Proof. This is a consequence of [ Reference Bourdon, Clark and Stankewicz8 , theorem 5·5].

Proof. See, for example, [ Reference Bourdon and Najman11 , lemma 2·5] and [ Reference Deligne and Rapoport19 , p. 274, proposition VI·3·2].

Remark 2·5. If $E/{\mathbb{Q}}(j(E))$ corresponds to an equation of the form $y^2=x^3+Ax+B$ and $P=(x,y) \in E$ , then we may take

\begin{equation*} \mathfrak{h}(P) = \begin{cases} x & AB \neq 0 \\ x^2 & B = 0 \\ x^3 & A = 0 \end{cases}. \end{equation*}

Note $AB \neq 0$ if E is non-CM. Thus by Lemma 2·4 we can compute the degree of a closed point on $X_1(N)$ associated to a non-CM elliptic curve by factoring division polynomials. See [ Reference Shimura45 , p. 107] for details, including a formulation of the Weber function which is more clearly model-independent.

Proof. The moduli interpretation ensures the map is defined over ${\mathbb{Q}}$ , and the degree calculation follows from [ Reference Diamond and Shurman23 , p·66].

Proof. Since $\mathcal{E}$ is rational, there exists $E_0\in \mathcal{E}$ with $j(E_0)\in {\mathbb{Q}}$ . By definition there exists an isogeny $\varphi\,:\, E \rightarrow E_0$ defined over $\overline{{\mathbb{Q}}}$ which we may assume is cyclic of degree N; see Lemma A·1 in [ Reference Cremona and Najman18 ]. Let C denote its kernel. Note that $F={\mathbb{Q}}(j(E),\mathfrak{h}(P))$ by Lemma 2·4 and there exists a Weierstrass equation of $E/F$ with $P \in E(F)$ . The proof strategy of [ Reference Clark14 , proposition 3·2] shows C is F-rational, as we will now explain. Suppose $\sigma(C) \neq C$ for some $\sigma \in {\mathrm{Gal}}_F$ , and consider the induced isogeny $E^{\sigma} \rightarrow (E/C)^{\sigma}$ . Since $j(E/C)=j(E_0) \in {\mathbb{Q}}$ , we see that $j((E/C)^{\sigma})=j(E_0)$ . Thus composition with an isomorphism to $E_0$ yields a cyclic N-isogeny $\psi\,:\, E \rightarrow E_0$ with kernel $\sigma(C)$ . But having two cyclic N-isogenies from E to $E_0$ with distinct kernels can happen only if E has complex multiplication (see the last paragraph of the proof of [ Reference Clark14 , proposition 3·2] for details). We have reached a contradiction.

Remark 3·3. Let $k \in{\mathbb{Z}}^{\geq 2}$ . Since $\deg(X_1(\ell^k) \rightarrow X_1(\ell))=\ell^{2k-2}$ for $\ell$ odd and $\deg(X_1(2^k) \rightarrow X_1(2))=2^{2k-3}$ , these lower bounds are best-possible whenever $d=0$ . In this case $E_0$ itself is a minimal torsion curve for $X_1(\ell^k)$ . This holds in certain cases; see, for example, Lemma 5·3. However, there are other $\overline{{\mathbb{Q}}}$ -isogeny classes for which these bounds can be further refined. See Proposition 6·1.

Proof. Let $E \in \mathcal{E}$ , and fix $P \in E$ of order $\ell^k$ . Define $F \,:\!=\, {\mathbb{Q}}(j(E),\mathfrak{h}(P))$ . By Lemma 3·1, there is a Weierstrass model of $E/F$ where $P \in E(F)$ and such that there exists an F-rational cyclic isogeny $\varphi\,:\, E \rightarrow E'$ with $j(E')\in{\mathbb{Q}}$ . By [ Reference Bourdon and Najman11 , lemma 4·6], we have $[F\,:\,{\mathbb{Q}}]$ is divisible by

\begin{equation*}\begin{cases}\ell^{\max(0,2k-2-d)} \text{ if $\ell$ is odd},\\\ell^{\max(0,2k-3-d)} \text{ if $\ell=2$},\end{cases}\end{equation*}

where $d={\mathrm{ord}}_{\ell}([{\mathrm{GL}}_2({\mathbb{Z}}_{\ell})\,:\, {\mathrm{im}} \rho_{E_0, \ell^{\infty}}])$ for any elliptic curve $E_0/{\mathbb{Q}}$ with $j(E_0)=j(E')$ .

