%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Sporadic points and complete classifications}\label{sec:sporadic}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

We treat the groups $\tg{6}{12}$, $\tg{3}{15}$, $\tg{3}{18}$ and $\tg{2}{26}$. Each of them is the
torsion group of an elliptic curve over a number field of degree $10$, but only of finitely many. For each of them
we give explicit elliptic curves over explicit decic fields with exactly this torsion group, and we determine all
the effective $\Q(\zeta_m)$-rational divisors of degree $d_m$ on $X_1(m,n)$ whose support contains no cusp. This shows
that the curves we give are, up to isomorphism, the only elliptic curves over decic fields with a subgroup isomorphic
to these groups, and it excludes the groups $\tg{3}{30}$, $\tg{3}{45}$, $\tg{3}{36}$, $\tg{6}{18}$ and $\torz{52}$.
\Cref{table:spor} lists the curves.

In all four cases the Jacobian of the modular curve $X$ is finite over $F=\Q(\zeta_m)$ by \cite[Theorem 4.1]{DS}
(for $X_1(2,26)$ see also \cite[Theorem 3.1(3)]{DEHMZ}), and $\gon_FX>d_m$. Hence, by \Cref{lem:closed}, distinct effective $F$-rational
divisors of degree $d_m$ on $X$ have distinct classes, and $X$ has only finitely many such divisors. We work at primes
$\mathfrak p$ of $F$ not dividing $n$, at which $X$ has good reduction, which are odd except in one step for
$X_1(3,15)$, so that reduction is injective on $J_X(F)$ by Katz's result. We identify the Galois orbits of cusps over $F$
with their reductions using the valuations of integral modular units, as in \cite[Sections 11, 12 and 13]{N69}. The arguments below then show that the class of the trace divisor $D$ is, after applying a level
automorphism, the class of a known effective divisor, which is then equal to $D$. By \Cref{lem:CM}, an elliptic curve
over a number field of degree $10$ whose torsion contains one of the four groups has no complex multiplication.
Finally, the following lemma shows that the pair $(K,E)$ is determined up to isomorphism.

\begin{lemma}\label{lem:spor-determined}
Let $m\mid n$ with $n\geq4$, let $F=\Q(\zeta_m)$ and $X=X_1(m,n)$ over $F$, and put $d_m=10/\varphi(m)$. Let $E_0$ be
an elliptic curve over a number field $K_0$ of degree $10$, and let $x_0=(E_0,P_0,Q_0)\in Y_1(m,n)(K_0)$ be a point
whose trace divisor $D_0$ is a closed point of degree $d_m$. Let $E$ be an elliptic curve over a number field $K$ of
degree $10$, and let $x=(E,P,Q)\in Y_1(m,n)(K)$ be a point whose trace divisor is $gD_0$ for a level automorphism $g$
of $X$ defined over $F$. Then $(K,E)$ is isomorphic to $(K_0,E_0)$.
\end{lemma}

\begin{proof}
As $D_0$ is a closed point of degree $d_m=[K_0:F]$, the residue field $F(x_0)\subseteq K_0$ of $x_0$ is $K_0$. The
automorphism $g$ is defined over $F$, so $g(x_0)\in Y_1(m,n)(K_0)$ also has residue field $K_0$, and its closed point
is $gD_0$. The trace divisor of $x$ is the sum of the $F$-conjugates of $x$, each counted $[K:F(x)]$ times
(\Cref{subsec:trace}), and as it is a closed point with multiplicity $1$, we have $F(x)=K$. So $x$ and $g(x_0)$ lie
above the same closed point of $X$, and there is an isomorphism $\sigma\colon K_0\to K$ over $F$ such that $x$ is the
image of $g(x_0)$ under $\sigma$. As $Y_1(m,n)$ is a fine moduli space and $g$ does not change the elliptic curve, $E\simeq E_0^\sigma$ over
$K$.
\end{proof}

For $b,c$ in a field we write
\begin{equation}\label{eq:spor-tate}
E(b,c):\quad y^2+(1-c)xy-by=x^3-bx^2
\end{equation}
for the curve in Tate normal form, on which $(0,0)$ has order $n$ when $(b,c)$ comes from a point of $Y_1(n)$, and
\begin{equation}\label{eq:spor-tate2}
E'(b,c):\quad y^2=x^3+cx^2+b(1-b-c)x,
\end{equation}
on which $(0,0)$ has order $2$ and $(b,b)$ has order $n$ when $(b,c)$ comes from a point of $Y_1(2,n)$.

\begin{table}[htbp]
\centering
\small
\renewcommand{\arraystretch}{1.1}
\begin{tabular}{c|l|c|l|l|l}
$G$ & $K$ & pairs & $E$ & \cite{x1mn} & \\\hline
$\tg{6}{12}$ & \lmfdbnf{10.0.4678854192.1} & $1$ & $E_{6,12}$ & \xpoint{6.12.10.a} & \Cref{prop:6x12}\\
$\tg{3}{15}$ & \lmfdbnf{10.0.306454347.1} & $1$ & $E_{3,15}$ & \xpoint{3.15.10.a} & \Cref{prop:3x15}\\
$\tg{3}{18}$ & \lmfdbnf{10.0.2571428187.1} & $2$ & $E_{3,18}$ & \xpoint{3.18.10.a} & \Cref{prop:3x18}\\
             &                             &     & $E'_{3,18}$ & \xpoint{3.18.10.b} & \\
$\tg{2}{26}$ & \lmfdbnf{10.6.396320664709.1} & $1$ & $E_{2,26}$ & \xpoint{2.26.10.a} & \Cref{prop:2x26}\\
             & \lmfdbnf{10.2.799905449.1} & $1$ & $E'_{2,26}$ & \xpoint{2.26.10.b} &
\end{tabular}
\caption{The elliptic curves $E$ over number fields $K$ of degree $10$ with $E(K)_\tors\simeq G$, for the four groups
$G$ of this section, the number of pairs $(K,E)$ up to isomorphism over each field, and the labels of the corresponding
points in \cite{x1mn}. These are all the pairs with $G\subseteq E(K)_\tors$. The curves $E_{3,18}$ and $E'_{3,18}$ are
$2$-isogenous. In all cases $j(E)$ has degree $5$ over $\Q$ and is not an algebraic integer, so $E$ has no complex
multiplication.}
\label{table:spor}
\end{table}
% The LMFDB labels are not in the Codex record: they were obtained on 2026-11-06 by looking up, in the LMFDB, the
% fields with the polredabs polynomials and discriminants of the census records (computed again with PARI's
% polredabs from the defining polynomials below). The degrees of j for E_{3,18}' and E_{2,26}, E_{2,26}' and the
% non-integrality were also rechecked then from the census models (Sage), see the report.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{The curve \texorpdfstring{$X_1(6,12)$}{X1(6,12)}}\label{subsec:spor-6x12}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Let $F=\Q(z)$ with $z^2-z+1=0$, so that $z$ is a primitive sixth root of unity, and let $X=X_1(6,12)$ over $F$. It is
a curve of genus $9$, $J_X(F)$ is finite \cite[Theorem 4.1]{DS}, and $\gon_FX\geq6$ \cite[Table 1]{NV}. We use the
plane model of $X$ from \cite{mdmagma} (see \githubbare{shared/mdmagma/models_X1_m_n/X1_6_12.txt}), with affine
coordinates $u,v$. On it the point $Q$ of order $12$ is the point $(0,0)$ of the curve $E(b,c)$ of
\eqref{eq:spor-tate}, where
\begin{equation}\label{eq:spor-rs}
r=\frac{u^2+1}{u^2-u},\qquad s=\frac{u^2+u+1}{u^2},\qquad b=rs(r-1),\qquad c=s(r-1).
\end{equation}
So $u$ is a coordinate on $X_1(12)\simeq\PP^1$, and $u:X\to\PP^1$ is the map forgetting $P$, which has degree $6$.

The level automorphisms of $X$ defined over $F$ are the maps
\begin{equation}\label{eq:spor-G612}
(P,Q)\longmapsto(\alpha P+2\beta Q,\ \gamma P+\delta Q),\qquad \alpha,\beta,\gamma\in\torz6,\ \
\delta\in\torz{12},\ \ \alpha\delta-2\beta\gamma\equiv1\ (\mathrm{mod}\ 6),
\end{equation}
which preserve the Weil pairing of $P$ and $2Q$. Taken modulo $\pm1$ they form a group $G_{6,12}$ of order $96$, as we
\gitlink{continuation_20261029/deps/full_aut_6_12.m}{compute}.

\begin{lemma}\label{lem:spor-612-jac}
Let $F=\Q(\zeta_6)$, let $X=X_1(6,12)$ over $F$, and let $\mathfrak p_7$ be a prime of $F$ above $7$, with residue
field $\F_7$.
\begin{enumerate}
\item[a)] The closed cusps of $X_{\F_7}$ have degrees $1^{16},2^8$, and each of them is the reduction of a Galois orbit of
cusps over $F$ of the same degree.
\item[b)] We have
\[
J_X(\F_7)\simeq\torz6\times\torz{12}\times(\torz{24})^2\times(\torz{48})^2,\qquad
\ClCusp_FX\simeq\torz6\times(\torz{12})^3\times(\torz{24})^2,
\]
so $\#\ClCusp_FX=5971968$, and reduction modulo $\mathfrak p_7$ is injective on $J_X(F)$.
\item[c)] The kernel of $A_{13}=T_{13}-14$ on $J_X(\F_7)$ contains the reduction of $\ClCusp_FX$, with quotient
isomorphic to $\torz2\times\torz2$. In particular $2J_X(F)\subseteq\ClCusp_FX$.
\item[d)] The non-cuspidal closed points of $X_{\F_7}$ of degrees $1,\dots,5$ number $0$, $0$, $96$, $468$ and
$3264$.
\end{enumerate}
\end{lemma}

