# Rank distribution in cubic twist families of elliptic curves

Anwesh Ray  Chennai Mathematical Institute, H1, SIPCOT IT Park, Kelambakkam,
Siruseri, Tamil Nadu 603103, India<EMAIL_ADDRESS>and  Pratiksha
Shingavekar  Chennai Mathematical Institute, H1, SIPCOT IT Park, Kelambakkam,
Siruseri, Tamil Nadu 603103, India<EMAIL_ADDRESS>
###### Abstract.

Let $a$ be an integer which is not of the form $n^{2}$ or $-3n^{2}$ for
$n\in\mathbb{Z}$. Let $E_{a}$ be the elliptic curve with rational $3$-isogeny
defined by $E_{a}\mathrel{\mathop{\mathchar 58\relax}}y^{2}=x^{3}+a$, and
$K\mathrel{\mathop{\mathchar 58\relax}}=\mathbb{Q}(\mu_{3})$. Assume that the
$3$-Selmer group of $E_{a}$ over $K$ vanishes. It is shown that there is an
explicit infinite set of cubefree integers $m$ such that the $3$-Selmer groups
over $K$ of $E_{m^{2}a}$ and $E_{m^{4}a}$ both vanish. In particular, the
ranks of these cubic twists are seen to be $0$ over $K$. Our results are
proven by studying stability properties of $3$-Selmer groups in cyclic cubic
extensions of $K$, via local and global Galois cohomological techniques.

###### 2020 Mathematics Subject Classification:

##  1\. Introduction

###  1.1. Motivation and historical background

Given an elliptic curve $E$ over $\mathbb{Q}$, and a subgroup $H$ of
$\operatorname{Aut}_{\bar{\mathbb{Q}}}(E)$, it is natural to consider the
twist family of elliptic curves $E_{\zeta}$ indexed by cohomology classes in

$\zeta\in H^{1}(\operatorname{Gal}(\bar{\mathbb{Q}}/\mathbb{Q}),H).$

When $H=\\{\pm 1\\}$, the associated twists are called quadratic twists and
are in bijection with $\mathbb{Q}^{\times}/(\mathbb{Q}^{\times})^{2}$.
Goldfeld’s conjecture predicts that the average rank of quadratic twists
$E_{m}$ over $\mathbb{Q}$ is $1/2$. In fact, the conjecture predicts that when
$m$ is ordered according to its absolute value, exactly half of the quadratic
twists have rank $0$ and the other half have rank $1$. The rank distribution
in quadratic twist families is studied for instance in [BM92, HB94, KMR13,
Smi17, KL19, BKOS19]. Set $K\mathrel{\mathop{\mathchar
58\relax}}=\mathbb{Q}(\mu_{3})$ and let $\mathcal{O}_{K}$ be its ring of
integers. We consider certain families of elliptic curves with rational
$3$-isogeny that have complex multiplication by $\mathcal{O}_{K}$. For a sixth
power free integer $a\neq 0$, set $E_{a}$ to be the elliptic curve

$E_{a}\mathrel{\mathop{\mathchar 58\relax}}y^{2}=x^{3}+a.$

Then for a cubefree integer $m$, we have the cubic twists $E_{m^{2}a}$ and
$E_{m^{4}a}$ of $E_{a}$. Let $\operatorname{Sel}_{3}(E_{a}/K)$ (resp.
$\Sh(E_{a}/K)$) denote the $3$-Selmer group (resp. Tate–Shafarevich group) of
$E_{a}$ over $K$. The Selmer group sits in a short exact sequence

$0\rightarrow
E_{a}(K)/3E_{a}(K)\rightarrow\operatorname{Sel}_{3}(E_{a}/K)\rightarrow\Sh(E_{a}/K)[3]\rightarrow
0.$

It is thus easy to see that $\operatorname{Sel}_{3}(E_{a}/K)=0$ if and only if

$E_{a}(K)[3]=0,\operatorname{rank}(E_{a}(K))=0\text{ and }\Sh(E_{a}/K)[3]=0.$

The main focus of this article is to study the following natural question.

######  Question 1.1.

Fix a sixth power free integer $a$ such that
$\operatorname{Sel}_{3}(E_{a}/K)=0$. What can one say about the family of
cubefree integers $m$ for which $\operatorname{Sel}_{3}(E_{m^{2}a}/K)=0$ and
$\operatorname{Sel}_{3}(E_{m^{4}a}/K)=0$?

We construct explicit infinite families of numbers $m$ for which the above
Selmer groups vanish.

###  1.2. Main results

We state the main results of this article. It is shown that there is an
explicit positive density of primes $\mathcal{Q}_{a}$ defined by local
congruence conditions, such that if $m$ is divisible only by primes
$\ell\in\mathcal{Q}_{a}$, then, $\operatorname{Sel}_{3}(E_{m^{2}a}/K)=0$ and
$\operatorname{Sel}_{3}(E_{m^{4}a}/K)=0$. In particular,

$\operatorname{rank}E_{m^{2}a}(K),\operatorname{rank}E_{m^{4}a}(K)=0\text{ and
}\Sh(E_{m^{2}a}/K)[3],\Sh(E_{m^{4}a}/K)[3]=0.$

When we specialize to the families $E_{\ell^{2}a}$ and $E_{\ell^{4}a}$, where
$\ell$ is a prime number, we find that our result holds for an explicit
positive density of primes $\ell$.

