1(40)}}\label{subsec:sporadic_40} \begin{proposition}\label{d8:prop:40} Every effective rational divisor of degree $8$ on $X_1(40)$ with non-cuspidal support is a translate under $\Delta(40)$ of the octic point of \Cref{d8:prop:realizations}. Hence, if $E$ is an elliptic curve over an octic field $K$ with a point of order $40$, then $(K,E)$ is isomorphic to the pair of \Cref{d8:prop:realizations}, and $E(K)_\tors\simeq\torz{40}$. In particular, $\torz{80}$ and $\tg{2}{40}$ are not subgroups of the torsion of an elliptic curve over an octic field. \end{proposition} \begin{proof} We work at $p=3$. By \cite[Table 2]{DEHMZ}, $J_1(40)(\Q)\simeq(\torz2)^3\times\torz8\times\torz{120}\times\torz{895440}$, and $\gon_\Q X_1(40)=12$ \cite[Theorem 4 and Table 1]{DvH}. We \gitlinkVIII{C40/fresh_sequential_run.m}{compute} $\#J_1(40)(\F_3)=31579088486400$, and that $H$ has order $6876979200=\#J_1(40)(\Q)$, so reduction is injective on $J_1(40)(\Q)$. The $21$ Galois orbits of cusps over $\Q$ have degrees $1^{12},2^2,4^6,8$, and the closed cusps of $X_1(40)_{\F_3}$ have degrees $1^{12},2^6,4^6$. The effective cuspidal divisors of degree $8$ over $\F_3$ give $194142$ classes, of which $113912$ are reductions of effective rational cuspidal divisors, and exactly these lie in $H$. So $\overline D$ is not cuspidal. The non-cuspidal closed points of degrees $1,\dots,8$ number $0,0,0,32,72,80,408,708$, with $4,9,10,51,96$ diamond orbits in degrees $4,\dots,8$, so the non-cuspidal part of $\overline D$ is a single closed point of degree $4,\dots,8$ or a sum of two quartic points. Allowing all the cuspidal parts, none of the single points gives a class in $H$, and of the $128$ sums of a representative quartic point and any quartic point, exactly one does. The octic point of \Cref{d8:prop:realizations} reduces to a sum of two distinct quartic points, and its translates give $4$ distinct classes, which contain this one. The rest of the proof is as explained at the beginning of the section. \end{proof} \subsection{The curve \texorpdfstring{$X_1(2,24)$}{X1(2,24)}}\label{subsec:sporadic_2_24} \Cref{d6:prop:2_24} shows that the non-cuspidal sextic points on $X_1(2,24)$ form at most two diamond orbits, and \Cref{d6:prop:2_24_points} shows that both occur. \begin{proposition}\label{d6:prop:2_24} Let $X=X_1(2,24)$. There are $\Q$-rational cuspidal divisors $L_1,L_2$ of degree $6$ on $X$ with the following property. If $D$ is an effective $\Q$-rational divisor of degree $6$ on $X$ with non-cuspidal support, then $\delta D\sim L_i$ for some $\delta\in\Delta(24)$ and some $i\in\{1,2\}$. Each of the linear systems $|L_1|$ and $|L_2|$ contains at most one effective divisor. \end{proposition} \begin{proof} The curve $X$ has genus $13$, its Jacobian has rank $0$ and $J(\Q)=\ClCusp_\Q X\simeq\torz2\times\torz4\times\torz4\times\torz{120}\times\torz{240}$ \cite[Theorem 3.1(3), Proposition 4.16 and Table 3]{DEHMZ}, and $\gon_{\F_5}X>6$ \cite[proof of Proposition 5.5]{DS}, so $\gon_\Q X>6$. We work at $p=5$. Let $D$ be an effective $\Q$-rational divisor of degree $6$ on $X$ with non-cuspidal support. If $D$ were not a single closed point of degree $6$, its support would contain a closed point of degree at most $3$, which would give an elliptic curve over a number field of degree at most $3$ with a subgroup isomorphic to $\tg{2}{24}$, which is impossible \cite{mazur,kamienny,DEHMZ}. So $D$ is a closed point of degree $6$, and hence the trace divisor of a point of $Y_1(2,24)(K)$, where $K$ is its residue field. In particular the totally cuspidal reduction of $D$ is excluded by \Cref{d6:prop:main}. As in \Cref{sec:noncyclic}, the orders of vanishing at the cusps of the units $b$, $b(1-b-c)$, $c^2-4b(1-b-c)$ and $x(jQ)$, $1\leq j\leq12$, do not change under reduction modulo $5$, and grouping the cusps by them gives $19$ blocks $B_1,\dots,B_{19}$, each of which is a sum of Galois orbits of cusps over $\Q$, of degrees $1^8,2^6,4^5$. Let $c_\ast$ be a rational cusp that forms a block by itself, and let $G\subseteq J(\F_5)$ be the subgroup generated by the classes $[\overline{B_j}-\deg(B_j)\overline{c_\ast}]$. We \gitlinkVI{fine_2_24_cusp_reduction.m}{compute} that $G$ has invariant factors $[2,4,4,120,240]$, so $G$ is the reduction of $J(\Q)$, and that $J(\F_5)/G\simeq(\torz2)^2\times\torz{24}\times\torz{48}$. The sets $\mathcal S_j$ defined after \Cref{d6:lem:classtest} have $1,5,23,75$ elements for $j=0,1,2,3$. Over $\F_5$ the numbers of non-cuspidal closed points of degrees $1,\dots,6$ are $0$, $0$, $16$, $96$, $672$, $2712$, so the non-cuspidal part $Z$ of the reduction of $D$ has type $[3]$, $[4]$, $[5]$, $[6]$ or $[3,3]$. Up to the diamond operators there are $4$, $32$, $168$, $684$ and $64$ such $Z$, respectively, and we compute that the condition $[Z-(\deg Z)\overline{c_\ast}]\bmod G\in\mathcal S_{6-\deg Z}$ holds for exactly two of them, both of type $[6]$. As $G$ is the reduction of $J(\Q)=\ClCusp_\Q X$, their classes $[Z-6\overline{c_\ast}]$ are the reductions of the classes $[L_1-6c_\ast]$ and $[L_2-6c_\ast]$ of rational cuspidal divisors $L_1$ and $L_2$ of degree $6$. So, after applying a diamond operator, $D$ reduces to one of these two points, and then $[D-L_i]$ is a rational torsion class with trivial reduction, so $D\sim L_i$. As $\gon_\Q X>6$, each of the linear systems $|L_1|$ and $|L_2|$ contains at most one effective divisor. \end{proof} \begin{proposition}\label{d6:prop:2_24_points} Let $K_{2,24}=\Q(w)$ with $w^6-2w^4+2w^2-3=0$, which is the field \lmfdbnf{6.2.231852.1}, and let $E_{2,24}=E'(b,c)$ as in \eqref{d6:eq:tate2} with \[ b=\tfrac{1}{32}\bigl(3w^5+7w^4-24w^3-15w^2+21w+18\bigr),\qquad c=\tfrac{1}{16}\bigl(-7w^5+5w^4+24w^3-5w^2-17w-2\bigr). \] \begin{enumerate} \item The point $Q=(b,b)$ has order $24$, and $E_{2,24}(K_{2,24})_\tors\simeq\tg{2}{24}$. \item The curve $X_1(2,24)$ has exactly $8$ non-cuspidal closed points of degree $6$. They form two orbits of length $4$ under $\Delta(24)$, and they all come from $E_{2,24}$. \item If $K$ is a sextic field and $E/K$ is an elliptic curve with $\tg{2}{24}\subseteq E(K)$, then $(K,E)\simeq(K_{2,24},E_{2,24})$. \end{enumerate} The curve $E_{2,24}$ has no complex multiplication, its $j$-invariant generates the cubic subfield of $K_{2,24}$, which has discriminant $-139$, and its conductor has norm $6$. \end{proposition} \begin{proof} Write $K=K_{2,24}$ and $E=E_{2,24}$. We \gitlinkVI{verify_sporadic_curves.m}{compute} (1) and the last sentence. Moreover $\#\Aut(K)=2$: as $h(t)=t^3-2t^2+2t-3$ is strictly increasing (its derivative $3t^2-4t+2$ has no real roots) and $h(0)<0$, it has exactly one real root, which is positive, so the defining polynomial $h(x^2)$ of $w$ has exactly two real roots $\pm w_0$. Hence, in a real embedding of $K$, every automorphism of $K$ sends $w$ to $\pm w$, and $w\mapsto-w$ is one. We prove (2). There are $32$ pairs $(P',Q')$ generating $E(K)_\tors\simeq\tg{2}{24}$ ($16$ choices of $Q'$ and $2$ of $P'\notin\dm{Q'}$), and two of them give the same point of $Y_1(2,24)$ if and only if they differ by an automorphism of $E$. As $\Aut(E)=\{\pm1\}$ and $(-P',-Q')=(P',-Q')$, they give $16$ distinct $K$-points of $Y_1(2,24)$. Each of them lies on a closed point of degree $6$, as $\tg{2}{24}\notin\Phi(d)$ for $d\in\{1,2,3\}$, whose residue field is isomorphic to $K$, and such a closed point has exactly $\#\Aut(K)=2$ points with values in $K$. Hence $E$ gives at least $8$ closed points of degree $6$. On the other hand, by \Cref{d6:prop:2_24}, for every non-cuspidal closed point $D$ of degree $6$ there are $\delta\in\Delta(24)$ and $i$ with $\delta D\sim L_i$, and $|L_i|$ contains at most one effective divisor, so there are at most $2\cdot\#\Delta(24)=8$ such points. Hence there are exactly $8$, they form two orbits of length $4$, and they all come from $E$, and the $16$ points above are all the $K$-points on these $8$ closed points. For (3), let $(K',E')$ be as in (3). It gives a non-cuspidal closed point of degree $6$ on $X_1(2,24)$, which is one of the $8$ points of (2), so its residue field is isomorphic to $K$. Choosing such an isomorphism, we obtain a $K$-point on one of these closed points, which is one of the $16$ points above, all of which come from $E$. As $X_1(2,24)$ is a fine moduli space, $(K',E')\simeq(K,E)$. \end{proof} \begin{remark} The sextic points on $X_1(37)$ and $X_1(2,24)$ are sporadic, since both curves have only finitely many points of degree at most $6$: for $X_1(37)$ this follows from $\gon_\Q X_1(37)=18$ \cite[Theorem 1]{DvH} and Frey's theorem \cite{frey}, and for $X_1(2,24)$ it follows from $\rk J_1(2,24)(\Q)=0$ and $\gon_\Q X_1(2,24)>6$ (see the proof of \Cref{d6:prop:2_24}) by \cite[Proposition 2.3]{DS}. The conductor of $E_{2,24}$ is very small, as for the curve with torsion $\tg{2}{18}$ over a quintic field in \cite{N5}. The points of degree $6$ that give the four sporadic groups in degree $6$ are listed in \cite{x1mn}, as the points \xpoint{1.25.6.a}, \xpoint{1.37.6.a}, \xpoint{4.12.6.a} and \xpoint{2.24.6.a}. \end{remark} \subsection{The curve \texorpdfstring{$X_1(2,26)$}{X1(2,26)}}\label{subsec:sporadic_2_26} \begin{proposition}\label{d9:prop:2_26}\label{d6:prop:2_26}\label{d7:prop:2_26}\label{d8:prop:2_26} The group $\tg{2}{26}$ is not a subgroup of the torsion of an elliptic curve over a number field of degree $d\in\{6,7,8\}$, and hence, by \Cref{d6:lem:4n}, neither is $\torz{52}$. Every closed point of degree $9$ of $Y_1(2,26)$ lies in the orbit of the point $x_{2,26}$ of \Cref{d9:prop:realizations} under the level automorphisms. This orbit has $12$ elements. \end{proposition} \begin{proof} Let $X=X_1(2,26)$, and let $D$ be the trace divisor of a point of $Y_1(2,26)$ over a number field of degree $d\in\{6,7,8\}$, or a closed point of degree $d=9$ of $Y_1(2,26)$. The group $J_X(\Q)\simeq\torz2\times\torz{14}\times\torz{266}\times\torz{3990}\times\torz{11970}$ is finite and generated by rational cuspidal divisors \cite[Theorem 3.1(3), Proposition 4.16 and Table 3]{DEHMZ}, and $\gon_\Q X\geq12$ \cite[Table 1]{NV}. The closed cusps of $X_{\F_3}$ have degrees $1^{18},3^6$, and the $18$ cusps of degree $1$ are reductions of rational cusps. We \gitlinkIX{C2x26/class_2_26_D9_lower8.m}{compute} that their differences generate a subgroup of $J_X(\F_3)\simeq\torz2\times\torz{14}\times\torz{70}\times\torz{1330}\times\torz{3990}\times\torz{11970}$ with the invariant factors of $J_X(\Q)$, which is therefore the reduction of $J_X(\Q)$, of index $350$. By \Cref{d9:table:2n} and \Cref{d6:prop:main}(a), the reduction of $D$ is $x+C$, with $x$ a non-cuspidal closed point of degree $e\in\{5,\dots,d\}$ and $C$ cuspidal of degree $d-e$. For $e=5,\dots,8$ there are $6$, $26$, $48$ and $159$ diamond orbit representatives $x$, and none of them, completed by any effective cuspidal divisor of degree at most $9-e$, has class, normalized