By [ Reference Bourdon and Najman11 , corollary 4·3 (2)], the curve E’ has a point of order $\ell$ over an extension $F'/F$ of degree dividing $\ell$ . In particular, there exists a closed point $x=[E',P'] \in X_1(\ell)$ such that ${\mathbb{Q}}(x) \subseteq F'$ . Hence

\begin{equation*}\deg(x) \mid \ell \cdot [F\,:\,{\mathbb{Q}}].\end{equation*}

Since $j(E')=j(E_0)$ , there exists a point $P_0\in E_0$ such that the closed point $x=[E_0,P_0]$ . If $\deg(x)$ is prime to $\ell$ , then $\deg(x) \mid [F\,:\,{\mathbb{Q}}]$ , and the conclusion follows since F is the residue field of the closed point associated to [E, P] by Lemma 2·4. So suppose $\deg(x)=\ell \cdot n_0$ . Note that $\ell \nmid n_0$ , since $\deg(X_1(\ell) \rightarrow X_1(1))\lt\ell^2$ , so $n_0 \mid [F\,:\,{\mathbb{Q}}]$ . If there exists $x_1\in X_1(\ell)$ associated to $E_0$ with $\deg(x_1) \mid n_0$ , then the conclusion follows with $x_1$ in place of x. So suppose not. By checking the possible images of the mod $\ell$ Galois representation associated to $E_0$ as in [ Reference González-Jiménez and Najman30 , theorem 5·6, Tables 1 and 2], we see that we must be in one of the following cases (see Section 2·1 for an explanation of subgroup labels used):

1. (i) $\ell=5$ , ${\mathrm{im}} \rho_{E_0,5}=5B.1.2, \,5B.1.3,\text{ or } 5B.4.2$ ;

2. (ii) $\ell=7$ , ${\mathrm{im}} \rho_{E_0,7}=7B.1.3, \, 7B.1.4,\text{ or }7B.6.3$ ;

3. (iii) $\ell=13$ , ${\mathrm{im}} \rho_{E_0,13}=13B.3.2, \, 13B.3.7,\, 13B.5.2 \text{ or }13B.4.2$ ;

4. (iv) $\ell=17$ , ${\mathrm{im}} \rho_{E_0,17}=17B.4.6$ ;

5. (v) $\ell=37$ , ${\mathrm{im}} \rho_{E_0,37}=37B.8.2$ .

In each case, there exists a ${\mathbb{Q}}$ -rational cyclic subgroup C of $E_0$ such that $E_0/C$ has mod $\ell$ image outside this list; see [ Reference Sutherland48 , theorem 3·32, Tables 3 and 4]. That is, in each case there exists a point on $X_1(\ell)$ associated to $E_0/C$ of degree dividing $n_0$ , and the result holds with $E_0/C$ in place of $E_0$ .

Remark 3·4. From the proof, we see that the statement of Proposition 3·2 holds for any $E_0/{\mathbb{Q}} \in \mathcal{E}$ which satisfies the following constraints:

1. (i) $\ell=5$ , ${\mathrm{im}} \rho_{E_0,5} \neq 5B.1.2, \,5B.1.3\text{ or } 5B.4.2$ ;

2. (ii) $\ell=7$ , ${\mathrm{im}} \rho_{E_0,7} \neq 7B.1.3, \, 7B.1.4\text{ or }7B.6.3$ ;

3. (iii) $\ell=13$ , ${\mathrm{im}} \rho_{E_0,13} \neq 13B.3.2, \, 13B.3.7,\, 13B.5.2 \text{ or }13B.4.2$ ;

4. (iv) $\ell=17$ , ${\mathrm{im}} \rho_{E_0,17} \neq 17B.4.6$ ;

5. (v) $\ell=37$ , ${\mathrm{im}} \rho_{E_0,37} \neq 37B.8.2$ ;

Proof. Let $\{P, Q\}$ be a basis for $E_1[\ell^{r+k}]$ , where $\ker(\varphi)=\langle \ell^k P \rangle$ . Then with respect to this basis, for any $\sigma \in {\mathrm{Gal}}_{F}$ , there exist $a, b, c ,d \in {\mathbb{Z}}$ such that

\begin{equation*} \rho_{E_1,\ell^{r+k}}(\sigma)=\begin{pmatrix}a & b \\\ell^{r} c & d\end{pmatrix}.\end{equation*}

One can check that $\{\varphi(P), \ell^{r} \varphi(Q)\}$ gives a basis for $E_2[\ell^k]$ . Moreover, for $\sigma \in {\mathrm{Gal}}_{F}$ , we have

\begin{align*}\sigma(\varphi(P))&=\varphi(\sigma(P))=\varphi(aP+\ell^{r} c Q)=a\varphi(P)+c \ell^{r} \varphi(Q),\\\sigma(\ell^{r} \varphi( Q))&=\ell^{r} \varphi(\sigma(Q))=\ell^{r}\varphi(bP+d Q)=\ell^{r}b\varphi(P)+d\ell^{r}\varphi( Q).\end{align*}

Thus

\begin{equation*} \rho_{E_2,\ell^{k}}(\sigma)=\begin{pmatrix}a &\ell^{r} b \\c & d\end{pmatrix},\end{equation*}

and ${\mathrm{im}} \rho_{E_2,\ell^k}$ can be deduced from ${\mathrm{im}} \rho_{E_1,\ell^{r+k}}$ .

