%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{The global method and diamond operators}\label{sec:global} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Most of the groups $\torz n$ with large $n$ are excluded by the global method of Derickx \cite[Proposition 4.3]{DN}, which we recall in \Cref{subsec:glob-method}. With $q=3$ it excludes the $45$ groups $\torz n$ of \Cref{table:global} (\Cref{prop:globaltable}), and with $q=2$ the groups $\torz{115}$, $\torz{121}$ and $\torz{135}$ (\Cref{subsec:glob-dyadic}). For $n=155$ the global method only shows that the trace divisor is invariant under a diamond operator whose order divides $10$, and we conclude by descending the point to a quadratic field (\Cref{subsec:glob-descent}). Finally, the groups $\torz{95}$ and $\torz{111}$ are excluded in \Cref{subsec:glob-95-111} with a diamond operator of order $3$, using that all the reductions modulo the primes above $2$ are bad. We use the following consequence of Kato's theorem, as in \cite[Section 4]{N69}. 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 \cite[Corollary 14.3]{kato}. Hence the group $(\dm a-1)J_1(n)(\Q)$ is finite if $\chi_f(a)=1$ for every newform $f$ of weight $2$ and level dividing $n$ with $L(f,1)=0$. We obtain this condition either from the LMFDB \cite{lmfdb}, where all the analytic ranks that we use are recorded as proved, or from the winding quotient (\Cref{subsec:hecke}). For a Galois orbit of even Dirichlet characters $\chi$ modulo $n$ we write $S_\chi$ for the plus quotient of the space of cuspidal modular symbols of weight $2$ and character $\chi$, and $W_\chi$ for its winding submodule, the Hecke module generated by the projection of the winding element. If $W_\chi=S_\chi$, then $L(f,1)\neq0$ for every newform $f$ of level dividing $n$ whose character lies in the orbit of $\chi$. Whenever we compute $W_\chi$ for all the orbits, we check that no orbit is missed by verifying that the sum of $\varphi(\ord\chi)\dim S_\chi$ over all of them is the genus of $X_1(n)$. For characters of large order we first check, as in the proof of \cite[Lemma 4.1]{N69}, whether the vectors $v,vT_2,\dots,vT_2^{k-1}$, where $v$ is the projection of the winding element and $k=\dim S_\chi$, are linearly independent modulo a prime of degree $1$ of the field of values of $\chi$ at which all the denominators are units. If they are, then $W_\chi=S_\chi$, and if not, we compute $W_\chi$ over the field of values of $\chi$. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \subsection{The global method}\label{subsec:glob-method} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% The main tool is the global method of Derickx \cite[Proposition 4.3]{DN}, which we recall for number fields of degree $10$, as in \cite[Section 4]{N5} and \cite[Section 4.1]{N69}. \begin{proposition}[Derickx {\cite[Proposition 4.3]{DN}}]\label{prop:global} Let $n\geq5$ 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 also that $n$ is odd and 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 $k_q=2(q+1)$ if not. If $10k_q<\gon_\Q X_1(n)$, then every effective $\Q$-rational divisor $D$ of degree $10$ on $X_1(n)$ whose support contains no cusp satisfies $\dm aD=D$. The inequality $10k_q<\gon_\Q X_1(n)$ holds as soon as \[ b(n):=\frac{325}{2^{15}}\cdot\frac{i(n)}{k_q}>10. \] \end{proposition} \begin{proof} For odd $q$ this is \cite[Proposition 4.3]{DN} with $H=\{\pm1\}$ and $d=10$. Its proof applies $A_q$, which kills the torsion of $J_1(n)(\Q)$ and the rational cuspidal divisors by \Cref{prop:tools}~b) and~c), to the class of $(\dm a-1)(D-10c_0)$, and $10k_q$ bounds the degree of the positive part of the resulting principal divisor. For $q=2$ the proof applies without change, by \Cref{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 (\Cref{subsec:hecke}). The last statement follows from Abramovich's bound (\Cref{subsec:hecke}), as $\gon_\Q X_1(n)\geq\gon_\C X_1(n)\geq\tfrac{325}{2^{15}}i(n)$. \end{proof} We apply \Cref{prop:global} to the trace divisor $D$ of a point of order $n$ over a number field of degree $10$ (\Cref{subsec:trace}). By \Cref{lem:CM} the support of $D$ consists of non-CM points, so if $\dm aD=D$, then by \Cref{lem:orbits} the order of $\dm a$ in $\Delta(n)$ divides $10$. For the $32$ levels of \Cref{table:global} that also occur in \cite[Table 1]{N69} we use the same elements $a$ as there. For $n=119$ the inequality $b(119)>10$ fails with $k_3=7$, and we use instead that $J_1(119)(\Q)$ is finite (see the proof of \Cref{prop:globaltable}). The other $11$ levels, all of them except $175$ divisible by $37$, are listed in \Cref{table:glob-new}. We \gitlink{computations/replay_new_certificates_v2.py}{check} the entries of \Cref{table:global} for the levels of \Cref{table:glob-new}. For the other levels the entries are those of \cite[Table 1]{N69}, apart from $k_3$ and $b(n)$ for $n=119$. They are recorded in \githubbare{exclusions/global3.json} and \githubbare{exclusions/global3_additional.json}, where for $n=133$ and $n=247$ the elements $a=27$ and $a=12$ are used instead, which generate the same subgroups of order $3$ of $\Delta(n)$ as $64$ and $144$. \begin{table}[tbp] \centering \footnotesize \begin{tabular}{rrrrrr} \toprule $n$ & $a$ & $\ord\dm a$ & $k_3$ & $i(n)$ & $b(n)$\\ \midrule $119$ & -- & -- & $4$ & $6912$ & $17.13$\\ $125$ & $3$ & $50$ & $7$ & $7500$ & $10.62$\\ $133$ & $64$ & $3$ & $8$ & $8640$ & $10.71$\\ $143$ & $67$ & $12$ & $8$ & $10080$ & $12.49$\\ $145$ & $3$ & $14$ & $7$ & $10080$ & $14.28$\\ $161$ & $3$ & $66$ & $7$ & $12672$ & $17.95$\\ $169$ & $3$ & $39$ & $7$ & $14196$ & $20.11$\\ $175$ & $18$ & $12$ & $8$ & $14400$ & $17.85$\\ $185$ & $2$ & $18$ & $8$ & $16416$ & $20.35$\\ $187$ & $122$ & $16$ & $8$ & $17280$ & $21.42$\\ $203$ & $6$ & $7$ & $8$ & $20160$ & $24.99$\\ $209$ & $34$ & $18$ & $8$ & $21600$ & $26.77$\\ $217$ & $3$ & $15$ & $7$ & $23040$ & $32.64$\\ $221$ & $3$ & $48$ & $7$ & $24192$ & $34.27$\\ $247$ & $144$ & $3$ & $8$ & $30240$ & $37.49$\\ $253$ & $3$ & $55$ & $7$ & $31680$ & $44.88$\\ $259$ & $3$ & $9$ & $7$ & $32832$ & $46.51$\\ $289$ & $3$ & $136$ & $7$ & $41616$ & $58.96$\\ $299$ & $3$ & $33$ & $7$ & $44352$ & $62.84$\\ $319$ & $3$ & $140$ & $7$ & $50400$ & $71.41$\\ $323$ & $101$ & $18$ & $8$ & $51840$ & $64.27$\\ $341$ & $56$ & $3$ & $8$ & $57600$ & $71.41$\\ $361$ & $3$ & $171$ & $7$ & $64980$ & $92.06$\\ $377$ & $5$ & $28$ & $8$ & $70560$ & $87.47$\\ $391$ & $3$ & $176$ & $7$ & $76032$ & $107.72$\\ $403$ & $118$ & $3$ & $8$ & $80640$ & $99.97$\\ $407$ & $10$ & $6$ & $8$ & $82080$ & $101.76$\\ $437$ & $3$ & $198$ & $7$ & $95040$ & $134.66$\\ $481$ & $3$ & $18$ & $7$ & $114912$ & $162.81$\\ $493$ & $3$ & $112$ & $7$ & $120960$ & $171.38$\\ $527$ & $3$ & $240$ & $7$ & $138240$ & $195.87$\\ $529$ & $3$ & $253$ & $7$ & $139656$ & $197.87$\\ $551$ & $12$ & $12$ & $8$ & $151200$ & $187.45$\\ $589$ & $3$ & $45$ & $7$ & $172800$ & $244.83$\\ $629$ & $6$ & $16$ & $8$ & $196992$ & $244.22$\\ $667$ & $3$ & $308$ & $7$ & $221760$ & $314.20$\\ $703$ & $3$ & $9$ & $7$ & $246240$ & $348.89$\\ $713$ & $2$ & $55$ & $8$ & $253440$ & $314.20$\\ $841$ & $12$ & $58$ & $8$ & $353220$ & $437.91$\\ $851$ & $3$ & $198$ & $7$ & $361152$ & $511.71$\\ $899$ & $3$ & $420$ & $7$ & $403200$ & $571.28$\\ $961$ & $3$ & $465$ & $7$ & $461280$ & $653.58$\\ $1073$ & $3$ & $252$ & $7$ & $574560$ & $814.08$\\ $1147$ & $3$ & $45$ & $7$ & $656640$ & $930.38$\\ $1369$ & $6$ & $74$ & $8$ & $936396$ & $1160.92$\\ \bottomrule \end{tabular} \medskip \caption{The levels $n$ of \Cref{prop:globaltable}, with $q=3$. The order of $\dm a$ is its order in $\Delta(n)$, and $b(n)$ is rounded down to two decimal places. For $n=119$ the group $J_1(119)(\Q)$ is finite, and the column $k_3$ gives the degree $4$ of $T_3$ instead (see the proof of \Cref{prop:globaltable}).} \label{table:global} \end{table} \begin{lemma}\label{lem:glob-finite} Let $n$ and $a$ be as in one of the rows of \Cref{table:global}. Then $(\dm a-1)J_1(n)(\Q)$ is finite. The group $J_1(119)(\Q)$ is also finite. \end{lemma} \begin{proof} An LMFDB search shows that every newform of weight $2$ and level dividing $119$ has analytic rank $0$ (see \githubbare{inputs/global_newforms_cache.json}), so $J_1(119)(\Q)$ is finite by Kato's theorem. For the other $31$ levels of \Cref{table:global} that occur in \cite[Table 1]{N69} with the same element $a$, the finiteness of $(\dm a-1)J_1(n)(\Q)$ is shown in the proof of \cite[Proposition 4.2]{N69}, in a part of that proof which does not depend on the degree. For $n=341$ and $n=403$, where \cite[Table 1]{N69} uses elements of order $2$, the computation of the winding submodules for the proof of \cite[Proposition 4.2]{N69} (\githubbare{inputs/global_archive/winding_probe_minimal.m}, with the logs \githubbare{inputs/global_archive/winding341.log} and \githubbare{inputs/global_archive/winding403.log}) shows that $W_\chi=S_\chi$ unless $\chi$ has order $1$ or $5$, respectively order $1$, $3$, $5$ or $15$, and it lists the elements of $(\torz n)^\times$ on which all these characters are trivial, among them $56$ and $118$. For the $11$ levels of \Cref{table:glob-new} we compute $W_\chi$ for every Galois orbit of even characters $\chi$ modulo $n$, with the scripts linked in \Cref{table:glob-new} (for $n=1147$ the computation is split between \githubbare{computations/winding1147.m} and \githubbare{computations/winding1147_resume.m}). In every case the sum of $\varphi(\ord\chi)\dim S_\chi$ is the genus of $X_1(n)$, and $W_\chi=S_\chi$ unless the order of $\chi$ is one of those listed in \Cref{table:glob-new}, and the same scripts check that the element $a$ of \Cref{table:global} lies in the kernel of these characters. (For $n=185$ only one of the two Galois orbits of even characters of order $4$, both of conductor $185$, has $W_\chi\neq S_\chi$, and it is trivial