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\section{The groups \texorpdfstring{$\tg mn$}{Z/mZ x Z/nZ} with \texorpdfstring{$m\geq3$}{m>=3}}\label{sec:mxn}
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Let $E$ be an elliptic curve over a number field $K$ of degree $10$ with $\tg mn\subseteq E(K)$ for some $m\geq3$.
By the Weil pairing $m\in\{3,4,6,11,22\}$, and for $m\in\{3,4\}$ the trace divisor is an effective
$\Q(\zeta_m)$-rational divisor of degree $d_m=5$ on $X_1(m,n)$ (\Cref{subsec:trace}). We treat here the group
$\tg{11}{11}$, for which $K=\Q(\zeta_{11})$, and the groups $\tg{3}{21}$, $\tg{3}{24}$, $\tg{3}{27}$, $\tg{3}{33}$,
$\tg{4}{16}$, $\tg{4}{20}$, $\tg{4}{24}$ and $\tg{4}{28}$. The remaining groups $\tg mn\in\mathcal M_{10}$ (\Cref{lem:minimal}) with
$m\geq 3$, namely $\tg{3}{30}$, $\tg{3}{36}$, $\tg{3}{39}$, $\tg{3}{45}$ and $\tg{6}{18}$, are ruled out in
\Cref{cor:3x30-3x45,cor:3x36-6x18,prop:78-3x39}. For $\tg{3}{21}$, $\tg{3}{24}$ and $\tg{4}{24}$ the Hecke sieve at
a single prime suffices, as in \cite[Section 12]{N69}. For $\tg{4}{16}$ we compare the reductions at two primes,
for $\tg{4}{20}$ we use stabilizers of level automorphisms, for $\tg{3}{27}$ and $\tg{3}{33}$ we push the trace
divisor forward to $X_1(27)$ and $X_1(33)$ (for $\tg{3}{33}$ together with a formal immersion argument at $2$ on
the part of the Jacobian on which a diamond operator acts by $-1$), and for $\tg{4}{28}$ we work on $X_1(2,28)$
over $\Q$.

Recall that by \cite[Theorem 4.1]{DS} the group $J_1(m,n)(\Q(\zeta_m))$ is finite for
$(m,n)\in\{(3,21),(3,24),(3,27),(4,16),(4,20),(4,24)\}$, and that in the Hecke sieve it suffices to test one point in
each orbit of the level automorphisms defined over $\Q(\zeta_m)$ (\Cref{subsec:sieve}). If $D$ is the trace
divisor of a point $x=(E,P,Q)$ and $g$ is a level automorphism defined over $\Q(\zeta_m)$, then $gD$ is the trace
divisor of $g(x)$, which corresponds to the same elliptic curve $E$.

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\subsection{The group \texorpdfstring{$\tg{11}{11}$}{Z/11Z x Z/11Z}}\label{subsec:mxn-11x11}

If $E[11]\subseteq E(K)$ with $[K:\Q]=10$, then $K=\Q(\zeta_{11})$ by the Weil pairing, and $E$ has a $K$-rational
point of order $11$. The modular curve $X_1(11)$ has genus $1$, and its Jacobian is the elliptic curve
\[
A\colon v^2+v=u^3-u^2
\]
(\lmfdbec{11}{a}{3}).

\begin{lemma}\label{lem:mxn-X111}
Let $K=\Q(\zeta_{11})$ and let $A$ be the elliptic curve $v^2+v=u^3-u^2$. Then $A(K)$ is finite.
\end{lemma}

\begin{proof}
Let $f$ be the newform of level $11$ attached to $A$. As in the proof of \cite[Lemma 5.1]{N69}, the restriction of
scalars of $A_K$ to $\Q$ is isogenous to a product of the abelian varieties attached to the Galois orbits of the
newforms associated with the twists $f\otimes\chi$, where $\chi$ runs over the characters of
$\Gal(K/\Q)\simeq(\torz{11})^\times$, so by Kato's theorem $A(K)$ is finite if $L(f\otimes\chi,1)\neq0$ for all
$\chi$. As $\chi$ has conductor $1$ or $11$, the newform associated with $f\otimes\chi$ has level $11$ or $121$, and
it has no complex multiplication, as $f$ has none ($j(A)=-2^{12}/11$ is not an integer). The data of the LMFDB
\cite{lmfdb} (see \githubbare{inputs/global_newforms_cache.json}) show that every newform of weight $2$ and level $11$
or $121$ without complex multiplication, with any character, has analytic rank $0$. Hence $A(K)$ is finite.
\end{proof}

\begin{proposition}\label{prop:11x11}
The group $\tg{11}{11}$ is not a subgroup of the torsion of an elliptic curve over a number field of degree $10$.
\end{proposition}

\begin{proof}
Suppose the opposite, that $E[11]\subseteq E(K)$ for an elliptic curve $E$ over a number field $K$ of degree $10$.
Then $K=\Q(\zeta_{11})$. Let $P\in E(K)$ be a point of order $11$, let $x\in Y_1(11)(K)$ be the point given by
$(E,P)$, and let $\mathfrak p$ be a prime of $K$ above $23$. As $23\equiv1\pmod{11}$, the prime $23$ splits completely
in $K$, so the residue field of $\mathfrak p$ is $\F_{23}$. Every elliptic curve over $\F_{23}$ has at most
$(\sqrt{23}+1)^2<34<121$ points, so by \Cref{lem:badred} and \Cref{lem:additive} the curve $E$ has multiplicative
reduction at $\mathfrak p$, and $x$ reduces modulo $\mathfrak p$ to a cusp. The $10$ cusps of $X_1(11)$ are defined
over $K$, and their reductions modulo $\mathfrak p$ are the $10$ cusps of $X_1(11)_{\F_{23}}$, so the reduction of $x$
is the reduction of a cusp $c\in X_1(11)(K)$. The class $[x-c]$ lies in $J_1(11)(K)\simeq A(K)$, which is finite by
\Cref{lem:mxn-X111}, and its reduction modulo $\mathfrak p$ is $0$, so $[x-c]=0$ by Katz's result. As $X_1(11)$ has genus $1$, it follows that
$x=c$, which is a contradiction.
\end{proof}

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\subsection{The group \texorpdfstring{$\tg{3}{21}$}{Z/3Z x Z/21Z}}\label{subsec:mxn-3x21}

\begin{lemma}\label{lem:mxn-321}
Let $F=\Q(\zeta_3)$, let $X=X_1(3,21)$ over $F$, and let $\mathfrak p$ be the prime of $F$ above $2$, with residue
field $\F_4$.
\begin{itemize}
\item[a)] The curve $X$ has genus $25$, and $J_X(F)$ is finite.
\item[b)] The cusps of $X$ form Galois orbits over $F$ of degrees $1^{24},6^4$, and the closed cusps of $X_{\F_4}$
have degrees $1^{24},3^8$. In particular, every closed cusp of $X_{\F_4}$ of degree $1$ is the reduction of an
$F$-rational cusp.
\item[c)] The numbers of non-cuspidal closed points of $X_{\F_4}$ of degrees $1,\dots,5$ are $0$, $0$, $0$, $144$ and $216$.
\item[d)] For every effective cuspidal divisor $\overline C$ of degree $5$ on $X_{\F_4}$ we have $\ell(2\overline C)=1$.
\end{itemize}
\end{lemma}

\begin{proof}
Part a) is \cite[Theorem 4.1]{DS}. The Galois orbits of the cusps and their reductions are
\gitlink{continuation_20261029/replay_cusp_tate.py}{computed} from the description of the cusps in
\Cref{subsec:modular}: the Frobenius at $\mathfrak p$ acts as $\sigma_4$, which has order $3$ in
$\Gal(\Q(\zeta_{21})/F)$, so each of the $4$ orbits of length $6$ splits into $2$ closed cusps of degree $3$. Part c) is
a direct \gitlink{computations/cont_local_3_21_q4.m}{computation}. For d), there are
$\binom{28}{5}+8\binom{25}{2}=100680$ effective cuspidal divisors $\overline C=\sum_Pm_PP$ of degree $5$ on $X_{\F_4}$. As
explained after \cite[Lemma 5.3]{N69}, $\ell(2\overline C)=1$ if and only if the $10$ linear conditions on
$H^0(X_{\F_4},\Omega^1)$ given by the vanishing of the first $2m_P$ coefficients of the local expansions at each $P$ are
linearly independent, and we \gitlink{computations/cont_double_cusp_3_21_D5_dt.m}{compute} that this holds for all
$100680$ divisors.
\end{proof}

\begin{proposition}\label{prop:3x21}
The group $\tg{3}{21}$ is not a subgroup of the torsion of an elliptic curve over a number field of degree $10$.
\end{proposition}

\begin{proof}
Suppose the opposite, let $F$, $X$ and $\mathfrak p$ be as in \Cref{lem:mxn-321}, and let $D$ be the trace divisor,
an effective $F$-rational divisor of degree $5$ on $X$ with non-cuspidal support. By \Cref{lem:mxn-321} a) and d) and
\Cref{lem:doubled}, the reduction of $D$ modulo $\mathfrak p$ is not entirely cuspidal. By \Cref{lem:mxn-321} c) it
is $D'+C'$, where either $D'$ is one of the $144$ non-cuspidal closed points of degree $4$ and $C'$ is a cusp of
degree $1$, or $D'$ is one of the $216$ non-cuspidal closed points of degree $5$ and $C'=0$. In both cases $C'$ is $0$ or
the reduction of an $F$-rational cusp by \Cref{lem:mxn-321} b), so $13$ is admissible for $C'$, and
$[A_{13}(D')]=0$ (\Cref{subsec:sieve}).

