\section{The remaining reductions}\label{sec:exceptions}

For the pairs $(d,p)$ of \Cref{table:local}, \Cref{prop:sieve} shows that the reduction $Z$ of a trace divisor is one of the divisors in \Cref{table:exceptions}, and the propositions of this section exclude them. Most of the cases are dealt with by \Cref{lem:stab}, in which case we give the element $h$, the reason why $y=0$ (if we use it), and the relevant inequality for the gonality, which always comes from \Cref{prop:gonality}. All the statements that $h$ fixes a closed point of $X_1(p)_{\F_2}$ are \gitlink{degree19/stabilizers.m}{checked} on the elliptic curves representing these points.

\subsection{The prime \texorpdfstring{$37$}{37}}

\begin{lemma}\label{lem:37twist}
Let $K$ be a number field of degree $9$, and let $E/K$ be an elliptic curve with $j(E)\in\{-7\cdot11^3,-7\cdot137^3\cdot2083^3\}$. Then $E(K)$ has no point of order $37$.
\end{lemma}

\begin{proof}
These are the $j$-invariants of the curves \lmfdbec{1225}{b}{2} and \lmfdbec{1225}{b}{1}, and as $j(E)\neq0,1728$, $E$ is a quadratic twist of one of them over $K$. Hence the $x$-coordinate of a point of order $37$ of $E(K)$ would be a root in $K$ of the $37$-division polynomial of one of these curves. But we \gitlink{degree09/P37/rational_isogeny_fibers.m}{compute} that the irreducible factors of these division polynomials have degrees $6,6,6,666$ and $18,222,222,222$, none of which divides $9$.
\end{proof}

\begin{proposition}\label{prop:37}
We have $37\notin S(9)$ and $37\notin S(11)$.
\end{proposition}

\begin{proof}
Let $d\in\{9,11\}$, and let $D$ be the trace divisor of a point of order $37$ over a number field $K$ of degree $d$. As $37\notin S(1)\cup S(3)$, $D$ is a closed point of degree $d$. By \Cref{prop:cusps,prop:sieve}, $Z$ is one of the \gitlink{audit11/stabilizers.m}{orbits} considered below. We \gitlink{audit11/rational_two.m}{compute} that the common kernel of $A_3$ and $A_5$ on $H_1(X_1(37),\Z)\otimes\F_2$, which contains $J_1(37)(\Q)[2]$ by \Cref{lem:torsion} b), is $0$. So $y=0$ in \Cref{lem:stab}, and we have $\gon_\Q X_1(37)=18$. As $37$ is not strange, $(h-1)J_1(37)(\Q)$ is finite for every $h\in\Delta(37)$ by \Cref{lem:finite}.

Suppose first that $Z=O_6$, one of $3$ closed points of degree $6$ fixed by $h=\dm8$, of order $6$. Then $c=d-6\in\{3,5\}$, and $18>d+c$, so $6\mid c$ by \Cref{lem:stab}, which is a contradiction.

Suppose now that $Z$ is one of the $2$ ordinary closed points of degree $9$, or the supersingular closed point of degree $9$, all of which are fixed by $h=\dm4$, of order $9$. Then $c\in\{0,2\}$ and $18>d+c$, so $hD=D$ and $9\mid c$. Hence $c=0$, $d=9$, and by \Cref{lem:orbits} the support of $D$ is a single orbit of $\dm4$, so the image of $D$ on $X_0(37)$ is a non-cuspidal rational point. The $j$-invariants of these points are $-7\cdot11^3$ and $-7\cdot137^3\cdot2083^3$ \cite[Section 5]{MSD}, which contradicts \Cref{lem:37twist}.

Suppose finally that $d=11$ and $Z=O_{10}$, so $c=1$. For one of the two orbits, $Z$ is fixed by $h=\dm 6$, of order $2$, and as $18>12$, \Cref{lem:stab} gives $2\mid 1$. The other orbit consists of the reductions of the closed points of degree $10$ in the supports of the trace $S$ of the point of order $37$ of \Cref{sec:lower} and of its images under $\Delta(37)$ \gitlink{audit11/decic37.m}{(we check this)}. Replacing $S$ by such an image we can assume $\overline S=Z$, so $D$ and $S+C$ both reduce to $Z+C'$. We \gitlink{audit11/decic37_jets.m}{compute} that for $t=\dm2-1$ the map $f_{t,11}$ is a formal immersion at $Z+\overline{c_1}$ for each of the $18$ rational cusps $c_1$. The group $tJ_1(37)(\Q)$ is finite, and as $J_1(37)(\Q)[2]=0$, reduction is injective on it, so $D=S+C$ by \Cref{lem:fi} b), which is a contradiction as $D$ contains no cusps.
\end{proof}

\subsection{The primes \texorpdfstring{$41$}{41} and \texorpdfstring{$43$}{43}}

\begin{proposition}\label{prop:41}
Let $d\in\{15,16\}$, and let $D$ be the trace divisor of a point of order $41$ over a number field of degree $d$. Then $Z\neq O_{14}$.
\end{proposition}

\begin{proof}
Here $O_{14}$ is one of the $10$ closed points of degree $14$ which are the reductions of the closed points of degree $14$ in the supports of the trace $S$ of an \gitlink{degree15/lift41_charzero.m}{explicit point} of degree $14$ on $X_1(41)$ and of its images under $\Delta(41)$, as we \gitlink{degree15/known41_lifting.m}{check}. Suppose that $Z=O_{14}$. Replacing $S$ by one of these images we can assume $\overline S=Z$, so $D$ and $S+C$ both reduce to $Z+C'$, where $c=d-14\in\{1,2\}$. As $J_1(41)(\Q)$ is finite (\Cref{lem:finite}), the tuple $t=(A_3,A_5,A_7)$ kills $J_1(41)(\Q)$ by \Cref{lem:torsion} b), and we compute that $f_{t,d}$ is a formal immersion at $Z+C'$ for all the $20$ (for $d=15$) and $210$ (for $d=16$, \gitlink{degree16/known41_lifting.m}{code}) possible $C'$. So $D=S+C$ by \Cref{lem:fi} b), which is impossible as $D$ contains no cusps and $C\neq0$.
\end{proof}

For $p=43$ and $d=16$ we need the following lemma.

