Selmer groups of twists of elliptic curves over K with K-rational torsion points

Abstract

We generalize a result of Frey on Selmer groups of twists of elliptic curves over $\mathbf Q$ with $\mathbf Q$-rational torsion points to elliptic curves defined over number fields of small degree K with a K-rational torsion point. We also provide examples of elliptic curves coming from Zywina that satisfy the conditions of our Corollary D.

1 Introduction

Let $\ell$ be an odd, rational prime and let E / K be an elliptic curve defined over a number field K. The K-rational points E(K) form a finitely generated group by the Mordell-Weil theorem. Recall from [1, Section X.4] that we have the following exact sequence

where ${{\mathrm{Sel}}}_{\ell }(E,K)$ denotes the $\ell$-Selmer group and is the $\ell$-Shafarevich-Tate group. If $K = \mathbf Q$, then Frey [2] provides explicit examples of quadratic twist of elliptic curves over $\mathbf Q$ with $\mathbf Q$-rational points of odd, prime order $\ell$ whose $\ell$-Selmer groups are non-trivial; a theorem of Mazur [3] implies that $\ell \in \left\{ 3,5,7 \right\}$.

Theorem 1.1

([2], Corollary) Suppose that $E/\mathbf Q$ is an elliptic curve with a $\mathbf Q$-rational torsion point P of odd prime order $\ell$, and suppose that P is not contained in the kernel of reduction modulo $\ell$; in particular, this means that E is not supersingular modulo $\ell$ if ${{\mathrm{ord}}}_{\ell }(j_E) \ge 0$. Let $\widetilde{S}_E$ be the subset of odd primes dividing the conductor N(E) of E defined by

$$\begin{aligned} \widetilde{S}_E&:= \left\{ p | N(E) : p \equiv -1 \pmod \ell ,\, \ell \not \mid {{\mathrm{ord}}}_p(\Delta _E) \right\} , \end{aligned}$$

where $j_E$ is the j-invariant of E and $\Delta _E$ is the discriminant of E. Suppose that $\widetilde{S}_E = {\varnothing }$. Suppose that $d\equiv 3 \pmod 4$ is a negative, square-free integer coprime to $\ell N(E)$ satisfying:

• 1. if ${{\mathrm{ord}}}_{\ell }(j_E) < 0$, then $\left( \frac{d}{\ell } \right) = -1$;

• 2. if p|N(E) is an odd prime, then

  $$\begin{aligned} \left( \frac{d}{p} \right) = \left\{ \begin{array}{llll} -1 &{} &{}\quad \text { if }{{\mathrm{ord}}}_p(j_E) \ge 0; \\ -1 &{} &{}\quad \text { if }{{\mathrm{ord}}}_p(j_E) < 0 \; \text { and }\; E/\mathbf Q _p \text { is a Tate curve}; \\ 1 &{} &{}\quad \text { otherwise}. \end{array}\right. \end{aligned}$$

Then we have that ${{\mathrm{Sel}}}_{\ell }(E^d,\mathbf Q )$ is non-trivial if and only if the $\ell$-torsion of the class group of $\mathbf Q (\sqrt{d})$ is non-trivial.

Remark 1.2

Frey actually proved a more explicit double divisibility statement [2, Theorem] concerning the $\ell$-Selmer group of $E^d$ and $\ell$-torsion of ray class groups, when $\widetilde{S}_E \ne \varnothing$.

In this paper, we generalize Frey’s results [2, Theorem, Corollary] to number fields K of small degree. We show that for specific quadratic twists $E^d$, the order of the $\ell$-torsion of some ray class group of $K(\sqrt{d})$ divides the order of ${{\mathrm{Sel}}}_{\ell }(E^d,K)$, and the order of ${{\mathrm{Sel}}}_{\ell }(E^d,K)$ divides the order of the $\ell$-torsion of a different ray class group of $K(\sqrt{d})$ times the degree of some maximal abelian extension of exponent $\ell$ with prescribed ramification and Galois conditions (cf. Theorems A, B for precise statements and Remark 3.1 for a colloquial statement). These results allow us to give explicit applications to elliptic curves defined over $\mathbf Q$ (cf. Corollaries C, D), and we provide explicit examples of elliptic curves over $\mathbf Q$ satisfying Corollary D in Sect. 6. Finally, in Corollary E, we generalize Theorem 1.1 to number fields of small degree.

Remark 1.3

Frey’s result [2, Theorem] has been used to produce twists of elliptic curves $E^d$ over $\mathbf Q$ such that is non-trivial. Most notably, Ono [4] utilized Frey’s result to prove that for a large class of $E/\mathbf Q$ with torsion subgroup over $\mathbf Q$ isomorphic to $\mathbf Z / \ell \mathbf Z$ where $\ell \in \left\{ 3,5,7 \right\}$, there are infinitely many negative square-free integers d for which

Later, Balog and Ono [5] used a new type of result concerning the non-triviality of class groups of imaginary quadratic fields (cf. [5, Theorem 1]) and Frey’s result to prove that for such $E/\mathbf Q$,

Frey’s idea was to obtain information about ${{\mathrm{Sel}}}_{\ell }(E^d,\mathbf Q )$ when $E(\mathbf Q )$ contains an element of order $\ell$. In particular, he studied the behavior of E over local fields $\mathbf Q _{\ell }$ and their algebraic closures $\overline{\mathbf{Q }}_{\ell }$. His work illustrated a deep relationship between $\ell$-ranks of Selmer groups and class groups of finite Galois extensions of exponent $\ell$. In this paper, we investigate the $\ell$-Selmer rank in families of quadratic twists of elliptic curves E / K with K-rational points of odd prime order $\ell$. We use Frey’s proof as a blueprint for our own, but the techniques we utilize come from class field theory. That being said, many of his arguments go through mutatis mutandis.

In order to state our results, we first need to recall some facts concerning prime torsion of elliptic curves defined over number fields of small degree. We give a succinct summary of these results and refer the reader to [6] for a more detailed synopsis. Let S(n) denote the set of primes that can arise as the order of a rational point on an elliptic curve defined over a number field of degree n and let Primes(n) denote the set of primes bounded by n. By Merel-Oesterlé’s bound, we know that

$$\begin{aligned} S(n) \subseteq {{\mathrm{\textsf {Primes}}}}((3^{n/2} + 1)^2). \end{aligned}$$

The exact value of the set S(n) is currently known for $n\le 5$, but reasonable good bounds on S(6) and S(7) are given in [7].

n	S(n)	References
1	${{\mathrm{\textsf {Primes}}}}(7)$	[3]
2	${{\mathrm{\textsf {Primes}}}}(13)$	[8]
3	${{\mathrm{\textsf {Primes}}}}(13)$	[9]
4	${{\mathrm{\textsf {Primes}}}}(17)$	[10]
5	${{\mathrm{\textsf {Primes}}}}(19)$	[11]
6	$\subseteq {{\mathrm{\textsf {Primes}}}}(19) \cup \left\{ 37,73 \right\}$	[7]

One can also consider the subset $S_\mathbf{Q }(n)\subseteq S(n)$ corresponding to primes that can arise as the order of a rational point on an elliptic curve $E_K = E \times _\mathbf{Q } K$ where E is defined over $\mathbf Q$ and K is a number field of degree n. From [12], it is known that

$$\begin{aligned} S_\mathbf{Q }(n) \subseteq {{\mathrm{\textsf {Primes}}}}(13) \cup \left\{ 37 \right\} \cup {{\mathrm{\textsf {Primes}}}}(2n+1), \end{aligned}$$

and [12, Corollary 1.1] states that for $1\le n \le 20$,

$$\begin{aligned} S_\mathbf{Q }(n) = \left\{ \begin{array}{ll} {{\mathrm{\textsf {Primes}}}}(7) &{}\quad \text { for }n= 1,2 \\ \left\{ 2,3,5,7,13 \right\} &{}\quad \text { for }n = 3,4 \\ {{\mathrm{\textsf {Primes}}}}(13) &{}\quad \text { for }n = 5,6,7 \\ {{\mathrm{\textsf {Primes}}}}(17) &{}\quad \text { for }n = 8\\ {{\mathrm{\textsf {Primes}}}}(19) &{}\quad \text { for }n = 9,10,11 \\ {{\mathrm{\textsf {Primes}}}}(19) \cup {37} &{}\quad \text { for }12\le n \le 20. \\ \end{array} \right. \end{aligned}$$

In this paper, we generalize the full double divisibility statement of [2, Theorem] to elliptic curves defined over small degree number fields K. We state explicit versions of our results, Theorems A, B and Corollary E, in Sect. 3 once we have established some notation.

1.1 Some remarks about the proofs

The problem of constructing elements in the Selmer group is a classical question with many avenues of approach. Frey’s condition that the elliptic curve E / K have a K-rational point of odd prime power order $\ell >3$ has two immediate consequences. First, the image of Galois under the mod $\ell$ representation is conjugate to

$$\begin{aligned} \left( \begin{array}{ll} 1 &{}\quad * \\ 0 &{}\quad * \end{array}\right) \subset {{\mathrm{GL}}}_2(\mathbf F _{\ell }), \end{aligned}$$

which will assist in our explicit description the Galois structure of splitting fields of $\ell$-covers of E / K and the splitting fields of elements in ${{\mathrm{Sel}}}_{\ell }(E^d,K)$. The second is that we can immediately identify a quotient of $H^1({{\mathrm{Gal}}}(\overline{K}/K),E(\overline{K})[\ell ])$, namely $H^1({{\mathrm{Gal}}}(\overline{K}/K),\mu _{\ell })$. Frey’s (and our) proof relies on an analysis of cocyles in $H^1({{\mathrm{Gal}}}(\overline{K}/K),E(\overline{K})[\ell ])$ and this fact will allow us to deduce local triviality in certain cases using Hilbert’s Theorem 90. A laborious aspect of our proofs is the case by case analysis of how primes $\mathfrak {p}$ dividing N(E) behave in the field $K(\sqrt{d})\cdot K(E[\ell ])$ where $d \in \mathcal {O}_K^{\times }/(\mathcal {O}_K^{\times })^2$ yields the quadratic twist $E^d$ of E and $K(E[\ell ])$ is the $\ell$-division field of E / K.

1.2 Organization of paper

In Sect. 2, we recall some classical facts from class field theory and algebraic number theory. In Sect. 3, we state our main results, Theorems A, B and Corollary E. In Sect. 4, we prove Theorem A, which yields a single divisibility statement. In Sect. 5, we prove the double divisibility statement of Theorem B by investigating the Galois structure of splitting fields of $\ell$-covers of E / K and the splitting fields of elements ${{\mathrm{Sel}}}_{\ell }(E^d,K)$. Finally in Sect. 6, we provide explicit examples of elliptic curves over $\mathbf Q$ coming from [13] that satisfy the Corollary D.

2 Background and notation

Let L / K be a Galois extension of K, with ring of integers $\mathcal {O}_L \text { and }\mathcal {O}_K.$ For any finite prime $\mathfrak {P}\in \mathcal {O}_L$ lying over a prime $\mathfrak {p}\in \mathcal {O}_K,$ let $D(\mathfrak {P})$ denote the decomposition group of $\mathfrak {P},$ let $I(\mathfrak {P})$ denote the inertia group of $\mathfrak {P}$ and let $\kappa ' := \mathcal {O}_L / \mathfrak {P}\text { and }\kappa = \mathcal {O}_K / \mathfrak {p}$ be the residue fields of characteristic $q = p^n$. In this note, we need a specific result concerning the Artin symbol and ramification theory for quadratic extensions L / K. For the definition of the Artin symbol $\left( \frac{L/K}{\mathfrak {p}} \right)$, we refer the reader to [14, Chapter IV].

Lemma 2.1

Let L / K be a quadratic extension, let $\mathfrak {p}$ be a prime ideal of $\mathcal {O}_K$, let $\mathfrak {P}$ denote some prime of $\mathcal {O}_L$ lying above $\mathfrak {p}$, and let $\langle \delta \rangle = {{\mathrm{Gal}}}(L/K).$ Then:

• 1. $\mathfrak {p}$ is unramified and splits completely in L $\Longleftrightarrow$ $\left( \frac{L/K}{\mathfrak {p}} \right) = {{\mathrm{id}}}$,

• 2. $\mathfrak {p}$ is unramified and non-split in L $\Longleftrightarrow$ $\left( \frac{L/K}{\mathfrak {p}} \right) = \delta$,

• 3. $\mathfrak {p}$ is ramified in L $\Longleftrightarrow$ $\mathfrak {p}| \Delta _{L/K}$ where $\Delta _{L/K}$ denotes the relative discriminant of L / K.

Proof

Part (3) follows directly from the definition of the Artin symbol. Since $\mathfrak {p}$ is unramified, we know that $|D(\mathfrak {P})| = f$ where f is the inertia degree of $\mathfrak {P}$ over $\mathfrak {p}$. A prime $\mathfrak {p}$ splits completely in L if and only if the ramification index e of $\mathfrak {P}$ above $\mathfrak {p}$ and the inertia degree f of $\mathfrak {P}$ above $\mathfrak {p}$ are equal to 1. Hence,

$$\begin{aligned} |D(\mathfrak {P})| = [\kappa ' : \kappa ] = 1 \Longleftrightarrow {{\mathrm{ord}}}\left( \frac{L/K}{\mathfrak {p}} \right) = 1 \Longleftrightarrow \left( \frac{L/K}{\mathfrak {p}} \right) = {{\mathrm{id}}}, \end{aligned}$$

which proves (1). For (2), our assumptions and the fundamental identity tell us that $e = 1$ and $g =1$ if and only if $f = 2$. Thus,

$$\begin{aligned} |D(\mathfrak {P})| = [\kappa ' : \kappa ] = 2 \Longleftrightarrow {{\mathrm{ord}}}\left( \frac{L/K}{\mathfrak {p}} \right) = 2 \Longleftrightarrow \left( \frac{L/K}{\mathfrak {p}} \right) = \delta . \end{aligned}$$

$\square$

In Theorem A, we use primitive Hecke characters to describe a subset of primes $\mathfrak {p}| N(E)$; we refer the reader to [14, Chapter VII, Section 6] for the definition of these characters.

