%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 10b_further.tex (tag fb): the groups Z/69Z, Z/78Z, Z/3Z x Z/39Z, Z/117Z,
% Z/90Z, Z/99Z and Z/105Z. Continues the section started in 10_further.tex.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\subsection{The group \texorpdfstring{$\torz{69}$}{Z/69Z}}\label{subsec:fb-69}

Over number fields of degree $7$, $8$ and $9$ the group $\torz{69}$ was excluded in
\cite[Proposition 9.9]{N69} with the unramified double cover $Y_{69}\to Z_{69}=X_0(69)$ of
\cite[Lemma 9.4]{N69}. Here $Y_{69}=X_1(69)/H$, where $H$ is the subgroup of order $11$ of $\Delta(69)$, and
we use the model of $Y_{69,\F_2}$ constructed in \cite[Section 9.2]{N69}. In degree $10$ the reduction of the
trace divisor modulo $2$ can also be a closed point of degree $10$, and for one of the two diamond orbits of such
points the double cover does not suffice. For this orbit we use the following quotient of $X_1(69)$ of degree $2$,
on which the images of a point and of its Hecke translates can be computed from the moduli interpretation.

\begin{lemma}\label{lem:fb-W69}
The diamond operator $\dm{22}$ has order $2$ in $\Delta(69)$ and fixes no geometric point of $X_1(69)$, in
characteristic $0$ or $2$. The curve $W_{69}:=X_1(69)/\dm{22}$ has genus $67$ and good reduction at $2$, and
$J(W_{69})(\Q)$ is finite. The curve $W_{69}$ parametrizes triples $(E,\pm P_{23},C_3)$, where $P_{23}$ is a point of
order $23$ and $C_3$ a cyclic subgroup of order $3$ of $E$, and the image of $(E,P)$ is $(E,\pm3P,\dm{23P})$. Let
$F_{23}$ be the function field of $X_1(23)_{\F_2}$, which has genus $12$, let $j\in F_{23}$ be the $j$-invariant,
and let $v$ be the coordinate on $X_0(3)_{\F_2}$ for which the pair $(E,C_3)$ is given by the curve
$y^2+xy+vy=x^3$ and the subgroup generated by $(0,0)$. Then $j=1/(v^3(v+1))$ on $X_0(3)_{\F_2}$, and the function
field of $W_{69,\F_2}$ is $F_{23}(v)$, where
\[
v^4+v^3+1/j=0 .
\]
\end{lemma}

\begin{proof}
We have $22^2\equiv1\pmod{69}$ and $22\not\equiv\pm1\pmod{69}$, so $\dm{22}$ has order $2$. As $22\equiv1\pmod 3$
and $22\equiv-1\pmod{23}$, and as $22^2\equiv1$, $22^2+22+1\equiv24$ and $22^2-22+1\equiv49$ modulo $69$,
\cite[Lemma 7.2]{N69} shows that $\dm{22}$ fixes no cusp and no non-cuspidal geometric point of $X_1(69)$, in
characteristic $0$ or $2$. Hence $\dm{22}$ acts freely on the smooth model of $X_1(69)$ over $\Z_{(2)}$, the
quotient is a smooth model of $W_{69}$ with special fibre $X_1(69)_{\F_2}/\dm{22}$, and as $X_1(69)$ has genus
$133$, the Riemann-Hurwitz formula $2\cdot133-2=2(2g(W_{69})-2)$ gives $g(W_{69})=67$. Let
$\pi:X_1(69)\to W_{69}$ be the quotient map. As $\pi_*\pi^*=2$ on $J(W_{69})$ and $J_1(69)(\Q)$ is finite
\cite[Theorem 3.1(2)]{DEHMZ}, the group $J(W_{69})(\Q)$ is finite.

The operator $\dm{22}$ maps $(E,P)$ to $(E,22P)$, so it fixes $23P$ and maps $3P$ to $-3P$. Hence $(E,P)$ and
$(E,P')$ have the same image on $W_{69}$ if and only if $P'=aP$ with $a\equiv\pm1\pmod 3$ and $a\equiv\pm1\pmod{23}$,
i.e. if and only if $\dm{23P'}=\dm{23P}$ and $3P'=\pm3P$. This proves the moduli description. Over $\F_2$ the
maps $(E,P)\mapsto(E,\pm3P)$ to $X_1(23)$ and $(E,P)\mapsto(E,\dm{23P})$ to $X_0(3)$ show that the function field of
$W_{69,\F_2}$ contains $F_{23}$ and $v$, and it has degree $4$ over $F_{23}$, as $X_1(69)\to X_1(23)$ has degree
$i(69)/i(23)=2112/264=8$. The curve $y^2+xy+vy=x^3$ over $\F_2(v)$ has the point $(0,0)$ of order $3$ and
$j$-invariant $1/(v^3(v+1))$, so $v$ is a root of $v^4+v^3+1/j$. We
\gitlink{next_cases_20261030/C69/quotient67/q67_profile_rr_v2.m}{compute} that this polynomial is irreducible over
$F_{23}$, so $F_{23}(v)$ is the function field of $W_{69,\F_2}$, and we also compute that it has genus $67$.
\end{proof}

\begin{lemma}\label{lem:fb-69red}
Let $E$ be an elliptic curve over a number field of degree $10$ with a point of order $69$, and let $D$ be the
trace divisor on $X_1(69)$ of this point. Then the reduction of $D$ modulo $2$ is $G+\overline C$, where $G$ is a
non-cuspidal closed point of $X_1(69)_{\F_2}$ of degree $e\in\{7,8,9,10\}$ and $\overline C$ is the reduction of an
effective rational cuspidal divisor $C$ of degree $10-e$. The non-cuspidal closed points of $X_1(69)_{\F_2}$ of degree
at most $10$ have degrees $7$, $8$, $9$ and $10$, and there are $66$, $44$, $44$ and $44$ of them.
\end{lemma}