Proof. Suppose there exists an isogeny $\varphi\,:\,E \rightarrow E'$ defined over $\overline{{\mathbb{Q}}}$ which is cyclic of order $\ell^r$ . Thus there exists a point $P\in E$ of order $\ell^r$ such that $E' \cong E/\langle P \rangle$ . Let $F\,:\!=\, {\mathbb{Q}}(\langle P \rangle)$ , and let $\psi\,:\,E \rightarrow E/\langle P \rangle$ be the induced F-rational isogeny. By Proposition 3·5 the $\ell$ -adic Galois representation of $E/\langle P \rangle$ , and thus the degrees of all closed points on $X_1(\ell^k)$ associated to $E/\langle P \rangle$ , is determined by ${\mathrm{im}} \rho_{E/F, \ell^{\infty}}$ . Since ${\mathrm{im}} \rho_{E/F, \ell^{\infty}}$ can be obtained from ${\mathrm{im}} \rho_{E/{\mathbb{Q}}, \ell^{\infty}}$ and the definition of F, the result follows.

Proof. Let d be the minimal degree of a point on $X_1(N)$ associated to $\mathcal{E}$ , and let $j_{min}$ be the j-invariant of a minimal torsion curve for $\mathcal{E}$ . Then $[{\mathbb{Q}}(j_{min})\,:\,{\mathbb{Q}}] \leq d.$ If $\mathcal{E}$ is CM, then the conclusion follows as there are only finitely many CM j-invariants in an extension of bounded degree (since there are only finitely many imaginary quadratic fields—and hence imaginary quadratic orders—of a given class number [ Reference Heilbronn33 ], this is a consequence of [ Reference Cox17 , theorem 11·1 proposition 13·2]). So suppose $\mathcal{E}$ is non-CM.

Let $E_{min}\in \mathcal{E}$ be an elliptic curve with $j(E_{min})=j_{min}$ . Since $\mathcal{E}$ is rational, there exists an elliptic curve $E_0/{\mathbb{Q}} \in \mathcal{E}$ with an isogeny $\varphi\,:\, E_0 \rightarrow E_{min}$ defined over $\overline{{\mathbb{Q}}}$ . We may assume $\varphi$ is cyclic of degree n by [ Reference Cremona and Najman18 , lemma A·1]. By [ Reference Clark14 , proposition 3·2] or [ Reference Cremona and Najman18 , corollary A·5], the field of moduli of this isogeny is

\begin{equation*}{\mathbb{Q}}(j_{min},j(E_0))={\mathbb{Q}}(j_{min}).\end{equation*}

Thus $\varphi$ can be defined over ${\mathbb{Q}}(j_{min})$ by replacing $E_0$ and $E_{min}$ with appropriate twists, and in particular a twist of $E_0$ defined over this field has a rational cyclic n-isogeny. Since isogenies are twist invariant (i.e., if $E_1, E_2$ are two elliptic curves over F with the same j-invariant, then $E_1$ has an F-rational cyclic n-isogeny if and only if $E_2$ does), it follows that $E_0$ attains a rational cyclic n-isogeny over ${\mathbb{Q}}(j_{min})$ , a number field of degree at most d.

By Serre’s open image theorem [ Reference Serre43 ], the image of the adelic Galois representation associated to $E_0/{\mathbb{Q}}$ has finite index in ${\mathrm{GL}}_2(\widehat{{\mathbb{Z}}})$ . For d fixed, it follows that there is a bound on the order of a cyclic subgroup of $E_0$ which can become F-rational over any number field F of degree at most d. Thus there are only finitely many choices for C such that $E_{min} \cong E_0/C$ over $\overline{{\mathbb{Q}}}$ .