Remark 2.2

Recall that there is a conductor-preserving correspondence between primitive Dirichlet characters of order $\ell$ and cyclic, degree $\ell$ number fields $F/\mathbf Q$. From [15, Theorem 3.7], the Dirichlet character $\chi$ corresponds to the fixed field F of $\ker \chi \subseteq \left( \mathbf Z / f_{\chi } \mathbf Z \right) ^{\times } = {{\mathrm{Gal}}}(\mathbf Q (\zeta _{f_{\chi }})/\mathbf Q )$. For any prime q,

$$\begin{aligned} \chi (q) = 0 \Longleftrightarrow q \text { ramifies in }F \quad \text { and }\quad \chi (q) = 1 \Longleftrightarrow q \text { splits in }F. \end{aligned}$$

By class field theory, any primitive Hecke character $\chi _{H}$ of K of order $\ell$ determines a cyclic extension N / K of degree $\ell$. Moreover, the set of primitive Hecke characters determining this cyclic extension equals $\{\chi _{H},\chi _{H}^2,\dots , \chi _{H}^{\ell - 1}\}$. These $\ell - 1$ Hecke characters have the same conductor $\mathfrak {f}$, and the determinant of L / K equals their product $\mathfrak {f}^{\ell - 1}$ by the Hasse conductor-discriminant theorem. Thus for any prime ideal $\mathfrak {q}$ of $\mathcal {O}_K$, we have that

$$\begin{aligned} \chi _{H}(\mathfrak {q}) = 0 \Longleftrightarrow \mathfrak {q}\text { ramifies in }N \quad \text { and }\quad \chi _{H}(\mathfrak {q}) = 1 \Longleftrightarrow \mathfrak {q}\text { splits in }N. \end{aligned}$$

2.1 Notation

We set the following notation:

$$\begin{aligned} K&:=\text { Galois number field,}\\ \ell&:= \text { odd, rational prime in } S(n) \backslash \left\{ 2,3 \right\} \text { such that }\ell \not \mid {{\mathrm{cl}}}(K) \text { and }\zeta _{\ell } \notin K, \\ L/K&:= \text { algebraic extension of }K, \\ \mathfrak {p}&:= \text { prime divisor of the rational prime }p \text { in } \mathcal {O}_K,\\ \mathfrak {P}&:= \text { prime divisor of }\mathfrak {p}\text { in } \mathcal {O}_L,\\ K_{\mathfrak {p}}&:= \text { completion of }K \text { with respect to }\mathfrak {p},\\ L_{\mathfrak {P}}&:= \text { completion of }L \text { with respect to }\mathfrak {P},\\ S&:= \text { finite set of primes of }\mathcal {O}_K ,\\ M/L&:= \text { Galois extension with abelian Galois group of exponent }\ell . \end{aligned}$$

More generally, lower case gothic font will denote a divisor of a rational prime of $\mathbf Q$, and similarly, upper case gothic font will denote a divisor of a prime of K.

Definition 2.3

M / L is said to be little ramified outside S if for primes $\mathfrak {p}\notin S$ and all $\mathfrak {P}_L|\mathfrak {p}$ one has

$$\begin{aligned} M\cdot L_{\mathfrak {P}}(\zeta _{\ell }) = L_{\mathfrak {P}}(\zeta _{\ell })(\root \ell \of {u_1},\ldots ,\root \ell \of {u_k}) \end{aligned}$$

with $k\in \mathbf N$ and ${{\mathrm{ord}}}_{\mathfrak {P}_L}(u_i) = 0$. Here $\zeta _{\ell }$ is a ${\ell }$th root of unity, $u_1,\dots , u_k$ are elements in $L_{\mathfrak {P}}(\zeta _{\ell })$, and ${{\mathrm{ord}}}_{\mathfrak {P}_L}$ is the normed valuation belonging to $\mathfrak {P}_L$.

If M / L little ramified outside S, then M / L is unramified at all divisors of primes $\mathfrak {p}\notin S \cup \left\{ \mathfrak {l} \right\}$.

2.2 Notation

We set the following notation, which comes directly from [2]:

$$\begin{aligned} L_S&:= \text { maximal abelian extension of exponent } \ell \text {of L which is} \\&\qquad \text { little ramified outside } S, \\ L_{S,u}&:= \text { maximal subfield of } L_{S} \text { which is unramified outside of } S, \\ H_S(L)&:= \text { Galois group of } L_{S}/L, \\ H_{S,u}(L)&:= \text { Galois group of } L_{S,u}/L ,\\ {{\mathrm{cl}}}_{S}(L)[{\ell }]&:= \text { order of } H_{S}(L), \\ {{\mathrm{cl}}}_{S,u}(L)[{\ell }]&:= \text { order of } H_{S,u}(L). \end{aligned}$$

Remark 2.4

If $S = \emptyset$, we see that ${{\mathrm{cl}}}_{\emptyset ,u}(L)$ is equal to the order of the subgroup of the divisor class group of L consisting of elements of order $\ell$ which we denote by ${{\mathrm{cl}}}(L)[\ell ]$.

Now assume that L / K is normal with cyclic Galois group generated by an element $\gamma$ of order $\ell -1$. Take an extension $\widetilde{\gamma }$ to $L(\zeta _{\ell })$. Let $\chi _{\ell }$ be the cyclotomic character induced by the action of ${{\mathrm{Gal}}}(L(\zeta _{\ell })/K)$ on $\langle \zeta _{\ell } \rangle$. Then $\chi _{\ell }(\widetilde{\gamma })$ is determined by

$$\begin{aligned} \widetilde{\gamma }(\zeta _{\ell }) = \zeta _{\ell }^{\chi _{\ell }(\widetilde{\gamma })}. \end{aligned}$$

Let M be normal over K containing L such that ${{\mathrm{Gal}}}(M/L)$ is abelian of exponent $\ell$. Then $\widetilde{\gamma }$ operates by conjugation on

$$\begin{aligned} {{\mathrm{Gal}}}(M(\zeta _{\ell })/L(\zeta _{\ell })) \cong {{\mathrm{Gal}}}(M/L), \end{aligned}$$

and this operation does not depend on choice of $\widetilde{\gamma }$. Hence the subgroup

$$\begin{aligned} H(\chi _{\ell }):= \left\{ \alpha \in {{\mathrm{Gal}}}(M/L) : \widetilde{\gamma }\alpha \widetilde{\gamma }^{-1}= \alpha ^{\chi _{\ell }(\widetilde{\gamma })} \right\} \subseteq {{\mathrm{Gal}}}(M/L) \end{aligned}$$

is well-defined. In the special case that $M = L_S$, we denote the order of $H_{S}(L)(\chi _{\ell })$ by ${{\mathrm{cl}}}_S(L)_{\ell }(\chi _{\ell })$.

Now we shall consider an elliptic curve E / K given by a Weierstrass equation $F(x,y) = 0$ with coefficients in $\mathcal {O}_K$ and minimal discriminant $\Delta _E$. For any extension L / K, we denote the L-rational points of E (including $\infty$) by E(L). Let $\chi _{H}$ be a primitive Hecke character of order $\ell$ and let

$$\begin{aligned} \widetilde{S}_E&:= \left\{ \mathfrak {p}| N(E) : \chi _{H}(\mathfrak {p}) \ne 0 ,{{\mathrm{ord}}}_{\mathfrak {p}}(\Delta _E)\not \equiv 0 \pmod \ell \right\} \\ S_E&:= \{\mathfrak {p}\in \widetilde{S}_E : {{\mathrm{ord}}}_{\mathfrak {p}}(j_E) < 0 \}. \end{aligned}$$

Let $d \in \mathcal {O}_K^{\times }/(\mathcal {O}_K^{\times })^2$ and denote the twist of E / K by $E^d/K$. Via the general theory of twists [1, Section X.2], we know that $E^d$ is isomorphic to E over $K(\sqrt{d})$ but not over K. Let $G_K := {{\mathrm{Gal}}}(\overline{K},K)$ denote the absolute Galois group. Let $\mathfrak {W}(E^d,K)[\ell ]$ be the set of elements of order $\ell$ in the kernel of

$$\begin{aligned} \rho :H^1(G_K,E^d(\overline{K})) \longrightarrow \bigoplus _{\mathfrak {p}\text { prime}} H^1({{\mathrm{Gal}}}(\overline{K_{\mathfrak {p}}}/K_{\mathfrak {p}}),E^d(\overline{K_{\mathfrak {p}}})). \end{aligned}$$

The group of elements of order $\ell$ in the Selmer group of $E^d$, denoted by ${{\mathrm{Sel}}}_{\ell }(E^d,K)$ is given as the pre-image of $\mathfrak {W}(E^d,K)[\ell ]$ by the map

$$\begin{aligned} \alpha :H^1(G_K,E^d(\overline{K})[\ell ]) \longrightarrow H^1(G_K,E^d(\overline{K})). \end{aligned}$$

There are two main cases we need to consider:

Case 1 Assume that ${{\mathrm{ord}}}_{\mathfrak {p}}(j_E) \ge 0$. Then there is a finite extension N / K such that E has good reduction modulo all $\mathfrak {P}_N | \mathfrak {p}$ i.e., we find an elliptic curve $\widetilde{E}$ such that $\tilde{E}$ modulo $\mathfrak {P}_N$ is an elliptic curve over the residue field of $\mathfrak {P}_N$. $\widetilde{E}(\overline{N_{\mathfrak {P}}})$ contains a subgroup $\widetilde{E}_-(N_{\mathfrak {P}})$ consisting of points $(\widetilde{x},\widetilde{y})$ with ${{\mathrm{ord}}}_{\mathfrak {P}_N}(\widetilde{x}) < 0$. $\widetilde{E}_-$ is the kernel of reduction modulo $\mathfrak {P}_N$, and ${{\mathrm{ord}}}_{\mathfrak {P}_N}(\widetilde{x}/\widetilde{y})$ is the level of $(\widetilde{x},\widetilde{y})$. For ease of notation, we say that a point $(x,y) \in E(\overline{N_{\mathfrak {P}}})$ is in the kernel of the reduction modulo $\mathfrak {P}_N$ if its image $(\widetilde{x},\widetilde{y}) \in \widetilde{E}_-(\overline{N_{\mathfrak {P}}})$.

Case 2 Assume that ${{\mathrm{ord}}}_{\mathfrak {p}}(j_E) < 0$. Then after an extension $L/K_{\mathfrak {p}}$ of degree $\le 2$, E becomes a Tate curve (via a theorem of Tate [1, Theorem C.14.1]); in particular, one has a Tate parametrization

$$\begin{aligned} \tau :\overline{L}^{\times }/\langle q \rangle \longrightarrow E(\overline{L}) \end{aligned}$$

where q is the $\mathfrak {p}$-adic period of E. One also has that

$$\begin{aligned} j_E = \frac{1}{q} + \sum _{i=0}^{\infty } a_iq^i \quad \text { with }a_i \in \mathbf Z \end{aligned}$$

and the points of order $\ell$ in $E(\overline{L})$ are of the form $\tau (\zeta _{\ell }^{\alpha }(q^{\beta /\ell }))$ where $\alpha ,\beta \in \left\{ 1,\dots , \ell -1 \right\}$.

Definition 2.5

If F / K is a number field and $\mathfrak {P}_F | \mathfrak {p}$ we say that a point $(x,y) \in E(F_{\mathfrak {P}})$ is in the connected component of the unity modulo $\mathfrak {P}_F$ if it is of the form $\tau (u)$ with u a $\mathfrak {P}_F$-adic unit, and (x, y) is in the kernel of the reduction modulo $\mathfrak {P}_F$ if $u-1 \in \mathfrak {P}_F$.

Remark 2.6

One should notice that if E is not a Tate curve over $K_{\mathfrak {p}}$ but over an extension of degree 2 of $K_{\mathfrak {p}}$, then for all points $P \in E(K_{\mathfrak {p}})$, 2P is in the connected component of unity modulo $\mathfrak {p}$.

3 Statement of results

As mentioned above, [2, Theorem] gives a double divisibility statement involving the $\ell$-torsion of the Selmer group. First, we generalize his single divisibility to elliptic curves E / K defined over number fields K of finite degree with K-rational points of odd, prime order $\ell$. Recall that S(n) is the set of primes that can arise as the order of a rational point on an elliptic curve defined over a number field of degree n.

Theorem A

Let K be a Galois number field and choose $\ell \in S(n)\backslash \left\{ 2,3 \right\}$ such that $\ell \not \mid {{\mathrm{cl}}}(K)$ and $\zeta _{\ell } \notin K$. Let E / K be an elliptic curve over K with a K-rational point P of order $\ell$; let $\chi _{H}$ denote a primitive Hecke character of K with order $\ell$; let $\mathfrak {q}$ denote a prime of $\mathcal {O}_K$ that lies above 2; and let $\mathfrak {l}$ denote a prime of $\mathcal {O}_K$ that lies above $\ell$. Suppose that P is not contained in the kernel of reduction modulo $\mathfrak {l}$; in particular, this means that E is not supersingular modulo $\mathfrak {l}$ if ${{\mathrm{ord}}}_{\mathfrak {l}}(j_E) \ge 0$. Let $S_E$ be the set of primes

$$\begin{aligned} S_E&:= \left\{ \mathfrak {p}| N(E){:} {{\mathrm{ord}}}_{\mathfrak {p}}(\Delta _{E}) \not \equiv 0 \, \pmod \ell , \, \chi _{H}(\mathfrak {p}) \ne 0, \text { and }{{\mathrm{ord}}}_{\mathfrak {p}}(j_{E}) < 0 \right\} . \end{aligned}$$

Suppose that $d \in \mathcal {O}_K^{\times }/(\mathcal {O}_K^{\times })^2$ is negative,¹ coprime to $\mathfrak {l}\cdot N(E)$, and satisfies the following divisibility and Artin symbol conditions where $\langle \delta \rangle = {{\mathrm{Gal}}}(K(\sqrt{d})/K)$:

• 1. if $\mathfrak {q}| N(E)$, then $\mathfrak {q}| \Delta _{K(\sqrt{d})/K}$;

• 2. if ${{\mathrm{ord}}}_{\mathfrak {l}}(j_{E}) < 0$, then $\left( \frac{K(\sqrt{d})/K}{\mathfrak {l}} \right) = \delta$;

• 3. if $\mathfrak {p}|N(E)$ is a prime of K with $\mathfrak {p}\notin S_E$, then

  • if ${{\mathrm{ord}}}_{\mathfrak {p}}(j_{E}) \ge 0$, then $\left( \frac{K(\sqrt{d})/K}{\mathfrak {p}} \right) = \delta$;

  • if ${{\mathrm{ord}}}_{\mathfrak {p}}(j_{E}) < 0 \text { and }E/K_{\mathfrak {p}} \text { is a Tate curve},$ then $\left( \frac{K(\sqrt{d})/K}{\mathfrak {p}} \right) = \delta$;

  • otherwise, $\left( \frac{K(\sqrt{d})/K}{\mathfrak {p}} \right) = {{\mathrm{id}}}$.