\begin{proof}
By \cite[Theorem 3.1(2)]{DEHMZ} the group $J_1(69)(\Q)$ is finite. As $i(69)=2112$, Abramovich's bound gives
$\gon_\C X_1(69)>20.94$, so $\gon_\Q X_1(69)>20$, and by \Cref{prop:main}~b), applied with $p=2$ and $d=10$, the
reduction of $D$ modulo $2$ is not entirely cuspidal. The numbers of non-cuspidal closed points of degree at most
$9$ are computed in the proof of \cite[Proposition 9.9]{N69}. We
\gitlink{computations/cont69_cover_all_decic_v2.m}{compute} that there are $20$ elliptic curves over $\F_{1024}$ with a
point of order $69$, all with cyclic group of order $966$. Each of them has $\varphi(69)/2=22$ points of order $69$
up to sign, which gives $440$ geometric points of $X_1(69)$ over $\F_{1024}$, and as there are no non-cuspidal
closed points of degree $1$, $2$ or $5$, these form $44$ closed points of degree $10$. As $2\cdot7>10$, the
non-cuspidal part of the reduction of $D$ is a single closed point $G$, with multiplicity $1$. The Galois orbits
of cusps of $X_1(69)$ over $\Q$ have degrees $1^{22},2^{11},22^2$, and the closed cusps of $X_1(69)_{\F_2}$ have
degrees $1^{22},2^{11},11^2,22$ (see the proof of \cite[Proposition 9.9]{N69}), so the closed cusps of degree $1$
and $2$ are reductions of Galois orbits of cusps of the same degree. As the cuspidal part of the reduction has
degree $10-e\leq3$, it is the reduction of an effective rational cuspidal divisor.
\end{proof}

\begin{proposition}\label{prop:69}
The group $\torz{69}$ 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 $69$ over a number field of degree $10$, and let $G$, $e$ and $C$
be as in \Cref{lem:fb-69red}. Let $P$ and $P_Y$ be the images of $G$ on $Z_{69,\F_2}$ and on $Y_{69,\F_2}$. As in
the proof of \cite[Proposition 9.9]{N69}, the groups $J(Y_{69})(\Q)$ and $J(Z_{69})(\Q)$ are finite, and as
$\dm5\in H$, the operator $A_5=T_5-6$ annihilates them (\Cref{prop:tools}) and annihilates the images of the
rational cuspidal divisors, such as $C$, as divisors (\Cref{lem:cuspHecke}). So $T_5P-6P$ is principal on
$Z_{69,\F_2}$ and $T_5P_Y-6P_Y$ is principal on $Y_{69,\F_2}$. The divisor $T_5P_Y$ is one of the effective divisors
on $Y_{69,\F_2}$ whose push-forward to $Z_{69,\F_2}$ is $T_5P$, which we call the lifts of $T_5P$. The non-trivial
automorphism of $Y_{69}$ over $Z_{69}$ commutes with $T_5$ and exchanges the two points of $Y_{69,\F_2}$ above $P$,
so we may take $P_Y$ to be a fixed one of them.

Suppose first that $e\leq9$. It is shown in the proof of \cite[Proposition 9.9]{N69} that the images of the points
$G$ on $Z_{69,\F_2}$ are $3$ closed points of degree $7$, $2$ of degree $8$ and $2$ of degree $9$, that the divisor
$T_5P-6P$ is not principal for two of those of degree $7$, both of degree $8$ and one of degree $9$, and that for the
remaining point of degree $7$ and the remaining one of degree $9$ the divisor $T-6P_Y$ is not principal for any of the
lifts $T$ of $T_5P$. These tests depend only on $G$ and not on $\overline C$, so they exclude these reductions.

Suppose now that $e=10$. Then the reduction of $D$ is $G$. The map $X_1(69)\to Z_{69}$ is the quotient by
$\Delta(69)$, and we \gitlink{computations/cont69_cover_all_decic_v2.m}{compute} that the $44$ closed points of degree
$10$ lie above $2$ closed points $P_1$ and $P_2$ of $Z_{69,\F_2}$, both of degree $10$, so they form $2$ diamond
orbits of $22$ points each. For both $i=1,2$, the divisor $T_5P_i-6P_i$ is principal, and $T_5P_i=Q_i+Q_i'$, where
$Q_i$ and $Q_i'$ are closed points of degree $10$ and $50$ which both split in $Y_{69}$. So there are $4$ lifts of
$T_5P_i$. For $P=P_1$ the divisor $T-6P_Y$ is not principal for any of them, so $G$ lies above $P_2$.

We \gitlink{next_cases_20261030/C69/direct_full/direct69_select_input_v2.m}{compute} that the $j$-invariants of the
points above $P_1$ and above $P_2$ have different minimal polynomials over $\F_2$, that those above $P_2$ are the
roots of
\[
j^{10}+j^8+j^7+j^6+j^5+j^4+j^3+j+1 ,
\]
and that for such a root $j_0$ there is exactly one elliptic curve $E_0$ over $\F_{1024}$ with $j(E_0)=j_0$ and a
point of order $69$, and that $E_0(\F_{1024})\simeq\torz{966}$. Let $P_0\in E_0(\F_{1024})$ be a point of order $69$ and
$G_0$ the closed point of $(E_0,P_0)$, which has degree $10$ and hence lies above $P_1$ or $P_2$, so it lies above
$P_2$. As the closed points above $P_2$ form one diamond orbit, $G=\dm aG_0$ for some $a\in(\torz{69})^\times$.