\begin{proof}
Parts a) and b) and the first four numbers in d) are in the proof of \cite[Proposition 12.7]{N69}, and reduction
modulo $\mathfrak p_7$ is injective on $J_X(F)=J_X(F)_\tors$ by Katz's result, as $7$ is odd, unramified in $F$ and
prime to $12$. We \gitlink{computations/cont_local_6_12_q7.m}{compute} the number $3264$ of closed points of degree
$5$. As $13\equiv1\pmod{12}$, the diamond operator $\dm{13}$ is trivial and $A_{13}=T_{13}-14$. We
\gitlink{computations/cont_diagonal_6_12_D5_v5.m}{compute} the kernel of $A_{13}$ on $J_X(\F_7)$, and that its
quotient by the reduction of $\ClCusp_FX$ is isomorphic to $\torz2\times\torz2$. As $13$ splits completely in $F$, $A_{13}$ annihilates
$J_X(F)$ by \Cref{prop:tools}~b), so the reduction of $J_X(F)$ lies in this kernel. As reduction is injective on
$J_X(F)$, it follows that $J_X(F)/\ClCusp_FX$ embeds into $\torz2\times\torz2$, which proves c).
\end{proof}

\begin{lemma}\label{lem:spor-612-sieve7}
Let $F=\Q(\zeta_6)$ and $X=X_1(6,12)$ over $F$. There are $3360$ explicit $F$-rational cuspidal divisors
$L_1,\dots,L_{3360}$ of degree $10$ on $X$ (not necessarily effective), given in
\githubbare{continuation_20261029/M6N12_D5_all_source_lifts.m}, such that for every effective $F$-rational divisor $D$ of
degree $5$ on $X$ whose support contains no cusp we have $2D\sim L_k$ for some $k$.
\end{lemma}

\begin{proof}
Let $D$ be such a divisor, let $\mathfrak p_7$ be as in \Cref{lem:spor-612-jac}, and fix an $F$-rational cusp $c_\ast$.
As $\gon_FX\geq6>5$, by \Cref{prop:main}~a) the reduction of $D$ modulo $\mathfrak p_7$ is not cuspidal, so by
\Cref{lem:spor-612-jac}~d) it is $x+C'$, where $x$ is a closed point of degree $3$, $4$ or $5$ and $C'$ is an effective
cuspidal divisor of degree $5-\deg x$. By \Cref{lem:spor-612-jac}~a) there are $\binom{17}{2}+8=144$ effective cuspidal
divisors of degree $2$ and $16$ of degree $1$, so there are
\[
96\cdot144+468\cdot16+3264=24576
\]
possibilities for $x+C'$. The class $[D-5c_\ast]$ lies in $J_X(F)$, so it is killed by $A_{13}$, and hence so is its
reduction $[x+C'-5\overline{c_\ast}]$. We \gitlink{computations/cont_diagonal_6_12_D5_v5.m}{compute} that this
holds for $3552$ of the $24576$ possibilities, and that they give $3360$ distinct classes
$2[x+C'-5\overline{c_\ast}]$. By \Cref{lem:spor-612-jac}~c) these classes lie in the reduction of $\ClCusp_FX$, and by
the injectivity of reduction we \gitlink{computations/cont_M6N12_D5_all_source_lifts.m}{obtain} $3360$ explicit
$F$-rational cuspidal divisors $L_1,\dots,L_{3360}$ of degree $10$ such that $2D\sim L_k$ for some $k$.
\end{proof}

\begin{lemma}\label{lem:spor-612-13}
Let $F=\Q(z)$ with $z^2-z+1=0$, let $X=X_1(6,12)$ over $F$, let $L_1,\dots,L_{3360}$ be as in
\Cref{lem:spor-612-sieve7}, and let $\mathfrak p_{13}=(13,z-4)$, a prime of $F$ above $13$, with residue field
$\F_{13}$. For exactly $192$ of the $3360$ indices $k$ the linear system $|\overline{L_k}|$ on $X_{\F_{13}}$ contains a
divisor of the form $2A$, where $A$ is effective and its support is not contained in the cusps, and for these
$\ell(L_k)=2$.
\end{lemma}

\begin{proof}
We \gitlink{computations/cont_RR612_norm13_shard0.m}{compute} all the effective $\F_{13}$-rational divisors in the
linear systems $|\overline{L_k}|$ (we have $\ell(\overline{L_k})=2$ for $2880$ of them and $\ell(\overline{L_k})=3$ for the
other $480$), and find that such a divisor $2A$ exists for exactly $192$ of the $3360$ classes, and that
$\ell(\overline{L_k})=2$ for all of these. For these $192$ divisors $L_k$ we have $\ell(L_k)\geq 10+1-9=2$ by the
Riemann-Roch theorem, and $\ell(L_k)\leq\ell(\overline{L_k})=2$ by semicontinuity \cite[Theorem III.12.8]{hartshorne},
so $\ell(L_k)=2$.
\end{proof}

\begin{lemma}\label{lem:spor-612-L8}
Let $F=\Q(\zeta_6)$, let $X=X_1(6,12)$ over $F$, and let $L_{8,1},\dots,L_{8,6}$ be the six effective cuspidal divisors
of degree $8$ with $\ell(L_{8,i})=2$ of the proof of \cite[Proposition 12.7]{N69}. Then $96$ of the $192$ divisors $L_k$
of \Cref{lem:spor-612-13} are linearly equivalent to divisors $L_{8,i}+2c$, where $c$ is an $F$-rational cusp, and no
effective $F$-rational divisor $D$ of degree $5$ on $X$ whose support contains no cusp satisfies
$2D\sim L_{8,i}+2c$.
\end{lemma}

\begin{proof}
We \gitlink{computations/cont_replay612_complete_v2.py}{find} that $96$ of the $192$ classes are the classes of the
divisors $L_{8,i}+2c$, where $c$ runs over the $16$ $F$-rational cusps. We verify these identities of classes with the
divisors of $141$ modular units. They span a sublattice of rank $23$ and index $5971968=\#\ClCusp_FX$ of the lattice
of $F$-rational cuspidal divisors of degree $0$, which is also the index of the sublattice of principal divisors, so
the two sublattices are equal. By \Cref{lem:spor-612-13} we have $\ell(L_{8,i}+2c)=2=\ell(L_{8,i})$, so every effective
divisor linearly equivalent to $L_{8,i}+2c$ is of the form $B+2c$ with $B\in|L_{8,i}|$. So $2D\sim L_{8,i}+2c$ would
imply that $c$ lies in the support of $D$, which is impossible.
\end{proof}

\begin{lemma}\label{lem:spor-612-double}
Let $F=\Q(\zeta_6)$ and $X=X_1(6,12)$ over $F$, and let $D$ be an effective $F$-rational divisor of degree $5$ on $X$
whose support contains no cusp. Then there is $g\in G_{6,12}$ such that $2gD\sim L$, where $L$ is the effective
cuspidal divisor of degree $10$ given in \githubbare{jobs/cont_reduce612_quintic_lattice_v2.log}.
\end{lemma}

\begin{proof}
By \Cref{lem:spor-612-sieve7} we have $2D\sim L_k$ for some $k$. Let $\mathfrak p_{13}$ be as in \Cref{lem:spor-612-13}.
As $\gon_FX\geq6>5$, by \Cref{prop:main}~a) the reduction of $D$ modulo $\mathfrak p_{13}$ is not cuspidal, so the
reduction of $2D$ is a divisor in $|\overline{L_k}|$ of the form $2A$, where $A$ is effective and its support is not
contained in the cusps. So $L_k$ is one of the $192$ divisors of \Cref{lem:spor-612-13}, and by \Cref{lem:spor-612-L8} it is not
linearly equivalent to any of the divisors $L_{8,i}+2c$. We
\gitlink{computations/cont_quintic612_class_orbits_v2.m}{compute} that the remaining $96$ classes are the classes of the
divisors $gL$, $g\in G_{6,12}$. As the elements of $G_{6,12}$ are automorphisms of $X$ over $F$ which map cusps to cusps,
$2D\sim gL$ is equivalent to $2g^{-1}D\sim L$.
\end{proof}

\begin{proposition}\label{prop:6x12}
Let $F=\Q(z)$ with $z^2-z+1=0$, let $w$ be a root of
\[
3w^5+(4-5z)w^4+(7-8z)w^3-7(z+1)w^2+6(z-1)w+2z-1,
\]
and put $K_{6,12}=F(w)$, which is the field \lmfdbnf{10.0.4678854192.1}. Let $E_{6,12}=E(b,c)$ be the curve
\eqref{eq:spor-tate} with $r$, $s$, $b$ and $c$ given by \eqref{eq:spor-rs} with $u=w$, i.e.
\[
E_{6,12}:\ y^2+(1+s-rs)xy+rs(1-r)y=x^3+rs(1-r)x^2,\qquad r=\frac{w^2+1}{w^2-w},\ \ s=\frac{w^2+w+1}{w^2}.
\]
\begin{enumerate}
\item[a)] We have $E_{6,12}(K_{6,12})_\tors\simeq\tg{6}{12}$. The point $Q=(0,0)$ has order $12$, the point $P$ with
\begin{align*}
2x(P)&=(25z+10)w^4+(45-20z)w^3+(79-17z)w^2+(22-109z)w+39z-46,\\
2y(P)&=(222z+237)w^4+(651-537z)w^3+(1012-601z)w^2\\
&\qquad-(1518z+223)w+808z-861
\end{align*}
has order $6$, and $e_6(P,2Q)=z$. The $j$-invariant of $E_{6,12}$ has degree $5$ over $\Q$ and is not an algebraic
integer.
\item[b)] Every effective $F$-rational divisor of degree $5$ on $X_1(6,12)$ whose support contains no cusp is one of the
$96$ closed points of degree $5$ in the orbit under $G_{6,12}$ of the trace divisor $D_{6,12}$ of $(E_{6,12},P,Q)$.
\item[c)] If $E$ is an elliptic curve over a number field $K$ of degree $10$ with $\tg{6}{12}\subseteq E(K)$, then
$(K,E)$ is isomorphic to $(K_{6,12},E_{6,12})$, and $E(K)_\tors\simeq\tg{6}{12}$. In particular, no group properly
containing $\tg{6}{12}$ is a subgroup of the torsion of an elliptic curve over a number field of degree $10$.
\end{enumerate}
\end{proposition}

\begin{proof}
We \gitlink{computations/cont_witness612.m}{compute} that the quintic polynomial is irreducible over $F$, that $Q$ and
$P$ have orders $12$ and $6$, and that $e_6(P,2Q)=z$, so $\tg{6}{12}\subseteq E_{6,12}(K_{6,12})$. The prime of
$K_{6,12}$ given by $z\mapsto14$, $w\mapsto8$ has residue field $\F_{61}$, and it is unramified, as both defining
polynomials have simple roots modulo $61$. The curve has good reduction $y^2-4xy+4y=x^3+4x^2$ there, with $72$ points
over $\F_{61}$. As $61$ is odd, reduction is injective on $E_{6,12}(K_{6,12})_\tors$, so its order divides $72$, which
proves the first claim of a). The same computation shows that the minimal polynomial of $j(E_{6,12})$ over $F$ has
rational, non-integral coefficients and degree $5$.