Then we have that the order of the $\ell$-torsion of the $S_E$-ray class group of $K(\sqrt{d})$ divides the order of ${{\mathrm{Sel}}}_{\ell }(E^d,K)$. More precisely, the single divisibility statement holds:

$$\begin{aligned} {{\mathrm{cl}}}_{S_E,u}(K(\sqrt{d}))[\ell ] \left| \# {{\mathrm{Sel}}}_{\ell }(E^d,K) \right. . \end{aligned}$$

(1)

We also prove a stronger, more explicit version of Theorem A in the form of a double divisibility statement, which completely generalizes [2, Theorem].

Theorem B

Let K be a Galois number field of degree $n\le 5$ such that $N_{K/\mathbf Q }(\mathfrak {q}) = 2 \text { for all }\mathfrak {q}|2$. Choose $\ell \in S(n)\backslash \left\{ 2,3 \right\}$ such that $\ell \not \mid {{\mathrm{cl}}}(K)$ and $\zeta _{\ell } \notin K$. Let E / K be an elliptic curve over K with a K-rational point P of order $\ell$; let $\chi _{H}$ denote a primitive Hecke character of K with order $\ell$; let $\mathfrak {q}$ denote a prime ideal of $\mathcal {O}_K$ that lies above 2; and let $\mathfrak {l}$ denote a prime ideal of $\mathcal {O}_K$ that lies above $\ell$. If $[K:\mathbf Q ] = 5$ and $\ell = 5$, then we must make the added assumption that $(\ell )\mathcal {O}_K$ is not totally ramified. Suppose that P is not contained in the kernel of reduction modulo $\mathfrak {l}$; in particular, this means that E is not supersingular modulo $\mathfrak {l}$ if ${{\mathrm{ord}}}_{\mathfrak {l}}(j_E) \ge 0$. Let $\widetilde{S}_E$ and $S_E$ be the sets of primes

$$\begin{aligned} \widetilde{S}_E&:= \left\{ \mathfrak {p}| N(E) {:} \chi _{H}(\mathfrak {p}) \ne 0,\, {{\mathrm{ord}}}_{\mathfrak {p}}(\Delta _{E}) \not \equiv 0\, \pmod \ell \right\} ,\\ S_E&:= \{\mathfrak {p}\in \widetilde{S}_E {:} {{\mathrm{ord}}}_{\mathfrak {p}}(j_{E}) < 0\}. \end{aligned}$$

Suppose that $d \in \mathcal {O}_K^{\times }/(\mathcal {O}_K^{\times })^2$ is negative, coprime to $\mathfrak {l}\cdot N(E)$, and satisfies the following divisibility and Artin symbol conditions where $\langle \delta \rangle = {{\mathrm{Gal}}}(K(\sqrt{d})/K)$:

• 1. if $\mathfrak {q}| N(E)$, then $\mathfrak {q}| \Delta _{K(\sqrt{d})/K}$;

• 2. if ${{\mathrm{ord}}}_{\mathfrak {l}}(j_{E}) < 0$, then $\left( \frac{K(\sqrt{d})/K}{\mathfrak {l}} \right) = \delta$;

• 3. if $\mathfrak {p}|N(E)$ is a prime of K with $\mathfrak {p}\notin S_E$, then

  • if ${{\mathrm{ord}}}_{\mathfrak {p}}(j_{E}) \ge 0$, then $\left( \frac{K(\sqrt{d})/K}{\mathfrak {p}} \right) = \delta$;

  • if ${{\mathrm{ord}}}_{\mathfrak {p}}(j_{E}) < 0 \text { and }E/K_{\mathfrak {p}} \text { is a Tate curve},$ then $\left( \frac{K(\sqrt{d})/K}{\mathfrak {p}} \right) = \delta$;

  • otherwise, $\left( \frac{K(\sqrt{d})/K}{\mathfrak {p}} \right) = {{\mathrm{id}}}$.

Then we have the following double divisibility

$$\begin{aligned} {{\mathrm{cl}}}_{S_E,u}(K(\sqrt{d}))[\ell ] \left| \# {{\mathrm{Sel}}}_{\ell }(E^d,K) \right| {{\mathrm{cl}}}_{\widetilde{S}_E,u}(K(\sqrt{d}))[\ell ]\cdot {{\mathrm{cl}}}_{S_E}(K')[\ell ](\chi _{\ell }), \end{aligned}$$

(2)

where $K'$ is the subfield of $K(\sqrt{d},\zeta _{\ell })$ of index 2 containing neither $\zeta _{\ell }$ nor $\sqrt{d}$.

Remark 3.1

In words, (2) states that the order of the $\ell$-torsion of the $S_E$-ray class group of $K(\sqrt{d})$ divides the order of ${{\mathrm{Sel}}}_{\ell }(E^d,K)$, and the order of ${{\mathrm{Sel}}}_{\ell }(E^d,K)$ divides the order of the $\ell$-torsion of the $\widetilde{S}_E$-ray class group of $K(\sqrt{d})$ times the degree of the maximal abelian extension $K''$ of $K'$ of exponent $\ell$ unramified outside of $S_E \cup \left\{ \mathfrak {l} \right\}$ such that the Galois group ${{\mathrm{Gal}}}(K'/K)$ acts on ${{\mathrm{Gal}}}(K''/K)$ by $\chi _{\ell }\varepsilon _d$, where $\varepsilon _d$ is the character prescribing the Galois action on $\sqrt{d}$.

Once we have proved Theorems A, B, we can immediately extend the divisibility statements (1), (2) to elliptic curves E defined over $\mathbf Q$ by considering the values of $S_\mathbf{Q }(n)$.

Corollary C

Let $E/\mathbf Q$ be an elliptic curve defined over $\mathbf Q$. For some Galois number field K, suppose that $E_K$ attains a K-rational point P of order $\ell$ where $\ell \in S_\mathbf{Q }(n)\backslash \left\{ 2,3 \right\}$ such that $\ell \not \mid {{\mathrm{cl}}}(K)$ and $\zeta _{\ell } \notin K$. In keeping with the notation and assumptions of Theorem A, we can produce examples of quadratic twists $E_K^d$ that satisfy the divisibility statement (1).

Corollary D

Let $E/\mathbf Q$ be an elliptic curve defined over $\mathbf Q$; let $E_K$ denote the base change of this curve to a Galois number field of degree $n\le 20$ such that $N_{K/\mathbf Q }(\mathfrak {q}) = 2 \text { for all }\mathfrak {q}|2$. Choose $\ell \in S_\mathbf{Q }(n)\backslash \left\{ 2,3 \right\}$ such that $\ell \not \mid {{\mathrm{cl}}}(K)$, $\zeta _{\ell } \notin K$, and the ramification index $e_{\mathfrak {l}}(K/\mathbf Q )$ satisfies $1>e_{\mathfrak {l}}(K/\mathbf Q )/(\ell -1) - 1$. Suppose that $E_K$ attains a K-rational point P of order $\ell$, then in keeping with the notation and assumptions of Theorem B, we can produce examples of quadratic twists $E_K^d$ that satisfy the double divisibility statement (2).

We can also generalize [2, Corollary], which we stated as Theorem 1.1.

Corollary E

Let $(E,\, \ell ,\, K,\, d)$ be as in Theorem B or in Corollary D. If $\widetilde{S}_E = \emptyset$, then ${{\mathrm{Sel}}}_{\ell }(E^d,K)$ is non-trivial if and only if the $\ell$-torsion of the class group of $K(\sqrt{d})$ is non-trivial, in particular

$$\begin{aligned} {{\mathrm{cl}}}(K(\sqrt{d}))[\ell ] \left| \# {{\mathrm{Sel}}}_{\ell }(E^d,K) \right| ({{\mathrm{cl}}}(K(\sqrt{d}))[\ell ])^2. \end{aligned}$$

Remark 3.2

In his Ph.D. thesis [16], Mailhot was able to recover and sharpen [2], Theorem] for elliptic curves defined over $\mathbf Q$ using purely cohomological methods. His refinement comes from prescribing a splitting behavior of primes above $K'$ instead of just a non-ramified condition. We remark that our methods and results are disjoint, however, we believe that [16, Corollary 2.17] can be generalized to elliptic curves defined over number fields K, using Theorem B.

4 Proof of Theorem A

In this section, we prove the divisibility statement (1). Before we proceed, we make a remark about some of the prime assumptions of Theorem A.

Remark 4.1

(Prime assumptions) If ${{\mathrm{ord}}}_{\mathfrak {p}}(j_{E}) < 0$, then we have that $E/{K_{\mathfrak {p}}}$ has a Tate parametrization. The second condition ${{\mathrm{ord}}}_{\mathfrak {p}}(\Delta _{E}) \not \equiv 0 \pmod \ell$ assists us in Lemma 4.2. In short, it allows us to understand ramification in the $\ell$-division field of $E_{K_{\mathfrak {p}}}$. The final condition $\chi _{H}(\mathfrak {p}) \ne 0$ is used in Lemma 4.3 and is an analogue of Frey’s condition that $p \equiv -1 \pmod \ell$. Moreover, this condition allows us to deduce, using Remark 2.2, that for a cyclic extension $M_2/K$ of degree $\ell$, $\mathfrak {p}$ is unramified in $M_2$.

The first step in the proof is to exhibit an element in ${{\mathrm{Sel}}}_{\ell }(E^d,K)$.

Lemma 4.2

Let $\ell > 3$ be a rational prime; let M / K be a non-abelian Galois extension of degree $2\ell$ containing $K(\sqrt{d})$ that is unramified over this field outside of $S_E$; let $\alpha$ be a generator of ${{\mathrm{Gal}}}(M/K(\sqrt{d}))$; and let $\phi$ the element in $H^1({{\mathrm{Gal}}}(M/K),E^d(M)[\ell ])$ determined by $\phi (\alpha ) = P$, where P is a K-rational point of order $\ell$. Then $\phi$ is an element of ${{\mathrm{Sel}}}_{\ell }(E^d,K)$.

Proof

First, we need to show that there exists some element

$$\begin{aligned} \phi \in H^1({{\mathrm{Gal}}}(M/K),E^d(M)[\ell ]) \end{aligned}$$

whose restriction $\overline{\phi }$ to ${{\mathrm{Gal}}}(M/K(\sqrt{d})) = \langle \alpha \rangle$ is given by $\overline{\phi }(\alpha ) = P$. We identify $E^d(M)[\ell ]$ with $E(M)[\ell ] = \langle P\rangle$. Since $E^d(K(\sqrt{d}))[\ell ] = \langle P \rangle$ and $\delta (P) = -P$ where $\langle \delta \rangle = {{\mathrm{Gal}}}(K(\sqrt{d})/K)$, we have invariance of $\phi$ under $\delta$ from the fact that $\delta \alpha \delta = \alpha ^{-1}$. Since

$$\begin{aligned} H^1({{\mathrm{Gal}}}(M/K),E^d(M)[\ell ]) = H^1({{\mathrm{Gal}}}(M/K(\sqrt{d})),E^d(M)[\ell ])^{\delta }, \end{aligned}$$

our assertions follows.

Hence it remains to show that $\overline{\phi }$ is locally trivial when regarded as an element of

$$\begin{aligned} H^1({{\mathrm{Gal}}}(M/K(\sqrt{d})),E^d(M)). \end{aligned}$$

We may restrict ourselves to primes $\mathfrak {P}_M | \mathfrak {l}\cdot N(E)$. By condition (1) of Theorem A, the divisors of $\mathfrak {q}$ are unramified in $M/K(\sqrt{d})$ if $\mathfrak {q}| N(E)$, and hence we may assume that $\mathfrak {P}_M \not \mid \mathfrak {q}$.

Assume that $\left( \frac{K(\sqrt{d})/K}{\mathfrak {p}} \right) = \delta$. In this case, $\mathfrak {P}_M$ is either fully ramified or decomposed (since M / K is non-abelian). So assume that $\mathfrak {P}_M$ is fully ramified and divides $\mathfrak {p}$. Then $\mathfrak {p}\in S_E$ and in particular $\mathfrak {p}\ne \mathfrak {l}$ and ${{\mathrm{ord}}}_{\mathfrak {p}}(\Delta _{E_K}) \ne 0 \pmod \ell$. We claim that $E^d/K_{\mathfrak {p}}(\sqrt{d})$ is a Tate curve and that P is contained in the connected component of the unity over $K_{\mathfrak {p}}(\sqrt{d})$ corresponding to an $\ell ^{\text {th}}$ root of unity $\zeta _{\ell }$ e.g., $P = \tau (\zeta _{\ell }^{\alpha })$ where $\tau$ is the Tate parametrization and $\alpha \in \left\{ 1,\ldots , \ell - 1 \right\}$.