\begin{lemma}\label{lem:43inj}
Reduction modulo $2$ is injective on $J_1(43)(\Q)_\tor$.
\end{lemma}

\begin{proof}
Let $Y=X_1(43)/\dm2$, where $\dm 2$ has order $7$, and let $f:X_1(43)\to Y$ be the quotient map. The curve $Y$ has genus $9$, and we \gitlink{degree16/torsion43_upper.m}{compute} that $J_Y$ contains an elliptic curve $A$ isogenous to \lmfdbec{43}{a}{1}, whose isogeny class consists of one curve with trivial rational torsion, so $A(\Q)[2]=0$, that $B=J_Y/A$ has rank $0$, and that the numbers of points of $B$ over $\F_3$, $\F_5$ and $\F_{23}$ have greatest common divisor $532=4\cdot 133$. Hence $J_Y(\Q)[2]$ injects into $B(\Q)_\tor$, which injects into $B(\F_3)$, $B(\F_5)$ and $B(\F_{23})$, so $\#J_Y(\Q)[2]\leq4$. Let $c_1,c_2,c_3$ be the rational cusps of $Y$. The divisors $38(c_2-c_1)$ and $38(c_3-c_1)$ are \gitlink{degree16/torsion43_Q_first.m}{principal}, and we \gitlink{degree16/torsion43_reduction.m}{compute} that the reductions of $x_1=19[c_2-c_1]$, $x_2=19[c_3-c_1]$ and $x_1+x_2$ are non-zero. So $J_Y(\Q)[2]=\{0,x_1,x_2,x_1+x_2\}$, and reduction is injective on it.

Now let $x\in J_1(43)(\Q)_\tor$ with $\overline x=0$. Then $2x=0$, so $f_*x\in J_Y(\Q)[2]$ has reduction $0$, and hence $f_*x=0$. We \gitlink{degree16/two43_h7.m}{compute} that $\dm 2$ acts trivially on the common kernel of $A_3$, $A_5$ and $A_7$ on $H_1(X_1(43),\Z)\otimes\F_2$, which contains $x$. Hence $0=f^*f_*x=\sum_{i=0}^6\dm2^ix=7x=x$.
\end{proof}

\begin{proposition}\label{prop:43}
Let $d\in\{13,16\}$, and let $D$ be the trace divisor of a point of order $43$ over a number field of degree $d$. Then $Z$ is not one of the divisors listed for $43$ in \Cref{table:exceptions}.
\end{proposition}

\begin{proof}
As $43$ is not strange, $(h-1)J_1(43)(\Q)$ is finite for every $h\in\Delta(43)$ by \Cref{lem:finite}. The $7$ closed points $O_{12}$ are the reductions of the closed points of degree $12$ in the supports of the trace $S$ of the point of order $43$ of \Cref{sec:lower} and its images under $\Delta(43)$, as we \gitlink{degree13/mixed43_lifting.m}{check}, and the $21$ closed points $O_{14}$ are, in the same way, the reductions of the trace $S$ of an explicit point of degree $14$ on $X_1(43)$ from the tables of \cite{vanHoeij} and its images (\gitlink{degree16/lift43_14.m}{check}). If $d=13$ and $Z=O_{12}$, or $d=16$ and $Z=O_{14}$, then replacing $S$ by one of its images we can assume $\overline S=Z$, so $D$ and $S+C$ both reduce to $Z+C'$, where $c=d-\deg Z\geq1$. For $d=13$ and $Z=O_{12}$ we \gitlink{degree13/mixed43_tangents.m}{compute} that $f_{t,13}$ is a formal immersion at $Z+\overline{c_1}$ for all the $21$ rational cusps $c_1$, with $t=(\dm{3}^{-1}-1)\dm{11}^{-1}A_{11}$, which kills $J_1(43)(\Q)$, and for $d=16$ and $Z=O_{14}$ we \gitlink{degree16/known43_cot.m}{compute} that $f_{t,16}$ is a formal immersion at all the $231$ divisors $Z+C'$, with $t$ combining several maps $\alpha A_q$, $q\in\{3,5,7\}$, with $\alpha J_1(43)(\Q)$ finite. In both cases \Cref{lem:fi} b) gives $D=S+C$, which is impossible as $D$ contains no cusps.

The remaining cases are for $d=16$. By \Cref{lem:43inj}, $y=0$ in \Cref{lem:stab}, and $\gon_\Q X_1(43)=24$. If $Z=O_{12}$ or $Z$ is one of the two orbits $O_{15}$, then $Z$ is fixed by $h=\dm{37}$, of order $3$, and $c\in\{4,1\}$, so as $24>16+c$, \Cref{lem:stab} gives $3\mid c$, which is a contradiction. If $Z\in\{SS_{14},2SS_7,2NC_7,SS_7+NC_7\}$, then $Z$ is fixed by $h=\dm2$, of order $7$, and $c=2$, so as $24>18$ we get $7\mid2$, which is again a contradiction.
\end{proof}

\subsection{The primes \texorpdfstring{$53$}{53} and \texorpdfstring{$61$}{61}}

\begin{proposition}\label{prop:53}
Let $13\leq d\leq 21$, and let $D$ be the trace divisor of a point of order $53$ over a number field of degree $d$. Then $Z\neq SS_{13}$.
\end{proposition}

\begin{proof}
There is a unique supersingular closed point $SS_{13}$ of degree $13$, and $h=\dm4$, of order $13$, acts simply transitively on its geometric points. As $53$ is not strange, $(h-1)J_1(53)(\Q)$ is finite, and we \gitlink{degree13/two53_verified.json}{compute} that $J_1(53)(\Q)[2]=0$, so $y=0$ in \Cref{lem:stab}.