Remark 4·2. The minimal torsion curve within a fixed geometric isogeny class may or may not be unique (up to isomorphism over $\overline{{\mathbb{Q}}}$ ). For example, let $E_0/{\mathbb{Q}}$ be the elliptic curve with LMFDB label 38.b2 and let $\mathcal{E}$ denote its geometric isogeny class. Since there exists $P_0 \in E_0({\mathbb{Q}})$ of order 5, the least degree of a point on $X_1(5)$ associated to $\mathcal{E}$ is 1, and so any minimal torsion curve for $X_1(5)$ must have j-invariant in ${\mathbb{Q}}$ . By Lemma 3·1, the only other elliptic curve $E' \in \mathcal{E}$ with j-invariant in ${\mathbb{Q}}$ has $j(E')=-{37966934881}/{4952198}$ . However, this curve does not give points on $X_1(5)$ of minimal degree. Thus $E_0$ is a unique minimal torsion curve for $X_1(5)$ , up to $\overline{{\mathbb{Q}}}$ -isomorphism.

On the other hand, let $E_0/{\mathbb{Q}}$ be the elliptic curve with LMFDB label 50.b1 and let $\mathcal{E}$ denote its geometric isogeny class. Then a similar argument shows that any elliptic curve ${\mathbb{Q}}$ -isogenous to $E_0$ is a minimal torsion curve for $X_1(3)$ , giving 4 distinct minimal torsion curves for this class up to $\overline{{\mathbb{Q}}}$ -isomorphism. Representatives for these curves can be found in the LMFDB isogeny class 50.b.

Proof. By Proposition 3·2, there exists $E_0/{\mathbb{Q}}\in \mathcal{E}$ and $x=[E_0,P_0] \in X_1(\ell)$ such that the degree of any point on $X_1(\ell^n)$ associated to $\mathcal{E}$ is divisible by

\begin{equation*}d_{\text{min}}(\ell^n)\,:\!=\, \begin{cases}\deg(x)\cdot \ell^{\max(0,2n-2-d)} \text{ if $\ell$ is odd},\\\deg(x) \cdot \ell^{\max(0,2n-3-d)} \text{ if $\ell=2$},\end{cases}\end{equation*}

where $d \,:\!=\, {\mathrm{ord}}_{\ell}([{\mathrm{GL}}_2({\mathbb{Z}}_{\ell})\,:\, {\mathrm{im}} \rho_{E_0, \ell^{\infty}}])$ .

We note ${\mathrm{im}} \rho_{E_0,\ell^{\infty}}$ has level $\ell^{k_0}$ for some $k_0 \in {\mathbb{Z}}^{\geq 0}$ by Serre’s open image theorem [ Reference Serre43 ]. Replacing $k_0$ with a larger integer if necessary, we may assume $2k_0-2-d \geq0$ if $\ell$ is odd and $2k_0-3-d \geq 0$ if $\ell=2$ . Now, let k be an integer with $k \geq k_0$ . Then by Proposition 3·2, there exists $\alpha_0 \in {\mathbb{Z}}^+$ such that the least degree of a closed point on $X_1(\ell^k)$ associated to $E_0 \in \mathcal{E}$ is

\begin{align*}d_{\text{min},E_0}(\ell^k)&= d_{\text{min}}(\ell^k) \cdot \alpha_0\\&= d_{\text{min}}(\ell^{k_0})\cdot \ell^{2(k-k_0)}\cdot \alpha_0.\end{align*}

Since $\deg(X_1(\ell^k) \rightarrow X_1(\ell^{k_0}))=\ell^{2(k-k_0)}$ , the assumptions concerning the level of ${\mathrm{im}} \rho_{E_0,\ell^{\infty}}$ imply

\begin{equation*}d_{\text{min},E_0}(\ell^{k_0})= d_{\text{min}}(\ell^{k_0}) \cdot \alpha_0.\end{equation*}

In particular, $\alpha_0$ does not depend on k.

Suppose first that $E_0$ is a minimal torsion curve for $X_1(\ell^k)$ for sufficiently large k. Then by Proposition 4·1 there exists a finite collection of elliptic curves $E_0, E_1, \ldots, E_s$ such that for any positive integer n a point of least degree on $X_1(\ell^n)$ coming from $\mathcal{E}$ can be associated to $E_i$ for some $0 \leq i \leq s$ . We may assume the isogeny $\varphi\,:\,E_i \rightarrow E_0$ is cyclic of degree $d_i$ by [ Reference Cremona and Najman18 , lemma A·1]. The result follows in this case with $C=\max\{d_0, d_1, \ldots, d_s\}$ .