The fact that $E^d/K_{\mathfrak {p}}(\sqrt{d})$ is a Tate curve follows since $\mathfrak {p}\in S_E$ and so ${{\mathrm{ord}}}_{\mathfrak {p}}(j_E) < 0$. Since ${{\mathrm{ord}}}_{\mathfrak {p}}(\Delta _{E}) \ne 0 \pmod \ell$, we know that adjoining $q^{1/\ell }$ to $K_{\mathfrak {p}}(\sqrt{d})$, where q is the $\mathfrak {p}$-adic period of E, is a non-trivial extension. Under the Tate parametrization $\tau$, we have that torsion points of order $\ell$ in $E^d(\overline{K_{\mathfrak {p}}(\sqrt{d})})[\ell ]$ are of the form $\tau (\zeta _{\ell }^{\alpha }q^{\beta /\ell })$ where $\alpha ,\beta \in \left\{ 1,\ldots , \ell - 1 \right\}$. Since P is a point of order $\ell$ defined over $K_{\mathfrak {p}}(\sqrt{d})$, we know that $\zeta _{\ell }^{\alpha } \in K_{\mathfrak {p}}(\sqrt{d})$ for some $\alpha \in \left\{ 1,\ldots , \ell -1 \right\}$ and that

$$\begin{aligned} \tau ^{-1}(P) = \zeta _{\ell }^{\alpha }q^{\beta /\ell } \in K_{\mathfrak {p}}(\sqrt{d}). \end{aligned}$$

In order for $\zeta _{\ell }^{\alpha }q^{\beta /\ell } \in K_{\mathfrak {p}}(\sqrt{d})$, we must have that $\beta = 0$ since q is not an $1/\ell ^{\text {th}}$ power. Thus, $\tau ^{-1}(P) = \zeta _{\ell }^{\alpha }$, and hence P is contained in the connected component of the unity over $K_{\mathfrak {p}}(\sqrt{d})$ corresponding to an $\ell ^{\text {th}}$ root of unity $\zeta _{\ell }$. Since $M_{\mathfrak {P}}/K_{\mathfrak {p}}(\sqrt{d})$ is cyclic of degree $\ell$, we have that $\zeta _{\ell } = \alpha x / x$ for some $x\in M_{\mathfrak {P}}$ by Hilbert’ Theorem 90, and therefore, $\overline{\phi }$ is trivial when considered in $H^1({{\mathrm{Gal}}}(M_{\mathfrak {P}}/K_{\mathfrak {p}}),E^d(M_{\mathfrak {P}}))$.

Next assume that $\left( \frac{K(\sqrt{d})/K}{\mathfrak {p}} \right) = {{\mathrm{id}}}$ and $\mathfrak {p}\ne \mathfrak {l}$. Then ${{\mathrm{ord}}}_{\mathfrak {p}}(j_{E}) < 0$ and E is a Tate curve over $K_{\mathfrak {p}}$, and so again P corresponds to some $\ell ^{\text {th}}$ root of unity $\zeta _{\ell }$ under the Tate parametrization of $E = E^d$ over $K_{\mathfrak {p}}(\zeta _{\ell })$ and hence $\overline{\phi }$ is split by $K_{\mathfrak {p}}(\zeta _{\ell })$ as seen above. But since the degree of $K_{\mathfrak {p}}(\zeta _{\ell })$ over $K_{\mathfrak {p}}$ is prime to $\ell$, $\overline{\phi }$ is split over $K_{\mathfrak {p}}$ already, and thus $\overline{\phi }$ is locally trivial.

There is one remaining case: $\mathfrak {p}= \mathfrak {l}$ and ${{\mathrm{ord}}}_{\mathfrak {l}}(j_{E}) \ge 0$. Let $\mathfrak {L}_M | \mathfrak {l}$. By the assumption, M / K is unramified at $\mathfrak {L}_M$, and we can find a normal extension N / K of degree prime to $\ell$ such that E has good reduction modulo all primes $\mathfrak {L}_N | \mathfrak {l}$. In particular, we may take $N = K(\zeta _{12},\root 12 \of {\mathfrak {l}})$. Now

$$\begin{aligned} H^1({{\mathrm{Gal}}}(M_{\mathfrak {L}}\cdot N/K_{\mathfrak {l}}\cdot N),E^d(M_{\mathfrak {L}}\cdot N)) = 0 \end{aligned}$$

since the reduction of $E^d$ modulo $\mathfrak {L}$ is good and $M_{\mathfrak {L}}N/K_{\mathfrak {l}}N$ is unramified, and hence it follows that

$$\begin{aligned} H^1({{\mathrm{Gal}}}(M_{\mathfrak {L}}/K_{\mathfrak {l}}),E^d(M_{\mathfrak {L}})) = 0. \end{aligned}$$

$\square$

Next, we look at the action of $\delta$ on $H_{S_E,u}(K(\sqrt{d}))$.

Lemma 4.3

The generator $\langle \delta \rangle = {{\mathrm{Gal}}}(K(\sqrt{d})/K)$ acts as $-{{\mathrm{id}}}$ on the Galois group $H_{S_E,u}(K(\sqrt{d}))$.

Proof

We may write

$$\begin{aligned} H_{S_E,u}(K(\sqrt{d})) = H^- \oplus H^+ \end{aligned}$$

where $H^-$ is the part where $\delta$ acts as $-{{\mathrm{id}}}$, and $H^+$ the part with $\delta = {{\mathrm{id}}}$. Let $\widetilde{M} := M_{S_E,u}^{H^-}$, which is the fixed field of $M_{S_E,u}$ by $H^-$. Assume that $M_1$ is a subfield of $\widetilde{M}$ that is cyclic over $K(\sqrt{d})$. Hence $M_1/K$ is cyclic of degree $2\cdot [M_1 {:} K(\sqrt{d})]$. Let $M_2$ be the cyclic extension of K with degree $[M_1:K(\sqrt{d})]$ contained in $M_1$. Then $M_2$ is unramifed outside of $S_E$. For $\mathfrak {p}\in S_E$, we have that $\chi _{H}(\mathfrak {p}) \ne 0$. Since $[M_2 {:} K] | \ell$ and $\ell \not \mid {{\mathrm{cl}}}(K)$, it follows that $M_2$ is not contained in the Hilbert class field of K and is unramified at all primes K. Thus, we have that $M_2 = K$, $M_1 = K(\sqrt{d})$ and hence $\widetilde{M} = K(\sqrt{d})$. $\square$

Proof of Theorem A

The divisibility of $\#{{\mathrm{Sel}}}_{\ell }(E^d,K)$ by ${{\mathrm{cl}}}_{S_E,u}(K(\sqrt{d}))[\ell ]$ follows from Lemmas 4.2, 4.3 since our element $\phi \in {{\mathrm{Sel}}}_{\ell }(E^d,K)$ is induced by $\alpha \in {{\mathrm{Gal}}}(M/K(\sqrt{d}))$ and the action of $\langle \delta \rangle$ on $H_{S_E,u}(K(\sqrt{d}))$ does not affect the order of $\alpha$ when considered as an element of $H_{S_E,u}(K(\sqrt{d})).$

5 Proof of Theorem B

Before we proceed with a proof of Theorem B, we wish to shed some light onto our assumptions. In general, our hypotheses allow us to control the ramification in cyclic extensions of $K(\sqrt{d})$.

Remark 5.1

(Field assumptions) We assume that our field K is a number field of degree $n\le 5$ such that $N_{K/\mathbf Q }(\mathfrak {q}) = 2 \text { for all }\mathfrak {q}|2$ and that for some $\ell \in S(n)\backslash \left\{ 2,3 \right\}$, $\ell \not \mid {{\mathrm{cl}}}(K)$ and $\zeta _{\ell } \notin K$. The degree and norm condition appear in Lemma 5.5 and allow us to deduce ramification conditions on prime divisors $\mathfrak {Q}_M|\mathfrak {q}$ where $M_1/K$ is cyclic. The condition that $\ell \not \mid {{\mathrm{cl}}}(K)$ implies that there does not exist an extension $M_2/K$ of degree $\ell$ contained in the Hilbert class field of K; once again this gives us a ramification consequence. The assumption that $\zeta _{\ell } \notin K$ is subtle, but it allows for more ramification possibilities since Kummer theory does not restrict cyclic extensions. The final condition that $e_{\mathfrak {l}}(K/\mathbf Q ) \ne 5$ when $[K{:}\mathbf Q ] = 5$ and $\ell = 5$ is due to a deep result of Katz [17] concerning the injectivity of $\ell$-torsion under the reduction map; the assumption $1>e_{\mathfrak {l}}(K/\mathbf Q )/(\ell -1) - 1$ from Theorem D is the general condition. This assumption allows us to use the fact that prime to 2 torsion will inject under the reduction map.

To prove Theorem B, it suffices to prove the divisibility statement

$$\begin{aligned} \# {{\mathrm{Sel}}}_{\ell }(E^d,K) \left| {{\mathrm{cl}}}_{\widetilde{S}_E,u}(K(\sqrt{d}))[\ell ]\cdot {{\mathrm{cl}}}_{S_E}(K')[\ell ](\chi _{\ell })\right. . \end{aligned}$$

To begin, we discuss the Galois structure of the $\ell$-division field of elliptic curves E / K from Theorem B.

5.1 Galois structure of splitting fields of $\ell$-covers of E

We want to determine the Galois group structure of splitting fields of elements in $H^1(G_K,E(\overline{K})[\ell ])$ for elliptic curves having a K-rational point P of order $\ell$. Recall that $\zeta _{\ell } \notin K$. Denote the $\ell$-division field by $K(E[\ell ])$; this is the field obtained by adjoining the x, y coordinates of all points of order $\ell$ of E to K. Then $K(E[\ell ])$ is a Galois extension of K containing $K(\zeta _{\ell })$, and it is cyclic over $K(\zeta _{\ell })$ of degree dividing $\ell$. From this point on, we shall abbreviate $E(\overline{K})[\ell ]$ with $E[\ell ]$, and similarly for $E^d[\ell ]$.

Lemma 5.2

The Galois group $K(E[\ell ])/K$ is generated by two elements $\overline{\gamma },\overline{\varepsilon }$ with $\overline{\gamma }^{\ell - 1} = {{\mathrm{id}}}$, $\overline{\varepsilon }^{\ell } = {{\mathrm{id}}}$, $\overline{ \gamma } | K(\zeta _{\ell })$ generates $K(\zeta _{\ell })/K$, and $\overline{\gamma }\overline{\varepsilon }\overline{\gamma }^{-1}= \overline{\varepsilon }^{\chi _{\ell }(\overline{\gamma })^{-1}}$.

Proof

Choose a base of the form $\left\{ P,Q \right\}$ of $E[\ell ]$ such that for $\sigma \in {{\mathrm{Gal}}}(K(E[\ell ])/K)$ the action of $\sigma$ on $E[\ell ]$ induces the matrix

$$\begin{aligned} \rho _{\sigma } = \left( \begin{array}{ll} 1 &{}\quad b\\ 0 &{}\quad a \end{array}\right) \in {{\mathrm{GL}}}_2(\mathbf F _{\ell }), \end{aligned}$$

with $a = \det (\rho _{\sigma }) \equiv \chi _{\ell }(\sigma ) \mod \ell$. Now we choose $\overline{\gamma }$ such that

$$\begin{aligned} \rho _{\overline{\gamma }} = \left( \begin{array}{ll} 1 &{}\quad 0\\ 0 &{}\quad w \end{array}\right) \in {{\mathrm{GL}}}_2(\mathbf F _{\ell }). \end{aligned}$$

with w a generator of $\left( \mathbf Z / \ell \mathbf Z \right) ^{\times }$. Also, we pick $\overline{\varepsilon } = {{\mathrm{id}}}$ if $K(E[\ell ]) = K(\zeta _{\ell })$. If $K(E[\ell ]) \ne K(\zeta _{\ell })$, we choose $\overline{\varepsilon }$ such that

$$\begin{aligned} \rho _{\overline{\varepsilon }} = \left( \begin{array}{ll} 1 &{}\quad 1 \\ 0 &{}\quad 1 \end{array}\right) \in {{\mathrm{GL}}}_2(\mathbf F _{\ell }). \end{aligned}$$

Then $\overline{\gamma }$ and $\overline{\varepsilon }$ generate ${{\mathrm{Gal}}}(K(E[\ell ])/K)$ and since

$$\begin{aligned} \left( \begin{array}{ll} 1 &{}\quad 0 \\ 0 &{}\quad w \end{array}\right) \left( \begin{array}{ll} 1 &{}\quad 1 \\ 0 &{}\quad 1 \end{array}\right) \left( \begin{array}{ll} 1 &{}\quad 0 \\ 0 &{}\quad w^{-1}\end{array}\right) = \left( \begin{array}{ll} 1 &{}\quad w^{-1}\\ 0 &{}\quad 1 \end{array}\right) = \left( \begin{array}{ll} 1 &{}\quad 1 \\ 0 &{}\quad 1 \end{array}\right) ^{w^{-1}} \end{aligned}$$

we have the relation $\overline{\gamma }\overline{\varepsilon }\overline{\gamma }^{-1}= \overline{\varepsilon }^{\chi _{\ell }(\overline{\gamma })^{-1}}$. $\square$

Remark 5.3

The choice of $\overline{\gamma }$ and $\overline{\varepsilon }$ is closely related to the choice of base $\left\{ P,Q \right\}$. In particular, we have $\overline{\varepsilon }(Q) = P + Q$ if $\overline{\varepsilon } \ne {{\mathrm{id}}}$ and $\overline{\gamma }(Q) = \chi _{\ell }(\overline{\gamma }) Q$.