We prove b). By \eqref{eq:spor-rs} the point $(E_{6,12},P,Q)$ of $X$ has $u$-coordinate $w$, so its residue field
is $F(w)=K_{6,12}$, and $D_{6,12}$ is a closed point of degree $5$. Let $D$ be as in b). By
\Cref{lem:spor-612-double}, applied to $D$ and to $D_{6,12}$, there are $g,g_0\in G_{6,12}$ with $2gD\sim L$ and
$2g_0D_{6,12}\sim L$. Let $\mathfrak p_{13}=(13,z-4)$ be the prime of $F$ above $13$ of \Cref{lem:spor-612-13}. We
\gitlink{additions/C6x12/square13.m}{compute} that $\ell(\overline L)=2$ on $X_{\F_{13}}$, and that exactly one of the
$14$ effective divisors in $|\overline L|$ is of the form $2A$ with $A$ effective and its support not contained in the
cusps (and $A$ is the sum of a closed point of degree $2$ and a closed point of degree $3$). By \Cref{prop:main}~a), as
$\gon_FX\geq6>5$, the reductions of $gD$ and of $g_0D_{6,12}$ modulo $\mathfrak p_{13}$ are not entirely cuspidal, and
twice each of them lies in $|\overline L|$, so both are equal to $A$. Hence $[gD-g_0D_{6,12}]\in J_X(F)$
reduces to $0$, so $gD\sim g_0D_{6,12}$ by Katz's result, as $13$ is odd, unramified in $F$ and prime to $12$, and
$gD=g_0D_{6,12}$ by \Cref{lem:closed}, as $\gon_FX\geq6>5$. So every $D$ as in b) is in the orbit of $D_{6,12}$ under
$G_{6,12}$. If $S\subseteq G_{6,12}$ is the stabilizer of $D_{6,12}$, then $\#S$ divides $\deg D_{6,12}=5$ by
\Cref{lem:orbits}, as $E_{6,12}$ has no complex multiplication by \Cref{lem:CM}, and $\#S$ divides $96$, so $S$ is
trivial and the orbit has $96$ elements. This proves b).

For c), let $E/K$ be as in c). By the Weil pairing $F\subseteq K$, and choosing $P',Q'\in E(K)$ of orders $6$ and $12$
with $e_6(P',2Q')=z$ we obtain a point of $Y_1(6,12)(K)$, whose trace divisor has degree $5$. By b) it is
$gD_{6,12}$ for some $g\in G_{6,12}$, so $(K,E)\simeq(K_{6,12},E_{6,12})$ by \Cref{lem:spor-determined}. The last
claim of c) follows from a).
\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{The curve \texorpdfstring{$X_1(3,15)$}{X1(3,15)}}\label{subsec:spor-3x15}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Let $F=\Q(z)$ with $z^2+z+1=0$, and let $X=X_1(3,15)$ over $F$. It is a curve of genus $9$, $J_X(F)$ is finite
\cite[Theorem 4.1]{DS}, and $\gon_FX\geq6$ \cite[Table 1]{NV}. As in \cite[Lemma 5.4]{N69} we use the normal form
$y^2+a_1xy+a_3y=x^3$, in which $P=(0,0)$ has order $3$, and we write $Q$ for the point of order $15$. The cusps are
defined over $\Q(\zeta_{15})=F(\zeta_5)$, and the $20$ Galois orbits of cusps over $F$ have degrees $1^{16},4^4$. We
identify them with their reductions at the primes of $F$ above $2$ and $7$ using the valuations of integral modular
units, as in the proof of \cite[Lemma 5.4]{N69}. Modulo the prime above $7$ these orbits stay irreducible, and modulo
the prime above $2$, whose residue field is $\F_4$, each orbit of degree $4$ splits into two closed cusps of degree $2$.

Put $Q_3=5Q$ and $Q_5=3Q$, of orders $3$ and $5$. The level automorphisms of $X$ defined over $F$ are the maps
\begin{equation}\label{eq:spor-G315}
(P,Q)\longmapsto\bigl(\alpha P+\beta Q_3,\ 2(\gamma P+\delta Q_3)+2\varepsilon Q_5\bigr),\qquad
\begin{pmatrix}\alpha&\beta\\\gamma&\delta\end{pmatrix}\in\SL_2(\F_3),\quad \varepsilon\in\F_5^\times,
\end{equation}
taken modulo $\pm1$, and they form a group $G_{3,15}\simeq(\SL_2(\F_3)\times\F_5^\times)/\dm{(-I,-1)}$ of order $48$
(see \cite[proof of Proposition 13.23]{N69}, where the representatives with $\varepsilon\in\{1,2\}$ are used).

\begin{lemma}\label{lem:spor-315-jac}
Let $F=\Q(\zeta_3)$ and $X=X_1(3,15)$ over $F$. Then
\[
J_X(F)=\ClCusp_FX\simeq\torz3\times\torz3\times\torz{24}\times\torz{96}\times\torz{96},
\]
and reduction modulo the prime of $F$ above $2$ is injective on $J_X(F)$.
\end{lemma}

\begin{proof}
By \cite[Lemma 5.4]{N69}, $\ClCusp_FX$ has these invariant factors, $J_X(F)/\ClCusp_FX$ has order dividing $3$, and
reduction modulo the prime $\mathfrak p$ of $F$ above $2$ is injective on $J_X(F)$. It remains to show that
$J_X(F)[3^\infty]\subseteq\ClCusp_FX$.

We \gitlink{computations/probe315_galois3.m}{compute} that $J_X(\F_4)\simeq(\torz3)^3\times\torz{24}\times(\torz{96})^2$,
so its $3$-primary part is $(\torz3)^6$, and that the classes $[\overline C-\deg(\overline C)\overline{c_\ast}]$ of the
closed cusps $\overline C$ of $X_{\F_4}$ generate a subgroup whose $3$-primary part is all of $(\torz3)^6$. Let
$L_0\subseteq F(\zeta_5)$ be the quadratic extension of $F$. As $4$ has order $2$ modulo $5$, the decomposition group
of a prime of $F(\zeta_5)$ above $\mathfrak p$ is $\Gal(F(\zeta_5)/L_0)$, so $\mathfrak p$ splits in $L_0$ into two
primes with residue field $\F_4$. Let $\lambda$ be one of them. As in the proof of \Cref{lem:doubled}, every closed cusp of
$X_{\F_4}$ is the reduction modulo $\lambda$ of an $L_0$-rational cuspidal divisor of the same degree. Let $C_1,\dots,C_r$
be such lifts of all the closed cusps, let $c_\ast$ be an $F$-rational cusp, and let $N$ be the prime-to-$3$ part of the
order of the subgroup of $J_X(L_0)$ generated by the classes $y_i=[C_i-\deg(C_i)c_\ast]$, which are torsion by the
Manin-Drinfeld theorem. The reductions of the classes $Ny_i$ generate the $3$-primary part of $J_X(\F_4)$.

Let $y\in J_X(F)[3^\infty]$. As reduction modulo $\lambda$ is injective on the torsion of odd order of $J_X(L_0)$ (the kernel of reduction is the
group of points of a formal group over a $2$-adic ring, which has no torsion of odd order), and the
$3$-primary part of $J_X(\F_4)$ is killed by $3$, we have $3y=0$, and $y=\sum_in_iNy_i$ for some integers $n_i$. Let
$\sigma$ be the non-trivial element of $\Gal(L_0/F)$. As $y$ is $F$-rational,
\[
2y=y+\sigma(y)=\sum_in_iN\,[C_i+\sigma(C_i)-2\deg(C_i)c_\ast],
\]
and the divisors $C_i+\sigma(C_i)$ are $F$-rational and cuspidal. So $2y\in\ClCusp_FX$, and as $3y=0$, also
$y=-2y\in\ClCusp_FX$.
\end{proof}

\begin{lemma}\label{lem:spor-315-classes}
Let $F=\Q(\zeta_3)$, let $X=X_1(3,15)$ over $F$, and fix an $F$-rational cusp $c_\ast$. Let $D$ be an effective $F$-rational
divisor of degree $5$ on $X$ whose support contains no cusp. Then $[D-5c_\ast]$ is one of $48$ explicit elements of
$\ClCusp_FX$.
\end{lemma}

\begin{proof}
We consider the reductions $D_2$ and $D_7$ of $D$ modulo the primes $\mathfrak p_2$ and $\mathfrak p_7$ of $F$ above $2$ and
$7$, with residue fields $\F_4$ and $\F_7$. We \gitlink{computations/diagonal315_D5.m}{compute} that the numbers of
non-cuspidal closed points of degrees $1,\dots,5$ are $0,0,0,102,192$ on $X_{\F_4}$ and $0,24,144,396,3312$ on $X_{\F_7}$.
Modulo $\mathfrak p_2$ we allow all the effective divisors of degree $5$: there are $22608$ cuspidal ones, $102\cdot16$
which are a point of degree $4$ plus one of the $16$ cusps of degree $1$, and the $192$ points of degree $5$, $24432$ in
all. Modulo $\mathfrak p_7$ the reduction is not cuspidal by \Cref{prop:main}~a), as $\gon_FX>5$. So $D_7$ is a point of
degree $2$, $3$, $4$ or $5$ plus an effective cuspidal divisor of the complementary degree, which gives $19584$, $19584$,
$6336$ and $3312$ possibilities, or a sum of two points of degree $2$ (possibly equal) and a cusp of degree $1$, which
gives $4800$ possibilities, or the sum of a point of degree $2$ and a point of degree $3$, which gives $3456$
possibilities. This makes $57072$ possibilities for $D_7$.

By \Cref{lem:spor-315-jac}, the class $y=[D-5c_\ast]$ lies in $\ClCusp_FX$, and its reductions modulo $\mathfrak p_2$ and
$\mathfrak p_7$ are $[D_2-5\overline{c_\ast}]$ and $[D_7-5\overline{c_\ast}]$. We
\gitlink{computations/diagonal315_D5_exact.m}{compute}, for all the pairs $(D_2,D_7)$ above, whether there is an
element of $\ClCusp_FX$ whose reductions are these two classes, and find that this leaves $240$ elements of
$\ClCusp_FX$ for $y$.