Write $Q_3=7Q$ and $Q_7=3Q$. The $72$ level 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 $\pm1$, are defined over $F$, and they have $2$ orbits on the non-cuspidal closed
points of $X_{\F_4}$ of degree $4$ and $3$ orbits on those of degree $5$. For the points of degree $4$ we computed in
the proof of \cite[Proposition 12.2]{N69} that $A_{13}x$ is not principal for a representative $x$ of each of the
$2$ orbits (see \githubbare{continuation_20261029/degree_independent_321/sieve_3_21_D4_p2.m}), and we
\gitlink{computations/cont_sieve_3_21_D5.m}{compute} that $A_{13}x$ is not principal for a representative $x$ of each
of the $3$ orbits of points of degree $5$. This is a contradiction.
\end{proof}

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\subsection{The groups \texorpdfstring{$\tg{3}{24}$ and $\tg{4}{24}$}{Z/3Z x Z/24Z and Z/4Z x Z/24Z}}\label{subsec:mxn-3x24}

\begin{lemma}\label{lem:mxn-324-424}
\begin{itemize}
\item[a)] Let $F=\Q(\zeta_3)$, $X=X_1(3,24)$ over $F$, and let $\mathfrak p$ be a prime of $F$ above $7$, with
residue field $\F_7$. Then $X$ has genus $25$, $J_X(F)$ is finite and $\gon_\C X>5.71$. The cusps of $X$ form Galois
orbits over $F$ of degrees $1^{24},2^4,4^4$, the closed cusps of $X_{\F_7}$ have degrees $1^{24},2^{12}$, and those of
degree $1$ are the reductions of the $F$-rational cusps. The numbers of non-cuspidal closed points of $X_{\F_7}$ of
degrees $1,\dots,5$ are $0$, $0$, $48$, $456$ and $3456$.
\item[b)] Let $F=\Q(i)$, $X=X_1(4,24)$ over $F$, and let $\mathfrak p$ be a prime of $F$ above $5$, with residue field
$\F_5$. Then $X$ has genus $33$, $J_X(F)$ is finite and $\gon_\C X>7.61$. The cusps of $X$ form Galois orbits over
$F$ of degrees $1^{16},2^{16},4^4$, the closed cusps of $X_{\F_5}$ have degrees $1^{16},2^{24}$, and those of degree
$1$ are the reductions of the $F$-rational cusps. The numbers of non-cuspidal closed points of $X_{\F_5}$ of
degrees $1,\dots,5$ are $0$, $0$, $0$, $96$ and $704$.
\end{itemize}
\end{lemma}

\begin{proof}
The finiteness of $J_X(F)$ is \cite[Theorem 4.1]{DS}, and the gonality bounds follow from Abramovich's bound, as
$i(3,24)=576$ and $i(4,24)=768$. The Galois orbits of the cusps and their reductions are
\gitlink{continuation_20261029/replay_cusp_tate.py}{computed} from the description of the cusps in
\Cref{subsec:modular}, and the numbers of points are computed
\gitlink{computations/cont_local_3_24_q7.m}{for $X_1(3,24)$} and \gitlink{computations/cont_local_4_24_q5.m}{for
$X_1(4,24)$} directly.
\end{proof}

\begin{proposition}\label{prop:3x24-4x24}
The groups $\tg{3}{24}$ and $\tg{4}{24}$ are not subgroups of the torsion of an elliptic curve over a number field
of degree $10$.
\end{proposition}

\begin{proof}
Suppose first that $\tg{3}{24}\subseteq E(K)$ for an elliptic curve $E$ over a number field $K$ of degree $10$, let $F$, $X$ and $\mathfrak p$ be as in \Cref{lem:mxn-324-424} a), and
let $D$ be the trace divisor, of degree $5$. As $\gon_FX\geq\gon_\C X>5$, \Cref{prop:main} a) shows that the
reduction of $D$ modulo $\mathfrak p$ is not entirely cuspidal, so it is $D'+C'$ of type $[3]$, with $C'$ effective
cuspidal of degree $2$, of type $[4]$, with $C'$ a cusp of degree $1$, or of type $[5]$, with $C'=0$. Write $R_3=8Q$
and $R_8=3Q$. The $48$ level 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 $1$, $13$
and $72$ orbits on the non-cuspidal closed points of $X_{\F_7}$ of degrees $3$, $4$ and $5$. In the types $[4]$ and
$[5]$ the prime $13$ is admissible for $C'$, and we compute that $A_{13}x$ is not principal for the representatives
$x$ of the $13$ orbits of \gitlink{computations/cont_sieve_3_24_D5_lower_supports.m}{degree $4$} and of the $72$
orbits of \gitlink{computations/cont_sieve_3_24_D5_shard0.m}{degree $5$} (the latter computation is split into the
six scripts \texttt{cont\_sieve\_3\_24\_D5\_shard}$i$, $i=0,\dots,5$). In the type
$[3]$ the cuspidal part $C'$ may contain cusps of degree $2$, which need not be reductions of $F$-rational cuspidal
divisors. But $31\equiv7\pmod{24}$, so $31$ is admissible for every $C'$ by \Cref{lem:cuspHecke}, and we
\gitlink{computations/cont_sieve_3_24_cubic31.m}{compute} that $A_{31}x$ is not principal for a representative $x$ of
the single orbit of points of degree $3$. This excludes all the possible reductions.

Suppose now that $\tg{4}{24}\subseteq E(K)$ for an elliptic curve $E$ over a number field $K$ of degree $10$, and let $F$, $X$ and $\mathfrak p$ be as in \Cref{lem:mxn-324-424} b).
As $\gon_FX>7.61>5$, the reduction of the trace divisor modulo $\mathfrak p$ is not entirely cuspidal by
\Cref{prop:main} a), so it is of type $[4]$, with $C'$ a cusp of degree $1$, or of type $[5]$, with $C'=0$, and in
both cases $13$ is admissible for $C'$. The $64$ level 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 the
points of degree $4$ and $11$ orbits on the points of degree $5$, and we compute that $A_{13}x$ is not principal for
the representatives $x$ of all of them (see \gitlink{computations/cont_sieve_4_24_D5_d4_r1.m}{this script} for the
first orbit of points of degree $4$ and \gitlink{computations/cont_sieve_4_24_D5_d5_r1.m}{this script} for the first
orbit of points of degree $5$, and the analogous scripts \texttt{cont\_sieve\_4\_24\_D5\_d4\_r}$i$ and
\texttt{cont\_sieve\_4\_24\_D5\_d5\_r}$i$ for the others).
\end{proof}

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\subsection{The group \texorpdfstring{$\tg{4}{16}$}{Z/4Z x Z/16Z}}\label{subsec:mxn-4x16}
For $\tg{4}{16}$ the Hecke sieve at one prime does not suffice, and we compare the reductions at two primes as in
\cite[Section 12.4]{N69}, using the following lemma.

\begin{lemma}[{\cite[Lemma 12.4]{N69}}]\label{lem:mxn-matched}
Let $X$ be a curve over a number field $F$ with $J_X(F)$ 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 $J_X(F)$, let $H_i\subseteq J_X(k_i)$ 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 $J_X(k_1)/H_1$ 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{lemma}\label{lem:mxn-416}
Let $F=\Q(i)$, let $X=X_1(4,16)$ over $F$, and let $\mathfrak p_5$, $\mathfrak p_3$ and $\mathfrak p_{13}$ be primes of
$F$ above $5$, $3$ and $13$, with residue fields $\F_5$, $\F_9$ and $\F_{13}$.
\begin{itemize}
\item[a)] The curve $X$ has genus $13$, $J_X(F)$ is finite and $\gon_FX\geq8$.
\item[b)] The cusps of $X$ form $24$ Galois orbits over $F$, of degrees $1^{16},2^4,4^4$. Modulo $\mathfrak p_5$ and
modulo $\mathfrak p_{13}$ each of them reduces to a single closed cusp of the same degree, and modulo $\mathfrak p_3$
the closed cusps have degrees $1^{24},2^8$.
\item[c)] The numbers of non-cuspidal closed points of degrees $1,\dots,5$ are $0$, $0$, $64$, $160$ and $512$ on $X_{\F_5}$ and
$0$, $0$, $320$, $1808$ and $11904$ on $X_{\F_9}$.
\item[d)] Let $H_5$ and $H_9$ be the reductions of $\ClCusp_F X$ modulo $\mathfrak p_5$ and $\mathfrak p_3$. Then
\begin{align*}
J_X(\F_5)&\simeq(\torz2)^6\times(\torz4)^2\times(\torz{40})^2\times(\torz{80})^2,\\
J_X(\F_9)&\simeq(\torz2)^8\times(\torz4)^2\times(\torz8)^2\times(\torz{80})^4,\\
H_5\simeq H_9\simeq\ClCusp_F X&\simeq(\torz2)^3\times(\torz4)^2\times\torz{20}\times(\torz{40})^2\times\torz{80},
\end{align*}
and $J_X(\F_5)/H_5\simeq(\torz2)^5$. In particular $2J_X(F)\subseteq\ClCusp_F X$.
\item[e)] The group $G$ of the $64$ level 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 $\pm1$, consists of automorphisms defined over $F$,
and it has $5$, $33$ and $186$ orbits on the non-cuspidal closed points of $X_{\F_9}$ of degrees $3$, $4$ and $5$.
\end{itemize}
\end{lemma}

