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\section{Further cyclic groups}\label{sec:further}
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We show that the groups $\torz n$ for
$$n\in\{57,63,65,69,78,81,87,90,99,105,117\}$$
and the group $\tg{3}{39}$ are not subgroups of the torsion of an elliptic curve over a number field of degree $10$.
For all of them we reduce modulo $2$ the trace divisor on $X_1(n)$, or its push-forward to a quotient of
$X_1(n)$, to $X_0(n)$ or to $X_1(n')$ for a divisor $n'$ of $n$. We exclude the reductions with non-cuspidal
support mostly with the Eichler-Shimura relation and Riemann-Roch computations over finite fields, often by
linear algebra in the canonical ring (\Cref{lem:mult}), and the entirely cuspidal reductions by gonality bounds
or by formal immersion at $2$. For $n\in\{57,65\}$ the Jacobian $J_1(n)$ has positive rank, and for these levels and
for $n\in\{99,105\}$ we use elements of the Hecke algebra whose image on $J_1(n)(\Q)$ is finite. The groups
$\torz n$ with $n\in\{57,63,65,69,81,87\}$ were excluded in some of the degrees $6$, $7$, $8$ and $9$ in
\cite{N69}, where the curves $X_1(39)$ and $X_1(45)$ were also studied, and we reuse the computations from there which
do not depend on the degree.

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\subsection{The group \texorpdfstring{$\torz{57}$}{Z/57Z}}\label{subsec:fa-57}

We largely follow the proof of \cite[Proposition 8.3]{N69}, which excludes $\torz{57}$ over nonic fields.
The Jacobians $J_1(57)$ and $J_1(65)$ have positive rank, and we use the following facts about them.

\begin{lemma}\label{lem:fa-MW}
Let $n\in\{57,65\}$, let $X$ be $X_1(n)$ or its quotient by a subgroup of $\Delta(n)$, and let $a_r$ be the
$r$-th Fourier coefficient of the newform \lmfdbnewform{57}{2}{a}{a} if $n=57$, and of
\lmfdbnewform{65}{2}{a}{a} if $n=65$. Then $(\dm a-1)J_X(\Q)$ is finite for every $a\in(\torz n)^\times$,
and $(T_r-a_r)J_X(\Q)$ is finite for $r\in\{2,5,7,11,13\}$ if $n=57$ and for $r\in\{2,3,7,11,17\}$ if
$n=65$, where
$$\begin{aligned}
(a_2,a_5,a_7,a_{11},a_{13})&=(-2,-3,-5,1,2)&&\text{for } n=57,\\
(a_2,a_3,a_7,a_{11},a_{17})&=(-1,-2,-4,2,2)&&\text{for } n=65.
\end{aligned}$$
\end{lemma}

\begin{proof}
By \cite[Lemma 9.1]{N69}, the newform $f$ above is the only newform of level dividing $n$ with $L(f,1)=0$, and
it has trivial character. By Kato's theorem all the isogeny factors of $J_1(n)$ except the one corresponding
to $f$ have finitely many rational points, and the operators $\dm a-1$ and $T_r-a_r$ act as $0$ on the factor
corresponding to $f$ (see the proofs of \cite[Lemma 9.1, Propositions 8.3 and 8.4]{N69}). The Jacobian of $X$
is a quotient of $J_1(n)$.
\end{proof}

We first restrict the possible reductions with a quotient of $X_1(57)$ of genus $9$. On it we use the model given
by $q$-expansions, which is the dual model of \Cref{subsec:hecke}. The diamond operator $\dm5$ acts on this
quotient as an involution, so it is the same on both models.

\begin{lemma}\label{lem:fa-57-Y}
Let $H\subseteq\Delta(57)$ be the subgroup of order $9$, which is generated by $\dm4$, let $Y_{57}:=X_1(57)/H$,
let $\pi\colon X_1(57)\to Y_{57}$ be the quotient map, and put $\sigma:=\dm5$ and $\alpha:=\sigma-1$. Then
$Y_{57}$ has genus $9$ and good reduction at $2$, the group $\alpha J_{Y_{57}}(\Q)$ is finite, and
$$J_{Y_{57}}(\F_2)\simeq\torz{105}\times\torz{210}.$$
The group $V:=\alpha J_{Y_{57}}(\F_2)$ is isomorphic to $(\torz7)^2$ and is generated by the images under
$\alpha$ of the classes of the cuspidal divisors of degree $0$ on $Y_{57,\F_2}$. The reduction modulo $2$ of
$\alpha J_{Y_{57}}(\Q)$ is contained in the kernel $V_0\simeq\torz7$ of the operator $T_5-\dm5-5$ of the dual
model on $V$.
\end{lemma}

\begin{proof}
The genus and the good reduction are \cite[Lemma 9.4]{N69}, and $\alpha J_{Y_{57}}(\Q)$ is finite by
\Cref{lem:fa-MW}. The group $J_{Y_{57}}(\F_2)$, the fact that $V\simeq(\torz7)^2$ is generated by the images of the
cuspidal classes, and the kernel $V_0\simeq\torz7$ of $T_5-\dm5-5$ on $V$ are computed in Step 1 of the proof of
\cite[Proposition 8.3]{N69}. By \Cref{prop:tools}~b), in the form for the dual model (\Cref{subsec:hecke}), the
operator $T_5-\dm5-5$ annihilates the finite group $\alpha J_{Y_{57}}(\Q)$, so the reduction of this group lies in
$V_0$.
\end{proof}

\begin{lemma}\label{lem:fa-57-filter}
Let $\pi$, $\alpha$ and $V_0$ be as in \Cref{lem:fa-57-Y}, and let $c_\ast$ be a rational cusp of $Y_{57}$. Let
$P$ be a non-cuspidal closed point of $X_1(57)_{\F_2}$ of degree $e\leq10$ such that
$\alpha[\pi_*P-e\,\overline{c_\ast}]\in V_0$. Then $e\in\{6,8,10\}$. The non-cuspidal closed points of
$X_1(57)_{\F_2}$ of degree $10$ form $12$ orbits under $\Delta(57)$, and if $e=10$, then $P$ lies in one of the $6$
orbits whose points $(E,R)$ have the property that $j(E/\dm R)$ is a root of one of the polynomials
$$x^{10}+x^7+1,\qquad x^{10}+x^9+x^7+x^3+1,\qquad x^{10}+x^7+x^6+x^3+1.$$
If $c$ is one of the $2$ closed cusps of $X_1(57)_{\F_2}$ of degree $9$, then
$\alpha[\pi_*c-9\,\overline{c_\ast}]\notin V_0$.
\end{lemma}

\begin{proof}
For two rational cusps $c_\ast$ and $c_\ast'$ of $Y_{57}$ the class $\alpha[c_\ast-c_\ast']$ lies in
$\alpha J_{Y_{57}}(\Q)$, so its reduction lies in $V_0$ by \Cref{lem:fa-57-Y}, and the conditions do not depend on
the choice of $c_\ast$. By the Hasse bound $e\geq6$, as $\#E(\F_{2^k})\leq44<57$ for every elliptic curve $E$
over $\F_{2^k}$ with $k\leq5$. For $e\leq9$ and for the cusps of degree $9$ the statement is shown in Step 1 of the proof of
\cite[Proposition 8.3]{N69}.