On the other hand, suppose $E_0$ is not a minimal torsion curve for $X_1(\ell^k)$ for all $k\geq k_0$ . Then there exists $E_1 \in \mathcal{E}$ which is a minimal torsion curve for some $X_1(\ell^{k_1})$ where $k_1 \geq k_0$ . Choose a Weierstrass equation for $E_1$ defined over $F\,:\!=\, {\mathbb{Q}}(j(E_1))$ . Replacing $k_1$ with a larger integer if necessary, we may assume that ${\mathrm{im}} \rho_{E_1,\ell^{\infty}}=\pi^{-1}({\mathrm{im}} \rho_{E_1, \ell^{k_1}})$ . By assumption, there exists a positive integer $\alpha_1$ with $\alpha_1\lt\alpha_0$ such that

\begin{equation*}d_{\text{min},E_1}(\ell^{k})=d_{\text{min}}(\ell^k)\cdot \alpha_1\end{equation*}

for all $k \geq k_1$ . If $E_1$ is a minimal torsion curve for $X_1(\ell^k)$ for sufficiently large k, then we are done as before. Continuing in this way produces a decreasing sequence of positive integers $\alpha_{0}\gt\alpha_1\gt\alpha_2 \ldots$ , so the process must stop after a finite number of steps.

Remark 4·4. It would be interesting to make the constant $C=C(\mathcal{E},\ell)$ explicit, perhaps in terms of invariants one could compute from the elliptic curve $E_0/{\mathbb{Q}}$ .

Proof. If $\ell \geq 11$ or $\ell=5$ , this follows from [ Reference Greenberg32 , theorems 1&2]. Note if $\ell \geq 11$ or if $\ell=5$ , the assumption in [ Reference Greenberg32 , theorems 1] holds, as explained in [ Reference Greenberg32 , remark 4·2·1, p. 1186-1187]. For $\ell=7$ , this follows from [ Reference Greenberg, Rubin, Silverberg and Stoll31 , theorem 5·5].

Proof. Suppose first that $\ell\gt5$ or that $\ell=5$ and $\mathcal{E}$ does not contain $E'/{\mathbb{Q}}$ with a rational cyclic 25-isogeny. By Proposition 3·2, there exists $E_0/{\mathbb{Q}}\in \mathcal{E}$ and $x=[E_0,P_0]\in X_1(\ell)$ such that

\begin{equation*}\deg(x)\cdot \ell^{\max(0,2k-2-d)}\end{equation*}

divides the degree of any point on $X_1(\ell^k)$ associated to $\mathcal{E}$ , where $d={\mathrm{ord}}_{\ell}([{\mathrm{GL}}_2({\mathbb{Z}}_{\ell})\,:$ ${\mathrm{im}} \rho_{E_0, \ell^{\infty}}])$ . By Theorem 5·2, we have $d=0$ . Since $\deg(X_1(\ell^k) \rightarrow X_1(\ell))=\ell^{2k-2}$ , lifts of x on $X_1(\ell^k)$ show this divisibility condition is best-possible with $j_{min}=j(E_0)$ .

Now, suppose $\ell=5$ and $\mathcal{E}$ contains $E'/{\mathbb{Q}}$ with a rational cyclic 25-isogeny. Then by Theorem 5·2, we have $ {\mathrm{ord}}_{5}([{\mathrm{GL}}_2({\mathbb{Z}}_{5})\,:\, {\mathrm{im}} \rho_{E', 5^{\infty}}])=1$ . By replacing E’ with a curve ${\mathbb{Q}}$ -isogenous, we may assume E’ has two independent 5-isogenies. The index of the image of the 5-adic Galois representation is unchanged by [ Reference Greenberg32 , proposition 2·1·1]. Then ${\mathrm{im}} \rho_{E',5}$ is one of the following, and we consider each case separately:

1. (1) 5Cs.1.1, 5Cs.1.3, or 5Cs.4.1: replacing E’ with a quadratic twist if necessary, we may assume ${\mathrm{im}} \rho_{E',5}=5Cs.1.1$ . By Proposition 3·2 and Remark 3·4, we see $5^{\max(0,2k-3)}$ divides the degree of any point on $X_1(5^k)$ associated to $\mathcal{E}$ . The curve E’ has a subgroup $C_1$ generated by a rational point P of order 5 and an independent rational cyclic subgroup $C_2$ of order 5. Then $E_2\,:\!=\, E'/C_2$ has a rational cyclic 25-isogeny, and the image of P is a point of order 5 in $E_2({\mathbb{Q}})$ lying in its kernel. Thus the image of the isogeny character $\chi\,:\, {\mathrm{Gal}}_{{\mathbb{Q}}} \rightarrow ({\mathbb{Z}}/25{\mathbb{Z}})^{\times}$ lands in the subgroup $\{a \,:\, a \equiv 1 \pmod{5}\}$ and so has order dividing 5. It follows that $E_2$ attains a rational point of order 25 in an extension of degree dividing 5. By Proposition 2·6, there are lifts of this point on $X_1(5^k)$ of degree at most $5^{2k-3}$ for all $k \geq 2$ . Thus the condition is best-possible for all k with $j_{min}=j(E_2)$ .