By \cite[Proposition 13.23 and its proof]{N69} there is a closed point $A_0$ of degree $4$ on $X$, coming from an
elliptic curve over an octic field, whose orbit under $G_{3,15}$ consists of $12$ closed points $A_1,\dots,A_{12}$ of
degree $4$ (see \githubbare{exclusions/315d5/M3N15_D4_exact_cusp_lifts.txt} for their classes). We
\gitlink{computations/diagonal315_D5_exact.m}{compute} that $192$ of
the $240$ elements above are the classes $[A_i+c-5c_\ast]$, where $c$ runs over the $16$ $F$-rational cusps. If $y$ is one of
them, then $D\sim A_i+c$, so $D=A_i+c$ by \Cref{lem:closed}, as $\gon_FX>5$. This is impossible, as the support of $D$
contains no cusp. So $y$ is one of the remaining $48$ elements.
\end{proof}

\begin{proposition}\label{prop:3x15}
Let $F=\Q(z)$ with $z^2+z+1=0$, let $w$ be a root of
\[
w^5+13(z+1)w^4-(24z+22)w^3+4(z+1)w^2+(z+4)w+z+1,
\]
and put $K_{3,15}=F(w)$, which is the field \lmfdbnf{10.0.306454347.1}. Let
\[
q=\frac{(293z+23)w^2+(105-238z)w+357-28z}{217w^2+(603z+271)w+283+86z}
\]
and
\[
E_{3,15}:\quad y^2+\bigl((1+q)w+2-q\bigr)xy+\bigl(qw^2+(1-q)w\bigr)y=x^3.
\]
\begin{enumerate}
\item[a)] We have $E_{3,15}(K_{3,15})_\tors\simeq\tg{3}{15}$. The points $P=(0,0)$ and $Q=(-w,w^2)$ have orders $3$ and
$15$, and $e_3(P,5Q)=z^2$. The $j$-invariant of $E_{3,15}$ has degree $5$ over $\Q$ and is not an algebraic integer.
\item[b)] Every effective $F$-rational divisor of degree $5$ on $X_1(3,15)$ whose support contains no cusp is one of the
$48$ closed points of degree $5$ in the orbit under $G_{3,15}$ of the trace divisor $D_{3,15}$ of $(E_{3,15},P,Q)$.
\item[c)] If $E$ is an elliptic curve over a number field $K$ of degree $10$ with $\tg{3}{15}\subseteq E(K)$, then
$(K,E)$ is isomorphic to $(K_{3,15},E_{3,15})$, and $E(K)_\tors\simeq\tg{3}{15}$.
\end{enumerate}
\end{proposition}

\begin{proof}
We \gitlink{computations/new315_witness.m}{compute} that the quintic polynomial is irreducible over $F$, that $P$ and $Q$
have orders $3$ and $15$, and that $e_3(P,5Q)=z^2$, so $\tg{3}{15}\subseteq E_{3,15}(K_{3,15})$. The prime of $K_{3,15}$
given by $z\mapsto6$, $w\mapsto28$ has residue field $\F_{43}$ and is unramified, and the curve has good reduction
$y^2+36xy+24y=x^3$ there, with group $\tg{3}{15}$. As reduction at an odd unramified prime is injective on the
torsion, this proves the first claim of a). The same computation gives the minimal polynomial of $j(E_{3,15})$ over $F$,
which has rational, non-integral coefficients and degree $5$. The computation uses an expression of $q$ as a
polynomial in $w$, and we \gitlink{computations/simplify315.py}{check} that it is equal to the formula above.

We prove b). As $j(E_{3,15})$ is a function on $X$ and has degree $5$ over $F$, the residue field of the point
$(E_{3,15},P,Q)$ is $F(j(E_{3,15}))=K_{3,15}$, so $D_{3,15}$ is a closed point of degree $5$. If $S\subseteq G_{3,15}$ is its
stabilizer, then $\#S$ divides $5$ by \Cref{lem:orbits}, as $E_{3,15}$ has no complex multiplication by \Cref{lem:CM},
and $\#S$ divides $48$. So $S$ is trivial, and the orbit of $D_{3,15}$ consists of $48$ distinct closed points of degree $5$. By
\Cref{lem:closed} their classes $[gD_{3,15}-5c_\ast]$ are distinct, and by \Cref{lem:spor-315-classes} they are among
the $48$ elements of $\ClCusp_FX$ of that lemma, so they are all of them. Now let $D$ be as in b). By
\Cref{lem:spor-315-classes}, $D\sim gD_{3,15}$ for some $g\in G_{3,15}$, and $D=gD_{3,15}$ by \Cref{lem:closed}.

Part c) follows from a), b) and \Cref{lem:spor-determined} as in the proof of \Cref{prop:6x12}~c), using the Weil
pairing $e_3(P',5Q')=z^2$.
\end{proof}

\begin{corollary}\label{cor:3x30-3x45}
The groups $\tg{3}{30}$ and $\tg{3}{45}$ are not subgroups of the torsion of an elliptic curve over a number field of
degree $10$.
\end{corollary}

\begin{proof}
Both groups contain $\tg{3}{15}$, so by \Cref{prop:3x15}~c) such a curve would have torsion group $\tg{3}{15}$.
\end{proof}

\begin{remark}\label{rem:spor-315}
The $5$ conjugates of $j(E_{3,15})$ give $5$ elliptic curves over $\Qbar$ which are pairwise not isomorphic, but by
\Cref{prop:3x15} they all come from the single pair $(K_{3,15},E_{3,15})$.
\end{remark}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{The curve \texorpdfstring{$X_1(3,18)$}{X1(3,18)}}\label{subsec:spor-3x18}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Let $F=\Q(\zeta_3)$ and $X=X_1(3,18)$ over $F$, a curve of genus $10$ with $J_X(F)$ finite \cite[Theorem 4.1]{DS} and
$\gon_FX\geq6$ \cite[Table 1]{NV}. We use the model of \cite[Section 12.4]{N69}, 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 $18$. We work
at primes $\mathfrak p_7$ and $\mathfrak p_{13}$ of $F$ above $7$ and $13$, with residue fields $\F_7$ and $\F_{13}$. As
in \cite[Section 12.4]{N69}, at both primes the closed cusps have degrees $1^{18},3^6$, every closed cusp is the
reduction of a Galois orbit of cusps over $F$ of the same degree, and we identify these orbits at the two primes using
the valuations of integral modular units. Reduction modulo $\mathfrak p_7$ and modulo $\mathfrak p_{13}$ is injective on
$J_X(F)$ by Katz's result.

The level automorphisms of $X$ defined over $F$ are the maps
\begin{equation}\label{eq:spor-G318}
(P,Q)\longmapsto(\alpha P+6\beta Q,\ \gamma P+\delta Q),\qquad \beta,\gamma\in\torz3,\quad \delta\in(\torz{18})^\times,
\quad \alpha\equiv\delta^{-1}\pmod3,
\end{equation}
taken modulo $\pm1$, so that we may take $\delta\in\{1,5,7\}$. They form a group $G_{3,18}$ of order $27$ (these are the
automorphisms of \cite[proof of Proposition 12.5]{N69}). Every element of $G_{3,18}$ fixes the point $9Q$ of order $2$.
If $g\neq1$ fixed a geometric non-cuspidal point $(E,P,Q)$ in characteristic $0$ or $p\geq5$, it would be induced by an
automorphism $u$ of $E$ whose image in $\Aut(E)/\{\pm1\}$ is non-trivial of order dividing $27$, hence of order $3$, and
such a $u$ fixes no non-zero point $T$ of order $2$: replacing $u$ by $-u$, which acts in the same way on $E[2]$, we
may assume that $u^2+u+1=0$, and then $uT=T$ gives $3T=0$. So $G_{3,18}$ acts freely on the non-cuspidal geometric points of $X$, also those with complex multiplication, and of
$X_{\F_7}$ and $X_{\F_{13}}$.

We also need an involution of $X$ which is not a level automorphism. For a point $(E,P,Q)$ of $Y_1(3,18)$ let
$\phi:E\to E'=E/\dm{9Q}$ and let $T'$ be the point of order $2$ in the kernel of the dual isogeny. Then
\begin{equation}\label{eq:spor-tau}
\tau(E,P,Q)=(E',\ \phi(P),\ 2\phi(Q)+T')
\end{equation}
defines an involution $\tau$ of $X$ over $F$: the point $2\phi(Q)+T'$ has order $18$, the Weil pairing is raised to
the power $2\deg\phi=4\equiv1\pmod3$, and $\tau^2(E,P,Q)=(E,-P,-Q)$.

\begin{lemma}\label{lem:spor-318-jac}
Let $F=\Q(\zeta_3)$, $X=X_1(3,18)$ over $F$, and let $\mathfrak p_7$ be a prime of $F$ above $7$.
\begin{enumerate}
\item[a)] We have $J_X(\F_7)\simeq(\torz3)^3\times(\torz9)^2\times(\torz{63})^2\times\torz{189}$ and
$\ClCusp_FX\simeq(\torz3)^4\times(\torz{63})^3$, of order $20253807$.
\item[b)] The common kernel of $A_{13}$ and $A_{19}$ on $J_X(\F_7)$ contains the reduction of $\ClCusp_FX$, with quotient
isomorphic to $(\torz3)^3$. In particular $3J_X(F)\subseteq\ClCusp_FX$.
\item[c)] The numbers of non-cuspidal closed points of degrees $1,\dots,5$ are $0,27,144,513,3672$ on $X_{\F_7}$ and
$0,54,846,7506,75384$ on $X_{\F_{13}}$. The numbers of orbits of $G_{3,18}$ on them in degrees $2,\dots,5$ are $1,8,19,136$
on $X_{\F_7}$ and $2,38,278,2792$ on $X_{\F_{13}}$.
\end{enumerate}
\end{lemma}

\begin{proof}
The invariants of $\ClCusp_FX$ are also given in \cite[Section 12.4]{N69}. We
\gitlink{computations/d31b_m318_paired_p7.m}{compute} $J_X(\F_7)$ and the reduction of $\ClCusp_FX$. As $13$ and $19$
are $\equiv1\pmod3$, the operators $A_{13}$ and $A_{19}$ annihilate $J_X(F)$ and the $F$-rational cuspidal divisors by
\Cref{prop:tools}. We \gitlink{computations/d31b_m318_kernel_structure_v2.m}{compute} their common kernel on
$J_X(\F_7)$, and as in the proof of \Cref{lem:spor-612-jac}~c) the injectivity of reduction gives $3J_X(F)\subseteq
\ClCusp_FX$. We \gitlink{computations/d31b_m3n18_frob_preflight.m}{compute} the numbers of closed points from the zeta
functions of $X_{\F_7}$ and $X_{\F_{13}}$, and the numbers of orbits by
\gitlink{computations/d31b_m318_paired_p7.m}{enumerating} the closed points at $7$ and
\gitlink{computations/d31b_m318_paired_p13.m}{at $13$} (in degree $5$ at $13$ see
\githubbare{computations/d31b_m318_p13_lean_s0.m}).
\end{proof}