Let $d \in \mathcal {O}_K^{\times }/(\mathcal {O}_K^{\times })^2$ be negative and relatively prime to $\mathfrak {l}\cdot N(E)$. We define $L_d$ to be the quadratic extension of $K(E[\ell ])$ given by the compositum $K(\sqrt{d})\cdot K(E[\ell ])$. The Galois group ${{\mathrm{Gal}}}(L_d/K)$ is generated by three elements $\delta , \gamma , \varepsilon$ with $\delta$ commuting with $\varepsilon$ and $\gamma$ and

$$\begin{aligned} \delta ^2&= {{\mathrm{id}}},&\delta (\sqrt{d}) = -\sqrt{d}, \\ \gamma ^{\ell - 1}&= {{\mathrm{id}}},&\gamma | K(E[\ell ]) = \overline{\gamma }, \\ \varepsilon ^{\ell }&= {{\mathrm{id}}},&\varepsilon | K(E[\ell ]) = \overline{\varepsilon }, \\ \gamma ^i\varepsilon ^j | K(\sqrt{d})&= {{\mathrm{id}}},&\gamma \varepsilon \gamma ^{-1}= \varepsilon ^{\chi _{\ell }(\gamma )^{-1}}. \end{aligned}$$

In particular, we have that $\delta$ operates as $-{{\mathrm{id}}}$ on $E^d[\ell ]$, the points of order $\ell$ of $E^d$. The fixed field of $\varepsilon$ is $K(\sqrt{d},\zeta _{\ell })$ and the fixed field of $\langle \varepsilon , \delta \gamma ^{(\ell - 1)/2} \rangle$ is $K'$ as defined in Theorem B. Thus, we have the following field diagram:

We now describe the elements in $H^1(G_K,E^d[\ell ])$. We have the exact inflation-restriction sequence

$$\begin{aligned} 0 \longrightarrow H^1({{\mathrm{Gal}}}(L_d/K),E^d[\ell ]) \overset{\text {inf.}}{\longrightarrow } H^1(G_K,E^d[\ell ]) \overset{\text {res.}}{\longrightarrow } H^1({{\mathrm{Gal}}}(\overline{K}/L_d),E^d[\ell ]) , \end{aligned}$$

where $H^1({{\mathrm{Gal}}}(\overline{K}/L_d),E^d[\ell ]) = {{\mathrm{Hom}}}_{{{\mathrm{Gal}}}(L_d/K)}({{\mathrm{Gal}}}(\overline{K}/L_d),E^d[\ell ]).$

Lemma 5.4

The group $H^1(G_K,E^d[\ell ])$ injects into ${{\mathrm{Hom}}}_{{{\mathrm{Gal}}}(L_d/K)}({{\mathrm{Gal}}}(\overline{K}/L_d),E^d[\ell ])$.

Proof

We need to show that $H^1({{\mathrm{Gal}}}(L_d/K),E^d[\ell ]) = 0$. If $\varepsilon = {{\mathrm{id}}}$, the degree of $L_d/K$ is prime to $\ell$, and the assertion follows. Now let $\varepsilon$ be of order $\ell$. Using the inflation-restriction sequence, one has that

$$\begin{aligned} H^1({{\mathrm{Gal}}}(L_d/K),E^d[\ell ]) = H^1(\langle \varepsilon \rangle ,E^d[\ell ])^{\langle \delta , \gamma \rangle }. \end{aligned}$$

Let $P_d,Q_d$ be the points of order $\ell$ of $E^d[\ell ]$ corresponding to $P,Q \in E[\ell ]$ . Then $P_d = \varepsilon Q_d - Q_d,$ and hence $H^1(\langle \varepsilon \rangle , E^d[\ell ])$ is generated by the class of cocycle $\psi$ which sends $\varepsilon$ to $Q_d$. Since $\delta \varepsilon \delta = \varepsilon$ and $\delta Q_d = -Q_d$, we have that $\psi \notin H^1(\langle \varepsilon \rangle ,E^d[\ell ])^{\langle \delta \rangle }$, and thus $H^1({{\mathrm{Gal}}}(L_d/K),E^d[\ell ]) = H^1(\langle \varepsilon \rangle ,E^d[\ell ])^{\langle \delta , \gamma \rangle } = 0$. $\square$

Take an element $\widetilde{\Phi } \in H^1(G_K,E^d[\ell ])$ with

$$\begin{aligned} {{\mathrm{res}}}\widetilde{\Phi } = \phi \in {{\mathrm{Hom}}}_{{{\mathrm{Gal}}}(L_d/K)}({{\mathrm{Gal}}}(\overline{K}/L_d),E^d[\ell ]) \end{aligned}$$

and denote by M the fixed field of the kernel of $\phi$. M / K is normal and ${{\mathrm{Gal}}}(M/L_{d})$ is possibly generated by two elements $\alpha _1,\alpha _2$ with $\alpha _i^{\ell } = {{\mathrm{id}}}$, which we may choose in such a way that

$$\begin{aligned} \phi (\alpha _1) = \mu _1P \quad \text { and }\quad \phi (\alpha _2) = \mu _2Q . \end{aligned}$$

We may also assume that $\mu _i = 1$ if $\alpha _i \ne {{\mathrm{id}}}$.

We extend $\delta ,\gamma ,\varepsilon \in {{\mathrm{Gal}}}(L_{d}/K)$ to elements $\widetilde{\delta }, \widetilde{\gamma }, \widetilde{\varepsilon } \in {{\mathrm{Gal}}}(M/K)$ and compute the actions of these elements on $\alpha _i$. We assume that $\widetilde{\delta }^2 = \widetilde{\gamma }^{\ell -1} = {{\mathrm{id}}}$. Since

$$\begin{aligned} \phi (\beta \alpha _i\beta ^{-1}) = \beta \phi (\alpha _i) \quad \forall \, \beta \in {{\mathrm{Gal}}}(M/K) \end{aligned}$$

via the fact that $\phi$ is a group homomorphism and the cocycle condition we get:

$$\begin{aligned} \widetilde{\delta }\alpha _i\widetilde{\delta }&= \alpha _i^{-1}&(\because \, \widetilde{\delta }| E^d[\ell ] = -{{\mathrm{id}}}) \\ \widetilde{\gamma }\alpha _1\widetilde{\gamma }^{-1}&= \alpha _1&(\because \,\widetilde{\gamma }P = P), \\ \widetilde{\gamma }\alpha _2\widetilde{\gamma }^{-1}&= \alpha _2^{\chi _{\ell }(\widetilde{\gamma })}&(\because \,\widetilde{\gamma }Q = \chi _{\ell }(\widetilde{\gamma })Q), \\ \widetilde{\varepsilon }\alpha _1\widetilde{\varepsilon }^{-1}&= \alpha _1&(\because \, \widetilde{\varepsilon }P = P), \\ \widetilde{\varepsilon }\alpha _2\widetilde{\varepsilon }^{-1}&= \alpha _1\alpha _2&\text { if }\, \varepsilon \ne {{\mathrm{id}}}\text { and }\alpha _2\ne {{\mathrm{id}}}(\because \text { then } \varepsilon \phi (\alpha ) = \\&&\varepsilon P = P+Q = \phi (\alpha _1\alpha _2); \text { necessarily } \alpha _1 \ne {{\mathrm{id}}}\\&&\text {in this case}). \end{aligned}$$

In particular, it follows that $\langle \alpha _1 \rangle$ is a normal subgroup of ${{\mathrm{Gal}}}(M/K)$ and that $\langle \alpha _2 \rangle$ is normal if either $\alpha _2 = {{\mathrm{id}}}$ or $\widetilde{\varepsilon } = {{\mathrm{id}}}$.

Now we distinguish between two cases:

Case 1 $\widetilde{\varepsilon } = {{\mathrm{id}}}$. In this case $\langle \alpha _1 \rangle$ and $\langle \alpha _2 \rangle$ are both normal in ${{\mathrm{Gal}}}(M/K)$ and hence

$$\begin{aligned} M_i := M^{\langle \alpha _i \rangle } \end{aligned}$$

are normal extensions of K. The Galois group of $M_2/K(\sqrt{d})$ is abelian and generated by the restriction of $\langle \widetilde{\gamma },\alpha _1 \rangle$ to $M_2$. Hence

$$\begin{aligned} \overline{M}_2 := M^{\langle \alpha _2,\widetilde{\gamma } \rangle } \end{aligned}$$

is Galois over K containing $K(\sqrt{d})$ and if $\alpha _1 \ne {{\mathrm{id}}}$, then ${{\mathrm{Gal}}}(\overline{M}_2/K)$ is non-abelian of order $2\ell$. Since

$$\begin{aligned} \widetilde{\delta }\widetilde{\gamma }^{(\ell -1)/2}\alpha _2(\widetilde{\delta } \widetilde{\gamma }^{(\ell -1)/2})^{-1}= \alpha _2, \end{aligned}$$

it follows that $M_1$ is abelian over $K'$ and hence

$$\begin{aligned} \overline{M}_1 := M^{\langle \alpha _1,\widetilde{\delta }\widetilde{\gamma }^{(\ell -1)/2}\rangle } \end{aligned}$$

is normal over K. Its Galois group is generated by

$$\begin{aligned} \overline{\alpha }_2 = \alpha _2 | \overline{M}_1 \quad \text { and }\quad \overline{\gamma } = \widetilde{\gamma } | \overline{M}_1, \end{aligned}$$

and its order is equal to $|\alpha _2|\cdot (\ell -1)$. Also one has the relation $\overline{\gamma }\overline{\alpha _2}\overline{\gamma }^{-1}= \overline{\alpha }_2^{\chi _{\ell }(\overline{\gamma })}.$ To summarize, we have that

$$\begin{aligned} \overline{M}_1(\phi )&:= M^{\langle \alpha _1,\widetilde{\delta }\widetilde{\gamma }^{(\ell -1)/2}\rangle }, \\ \overline{M}_2(\phi )&:= M^{\langle \alpha _2,\widetilde{\gamma } \rangle }. \end{aligned}$$

Case 2 $|\widetilde{\varepsilon } | = \ell$. In this case, we may assume that $\alpha _1 \ne {{\mathrm{id}}}$ for $\alpha _1 = {{\mathrm{id}}}$ implies that $\alpha _2 = {{\mathrm{id}}}$, as well.

Subcase (i) $\alpha _2 = {{\mathrm{id}}}$. We assert that ${{\mathrm{Gal}}}(M/K(\zeta _{\ell },\sqrt{d}))$ is not cyclic. Otherwise $\widetilde{\varepsilon }$ would be an element of order $\ell ^2$ with $\widetilde{\varepsilon }^{\,\ell } = \alpha _1$ (without lose of generality). So $\widetilde{\delta }\widetilde{\varepsilon }^{\,\ell }\widetilde{\delta } = \widetilde{\varepsilon }^{-\ell }$ and hence

$$\begin{aligned} \widetilde{\delta }\widetilde{\varepsilon }\widetilde{\delta } = \widetilde{\varepsilon }^{\,k} \quad \text {with }k\equiv -1\mod \ell . \end{aligned}$$

But since $\delta \varepsilon \delta = \varepsilon$, we would get $\delta \widetilde{\varepsilon }\delta = \widetilde{\varepsilon }\cdot (\widetilde{\varepsilon }^{\ell })^n = \widetilde{\varepsilon }^{\,1+\ell ^n}$ which gives a contradiction. Hence, we can choose $\widetilde{\varepsilon }$ so that

$$\begin{aligned} \widetilde{\varepsilon }^{\,\ell } = \widetilde{\alpha }_1^{\,\ell } = {{\mathrm{id}}}\quad \text { and }\quad \widetilde{\delta }\widetilde{\varepsilon }\widetilde{\delta } = \widetilde{\varepsilon }, \end{aligned}$$

which determines $\widetilde{\varepsilon }$ uniquely. Thus, $\overline{M}_2 := M^{\langle \widetilde{\varepsilon } , \widetilde{\gamma }\rangle }$ is normal over K and contains $K(\sqrt{d})$ and its Galois group is dihedral of order $2\ell$ and generated by $\langle \alpha _1,\widetilde{\delta } \rangle$. To summarize, we say that

$$\begin{aligned} \overline{M}_1(\phi )&:= M^{\langle \alpha _1,\widetilde{\delta }\widetilde{\gamma }^{(\ell -1)/2}\rangle }, \\ \overline{M}_2(\phi )&:= M^{\langle \widetilde{\varepsilon } , \widetilde{\gamma }\rangle }. \end{aligned}$$

Subcase (ii) $\alpha _2 \ne {{\mathrm{id}}}$. We have that $M_1 := M^{\langle \alpha _1\rangle }$ is normal over K and of degree $\ell$ over $L_d$. Since $\widetilde{\delta }\alpha _2\widetilde{\delta } = \alpha _2^{-1}$, we conclude as above that $\varepsilon$ has an extension $\widetilde{\varepsilon }$ to $M_1$ of order $\ell$ with $\widetilde{\delta }\widetilde{\varepsilon }\widetilde{\delta } = \widetilde{\varepsilon }$. Since $\widetilde{\delta }\widetilde{\gamma }^{(\ell -1)/2}$ acts trivially on $\alpha _2$ and $\widetilde{\varepsilon }$ acts trivially on $\alpha _2 | M_1$, we have that $\langle \widetilde{\delta }\widetilde{\gamma }^{(\ell -1)/2},\widetilde{\varepsilon } \rangle$ is a normal subgroup of ${{\mathrm{Gal}}}(M_1/K)$. Hence

$$\begin{aligned} \overline{M}_1 := M_1^{\langle \widetilde{\delta }\widetilde{\gamma }^{(\ell -1)/2},\widetilde{\varepsilon } \rangle } \end{aligned}$$

is normal over K containing $K'$, and its Galois group over $K'$ is generated by $\overline{\alpha }_2 =\alpha _2 | \overline{M}_1$, which is of order $\ell$ and satisfies the relation

$$\begin{aligned} \overline{\gamma }\overline{\alpha }_2\overline{\gamma }^{-1}= \overline{\alpha }_2^{\chi _{\ell }(\overline{\gamma })} \quad \text {with }\overline{\gamma } = \widetilde{\gamma } | K'. \end{aligned}$$

In order to simplify notation, we define $\overline{M}_2(\phi ) := K(\sqrt{d})$ if either $\varepsilon \ne {{\mathrm{id}}}$ or $\alpha _2 \ne {{\mathrm{id}}}$. To summarize, we say that

$$\begin{aligned} \overline{M}_1(\phi )&:= M_1^{\langle \widetilde{\delta }\widetilde{\gamma }^{(\ell -1)/2},\widetilde{\varepsilon } \rangle }, \\ \overline{M}_2(\phi )&:= K(\sqrt{d}). \end{aligned}$$

Hence for a given

$$\begin{aligned} \widetilde{\Phi } \in H^1(G_K,E^d[\ell ]) \end{aligned}$$

we have a field $M = M(\phi )$ which determines $\langle \phi \rangle$ completely where $\phi = {{\mathrm{res}}}(\widetilde{\Phi })$. We want to study the information we attain from the pair $(\overline{M}_1(\phi ),\overline{M}_2(\phi ))$. If $\varepsilon = {{\mathrm{id}}}$ or $\alpha _2 = {{\mathrm{id}}}$, then we get back $M(\phi )=M$ from $(\overline{M}_1(\phi ),\overline{M}_2(\phi ))$. In these cases, we shall say that $\phi$ is of first type. What happens if $\varepsilon \ne {{\mathrm{id}}}$ and $\alpha _2 \ne {{\mathrm{id}}}$?