2. (2) 5Cs: by Proposition 3·2 and Remark 3·4, we see $2\cdot 5^{2k-3}$ divides the degree of any point on $X_1(5^k)$ associated to $\mathcal{E}$ . Note E’ has two independent 5-isogenies with kernels $C_1$ and $C_2$ . Then $E_2\,:\!=\, E'/C_2$ has a rational cyclic 25-isogeny. By Proposition 2·3, the curve $E_2$ gives a closed point on $X_1(5)$ in degree dividing 2 and a closed point on $X_1(25)$ in degree dividing 10. Again by considering lifts of this point with degree bounds given by Proposition 2·6, we see the divisibility condition is best-possible for all k with $j_{min}=j(E_2)$ .

Proof. Suppose $3^{\max(0,2k-1-d)} \nmid [F\,:\,{\mathbb{Q}}]$ , so in particular $2k-1-d\gt0$ . Let $\{P,Q\}$ be a basis of $E[3^k]$ . Replacing F with at worst a quadratic extension $L/F$ , we may view $\varphi$ as an L-isogeny from E to $E_0/L$ . Then ${\mathrm{im}} \rho_{E/L,3^{k}}$ is contained in

\begin{equation*}H \,:\!=\, \left\{\begin{pmatrix}1 & x \\0 & y\end{pmatrix}|\, x \in {\mathbb{Z}}/3^k{\mathbb{Z}}, y \in ({\mathbb{Z}}/3^k{\mathbb{Z}})^{\times} \right\},\end{equation*}

which has order $3^k \cdot \varphi(3^k)=3^{2k-1}\cdot 2$ . If ${\mathrm{ord}}_{3}(\#{\mathrm{im}} \rho_{E/L,{3}^{k}})\lt{2k-1}$ , then the index of the mod $3^k$ Galois representation of $E/L$ is divisible by $3^{2k-1}$ . Thus

\begin{equation*}3^{2k-1} \mid [{\mathrm{GL}}_2({\mathbb{Z}}_{3})\,:\,{\mathrm{im}} \rho_{E/L,3^{\infty}}].\end{equation*}

By [ Reference Bourdon and Najman11 , lemma 4·5], we have $3^{2k-1} \mid [{\mathrm{GL}}_2({\mathbb{Z}}_{3})\,:\, {\mathrm{im}} \rho_{E_0/{\mathbb{Q}}, 3^{\infty}}] \cdot[L \cap {\mathbb{Q}}(E_0[3^{\infty}])\,:\,{\mathbb{Q}}]$ . Since

\begin{equation*}{\mathrm{ord}}_{3}([{\mathrm{GL}}_2({\mathbb{Z}}_{3})\,:\, {\mathrm{im}} \rho_{E_0/{\mathbb{Q}}, 3^{\infty}}])=d,\end{equation*}

it follows that $3^{2k-1-d} \mid [L \cap {\mathbb{Q}}(E_0[3^{\infty}])\,:\,{\mathbb{Q}}]$ . Since L is at most a quadratic extension of F, then $3^{2k-1-d} \mid [F\,:\,{\mathbb{Q}}]$ , contradicting our assumption. So we may assume ${\mathrm{ord}}_{3}(\#{\mathrm{im}} \rho_{E/L,3^{k}})={2k-1}$ .

Note ${\mathrm{im}} \rho_{E/F,3^{k}}$ contained in H as well, and since $L/F$ is at worst a quadratic extension, we have ${\mathrm{ord}}_{3}(\#{\mathrm{im}} \rho_{E/F,3^{k}})={2k-1}$ . If ${\mathrm{im}} \rho_{E/F,3^{k}}$ is properly contained in H, then $\#{\mathrm{im}} \rho_{E/F,3^{k}}=3^{2k-1}$ . Since ${\mathbb{Q}}(\zeta_{3^k}) \subseteq F(E[3^k])$ , it follows that 2 must divide $[F\,:\,{\mathbb{Q}}]$ . This contradicts F having odd degree. Hence ${\mathrm{im}} \rho_{E/F,3^{k}}=H$ .