Let $\pi:X_1(69)\to W_{69}$ be the quotient map of \Cref{lem:fb-W69}, let $P_W$ be the image of $G_0$, and put
$A_5=T_5-1-5\dm5$ on $W_{69}$. As $J(W_{69})(\Q)$ is finite (\Cref{lem:fb-W69}), $A_5\pi_*D$ is principal
(\Cref{prop:tools}), and so is its reduction $A_5\pi_*G=\dm aA_5\pi_*G_0$. We compute below that $P_W$ has degree
$10$, so $\pi_*G_0=P_W$ and $A_5P_W$ is principal. We
\gitlink{next_cases_20261030/C69/quotient67/q67_profile_rr_v2.m}{compute} $A_5P_W$ on the model of
\Cref{lem:fb-W69} as follows. The image of $(E_0,\pm3P_0)$ on $X_1(23)_{\F_2}$ is a closed point of degree $10$, at
which the reduction of $v^4+v^3+1/j$ is the product of a linear and an irreducible cubic polynomial over $\F_{1024}$.
The subgroup $\dm{23P_0}$ is defined over $\F_{1024}$, so $P_W$ is the unique point of degree $10$ of $W_{69,\F_2}$
above this point, and in the same way we obtain $\dm5P_W$ from $(E_0,5P_0)$. The six subgroups of order $5$ of
$E_0$ form two orbits under the Frobenius, of lengths $1$ and $5$. For a subgroup $S$ in each of them, the image of
$(E_0/S,\pm3P_0\bmod S)$ on $X_1(23)_{\F_2}$ is a closed point of degree $10$, respectively $50$, and at both of
them the quartic again factors as a linear times an irreducible cubic polynomial over the residue field. So
$T_5P_W=Q+Q'$, where $Q$ and $Q'$ are the unique points of $W_{69,\F_2}$ of degree $10$ and $50$ above these two
points. We compute that
\[
\ell\bigl(T_5P_W-P_W-5\dm5P_W\bigr)=0 ,
\]
so $A_5P_W$ is not principal, which is a contradiction.
\end{proof}

\subsection{The curve \texorpdfstring{$X_1(39)$}{X1(39)}}\label{subsec:fb-39}

Over number fields of degree $10$ there is an elliptic curve with a point of order $39$, and we show that it is the
only one up to isomorphism. This excludes $\torz{78}$, $\tg{3}{39}$ and $\torz{117}$, as in
\cite[Proposition 13.14]{N69} over octic fields.

Let $X=X_1(39)$. By \cite[Theorem 3.1(2), Corollary 4.14 and Table 2]{DEHMZ},
$J_X(\Q)\simeq\torz7\times\torz{31122}\times\torz{3236688}$, reduction modulo $2$ is injective on $J_X(\Q)$
\cite[Lemma 13.12]{N69}, and $\gon_\Q X=14$ \cite[Theorem 4 and Table 1]{DvH}. The Galois orbits of cusps over $\Q$ and
the closed cusps of $X_{\F_2}$ both have degrees $1^{12},2^6,12^2$, so every effective cuspidal divisor on $X_{\F_2}$
is the reduction of an effective rational cuspidal divisor of the same degree (see \cite[Section 13.8]{N69}). We
denote by $D_8$ the closed point of degree $8$ of $X$ given by the octic point of level $39$ of
\cite[Proposition 13.3]{N69}, and by $D_9$ the closed point of degree $9$ given by the point $x_{39}$ of
\cite[Proposition 13.5]{N69}. The point of degree $10$ is the following one.

\begin{lemma}\label{lem:fb-x39}
Let $(K_{39},E_{39})$ be the pair of \Cref{prop:explicit_cyclic} with $N=39$. The trace divisor $D_{10}$ of
$(E_{39},(0,0))$ on $X_1(39)$ is a closed point of degree $10$ (the point \xpoint{1.39.10.a} of \cite{x1mn}), and its
reduction modulo $2$ is a closed point of degree $10$.
\end{lemma}

\begin{proof}
We \gitlink{computations/d31a_78_independent_genuine.py}{check}, with the model of $(K_{39},E_{39})$ used in the proof
of \Cref{prop:explicit_cyclic}, that the reduction of $D_{10}$ modulo $2$ is a closed point of degree $10$. (Another
model of the pair replaces $D_{10}$ by a diamond translate, as $E_{39}(K_{39})_\tors\simeq\torz{39}$.) As every closed
point in the support of $D_{10}$ reduces to an effective divisor of the same degree, $D_{10}$ is a closed point of
degree $10$.
\end{proof}

\begin{lemma}\label{lem:fb-39}
Let $X=X_1(39)$.
\begin{itemize}
\item[a)] The non-cuspidal closed points of $X_{\F_2}$ of degrees $1,\dots,10$ number $0$, $0$, $0$, $0$, $0$, $12$, $0$, $49$, $52$ and $60$.
      Those of degree $6$ form one diamond orbit, and those of degree $8$, $9$ and $10$ form $5$ diamond orbits each.
\item[b)] Let $P$ be a non-cuspidal closed point of $X_{\F_2}$ of degree at most $10$ such that $A_5P$ is principal.
      Then $P$ is the reduction of $D_8$, or the reduction of a diamond translate of $D_9$ or of $D_{10}$.
\end{itemize}
\end{lemma}

\begin{proof}
The statements about the points of degree at most $8$ are \cite[Section 13.8]{N69} and the computation in the proof
of \cite[Proposition 13.14]{N69}: $A_5P$ is not principal for the points of degree $6$ and for those in $4$ of the
$5$ orbits of degree $8$, and the fifth orbit of degree $8$ consists of the reduction of $D_8$ alone. We
\gitlink{computations/d31a_78_source.m}{compute} that $39$ elliptic curves over $\F_{512}$ have a point of order $39$,
$36$ of them with cyclic group of order $546$ and $3$ with cyclic group of order $468$, and that $50$ elliptic curves
over $\F_{1024}$ have a point of order $39$, all with cyclic group of order $1014$. Their points of order $39$ give
$52$ closed points of degree $9$ and $60$ of degree $10$, each forming $5$ diamond orbits. As $A_5$ commutes with
the diamond operators, it suffices to test one point in each orbit, and we
\gitlink{computations/d31a_78_A5.m}{compute} that $\ell(A_5P)=0$ for the representatives of $4$ of the $5$ orbits in
each of the degrees $9$ and $10$. We \gitlink{computations/d31a_78_independent_genuine.py}{check} that the fifth
orbit of degree $9$ contains the reduction of $D_9$, and the fifth orbit of degree $10$ that of $D_{10}$.
\end{proof}