For a prime number $\ell$, set $\mathbb{F}_{\ell}\mathrel{\mathop{\mathchar
58\relax}}=\mathbb{Z}/\ell\mathbb{Z}$. If $E_{a}$ has good reduction at
$\ell$, we set $\widetilde{E}_{a}(\mathbb{F}_{\ell})$ to denote the group of
$\mathbb{F}_{\ell}$-points on the reduced curve of $E_{a}$ at $\ell$. More
generally, if $m=\ell_{1}\dots\ell_{k}$, for a fixed value of $k$ and
$\ell_{i}$ prime, then there is a positive density of such numbers for which
our result holds.

######  Theorem A.

Let $a$ be an integer and $E_{a}$ be the elliptic curve defined by
$y^{2}=x^{3}+a$. Assume that the following conditions are satisfied

  1. (1)

$a$ is not of the form $n^{2}$ or $-3n^{2}$ for $n\in\mathbb{Z}$,

  2. (2)

$\operatorname{Sel}_{3}(E_{a}/K)=0$.

Set $\mathcal{Q}_{a}$ to denote the set of primes $\ell\nmid a$ such that the
following conditions are satisfied

  1. (1)

$\ell\equiv 1\pmod{18}$,

  2. (2)

$\ell$ is a cube modulo $q$ for each prime $q$ that divides $a$.

  3. (3)

$\widetilde{E}_{a}(\mathbb{F}_{\ell})[3]=0$.

Then, the set $\mathcal{Q}_{a}$ is infinite with natural density given by

$\mathfrak{d}(\mathcal{Q}_{a})=\frac{1}{8\times 3^{s+1}},$

where $s$ is the number of prime divisors $q$ of $a$ such that $q\equiv
1\pmod{3}$. Moreover, if $m$ is any cubefree integer divisible only by primes
in $\mathcal{Q}_{a}$, then,

$\operatorname{Sel}_{3}(E_{m^{2}a}/K)=0\text{ and
}\operatorname{Sel}_{3}(E_{m^{4}a}/K)=0.$

For integers $a\in[-20,20]$, it is shown that
$\operatorname{Sel}_{3}(E_{a}/K)=0$ for

$a=-17,-16,-14,-10,-9,-8,-6,-5,-1,6,7,8,13,14,20,$

and thus Theorem A applies to these values of $a$. The values of
$\mathfrak{d}(\mathcal{Q}_{a})$ are given in Table 1. One of the main features
of our result is that the set of primes $\mathcal{Q}_{a}$ can be made
explicit. For instance, when $a=-1$, we have that

$\mathcal{Q}_{-1}\cap[1,1000]=\\{19,127,163,199,271,307,379,487,523,631,739,811,883,919,991\\}.$

For further details on numerical computations, we refer to sections 4.3 and
4.4.

###  1.3. Methodology

The study of Selmer ranks in twist families is related to the study the rank
growth and stability of elliptic curves defined over a number field in a
$\mathbb{Z}/p\mathbb{Z}$-extension for an odd prime $p$. Let $K$ be a number
field containing the $p$-th roots of unity and $E$ be an elliptic curve over
$K$ such that the $p$-Selmer group $\operatorname{Sel}_{p}(E/K)=0$. Let $L/K$
be a cyclic $p$-extension. Then, it is shown that under certain conditions on
$E$ and $L$, the Selmer group $\operatorname{Sel}_{p}(E/L)=0$ as well. This
strategy has been adapted from the technique of Mazur–Rubin [MR07, MRS07],
Brau [Bra14] and Česnavičius [Č17]. In keeping with these works, we study
stability questions in prime cyclic extensions for any prime number $p\geq 3$,
cf. Theorem 3.12 and Theorem 3.15 for further details.

###  1.4. Related work

In recent years, there has been an increasing interest in rank distribution
questions for families of elliptic curves from the perspective of _arithmetic
statistics_. These techniques involve the parameterization of Selmer groups of
elliptic curves and abelian varieties by integral forms in many variables. The
distribution properties of such integral forms can be studied via techniques
from analytic number theory and the geometry of numbers, and has gained
momentum in works of Bhargava–Shankar [BS15a, BS15b, BS13],
Bhargava–Elkies–Shnidman [BES20], Alpöge–Bhargava–Shnidman [ABS22],
Shnidman–Weiss [SW21, SW23] etc. Our techniques on the other hand rely of
local and global arguments in Galois cohomology (in keeping with those used by
Mazur–Rubin, Brau, Česnavičius), and one of the consequences is that we obtain
very explicit descriptions for the twist families for which the Selmer group
group vanishes. It is conceivable that our results should generalize to the
context of abelian varieties with complex multiplication. For instance,
Jacobians of twist families of bicyclic trigonal curves $y^{3}=f(x^{2})$, cf.
[SW21] for details.

###  1.5. Organization

Including the introduction, the article consists of four sections. In section
2, we set basic notation and recall definitions of Selmer groups and their
properties. In section 3, we study the stability of $p$-Selmer groups in
$p$-cyclic extensions of number fields, leveraging the technique of
Mazur–Rubin, Brau and Česnavičius. We prove our main results for cubic twist
families in section 4. These results are backed up by explicit numerical
computations in subsections 4.3 and 4.4.

##  2\. Notation and preliminaries

In this section, we introduce certain preliminary notions and set up relevant
notation. Given an abelian group $M$, set $\\#M$ to denote its size. Set
$M[n]$ to denote its $n$-torsion subgroup. Given a prime number $p$, set
$M[p^{\infty}]\mathrel{\mathop{\mathchar 58\relax}}=\bigcup_{k}M[p^{k}]$. The
$p$-adic Tate module associated to $M$ is the inverse limit

$\operatorname{T}_{p}(M)\mathrel{\mathop{\mathchar
58\relax}}=\varprojlim_{n}M[p^{n}]$

with respect to natural multiplication by $p$ maps $M[p^{k+1}]\rightarrow
M[p^{k}]$.