Suppose that $Z=SS_{13}$, so $c=d-13$. If $d=13$, then as $\gon_\Q X_1(53)>13$, \Cref{lem:stab} gives $hD=D$, and by \Cref{lem:orbits} $D$ is a single orbit of $\dm4$, so its image on $X_0(53)$ is a non-cuspidal rational point, which does not exist \cite[Theorem 1]{mazur78}. If $14\leq d\leq 16$, then $d+c\leq 19<20\leq\gon_\Q X_1(53)$, and \Cref{lem:stab} gives $13\mid c$, which is a contradiction. For $17\leq d\leq 21$ we use \Cref{lem:relD} with $L=C+\pi^*Y$, where $\pi:X_1(53)\to Y=X_1(53)/\dm4$ is the quotient map, and $t=((h-1)A_3,(h-1)A_5)$. Here $t$ kills $J_1(53)(\Q)$ and $\pi^*J_Y$, and $t[C+\pi^*E_0-dc_0]=0$ for every point $E_0$ of $Y$, as $h-1$ kills $\pi^*E_0$ and $A_q$ kills rational cuspidal divisors. We \gitlink{degree17/relative53.m}{compute} that the rank condition holds for all the possible $C'$, so $D$ lies on $L$ and contains $C\neq0$, which is a contradiction.
\end{proof}

\begin{lemma}\label{lem:W61}
\begin{itemize}
\item[a)] We have $J_1(61)(\Q)[2]=0$.
\item[b)] Let $W=X_1(61)/\dm4$, where $\dm4$ has order $15$, and let $s$ be the automorphism of $W$ induced by $\dm2$. Then $W$ has genus $8$, $s$ is an involution with $W/\dm s=X_0(61)$, $(s-1)J_W(\Q)$ is finite, $J_W(\Q)[2]=0$, and $W$ has no map of degree $2$ or $3$ to $\PP^1$ defined over $\Q$.
\end{itemize}
\end{lemma}

\begin{proof}
We \gitlink{degree15/quotient61/verify_sage.py}{compute} a), and that $W$ has \gitlink{degree15/quotient61/verify_small_quotient.m}{genus $8$}. The group $\Delta(61)/\dm4$ has order $2$ and is generated by the image of $\dm2$, so $s$ is an involution and $W/\dm s=X_1(61)/\Delta(61)=X_0(61)$. The characters trivial on $\dm4$ have order at most $2$, and the strange characters modulo $61$ have order $6$, so $(s-1)J_W(\Q)$ is finite by \Cref{lem:finite}. The pullback $J_W\to J_1(61)$ is injective on $2$-torsion, as its composition with the pushforward is multiplication by $15$, so $J_W(\Q)[2]=0$ by a).

The curve $X_0(61)$ has genus $4$ and no map of degree at most $3$ to $\PP^1$ over $\Q$: its canonical model is a complete intersection of a quadric and a cubic in $\PP^3$, and the discriminant of the quadric is not a square, so the two $g^1_3$'s are conjugate over $\Q(i)$ (see \cite{HS,jeon} and \gitlink{degree15/quotient61/verify_sage.py}{our check}). Suppose that $u:W\to\PP^1$ has degree $e\in\{2,3\}$. As $g(W)=8>(e-1)^2$, the Castelnuovo-Severi inequality implies that $u$ is the unique map of degree $e$ to $\PP^1$ up to automorphisms of $\PP^1$, so $u\circ s=\beta\circ u$ for some $\beta\in\mathrm{PGL}_2(\Q)$ with $\beta^2=1$. If $\beta=1$, then $u$ factors through $W\to X_0(61)$, which is impossible as $e\leq 3$. Otherwise $u$ induces a map $X_0(61)\to\PP^1/\dm\beta\simeq\PP^1$ of degree $e$, which is again impossible.
\end{proof}

\begin{proposition}\label{prop:61}
Let $d\in\{15,\dots,19,21,22\}$, and let $D$ be the trace divisor of a point of order $61$ over a number field of degree $d$. Then $Z\notin\{SS_{15},SS_{20},O_{20}\}$.
\end{proposition}

\begin{proof}
There is a unique supersingular closed point $SS_{15}$ of degree $15$, and it is \gitlink{degree15/quotient61/verify_reduction.m}{fixed} by all of $\Delta(61)$. Let $h=\dm3=\dm2^6$, of order $5$. As the strange characters modulo $61$ have order $6$, $(h-1)J_1(61)(\Q)$ is finite by \Cref{lem:finite}, and $J_1(61)(\Q)[2]=0$ by \Cref{lem:W61} a), so $y=0$ in \Cref{lem:stab}. Note that $\gon_\Q X_1(61)>18$ by \Cref{prop:gonality} a).

Suppose that $Z=SS_{15}$. If $d=16$, then $c=1$, and \Cref{lem:stab} gives $5\mid 1$. If $d\in\{17,18,19,21,22\}$, we use \Cref{lem:relD} with $L=C+\pi^*Y^{(3)}$, where $\pi:X_1(61)\to Y=X_1(61)/\dm h$ is the quotient map, and $t=((h-1)A_3,(h-1)A_5)$. Here $t$ kills $J_1(61)(\Q)$ and $\pi^*J_Y$, and $t[C+\pi^*E_0-dc_0]=0$ for every point $E_0$ of $Y^{(3)}$, as $h-1$ kills $\pi^*E_0$ and $A_q$ kills rational cuspidal divisors. We \gitlink{degree17/relative61.m}{compute} that the rank condition holds for all possible $C'$, so $D$ contains $C\neq0$, which is impossible. In the same way, for $d\in\{21,22\}$ and $Z\in\{SS_{20},O_{20}\}$, where $O_{20}$ is an orbit of $6$ closed points which are pullbacks from $Y$, we use $L=C+\pi^*Y^{(4)}$ (\gitlink{degree21/relative61_ss20.m}{$SS_{20}$}, \gitlink{degree21/relative61_o20.m}{$O_{20}$}).

It remains to deal with $d=15$ and $Z=SS_{15}$, so $c=0$. By \Cref{lem:stab} $hD=D$, so by \Cref{lem:orbits} $D$ is a pullback from $Y$, $D=\pi^*D_Y$, and $D_Y$ is a closed point of degree $3$. Let $W$ and $s$ be as in \Cref{lem:W61}. The image $D_W$ of $D_Y$ on $W$ is a closed point of degree $3$, as a rational point would map to a non-cuspidal rational point on $X_0(61)$. Its reduction is fixed by $s$, as $\overline D=SS_{15}$ is fixed by $\Delta(61)$. By \Cref{lem:W61} b) and \Cref{lem:stabgen} (with $C=0$ and condition i)), $sD_W\sim D_W$, and as $\gon_\Q W\geq 4$, $sD_W=D_W$. As $2$ and $3$ are coprime, $s$ fixes each of the three geometric points of $D_W$. If $w$ is one of them and $x$ is a point of $D$ above it, then $\dm2x=gx$ for some $g\in\dm4$, so $x$ is fixed by $\dm2g^{-1}\neq1$. This contradicts \Cref{lem:CM}, as $x$ has degree $15<20$.
\end{proof}