at a rational cusp, in the reduction of $J_X(\Q)$. This proves the first assertion, and for $d=9$ it shows that $e=9$. The $2328$ closed points of degree $9$ form $66$ orbits under the $36$ level automorphisms, and the three maps to $X_1(26)$ which keep $Q$, $P+Q$ and $P+14Q$ leave two of them, of which one has class outside the reduction of $J_X(\Q)$. The other is the orbit of the reduction of $x_{2,26}$, which has degree $9$ as the field of definition of $x_{2,26}$ is inert at $3$. So after applying a level automorphism, $D$ and $x_{2,26}$ have the same reduction, and $D=x_{2,26}$ by injectivity and \Cref{d9:lem:closed}. The stabilizer of the reduction has order $3$, and every element of it fixes $x_{2,26}$ by \Cref{d9:lem:stab}, so the orbit of $x_{2,26}$ has $36/3=12$ elements. \end{proof} \subsection{The curve \texorpdfstring{$X_1(3,15)$}{X1(3,15)}}\label{subsec:sporadic_3_15} \begin{proposition}\label{d8:prop:3_15} Let $z^2+z+1=0$, $F=\Q(z)$ and $K=F(w)$ with $w^4-5w^3+8w^2-6w+3=0$, which is the octic field \lmfdbnf{8.0.2313441.1}. Let \begin{align*} a_1&=\tfrac16\bigl((-7z-8)w^3+(23z+34)w^2+(-8z-43)w\bigr)+\tfrac12(-3z+10),\\ a_3&=\tfrac1{12}\bigl((17z+13)w^3+(-82z-71)w^2+(106z+119)w\bigr)-2z-5, \end{align*} and let $E:y^2+a_1xy+a_3y=x^3$, which has $j(E)=1331/8$. Then $E(K)_\tors\simeq\tg{3}{15}$. If $E'$ is an elliptic curve over an octic field $K'$ with $\tg{3}{15}\subseteq E'(K')$, then $(K',E')\simeq(K,E)$ and $E'(K')_\tors\simeq\tg{3}{15}$. In particular, $\tg{3}{30}$ is not a subgroup of the torsion of an elliptic curve over an octic field. \end{proposition} \begin{proof} The point $P=(0,0)$ has order $3$, and the point $Q=(x_Q,y_Q)$ with \begin{align*} x_Q&=\tfrac16\bigl((-z+1)w^3+(8z-2)w^2+(-17z-1)w\bigr)+\tfrac{3z}2,\\ y_Q&=\tfrac14\bigl((z+2)w^3+(-4z-8)w^2+(3z+6)w+2z+3\bigr) \end{align*} has order $15$, with $e_3(P,5Q)=z^2$, as we \gitlinkVIII{C3x15/realization_3_15_direct_v4_root.m}{compute}. The discriminant of $w^4-5w^3+8w^2-6w+3$ is $-507=-3\cdot13^2$, so $43$ is unramified in $K$. At the prime of $K$ given by $z\mapsto6$, $w\mapsto4$, of residue field $\F_{43}$, the curve has good reduction with group $\tg{3}{15}$. So $E(K)_\tors\simeq\tg{3}{15}$. Let $X=X_1(3,15)$ over $F$, and let $D_0$ be the trace divisor on $X$ of the point $(E,P,Q)$. The invariant $-y_Q^2/x_Q^3$ of the triple has degree $4$ over $F$, so $D_0$ is a closed point of degree $4$ with residue field $K$. Now let $D$ be the trace divisor of any point of $X$ over an octic field. We \gitlinkVIII{C3x15/diagonal_3_15_D4.m}{compute} all the possible reductions of $D$ at the primes above $2$ and $7$: $5000$ effective cuspidal divisors and $102$ quartic points over $\F_4$, and over $\F_7$, where the entirely cuspidal reduction is excluded by \Cref{d8:prop:main}(a) as $\gon_FX\geq6$ \cite[Table 1]{NV}, $3264$ quadratic points with a cuspidal part, $2304$ cubic points with a cusp, $396$ quartic points and $300$ sums of two quadratic points. Fix an $F$-rational cusp $c_\ast$. By \Cref{d8:lem:3_15_C}, $3[D-4c_\ast]\in\ClCusp_F X$, and comparing the two reductions inside $\ClCusp_F X$ leaves $60$ pairs, all of whose classes over $\F_4$ lie in the reduction of $\ClCusp_F X$. As reduction above $2$ is injective on $J_X(F)$ by \Cref{d8:lem:3_15_C}, it follows that $[D-4c_\ast]\in\ClCusp_F X$, and comparing the classes $[D-4c_\ast]$ themselves leaves $12$ elements of $\ClCusp_F X$. We \gitlinkVIII{C3x15/certify_exact_3_15_D4_class_orbit_v2.m}{check} that these $12$ classes form one orbit under the $48$ automorphisms $(P,Q)\mapsto\bigl(aP+bQ_3,\ 2(cP+dQ_3)+2uQ_5\bigr)$ of $X$ over $F$, where $Q_3=5Q$, $Q_5=3Q$, $\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\SL_2(\F_3)$ and $u\in\{1,2\}$. So $[D-4c_\ast]=[gD_0-4c_\ast]$ for an automorphism $g$, and as $\gon_FX>4$, $D=gD_0$ by \Cref{d8:lem:closed}. The automorphisms only change the level structure, so the elliptic curve corresponding to $D$ is $E$, over a field isomorphic to $K$. \end{proof} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Proof of the main theorems}\label{sec:proof} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \Cref{d6:table:summary,d7:table:summary,d8:table:summary,d9:table:summary} list, for each degree $d$ and each of the groups of $\mathcal M_d$, where it is shown that it is not a subgroup of the torsion of an elliptic curve over a number field of degree $d$. \begin{table}[H] \centering \footnotesize \begin{tabular}{@{}l@{\quad}l@{}} \toprule $G\in\mathcal M_6$ & treated in\\ \midrule $\torz{37n}$, $n=2,3,5,7,11,13,17,19,37$ & \Cref{d6:prop:37}\\ $\torz n$ for the $11$ levels $n$ of \Cref{table:global} with $d=6$ & \Cref{prop:global} (global method)\\ $\torz n$, $n=77,85,91,95,133,247$ & \Cref{prop:77_99,d6:prop:large} (diamond operators)\\ $\torz n$, $n=55,75$, & \Cref{d6:prop:rank0_cases}\\ \quad and $\tg{4}{24}$, $\tg{7}{7}$, $\tg{9}{9}$ & \\ $\torz{81}$, $\tg{3}{27}$ & \Cref{prop:81,d6:prop:3_27}\\ $\torz{45}$, $\tg{3}{15}$, $\tg{3}{21}$ & \Cref{prop:45,d6:prop:doubled_cases,d6:prop:3_21}\\ $\torz{54}$ & \Cref{d6:prop:54}\\ $\torz n$, $n=33,34,40,42$ & \Cref{d6:prop:sieve,d7:prop:sieve,d7:prop:40,d7:prop:42} (Hecke sieve)\\ $\torz n$, $n=35,38,49,50$ & \Cref{prop:35,prop:38,d6:prop:49}\\ $\torz{39}$, $\tg{2}{2n}$, $n=11,13,14$ & \Cref{d7:prop:39,d6:prop:2_22,d6:prop:2_26,d6:prop:2_28}\\ $\tg{3}{24}$, $\tg{4}{16}$, & \Cref{d6:prop:3_24,d6:prop:sieve_m},\\ \quad $\tg{4}{20}$, $\tg{6}{12}$ & \quad \Cref{d6:prop:4_20,d6:prop:6_12}\\ $\torz{32}$, $\torz{36}$, $\tg{2}{30}$, & \Cref{d6:prop:32,d6:prop:36},\\ \quad $\tg{3}{18}$ & \quad \Cref{d6:prop:2_30,d6:prop:3_18}\\ $\torz{51}$, $\torz{63}$ & \Cref{prop:51,prop:63}\\ & \quad (formal immersion at $2$)\\ $\torz{57}$, $\torz{65}$ & \Cref{prop:57,prop:65}\\ $\torz n$, $n=44,52,56,60$ & \Cref{d6:lem:4n}\\ $\torz{48}$ & \Cref{d6:prop:48}\\ \bottomrule \end{tabular} \medskip \caption{Treatment of the $66$ groups of $\mathcal M_6$.} \label{d6:table:summary} \end{table} \begin{table}[H] \centering \footnotesize \begin{tabular}{@{}l@{\quad}l@{}} \toprule $G\in\mathcal M_7$ & treated in\\ \midrule $\torz n$ for the $20$ levels $n$ of \Cref{table:global} with $d=7$ & \Cref{prop:global} (global method)\\ $\torz n$, $n=77,85,91,95,99$ & \Cref{prop:77_99} (diamond operators at $2$)\\ $\torz{81}$ & \Cref{prop:81}\\ $\torz n$, $n=32,34,35,38,40,42,46,49$ & \Cref{d6:prop:32,d7:prop:sieve,prop:35,prop:38},\\ & \quad \Cref{d7:prop:40,d7:prop:42,d7:prop:49} (Hecke sieve)\\ $\torz n$, $n=39,45,51,55$ & \Cref{d7:prop:39,prop:45,prop:51,prop:55}\\ $\torz n$, $n=57,63,65,69$ & \Cref{prop:57,prop:63,prop:65,prop:69}\\ $\torz{54}$ & \Cref{d7:prop:54}\\ $\tg{2}{2n}$, $n=11,12,13,14,15$ & \Cref{d7:prop:2_22,d7:prop:2_24,d7:prop:2_26}\\ & \quad and \Cref{d7:prop:2_28,d7:prop:2_30}\\ $\torz n$, $n=44,48,52,56,60$ & \Cref{d7:lem:4n}\\ $\torz{50}$, $\torz{75}$ & \Cref{d7:prop:25}\\ $\torz{66}$ & \Cref{d7:prop:33}\\ $\torz{72}$, $\tg{2}{36}$ & \Cref{d7:prop:36}\\ \bottomrule \end{tabular} \medskip \caption{Treatment of the $58$ groups of $\mathcal M_7$.