Fix a number field $K$ and denote by $\Sigma_{K}$ the set of all
nonarchimedian primes $v$ of $K$. Given a prime $v$, we shall let $\ell$ be
the rational prime such that $v|\ell$. Set $\mathbb{F}_{\ell}$ to denote the
finite field with $\ell$ elements. For $v\in\Sigma_{K}$, set $K_{v}$ to denote
the completion of $K$ at $v$. Take $\mathcal{O}_{K_{v}}$ to be the valuation
ring of $K_{v}$ and $\pi_{v}$ a choice of uniformizer of
$\mathcal{O}_{K_{v}}$. Let $k_{v}$ be the residue field
$\mathcal{O}_{K_{v}}/\pi_{v}$. Denote by $i_{v}\mathrel{\mathop{\mathchar
58\relax}}K\hookrightarrow K_{v}$ the natural inclusion of $K$ into $K_{v}$.
Let $\bar{K}$ (resp. $\bar{K}_{v}$) be an algebraic closure of $K$ (resp.
$K_{v}$) and $\operatorname{G}_{K}$ (resp. $\operatorname{G}_{K_{v}}$) denote
the absolute Galois group of $K$ (resp. $K_{v}$). Let $\operatorname{I}_{v}$
denote the inertia subgroup of $\operatorname{G}_{K_{v}}$. At each prime
$v\in\Sigma_{K}$, choose an embedding $\iota_{v}\mathrel{\mathop{\mathchar
58\relax}}\bar{K}\hookrightarrow\bar{K}_{v}$ that extends $i_{v}$. Note that
$\iota_{v}$ induces an inclusion $\iota_{v}^{*}\mathrel{\mathop{\mathchar
58\relax}}\operatorname{G}_{K_{v}}\hookrightarrow\operatorname{G}_{K}$. Let
$M$ be a $\operatorname{G}_{K}$-module and $i\in\mathbb{Z}_{\geq 0}$. Take
$H^{i}(K,M)$ (resp. $H^{i}(K_{v},M)$) to denote the cohomology group
$H^{i}(\operatorname{G}_{K},M)$ (resp. $H^{i}(\operatorname{G}_{K_{v}},M)$).
The restriction map induced by $\iota_{v}^{*}$ is denoted

$\operatorname{res}_{v}\mathrel{\mathop{\mathchar
58\relax}}H^{i}(K,M)\rightarrow H^{i}(K_{v},M).$

Let $E$ be an elliptic curve over $K$ and $n$ be a positive integer. Given a
prime $v\in\Sigma_{K}$ at which $E$ has good reduction, set $\widetilde{E}$ to
denote the reduction of $E$ at $v$. Take $\widetilde{E}(k_{v})$ to denote the
group of $k_{v}$-points on $\widetilde{E}$. Set $E[n]$ to denote the
$n$-torsion subgroup if $E(\bar{K})$. For $\mathcal{F}\in\\{K,K_{v}\\}$, we
have the exact sequence of $\operatorname{G}_{\mathcal{F}}$-modules

$0\rightarrow E[n]\rightarrow E\xrightarrow{\times n}E\rightarrow 0,$

from which we arrive at the following Kummer sequence

(2.1) $0\rightarrow
E(\mathcal{F})/nE(\mathcal{F})\xrightarrow{\delta_{E,\mathcal{F}}}H^{1}(\mathcal{F},E[n])\rightarrow
H^{1}(\mathcal{F},E)[n]\rightarrow 0.$

######  Definition 2.1.

Let $n$ be an odd natural number. With respect to notation above, the
$n$-Selmer group of $E$ over $K$ is defined as follows

$\operatorname{Sel}_{n}(E/K)\mathrel{\mathop{\mathchar
58\relax}}=\operatorname{ker}\left\\{H^{1}(K,E[n])\xrightarrow{\Phi}\prod_{v\in\Sigma_{K}}H^{1}(K_{v},E)[n]\right\\}.$

Here the map $\Phi$ is the product of maps $\Phi_{v}$, given by the composite

$H^{1}(K,E[n])\xrightarrow{\operatorname{res}_{v}}H^{1}(K_{v},E[n])\rightarrow
H^{1}(K_{v},E)[n].$

Let $\Sigma$ be a finite set of primes that contains the primes of $K$ that
lie above the primes that divide $n$ and the primes at which $E$ has bad
reduction. Let $K_{\Sigma}$ be the maximal algebraic extension of $K$ in which
the primes $\ell\notin\Sigma$ are unramified. Set
$\operatorname{G}_{K,\Sigma}$ to denote the Galois group
$\operatorname{Gal}(K_{\Sigma}/K)$. Given a
$\operatorname{G}_{K,\Sigma}$-module $M$, set $H^{i}(K_{\Sigma}/K,M)$ to
denote $H^{i}(\operatorname{G}_{K,\Sigma},M)$. The Selmer group
$\operatorname{Sel}_{n}(E/K)$ can also be described by finitely many local
conditions

$\operatorname{Sel}_{n}(E/K)=\operatorname{ker}\left\\{H^{1}(K_{\Sigma}/K,E[n])\xrightarrow{\Phi_{\Sigma}}\bigoplus_{v\in\Sigma}H^{1}(K_{v},E)[n]\right\\},$

where the map $\Phi_{\Sigma}\mathrel{\mathop{\mathchar
58\relax}}=\bigoplus_{v\in\Sigma}\Phi_{v}$, cf. [CR00, Chapter 1]. Given a
prime $p$, the natural map $E[p^{k}]\hookrightarrow E[p^{k+1}]$ induces a map
at the level of $p$-primary Selmer groups