If ${\mathrm{im}} \rho_{E/F,3^{\infty}} \neq \pi^{-1}({\mathrm{im}} \rho_{E/F,3^k})$ , then $[F(E[3^{k+1}])\,:\,F(E[3^k])]$ divides $3^3$ ; see, for example, [ Reference Bourdon, Ejder, Liu, Odumodu and Viray9 , proposition 3·5]. Since $\# {\mathrm{Gal}}(F(E[3^k])/F) =3^{2k-1}\cdot 2$ , we have

\begin{equation*}\# {\mathrm{Gal}}(F(E[3^{k+1}])/F) \mid 3^3\cdot 3^{2k-1}\cdot 2=3^{2k+2}\cdot 2.\end{equation*}

It follows that $\# {\mathrm{Gal}}(L(E[3^{k+1}])/L) \mid 3^{2k+2}\cdot 2$ , and so the index of the 3-adic Galois representation of $E/L$ is divisible by at least $3^{2k-1}\cdot 8$ . By [ Reference Bourdon and Najman11 , lemma 4·5], we have $3^{2k-1}\cdot 8 \mid [{\mathrm{GL}}_2({\mathbb{Z}}_{3})\,:\, {\mathrm{im}} \rho_{E_0/{\mathbb{Q}}, 3^{\infty}}] \cdot[L \cap {\mathbb{Q}}(E_0[3^{\infty}])\,:\,{\mathbb{Q}}]$ . It follows that $3^{2k-1-d} \mid [L \cap {\mathbb{Q}}(E_0[3^{\infty}])\,:\,{\mathbb{Q}}]$ . Since L is at worst a quadratic extension of F, then $3^{2k-1-d} \mid [F\,:\,{\mathbb{Q}}]$ , contradicting our assumption. Thus, ${\mathrm{im}} \rho_{E/F,3^{\infty}} = \pi^{-1}({\mathrm{im}} \rho_{E/F,3^k})$ .

Proof. Suppose $3^{\max(0,2k-1-d)} \nmid [F\,:\,{\mathbb{Q}}]$ , so $2k-1-d\gt0$ . First, suppose for the sake of contradiction that $r\gt k$ . Then

\begin{equation*}{\mathrm{im}} \rho_{E,3^{r}} \neq \pi^{-1}({\mathrm{im}} \rho_{E,3^k}),\end{equation*}

and by Lemma 6·4 we have $3^{2k-1-d} \mid [F\,:\,{\mathbb{Q}}]$ . We have reached a contradiction. So $r \leq k$ .

By assumption there exists $R \in E$ of order $3^r$ such that $\ker(\varphi) = \langle R \rangle$ . With respect to the basis $\{3^{k-r}P, Q\}$ of $E[3^r]$ , by Lemma 6·4 we may assume

\begin{equation*}{\mathrm{im}} \rho_{E/F, 3^r} =\left\{\begin{pmatrix}1 & x \\0 & y\end{pmatrix}|\, x \in {\mathbb{Z}}/3^r{\mathbb{Z}}, y \in ({\mathbb{Z}}/3^r{\mathbb{Z}})^{\times} \right\}.\end{equation*}

With respect to this basis, $R=\alpha 3^{k-r}P+ \beta Q$ for some $\alpha, \beta \in {\mathbb{Z}}/3^r{\mathbb{Z}}$ . We will show $\beta=0$ , from which we may conclude that $R \in \langle P \rangle$ .

Since R has order $3^r$ , we must have $3 \nmid \alpha$ or $3 \nmid \beta$ . First suppose $3 \nmid \alpha$ . The F-rationality of $\langle R \rangle$ and description of ${\mathrm{im}} \rho_{E/F, 3^r}$ as above implies there exists $\sigma \in {\mathrm{Gal}}_F$ and $\gamma_1 \in ({\mathbb{Z}}/3^r{\mathbb{Z}})^{\times}$ such that

\begin{equation*}\sigma(R)=\begin{pmatrix}1 & 0 \\0 & 2\end{pmatrix}\begin{pmatrix}\alpha \\\beta\end{pmatrix}=\begin{pmatrix}\alpha \\2\beta\end{pmatrix}=\begin{pmatrix}\gamma_1 \alpha \\\gamma_1 \beta\end{pmatrix}.\end{equation*}

So $\gamma_1 =1$ , which implies $\beta =0$ . Now suppose $3 \nmid \beta$ . As before, there must exist $\gamma_2 \in ({\mathbb{Z}}/3^r{\mathbb{Z}})^{\times}$ such that

\begin{equation*}\begin{pmatrix}1 & 1 \\0 & 1\end{pmatrix}\begin{pmatrix}\alpha \\\beta\end{pmatrix}=\begin{pmatrix}\alpha+\beta \\\beta\end{pmatrix}=\begin{pmatrix}\gamma_2 \alpha \\\gamma_2 \beta\end{pmatrix}.\end{equation*}

Again $\gamma_2=1$ and $\beta =0$ .