\begin{proof}
Part a) is \cite[Theorem 4.1]{DS} and \cite[Table 1]{NV}. In b), $\sigma_5$ and $\sigma_{13}$ generate
$\Gal(\Q(\zeta_{16})/F)$, while the Frobenius $\sigma_9$ at $\mathfrak p_3$ has order $2$, which splits each of the
$4$ orbits of degree $2$ into $2$ cusps of degree $1$ and each of the $4$ orbits of degree $4$ into $2$ cusps of
degree $2$. We identify the reductions of each Galois orbit at the three 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 numbers of points in c), the
groups in d) and the orbits in e) are \gitlink{computations/cont_diagonal_4_16_D5_v2.m}{computed} directly (see also
\githubbare{computations/cont_local_4_16_q5.m}). As $5$ is odd and unramified in $F$, reduction modulo $\mathfrak p_5$
is injective on $J_X(F)$, and the last claim of d) follows, as does $H_5\simeq\ClCusp_F X$. Similarly
$H_9\simeq\ClCusp_F X$.
\end{proof}

\begin{proposition}\label{prop:4x16}
The group $\tg{4}{16}$ is not a subgroup of the torsion of an elliptic curve over a number field of degree $10$.
\end{proposition}

\begin{proof}
Suppose the opposite, let $F$, $X$, $\mathfrak p_5$, $\mathfrak p_3$, $\mathfrak p_{13}$, $H_5$, $H_9$ and $G$ be as
in \Cref{lem:mxn-416}, and let $D$ be the trace divisor, of degree $5$. As $\gon_FX\geq8>5$, by \Cref{prop:main} a)
the reductions of $D$ modulo $\mathfrak p_5$ and modulo $\mathfrak p_3$ are not entirely cuspidal. By
\Cref{lem:mxn-416} c), modulo $\mathfrak p_5$ the reduction of $D$ is a closed point of degree $3$ plus an effective
cuspidal divisor of degree $2$, a point of degree $4$ plus one of the $16$ cusps of degree $1$, or a point of degree
$5$, which gives
\[
64\Bigl(\tbinom{17}{2}+4\Bigr)+160\cdot16+512=12032
\]
possibilities. Modulo $\mathfrak p_3$ the reduction has the same types. Replacing $D$ by $gD$ for a suitable $g\in G$,
we may assume that its non-cuspidal part modulo $\mathfrak p_3$ is one of the representatives of the orbits of
\Cref{lem:mxn-416} e), and this does not affect the set of the $12032$ possibilities modulo $\mathfrak p_5$, which is
stable under $G$. Allowing all the cuspidal parts, this gives
\[
5\Bigl(\tbinom{25}{2}+8\Bigr)+33\cdot24+186=2518
\]
possibilities for the reduction $\overline D_3$ of $D$ modulo $\mathfrak p_3$.

Fix an $F$-rational cusp $c_0$, and let $\overline D_5$ be the reduction of $D$ modulo $\mathfrak p_5$. By
\Cref{lem:mxn-matched} with $\mathfrak p_1=\mathfrak p_5$, $\mathfrak p_2=\mathfrak p_3$ and $e=2$ we have
$\psi(2[\overline D_5-5\overline{c_0}])=2[\overline D_3-5\overline{c_0}]$. We
\gitlink{computations/cont_diagonal_4_16_D5_v2.m}{compute} that only $158$ of the $2518$ possibilities for
$\overline D_3$ satisfy this relation for one of the $12032$ possibilities for $\overline D_5$: $157$ consisting of a
point of degree $3$ and an effective cuspidal divisor of degree $2$, and one point of degree $5$. In each of these
$158$ cases the class $2[D-5c_0]$, which lies in $\ClCusp_F X$ by \Cref{lem:mxn-416} d), is determined by its
reduction modulo $\mathfrak p_3$, so $2D\sim L_v$, where
\[
L_v=10c_0+\sum_{j=1}^{24}v_j\bigl(C_j-\deg(C_j)c_0\bigr)
\]
for an explicit $v\in\Z^{24}$, the $C_j$ being the $24$ Galois orbits of cusps over $F$. Reducing modulo $\mathfrak
p_{13}$, the divisor $2\overline D_{13}$, where $\overline D_{13}$ is the reduction of $D$, is an effective
$\F_{13}$-rational divisor in the linear system $|\overline{L_v}|$ all of whose multiplicities are even. We
\gitlink{computations/cont_M4N16_D5_RR13.m}{compute} that $\ell(\overline{L_v})=1$ in $150$ of the $158$ cases and
$\ell(\overline{L_v})=2$ in the other $8$, and that each of the $150+8\cdot14=262$ effective $\F_{13}$-rational divisors
in these linear systems has a point of odd multiplicity. This is a contradiction.
\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{The group \texorpdfstring{$\tg{4}{20}$}{Z/4Z x Z/20Z}}\label{subsec:mxn-4x20}
For $\tg{4}{20}$ the Hecke sieve leaves $4$ possible reductions whose non-cuspidal parts are fixed by an involution,
and we exclude these with \Cref{lem:section} and \Cref{lem:stab}.

\begin{lemma}\label{lem:mxn-420}
Let $F=\Q(i)$, let $X=X_1(4,20)$ over $F$, and let $\mathfrak p$ be the prime of $F$ above $3$, with residue field
$\F_9$.
\begin{itemize}
\item[a)] The curve $X$ has genus $25$, $J_X(F)$ is finite and $\gon_\C X>5.71$.
\item[b)] The cusps of $X$ form Galois orbits over $F$ of degrees $1^{24},4^6$, the closed cusps of $X_{\F_9}$ have
degrees $1^{24},2^{12}$, and those of degree $1$ are the reductions of the $F$-rational cusps.
\item[c)] The numbers of non-cuspidal closed points of $X_{\F_9}$ of degrees $1,\dots,5$ are $0$, $96$, $320$, $888$ and $11904$.
\item[d)] Let $G$ be the group of the $96$ level 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
$(-I,-1)$. Its elements are defined over $F$, and it has $2$, $4$, $12$ and $124$ orbits on the non-cuspidal closed
points of $X_{\F_9}$ of degrees $2$, $3$, $4$ and $5$.
\end{itemize}
\end{lemma}

\begin{proof}
The finiteness of $J_X(F)$ is \cite[Theorem 4.1]{DS}, and the gonality bound follows from Abramovich's bound, as
$i(4,20)=576$. The Galois orbits of the cusps and their reductions are
\gitlink{continuation_20261029/replay_cusp_tate420.py}{computed} from the description of the cusps in
\Cref{subsec:modular}, and the numbers of points are \gitlink{computations/cont_local_4_20_q9.m}{computed}
directly. The orbits of $G$ are computed in the scripts linked in the proof of \Cref{prop:4x20}.
\end{proof}

\begin{proposition}\label{prop:4x20}
The group $\tg{4}{20}$ is not a subgroup of the torsion of an elliptic curve over a number field of degree $10$.
\end{proposition}