Let $e=10$. The point $P$ corresponds to a pair $(E,R)$, where $E$ is an elliptic curve over $\F_{1024}$ and
$R\in E(\F_{1024})$ has order $57$, and in the dual model $\pi_*P$ lies above $j(E/\dm R)$, which does not change
under $\Delta(57)$. We \gitlink{continuation_20261030b/C57/cont57_census10.m}{compute} that $60$ of the $2053$
elliptic curves over $\F_{1024}$, up to isomorphism, have a point of order $57$, and that their $j$-invariants are
the roots of $6$ irreducible polynomials of degree $10$ over $\F_2$, and we
\gitlink{continuation_20261030b/C57/cont57_dual_match.m}{compute} that $40$ of these curves have
$E(\F_{1024})\simeq\torz{1026}$ and $20$ have $E(\F_{1024})\simeq\tg{3}{342}$, and that their $j$-invariants form $4$
and $2$ Frobenius orbits respectively. On a curve of the first kind the points of order $57$ generate one cyclic
subgroup, and on a curve of the second kind they generate $4$. Hence the non-cuspidal closed points of
$X_1(57)_{\F_2}$ of degree $10$ form $4+2\cdot4=12$ orbits under $\Delta(57)$. As
$\#(E/\dm R)(\F_{1024})=\#E(\F_{1024})$ and $57$ is squarefree, the curve $E/\dm R$ also has a point of order $57$
over $\F_{1024}$, so $\pi_*P$ is a closed point of degree $10$ above a root of one of the $6$ polynomials. There are
$48$ such closed points on $Y_{57,\F_2}$, and we \gitlink{continuation_20261030b/C57/cont57_quotient10.m}{find} that
the condition $\alpha[\pi_*P-10\,\overline{c_\ast}]\in V_0$ holds for $6$ of them, which lie above the roots of the
$3$ polynomials in the statement. The second computation above also gives $j(E/\dm R)$ for a point $(E,R)$ in each
of the $12$ orbits, and it is a root of one of these $3$ polynomials for $6$ of the $12$ orbits. So $P$ lies in one
of these $6$ orbits.
\end{proof}

\begin{lemma}\label{lem:fa-57-B}
Let $P$ be a non-cuspidal closed point of $X_1(57)_{\F_2}$ of degree $10$ in one of the $6$ orbits of
\Cref{lem:fa-57-filter}. Put $\sigma:=\dm5$ and $T:=T_5P$. Then the divisor
$$B:=\sigma T-T+P+4\sigma P-5\sigma^2P$$
on $X_1(57)_{\F_2}$ is not principal.
\end{lemma}

\begin{proof}
As the divisor $B$ for $\dm aP$ is $\dm aB$, it suffices to show that $B$ is not principal for a representative $P$
of each of the $6$ orbits.

For each representative we compute $T$ from the $6$ isogenies of degree $5$ of the corresponding elliptic curve,
carrying the level structure through them, and find that $T$ is the sum of $2$ closed points, of degrees $10$
and $50$, with multiplicity $1$. So the positive and the negative part of $B$ have degree $110$. As in Step 3 of
the proof of \cite[Proposition 8.3]{N69}, the points $\sigma P$, $\sigma^2P$ and the points of $T$ are computed on
the moduli model, so the identification of the diamond operators of the moduli model and the canonical model
plays no role. We apply \Cref{lem:mult} with $\mathcal L=\mathcal M=2K$ on $X_1(57)_{\F_2}$, which has genus
$85$, as in \cite[Section 8]{N69}. Put $R_i:=H^0(X_1(57)_{\F_2},iK)$. We compute the canonical model from an
integral basis of $S_2(\Gamma_1(57))$ and its $q$-expansions up to $q^{1100}$, and find $\dim R_1=85$,
$\dim R_2=252$ and $\dim R_4=588$, where $R_2$ and $R_4$ are spanned by products of the elements of the bases of
$R_1$ and $R_2$, and the identities between these products are checked beyond the Sturm bounds $480$ and $960$.
We obtain local parameters at the points in the support of $B$ from the deformation $a_6\mapsto a_6+t$ of the
ordinary curves $y^2+xy=x^3+a_2x^2+a_6$, by lifting the point of order $57$ by Hensel's lemma, and the canonical
coordinates from the modular units of \cite[Sections 2 and 2.1]{DvH} (see
\githubbare{continuation_20261030b/C57/cont57_good10_1.m}). For each of the $6$ representatives we
\gitlink{continuation_20261030b/C57/cont57_RR_v2.m}{compute} that the vanishing conditions at the positive and at
the negative part of $B$ both have rank $110$, that multiplication by the chosen section $F\in R_2$ has rank $252$,
and that the map of \Cref{lem:mult} has rank $252$, so it is injective and $\ell(B)=0$.
\end{proof}

\begin{lemma}\label{lem:fa-57-FI}
For $r\in\{2,5,7,11,13\}$ and $\ell\in\{5,7,11,13\}$ let
$$\Phi_{r,\ell}:=(T_r-a_r)(T_\ell-\dm\ell-\ell)$$
on the dual model of $X_1(57)$, with $a_r$ as in \Cref{lem:fa-MW}, and let $c_\ast$ be a rational cusp of
$X_1(57)$. Then each $\Phi_{r,\ell}$ annihilates $J_1(57)(\Q)$, and the morphism
$$f\colon X_1(57)^{(10)}\to J_1(57)^{20},\qquad B\mapsto\bigl(\Phi_{r,\ell}[B-10c_\ast]\bigr)_{r,\ell},$$
is a formal immersion modulo $2$ at every effective divisor of degree $10$ on $X_1(57)_{\F_2}$ supported on the
$18$ closed cusps of degree $1$ and the $9$ closed cusps of degree $2$.
\end{lemma}

\begin{proof}
The operator $T_r-a_r$ has finite image on $J_1(57)(\Q)$ by \Cref{lem:fa-MW}, and $T_\ell-\dm\ell-\ell$
annihilates $J_1(57)(\Q)_\tors$ by \Cref{prop:tools}~b) in the form for the dual model (\Cref{subsec:hecke}), so
$\Phi_{r,\ell}$ annihilates $J_1(57)(\Q)$. The image of the cotangent map of $f$ at an effective cuspidal divisor
$\overline C$ is the image of the restriction map from the space
$$W:=\sum_{r,\ell}\Phi_{r,\ell}^*H^0(X_1(57)_{\F_2},\Omega^1)$$
to $H^0(\overline C,\Omega^1|_{\overline C})$, so $f$ is a formal immersion at $\overline C$ if this restriction
has rank $10$. As in Step 2 of the proof of \cite[Proposition 8.3]{N69}, we compute $W$ modulo $2$ from the
action of the Hecke operators on an integral basis of $q$-expansions, and find $\dim W=82$. We use the
$q$-expansions as local expansions at the rational cusps, and at the cusps of degree $2$ we use local parameters
obtained from the Tate curve $E(q^{19})$ with the point $\zeta_3q$, which we check up to order $20$. As
$\Phi_{r,\ell}$ commutes with $\dm a$, we have $f(\dm aB)=\dm af(B)+f(10\dm ac_\ast)$, so $f$ is a formal immersion
at $\overline C$ if and only if it is one at $\dm a\overline C$. As $57$ is squarefree, $\Delta(57)$, of order $18$,
acts freely on the cusps \cite[Lemma 7.2]{N69}, so it acts simply transitively on the $18$ rational cusps. So, if $\overline C$ contains a rational cusp, we may assume that a rational cusp of maximal
multiplicity $n_0$ in $\overline C$ is the reduction of $c_\ast$. The other rational cusps then have multiplicity
at most $\min(n_0,5)$, and the cusps of degree $2$ at most $5$. For $n_0=1,\dots,10$ this gives
$$598417,\ 1165590,\ 455634,\ 124979,\ 29835,\ 6267,\ 1122,\ 162,\ 17,\ 1$$
divisors, and there are $\binom{13}{8}=1287$ divisors supported on the cusps of degree $2$. We
\gitlink{continuation_20261030b/C57/cont57_cusp_FI.py}{check} that the restriction of $W$ to each of these
$2383311$ divisors has rank $10$.
\end{proof}

\begin{proposition}\label{prop:57}
The group $\torz{57}$ is not a subgroup of the torsion of an elliptic curve over a number field of degree $10$.
\end{proposition}