\begin{proposition}\label{prop:78-3x39}
Every effective rational divisor of degree $10$ on $X_1(39)$ with non-cuspidal support is a diamond translate of
$D_{10}$. Hence, if $E$ is an elliptic curve over a number field $K$ of degree $10$ with a point of order $39$,
then $(K,E)$ is isomorphic to the pair $(K_{39},E_{39})$ of \Cref{prop:explicit_cyclic}, and $E(K)_\tors\simeq\torz{39}$. In
particular, $\torz{78}$ and $\tg{3}{39}$ are not subgroups of the torsion of an elliptic curve over a number field
of degree $10$.
\end{proposition}

\begin{proof}
Let $D$ be an effective rational divisor of degree $10$ on $X=X_1(39)$ with non-cuspidal support. We first note
that $D$ is equal to every effective rational divisor $B$ of degree $10$ with the same reduction modulo $2$: the
class $[D-B]\in J_X(\Q)$ reduces to $0$, so $D\sim B$ by injectivity, and $D=B$ by \Cref{lem:closed}, as
$\gon_\Q X=14>10$.

Suppose that the reduction of $D$ is entirely cuspidal. It is the reduction of an effective rational cuspidal
divisor $C$ of degree $10$, so $D=C$, which is impossible as the support of $D$ contains no cusp. So by
\Cref{lem:fb-39}~a) the reduction of $D$ is $P+\overline C$, where $P$ is a non-cuspidal closed point of degree $6$,
$8$, $9$ or $10$ (as $2\cdot6>10$) and $\overline C$ is the reduction of an effective rational cuspidal divisor $C$ of
degree $10-\deg P$. As $J_X(\Q)$ is finite, $A_5D$ is principal (\Cref{prop:tools}), and $A_5C=0$ as a divisor, so
$A_5P$ is principal. By \Cref{lem:fb-39}~b), the reduction of $D$ is the reduction of $D_8+C$, of $hD_9+C$ or of
$hD_{10}$ for a diamond operator $h$. These are effective rational divisors of degree $10$, so $D$ is equal to one of
them, and as the support of $D$ contains no cusp, $D=hD_{10}$.

Now let $E$ be an elliptic curve over a number field $K$ of degree $10$ with a point $P$ of order $39$, and let $D$
be the trace divisor of $(E,P)$ (\Cref{subsec:trace}). By the first part $D=hD_{10}$ for a diamond operator $h$, so
$(K,E)\simeq(K_{39},E_{39})$ by \Cref{lem:spor-determined}, applied with $m=1$ and the point $(E_{39},(0,0))$, whose
trace divisor $D_{10}$ is a closed point of degree $10$ by \Cref{lem:fb-x39}. Hence $E(K)_\tors\simeq\torz{39}$ by
\Cref{prop:explicit_cyclic}. If $E(K)$ contained $\torz{78}$ or $\tg{3}{39}$, it would contain
a point of order $39$, and $E(K)_\tors\simeq\torz{39}$ would contain a point of order $78$, respectively two
independent points of order $3$, which is impossible.
\end{proof}

\begin{proposition}\label{prop:117}
The group $\torz{117}$ is not a subgroup of the torsion of an elliptic curve over a number field of degree $10$.
\end{proposition}

\begin{proof}
Let $E$ be an elliptic curve over a number field $K$ of degree $10$ with a point $Q$ of order $117$. Then $3Q$ has
order $39$, so $E(K)_\tors\simeq\torz{39}$ by \Cref{prop:78-3x39}, which contains no point of order $117$.
\end{proof}

\subsection{The group \texorpdfstring{$\torz{90}$}{Z/90Z}}\label{subsec:fb-90}

The group $\torz{45}$ occurs over number fields of degree $10$, so we cannot exclude $\torz{90}$ on $X_1(45)$ alone.
As in the proof of \cite[Proposition 13.17]{N69}, where $\torz{78}$ is excluded over nonic fields, we use the
isogeny of degree $2$ given by the point of order $2$, and show that it would give an isogeny of degree $2$ from the
curve to one of its Galois conjugates, which is impossible by \Cref{lem:sw-twoisog}, as in the proofs of
\Cref{prop:66,prop:74}.

\begin{lemma}\label{lem:fb-45}
Let $X=X_1(45)$, which has genus $41$.
\begin{itemize}
\item[a)] The group $J_X(\Q)$ is finite, reduction modulo $2$ is injective on it, and $\gon_\Q X=18$.
\item[b)] The Galois orbits of cusps of $X$ over $\Q$ have degrees $1^{12},2^4,4^3,6^2,8,12$, and the closed cusps of
      $X_{\F_2}$ have degrees $1^{12},2^4,4^5,6^2,12$. In particular, every effective cuspidal divisor of degree
      at most $10$ on $X_{\F_2}$ which contains no closed cusp of degree $4$ is the reduction of an effective rational
      cuspidal divisor of the same degree.
\item[c)] The non-cuspidal closed points of $X_{\F_2}$ of degree at most $10$ are $60$ points of degree $8$, $60$
      points of degree $9$ and $108$ points of degree $10$. For every point $P$ of degree $8$ or $9$ there is a prime
      $q\nmid90$ such that $A_qP$ is not principal, and $\ell(P)=1$ for every point $P$ of degree $10$.
\item[d)] The space $U:=(\Cart-1)H^0(X_{\F_2},\Omega^1)$ has dimension $39$, and its restriction to every
      effective cuspidal divisor of degree $10$ on $X_{\F_2}$ which contains a closed cusp of degree $4$ is
      surjective.
\end{itemize}
\end{lemma}

\begin{proof}
a) We have $\rk J_X(\Q)=0$ \cite[Theorem 3.1(2)]{DEHMZ}, the injectivity of reduction modulo $2$ is shown at the end
of the proof of \cite[Proposition 7.6]{N69}, and $\gon_\Q X=18$ by \cite[Theorem D]{Varivoda26}.