\subsection{The prime \texorpdfstring{$73$}{73}}

\begin{lemma}\label{lem:Y73}
Let $Y=X_1(73)/\dm2$, where $\dm2$ has order $9$. Then every non-cuspidal point of $Y$ has degree at least $3$.
\end{lemma}

\begin{proof}
A non-cuspidal point of $Y$ of degree at most $2$ maps to a non-cuspidal point of degree at most $2$ on $X_0(73)$. There are no non-cuspidal rational points on $X_0(73)$ \cite[Theorem 1]{mazur78}, and the quadratic points were determined by Box \cite[Section 4.7]{Box}. They have $11$ different $j$-invariants ($3$ of them quadratic and not CM, and $8$ rational CM $j$-invariants). For each of them we \gitlink{degree18/quotient_quadratic.m}{check}, using the action of Frobenius on $E[73]$ at a prime which splits in the corresponding quadratic field, that no point of $Y$ of degree at most $2$ lies above it.
\end{proof}

\begin{proposition}\label{prop:73}
Let $9\leq d\leq 22$, and let $D$ be the trace divisor of a point of order $73$ over a number field of degree $d$. Then $Z$ is not a non-zero sum of points $SS_6$, $NC_9$ and $SS_{18}$.
\end{proposition}

\begin{proof}
Let $h=\dm2$, of order $9$. It fixes each of the $4$ supersingular closed points of degree $6$ and the $4$ non-rational cusps (of degree $9$), and the points of degree $18$, and the subgroup $\dm8$ of order $3$ fixes the geometric points of degree $6$. As $73$ is not strange, $(h-1)J_1(73)(\Q)$ is finite. We \gitlink{audit11/rational_two.m}{compute} that $h$ acts trivially on the common kernel of $A_3$ and $A_5$ on $H_1(X_1(73),\Z)\otimes\F_2$, which contains $J_1(73)(\Q)[2]$, so $y=0$ in \Cref{lem:stab} (condition ii) of \Cref{lem:stabgen}), and $\gon_\Q X_1(73)>26$. As $d<24=(73-1)/3$, the orbits of $\dm 2$ on the support of $D$ have length $9$ (\Cref{lem:orbits}).

Suppose that $Z$ is a non-zero sum of points $SS_6$, $NC_9$ and $SS_{18}$, and let $e=\deg Z$, so $c=d-e$ and $e\in\{6,9,12,15,18,21\}$. Suppose first that $d+c=2d-e\leq 26$. Then \Cref{lem:stab} gives $hD=D$ and $9\mid c$, and as the orbits of $h$ have length $9$, $9$ divides $d$. Hence $d\in\{9,18\}$, and by \Cref{lem:orbits} $D$ is a pullback from $Y=X_1(73)/\dm2$, of an effective divisor of degree $d/9\leq2$ with no cusps in its support. This contradicts \Cref{lem:Y73}.

Suppose now that $2d-e>26$. Then $d\geq 17$, and $Z$ and $d$ are as in \Cref{table:73}. In these cases we use \Cref{lem:relD} with $t=((h-1)A_3,(h-1)A_5)$ and
$$L=Z_0+R+\pi^*Y^{(r)},$$
where $\pi:X_1(73)\to Y$ is the quotient map, $Z_0$ is the divisor of degree $6k$ whose points are the unique lifts of the geometric points of $SS_6$ in $Z$ that are fixed by $\dm8$ (here $k$ is the number of points $SS_6$ in $Z$), $C'=\overline{R}+\pi^*E'$ with $E'$ a sum of $r'$ cusps of $Y$, $R$ has no full orbit of $\dm2$ in its support, and $r=r'$ plus the number of points $NC_9$ in $Z$ (which are pullbacks of cusps of $Y$). As $h$ commutes with $\dm 8$ and permutes the geometric points of each point $SS_6$, it fixes $Z_0$. So $t$ kills $J_1(73)(\Q)$ and $\pi^*J_Y$, and $t[Z_0+R+\pi^*E_0-dc_0]=0$ for every point $E_0$ of $Y^{(r)}$, as $h-1$ kills $Z_0$ and $\pi^*E_0$ and $A_q$ kills the rational cuspidal divisor $(h-1)(R-dc_0)$. We \gitlink{degree17/relative73.m}{compute} (\gitlink{degree18/relative73_nc.m}{$NC_9$}, \gitlink{degree21/relative73_multiple.m}{two or three non-ordinary points}) that the rank condition holds in all cases, so $D$ lies on $L$. If $R\neq0$, then $D$ contains a cusp, which is impossible. If $R=0$ and $k>0$, then $D$ contains a point fixed by $\dm8$, which contradicts \Cref{lem:CM}. If $R=0$ and $k=0$, then $Z=NC_9$, $d=18$, and $D$ is a pullback from $Y$ of an effective divisor of degree $2$ with no cusps in its support, which contradicts \Cref{lem:Y73}.
\end{proof}

\begin{table}[H]
\centering
\begin{tabular}{rl}
\toprule
$d$ & $Z$\\
\midrule
$17$ & $SS_6$\\
$18$ & $SS_6$, $NC_9$\\
$19$ & $SS_6$, $NC_9$\\
$20$ & $SS_6$, $NC_9$, $2SS_6$\\
$21$ & $SS_6$, $NC_9$, $2SS_6$, $SS_6+NC_9$\\
$22$ & $SS_6$, $NC_9$, $2SS_6$, $SS_6+NC_9$\\
\bottomrule
\end{tabular}
\medskip
\caption{The reductions at $73$ for which we use \Cref{lem:relD}.}
\label{table:73}
\end{table}

\subsection{The primes \texorpdfstring{$89$}{89} and \texorpdfstring{$97$}{97}}

\begin{proposition}\label{prop:89}
Let $11\leq d\leq 22$, and let $D$ be the trace divisor of a point of order $89$ over a number field of degree $d$. Then $Z\notin\{NC_{11},2NC_{11},SS_{22}\}$.
\end{proposition}

\begin{proof}
Let $h=\dm2$, of order $11$, which fixes the $4$ non-rational cusps. As $89$ is not strange, $(g-1)J_1(89)(\Q)$ is finite for every $g\in\Delta(89)$. We \gitlink{degree16/certify_anti89.m}{compute} that $A_3$ has odd degree on $(h-1)J_1(89)$, so $y=0$ in \Cref{lem:stab} (condition iii) of \Cref{lem:stabgen}), and $\gon_\Q X_1(89)>39$. If $Z=NC_{11}$, then $c=d-11$ and $d+c\leq 33$, so \Cref{lem:stab} gives $hD=D$ and $11\mid c$. If $c=0$, then $d=11$, $D$ is a single orbit of $\dm 2$ by \Cref{lem:orbits}, and its image on $X_0(89)$ is a non-cuspidal rational point, which does not exist \cite[Theorem 1]{mazur78}. So $c=11$ and $d=22$.