} \label{d7:table:summary} \end{table} \begin{table}[H] \centering \footnotesize \begin{tabular}{@{}l@{\quad}l@{}} \toprule $G\in\mathcal M_8$ & treated in\\ \midrule $\torz n$ for the $20$ levels $n$ of \Cref{table:global} with $d=8$ & \Cref{prop:global} (global method)\\ $\torz n$, $n=81,85,91,95$ & \Cref{prop:81,prop:77_99,d8:prop:91}\\ $\torz n$, $n=38,44,46,50$ & \Cref{prop:38,d8:prop:44}\\ $\torz n$, $n=45,49,51,55,63,75,77$ & \Cref{prop:45,d8:prop:49,prop:51},\\ & \quad \Cref{prop:55,prop:63,d8:prop:75,d8:prop:77}\\ $\torz n$, $n=57,65,69$ & \Cref{prop:57,prop:65,prop:69}\\ $\torz{54}$ & \Cref{d8:prop:54}\\ $\torz{48}$, $\tg{2}{2n}$, $n=13,14,15,16,18,21$ & \Cref{d8:prop:48,d8:prop:2_26,d8:prop:2_28},\\ & \quad \Cref{d8:prop:2_30,d8:prop:2_32,d8:prop:2_36,d8:prop:2_42}\\ $\torz n$, $n=52,56,60,64,72,84$ & \Cref{d8:lem:4n}\\ $\tg{3}{3n}$, $n=6,7,8,9$, $\tg{4}{4n}$, $n=4,5,6$, & \Cref{d8:prop:local,d8:prop:3_21,d8:prop:3_24},\\ \quad $\tg{5}{5n}$, $n=3,4,5$, & \quad \Cref{d8:prop:3_18,d8:prop:4_16,d8:prop:6_12,d8:prop:10_10},\\ \quad $\tg{6}{12}$, $\tg88$, $\tg{10}{10}$ & \quad \Cref{d8:prop:5_25,d8:prop:4_20,d8:prop:8_8}\\ $\torz{66}$, $\torz{99}$, $\tg{3}{33}$ & \Cref{d8:prop:33}\\ $\torz{68}$, $\tg{2}{34}$ & \Cref{d8:prop:34}\\ $\torz{70}$, $\torz{105}$, $\torz{175}$, $\tg{5}{35}$ & \Cref{d8:prop:35}\\ $\torz{78}$, $\torz{117}$, $\tg{3}{39}$ & \Cref{d8:prop:39}\\ $\torz{80}$, $\tg{2}{40}$ & \Cref{d8:prop:40}\\ $\tg{3}{30}$ & \Cref{d8:prop:3_15}\\ \bottomrule \end{tabular} \medskip \caption{Treatment of the $80$ groups of $\mathcal M_8$. The groups in the last six rows contain $\torz{33}$, $\torz{34}$, $\torz{35}$, $\torz{39}$, $\torz{40}$ or $\tg{3}{15}$, and they are ruled out by the determination of all the elliptic curves over octic fields containing these groups.} \label{d8:table:summary} \end{table} \begin{table}[H] \centering \footnotesize \begin{tabular}{@{}l@{\quad}l@{}} \toprule $G\in\mathcal M_9$ & treated in\\ \midrule $\torz n$ for the $35$ levels $n$ of \Cref{table:global} with $d=9$ & \Cref{prop:global} (global method)\\ $\torz n$, $n\in\{81,95,99,133,247\}$ & \Cref{prop:81,prop:77_99,d9:prop:133,d9:prop:247}\\ $\torz n$, $n\in\{35,38,44,46,50,58,62\}$ & \Cref{prop:35,prop:38,d9:prop:44,d9:prop:58_62}\\ $\torz{48}$ & \Cref{d9:prop:48}\\ $\torz n$, $n\in\{45,49,51,55,63,69,75,77\}$ & \Cref{prop:45,d9:prop:49,prop:51,prop:55},\\ & \quad \Cref{prop:63,prop:69,d9:prop:75,d9:prop:77}\\ $\torz n$, $n\in\{57,65,85,87,91,93\}$ & \Cref{d9:prop:57,d9:prop:65,d9:prop:85_91_93,d9:prop:87}\\ $\torz{54}$ & \Cref{d9:prop:54}\\ $\torz{66}$, $\torz{78}$, $\torz{117}$ & \Cref{d9:prop:33,d9:prop:78,d9:prop:117}\\ $\tg{2}{2n}$, $n\in\{14,15,16,17,18,20,21\}$, & \Cref{d9:prop:2_28,d9:prop:2_30,d9:prop:2_32_2_34_2_40},\\ \quad and $\torz{52}$ & \quad \Cref{d9:prop:2_36,d9:prop:2_42,d9:prop:52}\\ $\torz{4n}$, $n\in\{14,15,16,17,18,20,21\}$ & \Cref{d9:lem:4n}\\ \bottomrule \end{tabular} \medskip \caption{Treatment of the $81$ groups of $\mathcal M_9$.} \label{d9:table:summary} \end{table} \begin{proof}[Proof of \Cref{thm:main}] Let $d\in\{6,7,8,9\}$. Every group in $\mathcal G_d$ occurs over a number field of degree $d$: the groups in $\Phi^\infty(d)$ by definition, and the sporadic groups by \Cref{d6:prop:37,d6:prop:examples,d6:prop:2_24_points} for $d=6$, by \Cref{d7:prop:realizations} for $d=7$, by \Cref{d8:prop:realizations,d8:prop:5_10,d8:prop:3_15} for $d=8$, and by \Cref{d9:prop:realizations} for $d=9$. Conversely, by \Cref{lem:minimal} every group in $\Phi(d)\setminus\mathcal G_d$ contains one of the groups of $\mathcal M_d$, and these are not subgroups of the torsion of an elliptic curve over a number field of degree $d$ by the results listed in \Cref{d6:table:summary,d7:table:summary,d8:table:summary,d9:table:summary}. Hence $\Phi(d)=\mathcal