Assume that

$$\begin{aligned} \phi \ne \psi \in H^1(G_K,E^d[\ell ]) \end{aligned}$$

have fields $M(\phi )$ and $M(\psi )$ with Galois groups $\langle \alpha _1,\alpha _2 \rangle$ and $\langle \beta _1 , \beta _2\rangle$ as above such that

$$\begin{aligned} M(\phi )^{\alpha _1} = M(\psi )^{\beta _1}. \end{aligned}$$

Let N be the composite of $M(\phi )$ and $M(\psi )$. Then the Galois group ${{\mathrm{Gal}}}(N/L_d)$ is generated by three elements $\langle \sigma _1,\sigma _2,\sigma _3 \rangle$, which we can choose in such a way that

$$\begin{aligned} \sigma _1 | M(\phi ) = \alpha _1 ,\quad \sigma _1 |M(\psi ) = \beta _1^{\lambda } \end{aligned}$$

$$\begin{aligned} \sigma _2 | M(\phi ) = \alpha _2 ,\quad \sigma _2 |M(\psi ) = \beta _2^{\lambda } \end{aligned}$$

where $\lambda \in \left\{ 1,\ldots , \ell -1 \right\}$. N is a splitting field of $\phi$ and $\psi$, and

$$\begin{aligned} (\phi - \lambda ^{\ell - 1}\psi )(\sigma _1) = (\phi - \lambda ^{\ell - 1}\psi )(\sigma _2) = 0. \end{aligned}$$

Hence the fixed field of the kernel of $\phi - \lambda \psi$ is a cyclic extension of $L_d$ which is normal over K, and $\phi - \lambda ^{-1}\psi$ is of first type.

Thus, $\overline{M}_1(\phi )$ determines $\langle \phi \rangle$ up to elements of first type, and in order to determine all elements in $H^1(G_K,E^d[\ell ])$, it is enough to determine all dihedral extensions of K of degree $2\ell$ containing $K(\sqrt{d})$ and all extensions $M_1$ of degree $\ell$ over $K'$ which are normal over K such that conjugation by $\overline{\gamma }$ on ${{\mathrm{Gal}}}(\overline{M}_1,K')$ is equal to $\chi _{\ell }(\overline{\gamma })$.

Therefore to prove the double divisibility, one has to show that for $\phi \in {{\mathrm{Sel}}}_{\ell }(E^d,K)$, the field $\overline{M}_2(\phi )$ is unramified over $K(\sqrt{d})$ outside $\widetilde{S}_E$, and $\overline{M}_1(\phi )$ is unramifed over $K'$ outside $S_E$ and little ramified at divisors of $\mathfrak {l}$.

5.2 Splitting fields of elements in ${{\mathrm{Sel}}}_{\ell }(E^d,K)$

We shall continue to use the assumptions and the notations of the Theorem B and Sect. 5.1.

Lemma 5.5

Let $\phi$ be an element in ${{\mathrm{Sel}}}_{\ell }(E^d,K)$. Then $\overline{M}_1(\phi ) =: \overline{M}_1$ is unramified at $\mathfrak {q}$ over $K'$ and $\overline{M}_2(\phi ) =: \overline{M}_2$ is unramified at $\mathfrak {q}$ over $K(\sqrt{d})$.

Proof

We first prove the latter statement. Since $\mathfrak {q}| \Delta _{K(\sqrt{d})/K}\,$, we have that $K(\sqrt{d})$ and $K'$ are ramified at $\mathfrak {q}$ over K. Hence the norm of $\mathfrak {Q}|\mathfrak {q}$ in $K(\sqrt{d})$ is equal to $\mathfrak {q}$, and by assumption the norm of $\mathfrak {Q}|2$ is equal to 2. Suppose that $K(\sqrt{d})$ had a cyclic extension of degree $\ell$ in which $\mathfrak {Q}$ is ramified. Then the completion $K(\sqrt{d})_{\mathfrak {Q}}$ admits a cyclic extension of degree $\ell$ ramified at $\mathfrak {Q}$. Since $\ell$ is odd and $\mathfrak {Q}$ has residue characteristic two, this extension is tamely ramified. By local class field theory, the tamely ramified cyclic extensions of a local field $K(\sqrt{d})_{\mathfrak {Q}}$ all have degree dividing $|\kappa ^{\times }|$, where $\kappa$ is the residue field. Since $\kappa = \mathbf F _2$, we have that there are no tamely ramified and ramified extensions of $K(\sqrt{d})_{\mathfrak {Q}}$. Thus, $K(\sqrt{d})$ has no cyclic extension of degree $\ell$ in which $\mathfrak {Q}$ ramifies, and hence $\overline{M}_2$ is unramified at $\mathfrak {q}$ over $K(\sqrt{d})$.

To prove the former statement, we shall utilize the proof of [2, Lemma 3] and look prime by prime. For $\ell = 5$, the same argument as above can be applied to $\mathfrak {Q}_{K'}|\mathfrak {q}$. For $\ell = 7$, there is only one extension $\mathfrak {Q}|\mathfrak {q}$ to $K'$ which is ramified of order 2 and has norm 8. Assume that $\mathfrak {Q}_{K'}$ is ramified in $\overline{M}_1/K'$ and let $\mathfrak {Q}_{\overline{M}_1}$ be the unique extension of $\mathfrak {Q}_{K'}$ to $\overline{M}_1$. Let $M_t$ be the subfield of $\overline{M}_1$ in which $\mathfrak {Q}_{\overline{M}_1}$ is tamely ramified. Then $M_t$ is a cyclic extension of degree 7 over $K(\zeta _{7} + \zeta _{7}^{-1})$, and $\overline{M}_1$ is the compositum of $M_t$ with $K'$ over $K(\zeta _{7} + \zeta _{7}^{-1})$. Thus, ${{\mathrm{Gal}}}(\overline{M}_1/K(\zeta _{7} + \zeta _{7}^{-1}))$ is abelian. But this contradicts the fact that

$$\begin{aligned} \overline{\gamma }^{3}\overline{\alpha }\overline{\gamma }^{3} = \overline{\alpha }^{\chi _{7}(\overline{\gamma }^{3})} = \overline{\alpha }^{-1}, \end{aligned}$$

where $\langle \overline{\alpha }\rangle = {{\mathrm{Gal}}}(\overline{M}_1/K')$ and $\langle \overline{\gamma }\rangle = {{\mathrm{Gal}}}(K'/K)$.

For $\ell = 11,13,19,37$, we can use the same proof as the first statement since

$$\begin{aligned} 11&\not \mid (2^5 - 1)&13&\not \mid (2^2 - 1)&37&\not \mid (2^{3} - 1)&37&\not \mid (2^{18} - 1) \\ 13&\not \mid (2^6 - 1)&19&\not \mid (2^9 - 1)&37&\not \mid (2^{9} - 1)&37&\not \mid (2^{2} - 1). \\ 13&\not \mid (2^3 - 1)&19&\not \mid (2^3 - 1)&37&\not \mid (2^{6} - 1)&\end{aligned}$$

For $\ell = 17$, there there is only one extension $\mathfrak {Q}|\mathfrak {q}$ to $K'$ which is ramified of order 2 and has norm $2^8$ (note that $17|(2^8 - 1)$). If we assume that $\mathfrak {Q}_{K'}$ is ramified in $\overline{M}_1/K'$, then we can use the above argument to construct the same contradiction. $\square$

Remark 5.6

Since $73 | (2^{36} - 1),$ $73 | (2^9 - 1)$, and $73 | (2^{18} - 1)$, we may not assume that there is a unique cyclic extension of $K'$ with degree 73 in which $\mathfrak {Q}$ is ramified, and hence the above argument does work for $\ell = 73$. This precludes us from extending Theorem B to number fields K of degree $\ge 6$.

Therefore, we can assume that $\mathfrak {p}\not \mid \mathfrak {q}\cdot \mathfrak {l}$, but $\mathfrak {p}| N(E)$.

Lemma 5.7

Let $\phi$ be an element in ${{\mathrm{Sel}}}_{\ell }(E^d,K)$. Then $\overline{M}_1/K'$ is unramified outside of $S_E \cup \left\{ \mathfrak {l} \right\}$ and $\overline{M}_2/K(\sqrt{d})$ is unramified outside $\widetilde{S}_E \cup \left\{ \mathfrak {l} \right\}$.

Proof

We have to test prime numbers $\mathfrak {p}\ne \mathfrak {l}$ that divide N(E).

• 1. If ${{\mathrm{ord}}}_{\mathfrak {p}}(j_E) \ge 0$, then it follows from Néron’s list of minimal models of elliptic curves with potentially good reduction that $\ell$ must be equal to 3 [18, p.124]. Since we only consider primes $\ell > 3$, we can exclude this case from consideration.

• 2. Now assume that ${{\mathrm{ord}}}_{\mathfrak {p}}(j_E) < 0$. We have two subcases:

  • (a) If ${{\mathrm{ord}}}_{\mathfrak {p}}(j_E) \equiv 0 \mod \ell$, we have that $\mathfrak {p}\notin S_E$ and so $E^d$ is not a Tate curve over $K_{\mathfrak {p}}.$ Moreover, $K_{\mathfrak {p}}(E[\ell ])$ is unramified over $K_{\mathfrak {p}}$ and hence $\overline{M_1}/K'$ and $\overline{M_2}/K(\sqrt{d})$ are unramified at all divisors of $\mathfrak {p}$ if and only if $M_1/L_d$ (resp. $M_2/L_d$) are unramified at all divisors of $\mathfrak {p}$. We now use the triviality of the $\phi \in {{\mathrm{Sel}}}_{\ell }(E^d,K)$ over $K_{\mathfrak {p}}$ from Lemma 4.2. Also recall that M is the fixed field of the kernel of $\phi$. We shall show that $\mathfrak {Q}_M$ is unramified over $L_d$. There is a $\widetilde{P} \in E^d(M_{\mathfrak {P}})$ where $\mathfrak {P}_M|\mathfrak {p}$ such that for all $\sigma \in D(\mathfrak {P}_M)$, we have $\sigma \widetilde{P} - \widetilde{P} = \phi (\sigma )$. Hence

    $$\begin{aligned} P' := \ell \cdot \widetilde{P} \in E^d(K_{\mathfrak {p}}) \end{aligned}$$

    and so $2 P'$ is in the connected component of unity modulo $\mathfrak {p}$ via Remark 2.6. Hence $\widetilde{P} = \widetilde{P}_1 + P_2$ with $P_2 \in E^d[\ell ]$ and $2\widetilde{P}_1$ in the component of the unity of $E \mod \mathfrak {P}_M$, so $\widetilde{P}_1$ corresponds to a $\mathfrak {P}_M$-adic unity u under the Tate parametrization. Now take

    $$\begin{aligned} \alpha \in \langle \alpha _1,\alpha _2 \rangle \cap I(\mathfrak {P}_M) \end{aligned}$$

    where $I(\mathfrak {P}_M)$ is the interia group of $\mathfrak {P}_M$. Then $2(\alpha \widetilde{P} - \widetilde{P})$ corresponds to $\alpha u / u$ and is an $\ell ^{\text {th}}$ root of unity. By Hilbert’s Theorem 90, we have that $\alpha = {{\mathrm{id}}}$, and thus, $\mathfrak {P}_M$ is unramified over $L_d$.

  • (b) If ${{\mathrm{ord}}}_{\mathfrak {p}}(j_E) \not \equiv 0\mod \ell$, then the values at the Hecke characters $\chi$ of order $\ell$ tell us that either E is a Tate curve over $K_{\mathfrak {p}}$ or that $\mathfrak {p}\in S_E$. Consider the former situation. Our assumptions from Theorem B tell us that $\mathfrak {q}$ is not completely decomposed in $K(\sqrt{d})$ and $K'$. Since

    $$\begin{aligned} K_{\mathfrak {p}}^{\times }/(K_{\mathfrak {p}}^{\times })^{\ell }\cong K_{\mathfrak {p}}(\sqrt{d})^{\times }/(K_{\mathfrak {p}}(\sqrt{d})^{\times })^{\ell } \cong K'^{\times }_{\mathfrak {P}}/(K'^{\times }_{\mathfrak {P}})^{\ell } \end{aligned}$$

    for all $\mathfrak {P}_{K'}|\mathfrak {p}$, we see that for all cyclic extensions $\overline{M}_1$ of $K'$ and $\overline{M}_2/K(\sqrt{d})$ of degree $\ell$ and divisors $\mathfrak {P}_{M_i}|\mathfrak {p}$, one has that ${{\mathrm{Gal}}}(\overline{M}_{i,{\mathfrak {P}_{M_i}}}/K_{\mathfrak {q}})$ is abelian of even order. But this implies that

    $$\begin{aligned} \overline{M}_{1,\mathfrak {P}} = K_{\mathfrak {p}}' \quad \text { and }\quad \overline{M}_{2,\mathfrak {P}} = K_{\mathfrak {P}}(\sqrt{d}), \end{aligned}$$

    which is absurd. Thus $\mathfrak {p}\in S_E$ and our lemma follows.$\square$

The next step is to describe the behavior of $\overline{M}_i$ at divisors of $\mathfrak {l}$.

Lemma 5.8

Assume that ${{\mathrm{ord}}}_{\mathfrak {l}}(j_E) < 0$ and $\phi \in {{\mathrm{Sel}}}_{\ell }(E^d,K)$. Then $\overline{M}_2/K(\sqrt{d})$ is unramified at $\mathfrak {l}$ and $\overline{M}_1/K'$ is little ramified at divisors of $\mathfrak {l}$.