It remains to deal with $d=22$ and $Z\in\{NC_{11},2NC_{11},SS_{22}\}$. The supersingular closed points of degree $22$ form $5$ orbits under $\Delta(89)$, of lengths $4,4,4,2,1$. For the first three, $\alpha A_3Z$ or $\alpha A_5Z$ is not principal (\Cref{prop:sieve}), and the other two are fixed by $h$. Hence, by \Cref{lem:stab} and \Cref{lem:orbits}, in all the remaining cases $D$ is a pullback from $Y=X_1(89)/\dm2$, of a divisor $D'$ of degree $2$. The curve $Y$ has genus $27$, and we \gitlink{degree22/check_quotient89.py}{compute} that $\#Y(\F_2)=8$ and $\#Y(\F_4)=38$, so $\gon_\Q Y\geq\gon_{\F_4}Y\geq8$.

If $Z$ contains a non-rational cusp, then $\overline{D'}$ is a sum of two cusps of $Y_{\F_2}$, which are defined over $\F_2$. Let $h'$ be a generator of $\Delta(89)/\dm2$. We \gitlink{degree22/quotient89_cusps.m}{compute} that $f_{t,2}$ on $Y^{(2)}$, with $t$ combining the maps $(h'-1)B_qA_q$ for several primes $q$, is a formal immersion at all the $36$ sums of two such cusps. By \Cref{lem:finite} and \Cref{lem:torsion} (note that $\dm2$ has odd order $11$), $t$ kills $J_Y(\Q)$ and all the cuspidal divisors on $Y$. So $f_{t,2}$ maps $D'$ and the cuspidal divisor lifting $\overline{D'}$ (which is defined over an unramified extension of $\Q_2$) to $0$, and they are equal by \Cref{lem:fi} a), which is a contradiction.

If $Z=SS_{22}$, let $\iota$ be the element of order $2$ of $\Delta(89)/\dm2$, which is cyclic of order $4$. It fixes $\overline{D'}$, and $(\iota-1)J_Y(\Q)$ is finite, so by \Cref{lem:stabgen} (with $C=0$), $2\iota D'\sim 2D'$. As $\gon_\Q Y>4$, it follows that $\iota D'=D'$. A point of $Y$ fixed by $\iota$ lies below a point of $X_1(89)$ fixed by a non-trivial element of $\Delta(89)$, which has degree at least $30$ by \Cref{lem:CM}, so it has degree at least $3$. Hence $\iota$ has no fixed points in the support of $D'$, so $D'=y+\iota(y)$ for a point $y$ of degree at most $2$, and the image of $D'$ on $Y/\dm\iota$ is twice a rational point. Its image on $X_0(89)$ is a non-cuspidal rational point, which does not exist.
\end{proof}

At $p\in\{97,101\}$ we also need operators that kill the part of $J_1(p)$ belonging to the strange characters. We \gitlink{degree20/strange_eigenvalues.m}{compute} that for these $p$ the newforms with a strange character form a single Galois orbit, and that their coefficients lie in the field of values of their characters. Hence each $T_n$ acts on the corresponding abelian subvariety $B_p\subseteq J_1(p)$ through $\Z[\Delta(p)]$, and we put $\beta_n=T_n-a_n$, where $a_n\in\Z[\Delta(p)]$ is such that $\beta_n$ kills $B_p$. For every $h\in\Delta(p)$, the endomorphism $(h-1)\beta_n$ kills $B_p$ and the part of $J_1(p)$ with trivial character, so $(h-1)\beta_nJ_1(p)(\Q)$ is finite by \Cref{lem:finite}.

\begin{lemma}\label{lem:W97}
Let $H_8\subseteq\Delta(97)$ be the subgroup of order $8$, let $W=X_1(97)/H_8$, and let $h'$ be an element of order $3$ of $\Delta(97)/H_8$. Then $W$ has genus $41$, $(h'-1)J_W(\Q)$ is finite, $A_3$ restricts to an isogeny of odd degree on $(h'-1)J_W$, and $\gon_\Q W\geq7$.
\end{lemma}

\begin{proof}
The Jacobian $J_W$ corresponds to the characters modulo $97$ which are trivial on $H_8$, i.e.\ of order dividing $6$, and the strange characters have order $12$, so $(h'-1)J_W(\Q)$ is finite by \Cref{lem:finite}. We \gitlink{degree16/certify_anti97.m}{compute} that $W$ has genus $41$ and that $A_3$ has odd degree on $(h'-1)J_W$. A map $W\to\PP^1$ of degree $k$ gives a map $X_1(97)\to\PP^1$ of degree $8k$, and $\gon_\Q X_1(97)\geq 49$ by \Cref{prop:gonality} a), so $\gon_\Q W\geq 49/8$.
\end{proof}

\begin{proposition}\label{prop:97}
Let $16\leq d\leq 22$, and let $D$ be the trace divisor of a point of order $97$ over a number field of degree $d$. Then $Z\neq SS_{16}$.
\end{proposition}

\begin{proof}
The two supersingular closed points of degree $16$ are fixed by the subgroup $H_4\subseteq\Delta(97)$ of order $4$, and their geometric points are fixed by the subgroup $H_3$ of order $3$. As the strange characters modulo $97$ have order $12$, they are trivial on $H_4$, so $(h-1)J_1(97)(\Q)$ is finite for $h\in H_4$ by \Cref{lem:finite}, and \Cref{prop:gonality} a) gives $\gon_\Q X_1(97)\geq49$.

Suppose that $Z=SS_{16}$, and let $h$ be a generator of $H_4$. If $16\leq d\leq 19$, then $c=d-16\leq 3$ and $2(d+c)\leq 44<49$, so \Cref{lem:stab} gives $hD=D$ and $4\mid c$, so $c=0$ and $d=16$. As $16<(97-1)/3$, $D$ is a pullback from $X_1(97)/H_4$ (\Cref{lem:orbits}). Let $W$ and $h'$ be as in \Cref{lem:W97}. The image $D_W$ of $D$ on $W$ is a closed point of degree $m\in\{2,4\}$, and its reduction is fixed by $h'$, as the geometric points of $SS_{16}$ are fixed by $H_3$, whose image in $\Delta(97)/H_8$ is generated by $h'$. By \Cref{lem:W97} and \Cref{lem:torsion} b), condition iii) of \Cref{lem:stabgen} holds for $X=W$, $s=h'$ and $\beta=A_3$, so \Cref{lem:stabgen} (with $C=0$) gives $h'D_W\sim D_W$, and as $\gon_\Q W\geq 7>m$, $h'D_W=D_W$. As $3\nmid m$, $h'$ fixes each geometric point of $D_W$, so the points of $D$ are fixed by non-trivial elements of $\Delta(97)$, which contradicts \Cref{lem:CM}.