$\operatorname{Sel}_{p^{n}}(E/K)\rightarrow\operatorname{Sel}_{p^{n+1}}(E/K).$

The $p$-primary Selmer group is then defined to be the direct limit

$\operatorname{Sel}_{p^{\infty}}(E/K)\mathrel{\mathop{\mathchar
58\relax}}=\varinjlim_{n}\operatorname{Sel}_{p^{n}}(E/K).$

Let $\Sh(E/K)$ denote the Tate–Shafarevich group of $E$ over $K$. The Selmer
group fits into a short exact sequence

(2.2) $0\rightarrow
E(K)/nE(K)\xrightarrow{\delta_{E,n}}\operatorname{Sel}_{n}(E/K)\rightarrow\Sh(E/K)[n]\rightarrow
0.$

Given a prime $p$, passing to the direct limit for integers $n$ of the form
$p^{k}$, we have the following short exact sequence

(2.3) $0\rightarrow
E(K)\otimes_{\mathbb{Z}}\mathbb{Q}_{p}/\mathbb{Z}_{p}\rightarrow\operatorname{Sel}_{p^{\infty}}(E/K)\rightarrow\Sh(E/K)[p^{\infty}]\rightarrow
0.$

As is well known, the Tate–Shafarevich group $\Sh(E/K)$ is conjectured to be
finite. If $\Sh(E/K)[p^{\infty}]$ is known to be finite, then from (2.3),

$\operatorname{corank}_{\mathbb{Z}_{p}}\left(\operatorname{Sel}_{p^{\infty}}(E/K)\right)=\operatorname{rank}E(K).$

For each integer $k\geq 1$, multiplication by $p$ induces a map of Galois
modules $E[p^{n+1}]\xrightarrow{\times p}E[p^{n}]$. The _compact Selmer group_
is defined to be the inverse limit

$\mathfrak{S}_{p}(E/K)\mathrel{\mathop{\mathchar
58\relax}}=\varprojlim_{n}\operatorname{Sel}_{p^{n}}(E/K)$

with respect to these maps. From (2.2), we obtain the following exact sequence

(2.4) $0\rightarrow
E(K)\otimes_{\mathbb{Z}}\mathbb{Z}_{p}\rightarrow\mathfrak{S}_{p}(E/K)\rightarrow\operatorname{T}_{p}\left(\Sh(E/K)\right)\rightarrow
0.$

If the $p$-primary Tate–Shafarevich group $\Sh(E/K)[p^{\infty}]$ is finite,
then its Tate–module vanishes. Thus, if $\Sh(E/K)[p^{\infty}]$ is finite,
then,

$\operatorname{rank}_{\mathbb{Z}_{p}}\left(\mathfrak{S}_{p}(E/K)\right)=\operatorname{rank}E(K).$

It follows from [CR00, Lemma 1.8] that there is a short exact sequence

(2.5) $0\rightarrow
E(K)[p^{\infty}]\rightarrow\mathfrak{S}_{p}(E/K)\rightarrow\operatorname{T}_{p}\left(\operatorname{Sel}_{p^{\infty}}(E/K)\right)\rightarrow
0,$

and thus,

(2.6)
$\operatorname{corank}_{\mathbb{Z}_{p}}\left(\operatorname{Sel}_{p^{\infty}}(E/K)\right)=\operatorname{rank}_{\mathbb{Z}_{p}}\left(\mathfrak{S}_{p}(E/K)\right).$

Note if $\operatorname{Sel}_{p}(E/K)=0$, then, $\Sh(E/K)[3^{\infty}]=0$,
$\operatorname{rank}E(K)=0$ and $E(K)[3^{\infty}]=0$. Thus it follows from
(2.5) that $\mathfrak{S}_{p}(E/K)=0$. It then follows from the
Cassels–Poitou–Tate exact sequence [CR00, Ch. 1 (3)] that the map
$\Phi_{\Sigma}$ is surjective, from which we obtain a short exact sequence

(2.7) $0\rightarrow\operatorname{Sel}_{n}(E/K)\rightarrow
H^{1}(K_{\Sigma}/K,E[n])\xrightarrow{\Phi_{\Sigma}}\bigoplus_{v\in\Sigma}H^{1}(K_{v},E)[n]\rightarrow
0.$

##  3\. Rank stability in prime cyclic extensions

Let $E$ be an elliptic curve over a number field $K$ and $p$ be an odd prime
number. Denote by $\Sigma_{p}$ the set of primes $v$ of $K$ that lie above
$p$. In this section, we assume that $\operatorname{Sel}_{p}(E/K)=0$. Note
that this is equivalent to requiring that $E(K)$ has rank $0$, and both $E(K)$
and $\Sh(E/K)$ have no nontrivial $p$-torsion.

###  3.1. Local contributions to the growth of Selmer groups

We take $L/K$ to be a finite Galois extension with Galois group
$G=\operatorname{Gal}(L/K)$ isomorphic to $\mathbb{Z}/p\mathbb{Z}$. Let
$\sigma$ be a generator of $G$. Set $S$ to be the set of primes
$v\in\Sigma_{K}$ such that either one of the following conditions are
satisfied

  * •

$v\in\Sigma_{p}$,

  * •

$E$ has bad reduction at $v$,

  * •

$v$ is ramified in $L$.