\begin{proof}
Let $X=X_1(57)$, let $D$ be the trace divisor of a point of order $57$ over a number field of degree $10$, and let
$\overline D$ be its reduction modulo $2$. Let $\pi$, $\sigma$, $\alpha$ and $V_0$ be as in \Cref{lem:fa-57-Y}, and
let $c_\ast$ be a rational cusp of $Y_{57}$. The Galois orbits of cusps of $X$ have degrees $1^{18},2^9,18^2$, and
the closed cusps of $X_{\F_2}$ have degrees $1^{18},2^9,9^2,18$ \cite[Section 9.1]{N69}. So the closed cusps of
degree $1$ and $2$ are reductions of Galois orbits of cusps of the same degree, and every effective cuspidal
divisor on $X_{\F_2}$ supported on them is the reduction of an effective rational cuspidal divisor. For every
effective rational cuspidal divisor $C$ of degree $k$ on $X$, the classes $\alpha[\pi_*D-10c_\ast]$ and
$\alpha[\pi_*C-kc_\ast]$ lie in $\alpha J_{Y_{57}}(\Q)$, so their reductions lie in $V_0$ by \Cref{lem:fa-57-Y}.

Suppose first that $\overline D$ contains a non-cuspidal closed point $P$, of degree $e$. By the Hasse bound
$e\geq6$ (see the proof of \Cref{lem:fa-57-filter}), so, as $2\cdot6>10$, we have $\overline D=P+\overline C$, where
$\overline C$ is cuspidal of degree $10-e\leq4$. So $\overline C$ is supported on the cusps of degree $1$ and $2$, and it is the reduction of an
effective rational cuspidal divisor $C$. Hence
$$\alpha[\pi_*P-e\,\overline{c_\ast}]=\alpha[\pi_*\overline D-10\,\overline{c_\ast}]-\alpha[\pi_*\overline C-(10-e)\,\overline{c_\ast}]\in V_0,$$
and by \Cref{lem:fa-57-filter} we have $e\in\{6,8,10\}$, and if $e=10$, then $P$ lies in one of the $6$ orbits of
that lemma. Let $c$ be a rational cusp of $X$. The class
$[\sigma D-D]=(\sigma-1)[D-10c]+10[\sigma c-c]$ is torsion by \Cref{lem:fa-MW} and by Manin-Drinfeld, so it is
annihilated by $A_5=T_5-5\sigma-1$ (\Cref{prop:tools}~b)). As $A_5C=0$ by \Cref{prop:tools}~c), we also have
$A_5(\sigma C-C)=\sigma A_5C-A_5C=0$. Reducing modulo $2$, we get that
$$A_5(\sigma P-P)=\sigma T-T+P+4\sigma P-5\sigma^2P,\qquad T=T_5P,$$
is principal on $X_{\F_2}$. If $e=10$, this contradicts \Cref{lem:fa-57-B}. If $e\in\{6,8\}$, then $P$ is one of
the $2$ non-cuspidal closed points of $X_{\F_2}$ of degree $6$, or lies in the single $\Delta(57)$-orbit of
non-cuspidal closed points of degree $8$ \cite[Section 9.1 and proof of Proposition 9.6]{N69}, and it is shown in
Step 3 of the proof of \cite[Proposition 8.3]{N69} that this divisor is not principal for these points. This
computation depends only on $P$, so we again get a contradiction.

Suppose next that $\overline D$ is cuspidal and contains one of the $2$ closed cusps $c$ of degree $9$. Then
$\overline D=c+\overline r$, where $r$ is a rational cusp, and as above
$\alpha[\pi_*c-9\,\overline{c_\ast}]\in V_0$, which contradicts \Cref{lem:fa-57-filter}.

Suppose finally that $\overline D$ is supported on the cusps of degree $1$ and $2$. Then $\overline D$ is the
reduction of an effective rational cuspidal divisor $C$ of degree $10$. Let $f$ be the morphism of
\Cref{lem:fa-57-FI}. As $[D-C]\in J_1(57)(\Q)$, we have $f(D)=f(C)$, and $D$ and $C$ have the same reduction
$\overline D$, at which $f$ is a formal immersion by \Cref{lem:fa-57-FI}. By \cite[Lemma 4.2]{DKSS} (which is due
to Parent \cite[Lemme 4.13]{parentformal}) we get $D=C$, which is a contradiction, as the support of $D$ contains
no cusp.
\end{proof}

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\subsection{The group \texorpdfstring{$\torz{65}$}{Z/65Z}}\label{subsec:fa-65}

We largely follow the proof of \cite[Proposition 8.4]{N69}, which excludes $\torz{65}$ over nonic fields. The
following lemma collects the facts about a quotient of genus $9$ which we use for a reduction containing a cusp of
degree $6$. All of them are shown in the proof of \cite[Proposition 8.4]{N69}.

\begin{lemma}\label{lem:fa-65-Y}
Let $Y_{65}$ be the quotient of $X_1(65)$ by the kernel of the character $\dm u\mapsto\left(\frac{65}{u}\right)$ of
$\Delta(65)$, let $\pi\colon X_1(65)\to Y_{65}$ be the quotient map, put $\alpha:=\dm3-1$, and let $c_\ast$ be a
rational cusp of $Y_{65}$. Then $Y_{65}$ has genus $9$ and good reduction at $2$, and
$J_{Y_{65}}(\F_2)\simeq\torz4\times\torz{16}\times\torz{336}$. For each of the $4$ closed cusps $c$ of degree $6$
of $X_1(65)_{\F_2}$ we have $\pi_*c=6c_Y$ for a closed cusp $c_Y$ of $Y_{65,\F_2}$ of degree $1$, and the class
$6\alpha[c_Y-\overline{c_\ast}]$ is not annihilated by $T_3-\dm3-3$, the annihilator of the dual model.
\end{lemma}

\begin{proof}
The genus and the remaining statements are shown in the proof of \cite[Proposition 8.4]{N69}, and $Y_{65}$ has
good reduction at $2$ by \Cref{lem:goodred}. There the classes are normalized at a fixed rational cusp. For another
rational cusp $c_\ast'$ the class $\alpha[c_\ast-c_\ast']$ lies in $\alpha J_{Y_{65}}(\Q)$, which is finite by
\Cref{lem:fa-MW}, so it is annihilated by $T_3-\dm3-3$ (\Cref{prop:tools}~b) and \Cref{subsec:hecke}), and the
last statement does not depend on the choice of $c_\ast$.
\end{proof}

\begin{lemma}\label{lem:fa-65-B}
Let $P$ be a non-cuspidal closed point of $X_1(65)_{\F_2}$ of degree $10$. Put $\sigma:=\dm3^{-1}$,
$\tau:=\dm3$ and $T:=T_3P$. Then the divisor
$$B:=\sigma T-T-\sigma P-2P+3\tau P$$
on $X_1(65)_{\F_2}$ is not principal.
\end{lemma}

\begin{proof}
We \gitlink{computations/contc65_source_census_v2.m}{compute} that $45$ of the $2053$ elliptic curves over
$\F_{1024}$, up to isomorphism, have a point of order $65$, all with $E(\F_{1024})\simeq\torz{1040}$, and that they
give $108$ non-cuspidal closed points of $X_1(65)_{\F_2}$ of degree $10$, which form $5$ orbits under
$\Delta(65)$, of lengths $24$, $24$, $24$, $12$ and $24$. (Each closed point in the orbit of length $12$ is
mapped to itself by a diamond operator of order $2$, which permutes its geometric points as a power of the
Frobenius.) As the divisor
$B$ for $\dm aP$ is $\dm aB$, it suffices to treat a representative $P$ of each orbit.