If $20\leq d\leq 22$, we use \Cref{lem:relD} with
$$L=Z_0+R+\pi^*Y^{(r)},$$
where $\pi:X_1(97)\to Y=X_1(97)/H_3$ is the quotient map, $Z_0$ is the divisor of degree $16$ whose points are the lifts of the geometric points of $SS_{16}$ fixed by $H_3$, $C'=\overline R+\pi^*E'$ with $E'$ a sum of $r$ cusps of $Y$, and $R$ has no full orbit of $H_3$ in its support. Here we use $t=(h_3-1)\beta_nA_q$, where $h_3$ generates $H_3$ and $q$ is odd. As $(h_3-1)\beta_nJ_1(97)(\Q)$ is finite, $t$ kills $J_1(97)(\Q)$ by \Cref{lem:torsion} b). It also kills $\pi^*J_Y$, and $t[Z_0+R+\pi^*E_0-dc_0]=0$ for every point $E_0$ of $Y^{(r)}$, as $h_3-1$ kills $Z_0$ and $\pi^*E_0$ and $A_q$ kills rational cuspidal divisors. We \gitlink{degree20/sharp97.m}{compute} that the rank condition holds. So $D$ contains $R$, which is impossible if $R\neq0$, or $Z_0$, which contradicts \Cref{lem:CM}.
\end{proof}

\subsection{Reductions supported on the rational cusps at \texorpdfstring{$53$}{53}, \texorpdfstring{$101$}{101}, \texorpdfstring{$103$}{103} and \texorpdfstring{$131$}{131}}

\begin{proposition}\label{prop:cusps2}
Let $(d,p)$ be one of $(19,53)$, $(20,53)$, $(21,53)$, $(20,101)$, $(21,101)$ and $(22,101)$, or let $p=103$ and $17\leq d\leq 22$, or $p=131$ and $18\leq d\leq 22$. Let $D$ be the trace divisor of a point of order $p$ over a number field of degree $d$. Then $Z\neq 0$.
\end{proposition}

\begin{proof}
Suppose that $Z=0$, i.e.\ that $\overline D=C'$ is supported on the rational cusps.

Let $p=53$, $h=\dm4$, of order $13$, and $\pi:X_1(53)\to Y=X_1(53)/\dm4$. For $d=19$ the criterion of \Cref{prop:cusps} on $X_H$ with $H$ of order $2$ fails at only $7$ multiplicity vectors, and at all the $20444$ divisors on $X_1(53)$ lying above them we \gitlink{degree19/full_cusps53.m}{compute} that $f_{t,19}$ is a formal immersion, with $t=((\dm{10}-1)A_3,(\dm{10}-1)A_5)$, which kills $J_1(53)(\Q)$. For $d\in\{20,21\}$ we work on $X_1(53)$ with $t=((h-1)A_3,(h-1)A_5)$, and we \gitlink{degree20/full_cusps53.m}{compute} that $f_{t,d}$ is a formal immersion at every sum of rational cusps of degree $d$ that does not contain a full orbit of $\dm4$, while at the others the rank condition of \Cref{lem:relD} holds with $L=R+\pi^*Y$, where $R\neq0$ is the rest of the divisor. Here $f_{t,d}$ vanishes on $L$, as $h-1$ kills $\pi^*J_Y$ and the pullbacks $\pi^*E_0$, and $A_q$ kills rational cuspidal divisors. In both cases \Cref{lem:fi} b) or \Cref{lem:relD} shows that $D$ contains a cusp, which is a contradiction.

For $p=101$ we use $t=(h-1)\beta_nA_q$ with $h$ a generator of $\Delta(101)$, $\beta_n$ as above, $n\in\{2,3,5,7\}$, and $q$ odd. As $(h-1)\beta_nJ_1(101)(\Q)$ is finite, $t$ kills $J_1(101)(\Q)$ by \Cref{lem:torsion} b), and we \gitlink{degree20/sharp101.m}{compute} that $f_{t,d}$ is a formal immersion at all sums of rational cusps of degree $d$ that do not contain all the $50$ rational cusps, which is all of them. So $D$ contains a cusp by \Cref{lem:fi} b), which is a contradiction.

Let $p=103$ (resp.\ $p=131$), let $H\subseteq\Delta(p)$ be the subgroup of order $3$ (resp.\ $5$), $Y=X_H$, and $\pi:X_1(p)\to Y$ the quotient map. Then $\rho:Y\to X_0(p)$ has degree $17$ (resp.\ $13$), and the rational cusps of $Y$ form one fibre $F$ of $\rho$. On $Y$, the criterion of \Cref{prop:cusps} holds at all sums of rational cusps of degree $d$ that do not contain $F$, so by \Cref{lem:fi} b) we can assume that $\overline{\pi_*D}$ contains $F$, and we write $\overline{\pi_*D}=F+\overline R$ for a rational cuspidal divisor $R$ on $Y$. We use \Cref{lem:relD} on $Y^{(d)}$ with $L=R+\rho^*X_0(p)$ and $t=(\dm a-1)A_3$, where $\dm a$ generates $\Delta(p)/H$. As $p$ is not strange, $(\dm a-1)J_Y(\Q)$ is finite by \Cref{lem:finite}, so $t$ kills $J_Y(\Q)$ by \Cref{lem:torsion} b). It also kills $\rho^*J_0(p)$, on which $\dm a$ acts trivially, and $t[R+F-dc_0]=0$, as $A_3$ kills rational cuspidal divisors, so $f_{t,d}$ vanishes on $L$. We \gitlink{degree19/verify_relative_cusps_intrinsic.py}{compute} that the rank condition holds. Hence $\pi_*D$ lies on $L$. If $R\neq0$, then $\pi_*D$ contains a cusp, which is impossible. If $R=0$, then $p=103$, $d=17$, and $\pi_*D$ is the pullback of a non-cuspidal rational point on $X_0(103)$, which does not exist.
\end{proof}