Proof

The assumptions tells us that $E/K_{\mathfrak {l}}$ is a Tate curve but that $E^d/K_{\mathfrak {l}}$ is not a Tate curve. Since $K_{\mathfrak {l}}(E[\ell ]) = K_{\mathfrak {l}}(\zeta _{\ell })$, the behavior of $\overline{M}_i$ at $\mathfrak {l}$ is determined by the behavior of M at $\mathfrak {l}$. Let $\mathfrak {L}_M|\mathfrak {l}$, let $I(\mathfrak {L}_M)$ be the inertia group of $\mathfrak {L}_M$, and let

$$\begin{aligned} \alpha \in \langle \alpha _1,\alpha _2 \rangle \cap I(\mathfrak {L}_M). \end{aligned}$$

As in the proof of Lemma 5.7, we can use the fact that $E^d/K_{\mathfrak {l}}$ is not a Tate curve to show that there is a $\widetilde{Q} \in E^d(M_{\mathfrak {L}})$ where $\mathfrak {L}_M|\mathfrak {l}$ and $\alpha \widetilde{Q} - \widetilde{Q} = \phi (\alpha )$. Hence $2 \widetilde{Q}$ is in the connected component of unity modulo $\mathfrak {L}_M$ via Remark 2.6. This implies that

$$\begin{aligned} M_{\mathfrak {L}_M} = M_{\mathfrak {L}_M}^{\langle \alpha \rangle }(\root \ell \of {u}) \end{aligned}$$

where u is a $\mathfrak {L}_M$-adic unit corresponding to $2\widetilde{Q}$ under the Tate parametrization. Moreover, $M_1/L_d$ is little ramified at $\mathfrak {l}$.

Now assume that $\alpha _2 = {{\mathrm{id}}}$ or $\varepsilon = {{\mathrm{id}}}$. Then $\overline{M}_2/K(\sqrt{d})$ is of degree $\ell$, and we have to show that $\overline{M}_2/K(\sqrt{d})$ is unramified at $\mathfrak {L}_{\overline{M}_2}|\mathfrak {l}$. We recall the choice of point Q. Since $\gamma Q = \chi _{\ell }(\gamma )Q$ where $\langle \gamma \rangle = {{\mathrm{Gal}}}(K(\zeta _{\ell })/K)$, it follows that Q is in the kernel of the reduction of E modulo all divisors of $\mathfrak {l}$, and hence $P + \lambda Q$ is not in this kernel where $\lambda \in \mathbf N$. For $\alpha \in I(\mathfrak {L}_M)$, we saw that $\sigma \widetilde{Q} - \widetilde{Q} = \phi (\sigma )$ is in the kernel of the reduction modulo $\mathfrak {L}_M$, and hence

$$\begin{aligned} \alpha _1\alpha _2^{\lambda } \notin I(\mathfrak {L}_M) \quad \forall \, \lambda \in \mathbf N \text { and }\mathfrak {L}_M | \mathfrak {l}. \end{aligned}$$

Thus, it follows that $M^{\langle \alpha _2 \rangle }/L_d$ is unramified at $\mathfrak {L}_M$ and $\overline{M}_2/K(\sqrt{d})$ is unramified at $\mathfrak {l}$. $\square$

Finally, we look at the case where ${{\mathrm{ord}}}_{\mathfrak {l}}(j_E) \ge 0$.

Lemma 5.9

Assume that E / K has a K-rational point P of order $\ell > 3$, that ${{\mathrm{ord}}}_{\mathfrak {l}}(j_E) \ge 0$, and that P is not contained in the kernel of reduction modulo $\mathfrak {l}$, in particular, this means that E is not supersingular modulo $\mathfrak {l}$. Let $\phi$ be an element in ${{\mathrm{Sel}}}_{\ell }(E^d,K)$ with corresponding fields $\overline{M}_1$ and $\overline{M}_2$. Then $\overline{M}_1/K'$ is little ramified at $\mathfrak {l}$, and $\overline{M}_2/K(\sqrt{d})$ is unramified at $\mathfrak {l}$.

Proof

Suppose that ${{\mathrm{ord}}}_{\mathfrak {l}}(j_E) \ge 0$, which implies that E has potentially good reduction at $\mathfrak {l}$. Since E / K has a K-rational point P of order $\ell >3$, we know that ${{\mathrm{Gal}}}(K(E[\ell ])/K(\zeta _{\ell }))$ is a subgroup of the additive group $\mathbf F _{\ell }^+$. We want to show that all divisors of $\mathfrak {l}$ are not ramified in $K(E[\ell ])/K(\zeta _{\ell })$. If E has good reduction over $K(\zeta _{\ell })$, then we are immediately done. If E does not have good reduction over $K(\zeta _{\ell })$, then there must exist some extension $N/K(\zeta _{\ell })$ such that $[N{:}K(\zeta _{\ell })] | 6$ and that E has good reduction at all divisors $\mathfrak {L}_N | \mathfrak {l}$; this divisibility condition is similar to the proof of [1, Proposition VII.5.4.c]. From our assumptions, it follows that $N_{\mathfrak {L}}$ contains $K(E[\ell ])$ and that $\langle Q \rangle$ is the subgroup of order $\ell$ of the kernel of reduction modulo $\mathfrak {L}_N$. Hence all divisors of $\mathfrak {l}$ are not ramified in $K(E[\ell ])/K(\zeta _{\ell })$, and we can prove the lemma by looking at the behavior of $\mathfrak {l}$ in $M/L_d$.

Assume that $\mathfrak {L}_M|\mathfrak {l}$ and let $I(\mathfrak {L}_M)$ be the inertia group of $\mathfrak {L}_M$. Suppose that $\alpha _1^{\mu }\alpha _2^{\lambda }\in I(\mathfrak {L}_M).$ There there is a $\widetilde{P} \in E(M_{\mathfrak {L}})$ with

$$\begin{aligned} (\alpha _1^{\mu }\alpha _2^{\lambda })\widetilde{P} - \widetilde{P} = \mu P + \lambda Q. \end{aligned}$$

But we know that for $\mu \ne 0$, the point $\mu P + \lambda Q$ is not in the kernel of reduction modulo $\mathfrak {L}_M$. Let $\widetilde{E}$ be a model of E over N having good reduction modulo $\mathfrak {L}_N|\mathfrak {l}$. Since $(I(\mathfrak {L}_M) - {{\mathrm{id}}})\widetilde{E}(N\cdot M_{\mathfrak {L}})$ is contained in this kernel, we must have that $\mu = 0$, and hence

$$\begin{aligned} I(\mathfrak {L}_M) \cap {{\mathrm{Gal}}}(M/L_d) \subseteq \langle \alpha _2 \rangle . \end{aligned}$$

Thus, $M^{\langle \alpha _2 \rangle }/L_d$ is unramified at $\mathfrak {L}_M$; moreover, $\overline{M}_2/K(\sqrt{d})$ is unramified above $\mathfrak {l}$.

Now assume that $I(\mathfrak {L}_M) = \langle \alpha _2 \rangle$. Then $Q = \alpha _2\widetilde{Q} - \widetilde{Q}$ and since $\langle \alpha _2 \rangle$ acts trivially on $\widetilde{E}(N\cdot M_{\mathfrak {L}})/\widetilde{E}_-(N\cdot M_{\mathfrak {L}})$, we may assume that $\widetilde{Q} \in \widetilde{E}_-(N\cdot M_{\mathfrak {L}})$ and hence $\ell \cdot \widetilde{Q} \in \widetilde{E}_-(N\cdot K_{\mathfrak {l}}).$ Since $\widetilde{E}$ has ordinary reduction modulo $\mathfrak {L}_M$, we have that $N\cdot K_{\mathfrak {l}}(\widetilde{Q})$ is little ramified at divisors of $\mathfrak {l}$. Thus, our lemma follows. $\square$

Lemmas 5.5, 5.7, 5.8 and 5.9 prove that for $\phi \in {{\mathrm{Sel}}}_{\ell }(E^d,K)$, the field $\overline{M}_2(\phi )$ is unramified over $K(\sqrt{d})$ outside $\widetilde{S}_E \cup \left\{ \mathfrak {l} \right\}$, and $\overline{M}_1(\phi )$ is unramifed over $K'$ outside $S_E$ and little ramified at divisors of $\mathfrak {l}$. Moreover, we have proved that

$$\begin{aligned} \# {{\mathrm{Sel}}}_{\ell }(E^d,K) \left| {{\mathrm{cl}}}_{\widetilde{S}_E,u}(K(\sqrt{d}))[\ell ]\cdot {{\mathrm{cl}}}_{S_E}(K')[\ell ](\chi _{\ell }) \right. , \end{aligned}$$

which completes the proof of Theorem B.

5.3 Proof of Corollary E

Since we have established our double divisibility statement (2), we can proceed with a proof of Corollary E. By the definitions established in Sect. 2, we have that

$$\begin{aligned} {{\mathrm{cl}}}_{\widetilde{S}_E,u}(K(\sqrt{d}))[\ell ]\cdot {{\mathrm{cl}}}_{S_E}(K')[\ell ](\chi _{\ell }) \left| {{\mathrm{cl}}}_{\emptyset ,u}(K(\sqrt{d}))[\ell ]\cdot {{\mathrm{cl}}}_{\emptyset }(K')[\ell ](\chi _{\ell })\cdot \varepsilon _S \right. \end{aligned}$$

where $\varepsilon _S$ is a number depending only on $\widetilde{S}_E$. Note that when $\widetilde{S}_E = \emptyset$, we have that $\varepsilon _S = 1$ and that ${{\mathrm{cl}}}_{\emptyset ,u}(K(\sqrt{d}))[\ell ] = {{\mathrm{cl}}}(K(\sqrt{d}))[\ell ]$ by Remark 2.4. Corollary E follows immediately from the following lemma.

Lemma 5.10

${{\mathrm{cl}}}_{\emptyset }(K')[\ell ](\chi _{\ell }) \, | \, {{\mathrm{cl}}}(K(\sqrt{d}))[\ell ]$.

Proof

Let M / K be a Galois extension containing $K'$ with $\langle \alpha \rangle = {{\mathrm{Gal}}}(M/K)$, with the relations

$$\begin{aligned} \alpha ^{\ell } = {{\mathrm{id}}}\quad \text { and }\quad \overline{\gamma }\overline{\alpha }\overline{\gamma }^{-1}= \alpha ^{\chi _{\ell }(\overline{\gamma })} \, {{\mathrm{where}}}\, \langle \overline{\gamma } \rangle = {{\mathrm{Gal}}}(K'/K). \end{aligned}$$

We assume that M is unramified outside $\mathfrak {l}$ and little ramified at $\mathfrak {l}$; hence

$$\begin{aligned} M(\zeta _{\ell }) = K'(\sqrt{d})(\root \ell \of {c}), \end{aligned}$$

with $c \in M(\sqrt{d})$ and the principal divisor of c is a $\ell ^{\text {th}}$ power. We want to extend c to an element of order $\ell$ in the divisor class group of $K(\sqrt{d})$.

Let $\widetilde{\gamma }$ be an extension of $\overline{\gamma }$ to ${{\mathrm{Gal}}}(M(\sqrt{d})/K)$ such that $\widetilde{\gamma }^{\ell - 1} = {{\mathrm{id}}}$, $\widetilde{\gamma } |{K(\zeta _{\ell })}$ generates ${{\mathrm{Gal}}}(K(\zeta _{\ell })/K)$, and $\widetilde{\gamma } |{K(\sqrt{d})} = {{\mathrm{id}}}$. Since $M(\sqrt{d}) / K$ is normal, we have $\widetilde{\gamma }(c) = c^i\cdot e^{\ell }$ with $1\le i \le \ell -1$ and $e \in K'(\sqrt{d})$. Hence,

$$\begin{aligned} \widetilde{\gamma }(\root \ell \of {c}) = (\root \ell \of {c})^i\cdot e \cdot \xi _{\widetilde{\gamma }} \end{aligned}$$

with $\xi _{\widetilde{\gamma }}^{\ell } = 1$. Let $\widetilde{\alpha }$ be an extension of $\alpha$ to $M(\sqrt{d})$ of order $\ell$. We can see that $i = 1$ since

$$\begin{aligned} \widetilde{\gamma }\widetilde{\alpha }(\root \ell \of {c}) = \xi _{\widetilde{\alpha }}^{\chi _{\ell }(\overline{\gamma })}\widetilde{\gamma }(\root \ell \of {c}) \end{aligned}$$

and

$$\begin{aligned} \widetilde{\alpha }^{\chi _{\ell }(\gamma )}\widetilde{\gamma }(\root \ell \of {c}) = \widetilde{\alpha }^{\chi _{\ell }(\gamma )}(\xi _{\widetilde{\gamma }}(\root \ell \of {c})^i \cdot e) = \xi _{\widetilde{\alpha }}^{i\cdot \chi _{\ell }(\overline{\gamma })}\cdot \widetilde{\gamma }(\root \ell \of {c}), \end{aligned}$$

and hence

$$\begin{aligned} M(\sqrt{d}) = K(\sqrt{d}, \root \ell \of {c},\zeta _{\ell }). \end{aligned}$$

There exists an element $\widetilde{c} = c^{\ell - 1}\cdot e'^{\ell } \in M(\sqrt{d})$ with $e' \in K'(\sqrt{d})$ such that the divisor of $\widetilde{c}$ is a $\ell$th power. However, since $\pm \widetilde{c}$ is not an $\ell ^{\text {th}}$ power in $K(\sqrt{d})$, it is an element of order $\ell$ in the divisor class group of $K(\sqrt{d})$. $\square$

6 Elliptic curves satisfying Corollary D

Let E be an elliptic curve over a number field K. In a recent work [13], Zywina has described all known, and conjecturally all, pairs $(E/\mathbf Q ,\ell )$ such that mod $\ell$ image of Galois, $\rho _{E,\ell }(G_\mathbf{Q })$, is non-surjective. Using Zywina’s classification, we can find elliptic curves $E/\mathbf Q$ that will satisfy the conditions of Corollary D. First, we present an example of this technique for the case when $\ell = 3$. We remark that this case does not apply to Corollary D; however, it best illustrates the technique.