We use the following formal argument, which replaces \Cref{lem:stab} when the gonality is too small for it. Let
$p\in\{7,13\}$, let $R=\Z_p$ be the completion of the ring of integers of $F$ at $\mathfrak p_p$, and let $h$ be an automorphism of $X$ over $F$ which maps cusps to cusps. We denote by $\alpha=h_*-1$ the induced endomorphism of
$J_X$. Recall that the cotangent space of $X^{(5)}_{\F_p}$ at an effective divisor $\overline D$ is
$H^0(\overline D,\Omega^1|_{\overline D})$, and that, after identifying the cotangent space of $J_X$ at $0$ with
$H^0(X_{\F_p},\Omega^1)$, the cotangent map at $\overline D$ of the map $X^{(5)}\to J_X$, $Z\mapsto[\alpha(Z)]=[hZ-Z]$, is
$\omega\mapsto((h^*-1)\omega)|_{\overline D}$.

\begin{lemma}\label{lem:spor-318-fibre}
Let $F=\Q(\zeta_3)$, $X=X_1(3,18)$ over $F$, $p\in\{7,13\}$, and let $h$ be an automorphism of $X$ over $F$ of order
$r\in\{2,3\}$ which maps cusps to cusps and extends to the smooth model of $X$ over $R=\Z_p$. Let $\overline N$ be a
non-cuspidal effective divisor on $X_{\F_p}$ with $h\overline N=\overline N$ such that $h$ acts freely on the geometric
points of $\overline N$, and let $\overline C$ be an effective cuspidal divisor on $X_{\F_p}$ which is the reduction of an
$F$-rational cuspidal divisor $C$, with $\deg\overline N+\deg\overline C=5$. Let $Y=X/\dm h$ and $\pi:X\to Y$ the quotient
map, and assume that
\begin{enumerate}
\item[a)] $\overline N=\pi^*(\overline M)$ for an effective divisor $\overline M$ on $Y_{\F_p}$ of degree $k=\deg\overline
N/r$, which is a sum of $k$ distinct points or a single closed point of degree $k$, and
\item[b)] the restriction map $(h^*-1)H^0(X_{\F_p},\Omega^1)\to H^0(\overline N+\overline C,\Omega^1|_{\overline
N+\overline C})$ has rank $5-k$.
\end{enumerate}
Then no effective $F$-rational divisor $D$ of degree $5$ on $X$ whose support contains no cusp reduces to $\overline
N+\overline C$ modulo $\mathfrak p_p$.
\end{lemma}

\begin{proof}
Suppose that $D$ reduces to $\overline N+\overline C$. As $h$ acts freely on the points of $\overline N$, it acts freely
on a neighbourhood of them in the smooth model of $X$ over $R$, so $\pi$ is \'etale there and $Y$ is smooth over $R$ at
the support of $\overline M$. Hence $\overline M$ lifts to an effective divisor $M$ on $Y$ over $R$ of the same shape. The
class
\[
y=[hD-D-hC+C]\in J_X(F)
\]
is torsion, as $J_X(F)$ is finite, and its reduction is $[h\overline N-\overline N]=0$. As $p$ is odd and unramified in
$F$, $y=0$ by Katz's result. So $D$ lies in the fibre $V=\{Z: [hZ-Z-hC+C]=0\}$ of the map $X^{(5)}_R\to J_{X,R}$ over
$0$. For every effective divisor $M'$ on $Y$ in a formal neighbourhood of $M$, the divisor $C+\pi^*M'$ lies in $V$, as
$\pi^*M'$ is $h$-invariant. These divisors form a closed formal subscheme $W$ of $V$ through $\overline N+\overline C$
which is smooth of relative dimension $k$ over $R$, as $M'$ is determined by $\pi^*M'$, and $\pi$ is \'etale at the
points of $\overline N$, whose support is disjoint from that of $\overline C$. On the other hand, by b) and the
description of the cotangent map above, the formal neighbourhood of $\overline N+\overline C$ in $V$ is contained in the
zero locus of $5-k$ functions on the formal neighbourhood of $\overline N+\overline C$ in $X^{(5)}_R$ (which is smooth of
relative dimension $5$ over $R$) with linearly independent differentials. This zero locus is formally smooth of relative
dimension $k$ over $R$ and integral, and it contains $W$, which has the same dimension, so it equals $W$. So $D=C+\pi^*M'$
for some $M'$, and the support of $D$ contains the cusps in the support of $C$, which is a contradiction.
\end{proof}

\begin{lemma}\label{lem:spor-318-cubic}
Let $F=\Q(\zeta_3)$, $X=X_1(3,18)$ over $F$, $p\in\{7,13\}$, and let $\mathfrak p_p$ be a prime of $F$ above $p$. Let
$\overline N$ be a non-cuspidal closed point of degree $3$ of $X_{\F_p}$ whose orbit under $G_{3,18}$ has length $9$, and
let $\overline C$ be an effective cuspidal divisor of degree $2$ on $X_{\F_p}$. Then no effective $F$-rational divisor
of degree $5$ on $X$ whose support contains no cusp reduces to $\overline N+\overline C$ modulo $\mathfrak p_p$.
\end{lemma}

\begin{proof}
As the closed cusps of $X_{\F_p}$ have degrees $1$ and $3$, the divisor $\overline C$ is a sum of cusps of degree $1$, and
it is the reduction of an $F$-rational cuspidal divisor, as every closed cusp is the reduction of a Galois orbit of
cusps over $F$ of the same degree. There are $\binom{19}{2}=171$ such divisors. If $D$ reduces to $\overline N+\overline
C$, then $gD$ reduces to $g\overline N+g\overline C$ for every $g\in G_{3,18}$, so we may assume that $\overline N$ is one
of the orbit representatives of \Cref{lem:spor-318-jac}~c). As $G_{3,18}$ acts freely on the non-cuspidal geometric
points of $X_{\F_p}$, the stabilizer of a closed point of degree $3$ acts freely on its $3$ geometric points, so the
orbits of these points have length $9$ or $27$, and by the numbers in \Cref{lem:spor-318-jac}~c), $4$ of the $8$ orbits
at $7$ and $10$ of the $38$ orbits at $13$ have length $9$. The stabilizer of $\overline N$ in $G_{3,18}$ has order $3$.
Let $h$ be a generator of it. Then $h$ permutes the $3$ geometric points of $\overline N$ freely, so $\overline N$ is the
fibre of $X\to X/\dm h$ over a point of degree $1$. As a level automorphism, $h$ maps cusps to cusps and extends to the smooth model of $X$ over $\Z_p$. We
\gitlink{computations/d31b_m318_relative_cubic_rank.m}{compute} that the assumption b) of \Cref{lem:spor-318-fibre},
with $r=3$ and $k=1$, holds in all the $4\cdot171=684$ cases at $7$, and
\gitlink{computations/d31b_m318_relative_cubic_rank_p13.m}{in all} the $10\cdot171=1710$ cases at $13$, so
\Cref{lem:spor-318-fibre} gives the claim.
\end{proof}

\begin{lemma}\label{lem:spor-318-quartic}
Let $F=\Q(\zeta_3)$, $X=X_1(3,18)$ over $F$, and let $\mathfrak p_7$ be a prime of $F$ above $7$. Let $\overline N_{12}$
and $\overline N_{17}$ be the representatives number $12$ and $17$ of the $19$ orbits of $G_{3,18}$ on the non-cuspidal
closed points of degree $4$ of $X_{\F_7}$ listed in \githubbare{continuation_20261031b/M3N18/paired_p7_degree4_data.m}.
Let $i\in\{12,17\}$, $g\in G_{3,18}$, and let $\overline c$ be a cusp of $X_{\F_7}$ of degree $1$. Then no effective
$F$-rational divisor of degree $5$ on $X$ whose support contains no cusp reduces to $g(\overline N_i+\overline c)$
modulo $\mathfrak p_7$.
\end{lemma}

\begin{proof}
Suppose that $D$ is such a divisor. Replacing $D$ by $g^{-1}D$, we may assume that $g=1$. The points $\overline N_{12}$ and $\overline N_{17}$ are fixed by the
involutions $h_{12}=g_{12}^{-1}\tau g_{12}$ and $h_{17}=g_{17}^{-1}\tau g_{17}$ respectively, where $g_{12}(P,Q)=(P+12Q,Q)$
and $g_{17}(P,Q)=(P+12Q,P+Q)$, which act on their residue field $\F_{7^4}$ as the square of the Frobenius, as we
\gitlink{computations/d31b_m318_quartic_phase_binding.m}{compute} (see also
\githubbare{computations/d31b_m318_tau_frame_p7.m} for $\tau$). As $\tau$ is defined by an isogeny of degree $2$, the
involutions $h_i$ map cusps to cusps and extend to the smooth model of $X$ over $\Z_7$. So $h_i$ acts freely on the
geometric points of $\overline N_i$, and $\overline N_i=\pi^*(\overline M_i)$ for a closed point $\overline M_i$ of
degree $2$ of $X/\dm{h_i}$, where $\pi$ is the quotient map. The cusp $\overline c$ is the reduction of an $F$-rational
cusp. The space $(h_i^*-1)H^0(X_{\F_7},\Omega^1)$ has dimension $6$, and we
\gitlink{computations/d31b_m318_relative_quartic_12.m}{compute} (and \gitlink{computations/d31b_m318_relative_quartic_17.m}{for
$\overline N_{17}$}) that its restriction to $\overline N_i+\overline c$ has rank $3$ for each of the $18$ cusps
$\overline c$ of degree $1$. So \Cref{lem:spor-318-fibre}, with $r=2$ and $k=2$, gives a contradiction.
\end{proof}