\subsection{The primes \texorpdfstring{$109$}{109}, \texorpdfstring{$113$}{113}, \texorpdfstring{$127$}{127} and \texorpdfstring{$137$}{137}}

\begin{proposition}\label{prop:109}
Let $12\leq d\leq 22$, and let $D$ be the trace divisor of a point of order $109$ over a number field of degree $d$. Then $Z\notin\{SS_9,NC_{18},2SS_9,SS_9+SS_{12}\}$.
\end{proposition}

\begin{proof}
All of these divisors are fixed by $h=\dm{23}$, of order $18$, $(h-1)J_1(109)(\Q)$ is finite as $109$ is not strange, and $\gon_\Q X_1(109)>58$. We only use $2y=0$ in \Cref{lem:stab}. If $Z=SS_9$ and $d\leq 19$, or if $Z$ is one of the others, then $2(d+c)\leq58$, so $hD=D$ and $18\mid c$. As $c\leq13$, this gives $c=0$, so $Z=NC_{18}$ or $2SS_9$ with $d=18$, or $Z=SS_9+SS_{12}$ with $d=21$. As $d<(109-1)/3$, the orbits of $h$ on the support of $D$ have length $18$ by \Cref{lem:orbits}. So the last case is impossible, and in the first two $D$ is a single orbit of $\dm{23}$, whose image on $X_0(109)$ is a non-cuspidal rational point, which does not exist. If $Z=SS_9$ and $d\geq 20$, we use \Cref{lem:relD} with $L=C+\pi^*Y$ or $L=R+\pi^*Y^{(2)}$, where $\pi:X_1(109)\to Y=X_1(109)/\dm{h^2}$, $\dm{h^2}$, of order $9$, acts freely on the geometric points of $SS_9$, and $R$ is a rational cuspidal divisor, and $t=(h^2-1)A_q$ (\gitlink{degree20/relative109.m}{code}). Here $t$ kills $J_1(109)(\Q)$ and $\pi^*J_Y$, and $f_{t,d}$ vanishes on $L$ as $h^2-1$ kills the pullbacks from $Y$ and $A_q$ kills rational cuspidal divisors. As the rank condition holds, $D$ contains a cusp, which is a contradiction.
\end{proof}

\begin{proposition}\label{prop:113}
Let $13\leq d\leq 22$, and let $D$ be the trace divisor of a point of order $113$ over a number field of degree $d$. Then $Z$ is not one of the divisors listed for $113$ in \Cref{table:exceptions}.
\end{proposition}

\begin{proof}
Here $O_8$ denotes one of the $7$ ordinary closed points of degree $8$ which are fixed by $h_8=\dm{40}$, of order $8$ (the other $28$ ordinary closed points of degree $8$, alone or together with non-ordinary points, are excluded by \Cref{prop:sieve}), and $\gon_\Q X_1(113)>63$. The element $h_2=\dm2$, of order $14$, fixes the closed points $SS_7$ and $NC_{14}$, and the involution $\iota=\dm{98}$ fixes the points $SS_7$ and $O_8$. As $113$ is not strange, $(g-1)J_1(113)(\Q)$ is finite for every $g\in\Delta(113)$. We only use $2y=0$ in \Cref{lem:stab}, and as $d<(113-1)/3$, \Cref{lem:orbits} b) applies to every subgroup of $\Delta(113)$ that fixes $D$.

If $Z\in\{2SS_7,3SS_7,NC_{14},SS_7+NC_{14}\}$, then $2(d+c)\leq 60$, so $h_2D=D$ and $14\mid c$, so $c=0$. Then $d\in\{14,21\}$, and as $14\nmid21$, $d=14$ and $D$ is a single orbit of $h_2$, whose image on $X_0(113)$ is a non-cuspidal rational point, which is impossible.

If $Z=O_8$ and $d\leq19$, or $Z=2O_8$, then $2(d+c)\leq 62$, so $h_8D=D$ and $8\mid c$. Hence $d=16$ and $c=0$ or $c=8$, and $D$ is a pullback of a divisor of degree $2$ from $X_1(113)/\dm{h_8}$. Its image on $X_0(113)$ is a non-cuspidal point of degree at most $2$, which has complex multiplication, as all the quadratic points on $X_0(113)$ are CM points \cite[Table 14]{AKMNOV} and there are no non-cuspidal rational points \cite[Theorem 1]{mazur78}. But then by \Cref{lem:silverberg} a point of order $113$ on $E$ is defined over a field of degree at least $112/6>16$, which is a contradiction. If $Z=O_8$ and $d\geq 20$, we use \Cref{lem:relD} with $L=R+\pi^*Y^{(r)}$, where $\pi$ is the quotient by $\dm{h_8}$, which is \'etale near $Z$, $R$ is a rational cuspidal divisor, and $t=((h_8-1)A_2,(h_8-1)A_3)$ (\gitlink{degree20/relative113_ordinary.m}{code}). Here $t$ kills $J_1(113)(\Q)$ by \Cref{lem:torsion} b), and $f_{t,d}$ vanishes on $L$ as $h_8-1$ kills the pullbacks and $A_q$ kills rational cuspidal divisors.

If $Z=O_8+SS_7$, then $2(d+c)\leq 58$, so $\iota D=D$ and $2\mid c=d-15$. So $d$ is odd, and the orbits of $\iota$ on the support of $D$ have length $2$ by \Cref{lem:orbits}, which is a contradiction. Finally, if $Z=O_8+NC_{14}$ and $d=22$, we use \Cref{lem:relD} with $t=((h_8-1)B_qA_q)_{q\in\{2,3\}}$ and $L=Z_1+\pi^*Y$, where $Z_1$ is the lift of $NC_{14}$ to a cuspidal divisor (\gitlink{degree22/relative113_nc.m}{code}). Here $t$ kills $J_1(113)(\Q)$ by \Cref{lem:torsion} b), and as $h_8-1$ kills the pullbacks and $B_qA_q$ kills all cuspidal divisors by \Cref{lem:torsion} c), $f_{t,22}$ vanishes on $L$. Hence $D$ contains a cusp, which is a contradiction.
\end{proof}

\begin{proposition}\label{prop:127}
Let $17\leq d\leq 22$, and let $D$ be the trace divisor of a point of order $127$ over a number field of degree $d$. Then $Z$ is not one of the divisors listed for $127$ in \Cref{table:exceptions}.
\end{proposition}

\begin{proof}
The element $h_7=\dm2$, of order $7$, fixes the $9$ non-rational cusps and the $48$ supersingular closed points of degree $14$, and the ordinary closed points $O_9$ are fixed by the subgroup $H_9$ of order $9$. As $127$ is not strange, $(g-1)J_1(127)(\Q)$ is finite for every $g\in\Delta(127)$. We have $\gon_\Q X_1(127)>79$, and $A_2$ kills $J_1(127)(\Q)_\tor$ by \Cref{lem:torsion} b). We only use $2y=0$ in \Cref{lem:stab}.