b) The degrees of the Galois orbits of cusps are given in \cite[Lemma 2.8]{DEHMZ}, and those of the closed cusps
of $X_{\F_2}$ in the proof of \cite[Proposition 7.6]{N69}. The cuspidal subscheme is finite \'etale over
$\Z[\tfrac1{45}]$, so distinct geometric cusps have distinct reductions. Hence the $12$ rational and the $4$
quadratic closed cusps of $X_{\F_2}$ are the reductions of the Galois orbits of degree $1$ and $2$, and as $X_{\F_2}$
has no closed cusps of degree $3$, each of the two Galois orbits of degree $6$ reduces to a closed cusp of degree
$6$. This proves the last statement, as a cuspidal divisor of degree at most $10$ on $X_{\F_2}$ which contains no
closed cusp of degree $4$ is supported on the closed cusps of degree $1$, $2$ and $6$.

c) The points of degree at most $9$ and the Hecke tests for them are in the proof of \cite[Proposition 7.6]{N69}:
the points of degree $8$ and $9$ form $5$ diamond orbits each, for a representative $P$ of each orbit of degree $8$
one of the classes $[A_qP]$ with $q\in\{7,11,13,17,19\}$ is non-zero, and for a representative of each orbit of
degree $9$ the divisor $A_7P$ is not principal. As $A_q$ commutes with the diamond operators, this holds for all the
points. We \gitlink{computations/d31a_90_source.m}{compute} that the points of order $45$ of the elliptic curves
over $\F_{1024}$ give $108$ closed points of degree $10$ of $X_{\F_2}$, which form $10$ orbits under the diamond
operators, and we \gitlink{additions/C90/ell45.m}{compute} that $\ell(P)=1$ for a representative $P$ of each orbit. As
the diamond operators are automorphisms of $X_{\F_2}$, it follows that $\ell(P)=1$ for all the $108$ points.

d) The dimension of $U$ is computed in the proof of \cite[Proposition 7.6]{N69}, together with the restrictions of
$U$ to the divisors $mc$ for every closed cusp $c$ of $X_{\F_2}$ and every $m$ with $m\deg c\leq8$. In an effective
cuspidal divisor of degree $10$ which contains a closed cusp of degree $4$, this cusp has multiplicity at most $2$,
and every other closed cusp $c$ occurs with a multiplicity $m$ with $m\deg c\leq6$. So the restriction of $U$ to
such a divisor is determined by these data. There are $94420$ such divisors, and we
\gitlink{computations/d31a_90_quartic_boundary_v2.py}{compute} that the restriction of $U$ to each of them has
rank $10$.
\end{proof}

\begin{lemma}\label{lem:fb-45red}
Let $E$ be an elliptic curve over a number field $K$ of degree $10$ with a point $P$ of order $45$, and let $D$ be
the trace divisor of $(E,P)$ on $X_1(45)$. Then $D$ is a closed point of degree $10$ with residue field $K$, the field
$K$ has a unique prime $\mathfrak p$ above $2$, which is unramified with residue field $\F_{1024}$, $E$ has good
reduction at $\mathfrak p$, and the reduction of $D$ modulo $2$ is a non-cuspidal closed point of degree $10$.
\end{lemma}

\begin{proof}
Let $X=X_1(45)$. Suppose first that the reduction $\overline D$ of $D$ is entirely cuspidal. If $\overline D$
contains no closed cusp of degree $4$, then by \Cref{lem:fb-45}~b) it is the reduction of an effective rational
cuspidal divisor $C$ of degree $10$, so $D\sim C$ by \Cref{lem:fb-45}~a), and $D=C$ by \Cref{lem:closed}, as
$\gon_\Q X=18>10$. This is impossible, as the support of $D$ contains no cusp. If $\overline D$ contains a closed cusp
of degree $4$, then \Cref{prop:cartier}, applied with $Y=X$, $r=1$ and $\psi_1=1$ (as $J_X(\Q)$ is finite), and
\Cref{lem:fb-45}~d) give a contradiction.

So $\overline D$ is not entirely cuspidal, and by \Cref{lem:fb-45}~c), as $2\cdot8>10$, it is $G+\overline C$ with
$G$ a non-cuspidal closed point of degree $8$, $9$ or $10$ and $\overline C$ cuspidal of degree at most $2$. If
$\deg G\leq9$, then by \Cref{lem:fb-45}~b) $\overline C$ is the reduction of an effective rational cuspidal divisor
$C$. As $J_X(\Q)$ is finite, $A_qD$ is principal for every prime $q\nmid90$ (\Cref{prop:tools}), and $A_qC=0$ as a
divisor, so $A_qG$ is principal, which contradicts \Cref{lem:fb-45}~c). Hence $\overline D=G$ is a closed point of
degree $10$ with multiplicity $1$. If $x\in X(K)$ is the point given by $(E,P)$, then $D$ is the sum of the
conjugates of $x$, each counted $[K:\Q(x)]$ times, so $[K:\Q(x)]=1$ and $D$ is a closed point with residue field
$K$. By \Cref{subsec:trace}, $G$ is the reduction of $x$ modulo a prime $\mathfrak p$ of $K$ above $2$ which is
unramified and has residue field $\kappa(G)=\F_{1024}$, so $\mathfrak p$ is the only prime of $K$ above $2$. As $G$
is not a cusp and $X$ is a fine moduli space, $E$ has good reduction at $\mathfrak p$.
\end{proof}

\begin{proposition}\label{prop:90}
The group $\torz{90}$ is not a subgroup of the torsion of an elliptic curve over a number field of degree $10$.
\end{proposition}

\begin{proof}
Let $E$ be an elliptic curve over a number field $K$ of degree $10$ with a point $Q$ of order $90$. Put $P=2Q$ and
$R=45Q$, let $\phi:E\to E'=E/\dm R$, which is defined over $K$, and put $P'=\phi(P)$, which has order $45$. Let $D$
and $D'$ be the trace divisors of $(E,P)$ and $(E',P')$ on $X=X_1(45)$. By \Cref{lem:fb-45red}, applied to both
of them, the prime $\mathfrak p$ of $K$ above $2$ is unique and unramified with residue field $\F_{1024}$, $E$ and $E'$
have good reduction at $\mathfrak p$, and $D$ and $D'$ are closed points of degree $10$ with residue field $K$ which
reduce to non-cuspidal closed points of degree $10$. As $\mathfrak p$ is unramified over $2$, the reduction of $E$ at
$\mathfrak p$ is ordinary by \Cref{lem:sw-twoisog}~a).