\begin{lemma}\label{lem:spor-318-classes}
Let $F=\Q(\zeta_3)$, $X=X_1(3,18)$ over $F$, and let $D$ be an effective $F$-rational divisor of degree $5$ on $X$ whose
support contains no cusp. Then there are $g\in G_{3,18}$ and $i\in\{1,2\}$ with $gD\sim A_i$, where $A_1$ and $A_2$ are
the $F$-rational cuspidal divisors of degree $5$ given in \githubbare{jobs/d31b_m318_short_cusp_divisors.log} (see also
\githubbare{continuation_20261031b/M3N18/endomorphism_gluing/short_cusp_divisors.m}).
\end{lemma}

\begin{proof}
Fix an $F$-rational cusp $c_\ast$. As $\gon_FX\geq6>5$, by \Cref{prop:main}~a) the reductions of $D$ modulo $\mathfrak
p_7$ and $\mathfrak p_{13}$ are not cuspidal. By \Cref{lem:spor-318-jac}~c), the non-cuspidal part of each of them is a
closed point of degree $2,\dots,5$ or a sum of two points of degree $2$ (possibly equal) or of a point of degree $2$ and a
point of degree $3$, and its cuspidal part is an effective cuspidal divisor of the complementary degree, which may contain
cusps of degree $3$. Applying an element of $G_{3,18}$, we may assume that at $\mathfrak p_7$ the first point of the
non-cuspidal part is one of the orbit representatives of \Cref{lem:spor-318-jac}~c), and we do the same separately at
$\mathfrak p_{13}$. This gives the following numbers of cases, according to the type of the non-cuspidal part.
\[
\begin{array}{c|rrrrrr|r}
 p&[2]&[3]&[4]&[5]&[2,2]&[2,3]&\text{total}\\\hline
 7&1146&1368&342&136&486&144&3622\\
 13&2292&6498&5004&2792&1944&1692&20222
\end{array}
\]

By the Hecke sieve (\Cref{subsec:sieve}), with $q=13$ and $q=19$, which are admissible for every cuspidal part as every
closed cusp is the reduction of an $F$-rational cuspidal divisor of the same degree, the class of the reduction modulo
$\mathfrak p_7$ of $D-5c_\ast$ lies in the common kernel of $A_{13}$ and $A_{19}$. By \Cref{lem:spor-318-cubic}, the
cases of a point of degree $3$ in an orbit of length $9$ with a cuspidal part, $4\cdot171=684$ at $\mathfrak p_7$ and
$10\cdot171=1710$ at $\mathfrak p_{13}$, do not occur. We
\gitlink{computations/d31b_m318_paired_three_relative_v2.py}{compute} that the Hecke sieve and these exclusions leave
$242$ of the $3622$ cases at $\mathfrak p_7$. Here, for the types $[2,2]$ and $[2,3]$, we use the classes of all the
closed points of degree $2$ and $3$ of $X_{\F_7}$ and $X_{\F_{13}}$, not only those of the orbit representatives, which
we \gitlink{independent_checks/recompute_318_classes.m}{compute} from the model of $X$.

By \Cref{lem:spor-318-jac}~b), the class $v=3[D-5c_\ast]$ lies in $\ClCusp_FX$, so its reductions modulo $\mathfrak p_7$ and
$\mathfrak p_{13}$ are the reductions of the same element of $\ClCusp_FX$. As we normalized the reductions by
$G_{3,18}$ at the two primes separately, we compare each case at $\mathfrak p_{13}$ with the $27$ translates of each of the
$242$ cases at $\mathfrak p_7$, computing the action of $G_{3,18}$ on the classes from its action on the cusps. We
\gitlink{computations/d31b_m318_paired_three_relative_v2.py}{find} that only $78$ of the cases at $\mathfrak p_{13}$ admit
such a match, with $53$ distinct non-cuspidal parts $\overline N$. For these we use the Hecke sieve at $\mathfrak p_{13}$
with $q=7$: as $7$ is admissible for every cuspidal part, $A_7\overline N$ has to be principal, and we
\gitlink{computations/d31b_m318_final_Hecke7_v2_s0.m}{compute} that it is not for $43$ of the $53$ divisors $\overline N$.
The remaining $10$ give $12$ cases at $\mathfrak p_{13}$. The cases at $\mathfrak p_7$ which match them are of type
$[4]$, with a point in the orbit of $\overline N_{12}$ or of $\overline N_{17}$ plus a cusp of degree $1$, which
\Cref{lem:spor-318-quartic} excludes, or one of two cases of type $[2,3]$.

There remain $4$ pairs of cases at $\mathfrak p_7$ and $\mathfrak p_{13}$, with $2$ cases of type $[2,3]$ at $\mathfrak
p_7$. For both of these, we \gitlink{computations/d31b_m318_direct_cusp_membership.py}{compute} that the class of the
reduction of $D-5c_\ast$ itself (not only its multiple by $3$) lies in the reduction of $\ClCusp_FX$, so by the
injectivity of reduction $[D-5c_\ast]\in\ClCusp_FX$. Comparing the classes $[D-5c_\ast]$ at the two primes, with the
action of $G_{3,18}$ on the cusps taken into account, two of the $4$ pairs are excluded, as their classes at $\mathfrak
p_{13}$ do not lie in the reduction of $\ClCusp_FX$, and the other two give the classes $[A_1-5c_\ast]$ and
$[A_2-5c_\ast]$. By injectivity, $gD\sim A_1$ or $gD\sim A_2$ for some $g\in G_{3,18}$.
\end{proof}

\begin{proposition}\label{prop:3x18}
Let $F=\Q(\zeta_3)$, let $u$ be a root of
\[
3u^5+(1+5\zeta_3)u^4+(6\zeta_3-3)u^3-(6\zeta_3+9)u^2-(5\zeta_3+4)u+3,
\]
and put $K_{3,18}=F(u)$, which is the field \lmfdbnf{10.0.2571428187.1}. Let
\[
19t=(4\zeta_3-25)u^4+(12-65\zeta_3)u^3+(20+31\zeta_3)u^2+(69+30\zeta_3)u+3\zeta_3-33,
\]
\[
E_{3,18}:\quad y^2+\bigl((u+2)t+1-u\bigr)xy+\bigl((u+1)t^2-ut\bigr)y=x^3,
\]
$P=(0,0)$, $Q=(-t,t^2)$, and let $E_{3,18}'=E_{3,18}/\dm{9Q}$.
\begin{enumerate}
\item[a)] The points $P$ and $Q$ have orders $3$ and $18$ and generate a subgroup isomorphic to $\tg{3}{18}$, and
$E_{3,18}(K_{3,18})_\tors\simeq E'_{3,18}(K_{3,18})_\tors\simeq\tg{3}{18}$. The $j$-invariants of $E_{3,18}$ and
$E'_{3,18}$ have degree $5$ over $\Q$, they are not algebraic integers, and their minimal polynomials over $\Q$ are
different.
\item[b)] Every effective $F$-rational divisor of degree $5$ on $X_1(3,18)$ whose support contains no cusp is one of the
$54$ closed points of degree $5$ in the two orbits under $G_{3,18}$ of the trace divisor $D_{3,18}$ of $(E_{3,18},P,Q)$ and of
$\tau(D_{3,18})$, each of which has $27$ elements.
\item[c)] If $E$ is an elliptic curve over a number field $K$ of degree $10$ with $\tg{3}{18}\subseteq E(K)$, then the
pair $(K,E)$ is isomorphic to $(K_{3,18},E_{3,18})$ or to $(K_{3,18},E'_{3,18})$, and $E(K)_\tors\simeq\tg{3}{18}$. These two pairs
are not isomorphic.
\end{enumerate}
\end{proposition}

\begin{proof}
We \gitlink{computations/d31b_m318_global_point13900_v2.m}{compute} that the quintic polynomial is irreducible over $F$,
that $P$ and $Q$ have orders $3$ and $18$, and that their Weil pairing is primitive, so $\tg{3}{18}\subseteq
E_{3,18}(K_{3,18})$. The image of $(P,Q)$ under the isogeny gives $\tg{3}{18}\subseteq E'_{3,18}(K_{3,18})$, as
in \eqref{eq:spor-tau}. We \gitlink{computations/d31b_m318_parent_nf_reductions.m}{compute} that an absolute defining
polynomial of $K_{3,18}$ is squarefree modulo $7$, with irreducible factors of degrees $2,2,3,3$, and that at the two primes of
$K_{3,18}$ of residue degree $2$ above $7$ both curves have good reduction with $54$ points over $\F_{49}$. As $7$ is odd and
unramified in $K_{3,18}$, the torsion of both curves has order dividing $54$. We
\gitlink{independent_checks/sporadic_check.m}{compute} the minimal polynomials of $j(E_{3,18})$ and $j(E'_{3,18})$,
which proves the rest of a).

We prove b). As $j(E_{3,18})$ has degree $5$ over $\Q$ and $[F:\Q]=2$, it has degree $5$ over $F$. The residue field of
the point $(E_{3,18},P,Q)$ contains $F(j(E_{3,18}))$ and is contained in $K_{3,18}$, so it is $K_{3,18}$, and $D_{3,18}$
is a closed point of degree $5$. As $\tau$ is an automorphism of $X$ over $F$ which maps cusps to cusps,
$\tau(D_{3,18})$ is a closed point of degree $5$ whose support contains no cusp, and its elliptic curve is
$E'_{3,18}$. By \Cref{lem:spor-318-classes} there are $g_1,g_2\in G_{3,18}$ and $i_1,i_2\in\{1,2\}$ with
$g_1D_{3,18}\sim A_{i_1}$ and $g_2\tau(D_{3,18})\sim A_{i_2}$. If $i_1=i_2$, then $g_1D_{3,18}=g_2\tau(D_{3,18})$ by
\Cref{lem:closed}, as $\gon_FX\geq6>5$, and as the level automorphisms do not change the elliptic curve,
$j(E_{3,18})$ and $j(E'_{3,18})$ would be conjugate, which contradicts a). So $\{i_1,i_2\}=\{1,2\}$. Now let $D$ be as
in b). By \Cref{lem:spor-318-classes} there are $g\in G_{3,18}$ and $i\in\{1,2\}$ with $gD\sim A_i$, so $gD$ is
linearly equivalent, and by \Cref{lem:closed} equal, to $g_1D_{3,18}$ or to $g_2\tau(D_{3,18})$. By \Cref{lem:orbits},
as $E_{3,18}$ and $E'_{3,18}$ have no complex multiplication by \Cref{lem:CM}, the stabilizers of $D_{3,18}$ and of
$\tau(D_{3,18})$ in $G_{3,18}$ have order dividing $\gcd(27,5)=1$, so both orbits have $27$ elements. As the level
automorphisms do not change the elliptic curve, and the $j$-invariants of $E_{3,18}$ and $E'_{3,18}$ are not conjugate
by a), the two orbits are distinct.