Suppose that $Z$ is a sum of points $NC_7$ and $SS_{14}$. Then $2(d+c)\leq 74$, so $h_7D=D$ and $7\mid c$. This is impossible unless $d=21$, in which case $D$ is a pullback from $Y=X_1(127)/\dm2$ (\Cref{lem:orbits}), of a divisor $D'$ of degree $3$. The curve $Y$ has genus $88$, and its points over $\F_2$ which are images of cusps are the $9$ rational cusps and the $9$ images of the non-rational cusps. Let $h'$ be a generator of $\Delta(127)/\dm2$. By \Cref{lem:finite} and \Cref{lem:torsion} (note that $\dm 2$ has odd order), the maps $(h'-1)B_qA_q$ with $q\in\{2,3\}$ kill $J_Y(\Q)$ and all the cuspidal divisors on $Y$.

If $Z$ consists of non-rational cusps, then $\overline{D'}$ is a sum of three of these $18$ points, and we \gitlink{degree21/quotient127_cusps.m}{compute} that $f_{t,3}$ on $Y^{(3)}$, with $t=((h'-1)B_qA_q)_{q\in\{2,3\}}$, is a formal immersion at all the $1140$ such sums. So $f_{t,3}$ maps $D'$ and the cuspidal divisor lifting $\overline{D'}$ (which is defined over an unramified extension of $\Q_2$) to $0$, and they are equal by \Cref{lem:fi} a), which is a contradiction. If $Z=SS_{14}$, so that $c=7$, then for $5$ of the $6$ orbits of points $SS_{14}$, $\alpha A_qZ$ is not principal for some $q\in\{3,5,7\}$, which contradicts \Cref{lem:hecke}. For the remaining orbit (of length $3$), $\overline{D'}=S+\overline{c_1}$, where $c_1$ is a rational cusp of $Y$ and $S$ is a supersingular point of degree $2$ fixed by the element $h_3'$ of order $3$ of $\Delta(127)/\dm2$. We use \Cref{lem:relD} on $Y^{(3)}$ with $L=S_0+c_1$, where $S_0$ is the unique lift of $S$ fixed by $h_3'$, and with $t$ combining the maps $(h_3'-1)A_q$ for several primes $q\in\{2,3,5\}$ (\gitlink{degree21/quotient127_ss.m}{code}). Here $t$ kills $J_Y(\Q)$ by \Cref{lem:finite} and \Cref{lem:torsion} b), and $t[S_0+c_1-3c_0]=0$, as $h_3'$ fixes $S_0$ and $A_q$ kills rational cuspidal divisors. As the rank condition holds, $D'=S_0+c_1$ contains a cusp, which is a contradiction. The case $Z=SS_{14}+NC_7$ is excluded by \Cref{prop:sieve}.

If $Z=O_9$ or $Z=2O_9$, then $2(d+c)\leq 72$, so $D$ is $H_9$-invariant and $9\mid c$. Hence $d=18$ and $D$ is a pullback of a divisor of degree $2$ from $X_1(127)/H_9$. Its image on $X_0(127)$ is a non-cuspidal point of degree at most $2$, which has complex multiplication, as all the quadratic points on $X_0(127)$ are CM points \cite[Table 16]{AKMNOV} and there are no non-cuspidal rational points \cite[Theorem 1]{mazur78}. By \Cref{lem:silverberg} a point of order $127$ on $E$ is then defined over a field of degree at least $126/6=21>18$, which is a contradiction.

Finally, if $Z=O_9+NC_7$, we use \Cref{lem:relD} with $L=C+\pi_9^*(X_1(127)/H_9)+\pi_7^*(X_1(127)/\dm2)$, where $\pi_9$ and $\pi_7$ are the quotient maps, which is smooth of dimension $2$ as $H_9$ and $\dm2$ act freely on the supports of $O_9$ and $NC_7$, and $t=(h_9-1)(h_7-1)A_q$, $q\in\{2,3\}$, where $h_9$ generates $H_9$ (\gitlink{degree17/relative127.m}{code}). Here $t$ kills $J_1(127)(\Q)$ and the pullbacks of the Jacobians of both quotients, and $f_{t,d}$ vanishes on $L$ as $A_q$ kills rational cuspidal divisors. So $D$ contains $C\neq0$, which is a contradiction.
\end{proof}

\begin{proposition}\label{prop:137}
Let $19\leq d\leq 22$, and let $D$ be the trace divisor of a point of order $137$ over a number field of degree $d$. Then $Z\notin\{SS_{17},O_{17}\}$.
\end{proposition}

\begin{proof}
The point $SS_{17}$ is fixed by $\dm2$, of order $34$, and $O_{17}$ is \gitlink{degree19/stabilizer137_ordinary.m}{fixed} by $\dm{81}$, of order $17$. As $137$ is not strange, $(g-1)J_1(137)(\Q)$ is finite for every $g\in\Delta(137)$. In both cases $c=d-17\in\{2,3,4,5\}$ and $2(d+c)\leq 54<\gon_\Q X_1(137)$, so \Cref{lem:stab} gives $34\mid c$ or $17\mid c$, which is a contradiction.
\end{proof}

\subsection{Proof of \texorpdfstring{\Cref{thm:main}}{Theorem 1.1}}
By \Cref{prop:lower} all the primes listed in \Cref{thm:main} are in $S(d)$. Conversely, let $9\leq d\leq 22$ and let $p\in S(d)$ be a prime which is not listed. By \Cref{prop:large,prop:medium,prop:127medium,prop:193}, $p$ is one of the primes of \Cref{table:local}. Let $D$ be the trace divisor of a point of order $p$ over a number field of degree $d$. By \Cref{prop:sieve}, $Z$ is one of the divisors listed in \Cref{table:exceptions}, and these are excluded by \Cref{prop:37,prop:41,prop:43,prop:53,prop:61,prop:73,prop:89,prop:97,prop:cusps2,prop:109,prop:113,prop:127,prop:137}.