\begin{proof}
Suppose the opposite, let $F$, $X$, $\mathfrak p$ and $G$ be as in \Cref{lem:mxn-420}, and let $D$ be the trace
divisor, of degree $5$. As $\gon_FX\geq\gon_\C X>5$ by \Cref{lem:mxn-420} a), the reduction of $D$ modulo $\mathfrak p$ is not
entirely cuspidal by \Cref{prop:main} a), so by \Cref{lem:mxn-420} c) it is $D'+C'$ with $D'$ of type $[2]$, $[3]$,
$[4]$, $[5]$, $[2,2]$ or $[2,3]$ and $C'$ effective cuspidal of degree $3$, $2$, $1$, $0$, $1$ or $0$, respectively. In
the type $[2,2]$ the two points may coincide. As $29\equiv9\pmod{20}$, the prime $29$ is admissible for every $C'$ by
\Cref{lem:cuspHecke}, and by \Cref{lem:mxn-420} b) every prime, in particular $13$, is admissible for $C'$ when $C'$ is
$0$ or a cusp of degree $1$. Replacing $D$ by $gD$ for some $g\in G$, we may assume that $D'$ is a representative of
an orbit of $G$, and in the types $[2,3]$ and $[2,2]$ that the quadratic point, respectively one of the two quadratic
points, is such a representative. This gives the numbers of cases in the following table, and we compute that $A_qD'$
is not principal for one of the listed primes $q$ in all but $4$ of them.
\begin{center}
\begin{tabular}{cccc}
\toprule
type of $D'$ & cases & $q$ & remaining\\
\midrule
$[2]$ & $2$ & $29$ & $0$\\
$[3]$ & $4$ & $29$ & $0$\\
$[4]$ & $12$ & $29$, $13$ & $2$\\
$[5]$ & $124$ & $13$ & $0$\\
$[2,2]$ & $2\cdot96=192$ & $29$, $13$ & $2$\\
$[2,3]$ & $2\cdot320=640$ & $13$ & $0$\\
\bottomrule
\end{tabular}
\end{center}
For the types $[2]$, $[3]$ and $[4]$ we \gitlink{computations/cont_sieve_4_20_lower.m}{compute} $A_{29}D'$, and for the
$4$ points of degree $4$ that remain we \gitlink{computations/cont_sieve_4_20_quartic_followup.m}{compute} $A_{13}D'$.
For the type $[5]$ we \gitlink{computations/cont_sieve_4_20_quintic_shard0.m}{compute} $A_{13}D'$ (the
computation is split into twelve scripts \texttt{cont\_sieve\_4\_20\_quintic\_shard}$i$, $i=0,\dots,11$). For the types
$[2,2]$ and $[2,3]$ we \gitlink{computations/cont_sieve_4_20_mixed_shard0.m}{compute} $A_{29}D'$ and $A_{13}D'$,
respectively (in eight scripts \texttt{cont\_sieve\_4\_20\_mixed\_shard}$i$, $i=0,\dots,7$), and for the $8$ divisors of type $[2,2]$ that remain we
\gitlink{computations/cont_sieve_4_20_pairs_followup.m}{compute} $A_{13}D'$.

In each of the $4$ remaining cases the reduction of $D$ is $B+\overline c$, where $B$ is a closed point of degree $4$ or
a sum of two closed points of degree $2$, and $\overline c$ is a closed cusp of degree $1$. We
\gitlink{computations/cont_sieve_4_20_pairs_followup.m}{compute} that the stabilizers in $G$ of the two
remaining points of degree $4$ have order $8$ and both contain the element with
$\begin{pmatrix}a&b\\c&d\end{pmatrix}=-I$ and $u=1$, and that the stabilizers of the two remaining divisors of type
$[2,2]$ have order $4$ and contain the elements with $u=1$ and
\[
\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}3&2\\2&3\end{pmatrix},\quad\text{respectively}\quad
\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}1&2\\0&1\end{pmatrix}.
\]
Let $g$ be this element in the case at hand, a level automorphism of order $2$ defined over $F$, which fixes $B$. We
\gitlink{additions/C4x20/rr420.m}{compute} that $\ell(B+\overline c_1+\overline c_2)=1$ on $X_{\F_9}$ for each of the $4$
divisors $B$ and all the closed cusps $\overline c_1,\overline c_2$ of degree $1$, possibly equal ($1200$ divisors in
total). As $g\overline c$ is again a closed cusp of degree $1$, we have $\ell(B+\overline c+g\overline c)=1$, and
\Cref{lem:section} with $r=2$ gives $g\overline c=\overline c$. So $g$ fixes the reduction $B+\overline c$ of $D$, and as
$J_X(F)$ is finite and $\gon_FX\geq\gon_\C X>5$, \Cref{lem:stab} a), with $C=0$, gives $gD=D$. By \Cref{lem:CM}, the
points in the support of $D$ correspond to elliptic curves without complex multiplication, so \Cref{lem:orbits},
applied to the group of order $2$ generated by $g$, shows that $\deg D$ is even. This contradicts $\deg D=5$.
\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{The group \texorpdfstring{$\tg{3}{27}$}{Z/3Z x Z/27Z}}\label{subsec:mxn-3x27}

\begin{lemma}\label{lem:mxn-327}
Let $F=\Q(\zeta_3)$, let $X=X_1(3,27)$ over $F$, let $\mathfrak p$ be the prime of $F$ above $2$, with residue field
$\F_4$, and let $\pi:X\to Y=X_1(27)_F$ be the map $(E,P,Q)\mapsto(E,Q)$.
\begin{itemize}
\item[a)] The curve $X$ has genus $46$, $\gon_\C X>9.64$, and $J_X(F)$ is finite of odd order.
\item[b)] Every Galois orbit of cusps of $X$ over $F$ reduces modulo $\mathfrak p$ to a single closed cusp of the same
degree. These degrees are $1^{27},3^6,9^3$.
\item[c)] The numbers of non-cuspidal closed points of $X_{\F_4}$ of degrees $1,\dots,5$ are $0$, $0$, $0$, $0$ and $162$.
\item[d)] The map $\pi$ sends the $162$ non-cuspidal closed points of degree $5$ of $X_{\F_4}$ onto $54$ closed points
of degree $5$ of $Y_{\F_4}$, which form $6$ orbits under the diamond operators.
\end{itemize}
\end{lemma}

\begin{proof}
For a), the gonality bound is Abramovich's bound with $i(3,27)=972$, and $J_X(F)$ is finite by
\cite[Theorem 4.1]{DS}. Over $F$ the curve $X$ is isomorphic to $X_H(81)$, the quotient of $X_1(81)$ by the subgroup
$H=\{1,28,55\}$ of the diamond operators \cite[Section 2]{DS}, and this isomorphism comes from the moduli
interpretation, so it is compatible with reduction at the primes of $F$ above $2$ and $37$. The Jacobian of $X_H(81)$
has dimension $46$, and we \gitlink{computations/d31b_m327_torsion_gcd.m}{compute} the characteristic polynomial of
Frobenius at $37$ on it and find that $\#J_X(\F_{37})$ is odd. As $37$ splits in $F$ and reduction modulo a
prime above $37$ is injective on $J_X(F)$ by Katz's result, $J_X(F)$ has odd order.

The group $\Gal(\Q(\zeta_{27})/F)$ is cyclic of order $9$, and it is generated by the Frobenius $\sigma_4$ at
$\mathfrak p$, as $4$ has order $9$ modulo $27$. So $\sigma_4$ acts transitively on each Galois orbit of cusps over
$F$, which proves the first claim of b). The degrees are
\gitlink{computations/d31b_m327_cusp_parity_v2.py}{computed} from the description of the cusps in
\Cref{subsec:modular}. We \gitlink{computations/d31b_m3n27_frob_preflight.m}{compute} c) from the Frobenius
polynomial of the Jacobian of $X_H(81)$ at $2$, and d) by \gitlink{computations/d31b_m3n27_physical_source.m}{running}
through all the elliptic curves over $\F_{4^5}$ with a subgroup isomorphic to $\tg{3}{27}$ and all their level
structures.
\end{proof}

\begin{proposition}\label{prop:3x27}
The group $\tg{3}{27}$ is not a subgroup of the torsion of an elliptic curve over a number field of degree $10$.
\end{proposition}

\begin{proof}
Suppose the opposite, let $F$, $X$, $\mathfrak p$, $\pi$ and $Y$ be as in \Cref{lem:mxn-327}, and let $D$ be the trace
divisor, an effective $F$-rational divisor of degree $5$ on $X$ with non-cuspidal support.

Suppose first that the reduction of $D$ modulo $\mathfrak p$ is an effective cuspidal divisor $\overline C$. By
\Cref{lem:mxn-327} b), $\overline C$ is the reduction of an effective $F$-rational cuspidal divisor $C$. The class
$[D-C]\in J_X(F)$ reduces to $0$, so $2[D-C]=0$ by Parent's result, and $[D-C]=0$ as $J_X(F)$ has odd order. So
$D\sim C$ and $D\neq C$, which contradicts $\gon_FX\geq\gon_\C X>5$ (\Cref{lem:closed} a)).