Denote by $S(L)$ the primes $w\in\Sigma_{L}$ such that $w|v$ for some prime
$v\in S$. Then there are exact sequences

$\begin{split}&0\rightarrow\operatorname{Sel}_{p}(E/K)\rightarrow
H^{1}(K_{S}/K,E[p])\xrightarrow{\Phi_{K}}\bigoplus_{v\in
S}H^{1}(K_{v},E)[p]\rightarrow 0,\\\
&0\rightarrow\operatorname{Sel}_{p}(E/L)^{G}\rightarrow
H^{1}(K_{S}/L,E[p])^{G}\xrightarrow{\Phi_{L}}\left(\bigoplus_{w\in
S(L)}H^{1}(L_{w},E)[p]\right)^{G}.\\\ \end{split}$

From the Cassels-Poitou-Tate sequence (cf. (2.7) and the discussion preceding
it), the map $\Phi_{K}$ is surjective. For $v\in S$, consider the restriction
map

$\gamma_{v}\mathrel{\mathop{\mathchar
58\relax}}H^{1}(K_{v},E)[p]\rightarrow\left(\bigoplus_{w|v}H^{1}(L_{w},E)[p]\right)^{G},$

where $w$ runs through all primes of $L$ that lie above $v$. Let

$\gamma\mathrel{\mathop{\mathchar 58\relax}}\bigoplus_{v\in
S}H^{1}(K_{v},E)[p]\rightarrow\left(\bigoplus_{w\in
S(L)}H^{1}(L_{w},E)[p]\right)^{G}$

denote the direct sum of maps $\bigoplus_{v\in S}\gamma_{v}$.

There is a natural restriction map

$\alpha\mathrel{\mathop{\mathchar
58\relax}}\operatorname{Sel}_{p}(E/K)\rightarrow\operatorname{Sel}_{p}(E/L)^{G}$

which fits into the fundamental diagram

(3.1)
${0=\operatorname{Sel}_{p}(E/K)}$${H^{1}(K_{S}/K,E[p])}$${\bigoplus_{v\in
S}H^{1}(K_{v},E)[p]}$${0}$${0}$${\operatorname{Sel}_{p}(E/L)^{G}}$${H^{1}(L_{S}/L,E[p])^{G}}$${\left(\bigoplus_{w\in
S(L)}H^{1}(L_{w},E)[p]\right)^{G},}$$\scriptstyle{\alpha}$$\scriptstyle{\Phi_{K}}$$\scriptstyle{\beta}$$\scriptstyle{\gamma}$

From the snake lemma one obtains an exact sequence

(3.2)
$0=\operatorname{ker}(\alpha)\rightarrow\operatorname{ker}(\beta)\rightarrow\operatorname{ker}(\gamma)\rightarrow\operatorname{cok}(\alpha)\rightarrow\operatorname{cok}(\beta)\rightarrow\operatorname{cok}(\gamma)$

The following basic result will be useful in many of our arguments.

######  Lemma 3.1.

Let $V$ be a finite dimensional $\mathbb{F}_{p}$-vector space and $G$ be a
finite $p$-group which acts on $V$. Then, the following implication holds

$V^{G}=0\Rightarrow V=0.$

###### Proof.

We have that

$\\#V=\sum_{i}\\#\mathcal{O}_{i},$

where the sum ranges over $G$-orbits of $V$. Suppose that $V^{G}=0$, then
$\\{0\\}$ is a $G$-orbit of $V$ which contains only $1$-element. Every other
$G$-orbit has $p^{n_{i}}$ elements with $n_{i}>0$. This implies that
$\\#V\equiv 1\mod{p}$, hence, $V=0$. ∎

######  Proposition 3.2.

Let $K$ be a number field and $E$ be an elliptic curve over $K$ such that
$\operatorname{Sel}_{p}(E/K)=0$. Let $L/K$ be a cyclic $p$-extension with
Galois group $G\mathrel{\mathop{\mathchar 58\relax}}=\operatorname{Gal}(L/K)$.
Assume moreover that for all $v\in S$, the map $\gamma_{v}$ is injective.
Then, we have that $\operatorname{Sel}_{p}(E/L)=0$. In particular, the rank of
$E(L)$ is $0$.

###### Proof.

Since it is assumed that $\operatorname{Sel}_{p}(E/K)=0$, we find that
$E(K)[p]=0$. Note that $E(K)[p]=E(L)[p]^{G}$. We deduce from Lemma 3.1 that
$E(L)[p]=0$. The restriction map $\beta$ in (3.1) sits in an inflation-
restriction sequence

$0\rightarrow H^{1}(L/K,E(L)[p])\rightarrow
H^{1}(K_{S}/K,E[p])\xrightarrow{\beta}H^{1}(K_{S}/L,E[p])^{G}\rightarrow
H^{2}(L/K,E(L)[p]).$

The groups $H^{i}(L/K,E(L)[p])=0$ and hence, $\beta$ is an isomorphism. The
exact sequence (3.2) then implies that

$\operatorname{cok}(\alpha)\simeq\operatorname{ker}(\gamma)\simeq\bigoplus_{v\in
S}\operatorname{ker}\gamma_{v}.$

Since it is assumed that $\gamma_{v}$ is injective for all $v\in S$, we deduce
that $\operatorname{cok}(\alpha)=0$. Since $\operatorname{Sel}_{p}(E/K)=0$, it
follows that $\operatorname{cok}(\alpha)=\operatorname{Sel}_{p}(E/L)^{G}=0$.
It then follows from Lemma 3.1 that $\operatorname{Sel}_{p}(E/L)=0$, and this
concludes the proof of the result. ∎

We analyze conditions for $\gamma_{v}$ to be injective for $v\in S$. Let $w|v$
be a chosen prime of $L$ and consider the restriction map

$\gamma_{v}^{\prime}\mathrel{\mathop{\mathchar
58\relax}}H^{1}(K_{v},E)[p]\rightarrow
H^{1}(L_{w},E)[p]^{\operatorname{Gal}(L_{w}/K_{v})}.$

Since $G$ permutes the primes $w$ of $L$ that lie above $v$ transitively, it
follows that

(3.3) $\operatorname{ker}\gamma_{v}=\operatorname{ker}\gamma_{v}^{\prime}.$

######  Lemma 3.3.

Let $v\in S$ and $w|v$ be a prime of $L$. Then, we have that

$\operatorname{ker}\gamma_{v}\simeq H^{1}(L_{w}/K_{v},E(L_{w}))[p].$

In particular, if $v$ splits in $L$, then, $\gamma_{v}$ is injective.