The reduction of $\phi$ at $\mathfrak p$ is an isogeny of degree $2$ of an ordinary elliptic curve, so its kernel is
either the connected subgroup of order $2$, in which case it is the Frobenius isogeny, or the unique \'etale
subgroup of order $2$, in which case it is the Verschiebung up to isomorphism. Hence, as in the proof of
\cite[Proposition 13.17]{N69}, the reduction of $(E',P')$ is the image of the reduction of $(E,P)$ under the
Frobenius, or under $\dm2$ composed with the inverse of the Frobenius. As closed points are stable under the
Frobenius, the reduction of $D'$ is that of $D$ or of $\dm2D$. These are effective rational divisors of degree $10$,
so the class of $D'-D$, respectively of $D'-\dm2D$, lies in $J_X(\Q)$ and reduces to $0$, and by \Cref{lem:fb-45}~a)
we get $D'\sim D$, respectively $D'\sim\dm2D$. The reductions of $D$ and $\dm2D$ are
non-cuspidal closed points of degree $10$, so $\ell(D)=\ell(\dm2D)=1$ by \Cref{lem:fb-45}~c) and semicontinuity
\cite[Theorem III.12.8]{hartshorne}, and hence $D'=D$, respectively $D'=\dm2D$.

Note that $\dm2D$ is the trace divisor of $(E,2P)$. So the $K$-point of $X$ given by $(E',P')$ and the one given
by $(E,P)$, respectively $(E,2P)$, lie on the same closed point with residue field $K$, and they differ by an
automorphism $\sigma$ of $K$. As $X$ is a fine moduli space, $E'\simeq E^\sigma$ over $K$, and $\sigma(\mathfrak p)=
\mathfrak p$, as $\mathfrak p$ is the only prime of $K$ above $2$. As $\torz{90}$ is not a subgroup of any of the
groups of \Cref{lem:CM}, the curve $E$ has no complex multiplication, so $\End_{\overline K}(E)=\Z$, and this
contradicts \Cref{lem:sw-twoisog}~b).
\end{proof}

\subsection{The groups \texorpdfstring{$\torz{99}$ and $\torz{105}$}{Z/99Z and Z/105Z}}\label{subsec:fb-99-105}

For $n=99$ and $n=105$ we do not show that $J_1(n)(\Q)$ is finite, and we only use that $(\dm a-1)J_1(n)(\Q)$ is
finite for a suitable $a$. The entirely cuspidal reductions modulo $2$ are then excluded by the following lemma, and
the non-cuspidal ones with Hecke operators on $X_1(33)$ and on a quotient of $X_1(99)$ of genus $29$, respectively on
$X_1(35)$.

\begin{lemma}\label{lem:fb-cusp}
Let $n$ be odd, let $X=X_1(n)$, and let $a\in(\torz n)^\times$ be such that $(\dm a-1)J_X(\Q)$ is finite,
$\gon_\C X>40$, and the order of $\dm a$ in $\Delta(n)$ does not divide $10$. Then no effective rational divisor of
degree $10$ on $X$ whose support contains no cusp reduces modulo $2$ to a cuspidal divisor.
\end{lemma}

\begin{proof}
Let $D$ be such a divisor, and suppose that its reduction modulo $2$ is cuspidal. The cusps of $X$ are defined over
$\Q(\zeta_n)$, which is unramified above $2$, and the cuspidal subscheme is finite \'etale over $\Z[\tfrac1n]$, so
the reduction of $D$ lifts to an effective cuspidal divisor $C$ of degree $10$ defined over $\Q(\zeta_n)$. The
operator $g=\dm a$ fixes the reduction of $D-C$, which is $0$. By \Cref{lem:stab}~c), applied with $p=2$ and
$d=e=10$, as $\gon_\C X>40$, we get $gD+C=D+gC$, and as the support of $D$ contains no cusp, $gD=D$. Every point
in the support of $D$ reduces to a cusp modulo a prime above $2$, so its $j$-invariant is not integral, and the
corresponding elliptic curve has no complex multiplication. By \Cref{lem:orbits}, applied to the group generated
by $g$, the order of $g$ in $\Delta(n)$ divides $\deg D=10$, which is a contradiction.
\end{proof}

\begin{lemma}\label{lem:fb-99}
Let $X=X_1(99)$, which has genus $281$.
\begin{itemize}
\item[a)] The group $(\dm a-1)J_X(\Q)$ is finite for every $a\in(\torz{99})^\times$.
\item[b)] The curve $X_{\F_2}$ has no non-cuspidal closed points of degree at most $9$, and $120$ of degree $10$,
      which form $4$ diamond orbits.
\item[c)] We have $\gon_\C X>42.84$.
\end{itemize}
\end{lemma}

\begin{proof}
a) We \gitlink{computations/cont_winding99.m}{compute}, for each Galois orbit of even Dirichlet characters modulo
$99$, the winding submodule of the space of cuspidal modular symbols of weight $2$ with that character. The dimensions
of these spaces, counted with the sizes of the orbits, add up to $281$. The winding submodule is the whole space for
every non-trivial character, and it has codimension $1$ for the trivial character. By Kato's theorem
\cite[Corollary 14.3]{kato}, every factor of $J_X$ with non-trivial character has finitely many rational points, and
$\dm a$ acts trivially on the factor with trivial character, so $(\dm a-1)J_X(\Q)$ is finite.

b) We \gitlink{computations/contc99_source_census.m}{compute} that no elliptic curve over $\F_{2^k}$ with $k\leq9$ has
a point of order $99$, and that $40$ elliptic curves over $\F_{1024}$ have one, all with cyclic group of order $990$.
Their points of order $99$ give $120$ closed points of degree $10$, which form $4$ orbits under $\Delta(99)$, a group
of order $30$.