So by \Cref{lem:mxn-327} c) the reduction of $D$ is one of the $162$ non-cuspidal closed points $x$ of degree $5$ of
$X_{\F_4}$, and the reduction of the effective $F$-rational divisor $\pi_*D$ of degree $5$ on $Y$ is the closed point
$N=\pi(x)$ of degree $5$. As $\pi^*:J_Y\to J_X$ has finite kernel and $J_X(F)$ is finite, $J_Y(F)$ is finite. Let $c_0$ be a rational cusp of $Y$. As $7$ splits in $F$ and $7\nmid 2\cdot27$, the
operator $A_7=T_7-7\dm7-1$ on $Y$ kills $J_Y(F)$ and $c_0$ (\Cref{prop:tools} b) and c)), so
$[A_7(\pi_*D)]=A_7[\pi_*D-5c_0]=0$, and $A_7N$ is principal on $Y_{\F_4}$. As $A_7$ commutes with the diamond
operators, by \Cref{lem:mxn-327} d) it suffices to check this for $6$ representatives $N$, and we
\gitlink{computations/d31b_m3n27_quotient_A7_v5.m}{compute} that $\ell(A_7N)=0$ for each of them. This is a
contradiction.
\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{The group \texorpdfstring{$\tg{3}{33}$}{Z/3Z x Z/33Z}}\label{subsec:mxn-3x33}
The group $J_1(3,33)(\Q(\zeta_3))$ is infinite, as over $\Q(\zeta_3)$ the curve $X_1(3,33)$ is isomorphic to a
quotient of $X_1(99)$ \cite[Section 2]{DS}, which maps to $X_0(99)$, and the elliptic curve \lmfdbec{99}{a}{2}, a
quotient of $J_0(99)$, has rank $1$. So \Cref{prop:main} and the Hecke sieve cannot be applied to $X_1(3,33)$
directly. Let $h=\dm{10}$, which is the involution $(E,P,Q)\mapsto(E,P,10Q)$ of $X_1(3,33)$, as $10\equiv1\pmod3$
and $10^2\equiv1\pmod{33}$. We use the part of the Jacobian on which $h$ acts by $-1$, which has finitely many points
over $\Q(\zeta_3)$ (\Cref{lem:mxn-333image}), and for the entirely cuspidal reduction modulo $2$ a formal immersion
argument over the unramified quadratic extension of $\Q_2$, which is a variant of \Cref{prop:cartier} with $\Cart^2$
in place of $\Cart$.

\begin{lemma}\label{lem:mxn-333image}
Let $F=\Q(\zeta_3)$, $X=X_1(3,33)$ and $h=\dm{10}$. Then $(h_*-1)J_X(F)$ is finite.
\end{lemma}

\begin{proof}
Over $F$ the curve $X$ is isomorphic to the quotient $X_H(99)$ of $X_1(99)$ by the subgroup
$H=\{1,34,67\}$ of the diamond operators \cite[Section 2]{DS}, and $h$ corresponds to the diamond operator $\dm{10}$ of $X_H(99)$. The image $B$ of
$\dm{10}_*-1$ on the Jacobian of $X_H(99)$ is an abelian variety over $\Q$, and the newforms occurring in it are those
whose character $\varepsilon$ satisfies $\varepsilon(34)=1$ and $\varepsilon(10)=-1$. Such characters factor through
$(\torz{33})^\times$, so their conductors divide $33$. We
\gitlink{computations/d31b_m333_antih10_winding.m}{compute} that they form two Galois orbits, of orders $2$ and $10$,
and that the corresponding spaces of cuspidal modular symbols with sign $+1$ have dimension $8$ over the field of
values of the character. This accounts for $8+4\cdot8=40=81-41$ dimensions, where $81$ and $41$ are the genera of
$X_H(99)$ and of its quotient by $\dm{10}$, which we also compute, so $B$ has dimension $40$. We also compute that in
both cases the winding element generates the whole space as a module over the Hecke algebra: the images of the winding
element under the products of $T_2,T_3,\dots,T_{23}$ span the space modulo a prime of degree $1$ of the field of
values, above $5$, respectively $11$. So $L(f,1)\neq0$ for every newform $f$ occurring in $B$.

Let $\chi_{-3}$ be the quadratic character of $F$. By the argument of the proof of \cite[Theorem 4.1]{DS}, $B(F)$ is
finite if $L(f,1)\neq0$ and $L(f\otimes\chi_{-3},1)\neq0$ for all such $f$, and as the conductor of $\varepsilon$
divides $33$ and $99=3\cdot33$, the twist $f\otimes\chi_{-3}$ again lies in $S_2(\Gamma_0(99),\varepsilon)$ (see
loc.\ cit.), so it again occurs in $B$. Hence $B(F)$ is finite, and so is $(h_*-1)J_X(F)$, as $(h_*-1)J_X$ corresponds
to $B$ under the isomorphism.
\end{proof}

Let $R$ be the ring of integers of the unramified quadratic extension of $\Q_2$, with residue field $k=\F_4$. For an
abelian scheme $A$ over $R$ of dimension $g$ we choose coordinates $z=(z_1,\dots,z_g)$ of its formal group. The points
of $A(R)$ which reduce to $0$ are those with coordinates in $2R$ (see the proof of \cite[Lemma 7.1]{N69}), and for such a
point $P$ we write $z(P)=2u$ with $u\in R^g$, and call the reduction $\tau(P):=\overline u$ the \textit{first order
tangent} of $P$. We view it as a tangent vector of $A_k$ at
$0$, which does not depend on the choice of coordinates. We write $V_4$ for the Verschiebung of $A_k$ relative to $k$. It
satisfies $V_4\circ\Frob_4=[4]$, where $\Frob_4$ is the Frobenius endomorphism of $A_k$, which raises the coordinates
to the $4$-th power. If $A$ is the Jacobian of a curve $\mathcal X$ over $R$, then $V_4^*=\Cart^2$ on
$H^0(A_k,\Omega^1)=H^0(\mathcal X_k,\Omega^1)$, where $\Cart$ is the Cartier operator \cite{oda}.

\begin{lemma}\label{lem:mxn-F4tangent}
Let $R$ be the ring of integers of the unramified quadratic extension of $\Q_2$, with residue field $k=\F_4$, let $A$
be an abelian scheme over $R$, and let $P\in A(R)$ be a torsion point which reduces to $0$.
\begin{itemize}
\item[a)] We have $2P=0$, and if $\tau(P)=0$, then $P=0$.
\item[b)] For every $\omega\in H^0(A_k,\Omega^1)$ we have $(V_4^*\omega-\omega)(\tau(P))=0$.
\end{itemize}
\end{lemma}

\begin{proof}
Multiplication by $2$ maps the points with coordinates in $2R$ to points with coordinates in $4R$, and a non-zero point
$v$ with coordinates in $4R$ is not killed by $2$: 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
$2s>s+1$. Multiplication by an odd integer is an automorphism of the formal group, so the points with coordinates in
$4R$ form a torsion-free group. As $2P$ has coordinates in $4R$ and is torsion, $2P=0$, and if $\tau(P)=0$, then $P$ has coordinates in $4R$, so $P=0$. This proves a).

For b), write $[2](z)=2z+\mathcal Q(z)+\mathcal S(z)$, where the coordinates of $\mathcal Q$ are homogeneous of degree
$2$ and $\mathcal S$ has only terms of degree at least $3$, all with coefficients in $R$. As $[2]=V\circ\Frob$ on
$A_k$, where $\Frob:A_k\to A_k^{(2)}$ is the relative Frobenius and $V$ the Verschiebung, the reduction of $\mathcal Q$
is $\overline{\mathcal Q}(z)=\overline M(z_1^2,\dots,z_g^2)^t$, where $\overline M$ is the matrix of the linear part of
$V$ (as in the proof of \cite[Lemma 7.1]{N69}). As $2P=0$, we have $0=[2](2u)\equiv4u+4\mathcal Q(u)\pmod8$, so
$\overline u=\overline M\,\overline u^{[2]}$, where $\overline u^{[2]}$ is obtained by squaring the coordinates of
$\overline u$. Applying the Frobenius of $k$ to this identity and using $\overline u^{[4]}=\overline u$, which holds
as $\overline u\in\F_4^g$, we get $\overline u^{[2]}=\overline M^{[2]}\,\overline u$, and hence
$\overline u=\overline M\,\overline M^{[2]}\,\overline u$. The matrix $\overline M^{[2]}$ is the matrix of the linear
part of $V^{(2)}:A_k^{(4)}\to A_k^{(2)}$, and $A_k^{(4)}=A_k$, so $\overline M\,\overline M^{[2]}$ is the matrix of the
tangent map of $V_4=V\circ V^{(2)}$ at $0$. Hence $\tau(P)$ is fixed by the tangent map of $V_4$, which is b).
\end{proof}

\begin{lemma}\label{lem:mxn-relformal}
Let $R$ be the ring of integers of the unramified quadratic extension of $\Q_2$, with residue field $k=\F_4$. Let
$\mathcal X$ be a smooth proper curve over $R$ with Jacobian $A$, let $h$ be an automorphism of $\mathcal X$ of
order $2$, and put $\alpha=h_*-1\in\End(A)$. Let $\Sigma\subseteq\mathcal X$ be a closed subscheme which is finite
\'etale over $R$ and stable under $h$, such that $h$ fixes no closed point of $\Sigma_k$. Let $\overline C$ be an
effective divisor of degree $d$ on $\mathcal X_k$ supported on $\Sigma_k$. For each pair $\{c,hc\}$ of closed points
of $\Sigma_k$ let $m_c$ be the smaller of the multiplicities of $c$ and $hc$ in $\overline C$, and put
$\overline B=\sum m_c(c+hc)$, the sum being over these pairs, $\overline C_0=\overline C-\overline B$ and
$k_0=\deg\overline B/2$. Let $C$ and $C_0$ be the effective divisors on $\mathcal X$ supported on $\Sigma$ which reduce
to $\overline C$ and $\overline C_0$ (they exist and are unique, as $\Sigma$ is finite \'etale over $R$), and put
\[
U=(\Cart^2-1)\,\alpha^*H^0(\mathcal X_k,\Omega^1)\subseteq H^0(\mathcal X_k,\Omega^1).
\]
Suppose that the restriction map $U\to H^0(\overline C,\Omega^1|_{\overline C})$ has rank $d-k_0$. If
$D\in\mathcal X^{(d)}(R)$ reduces to $\overline C$ and $\alpha[D-C]$ is torsion, then $C_0\leq D$.
\end{lemma}