###### Proof.

From the inflation-restriction sequence, we find that

$\operatorname{ker}\gamma_{v}^{\prime}\simeq H^{1}(L_{w}/K_{v},E(L_{w}))[p].$

Since
$\operatorname{ker}\gamma_{v}\simeq\operatorname{ker}\gamma_{v}^{\prime}$, the
result follows. ∎

Since $\gamma_{v}$ is injective when $v$ splits in $L$, we consider two cases

  1. (1)

$v\nmid p$ and $v$ is inert or ramified in $L$,

  2. (2)

$v|p$ and $v$ is inert or ramified in $L$.

We note that in the above cases, there is a unique prime $w|v$ and hence,
$\gamma_{v}^{\prime}=\gamma_{v}$.

######  Lemma 3.4.

For $v\in S\setminus\Sigma_{p}$ assume that _either_ of the following
conditions is satisfied

  1. (1)

$v$ is a prime of bad reduction and is split in $L$,

  2. (2)

$v$ is a prime of good reduction, $v$ is ramified in $L$ and
$\widetilde{E}(k_{v})[p]=0$.

Then, $\gamma_{v}$ is injective.

We remark that a prime $v\in S\setminus\Sigma_{p}$ at which $E$ has good
reduction is necessarily ramified in $L$.

###### Proof.

The result follows from [PR24, Proposition 2.8]. ∎

Throughout the rest of this section, we shall study the cases when $v|p$. This
analysis is based on unpublished work of Brau [Bra14], results of Mazur–Rubin
[MR07] and Mazur–Rubin–Silverberg [MRS07]. For a prime $v$ of a number field
$F$ and a $\operatorname{G}_{F_{v}}$-module $V$, set

$H^{1}_{\operatorname{nr}}(F_{v},V)\mathrel{\mathop{\mathchar
58\relax}}=\operatorname{image}\left\\{\operatorname{inf}\mathrel{\mathop{\mathchar
58\relax}}H^{1}(\operatorname{G}_{F_{v}}/I_{v},V^{\operatorname{I}_{v}})\longrightarrow
H^{1}(K_{v},V)\right\\}.$

######  Definition 3.5.

Let $p$ be an odd prime and $F$ be a number field. For a continuous
$p$-primary $\operatorname{G}_{F}$-module $V$, a _Selmer structure_
$\mathcal{F}$ on $V$ is a collection of cohomology classes

$H^{1}_{\mathcal{F}}(F_{v},V)\subseteq H^{1}(F_{v},V)$

for $v\in\Sigma_{F}$ such that for all but finitely many primes $v$,

$H^{1}_{\mathcal{F}}(F_{v},V)=H^{1}_{\operatorname{nr}}(F_{v},V).$

Associated to $\mathcal{F}$ is the Selmer group

$H^{1}_{\mathcal{F}}(F,V)\mathrel{\mathop{\mathchar
58\relax}}=\operatorname{ker}\left\\{H^{1}(F,V)\longrightarrow\prod_{v\in\Sigma_{F}}\left(\frac{H^{1}(F_{v},V)}{H^{1}_{\mathcal{F}}(F_{v},V)}\right)\right\\}.$

Given two Selmer structures $\mathcal{F}$ and $\mathcal{G}$, set

$\begin{split}&H^{1}_{\mathcal{F}+\mathcal{G}}(F_{v},V)\mathrel{\mathop{\mathchar
58\relax}}=H^{1}_{\mathcal{F}}(F_{v},V)+H^{1}_{\mathcal{G}}(F_{v},V),\\\
&H^{1}_{\mathcal{F}\cap\mathcal{G}}(F_{v},V)\mathrel{\mathop{\mathchar
58\relax}}=H^{1}_{\mathcal{F}}(F_{v},V)\cap
H^{1}_{\mathcal{G}}(F_{v},V).\end{split}$

We would like to draw attention to two Selmer conditions $\mathcal{F}$ and
$\mathcal{A}$ associated to the $\operatorname{G}_{K}$-module $E[p]$. For each
prime $v\in\Sigma_{K}$, set

$H^{1}_{\mathcal{F}}(K_{v},E[p])\mathrel{\mathop{\mathchar
58\relax}}=\operatorname{image}\left\\{\frac{E(K_{v})}{pE(K_{v})}\xrightarrow{\delta_{E,K_{v}}}H^{1}(K_{v},E[p])\right\\},$

where $\delta_{E,K_{v}}$ is the local Kummer map (2.1). Observe that
$\operatorname{Sel}_{p}(E/K)=H^{1}_{\mathcal{F}}(K,E[p])$. Let $L/K$ be a
$G$-extension where $G\simeq\mathbb{Z}/p\mathbb{Z}$. Set
$\operatorname{Res}_{K}^{L}E$ to denote the Weil–restriction of scalars of $E$
from $L$ to $K$. For every $K$-algebra $X$, we have that