on $2$.) Hence $(\dm a-1)J_1(n)(\Q)$ is finite. \end{proof} \begin{table}[tbp] \centering \begin{tabular}{rrl} \toprule $n$ & $g(X_1(n))$ & orders of the $\chi$ with $W_\chi\neq S_\chi$\\ \midrule \gitlink{computations/winding175.m}{$175$} & $1033$ & $1$, $5$\\ \gitlink{computations/winding185.m}{$185$} & $1225$ & $1$, $4$\\ \gitlink{computations/winding259.m}{$259$} & $2521$ & $1$\\ \gitlink{computations/winding407.m}{$407$} & $6481$ & $1$, $6$\\ \gitlink{computations/winding481.m}{$481$} & $9145$ & $1$\\ \gitlink{computations/winding629.m}{$629$} & $15841$ & $1$, $9$\\ \gitlink{computations/winding703.m}{$703$} & $19873$ & $1$\\ \gitlink{computations/winding851.m}{$851$} & $29305$ & $1$\\ \gitlink{computations/winding1073.m}{$1073$} & $46873$ & $1$\\ \gitlink{computations/winding1147_resume.m}{$1147$} & $53641$ & $1$\\ \gitlink{computations/winding1369.m}{$1369$} & $77044$ & $1$, $3$, $9$\\ \bottomrule \end{tabular} \medskip \caption{The $11$ levels of \Cref{table:global} that do not occur in \cite[Table 1]{N69}, the genus of $X_1(n)$, and the orders of the even characters $\chi$ modulo $n$ whose winding submodule is not the whole space. Each level is a link to the script that computes the winding submodules.} \label{table:glob-new} \end{table} \begin{proposition}\label{prop:globaltable} Let $n$ be one of the $45$ levels 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 $10$. \end{proposition} \begin{proof} Let $D$ be the trace divisor of a point of order $n$ over a number field of degree $10$. By \Cref{lem:CM} the support of $D$ consists of non-CM points. Suppose first that $n=119$. As $J_1(119)(\Q)$ is finite by \Cref{lem:glob-finite}, the divisor $A_3D$ is principal by \Cref{prop:tools}~d). Its positive part is contained in $T_3D$, which has degree $40$, and $\gon_\Q X_1(119)\geq\tfrac{325}{2^{15}}\cdot6912>68$ by Abramovich's bound, so $A_3D=0$. This contradicts \Cref{lem:tree}. Suppose now that $n\neq119$, and take $q=3$ and $a$ as in \Cref{table:global}. By \Cref{lem:glob-finite} the hypotheses of \Cref{prop:global} are satisfied, and by \Cref{table:global} we have $b(n)>10$, so $\dm aD=D$. By \Cref{lem:orbits} the order of $\dm a$ in $\Delta(n)$ divides $10$, which contradicts \Cref{table:global}. \end{proof} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \subsection{The global method at the prime \texorpdfstring{$2$}{2}}\label{subsec:glob-dyadic} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% For $n=115$, $121$ and $135$ we use \Cref{prop:global} with $q=2$. Note that for $n=115$ this is the argument of \cite[Proposition 4.2]{N69} in degree $9$, which also works in degree $10$. For $n=121$ the choice $q=3$ and $a=2$ gives only $b(121)=9.0007\ldots$ (see \cite[Table 1]{N69}), and $3$ divides $135$. \begin{lemma}\label{lem:glob-odd} Let $(n,a,p)$ be one of the triples $(115,16,3)$, $(121,2,3)$ and $(135,16,31)$, and let $B_n:=(\dm a-1)J_1(n)$. Then $B_n(\Q)$ is finite and $\#B_n(\F_p)$ is odd. In particular $(\dm a-1)J_1(n)(\Q)$ is finite of odd order. \end{lemma} \begin{proof} The abelian variety $B_n$ is isogenous to the product of the parts of $J_1(n)$ with characters $\chi$ such that $\chi(a)\neq1$. These are the characters of order $11$, $22$ and $44$ for $n=115$, of order $5$, $11$ and $55$ for $n=121$, and of order $3$, $6$, $9$, $12$, $18$ and $36$ for $n=135$, and $B_n$ has dimension $400$, $520$ and $480$, respectively. For $n=115$ the group $B_n(\Q)$ is finite by \cite[Lemma 4.1]{N69}. For $n=121$ the only newform of weight $2$ and level dividing $121$ with positive analytic rank is \lmfdbnewform{121}{2}{a}{b}, which has trivial character (see \githubbare{inputs/global_newforms_cache.json}). For $n=135$ we \gitlink{computations/cont_winding135.m}{compute} that $W_\chi=S_\chi$ for every non-trivial even character $\chi$ modulo $135$, the sum of $\varphi(\ord\chi)\dim S_\chi$ over all the orbits being $529=g(X_1(135))$. Hence $B_n(\Q)$ is finite in all three cases. For every Galois orbit of characters $\chi$ with $\chi(a)\neq1$ we compute the norm of $\det\bigl((1+p\chi(p))-T_p\bigr)$ on $S_\chi$. By the Eichler-Shimura relation it is a power of the number of $\F_p$-points of the part of $J_1(n)$ with characters in the orbit of $\chi$, and we find that all these norms are odd. For $n=115$ this is the computation in the proof of \cite[Proposition 4.2]{N69}, and for $n=121$ and $n=135$ we compute it with \githubbare{computations/image121_no2_v2.m} and \githubbare{computations/cont_image135_a16_no2.m}. As the number of points over a finite field is invariant under isogeny, $\#B_n(\F_p)$ is odd. As $p$ is odd and $B_n$ has good reduction at $p$, reduction modulo $p$ is injective on $B_n(\Q)_\tors$ by Katz's result, so $B_n(\Q)$ has odd order, and so has its subgroup $(\dm a-1)J_1(n)(\Q)$. \end{proof} \begin{proposition}\label{prop:115-121} The groups $\torz{115}$ and $\torz{121}$ are not subgroups of the torsion of an elliptic curve over a number field of degree $10$. \end{proposition} \begin{proof} Let $n\in\{115,121\}$, put $a=16$ if $n=115$ and $a=2$ if $n=121$, and let $D$ be the trace divisor of a point of order $n$ over a number field of degree $10$. By \Cref{lem:glob-odd} we can apply \Cref{prop:global} with $q=2$. As $16\not\equiv\pm2^{\pm1}\pmod{115}$ we have $k_2=6$ for $n=115$, and $k_2=5$ for $n=121$. As $i(115)=6336$ and $i(121)=7260$, we get $b(115)=\tfrac{10725}{1024}>10.47$ and $b(121)=\tfrac{117975}{8192}>14.40$, so $\dm aD=D$. By \Cref{lem:CM} the support of $D$ consists of non-CM points, so by \Cref{lem:orbits} the order of $\dm a$ in $\Delta(n)$ divides $10$. But $\dm{16}$ has order $11$ in $\Delta(115)$ and $\dm2$ has order $55$ in $\Delta(121)$, which is a contradiction. \end{proof} \begin{proposition}\label{prop:135} The group $\torz{135}$ is not a subgroup of the torsion of an elliptic curve over a number field of degree $10$. \end{proposition} \begin{proof} Let $D$ be the trace divisor of a point of order $135$ over a number field of degree $10$. By \Cref{lem:glob-odd} we can apply \Cref{prop:global} with $q=2$ and $a=16$. As $16\not\equiv\pm2^{\pm1}\pmod{135}$ we have $k_2=6$, and as $i(135)=7776$, we get $b(135)=\tfrac{26325}{2048}>12.85$, so $\dm{16}D=D$. By \Cref{lem:CM} the support of $D$ consists of non-CM points, and by \Cref{lem:orbits} the order of $\dm{16}$ in $\Delta(135)$ divides $10$. We \gitlink{computations/cont_image135_a16_no2.m}{compute} that this order is $9$, which is a contradiction. \end{proof} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \subsection{The