Proof. By [ Reference Cremona and Najman18 , lemma A·1], we may assume $\varphi$ is cyclic and generated by a point of order $3^r\cdot n$ where $3 \nmid n$ . If $n\gt1$ , then by replacing E with an n-isogenous curve if necessary, we may assume $\varphi$ has degree $3^r$ . If ${\mathrm{im}} \rho_{E_0, 3^{\infty}}=9.36.0.6$ or $9.36.0.8$ , then ${\mathrm{ord}}_{3}([{\mathrm{GL}}_2({\mathbb{Z}}_{3})\,:\, {\mathrm{im}} \rho_{E_0, 3^{\infty}}])=2$ . Suppose for the sake of contradiction that $3^{2k-3} \nmid [F\,:\,{\mathbb{Q}}]$ . By Lemma 6·5, we have $r \leq k$ and $\ker(\varphi) \subseteq \langle P \rangle$ .

First, suppose $r=k-1$ or k. Set $t=r+2$ , and let $\{R,S\}$ be a basis of $E[3^{t}]$ such that $3^{t-k}R = P$ and $3^{t-k}S=Q$ . Then we will show $\{\varphi(R), 3^{k-2}\varphi(Q)\}$ is a basis of E’[ Reference Bourdon, Ejder, Liu, Odumodu and Viray9 ]. Suppose there exist $\alpha, \beta\in {\mathbb{Z}}/9{\mathbb{Z}}$ such that $\alpha \varphi(R)=\beta 3^{k-2}\varphi(Q)$ . This implies $\alpha R - \beta 3^{k-2} Q \in \ker(\varphi) \subseteq \langle P \rangle = \langle 3^{t-k}R \rangle$ . Thus $\beta 3^{t-s}S$ is in the cyclic subgroup generated by R, and so $3^2 \mid \beta$ since $\{R,S\}$ is a basis of $E[3^{t}]$ . Thus $\alpha \varphi(R)=\beta 3^{k-2}\varphi(Q)=\mathcal{O}$ , as desired.

By Lemma 6·4 with respect to the basis $\{P,Q\}$ , there exists $\sigma \in {\mathrm{Gal}}_F$ such that

\begin{equation*}\rho_{E/F,3^k}(\sigma)=\begin{pmatrix}1 & 0 \\0 & 4\end{pmatrix}.\end{equation*}

Also by Lemma 6·4 we have ${\mathrm{im}} \rho_{E/F,3^{\infty}} = \pi^{-1}({\mathrm{im}} \rho_{E/F,3^k})$ , so with respect to the basis $\{R,S\}$ of $E[3^{t}]$ , we know there exists $\sigma' \in {\mathrm{Gal}}_F$ such that

\begin{equation*}\rho_{E/F,3^{t}}(\sigma')=\begin{pmatrix}1 & 0 \\0 & 4\end{pmatrix}.\end{equation*}

Under the basis $\{\varphi(R), 3^{k-2}\varphi(Q)\}$ of E ^′[ Reference Bourdon, Ejder, Liu, Odumodu and Viray9 ],

\begin{align*}\sigma'(\varphi(R))&=\varphi(\sigma'(R))=\varphi(R)\end{align*}

\begin{align*}\sigma'(3^{k-2}\varphi(Q))&=3^{k-2}\varphi(\sigma'(Q))=4\cdot 3^{k-2}\varphi(Q).\end{align*}

So

\begin{equation*}\rho_{E'/F, 9}(\sigma')=\begin{pmatrix}1 & 0 \\0 & 4\end{pmatrix}.\end{equation*}

After at worst a quadratic extension $L/F$ , we have $E'/L \cong_L E_0/L$ . Since the matrix above has order 3,

\begin{equation*}\rho_{E_0/L, 9}(\sigma')=\begin{pmatrix}1 & 0 \\0 & 4\end{pmatrix}.\end{equation*}

This means that the group generated by $\rho_{E_0/L, 9}(\sigma')$ is conjugate to an order 3 subgroup of 9.36.0.6 or 9.36.0.8 mod 9, and a Magma computation shows no such subgroup exists.

Now, suppose $r \leq k-2$ . Then $\{3^{k-r-2} \varphi(P), 3^{k-2}\varphi(Q)\}$ is a basis of E’[ Reference Bourdon, Ejder, Liu, Odumodu and Viray9 ] since $\ker(\varphi) \subseteq \langle P \rangle$ . Since $P \in E(F)$ , we have $3^{k-r-2} \varphi(P) \in E'(F)$ . Moreover, for $\sigma \in {\mathrm{Gal}}_F$ as above,

\begin{align*}\sigma(3^{k-2}\varphi(Q))&=3^{k-2}\varphi(\sigma(Q))=4 \cdot 3^{k-2}\varphi(Q),\end{align*}

so

\begin{equation*}\begin{pmatrix}1 & 0 \\0 & 4\end{pmatrix}\in {\mathrm{im}} \rho_{E'/F,9}.\end{equation*}

We reach a contradiction as before.