Part c) follows from a), b) and \Cref{lem:spor-determined}, applied to the point $(E_{3,18},P,Q)$ and to its image
under $\tau$, as in the proof of \Cref{prop:6x12}~c). As the minimal polynomials of $j(E_{3,18})$ and
$j(E'_{3,18})$ are different, no field automorphism maps one to the other, so the two pairs are not isomorphic.
\end{proof}

\begin{corollary}\label{cor:3x36-6x18}
The groups $\tg{3}{36}$ and $\tg{6}{18}$ are not subgroups of the torsion of an elliptic curve over a number field of
degree $10$.
\end{corollary}

\begin{proof}
Both groups contain $\tg{3}{18}$, so by \Cref{prop:3x18}~c) the torsion group of such a curve would be $\tg{3}{18}$, of
order $54$.
\end{proof}

\begin{remark}\label{rem:spor-318}
For the curve $X_1(3,18)$ over sextic and octic fields, where the trace divisor has degree $3$ or $4$, the stabilizer
argument of \Cref{lem:stab}~a) was enough to deal with the points of degree $3$ fixed by an element of order $3$ of
$G_{3,18}$ (see the proof of \cite[Proposition 12.5]{N69}). Here it would require $\gon_FX>7$, which is why we use
\Cref{lem:spor-318-fibre} instead. Note also that the field $K_{3,18}$ has an automorphism which acts non-trivially on $F$
(we \gitlink{independent_checks/aut_318_check.m}{compute} that $\Aut(K_{3,18})$ has order $2$ and that its
non-trivial element does not fix $\zeta_3$), so the last claim of \Cref{prop:3x18}~c) does not follow from b) alone.
\end{remark}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{The curve \texorpdfstring{$X_1(2,26)$}{X1(2,26)}}\label{subsec:spor-2x26}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Let $X=X_1(2,26)$ over $\Q$, a curve of genus $25$. We use the normal form \eqref{eq:spor-tate2}, with $P=(0,0)$ and
$Q=(b,b)$. By \cite[Theorem 3.1(3), Proposition 4.16 and Table 3]{DEHMZ},
$J_X(\Q)\simeq\torz2\times\torz{14}\times\torz{266}\times\torz{3990}\times\torz{11970}$ is generated by rational cuspidal
divisor classes, and $\gon_\Q X\geq12$ \cite[Table 1]{NV}. As in the proof of \cite[Proposition 13.22]{N69}, the closed cusps
of $X_{\F_3}$ have degrees $1^{18},3^6$, the $18$ cusps of degree $1$ are the reductions of the $18$ rational cusps, and
their differences generate the reduction of $J_X(\Q)$, which is a subgroup of index $350$ of
\[
J_X(\F_3)\simeq\torz2\times\torz{14}\times\torz{70}\times\torz{1330}\times\torz{3990}\times\torz{11970}
\]
Reduction modulo $3$ is injective on $J_X(\Q)$ by Katz's
result. The level automorphisms of $X$ permute the three points of order $2$ and act by $\Delta(13)$ on the point of order
$13$, and they form a group $G_{2,26}\simeq S_3\times\Delta(13)$ of order $36$ (see \cite[Section 11]{N69}). They are
defined over $\Q$.

\begin{lemma}\label{lem:spor-226-j0}
Let $K$ be a number field of degree at most $10$ and $E/K$ an elliptic curve with $j(E)=0$. Then $\tg{2}{26}\not\subseteq E(K)$.
\end{lemma}

\begin{proof}
Suppose the opposite. Then $E$ has a model $y^2=x^3-\lambda^3$ with $\lambda\in K^\times$, and as the other two roots
$\lambda\zeta_3^{\pm1}$ of $x^3-\lambda^3$ lie in $K$, we have $F=\Q(\zeta_3)\subseteq K$, so $[K:F]\leq5$. The curve $E$ is the
quadratic twist by $\lambda$ of $E_0:y^2=x^3-1$, so the $x$-coordinate of a point of order $13$ in $E(K)$, divided by $\lambda$,
is the $x$-coordinate of a point of order $13$ of $E_0$, and lies in $K$. So the $13$-division polynomial of $E_0$ has a root of
degree at most $5$ over $F$. We \gitlink{computations/late226_j0_twists_v2.m}{compute} that its irreducible factors over $F$
have degrees $6,6,72$, which is a contradiction.
\end{proof}

\begin{lemma}\label{lem:spor-226-red}
Let $X=X_1(2,26)$ and let $D$ be an effective rational divisor of degree $10$ on $X$ whose support contains no cusp. Then
there is $g\in G_{2,26}$ such that the reduction of $gD$ modulo $3$ is one of the following divisors, where
$\overline{N_9}$, $\overline{N_5}$, $\overline{N_5'}$ and $\overline{N_{10}}$ are closed points of $X_{\F_3}$ of degrees
$9$, $5$, $5$ and $10$ determined in the proof:
\begin{enumerate}
\item[a)] $\overline{N_9}+\overline c$, where $\overline c$ is a cusp of degree $1$,
\item[b)] $\overline{N_5}+\overline{N_5'}$,
\item[c)] $\overline{N_{10}}$.
\end{enumerate}
\end{lemma}

\begin{proof}
As $\gon_\Q X\geq12>10$, by \Cref{prop:main}~a) the reduction of $D$ modulo $3$ is not cuspidal. We
\gitlink{computations/late226_source.m}{compute} that the numbers of non-cuspidal closed points of $X_{\F_3}$ of degrees
$1,\dots,10$ are $0,0,0,0,36,122,288,936,2328,5382$ (the first nine are also in \cite[Table 4]{N69}). So the reduction of $D$ is
$x+C'$, where $x$ is a closed point of degree $e\in\{5,\dots,10\}$ or a sum of two closed points of degree $5$, and $C'$ is an
effective cuspidal divisor of degree $10-\deg x$, which may contain cusps of degree $3$. The $36$ points of degree $5$ form a single
orbit under $G_{2,26}$, and the points of degrees $6,\dots,10$ form $7$, $8$, $29$, $66$ and $154$ orbits. Applying an
element of $G_{2,26}$, we may assume that $x$, or its first summand of degree $5$, is one of these $265$ orbit
representatives, which gives $265+36=301$ possibilities for $x$.

By \Cref{lem:classtest}, applied with the reduction of $J_X(\Q)$, the class $[x+C'-10\overline{c_\ast}]$ lies in the
reduction of $J_X(\Q)$, so a possibility for $x$ can only occur if this holds for some effective cuspidal divisor $C'$
of degree $10-\deg x$ on $X_{\F_3}$. To speed up the
computation, we first \gitlink{computations/late226_filter26.m}{apply} the analogous condition \eqref{eq:pushclass} to the
push-forwards of $x$ by the four maps $X\to X_1(26)$ of \cite[Section 11]{N69}, which keep $Q$ or $P+Q$, or divide $E$ by
$\dm P$ or $\dm{P+13Q}$ and keep the image of $Q$, with the reduction of $J_1(26)(\Q)$, which is finite and generated by the rational
cuspidal divisor classes \cite[Theorem 3.1(2), Corollary 4.14 and Table 2]{DEHMZ}. This leaves $32$ of the $301$
possibilities. For these we \gitlink{computations/late226_picard_s0.m}{test} the condition on $X$ itself, allowing all the
effective cuspidal divisors $C'$ over $\F_3$, and find that only the three divisors in the statement satisfy it (see
\githubbare{continuation_20261029b/C226/picard_template.m} and the logs of the eight runs).
\end{proof}

\begin{lemma}\label{lem:spor-226-nine}
Let $X=X_1(2,26)$. No effective rational divisor $D$ of degree $10$ on $X$ whose support contains no cusp reduces modulo $3$
to the divisor $\overline{N_9}+\overline c$ of \Cref{lem:spor-226-red}~a).
\end{lemma}

\begin{proof}
Suppose the opposite. Let $g\in G_{2,26}$ be the level automorphism $(P,Q)\mapsto(13Q,P+17Q)$. As $g^3(P,Q)=(-P,-Q)$, it
has order $3$ in $G_{2,26}$, and we \gitlink{computations/late226_stabilizer_replay.py}{compute} that it fixes
$\overline{N_9}$, on which it acts as the cube of the Frobenius. Let $c$ be the rational cusp which reduces to $\overline c$. By
\Cref{lem:stab}~a), with $p=3$, $C=c$ and $\gon_\Q X\geq12>11$, we get $gD=D$.