$(\operatorname{Res}_{K}^{L}E)(X)=E(L\otimes_{K}X).$

Let $A$ denote the kernel of the norm map
$\operatorname{Res}_{K}^{L}E\xrightarrow{N_{L/K}}E$. Let $\mathcal{O}$ be the
ring of integers of $\mathbb{Q}(\mu_{p})$ and $\mathfrak{p}$ be the ideal of
$\mathcal{O}$ that lies above $p$. The natural inclusion
$\mathbb{Z}[G]\hookrightarrow\operatorname{End}_{K}(\operatorname{Res}_{K}^{L}E)$
induces a natural map $\mathbb{Z}[G]\rightarrow\operatorname{End}_{K}(A)$,
which factors as

$\mathbb{Z}[G]\twoheadrightarrow\mathcal{O}\hookrightarrow\operatorname{End}_{K}(A).$

There is a canonical $\operatorname{G}_{K}$-isomorphism
$A[\mathfrak{p}]\xrightarrow{\sim}E[p]$, cf. [MR07, Proposition 4.1]. Let
$\pi$ be a generator of $\mathfrak{p}$ and consider the composed map

$\lambda_{A,v}\mathrel{\mathop{\mathchar 58\relax}}\frac{A(K_{v})}{\pi
A(K_{v})}\hookrightarrow
H^{1}(K_{v},A[\mathfrak{p}])\xrightarrow{\sim}H^{1}(K_{v},E[p]).$

Let $H^{1}_{\mathcal{A}}(K_{v},E[p])$ be the image of this composed map. This
defines $\mathcal{A}$, which is referred to as the _twisted Selmer condition_
with respect to $L/K$.

######  Lemma 3.6.

For every prime $v\in S$, we have that

$\operatorname{ker}\gamma_{v}\simeq\frac{H^{1}_{\mathcal{F}+\mathcal{A}}(K_{v},E[p])}{H^{1}_{\mathcal{A}}(K_{v},E[p])}\simeq\frac{H^{1}_{\mathcal{F}}(K_{v},E[p])}{H^{1}_{\mathcal{F}\cap\mathcal{A}}(K_{v},E[p])}.$

###### Proof.

The result above is [Bra14, Lemma 5.1], and follows from Proposition 4.4 of
_loc. cit_. ∎

######  Lemma 3.7.

Let $v\in\Sigma_{p}$ be a prime at which $E$ has good ordinary reduction, and
assume that $v$ is ramified in $L$. Then if $\widetilde{E}(k_{v})[p]=0$, the
map $\gamma_{v}$ is injective.

###### Proof.

Lemma 3.6 asserts that

(3.4)
$\operatorname{ker}\gamma_{v}\simeq\frac{H^{1}_{\mathcal{F}}(K_{v},E[p])}{H^{1}_{\mathcal{F}\cap\mathcal{A}}(K_{v},E[p])}.$

It follows from [MR07, Proposition 5.2] that

(3.5)
$\frac{H^{1}_{\mathcal{F}}(K_{v},E[p])}{H^{1}_{\mathcal{F}\cap\mathcal{A}}(K_{v},E[p])}\simeq\frac{E(K_{v})}{\left(E(K_{v})\cap
N_{L_{w}/K_{v}}E(L_{w})\right)}.$

Denote by $E_{1}(K_{v})$ (resp. $E_{1}(L_{w})$) denote the kernel of the
reduction map $E(K_{v})\rightarrow\widetilde{E}(k_{v})$ (resp.
$E(L_{w})\rightarrow\widetilde{E}(k_{w})$). Since $\widetilde{E}(k_{v})[p]=0$
and $L/K$ is a $p$-extension, it follows that $\widetilde{E}(k_{w})[p]=0$ (cf.
Lemma 3.1). Therefore, we find that

$\begin{split}E(K_{v})/pE(K_{v})\simeq E_{1}(K_{v})/pE_{1}(K_{v}),\\\
E(L_{w})/pE(L_{w})\simeq E_{1}(L_{w})/pE_{1}(L_{w}).\\\ \end{split}$

Hence it is easy to see that

$\frac{E(K_{v})}{\left(E(K_{v})\cap
N_{L_{w}/K_{v}}E(L_{w})\right)}\simeq\frac{E_{1}(K_{v})}{\left(E_{1}(K_{v})\cap
N_{L_{w}/K_{v}}E_{1}(L_{w})\right)}.$

From [MR07, Proposition B.2] it follows that

$\frac{E_{1}(K_{v})}{\left(E_{1}(K_{v})\cap
N_{L_{w}/K_{v}}E_{1}(L_{w})\right)}=0.$

Combining this with (3.4) and (3.5), we conclude that $\gamma_{v}$ is
injective. ∎

######  Lemma 3.8.

Let $v\in\Sigma_{p}$ be a prime at which $E$ has good reduction and assume
that $v$ is inert in $L$. Then, $\gamma_{v}$ is injective.

###### Proof.

From the proof of Lemma 3.7, we find that

$\operatorname{ker}\gamma_{v}\simeq\frac{E_{1}(K_{v})}{\left(E_{1}(K_{v})\cap
N_{L_{w}/K_{v}}E_{1}(L_{w})\right)}.$

The norm map $N_{L_{w}/K_{v}}\mathrel{\mathop{\mathchar
58\relax}}E_{1}(L_{w})\rightarrow E_{1}(K_{v})$ is surjective when
$L_{w}/K_{v}$ is unramified (cf. [Haz74, Proposition 3.1]). Therefore,
$\gamma_{v}$ is injective and this proves the result. ∎