G_d$. \end{proof} \begin{proof}[Proof of \Cref{thm:sporadic}] For $d=6$ the theorem follows from \Cref{d6:prop:37,d6:prop:2_24_points}, for $d=7$ from \Cref{d7:prop:25,d7:prop:33,d7:prop:36}, and for $d=8$ from \Cref{d8:prop:33,d8:prop:34,d8:prop:35,d8:prop:39,d8:prop:40,d8:prop:3_15}, with the curves of \Cref{d7:prop:realizations,d8:prop:realizations}. For $d=9$ and $G=\torz{33}$ it is \Cref{d9:prop:33}. Finally, let $d=9$ and $\tg{2}{26}\subseteq E(K)$. By \Cref{d9:rem:exactdegree} the corresponding point of $X_1(2,26)$ has degree $9$, so by \Cref{d9:prop:2_26} it lies in the orbit of $x_{2,26}$ under the level automorphisms, which do not change the elliptic curve. As $X_1(2,26)$ is a fine moduli space, $(K,E)$ is isomorphic to $(K_{2,26},E_{2,26})$, and $E(K)_\tors\simeq\tg{2}{26}$ by \Cref{d9:prop:realizations}. \end{proof} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{The computations}\label{sec:computations}\label{d6:sec:computations}\label{d7:sec:computations}\label{d8:sec:computations}\label{d9:sec:computations} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Our computations were performed in Magma V2.30-1 \cite{magma}, apart from some auxiliary ones in Python in all four degrees (such as the lists of groups, the values of Abramovich's bound, the queries to the LMFDB and the replay of data exported from Magma), in PARI/GP in degrees $8$ and $9$, and in Sage in degree $9$. They were run on the \textit{Mordell} server at the Department of Mathematics, University of Zagreb, with an AMD EPYC 9175F CPU and $384$ GB of RAM. We used the package \texttt{mdmagma} \cite{mdmagma} for the moduli-theoretic operations, Sutherland's optimized models of $X_1(n)$ \cite{sutherland,sutherlandtables}, the Derickx-Sutherland models of $X_1(m,n)$ \cite{DSmodels}, the low degree points of van Hoeij \cite{vanHoeij}, Sutherland's package \cite{sutherlandmagma} for independent counts of points on modular curves over finite fields, and the LMFDB \cite{lmfdb} for the analytic ranks of newforms. The Hecke operators on closed points of $X_1(m,n)_{\F_p}$ were computed with the routines of \cite[\S 10]{N5}, extended to $m\geq3$ by carrying both level points through the isogenies. In degree $6$ these extensions were tested against the full computation of the isogenies on sample points. In degree $9$ the divisors $T_qP$ were computed both by our own routines and by Magma's isogeny functions. The computations took about $11$, $19$, $39$ and $38$ hours of CPU time in total for the degrees $6$, $7$, $8$ and $9$. For degree $7$ this is the time of the computations recorded in the repository. These figures do not include the reruns made for the repositories, which took about $15$ hours in degree $8$ and $3$ hours in degree $9$, nor the independent recomputations described below. The figure for degree $9$ includes the computations for the prime orders, which this paper does not use. The completeness of the enumeration of closed points of small degree, which every sieve computation assumes, was checked independently of the models by counting the isomorphism classes of elliptic curves with the required level structure over $\F_{p^e}$, using $\#Y_1(m,n)(\F_{p^e})=\sum_{k\mid e}k\cdot n_k$, where $n_k$ is the number of non-cuspidal closed points of degree $k$ (see \githubbareVI{independent_counts.m}, \githubbareVII{independent_counts_fast.m}, the \gitlinkVIII{common/independent_counts.m}{octic} and the \gitlinkIX{C38/independent_counts_D9.m}{nonic} code, and the other scripts listed in the \texttt{README} files). For $X_1(10,10)$ in \Cref{d8:prop:10_10} we \gitlinkVIII{C10x10/count_full10_points_root.m}{count} the