For each representative we \gitlink{computations/contc65_ordinary10_9.m}{compute} $T$ from the $4$ isogenies of
degree $3$ of the corresponding elliptic curve, and find that $T$ is the sum of $2$ closed points of degree $20$
with multiplicity $1$. So the positive and the negative part of $B$ have degree $70$. As in the proof of
\cite[Proposition 8.4]{N69}, we compute $X_1(65)$, of genus $121$, through its canonical forms and the modular
units of \cite[Sections 2 and 2.1]{DvH}, and apply \Cref{lem:mult} with $\mathcal L=\mathcal M=K$. The space
$H^0(X_1(65)_{\F_2},2K)$ has dimension $360$ and is spanned by the products of the canonical forms, which we check
beyond the Sturm bound $672$. We obtain local parameters from the deformation $a_6\mapsto a_6+t$ of the ordinary
curves $y^2+xy=x^3+a_2x^2+a_6$ by lifting the point of order $65$ by Hensel's lemma. As the identification of the
diamond operators of the moduli model with those of the canonical model is a matter of convention
(\Cref{subsec:hecke}), we test, as in \cite[Section 8]{N69}, both $B$ and the divisor
$\sigma T-T+2P+\tau P-3\sigma P$ obtained from the annihilator of the dual model, for each of the $24$ simultaneous
diamond translates of the points of $T$. For each of these $5\cdot24\cdot2=240$ divisors we
\gitlink{computations/contc65_R1_sieve.py}{check} that the map of \Cref{lem:mult}, a $121\times309$ matrix, has rank
$121$, so it is injective and the divisor is not principal.
\end{proof}

\begin{lemma}\label{lem:fa-65-FI}
For $r\in\{2,3,7,11,17\}$ and $\ell\in\{3,7,11,17\}$ let
$$\Phi_{r,\ell}:=(T_r-a_r)(T_\ell-\dm\ell-\ell)$$
on the dual model of $X_1(65)$, with $a_r$ as in \Cref{lem:fa-MW}, and let $c_\ast$ be a rational cusp of
$X_1(65)$. Then each $\Phi_{r,\ell}$ annihilates $J_1(65)(\Q)$, and the morphism
$$f\colon X_1(65)^{(10)}\to J_1(65)^{20},\qquad B\mapsto\bigl(\Phi_{r,\ell}[B-10c_\ast]\bigr)_{r,\ell},$$
is a formal immersion modulo $2$ at every effective divisor of degree $10$ on $X_1(65)_{\F_2}$ supported on the
$24$ closed cusps of degree $1$ and the $6$ closed cusps of degree $4$.
\end{lemma}

\begin{proof}
This is proved in the same way as \Cref{lem:fa-57-FI}. The space
$$W:=\sum_{r,\ell}\Phi_{r,\ell}^*H^0(X_1(65)_{\F_2},\Omega^1)$$
has dimension $106$, and at the cusps of degree $4$ we use local parameters obtained from the Tate curve
$E(q^{13})$ with the point $\zeta_5q$ and its diamond translates. As $4\nmid10$, every effective divisor of degree
$10$ supported on these cusps contains a rational cusp. As $65$ is squarefree, $\Delta(65)$, of order $24$, acts
freely on the cusps \cite[Lemma 7.2]{N69}, so it acts simply transitively on the $24$ rational cusps, and we may
assume that a rational cusp of maximal multiplicity $n_0$ is the reduction of
$c_\ast$. The other rational cusps then have multiplicity at most $\min(n_0,5)$, and the cusps of degree $4$ at
most $2$. For $n_0=1,\dots,10$ this gives
$$1019567,\ 4152510,\ 1521680,\ 377867,\ 80868,\ 14956,\ 2300,\ 276,\ 23,\ 1$$
divisors, $7170048$ in total, and we \gitlink{computations/contc65_cusp_FI_n1.py}{check} that the restriction of
$W$ to each of them has rank $10$ (see also \githubbare{computations/contc65_cusp_FI_common.py}).
\end{proof}

\begin{proposition}\label{prop:65}
The group $\torz{65}$ is not a subgroup of the torsion of an elliptic curve over a number field of degree $10$.
\end{proposition}

\begin{proof}
Let $X=X_1(65)$, let $D$ be the trace divisor of a point of order $65$ over a number field of degree $10$, and let
$\overline D$ be its reduction modulo $2$. The Galois orbits of cusps of $X$ have degrees $1^{24},4^6,12^2,24$,
and the closed cusps of $X_{\F_2}$ have degrees $1^{24},4^6,6^4,12^2$ \cite[Section 9.1]{N69}. So the closed
cusps of degree $1$ and $4$ are reductions of Galois orbits of cusps of the same degree, and every effective
cuspidal divisor on $X_{\F_2}$ supported on them is the reduction of an effective rational cuspidal divisor. By
the Hasse bound and \cite[proof of Proposition 8.4]{N69}, the non-cuspidal closed points of $X_{\F_2}$ of degree at
most $9$ have degrees $6$, $7$, $8$ and $9$.

Suppose first that $\overline D$ contains a non-cuspidal closed point $P$, of degree $e$. Then $e\geq6$, and as
$2\cdot6>10$ we have $\overline D=P+\overline C$, where $\overline C$ is cuspidal of degree $10-e\leq4$, so
$\overline C$ is the reduction of an effective rational cuspidal divisor $C$. Put $\sigma=\dm3^{-1}$, $\tau=\dm3$ and $A_3=T_3-1-3\tau$. As in the
proof of \Cref{prop:57}, the class $[\sigma D-D]$ is torsion by \Cref{lem:fa-MW} and by Manin-Drinfeld, so it is
annihilated by $A_3$, and $A_3(\sigma C-C)=0$. Reducing modulo $2$, we get that
$$A_3(\sigma P-P)=\sigma T-T-\sigma P-2P+3\tau P,\qquad T=T_3P,$$
is principal on $X_{\F_2}$. For $e=10$ this contradicts \Cref{lem:fa-65-B}. For $e\leq9$ it is shown in the proof
of \cite[Proposition 8.4]{N69}, for a representative $P$ of each $\Delta(65)$-orbit and hence for all $P$, that this
divisor is not principal (for $e=6$ it is shown for $A_3(\tau P-P)=-\tau A_3(\sigma P-P)$). This computation
depends only on $P$, so we again get a contradiction.

Suppose next that $\overline D$ is cuspidal and contains one of the $4$ closed cusps $c$ of degree $6$. Then
$\overline D=c+\overline C$, where $\overline C$ has degree $4$ and is the reduction of an effective rational
cuspidal divisor $C$. Let $\pi$, $\alpha$, $c_\ast$ and $c_Y$ be as in \Cref{lem:fa-65-Y}. The classes
$\alpha[\pi_*D-10c_\ast]$ and $\alpha[\pi_*C-4c_\ast]$ lie in $\alpha J_{Y_{65}}(\Q)$, which is finite by
\Cref{lem:fa-MW}, so they are annihilated by $T_3-\dm3-3$ (\Cref{prop:tools}~b) and \Cref{subsec:hecke}). The
reduction of their difference is $\alpha[\pi_*c-6\,\overline{c_\ast}]=6\alpha[c_Y-\overline{c_\ast}]$, which is
then also annihilated by $T_3-\dm3-3$, and this contradicts \Cref{lem:fa-65-Y}.