\begin{proof}
Note first that $h^*$ commutes with $\Cart$, so $U=\alpha^*(\Cart^2-1)H^0(\mathcal X_k,\Omega^1)\subseteq
\alpha^*H^0(\mathcal X_k,\Omega^1)$. Choose one point $c$ in each pair $\{c,hc\}$ and put $\overline G=\sum m_cc$. Then
$\overline B=\overline G+h\overline G$ and $\deg\overline G=k_0$. Let $G_0$ be the effective divisor supported on
$\Sigma$ which reduces to $\overline G$. Consider the morphisms
\[
f:\mathcal X^{(d)}\to A,\quad E\mapsto\alpha[E-C],\qquad\qquad \iota:\mathcal X^{(k_0)}\to\mathcal X^{(d)},\quad
G\mapsto C_0+G+hG.
\]
As $C=C_0+G_0+hG_0$ and $h_*(G+hG)=G+hG$, we have $f\circ\iota=0$. Let $\mathcal W$ and $\mathcal T$ be the formal
completions of $\mathcal X^{(d)}$ at $\overline C$ and of $\mathcal X^{(k_0)}$ at $\overline G$, which are formally smooth
over $R$ of relative dimensions $d$ and $k_0$.

We first show that $\iota$ induces a closed immersion $\mathcal T\to\mathcal W$. We may check this after an unramified
extension of $R$, so we may assume that the points of $\Sigma_k$ are rational. Then $\mathcal W$ is the product, over
the points $P$ in the support of $\overline C$, of the formal completions of the symmetric powers of $\mathcal X$ at
$n_P P$, where $n_P$ is the multiplicity of $P$ in $\overline C$, and the latter are described by the coefficients of a
monic polynomial of degree $n_P$ in a local parameter $t_P$ at $P$. We choose $t_{hc}=t_c\circ h^{-1}$. For a pair
$\{c,hc\}$, the polynomials of $\iota(G)$ at $c$ and at $hc$ are $\gamma_c\cdot\gamma$ and $\gamma_{hc}\cdot\gamma$,
where $\gamma$ is the polynomial of $G$ at $c$ and $\gamma_c,\gamma_{hc}$ are those of $C_0$ at $c$ and $hc$. By the
choice of $m_c$, one of $\gamma_c$ and $\gamma_{hc}$ is equal to $1$, so the coefficients of $\gamma$ are among the
coordinates of $\iota(G)$. This proves the claim. In particular, the tangent space $\mathcal T_0$ of $\iota(\mathcal T)_k$
at $\overline C$ has dimension $k_0$.

The cotangent space of $\mathcal W_k$ at $\overline C$ is $H^0(\overline C,\Omega^1|_{\overline C})$, and the
cotangent map of $f$ at $\overline C$ is $\omega\mapsto\alpha^*\omega|_{\overline C}$ (see the proof of
\cite[Proposition 7.3]{N69}). As $f\circ\iota=0$, the space $\mathcal T_0$ is contained in the annihilator of the
restriction of $\alpha^*H^0(\mathcal X_k,\Omega^1)$ to $\overline C$, hence in the annihilator of the restriction of $U$.
By assumption the latter has dimension $d-(d-k_0)=k_0$, so it is equal to $\mathcal T_0$.

Let $s=(s_1,\dots,s_d)$ be coordinates of $\mathcal W$ over $R$ with $s(C)=0$, choose coordinates $z$ of the formal
group of $A$ as above, and write $z(f(E))=\Phi(s(E))$ with $\Phi\in R[[s]]^{\dim A}$ and $\Phi(0)=0$. As $D$ reduces to $\overline C$, we have $s(D)=2v$ with $v\in R^d$, and $\overline v$
is a tangent vector of $\mathcal W_k$ at $\overline C$. The point $f(D)$ is torsion and reduces to $0$, and
$z(f(D))=\Phi(2v)\equiv2Jv\pmod4$, where $J$ is the Jacobian matrix of $\Phi$ at $0$, so $\tau(f(D))=df(\overline v)$.
By \Cref{lem:mxn-F4tangent} b), every $\omega\in H^0(\mathcal X_k,\Omega^1)$ satisfies
\[
0=\bigl((\Cart^2-1)\omega\bigr)\bigl(df(\overline v)\bigr)=\bigl(\alpha^*(\Cart^2-1)\omega\bigr)|_{\overline C}(\overline v),
\]
so $\overline v$ is annihilated by the restriction of $U$, and $\overline v\in\mathcal T_0$. Let $t$ be coordinates of
$\mathcal T$ over $R$ with $t(G_0)=0$. As $\overline v$ lies in the image of the tangent map of $\iota$, there is $w\in
R^{k_0}$ such that the point $G'\in\mathcal T(R)$ with $t(G')=2w$ satisfies $s(\iota(G'))\equiv2v=s(D)\pmod4$. Then
$z(f(D))=\Phi(s(D))\equiv\Phi(s(\iota(G')))=z(f(\iota(G')))=0\pmod4$, as $\Phi$ has coefficients in $R$, so
$\tau(f(D))=0$, and $f(D)=0$ by \Cref{lem:mxn-F4tangent} a).

Finally, the restriction of $\alpha^*H^0(\mathcal X_k,\Omega^1)$ to $\overline C$ has rank at least $d-k_0$, so we can
choose $d-k_0$ linear combinations $\phi_1,\dots,\phi_{d-k_0}$ of the coordinates of $\Phi$ whose differentials at $0$ are
linearly independent modulo $2$. By the implicit function theorem for formal power series, the closed formal
subscheme $\mathcal Z\subseteq\mathcal W$ defined by $\phi_1=\dots=\phi_{d-k_0}=0$ is formally smooth over $R$ of
relative dimension $k_0$. It contains $\iota(\mathcal T)$, which is also formally smooth of relative dimension $k_0$, and
the tangent spaces of $\mathcal Z_k$ and $\iota(\mathcal T)_k$ at $\overline C$ coincide. So $\iota(\mathcal T)=\mathcal
Z$, as a surjective homomorphism between power series rings over $R$ in $k_0$ variables is an isomorphism. As
$f(D)=0$, we have $D\in\mathcal Z(R)=\iota(\mathcal T(R))$, so $D=C_0+G+hG$ for some $G$, and $C_0\leq D$.
\end{proof}

\begin{lemma}\label{lem:mxn-333}
Let $F=\Q(\zeta_3)$, let $X=X_1(3,33)$ over $F$, let $h=\dm{10}$, let $\mathfrak p$ be the prime of $F$ above $2$, with
residue field $\F_4$, and let $\pi:X\to Y=X_1(33)_F$ be the map $(E,P,Q)\mapsto(E,Q)$.
\begin{itemize}
\item[a)] The numbers of non-cuspidal closed points of $X_{\F_4}$ of degrees $1,\dots,5$ are $0$, $0$, $0$, $0$ and $264$, and the
closed cusps of $X_{\F_4}$ have degrees $1^{40},5^8$. The involution $h$ fixes no closed cusp of $X_{\F_4}$, and it
interchanges the cusps of degree $1$ in $20$ pairs and those of degree $5$ in $4$ pairs.
\item[b)] Let $U=(\Cart^2-1)(h^*-1)H^0(X_{\F_4},\Omega^1)$. Then $\dim U=36$. For every effective cuspidal divisor
$\overline C$ of degree $5$ on $X_{\F_4}$, the restriction map $U\to H^0(\overline C,\Omega^1|_{\overline C})$ has rank
$5-k_0$, where $k_0$ is defined as in \Cref{lem:mxn-relformal}.
\item[c)] The map $\pi$ sends the $264$ non-cuspidal closed points of degree $5$ of $X_{\F_4}$ onto $88$ closed points
of degree $5$ of $Y_{\F_4}$, which form $12$ orbits under the diamond operators.
\end{itemize}
\end{lemma}