elliptic curves with full level $10$ structure. In the sieve computations on the curves $X_1(n)$ the degrees of the closed cusps over $\F_p$, on which the admissibility of the primes $q$ depends, are computed from the models and checked against the prediction of \cite[Lemma 2.8]{DEHMZ}. Several results were also checked by computations that their proofs do not need. They are listed, result by result and with their logs, in the section ``Further checks'' of the \texttt{README} files of the septic, octic and nonic repositories, and for degree $6$ in the \texttt{README} of the sextic repository. Most of them are the original computations in the smaller degrees, which are covered by the computation in the largest degree (see \Cref{subsec:sieve_method} and \Cref{rem:smaller}), for instance the Hecke sieve in degree $6$ on $X_1(32)$, $X_1(34)$, $X_1(39)$, $X_1(40)$, $X_1(2,22)$ and $X_1(2,28)$, the computations on the quotient $Y_{51}$ of \Cref{d6:lem:quotients_51_63} in degrees $6$ and $7$, and those for $\torz{65}$ in degree $6$ on a quotient of $X_1(65)$ of genus $41$. Most of the computations of ranks of restriction maps, of ranks of the matrices in \Cref{d9:lem:mult} and of classes in finite groups in degrees $8$ and $9$ were done, or replayed, by independent programs, using integer and binary linear algebra, from the data exported by the Magma computations. Other results were checked by a second method, such as the finiteness of $J_1(69)(\Q)$ and $J_1(81)(\Q)$ by winding computations, independently of \cite{DEHMZ}, the spaces $\Omega(jP)$ used for \Cref{prop:77,d8:prop:3_21}, and $\ell(B)=0$ in the proof of \Cref{d9:prop:2_42}, by lattice reduction. We checked the computations on which the proofs depend most heavily a second time, with code written independently of the first implementation and, where models of the curves are involved, with different models. In all four degrees these are the list of \Cref{lem:minimal}, the explicit examples of \Cref{sec:sporadic} and the analytic ranks and character values taken from the LMFDB (in degree $9$ those at the levels up to $99$ and at $115$, $133$ and $247$, and those for the curves $X_1(2,2n)$). In degree $6$ they also include the $2$-torsion computations of \Cref{lem:J49} and \Cref{d6:lem:57two}, the doubled cuspidal computations for $X_1(45)$ and $X_1(3,21)$, and the Cartier operator computations on the quotients of \Cref{d6:lem:quotients_51_63} and on the quotient of $X_1(65)$ mentioned above. In degree $7$ they include the numbers of non-cuspidal closed points over $\F_2$ and $\F_3$ and the septic points on $X_1(25)$ and $X_1(33)$. In degree $8$ they include all the numbers of non-cuspidal closed points and their orbits, the orders of the groups $J(\F_p)$, the octic points on $X_1(33)$, $X_1(34)$, $X_1(35)$, $X_1(39)$ and $X_1(40)$, the quartic points on $X_1(3,15)$ over $\Q(\zeta_3)$, and the formal immersion criteria in \Cref{prop:45,prop:63,prop:49,d8:prop:3_21,d8:prop:5_25} (for $X_1(51)$ our models were too large for this). In degree $9$ they include $47$ of the lists of numbers of non-cuspidal closed points that we use, by an enumeration of all elliptic curves over $\F_{p^e}$ for $p\leq5$ and $e\leq9$, and the list in \Cref{d9:lem:CM}(3), which we also derived from the results of Bourdon and Pollack \cite{BP}. In all cases the results agree. These computations are in the folders \githubdirVI{independent}, \githubdirVII{independent}, \githubdirVIII{independent} and \githubdirIX{independent} of the four repositories. \bibliographystyle{amsalpha} \bibliography{torsion69} \end{document}