Suppose finally that $\overline D$ is cuspidal and does not contain a cusp of degree $6$. As it has degree $10$, it
is supported on the cusps of degree $1$ and $4$, so it is the reduction of an effective rational cuspidal divisor
$C$. As in the proof of \Cref{prop:57}, the morphism $f$ of \Cref{lem:fa-65-FI} satisfies $f(D)=f(C)$, and it is a
formal immersion at $\overline D$, so $D=C$ by \cite[Lemma 4.2]{DKSS}, which is a contradiction.
\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{The group \texorpdfstring{$\torz{63}$}{Z/63Z}}\label{subsec:fa-63}

We work on the quotient $Y_{63}$ of \cite[Lemma 7.8]{N69}, as in \cite[Proposition 7.12]{N69}. In degree $10$ the
cuspidal part of a reduction can contain a closed cusp of degree $3$ which is not the reduction of a rational
cuspidal divisor, and we treat it separately.

\begin{lemma}\label{lem:fa-63-Y}
Let $Y_{63}:=X_1(63)/\dm8$ and let $\pi\colon X_1(63)\to Y_{63}$ be the quotient map.
\begin{itemize}
\item[a)] The diamond operator $\dm8$ has order $2$ and acts freely on $X_1(63)$, also in characteristic $2$, the
curve $Y_{63}$ has genus $49$ and good reduction at $2$, and $J_{Y_{63}}(\Q)$ is finite.
\item[b)] The Galois orbits of cusps of $Y_{63}$ have degrees $1^9,2^3,3^6,6,9$, and the closed cusps of
$Y_{63,\F_2}$ have degrees $1^9,2^3,3^9,6$. The closed cusps of degree $1$ and $2$, and $6$ of the closed cusps of
degree $3$, are reductions of Galois orbits of cusps of the same degree. The other $3$ closed cusps of degree $3$,
which we denote by $C_1$, $C_2$ and $C_3$, are the closed cusps of degree $3$ at which $j$ has a simple pole, and
$\sigma_5$ acts on them as $\dm5$.
\item[c)] Let $P$ be a non-cuspidal closed point of $X_1(63)_{\F_2}$ of degree $e\leq10$. Then $e\in\{7,8,9,10\}$,
and $\pi_*P$ is a closed point of $Y_{63,\F_2}$ of degree $e$. For $e=7,8,9,10$ these closed points of
$Y_{63,\F_2}$ are $27$, $27$, $36$ and $27$ points, which form $3$, $3$, $4$ and $3$ orbits under the diamond
operators $\dm2$ and $\dm5$.
\end{itemize}
\end{lemma}

\begin{proof}
Part a) is \cite[Lemma 7.8]{N69}. We \gitlink{continuation_20261031a/C63/inputs/C63/cusp63.py}{compute} the
degrees in b) from the description of the cusps of $X_1(63)$, and we check there that the closed cusps of degree at
most $2$ are reductions of Galois orbits of cusps of the same degree (see also \cite[proof of Proposition
7.12]{N69}). Comparing the two lists of degrees, each Galois orbit of degree $3$ reduces to a closed cusp of degree
$3$, and the orbit of degree $9$ splits into the other $3$ closed cusps of degree $3$. The cusps of $X_1(63)$ at which
$j$ has a simple pole are the $18$ cusps $(\mathrm{Tate}(q),\zeta)$, with $\zeta$ a primitive $63$-rd root of unity
taken up to sign. As $\dm8$ acts freely, their images are the cusps of $Y_{63}$ at which $j$ has a simple pole, and
they form the Galois orbit of degree $9$. Both $\sigma_5$ and $\dm5$ map $(\mathrm{Tate}(q),\zeta)$ to
$(\mathrm{Tate}(q),\zeta^5)$, so they act in the same way on these cusps.

For $e\leq9$ part c) is shown in the proof of \cite[Proposition 7.12]{N69}, where it is also shown that there are
no such points of degree at most $6$. Let $e=10$. The function field of $Y_{63}$ is obtained by adjoining to the function field $\Q(t)$ of
$X_1(9)$ a root $u$ of the $7$-division polynomial of the universal elliptic curve $E_t$ with a point of order $9$ in
Tate normal form \cite[Lemma 7.8]{N69}. We \gitlink{computations/d31a_63_good10.m}{enumerate} the pairs $(t,u)$
over $\F_{1024}$ for which $E_t$ is an elliptic curve with $63\mid\#E_t(\F_{1024})$ and $u$ is the $x$-coordinate
of a point of order $7$ of $E_t(\F_{1024})$, which give the images on $Y_{63}$ of all the non-cuspidal points of
$X_1(63)$ over $\F_{1024}$. There are $270$ of them, and they form $27$ Frobenius orbits of length $10$, which form
$3$ orbits under $\dm2$ and $\dm5$.
\end{proof}

\begin{lemma}\label{lem:fa-63-A5}
Let $Y_{63}$ and $C_1,C_2,C_3$ be as in \Cref{lem:fa-63-Y}, and put $A_5:=T_5-1-5\dm5$ on $Y_{63}$.
\begin{itemize}
\item[a)] For every non-cuspidal closed point $P$ of $Y_{63,\F_2}$ of degree $10$ as in \Cref{lem:fa-63-Y}~c), the
divisor $A_5P$ is not principal.
\item[b)] For every non-cuspidal closed point $P$ of $Y_{63,\F_2}$ of degree $7$ as in \Cref{lem:fa-63-Y}~c), and
all $i\neq j$ in $\{1,2,3\}$, the divisor $A_5P+4(C_i-C_j)$ is not principal.
\end{itemize}
\end{lemma}

\begin{proof}
The diamond operators commute with $A_5$ and permute $C_1$, $C_2$ and $C_3$, so it suffices to consider a
representative $P$ of each of the $3$ orbits in each degree, together with all the ordered pairs $(i,j)$. We use
the model of $Y_{63,\F_2}$ described in the proof of \Cref{lem:fa-63-Y}, and compute $T_5P$ from the $6$ isogenies
of degree $5$ of the corresponding elliptic curve, carrying the points of order $9$ and $7$ through them. For $P$ of
degree $10$, $T_5P$ is the sum of $2$ closed points of degree $30$, and we
\gitlink{computations/d31a_63_good10.m}{compute} that $\ell(A_5P)=0$ for the $3$ representatives. For $P$ of degree
$7$, $T_5P$ is a closed point of degree $42$, and we \gitlink{computations/d31a_63_good7.m}{compute} that
$\ell(A_5P+4(C_i-C_j))=0$ in all the $3\cdot6=18$ cases.
\end{proof}

\begin{lemma}\label{lem:fa-63-cusp}
Let $Y_{63}$ be as in \Cref{lem:fa-63-Y}, and put $U:=(\Cart-1)H^0(Y_{63,\F_2},\Omega^1)$. Then $\dim U=45$, and the
restriction map $U\to H^0(\overline C,\Omega^1|_{\overline C})$ is surjective for every effective cuspidal divisor
$\overline C$ of degree $10$ on $Y_{63,\F_2}$.
\end{lemma}

\begin{proof}
The dimension of $U$ is computed in the proof of \cite[Proposition 7.12]{N69}. By \Cref{lem:fa-63-Y}~b) there are
$[x^{10}](1-x)^{-9}(1-x^2)^{-3}(1-x^3)^{-9}(1-x^6)^{-1}=239238$ effective cuspidal divisors of degree $10$ on
$Y_{63,\F_2}$. For a closed cusp $c$ and $m\deg c\leq9$ we use the subspaces $\Omega(mc)$ from the computation for
degree $9$ in the proof of \cite[Proposition 7.12]{N69}, and we \gitlink{computations/d31a_63_high_1.m}{compute}
the subspaces $\Omega(10c)$ for the $9$ cusps $c$ of degree $1$ and $\Omega(5c)$ for the $3$ cusps $c$ of degree
$2$. Using these, we \gitlink{computations/d31a_63_existing_boundary.py}{check} that
$\dim U-\dim\bigl(U\cap\Omega(\overline C)\bigr)=10$ for each of the $239238$ divisors $\overline C$.
\end{proof}

\begin{proposition}\label{prop:63}
The group $\torz{63}$ is not a subgroup of the torsion of an elliptic curve over a number field of degree $10$.
\end{proposition}

\begin{proof}
Let $D$ be the trace divisor of a point of order $63$ over a number field of degree $10$, let $\pi$, $C_1$, $C_2$
and $C_3$ be as in \Cref{lem:fa-63-Y}, and let $c_\ast$ be a rational cusp of $Y_{63}$. As $J_{Y_{63}}(\Q)$ is
finite, the operator $A_5=T_5-1-5\dm5$ annihilates it (\Cref{prop:tools}~b)), and $A_5c_\ast=0$
(\Cref{prop:tools}~c)), so $A_5\pi_*D$ is principal, as its class is $A_5[\pi_*D-10c_\ast]=0$.