\begin{proof}
By \cite[Lemma 7.2]{N69}, the diamond operator $\dm{10}$ of $X_1(33)$ fixes no geometric cusp in characteristic $2$,
as $10\equiv1\pmod3$ and $10\equiv-1\pmod{11}$, and as $\pi\circ h=\dm{10}\circ\pi$, the involution $h$ fixes no
geometric cusp of $X_{\F_4}$. As $h$ commutes with the Frobenius, it permutes the closed cusps of each degree, and it
fixes none of them, as an involution of the $5$ geometric points of a closed cusp of degree $5$ has a fixed point. This
gives the pairs in a), once we know the degrees of the closed cusps, which we
\gitlink{independent_checks/cusp_degrees.py}{compute} from the description of the Galois action on the cusps. We
\gitlink{computations/d31b_m3n33_frob_preflight.m}{compute} the numbers of points in a) from the Frobenius
polynomial of the Jacobian of $X_H(99)$ at $2$, with $H=\{1,34,67\}$ as in the proof of \Cref{lem:mxn-333image}. For b) we \gitlink{computations/d31b_m333_kummer_frame.m}{compute} a model of $X_{\F_4}$ as a
cyclic cubic cover of $X_1(33)_{\F_4}$: if $E$ is written as $y^2+A_1xy+A_3y=x^3$ with the point $11Q$ of order $3$ at
$(0,0)$, then adjoining a cube root $r$ of $A_1^3-27A_3$ to the function field of $X_1(33)_{\F_4}$ gives the
$x$-coordinate of a point of order $3$ independent of $11Q$, and we check that the resulting curve has genus $81$, the
genus of $X$. On this model we compute a basis of $H^0(X_{\F_4},\Omega^1)$, the matrices of $\Cart$ and $h^*$ in this
basis, and the space $U$. We then \gitlink{computations/d31b_m333_analytic_jets_v4.m}{compute} the first $5$
coefficients of the local expansions of the elements of $U$ at the $40$ cusps of degree $1$ and the first coefficient at
the $8$ cusps of degree $5$. There are $\binom{44}{5}=1086008$ effective divisors of degree $5$ supported on the
cusps of degree $1$, and $864008$, $213600$ and $8400$ of them have $k_0=0$, $1$ and $2$, respectively. We
\gitlink{computations/d31b_m333_prefix_ranks.py}{compute} that the restriction of $U$ to each of them has rank
$5-k_0$, and that the restriction to each of the $8$ cusps of degree $5$, for which $k_0=0$, has rank $5$. Part c) is
\gitlink{computations/d31b_m3n33_physical_source.m}{computed} by running through all the elliptic curves over
$\F_{4^5}$ with a subgroup isomorphic to $\tg{3}{33}$ and all their level structures.
\end{proof}

\begin{proposition}\label{prop:3x33}
The group $\tg{3}{33}$ is not a subgroup of the torsion of an elliptic curve over a number field of degree $10$.
\end{proposition}

\begin{proof}
Suppose the opposite, let $F$, $X$, $h$, $\mathfrak p$, $\pi$ and $Y$ be as in \Cref{lem:mxn-333}, and let $D$ be the
trace divisor, an effective $F$-rational divisor of degree $5$ on $X$ with non-cuspidal support. By
\Cref{lem:mxn-333} a), the reduction of $D$ modulo $\mathfrak p$ is either an effective cuspidal divisor or a
non-cuspidal closed point of degree $5$.

Suppose first that $D$ reduces to an effective cuspidal divisor $\overline C$. Let $R$ be the completion of the ring of
integers of $F$ at $\mathfrak p$, the ring of integers of the unramified quadratic extension of $\Q_2$, and let $\mathcal X$
be the smooth model of $X$ over $R$, to which $h$ extends. The cuspidal subscheme $\Sigma$ of $\mathcal X$ is finite
\'etale over $R$ and stable under $h$, and $h$ fixes no closed point of $\Sigma_{\F_4}$ by \Cref{lem:mxn-333} a). Let $C$
be the effective divisor supported on $\Sigma$ which reduces to $\overline C$, and let $c_0$ be an $F$-rational cusp.
The class $\alpha[D-5c_0]$, where $\alpha=h_*-1$, is torsion by \Cref{lem:mxn-333image}, and $[C-5c_0]$ is torsion by
the Manin-Drinfeld theorem, so $\alpha[D-C]$ is torsion. As $D\in X^{(5)}(F)\subseteq\mathcal X^{(5)}(R)$ and
$(h_*-1)^*=h^*-1$ on differentials, \Cref{lem:mxn-relformal} and \Cref{lem:mxn-333} b) give $C_0\leq D$. As $\deg C_0
=5-2k_0$ is odd, $C_0\neq0$, which contradicts the fact that the support of $D$ contains no cusp.

So the reduction of $D$ is one of the $264$ non-cuspidal closed points $x$ of degree $5$ of $X_{\F_4}$, and the
reduction of the effective $F$-rational divisor $M=\pi_*D$ of degree $5$ on $Y$ is the closed point $N=\pi(x)$ of degree
$5$. Let $\sigma$ be the non-trivial automorphism of $F$. Then $M+M^\sigma$ is an effective rational divisor of degree
$10$ on $X_1(33)$, and as $2$ is inert in $F$, its reduction modulo $2$ is the divisor on $X_1(33)_{\F_2}$ whose base
change to $\F_4$ is $N+N^\sigma$, where $N^\sigma$ is the image of $N$ under the Frobenius of $\F_4$. The group
$J_1(33)(\Q)$ is finite \cite[Theorem 3.1(2)]{DEHMZ}, so by \Cref{prop:tools} d) the divisor $A_5(M+M^\sigma)$ is
principal, and hence so is $A_5(N+N^\sigma)$. As $A_5$ commutes with the diamond operators, by \Cref{lem:mxn-333} c)
it suffices to consider $12$ representatives $N$, and we
\gitlink{computations/d31b_m333_rational_trace_A5.m}{compute} that $\ell(A_5(N+N^\sigma))=0$ for $9$ of them.

For the remaining $3$ representatives we use the other elliptic curves in the isogeny class. Let $(E,P,Q)$ be the
point of $Y_1(3,33)(K)$ given by $E$. As the reduction of $D$ modulo $\mathfrak p$ is a closed point of degree
$5=[K:F]$, there is a unique prime $\mathfrak P$ of $K$ above $\mathfrak p$, its residue field is $\F_{4^5}$, and $E$
has good reduction at $\mathfrak P$ (\Cref{subsec:trace}). As $E[3]\subseteq E(K)$, for each of the $3$ subgroups $W\subseteq E[3]$ of order $3$
different from the subgroup generated by $11Q$, the isogeny $\phi_W:E\to E/W$ is defined over $K$ and $\phi_W(Q)$ has
order $33$. So $y_W=(E/W,\phi_W(Q))$ is a point of $Y_1(33)(K)$, and the argument above, applied to its trace divisor
$M_W$ over $F$, shows that the reduction of $A_5(M_W+M_W^\sigma)$ modulo $2$ is principal. This reduction is
determined by the reduction of $y_W$ modulo $\mathfrak P$, which depends only on the reduction of $(E,Q)$ and of $W$, and
the construction commutes with the diamond operators. We
\gitlink{computations/d31b_m333_rational_trace_3isog_A5.m}{compute} that for each of the $3$ remaining
representatives $N$, the dimensions of the Riemann-Roch spaces of these reductions, for the $3$ subgroups $W$, are
$(0,0,1)$, $(1,0,0)$ and $(1,0,0)$, respectively. So in each case one of these divisors is not principal, which is a contradiction.
\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{The group \texorpdfstring{$\tg{4}{28}$}{Z/4Z x Z/28Z}}\label{subsec:mxn-4x28}
The group $J_1(4,28)(\Q(i))$ is infinite, as over $\Q(i)$ the curve $X_1(4,28)$ is isomorphic to a quotient of
$X_1(112)$ \cite[Section 2]{DS}, which maps to $X_0(112)$, and the elliptic curve \lmfdbec{112}{a}{2}, a quotient of
$J_0(112)$, has rank $1$. So we work instead with the curve $X_1(2,28)$ over $\Q$, whose Jacobian has rank $0$, and
with four maps from it to $X_1(28)$. The additional $4$-torsion is used twice: through the residue fields of the
possible reductions, and through a $2$-isogeny.

\begin{lemma}\label{lem:mxn-428isog}
Let $L$ be a field of characteristic different from $2$ and $7$, let $E/L$ be an elliptic curve, and let $P_2,Q,U\in E(L)$
be points such that $P_2$ has order $2$, $Q$ has order $28$, $P_2\neq14Q$ and $2U=P_2$. Let $\phi:E\to E'$ be the
isogeny with kernel $\{0,P_2\}$. Then $\phi(U)$ has order $2$, $\phi(Q)$ has order $28$ and $\phi(U)\neq14\phi(Q)$, so
$(E',\phi(U),\phi(Q))$ is a point of $Y_1(2,28)(L)$.
\end{lemma}

\begin{proof}
As $P_2\neq14Q$, the kernel of $\phi$ meets the group generated by $Q$ trivially, so $\phi(Q)$ has order $28$. We have
$2\phi(U)=\phi(P_2)=0$, and $\phi(U)\neq0$ as $U\notin\{0,P_2\}$. The dual isogeny $\widehat\phi$ maps $\phi(U)$ to
$2U=P_2\neq0$ and $14\phi(Q)$ to $28Q=0$, so $\phi(U)\neq14\phi(Q)$.
\end{proof}