group \texorpdfstring{$\torz{155}$}{Z/155Z}}\label{subsec:glob-descent} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% For $n=155$ we \gitlink{computations/cont_winding155.m}{compute} that every element $a$ for which $(\dm a-1)J_1(155)(\Q)$ is known to be finite lies in the subgroup of order $5$ of $\Delta(155)$ generated by $\dm{16}$, so the global method only shows that the trace divisor is invariant under $\dm{16}$. We use that a closed point which is mapped to itself by a diamond operator has an image on a modular curve of smaller level which is defined over a smaller field. \begin{lemma}\label{lem:glob-descent} Let $n\geq5$, let $a\in(\torz n)^\times$, and let $r$ be the order of $\dm a$ in $\Delta(n)$. Let $x$ be a closed point of $Y_1(n)$ whose geometric points have no complex multiplication, and suppose that $\dm a$ maps $x$ to itself. Then $\dm a$ induces an automorphism $\phi$ of the residue field $\Q(x)$ of $x$ of order $r$. If $\pi:X_1(n)\to Z$ is a morphism of curves over $\Q$ with $\pi\circ\dm a=\pi$, then the residue field of $\pi(x)$ is contained in the fixed field $\Q(x)^\phi$, which has degree $[\Q(x):\Q]/r$. \end{lemma} \begin{proof} As $\dm a$ is an automorphism of $X_1(n)$ over $\Q$ which maps the closed point $x$ to itself, it induces an automorphism $\phi$ of $\Q(x)$, and it acts on the geometric points of $x$, which correspond to the embeddings $\iota:\Q(x)\to\Qbar$, by $\iota\mapsto\iota\circ\phi$ (or $\iota\mapsto\iota\circ\phi^{-1}$, depending on the convention). If $\phi^k=1$ for some $014$. So $\dm{16}D=D$ by \Cref{prop:global}, and $\dm{16}$ maps $x$ to itself. By \Cref{lem:CM}, the geometric points of $x$ have no complex multiplication, so by \Cref{lem:glob-descent} the degree of $K$ is divisible by $5$, and $\dm{16}$ induces an automorphism $\phi$ of $K$ of order $5$. As $31$ is not in $S(5)$ \cite[Theorem 1.2]{DKSS}, we have $[K:\Q]\neq5$. Hence $[K:\Q]=10$, and $F:=K^\phi$ is a quadratic field. The residue fields of $K$ at the primes above $2$ have at most $2^{10}$ elements, so by \Cref{lem:glob-155-F2}, \Cref{lem:badred} and \Cref{lem:additive} the curve given by $x$ has multiplicative reduction at every prime of $K$ above $2$, and $j(x)$ has negative valuation at each of them. As $16\equiv2^4\pmod{31}$, the composition of $X_1(155)\to X_1(31)$, $(E,P)\mapsto(E,5P)$, with the quotient map $X_1(31)\to Y_{31}$ is invariant under $\dm{16}$, so by \Cref{lem:glob-descent} the image $y$ of $x$ on $Y_{31}$ is defined over $F$. In particular $j(y)=j(x)\in F$, and it has negative valuation at every prime of $F$ above $2$. Let $D_Y$ be the sum of the images of $y$ under the two embeddings of $F$ into $\Qbar$. It is an effective $\Q$-rational divisor of degree $2$ on $Y_{31}$ whose support contains no cusp, and it reduces to a cuspidal divisor modulo $2$. But $J_{Y_{31}}(\Q)$ is finite and $\gon_\Q Y_{31}\geq5>4$ by \Cref{lem:glob-Y31}, so this contradicts \Cref{prop:main}~b), applied with $X=Y_{31}$, $p=2$ and $d=2$. \end{proof} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \subsection{The groups \texorpdfstring{$\torz{95}$ and $\torz{111}$}{Z/95Z and Z/111Z}}\label{subsec:glob-95-111} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% For $n=95$ and $n=111$ we use, as in \cite[Section 