Suppose first that the reduction of $D$ modulo $2$ is not entirely cuspidal. By \Cref{lem:fa-63-Y}~c) its
non-cuspidal part is a single closed point of degree $e\in\{7,8,9,10\}$, with multiplicity $1$ as $2\cdot7>10$, and
the reduction of $\pi_*D$ is $P+\overline C$, where $P$ is one of the closed points of $Y_{63,\F_2}$ of degree $e$ of
\Cref{lem:fa-63-Y}~c) and $\overline C$ is cuspidal of degree $10-e\leq3$. If $\overline C$ is the reduction of an effective rational
cuspidal divisor $C$, then $A_5C=0$ by \Cref{prop:tools}~c), so $A_5P$ is principal. For $e=10$ this contradicts
\Cref{lem:fa-63-A5}~a), and for $e\leq9$ it is shown in the proof of \cite[Proposition 7.12]{N69} that $A_5P$ is not
principal for a representative $P$ of each orbit, and hence for all of them. Otherwise, by \Cref{lem:fa-63-Y}~b),
$e=7$ and $\overline C=C_i$ for some $i$. By \Cref{lem:cuspHecke}, as $\sigma_5$ acts on $C_i$ as $\dm5$, we have
$$A_5C_i=\sigma_5C_i-C_i+5\dm5\bigl(\dm5^{-1}C_i-C_i\bigr)=4(C_i-\sigma_5C_i),$$
and $\sigma_5C_i=C_j$ for some $j$. If $j=i$, then $A_5P$ is principal, which we have just excluded. If $j\neq i$,
then $A_5P+4(C_i-C_j)$ is principal, which contradicts \Cref{lem:fa-63-A5}~b).

So the reduction of $\pi_*D$ is entirely cuspidal. By \Cref{lem:fa-63-Y}~a) the curve $Y_{63}$ satisfies the
assumptions of \Cref{prop:cartier}, and $J_{Y_{63}}(\Q)$ is finite, so we may take $r=1$ and $\psi_1=1$. By
\Cref{lem:fa-63-cusp} the restriction of $U$ to the reduction of $\pi_*D$ is surjective, and \Cref{prop:cartier}
gives a contradiction, as $\pi_*D$ is an effective rational divisor of degree $10$ with non-cuspidal support.
\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{The group \texorpdfstring{$\torz{81}$}{Z/81Z}}\label{subsec:fa-81}

In degrees $6$, $7$, $8$ and $9$ the group $\torz{81}$ is excluded in \cite[Proposition 4.8]{N69} by showing that
$T_2D=D+2\dm2D$ for the trace divisor $D$, which follows from $T_2D\sim D+2\dm2D$ as both sides have degree
$3d<\gon_\C X_1(81)$. In degree $10$ both sides have degree $30$, and we replace the gonality bound by a
Riemann-Roch computation over $\F_2$ on a quotient of $X_1(81)$ of genus $46$, after excluding some of the
reductions with $X_0(81)$.

\begin{lemma}\label{lem:fa-81-MW}
Let $Y_{81}:=X_1(81)/\dm{28}$, where $\dm{28}$ has order $3$. Then $Y_{81}$ has good reduction at $2$. The groups
$J_1(81)(\Q)$, $J_0(81)(\Q)$ and $J_{Y_{81}}(\Q)$ are finite and have no non-trivial $2$-torsion, reduction modulo
$2$ is injective on each of them, and $A_2:=T_2-1-2\dm2$ annihilates $J_{Y_{81}}(\Q)$.
\end{lemma}

\begin{proof}
The curve $Y_{81}$ has good reduction at $2$ by \Cref{lem:goodred}. The group $J_1(81)(\Q)$ is finite by
\cite[Theorem 3.1(2)]{DEHMZ}, and $J_1(81)(\Q)[2]=0$ by \cite[proof of Proposition 4.8]{N69}. Let $\rho$ be one of
the natural maps $X_1(81)\to X_0(81)$ and $X_1(81)\to Y_{81}$, of degrees $27$ and $3$. As $\rho_*\rho^*$ is
multiplication by the odd integer $\deg\rho$ on the Jacobian of the target, the kernel of $\rho^*$ on its rational
points is finite of odd order, so the Jacobian of the target has finitely many rational points and no non-trivial
rational $2$-torsion. By Parent's result (\Cref{subsec:hecke}) the kernel of reduction modulo $2$ on these finite
groups has exponent at most $2$, so it is trivial. By the Eichler-Shimura relation \cite[(10.2.3)]{DiamondIm} at
$2$ we have $T_2=F+\dm2V$ on the special fibre of the N\'eron model of $J_{Y_{81}}$ at $2$, and on
$J_{Y_{81}}(\F_2)$ the Frobenius $F$ is the identity, so $V=VF=2$. Hence $A_2$ annihilates $J_{Y_{81}}(\F_2)$,
and as reduction is injective on $J_{Y_{81}}(\Q)$ and commutes with $T_2$ and $\dm2$, it annihilates
$J_{Y_{81}}(\Q)$, as in \cite[proof of Proposition 4.8]{N69}.
\end{proof}

\begin{lemma}\label{lem:fa-81-points}
The non-cuspidal closed points of $X_1(81)_{\F_2}$ of degree at most $10$ are $135$ points of degree $9$ and $135$
points of degree $10$, and the points of each degree form $5$ orbits under $\Delta(81)$. Every Galois orbit of
cusps of $X_1(81)$ reduces to a single closed cusp of $X_1(81)_{\F_2}$ of the same degree.
\end{lemma}

\begin{proof}
We \gitlink{computations/b81_source_census.m}{compute} that no elliptic curve over $\F_{2^e}$ with $e\leq8$ has a point
of order $81$, that over $\F_{512}$ there are $45$ such curves, up to isomorphism, all with
$E(\F_{512})\simeq\torz{486}$, and that over $\F_{1024}$ there are $30$, of which $20$ have
$E(\F_{1024})\simeq\torz{972}$ and $10$ have $E(\F_{1024})\simeq\tg{3}{324}$. Their points of order $81$ give the
closed points and the orbits in the statement. The cusps are defined over $\Q(\zeta_{81})$ and the cuspidal
subscheme is finite \'etale over $\Z[\tfrac1{81}]$. As $2$ generates $(\torz{81})^\times$, the Frobenius at $2$
generates $\Gal(\Q(\zeta_{81})/\Q)$, so every Galois orbit of cusps reduces to a single closed cusp.
\end{proof}

\begin{lemma}\label{lem:fa-81-X0}
Let $\pi_0\colon X_1(81)\to X_0(81)$ be the natural map, let $c_\ast$ be a rational cusp of $X_0(81)$, and let $G$
be the subgroup of odd order of $J_0(81)(\F_2)$. Then $J_0(81)(\F_2)\simeq\torz3\times\torz{18}$, and for every
non-cuspidal closed point $P$ of $X_1(81)_{\F_2}$ of degree $e\leq10$, the push-forward $\pi_{0*}P$ is a closed point
of degree $e$. For $P$ in $2$ of the $5$ orbits of \Cref{lem:fa-81-points} of degree $9$, and in $3$ of the $5$
orbits of degree $10$, we have $[\pi_{0*}P-e\,\overline{c_\ast}]\notin G$.
\end{lemma}

\begin{proof}
The diamond operators act trivially on $X_0(81)$, so $\pi_{0*}P$ depends only on the orbit of $P$. We
\gitlink{computations/b81_x0_setup.m}{compute} $J_0(81)(\F_2)$ with Magma's proven class group algorithm on a model of
$X_0(81)_{\F_2}$, which has genus $4$. For a representative $P$ of each of the $10$ orbits we
\gitlink{computations/b81_x0_filter.m}{compute} $\pi_{0*}P$ from the $j$-invariants of the corresponding elliptic
curve and of its quotient by the cyclic subgroup of order $81$, find that it is a closed point of degree $e$, and
compute its class.
\end{proof}