If $g$ fixed a geometric point $(E,P,Q)$ in the support of $D$, there would be an automorphism $u$ of $E$ with
$u(P,Q)=\pm g(P,Q)$, whose image in $\Aut(E)/\{\pm1\}$ has order $3$, so $j(E)=0$. All the geometric points in the support of
$D$ are conjugate to a non-cuspidal point defined over a number field of degree at most $10$, over which, as $X$ is a fine
moduli space, its elliptic curve has a subgroup isomorphic to $\tg{2}{26}$, so this is impossible by \Cref{lem:spor-226-j0}. So $g$ acts freely on the geometric points in the support of
$D$, and as $gD=D$, these points form orbits of length $3$ on which the coefficients of $D$ are constant. Hence $3\mid\deg
D=10$, which is a contradiction.
\end{proof}

Let $K_{2,26}=\Q(a_1)$ and $K'_{2,26}=\Q(a_2)$, where $a_1$ and $a_2$ are roots of
\begin{align*}
F_1(x)&=2x^{10}+11x^9+17x^8-7x^7-42x^6-20x^5+9x^4+9x^3+10x^2+6x+1,\\
F_2(x)&=x^{10}+3x^9-3x^8-23x^7-26x^6+3x^5+13x^4-2x^3-9x^2-5x-1,
\end{align*}
which are the fields \lmfdbnf{10.6.396320664709.1} and \lmfdbnf{10.2.799905449.1}. Put
\[
b_1=\frac{B_1(a_1)}{406832},\qquad c_1=\frac{C_1(a_1)}{203416},\qquad b_2=\frac{B_2(a_2)}{784},\qquad c_2=\frac{C_2(a_2)}{1960},
\]
where the coefficients of the polynomials $B_i$ and $C_i$ are given in \Cref{table:spor-226}, and let $E_{2,26}=E'(b_1,c_1)$
over $K_{2,26}$ and $E'_{2,26}=E'(b_2,c_2)$ over $K'_{2,26}$ be the curves \eqref{eq:spor-tate2}, with $P=(0,0)$ and
$Q=(b_i,b_i)$.

\begin{table}[htbp]
\centering
\small
\begin{tabular}{r|rr|rr}
$j$&$[x^j]B_1$&$[x^j]C_1$&$[x^j]B_2$&$[x^j]C_2$\\\hline
0&$-621631$&$129155$&$-27507$&$90619$\\
1&$-2586377$&$551533$&$-95105$&$307177$\\
2&$-2555940$&$1269576$&$-100181$&$325777$\\
3&$-4632176$&$2528052$&$101187$&$-321491$\\
4&$2207386$&$1757370$&$190643$&$-622235$\\
5&$13819447$&$-2078127$&$-236919$&$755143$\\
6&$4462921$&$-1932589$&$-349828$&$1128928$\\
7&$-4689287$&$-133261$&$-76640$&$249316$\\
8&$-3688261$&$347217$&$41312$&$-133040$\\
9&$-721870$&$92982$&$17265$&$-55877$
\end{tabular}
\caption{The coefficients of the polynomials $B_1,C_1,B_2,C_2$ defining the curves $E_{2,26}$ and $E'_{2,26}$.}
\label{table:spor-226}
\end{table}

\begin{proposition}\label{prop:2x26}
Let $K_{2,26}$, $K'_{2,26}$, $E_{2,26}$ and $E'_{2,26}$ be as above.
\begin{enumerate}
\item[a)] We have $E_{2,26}(K_{2,26})_\tors\simeq E'_{2,26}(K'_{2,26})_\tors\simeq\tg{2}{26}$, generated by $P$ and $Q$. The
$j$-invariants of $E_{2,26}$ and $E'_{2,26}$ have degree $5$ over $\Q$, they are not algebraic integers, and their minimal
polynomials over $\Q$ are different.
\item[b)] Every effective rational divisor of degree $10$ on $X_1(2,26)$ whose support contains no cusp is one of the $36$
closed points of degree $10$ in the orbits under $G_{2,26}$ of the trace divisors $D_{2,26}$ and $D'_{2,26}$ of
$(E_{2,26},P,Q)$ and $(E'_{2,26},P,Q)$. Each of the two orbits has $18$ elements.
\item[c)] If $E$ is an elliptic curve over a number field $K$ of degree $10$ with $\tg{2}{26}\subseteq E(K)$, then the pair
$(K,E)$ is isomorphic to $(K_{2,26},E_{2,26})$ or to $(K'_{2,26},E'_{2,26})$, and $E(K)_\tors\simeq\tg{2}{26}$.
\end{enumerate}
\end{proposition}

\begin{proof}
We \gitlink{computations/resolve226_complete_native.m}{compute} that $F_1$ and $F_2$ are irreducible, that $P$ and $Q$ have
orders $2$ and $26$ on both curves, and that $P\neq13Q$, so $\tg{2}{26}\subseteq E_{2,26}(K_{2,26})$ and
$\tg{2}{26}\subseteq E'_{2,26}(K'_{2,26})$. At the prime of $K_{2,26}$ given by $a_1\mapsto23\in\F_{41}$ and at the prime of
$K'_{2,26}$ given by $a_2\mapsto39\in\F_{43}$, which are unramified as the defining polynomials are separable modulo $41$ and
$43$, the curves have good reduction, with $b_1\mapsto6$, $c_1\mapsto25$, respectively $b_2\mapsto21$, $c_2\mapsto11$, and
$52$ points. So both torsion groups have order $52$. We \gitlink{computations/resolve226_bind_classes.py}{compute} the
minimal polynomials of the $j$-invariants, which proves the rest of a).

We \gitlink{computations/resolve226_coordinate_fields.py}{compute} that $\Q(b_1,c_1)=K_{2,26}$ and $\Q(b_2,c_2)=K'_{2,26}$,
so $D_{2,26}$ and $D'_{2,26}$ are closed points of degree $10$ with these residue fields. Modulo $3$, $F_1$ and $F_2$ factor
into irreducible factors of degrees $5,5$ and $10$, and we \gitlink{computations/resolve226_bind_classes.py}{compute}, by
evaluating $b_i$ and $c_i$ at the roots of these factors, that $D_{2,26}$ reduces to the divisor of
\Cref{lem:spor-226-red}~b) and $D'_{2,26}$ to the divisor of \Cref{lem:spor-226-red}~c).

Now let $D$ be as in b). By \Cref{lem:spor-226-red,lem:spor-226-nine}, there is $g\in G_{2,26}$ such that $gD$ and
$D_{2,26}$, or $gD$ and $D'_{2,26}$, have the same reduction modulo $3$. Then $[gD-D_{2,26}]$, respectively
$[gD-D'_{2,26}]$, lies in $J_X(\Q)$ and reduces to $0$, so it is $0$, and $gD=D_{2,26}$, respectively $gD=D'_{2,26}$, by
\Cref{lem:closed}, as $\gon_\Q X\geq 12>10$. The two orbits are distinct, as the $j$-invariants are not conjugate. We
\gitlink{computations/late226_two_classes_stabilizers.py}{compute} that the reductions of $D_{2,26}$ and $D'_{2,26}$ are fixed
by the elements $(P,Q)\mapsto(P,P+5Q)$ and $(P,Q)\mapsto(13Q,P+18Q)$ of $G_{2,26}$, respectively, which have order $2$. By
\Cref{lem:stab}~a), with $C=0$, these elements fix $D_{2,26}$ and $D'_{2,26}$. On the other hand, by \Cref{lem:orbits},
the stabilizers have order dividing $\gcd(36,10)=2$, as the curves have no complex multiplication by \Cref{lem:CM}. So
both stabilizers have order $2$ and both orbits have $18$ elements. This proves b), and c) follows from a), b) and
\Cref{lem:spor-determined}, applied to the points $(E_{2,26},P,Q)$ and $(E'_{2,26},P,Q)$, as in the proof of
\Cref{prop:6x12}~c).
\end{proof}

\begin{proposition}\label{prop:52}
The group $\torz{52}$ 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, and let $E_0$ be an elliptic curve over a number field $K$ of degree $10$ with a point $R$ of order
$52$. Let $\phi:E_0\to E=E_0/\dm{26R}$. As in \Cref{lem:isogeny} (or the proof of \cite[Proposition 11.16]{N69}), the
point $\phi(R)$ has order $26$ and is independent of the generator of the kernel of the dual isogeny $\widehat\phi$, as
$\widehat\phi(13\phi(R))=26R\neq0$, so $\tg{2}{26}\subseteq E(K)$. By \Cref{prop:2x26}~c), after replacing $(K,E_0,E)$ by an
isomorphic triple, we may assume that $(K,E)$ is $(K_{2,26},E_{2,26})$ or $(K'_{2,26},E'_{2,26})$. Then $E_0\simeq
E/\ker\widehat\phi$, where $\ker\widehat\phi$ is one of the three subgroups of order $2$ of $E$, and the kernel $\dm{26R}$ of
the dual of $E\to E/\ker\widehat\phi$ is generated by the point $26R=2\cdot13R$, which is divisible by $2$ in $E_0(K)$.

Let $r_0=0$, $r_1=x(13Q)$ and $r_2=-c_i-r_1$ be the $x$-coordinates of the points of order $2$ of $E'(b_i,c_i)$, let
$\{s,t,w\}=\{0,1,2\}$, and put $u=r_t-r_s$ and $v=r_w-r_s$. Moving $(r_s,0)$ to $(0,0)$, the curve becomes
$y^2=x(x-u)(x-v)$, and its quotient by $\dm{(0,0)}$ is
\[
y^2=x^3+2(u+v)x^2+(u-v)^2x,
\]
on which the kernel of the dual isogeny is generated by $(0,0)$. A point $T$ with $2T=(0,0)$ satisfies $x(T)^2=(u-v)^2$, and
$y(T)^2=4(u-v)^2u$ if $x(T)=u-v$, and $y(T)^2=4(u-v)^2v$ if $x(T)=v-u$. So $(0,0)$ is divisible by $2$ over $K$ only if $u$ or
$v$ is a square in $K$. We \gitlink{computations/resolve226_complete_native.m}{compute} that none of the $6$ differences
$r_t-r_s$, $s\neq t$, is a square in $K_{2,26}$, respectively $K'_{2,26}$: their residues at the degree $1$ primes given in the
following table are non-zero non-squares.
\[
\begin{array}{c|c|cc}
(s,t) & K_{2,26}:\ p=41,\ a_1\mapsto23 & \multicolumn{2}{c}{K'_{2,26}:\ p=43}\\
 & r_t-r_s & a_2\mapsto & r_t-r_s\\\hline
(0,1)&38&16&33\\
(0,2)&19&39&18\\
(1,0)&3&39&29\\
(1,2)&22&16&19\\
(2,0)&22&16&34\\
(2,1)&19&39&39
\end{array}
\]
As a square in $K$ which is a unit at such a prime has a square residue, this is a contradiction.
\end{proof}

\begin{remark}\label{rem:spor-census}
The points of $X_1(m,n)$ of degree $d_m$ over $\Q(\zeta_m)$ found in this section are sporadic, as by the finiteness of
$J_X(\Q(\zeta_m))$ and $\gon_{\Q(\zeta_m)}X>d_m$ the curve $X$ has only finitely many points of degree at most $d_m$ over
$\Q(\zeta_m)$ (see the beginning of the section). They are listed in \cite{x1mn} as the points \xpoint{6.12.10.a},
\xpoint{3.15.10.a}, \xpoint{3.18.10.a}, \xpoint{3.18.10.b}, \xpoint{2.26.10.a} and \xpoint{2.26.10.b}, with one
representative for each pair $(K,E)$.
\end{remark}