c) As $i(99)=4320$, Abramovich's bound gives $\gon_\C X\geq\tfrac{325}{2^{15}}\cdot4320>42.84$.
\end{proof}

For the remaining orbit of $\torz{99}$ we use the following quotient of $X_1(99)$.

\begin{lemma}\label{lem:fb-V99}
Let $H_{99}\subseteq\Delta(99)$ be the subgroup of the $\dm a$ with $a\equiv\pm1\pmod 9$, which has order $10$, let
$V_{99}=X_1(99)/H_{99}$, and let $\pi:X_1(99)\to V_{99}$ be the quotient map.
\begin{itemize}
\item[a)] The curve $V_{99}$ parametrizes triples $(E,\pm P_9,C_{11})$, where $P_9$ is a point of order $9$ and
      $C_{11}$ a cyclic subgroup of order $11$ of $E$, and $\pi(E,P)=(E,\pm11P,\dm{9P})$. It is the normalization of
      $X_1(9)\times_{X(1)}X_0(11)$, it has good reduction at $2$, and every $\dm a$ induces an automorphism
      $\delta_a$ of $V_{99}$, which depends only on $a$ modulo $9$ up to sign. In particular $\delta_{10}=1$.
\item[b)] The group $(\delta_2-1)J(V_{99})(\Q)$ is finite.
\item[c)] Let $t$ be the coordinate on $X_1(9)$ for which the pair $(E,P_9)$ is given by the curve
      \[
      E_t:y^2+(1-c)xy-by=x^3-bx^2,\qquad c=t^2(t-1),\quad b=c(t^2-t+1),
      \]
      and the point $(0,0)$. Then the function field of $V_{99,\F_2}$ is generated over $\F_2(t)$ by the function
      field of $X_0(11)_{\F_2}$, it has degree $12$ over $\F_2(t)$, and the curve $V_{99}$ has genus $29$.
\end{itemize}
\end{lemma}

\begin{proof}
a) Two points $(E,P)$ and $(E,P')$ of $X_1(99)$ satisfy $11P'=\pm11P$ and $\dm{9P'}=\dm{9P}$ if and only if $P'=aP$
with $a\equiv\pm1\pmod9$, i.e. if and only if $(E,P')=\dm a(E,P)$ with $\dm a\in H_{99}$. This proves the moduli
description, and hence $V_{99}$ is the normalization of $X_1(9)\times_{X(1)}X_0(11)$. It has good reduction at $2$
by \Cref{lem:goodred}, as its genus is positive by c). As $\Delta(99)$ is abelian, every $\dm a$ induces an
automorphism $\delta_a$ of $V_{99}$, which acts by $(E,\pm P_9,C_{11})\mapsto(E,\pm aP_9,C_{11})$. As
$10\equiv1\pmod9$, we have $\delta_{10}=1$.

b) We have $\pi_*\pi^*=10$ on $J(V_{99})$ and $\pi_*\dm2=\delta_2\pi_*$. So for $y\in J(V_{99})(\Q)$ we get
$10(\delta_2-1)y=\pi_*\bigl((\dm2-1)\pi^*y\bigr)$, which lies in the finite group $\pi_*\bigl((\dm2-1)J_1(99)(\Q)\bigr)$
by \Cref{lem:fb-99}~a). As $J(V_{99})(\Q)$ is finitely generated, $(\delta_2-1)J(V_{99})(\Q)$ is finite.

c) The map $V_{99}\to X_1(9)$ has degree $i(99)/(10\,i(9))=4320/360=12$, which is also the degree of
$X_0(11)\to X(1)$. Over $\F_2$ the maps $(E,\pm P_9,C_{11})\mapsto(E,\pm P_9)$ and $(E,\pm P_9,C_{11})\mapsto(E,C_{11})$
to $X_1(9)$ and $X_0(11)$ show that the function field of $V_{99,\F_2}$ contains $\F_2(t)$ and the function field of
$X_0(11)_{\F_2}$. We \gitlink{computations/contc99_independent_Y29_model_v5.m}{compute} that $(0,0)$ has order $9$ on
$E_t$ over $\F_2(t)$, and, eliminating the $j$-invariant between $E_t$ and the $j$-map of a model of
$X_0(11)_{\F_2}$, that the compositum of these two fields is given by an irreducible polynomial of degree $12$ over
$\F_2(t)$. So it is the function field of $V_{99,\F_2}$, and we compute that it has genus $29$. On this model we also
express $j(E/C_{11})$, using the Atkin-Lehner involution of $X_0(11)$. The image on $V_{99,\F_2}$ of a point $(E,P)$
of $X_1(99)_{\F_2}$ is then determined by the value of $t$ at $(E,\pm11P)$ and by the point $(E,\dm{9P})$ of
$X_0(11)_{\F_2}$, which we compute from $j(E)$ and $j(E/\dm{9P})$.
\end{proof}

\begin{proposition}\label{prop:99}
The group $\torz{99}$ 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 $99$ over a number field of degree $10$. As
$2^{15}\equiv-1\pmod{99}$ and $2,2^3,2^5\not\equiv\pm1\pmod{99}$, the operator $\dm2$ has order $15$ in
$\Delta(99)$. By \Cref{lem:fb-99} and \Cref{lem:fb-cusp}, applied with $a=2$, the reduction $\overline D$ of $D$
modulo $2$ is not entirely cuspidal, so by \Cref{lem:fb-99}~b) it is a non-cuspidal closed point $G$ of degree $10$.
Then $G=\dm aG_i$ for some $a$ and $i$, where $G_1,\dots,G_4$ are representatives of the $4$ diamond orbits of such
points.