To exclude the remaining reductions we use the following model of $Y_{81}$. Let $(E_{27},R)$ be the universal
elliptic curve over $X_1(27)$ with a point $R$ of order $27$, in Tate normal form, put $S:=9R$, which has order $3$,
and let $\phi\colon E_{27}\to E_{27}/\dm S$ be the quotient map and $\widehat\phi$ its dual. For a point $P$ of
order $81$, let $\phi_P$ be the quotient by $\dm{27P}$. The map $(E,P)\mapsto(E,3P,\phi_P(P))$ is invariant under
$P\mapsto28P$, as $\phi_P(27P)=0$, and $\widehat{\phi_P}(\phi_P(P))=3P$. So $Y_{81}$ parametrizes the points $V$ of
$E_{27}/\dm S$ with $\widehat\phi(V)=R$, and its function field is obtained from that of $X_1(27)$ by adjoining $x(V)$,
which is a root of a cubic polynomial.

\begin{lemma}\label{lem:fa-81-model}
Let $(E_{27},R)$, $S$ and $V$ be as above, let $m$ be the slope of the tangent line to $E_{27}$ at $S$, and put
$$f:=\bigl(y(R)-y(S)\bigr)-m\bigl(x(R)-x(S)\bigr).$$
Then the function field of $Y_{81,\F_4}$ is the extension of the function field of $X_1(27)_{\F_4}$ generated by a
root $w$ of $w^3=f$, and $x(V)=a+Aw+Bw^2$ for functions $a$, $A$, $B$ on $X_1(27)_{\F_4}$. The map
$Y_{81,\F_4}\to X_1(27)_{\F_4}$ is ramified at $9$ points and at no other point. These $9$ points are the images of
the closed cusps of $X_1(81)_{\F_2}$ of degree $1$, and $Y_{81}$ has genus $46$.
\end{lemma}

\begin{proof}
We \gitlink{computations/b81_y46_kummer_model_v2.m}{compute} a model of $X_1(27)_{\F_4}$, which has genus $13$, the
cubic polynomial of $x(V)$, which is irreducible, and functions $a$, $A$, $B$ such that $a+Aw+Bw^2$ is a root of it.
The function $f$ has valuation $2$ modulo $3$ at $9$ places, which are cusps, and valuation divisible by $3$ at all
the other places, so the extension is ramified at these $9$ places, and the Riemann-Hurwitz formula gives
$g(Y_{81})=1+3\cdot12+9=46$. By \Cref{lem:fa-81-points}, the closed cusps of $X_1(81)_{\F_2}$ of degree $1$ are
the reductions of the $27$ rational cusps $(\mathrm{Tate}(q),q^{b/81})$, with $b\in(\torz{81})^\times/\{\pm1\}$. At
these cusps $j$ has a pole of order $81$, and at their images $(\mathrm{Tate}(q),q^{b/27})$ on $X_1(27)$ a pole of
order $27$, so the map $X_1(81)\to X_1(27)$ has ramification index $3$ at them. As $\dm{28}$ maps the cusp with label
$b$ to the one with label $28b\not\equiv\pm b\pmod{81}$, the map $X_1(81)\to Y_{81}$ is unramified at them, so
$Y_{81}\to X_1(27)$ is ramified at their images, which are therefore the $9$ ramified cusps.
\end{proof}

\begin{lemma}\label{lem:fa-81-RR}
Let $\pi\colon X_1(81)\to Y_{81}$ be the quotient map and $\delta:=\dm2$. Let $P$ be a non-cuspidal closed point of
$X_1(81)_{\F_2}$ of degree $e\in\{9,10\}$ in one of the $5$ orbits of \Cref{lem:fa-81-points} which are not excluded
in \Cref{lem:fa-81-X0}. If $e=9$, let $c$ be a closed cusp of $X_1(81)_{\F_2}$ of degree $1$, and if $e=10$, put
$c=0$. Then $L:=\pi_*(P+c)$ satisfies
$$\ell(L+2\delta L)=1$$
on $Y_{81,\F_2}$.
\end{lemma}

\begin{proof}
For $a\in(\torz{81})^\times$ we have $\pi_*(\dm aP+c)=\dm a\pi_*(P+\dm a^{-1}c)$, so it suffices to consider a
representative $P$ of each of the $5$ orbits, $3$ of degree $9$ and $2$ of degree $10$. By \Cref{lem:fa-81-model}
there are $9$ possibilities for $\pi_*c$ if $e=9$, so there are $3\cdot9+2=29$ cases. As $\ell$ does not change
under base change, we compute $\ell(L+2\delta L)$ on $Y_{81,\F_4}$, with the model of \Cref{lem:fa-81-model}. Let
$M$ be the push-forward of $\pi_*P+2\delta\pi_*P$ to $X_1(27)_{\F_4}$, and let $L'$ be the space of the functions on
$Y_{81,\F_4}$ whose poles are bounded by $L+2\delta L$ at the cusps and by the pull-back of $M$ at the other points.
Under the automorphisms $w\mapsto\zeta w$, with $\zeta^3=1$, of $Y_{81,\F_4}$ over $X_1(27)_{\F_4}$, the space $L'$
decomposes as $\bigoplus_{k=0}^2 V_k\,w^k$, where $V_k$ is a Riemann-Roch space on $X_1(27)_{\F_4}$ determined by $M$,
by the poles allowed at the cusps and by the valuations of $f$. Above every point $p$ in the support of $M$ only one of
the $3$ points of $Y_{81,\F_4}$ occurs in $L+2\delta L$, and $L(L+2\delta L)$ is the subspace of $L'$ of the
functions without poles at the other $2$ points. We determine the point which occurs by the formula for $x(V)$ and a
computation of the isogenies of degree $3$ of the corresponding elliptic curve. We
\gitlink{computations/b81_rr_case1.m}{compute} that $V_0$, $V_1$ and $V_2$ have dimensions $15$, $11$ and $13$ in the
$27$ cases with $e=9$, and $18$, $12$ and $15$ in the $2$ cases with $e=10$, and that the conditions at the other
points have rank $38$ and $44$ respectively. Hence $\ell(L+2\delta L)=39-38=45-44=1$.
\end{proof}

\begin{proposition}\label{prop:81}
The group $\torz{81}$ is not a subgroup of the torsion of an elliptic curve over a number field of degree $10$.
\end{proposition}

\begin{proof}
Let $D$ be the trace divisor of a point of order $81$ over a number field of degree $10$, and let $\overline D$ be
its reduction modulo $2$. The group $J_1(81)(\Q)$ is finite by \Cref{lem:fa-81-MW}, and
$\gon_\C X_1(81)\geq\tfrac{325}{2^{15}}\cdot2916>28.92>20$, as $i(81)=2916$. So $\overline D$ is not entirely
cuspidal by \Cref{prop:main}~b). By \Cref{lem:fa-81-points}, as $2\cdot9>10$, we have $\overline D=P+\overline c$,
where $P$ is a non-cuspidal closed point of degree $e\in\{9,10\}$, and $c$ is a rational cusp if $e=9$, while $c=0$
if $e=10$.