4.2]{N69}, that the elliptic curve has bad reduction at all the primes above $2$, together with a diamond operator of order $3$ and \Cref{lem:stab}~c). \begin{lemma}\label{lem:glob-95-111} Let $n\in\{95,111\}$, and let $h=\dm{11}$ if $n=95$ and $h=\dm{10}$ if $n=111$. \begin{enumerate} \item[a)] For $1\leq f\leq10$ no elliptic curve over $\F_{2^f}$ has a point of order $n$. \item[b)] The diamond operator $h$ has order $3$ in $\Delta(n)$, and $(h-1)J_1(n)(\Q)$ is finite. \end{enumerate} \end{lemma} \begin{proof} a) For $n=95$ and $f\leq7$, and for $n=111$ and $f\leq6$ or $f=8$, the Hasse interval $\bigl[2^f+1-2^{f/2+1},2^f+1+2^{f/2+1}\bigr]$ contains no multiple of $n$. The remaining multiples of $n$ in the Hasse intervals are $285$, $475$ and $1045$ for $n=95$ and $f=8,9,10$, and $111$, $555$ and $999$ for $n=111$ and $f=7,9,10$, which correspond to the traces of Frobenius $-28$, $38$, $-20$, respectively $18$, $-42$, $26$. These are even, and an elliptic curve over $\F_{2^f}$ with even trace of Frobenius is supersingular, so its trace is $0$ or $\pm2^{(f+1)/2}$ for odd $f$, and $0$, $\pm2^{f/2}$ or $\pm2^{f/2+1}$ for even $f$ \cite[Theorem 4.1]{waterhouse}. None of the six traces above is of this form. b) The group $\Delta(n)$ is cyclic of order $36$, generated by $\dm2$, and $11\equiv2^{12}\pmod{95}$ and $10\equiv2^{24}\pmod{111}$. So $h$ has order $3$, and a character of $\Delta(n)$ is non-trivial on $h$ if and only if its order is $9$, $18$ or $36$. We compute, with \githubbare{computations/next95_winding.m} and \githubbare{computations/winding111.m}, that $W_\chi=S_\chi$ for every Galois orbit of even characters $\chi$ modulo $n$ except the characters of order $12$ for $n=95$, and of order $1$ and $6$ for $n=111$, the sums of $\varphi(\ord\chi)\dim S_\chi$ being $289=g(X_1(95))$ and $385=g(X_1(111))$. These characters are trivial on $h$, so $(h-1)J_1(n)(\Q)$ is finite. \end{proof} \begin{proposition}\label{prop:95-111} The groups $\torz{95}$ and $\torz{111}$ are not subgroups of the torsion of an elliptic curve over a number field of degree $10$. \end{proposition} \begin{proof} Let $n\in\{95,111\}$, let $h$ be as in \Cref{lem:glob-95-111}, let $E$ be an elliptic curve over a number field $K$ of degree $10$ with a point of order $n$, and let $D$ be its trace divisor. The residue fields of $K$ at the primes above $2$ are $\F_{2^f}$ with $f\leq10$, so by \Cref{lem:glob-95-111}~a) and \Cref{lem:badred} the curve $E$ has bad reduction at every prime of $K$ above $2$. By Abramovich's bound, $\gon_\C X_1(95)\geq\tfrac{325}{2^{15}}\cdot4320=\tfrac{43875}{1024}>42$ and $\gon_\C X_1(111)\geq\tfrac{325}{2^{15}}\cdot5472=\tfrac{55575}{1024}>54$, so in both cases $\gon_\C X_1(n)>40$. By \Cref{lem:additive} the reduction of $D$ modulo a prime above $2$ is cuspidal, and as the cusps are defined over $\Q(\zeta_n)$, which is unramified above $2$, it lifts to an effective cuspidal divisor $C$ of degree $10$ over a number field unramified above $2$. By \Cref{lem:glob-95-111}~b), \Cref{lem:stab}~c), with $g=h$, $p=2$ and $d=e=10$, gives $hD=D$, as the reduction of $D-C$ is $0$. By \Cref{lem:CM} the support of $D$ consists of non-CM points, so by \Cref{lem:orbits} the order $3$ of $h$ divides $10$, which is a contradiction. \end{proof}