Let $E/\mathbf Q$ be a non-CM elliptic curve over $\mathbf Q$ such that $\rho _{E,3}(G_\mathbf{Q })$ is conjugate to

$$\begin{aligned} B(3):= \left( \begin{array}{ll} * &{}\quad * \\ 0 &{}\quad * \end{array}\right) \subset {{\mathrm{GL}}}_2(\mathbf F _{3}) . \end{aligned}$$

We can use Galois theory to prove the following result:

Proposition 6.1

Let $E/\mathbf Q$ have mod 3 image of Galois conjugate to B(3). Then $\mathbf Q (E[3]) = \mathbf Q (x(E[3])) \cdot K$ where K is an explicitly computable quadratic extension.

Before we prove Proposition 6.1, we prove the following lemma which tells us over which extension E obtains a 3-torsion point.

Lemma 6.2

For $E/\mathbf Q$ from Proposition 6.1, there exists some quadratic extension K such that E has a K-rational 3-torsion point. In particular, $E(K)[3] = \langle P \rangle$.

Proof

Let $E:y^2 = x^3 - Ax - B$ for $A,B\in \mathbf Q$. Via the Weil-pairing, we know that $\mathbf Q (\zeta _3) \subseteq \mathbf Q (E[3])$. It is also a well known fact that $B(3) \cong S_3 \times \mathbf Z / 2 \mathbf Z$. Combining these results with our assumptions, we have the following diagram of Galois sub-fields of $\mathbf Q (E[3])$:

where the extension $\mathbf Q (x(E[3]))$ is the index 2 sub-field of $\mathbf Q (E[3])$ generated by the x-coordinates of points in $E(\overline{\mathbf{Q }})[3]$. Recall that the roots of the 3-division polynomial

$$\begin{aligned} \psi _3(x) = 3x^4 + 6Ax^2 + 12Bx -A^2 \end{aligned}$$

correspond to x-coordinates of $E(\overline{\mathbf{Q }})[3]$. In particular, $\psi _3(x)$ is the minimal polynomial of the degree 6, Galois extension $\mathbf Q (x(E[3]))$.

Since $S_4$ does not contain any transitive subgroups of order 6, we know that $\psi _3(x)$ must have a linear factor, so we write $\psi _3(x) = (x-\alpha )g(x)$ where $\alpha \in \mathbf Q$ and g(x) is an irreducible cubic. This implies that there exists some $P\in E(\overline{\mathbf{Q }})[3]$ with $\mathbf Q$-rational x-coordinate given by $\alpha$. Moreover, we see that there is a 3-torsion point

$$\begin{aligned} P = (\alpha , \sqrt{f(\alpha )}). \end{aligned}$$

that is defined over the quadratic extension $\mathbf Q (\sqrt{f(\alpha )})$. $\square$

Remark 6.3

From the above proof, one can easily see that ${{\mathrm{Gal}}}(\mathbf Q (x(E[3]))/\mathbf Q ) \cong S_3$. Indeed, since $\mathbf Q (x(E[3]))$ is Galois, we showed that the Galois group of $\psi _3(x)$ is actually the Galois group of the cubic g(x). Since $[\mathbf Q (x(E[3])) : \mathbf Q ] = 6$, we know g(x) must be an irreducible cubic with non-square discriminant, which immediately implies our claim.

Proof of Proposition 6.1

(Proof of Proposition 6.1) Let K denote the quadratic extension from Lemma 6.2. It is clear that $K \subset \mathbf Q (E[3])$ and that $K \nsubseteq \mathbf Q (x(E[3]))$, so we have $\mathbf Q (E(3))$ is the compositum of $\mathbf Q (x(E[3]))$ and K. $\square$

The idea behind finding elliptic curves over $\mathbf Q$ such that $E(\mathbf Q )[\ell ] = \left\{ \mathcal {O} \right\}$ and $E(K)[\ell ] = \langle P \rangle$ is to consider $E/\mathbf Q$ with $\rho _{E,\ell }(G_\mathbf{Q })$ conjugate to a subgroup H such that

$$\begin{aligned} \left( \begin{array}{ll} 1 &{}\quad * \\ 0 &{}\quad * \end{array}\right) \subsetneq H \subseteq \left( \begin{array}{ll} * &{}\quad * \\ 0 &{}\quad * \end{array}\right) =: B(\ell ) . \end{aligned}$$

We can see that E will attain an $\ell$ torsion point over an extension K where the degree of $K/\mathbf Q$ is determined the cardinality of the upper left entry. For $\ell = 3$, we saw that $H = B(3)$ and thus the upper left entry has order 2, which gives a less explicit proof of Proposition 6.1.

Let $\ell \in \left\{ 5,13 \right\}$. Below, we provide examples of elliptic curves $E/\mathbf Q$ that do not have a $\mathbf Q$-rational point of order $\ell$ but attain a K-rational point P of order $\ell$ over some extension of small degree K that satisfies the conditions of Corollary D. The final step in our verification is showing P is not contained in the kernel of reduction modulo $\mathfrak {l}$; in particular, this means that E / K is not supersingular modulo $\mathfrak {l}$ if ${{\mathrm{ord}}}_{\mathfrak {l}}(j_E) \ge 0$. This condition is computable via the Magma command IsSupersingular.

In order to conduct a thorough search, we consider all subgroups H which can occur as an image of Galois for a non-CM $E/\mathbf Q$ and satisfy the above containment. In particular, we run through a large list elliptic curves $E/\mathbf Q$ with prescribed non-surjective mod $\ell$ image of Galois coming from the modular curves $X_H$ of Zywina [13]. Since this list is comprehensive, we also give examples of elliptic curves over $\mathbf Q$ that do not satisfy and potentially satisfy Corollary D, modulo some computations.

For $\ell = 5$, we only have one example.

Example 6.3.1

($\ell = 5$) Let $E/\mathbf Q$ be the elliptic curve

$$\begin{aligned} E:y^2 = f(x) = x^3 - \frac{185193}{185193}x + \frac{185193}{149}. \end{aligned}$$

E has mod 5 image of Galois conjugate to $B(5) \subset {{\mathrm{GL}}}_2(\mathbf F _5)$, and hence E attains a K-rational point of order 5 over a bi-quadratic extension K of $\mathbf Q$. The first quadratic extension $L/\mathbf Q$ is given by adjoining the quadratic root $\alpha$ of the 5-division polynomial $\psi _5$, and then the second quadratic is given by adjoining the square root of the $f(\alpha )$. For E defined above, we compute that ${{\mathrm{cl}}}(K) = 8$, $\zeta _5\notin K$, 2 is ramified in $\mathcal {O}_K$, and that E / K is not supersingular modulo $\mathfrak {l}$ if ${{\mathrm{ord}}}_{\mathfrak {l}}(j_E) \ge 0$ where $\mathfrak {l}| 5$. Therefore, the elliptic curve E and the number field K satisfy the conditions of Corollary D.

For $\ell = 7$, we have two possibilities.

Potential example 6.3.2 ($\ell = 7$) Let $E/\mathbf Q$ be the elliptic curve

$$\begin{aligned} E:y^2 = f(x) = x^3 - \frac{81469949623875}{3017401762489}x + \frac{162939899247750}{3017401762489}, \end{aligned}$$

which has mod 7 image conjugate to B(7). E attains a K-rational point of order 7 over an extension K of degree 6. The extension K is given by first adjoining the root $\alpha$ of the cubic factor of $\psi _7$ and then adjoining the square root of $f(\alpha )$. We verify almost all of the conditions from Corollary D for E and K; however, we are not able to verify that $7 \not \mid {{\mathrm{cl}}}(K)$.

Non-example 6.3.3 ($\ell = 7$) Suppose that $E/\mathbf Q$ has $\rho _{E,7}(G_\mathbf{Q })$ conjugate to

$$\begin{aligned} H:= \left( \begin{array}{ll} a^2 &{}\quad * \\ 0 &{}\quad * \end{array}\right) \quad {{\mathrm{where}}}a\in \mathbf F _7 . \end{aligned}$$

Since $\#(\mathbf F _7^{\times })^2 = 3$, we have that E attains a K-rational point of order 7 over a cubic extension K. Moreover, this extension is given adjoining the root of the cubic factor of the 7-division polynomial $\psi _7$. In our search, we find that all E / K are supersingular modulo $\mathfrak {l}$ if ${{\mathrm{ord}}}_{\mathfrak {l}}(j_E) \ge 0$ where $\mathfrak {l}| 7$.

For $\ell = 11$, there do not exist any subgroups coming from [13] that have our desired condition. For $\ell = 13$, we find a few examples of curves satisfying Corollary D.

Example 6.3.4

($\ell = 13$) Suppose that $E/\mathbf Q$ has $\rho _{E,13}(G_\mathbf{Q })$ conjugate to

$$\begin{aligned} H = \left( \begin{array}{ll} a^3 &{}\quad * \\ 0 &{}\quad * \end{array}\right) \quad {{\mathrm{where}}}a\in \mathbf F _{13}, \end{aligned}$$

then E attains a K-rational point of order 13 over a bi-quadratic extension $K/\mathbf Q$ since $\#(\mathbf F _{13}^{\times })^3 = 4$. As an example, consider the elliptic curve

$$\begin{aligned} E:y^2 = x^3 - \frac{2248091}{180353}x + \frac{4496182}{180353}, \end{aligned}$$

which has mod 13 image conjugate to H. E attains a K-rational point of order 13 over a bi-quadratic extension K of $\mathbf Q$. The first quadratic extension $L/\mathbf Q$ is given by adjoining a quadratic root $\alpha$ of the 13-division polynomial $\psi _{13}$, and then the second quadratic is given by adjoining the square root of the $f(\alpha )$. We compute that ${{\mathrm{cl}}}(K) = 2$, $\zeta _{13}\notin K$, (2) splits in $\mathcal {O}_K$, and E / K is not supersingular modulo $\mathfrak {l}$ if ${{\mathrm{ord}}}_{\mathfrak {l}}(j_E) \ge 0$ where $\mathfrak {l}| 13$. Therefore, the elliptic curve E and the number field K satisfy the conditions of Corollary D.

Example 6.3.5

($\ell = 13$) Suppose that $E/\mathbf Q$ has $\rho _{E,13}(G_\mathbf{Q })$ conjugate to

$$\begin{aligned} H := \left( \begin{array}{ll} a^4 &{}\quad * \\ 0 &{}\quad * \end{array}\right) \quad {{\mathrm{where}}}a\in \mathbf F _{13} . \end{aligned}$$

Since $\#(\mathbf F _{13}^{\times })^4 = 3$, E attains a K-rational point of order 13 over cubic extension $K/\mathbf Q$. For example, consider the elliptic curve

$$\begin{aligned} E:y^2 = x^3 + 13674069x + 324405221670. \end{aligned}$$

Using [13], E has mod 13 image conjugate to H. Now let $K/\mathbf Q$ denote the number field defined by the cubic factor of $\psi _{13}$. For notational purposes, we shall write $K =\mathbf Q (\alpha )$ where $\alpha$ is the primitive element of K. By base changing to K, we find that $E_K = E\times _\mathbf{Q } K$ has K-rational 13-torsion point. We also compute that ${{\mathrm{cl}}}(F) = 1$, 2 splits in $\mathcal {O}_K$, $\zeta _{13} \notin K$, and that E / K is not supersingular modulo $\mathfrak {l}$ if ${{\mathrm{ord}}}_{\mathfrak {l}}(j_E) \ge 0$ where $\mathfrak {l}| 13$. Therefore, the elliptic curve E and the number field K satisfy the conditions of Corollary D.

Example 6.3.6

($\ell = 13$) Suppose that an elliptic curve has $\rho _{E,13}(G_\mathbf{Q })$ conjugate to

$$\begin{aligned} \left( \begin{array}{ll} a^2 &{}\quad * \\ 0 &{}\quad * \end{array}\right) \quad {{\mathrm{where}}}a\in \mathbf F _{13} . \end{aligned}$$

Since $\#(\mathbf F _{13}^{\times })^2 = 6$, E will attain a K-rational point of order 13 over an extension of degree 6. As an example, consider the elliptic curve

$$\begin{aligned} E:y^2 = x^3 - \frac{12096}{529}x + \frac{24192}{529}, \end{aligned}$$

which satisfies the above property. E attains a K-rational point of order 13 over a sextic extension K of $\mathbf Q$. The first cubic extension $L/\mathbf Q$ is given by adjoining a cubic root $\alpha$ of the 13-division polynomial $\psi _{13}$, and then the second quadratic is given by adjoining the square root of the $f(\alpha )$. We also compute that ${{\mathrm{cl}}}(F) = 4$, 2 splits in $\mathcal {O}_K$, $\zeta _{13} \notin K$, and that E / K is not supersingular modulo $\mathfrak {l}$ if ${{\mathrm{ord}}}_{\mathfrak {l}}(j_E) \ge 0$ where $\mathfrak {l}| 13$. Therefore, the elliptic curve E and number field K satisfy the conditions of Corollary D.

Potential example 6.3.7 ($\ell = 13$) Suppose that an elliptic curve $E/\mathbf Q$ has mod 13 image conjugate to B(13). The curve E will attain a K-rational point of order 13 over an extension of degree 12. The difficultly in verifying the conditions of Corollary D is computing the class number and ramification indicies for the duodecic extension K.

Finally for $\ell = 37$, there is only one $E/\overline{\mathbf{Q }}$ that we need to consider.

Potential example 6.3.8 ($\ell = 37$) Suppose that $E/\mathbf Q$ is the elliptic curve with j-invariant $-7\cdot 11^3$, which has affine equation

$$\begin{aligned} E:y^2 = x^3 - \frac{251559}{11045}x + \frac{503118}{11045}. \end{aligned}$$

From [13, Theorem 1.10.(ii)], we know that the mod 37 image of E is conjugate to

$$\begin{aligned} H := \left( \begin{array}{ll} a^3 &{}\quad * \\ 0 &{}\quad * \end{array}\right) \quad {{\mathrm{where}}}a\in \mathbf F _{37} . \end{aligned}$$

Since $\#(\mathbf F _{37}^{\times })^3 = 12$, E attains a K-rational point of order 37 over a duodecic extension $K/\mathbf Q$.

As before, the difficultly in verifying the conditions of Corollary D is computing the class number and ramification indicies for the duodecic extension K.