\begin{lemma}\label{lem:mxn-428}
Let $Y=X_1(2,28)$ and $Z=X_1(28)$.
\begin{itemize}
\item[a)] The curve $Y$ has genus $25$, $J_Y(\Q)$ is finite and $\gon_\Q Y\geq12$.
\item[b)] The group $J_Z(\Q)$ is equal to $\ClCusp_\Q Z$, and
$J_Z(\Q)\simeq\torz2\times\torz4\times\torz{12}\times\torz{936}$. Its reduction modulo $3$ has index $4$ in
$J_Z(\F_3)\simeq\torz4\times\torz4\times\torz{24}\times\torz{936}$.
\item[c)] The closed cusps of $Y_{\F_3}$ have degrees $1^{12},2^6,3^4,6^2$, and those of degrees $1$ and $2$ are the
reductions of Galois orbits of cusps of $Y$ over $\Q$ of the same degrees.
\item[d)] The curve $Y_{\F_3}$ has no non-cuspidal closed points of degree at most $4$. The numbers of its non-cuspidal
closed points of degrees $5$, $6$, $8$ and $10$ are $48$, $64$, $606$ and $6384$, and they form $2$, $5$, $29$ and $272$
orbits under the group of the $24$ automorphisms $(P_2,Q)\mapsto(P_2+14bQ,\ cP_2+aQ)$ of $Y$, with $b,c\in\{0,1\}$ and
$a\in\{1,3,5,9,11,13\}$, which are defined over $\Q$.
\end{itemize}
\end{lemma}

\begin{proof}
Part a) is \cite[Theorem 3.1(3)]{DEHMZ} and \cite[Table 1]{NV}, and b) is \cite[Theorem 3.1(2), Corollary 4.14 and
Table 2]{DEHMZ} together with a \gitlink{computations/d31b_m428_Y_fourmap.m}{computation} of $J_Z(\F_3)$ and of the
reduction of $J_Z(\Q)$. The degrees in c) are \gitlink{computations/d31b_Y228_AF_cusp_signature_probe.m}{computed}
directly. We \gitlink{independent_checks/cusp_degrees.py}{compute} that the cusps of $Y$ form Galois orbits over $\Q$
of degrees $1^{12},2^6,6^4$, and that modulo $3$ the orbits of degrees $1$ and $2$ stay irreducible, while those of degree
$6$ give closed cusps of degree $3$ or $6$. So the $18$ closed cusps of $Y_{\F_3}$ of degrees $1$ and $2$ are the reductions
of the $18$ Galois orbits of degrees $1$ and $2$.
An elliptic curve over $\F_{3^j}$ with a subgroup isomorphic to $\tg{2}{28}$ has a number of points divisible by $56$,
which is impossible for $j\leq4$ by the Hasse bound, as $(3^{j/2}+1)^2<56$ for $j\leq3$ and $(9-1)^2=64>56$ and
$(9+1)^2=100<112$. The numbers of points and orbits in d) are \gitlink{computations/d31b_m428_Y_fourmap.m}{computed}
directly.
\end{proof}

\begin{proposition}\label{prop:4x28}
The group $\tg{4}{28}$ is not a subgroup of the torsion of an elliptic curve over a number field of degree $10$.
\end{proposition}

\begin{proof}
Suppose the opposite, that $\tg{4}{28}\subseteq E(K)$ for an elliptic curve $E$ over a number field $K$ of degree $10$.
Then $\Q(i)\subseteq K$ by the Weil pairing. Let $Y$ and $Z$ be as in \Cref{lem:mxn-428}, choose $P_2,Q\in E(K)$ of orders
$2$ and $28$ with $P_2\neq14Q$, and let $D$ be the trace divisor of the point $(E,P_2,Q)\in Y_1(2,28)(K)$, an effective
rational divisor of degree $10$ on $Y$ with non-cuspidal support. As $J_Y(\Q)$ is finite and $\gon_\Q Y\geq12>10$, the
reduction of $D$ modulo $3$ is not entirely cuspidal by \Cref{prop:main} a). Write it as $D'+C'$ as in
\Cref{subsec:sieve}.

Let $\mathfrak P$ be a prime of $K$ above $3$ at which $E$ has good reduction, with ramification index $e(\mathfrak
P)$ and residue degree $f(\mathfrak P)$. As $3$ is inert in $\Q(i)\subseteq K$, $f(\mathfrak P)$ is even. The reduction
$\overline x$ of $(E,P_2,Q)$ modulo $\mathfrak P$ is a non-cuspidal closed point of $Y_{\F_3}$ whose degree divides
$f(\mathfrak P)$, and it occurs in the reduction of $D$ with multiplicity at least $e(\mathfrak P)f(\mathfrak
P)/\deg\overline x$ (\Cref{subsec:trace}).
By \Cref{lem:mxn-428} d), $\deg\overline x\geq5$. As the numbers $e(\mathfrak P)f(\mathfrak P)$ add up to $10$, there is
exactly one such $\mathfrak P$, with $e(\mathfrak P)=1$ and $f(\mathfrak P)\in\{6,8,10\}$, and $D'$ is a closed point
of degree $6$, $8$ or $10$, with $C'$ of degree $4$, $2$ or $0$, respectively, or twice a closed point of degree $5$,
with $C'=0$. If $\deg D'\in\{6,8,10\}$, then the residue field of $D'$ is the residue field of $\mathfrak P$, and as
$Y$ is a fine moduli space, the elliptic curve of the point $D'$ is the reduction of $E$ modulo $\mathfrak P$, which has
full $4$-torsion over this field, as $E[4]\subseteq E(K)$. Writing this curve as $y^2=(x-e_1)(x-e_2)(x-e_3)$, this implies
that all the differences $e_i-e_j$ are squares in the residue field, by the usual description of the rational points
divisible by $2$ on an elliptic curve with full rational $2$-torsion over a field of characteristic different from $2$
(see \cite[Proposition X.1.4 and its proof]{silverman}). This
condition depends only on the orbit of the point under the $24$ automorphisms of \Cref{lem:mxn-428} d), and it holds
for $3$ of the $5$ orbits of points of degree $6$, for $11$ of the $29$ orbits of degree $8$ and for $158$ of the $272$
orbits of degree $10$. Replacing $(P_2,Q)$ by its image under one of these automorphisms, which gives another point of
$Y_1(2,28)(K)$ with the same $E$, we may assume that the closed point under $D'$ is one of the orbit representatives.
This leaves $2+3+11+158=174$ possibilities for this point.

We now use the four maps $f_j:Y\to Z$,
\begin{align*}
f_1(E,P_2,Q)&=(E,Q), & f_3(E,P_2,Q)&=(E/\langle P_2\rangle,\phi_3(Q)),\\
f_2(E,P_2,Q)&=(E,P_2+Q), & f_4(E,P_2,Q)&=(E/\langle P_2+14Q\rangle,\phi_4(Q)),
\end{align*}
where $\langle T\rangle$ is the subgroup generated by $T$ and $\phi_3$, $\phi_4$ are the quotient isogenies. They are
defined over $\Q$, map cusps to cusps and non-cusps to non-cusps, and extend over $\Z_{(3)}$. By \Cref{lem:mxn-428} b),
\eqref{eq:pushclass} applied to each $f_j$, with $G$ the reduction of $J_Z(\Q)$, gives four necessary conditions on
$D'$, and we \gitlink{computations/d31b_m428_Y_fourmap.m}{compute} that only $19$ of the $174$ possibilities satisfy
all four of them: $3$ in which $D'$ is a closed point of degree $8$, and $16$ in which it is a closed point of degree
$10$.

As $J_Y(\Q)$ is finite, $[A_5(D)]=0$ by \Cref{prop:tools} d), and by \Cref{lem:mxn-428} c) and \Cref{lem:cuspHecke},
$A_5(C')=0$ whenever $C'$ has degree at most $2$. So $A_5(D')$ is principal (\Cref{subsec:sieve}), and we
\gitlink{computations/d31b_Y228_A5_v5_s0.m}{compute} (in six scripts \texttt{d31b\_Y228\_A5\_v5\_s}$i$, $i=0,\dots,5$, with the
common part \githubbare{computations/d31b_Y228_A5_common_v5.m}) that this excludes $15$ of the $19$ possibilities. The
remaining $4$ are closed points of degree $10$, with $C'=0$.

In these $4$ cases we use the additional $4$-torsion once more. In each of them $\mathfrak P$ is the only prime of
$K$ above $3$, with residue field $\F_{3^{10}}$. As $E[4]\subseteq E(K)$, all the $4$ points $U$ with $2U=P_2$ are
defined over $K$, and as reduction modulo $\mathfrak P$ is injective on $E[4]$, we may choose $U$ such that its
reduction is the point $\overline U$ with $2\overline U=\overline{P_2}$ used in the computation below. By
\Cref{lem:mxn-428isog}, $x_1=(E/\langle P_2\rangle,\phi_3(U),\phi_3(Q))$ is again a point of $Y_1(2,28)(K)$, and its trace
divisor $D_1$ is an effective rational divisor of degree $10$ on $Y$ with non-cuspidal support. As $E/\langle P_2\rangle$
has good reduction at $\mathfrak P$, the reduction of $D_1$ modulo $3$ is $(10/\deg\overline{x_1})\,\overline{x_1}$
(\Cref{subsec:trace}), where
$\overline{x_1}$ is the closed point of $Y_{\F_3}$ given by the reduction of $x_1$ modulo $\mathfrak P$, and as above
$A_5$ of this divisor has to be principal. We \gitlink{computations/d31b_m428_neighbors_A5.m}{compute} that it is not,
in each of the $4$ cases. This is a contradiction.
\end{proof}