Let $\pi_0$, $c_\ast$ and $G$ be as in \Cref{lem:fa-81-X0}. The class $[\pi_{0*}D-10c_\ast]$ lies in
$J_0(81)(\Q)$, and so does $[\pi_{0*}c-c_\ast]$ if $e=9$. As $J_0(81)(\Q)$ is finite of odd order by
\Cref{lem:fa-81-MW}, the reductions of these classes lie in $G$, and hence $[\pi_{0*}P-e\,\overline{c_\ast}]\in G$. By \Cref{lem:fa-81-X0}, $P$ lies in one of the $5$
orbits of \Cref{lem:fa-81-RR}.

Let $\pi$ and $\delta$ be as in \Cref{lem:fa-81-RR}, and let $y_\ast$ be a rational cusp of $Y_{81}$. By
\Cref{lem:cuspHecke} with $q=2$ we have $T_2y_\ast=y_\ast+2\delta y_\ast$, so $A_2y_\ast=0$, and as $A_2$ annihilates
$J_{Y_{81}}(\Q)$ by \Cref{lem:fa-81-MW}, we get $[A_2\pi_*D]=A_2[\pi_*D-10y_\ast]=0$, so
$$T_2\pi_*D\sim\pi_*D+2\delta\pi_*D.$$
Both sides are effective of degree $30$. The divisor $\pi_*D+2\delta\pi_*D$ extends to an effective relative divisor
on the smooth model of $Y_{81}$ over $\Z_{(2)}$, whose special fibre is $L+2\delta L$ with $L=\pi_*(P+\overline c)$,
so by the semicontinuity theorem \cite[Theorem III.12.8]{hartshorne} and \Cref{lem:fa-81-RR},
$\ell(\pi_*D+2\delta\pi_*D)\leq1$. Hence $\pi_*D+2\delta\pi_*D$ is the only effective divisor in its linear
equivalence class, and $T_2\pi_*D=\pi_*D+2\delta\pi_*D$.

So the finite non-empty set of $j$-invariants of the points in the support of $\pi_*D$ is closed under cyclic
isogenies of degree $2$, as $T_2$ maps a point to the sum of the points obtained from its $3$ isogenies of degree
$2$, and $\delta$ does not change the $j$-invariant. But for $E_\tau=\C/(\Z+\tau\Z)$ the map $E_\tau\to E_{2\tau}$,
$z\mapsto2z$, is a cyclic isogeny of degree $2$, and $j(2^k\tau)=q^{-2^k}+744+O(q^{2^k})$, with $q=e^{2\pi i\tau}$,
is unbounded as $k\to\infty$, which is a contradiction.
\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{The group \texorpdfstring{$\torz{87}$}{Z/87Z}}\label{subsec:fa-87}

Over nonic fields the group $\torz{87}$ is excluded in \cite[Proposition 8.2]{N69} with a quotient of $X_1(87)$ of
genus $33$. In degree $10$ we work on $X_0(87)$, which has genus $9$. As $87$ is squarefree, the $4$ cusps of
$X_0(87)$ are rational.

\begin{lemma}\label{lem:fa-87-points}
Let $\pi_0\colon X_1(87)\to X_0(87)$ be the natural map, and let $P$ be a non-cuspidal closed point of
$X_1(87)_{\F_2}$ of degree $e\leq10$. Then $e\in\{9,10\}$ and $\pi_{0*}P$ is a closed point of $X_0(87)_{\F_2}$ of
degree $e$. These closed points of $X_0(87)_{\F_2}$ are $4$ points of degree $9$ and $8$ points of degree $10$.
\end{lemma}

\begin{proof}
No elliptic curve over $\F_{2^e}$ with $e\leq8$ has a point of order $87$ \cite[proof of Proposition 8.2]{N69}. We
\gitlink{computations/b87_base_test_v3.m}{compute} that over $\F_{512}$ there are $36$ elliptic curves with a point
of order $87$, up to isomorphism, all with $E(\F_{512})\simeq\torz{522}$, and that over $\F_{1024}$ there are $50$,
of which $40$ have $E(\F_{1024})\simeq\torz{1044}$ and $10$ have $E(\F_{1024})\simeq\tg{3}{348}$. The subgroup of
order $29$ is unique, and on the curves of the last kind each of the $4$ subgroups of order $3$ is generated by a
rational point. So the points of order $87$ generate $36$ and $40+10\cdot4=80$ cyclic subgroups of order $87$,
respectively, and the corresponding points of $X_0(87)$ form $4$ Frobenius orbits of length $9$ and $8$ Frobenius
orbits of length $10$.
\end{proof}

\begin{lemma}\label{lem:fa-87-RR}
Let $P$ be one of the $12$ closed points of $X_0(87)_{\F_2}$ of \Cref{lem:fa-87-points}, of degree $e$. Then the
divisor $T_5P-6P$ on $X_0(87)_{\F_2}$ is not principal.
\end{lemma}

\begin{proof}
We obtain a model of $X_0(87)_{\F_2}$ as the normalization of the fibre product of $X_0(29)_{\F_2}$, which has genus
$2$, and $X_0(3)_{\F_2}$ over the $j$-line, by adjoining to the function field of $X_0(29)_{\F_2}$ a root $v$ of
$v^4+v^3+1/j=0$. Here $j=1/(v^3(v+1))$ is the $j$-invariant of the curve $y^2+xy+vy=x^3$, on which $(0,0)$ has
order $3$. The point $P$ is determined by the kernel of the isogeny of degree $29$ and by the point of order $3$ of
the corresponding elliptic curve. We \gitlink{computations/b87_base_test_v3.m}{compute} $T_5P$ from the $6$ cyclic
subgroups of order $5$, carrying the subgroup of order $87$ through the isogenies. For $e=9$ it is a closed point of
degree $54$, and for $e=10$ it is the sum of closed points of degrees $10$ and $50$, with multiplicity $1$. We then
compute that $\ell(T_5P-6P)=0$ for all the $12$ points $P$.
\end{proof}

\begin{proposition}\label{prop:87}
The group $\torz{87}$ is not a subgroup of the torsion of an elliptic curve over a number field of degree $10$.
\end{proposition}

\begin{proof}
Let $D$ be the trace divisor of a point of order $87$ over a number field of degree $10$, and let $\overline D$ be
its reduction modulo $2$. The group $J_1(87)(\Q)$ is finite \cite[Theorem 3.1(2)]{DEHMZ}, and
$\gon_\C X_1(87)\geq\tfrac{325}{2^{15}}\cdot3360>33>20$, as $i(87)=3360$. So $\overline D$ is not entirely cuspidal
by \Cref{prop:main}~b). By \Cref{lem:fa-87-points}, as $2\cdot9>10$, we have $\overline D=P+\overline C$, where $P$
is a non-cuspidal closed point of degree $e\in\{9,10\}$ and $\overline C$ is cuspidal of degree $10-e\leq1$.

Let $\pi_0$ be as in \Cref{lem:fa-87-points}. The group $J_0(87)(\Q)$ is finite, as $J_0(87)$ is a quotient of
$J_1(87)$. As the diamond operators act trivially on $X_0(87)$, we have $A_5=T_5-6$ on $X_0(87)$, which annihilates
$J_0(87)(\Q)$ by \Cref{prop:tools}~b). As all the cusps of $X_0(87)$ are rational, \Cref{lem:cuspHecke} shows that
$A_5$ annihilates every cuspidal divisor on $X_0(87)$ and on $X_0(87)_{\F_2}$. Hence, for a cusp $c_\ast$ of
$X_0(87)$, the class of $A_5\pi_{0*}D$ is $A_5[\pi_{0*}D-10c_\ast]=0$, and reducing modulo $2$ we get that $A_5\pi_{0*}P=T_5\pi_{0*}P-6\pi_{0*}P$ is principal on $X_0(87)_{\F_2}$. As
$\pi_{0*}P$ is one of the $12$ points of \Cref{lem:fa-87-points}, this contradicts \Cref{lem:fa-87-RR}.
\end{proof}