We first use the map $\pi_{33}:X_1(99)\to X_1(33)$, $(E,P)\mapsto(E,3P)$, which commutes with the diamond operators.
We have $\rk J_1(33)(\Q)=0$ \cite[Theorem 3.1(2)]{DEHMZ}, so for $A_5=T_5-1-5\dm5$ on
$X_1(33)$ the divisor $A_5\pi_{33*}D$ is principal (\Cref{prop:tools}~d)), and so is its reduction
$\dm aA_5\pi_{33*}G_i$. We \gitlink{computations/contc99_quotient33_RR.m}{compute} that for every $i$ the image
$P_i=\pi_{33*}G_i$ is a closed point of degree $10$, that the six subgroups of order $5$ form Frobenius orbits of
lengths $1$, $1$, $2$ and $2$, which give $T_5P_i$ as a sum of closed points of degrees $10$, $10$, $20$ and $20$, each
with multiplicity $1$, and that $\ell(A_5P_i)=0$ for $i=1,2,4$. So $G=\dm aG_3$.

Let $\pi:X_1(99)\to V_{99}$ be as in \Cref{lem:fb-V99}, let $c_0$ be the image of the standard cusp, which is
rational, and put $A_5=T_5-1-5\delta_5$ on $V_{99}$. We have
\[
(\delta_2-1)\pi_*D=(\delta_2-1)(\pi_*D-10c_0)+10(\delta_2c_0-c_0).
\]
The class of the first term lies in the finite group $(\delta_2-1)J(V_{99})(\Q)$ (\Cref{lem:fb-V99}~b)), and the
second term is a rational cuspidal divisor of degree $0$, whose class is torsion by the Manin-Drinfeld theorem. So
the class of $(\delta_2-1)\pi_*D$ is a rational torsion point of $J(V_{99})$, and it is annihilated by $A_5$
(\Cref{prop:tools}). Hence the reduction $(\delta_2-1)A_5\pi_*G$ of $A_5(\delta_2-1)\pi_*D$ is principal. As the
operators $\delta_b$ are automorphisms of $V_{99}$ which commute with $A_5$, it follows that
$B=(\delta_2-1)A_5P_V$ is principal, where $P_V=\pi_*G_3$. Writing $T=T_5P_V$, and using $\delta_{10}=1$, we have
\[
B=\delta_2T-T-\delta_2P_V+5\delta_5P_V-4P_V .
\]
We \gitlink{computations/contc99_independent_Y29_RR.m}{compute} $B$ on the model of \Cref{lem:fb-V99}~c), carrying the
level structure of a point of $G_3$ through the six isogenies of degree $5$: the point $P_V$ has degree $10$, $T$ and
$\delta_2T$ are sums of closed points of degrees $10$, $10$, $20$ and $20$, each with multiplicity $1$, and the
positive and negative parts of $B$ have degree $110$. We compute that $\ell(B)=0$, so $B$ is not principal, which is
a contradiction.
\end{proof}

\begin{lemma}\label{lem:fb-105}
Let $X=X_1(105)$, which has genus $289$.
\begin{itemize}
\item[a)] The group $(\dm{22}-1)J_X(\Q)$ is finite.
\item[b)] The curve $X_{\F_2}$ has no non-cuspidal closed points of degree at most $9$, and $144$ of degree $10$,
      which form $6$ diamond orbits.
\item[c)] We have $\gon_\C X>45.70$.
\end{itemize}
\end{lemma}

\begin{proof}
a) We \gitlink{computations/cont_winding105.m}{compute} the winding submodules as in the proof of
\Cref{lem:fb-99}. The dimensions add up to $289$, and the winding submodule is the whole space for every Galois orbit
of even characters except one, which consists of characters of order $6$ and conductor $21$, and for which it has
codimension $1$. As $22\equiv1\pmod{21}$, these characters are trivial at $22$, so $\dm{22}$ acts trivially on the
corresponding factor of $J_X$, and by Kato's theorem \cite[Corollary 14.3]{kato} the group $(\dm{22}-1)J_X(\Q)$ is
finite.

b) We \gitlink{computations/contc105_source_census.m}{compute} that no elliptic curve over $\F_{2^k}$ with $k\leq9$ has
a point of order $105$, and that $60$ elliptic curves over $\F_{1024}$ have one, all with cyclic group of order
$1050$. Their points of order $105$ give $144$ closed points of degree $10$, which form $6$ orbits under
$\Delta(105)$, a group of order $24$.

c) As $i(105)=4608$, Abramovich's bound gives $\gon_\C X\geq\tfrac{325}{2^{15}}\cdot4608>45.70$.
\end{proof}

\begin{proposition}\label{prop:105}
The group $\torz{105}$ 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 $105$ over a number field of degree $10$. We have
$22^2\equiv64$ and $22^4\equiv1$ modulo $105$, and $64\not\equiv-1\pmod{105}$, so $\dm{22}$ has order $4$ in
$\Delta(105)$. By \Cref{lem:fb-105} and \Cref{lem:fb-cusp}, applied with $a=22$, the reduction of $D$ modulo $2$ is
not entirely cuspidal, so by \Cref{lem:fb-105}~b) it is a non-cuspidal closed point $G$ of degree $10$, and
$G=\dm aG_i$ for a representative $G_i$ of one of the $6$ diamond orbits.

Let $\pi_{35}:X_1(105)\to X_1(35)$ be the map $(E,P)\mapsto(E,3P)$, which commutes with the diamond operators.
We have $\rk J_1(35)(\Q)=0$ \cite[Theorem 3.1(2)]{DEHMZ}, so for $A_3=T_3-1-3\dm3$ on $X_1(35)$
the divisor $A_3\pi_{35*}D$ is principal (\Cref{prop:tools}~d)), and so is its reduction $\dm aA_3\pi_{35*}G_i$. We
\gitlink{computations/contc105_quotient35_RR.m}{compute} that for each of the $6$ representatives the image
$P_i=\pi_{35*}G_i$ is a closed point of degree $10$, that the four subgroups of order $3$ form Frobenius orbits of
lengths $1$ and $3$, which give $T_3P_i$ as a sum of closed points of degrees $10$ and $30$, each with multiplicity
$1$, and that
\[
\ell\bigl(T_3P_i-P_i-3\dm3P_i\bigr)=0 .
\]
So $A_3P_i$ is not principal for any $i$, which is a contradiction.
\end{proof}