Next, we deal with the case when $v\in\Sigma_{p}$ is a prime at which $E$ has
good supersingular reduction. Let $\mathcal{K}$ be a finite extension of
$\mathbb{Q}_{p}$ with valuation ring $\mathcal{O}_{\mathcal{K}}$, maximal
ideal $\mathfrak{m}_{\mathcal{K}}$ and residue field
$k=\mathcal{O}_{\mathcal{K}}/\mathfrak{m}_{\mathcal{K}}$. Let
$\mathcal{F}(X,Y)\in\mathcal{O}\llbracket X,Y\rrbracket$ be a formal group law
and $\mathcal{F}(\mathfrak{m}_{\mathcal{K}})$ be the induced abelian group.
The group operation is defined as follows

$x\oplus_{\mathcal{F}}y\mathrel{\mathop{\mathchar 58\relax}}=\mathcal{F}(x,y)$

for $x,y\in\mathfrak{m}_{\mathcal{K}}$. Let $\mathcal{L}/\mathcal{K}$ be a
finite Galois extension with Galois group
$\mathcal{G}=\operatorname{Gal}(\mathcal{L}/\mathcal{K})$. Writing
$\mathcal{G}=\\{\sigma_{1},\dots,\sigma_{n}\\}$, consider the induced norm map

$N_{\mathcal{L}/\mathcal{K}}\mathrel{\mathop{\mathchar
58\relax}}\mathcal{F}(\mathfrak{m}_{\mathcal{L}})\rightarrow\mathcal{F}(\mathfrak{m}_{\mathcal{K}}),$

defined by

$N_{\mathcal{L}/\mathcal{K}}(x)\mathrel{\mathop{\mathchar
58\relax}}=\sigma_{1}(x)\oplus_{\mathcal{F}}\sigma_{2}(x)\oplus_{\mathcal{F}}\dots\oplus_{\mathcal{F}}\sigma_{n}(x).$

Assume that $\mathcal{G}\simeq\mathbb{Z}/p\mathbb{Z}$ and $\sigma$ be a
generator of $\mathcal{G}$. Let $v_{\mathcal{L}}$ (resp. $v_{\mathcal{K}}$)
denote the normalized valuation of $\mathcal{O}_{\mathcal{L}}$ (resp.
$\mathcal{O}_{\mathcal{K}}$). Assume that $\mathcal{L}/\mathcal{K}$ is a
ramified extension. Set $\pi_{\mathcal{L}}$ (resp. $\pi_{\mathcal{K}}$) to
denote a uniformizer of $\mathcal{O}_{\mathcal{L}}$ (resp.
$\mathcal{O}_{\mathcal{K}}$). Setting

(3.6) $t\mathrel{\mathop{\mathchar
58\relax}}=v_{\mathcal{L}}\left(\sigma(\pi_{\mathcal{L}})-\pi_{\mathcal{L}}\right)-1,$

we note that $t\geq 1$ since $\mathcal{L}/\mathcal{K}$ must be a wildly
ramified extension. On the other hand, suppose $\mathcal{L}/\mathcal{K}$ is
unramified, then, $t\mathrel{\mathop{\mathchar 58\relax}}=-1$.

Suppose now that $L/K$ is a $\mathbb{Z}/p\mathbb{Z}$-extension of number
fields, $v\in\Sigma_{K}$ and $w|v$ is a prime of $L$. Let
$t_{v}\mathrel{\mathop{\mathchar 58\relax}}=t$ be the invariant defined above
(3.6) for $\mathcal{L}\mathrel{\mathop{\mathchar 58\relax}}=L_{w}$ and
$\mathcal{K}\mathrel{\mathop{\mathchar 58\relax}}=K_{v}$.

######  Proposition 3.9.

Let $E$ be an elliptic curve with good supersingular reduction at a prime
$v\in\Sigma_{p}$. Suppose that $v$ is ramified in $L$. Then, $\gamma_{v}$ is
injective if and only if $t_{v}=1$.

###### Proof.

The result is [Bra14, Proposition 5.10]; we give a brief sketch. Let $w$ be
the prime of $L$ that lies above $v$. We identify
$\operatorname{Gal}(L_{w}/K_{v})$ with $G$. Since $E$ has good supersingular
reduction at $v$, we find that

$a_{v}(E)\mathrel{\mathop{\mathchar
58\relax}}=\operatorname{Norm}_{K/\mathbb{Q}}(v)+1-\\#\widetilde{E}(k_{v})$

is divisible by $p$. This implies that $\widetilde{E}(k_{v})[p]=0$, in other
words,

$\left(\widetilde{E}(k_{w})[p]\right)^{G}=0.$

It follows from Lemma 3.1 that $\widetilde{E}(k_{w})[p]=0$. Let $E_{1}$ denote
the formal group of $E$. Consider the short exact sequence of $G$-modules

$0\rightarrow E_{1}(L_{w})\rightarrow
E(L_{w})\rightarrow\widetilde{E}(k_{w})\rightarrow 0$

and the associated long exact sequence in cohomology

$\widetilde{E}(k_{v})\rightarrow H^{1}(G,E_{1}(L_{w}))\rightarrow
H^{1}(G,E(L_{w}))\rightarrow H^{1}(G,\widetilde{E}(k_{w})).$

Since $p\nmid\widetilde{E}(k_{w})$, it follows from the restriction-
corestriction sequence that

$H^{1}(G,\widetilde{E}(k_{w}))=0.$

On the other hand, $H^{1}(G,E_{1}(L_{w}))$ is a $p$-primary group, and
$p\nmid\\#\widetilde{E}(k_{v})$. Hence, there is an isomorphism

$H^{1}(G,E(L_{w}))\simeq H^{1}(G,E_{1}(L_{w})).$

The proof of [Bra14, Proposition 3.6] shows that the Herbrand-quotient of
$E_{1}(L_{w})$ is $1$, and hence,

$H^{1}(G,E_{1}(L_{w}))\simeq H^{2}(G,E_{1}(L_{w}))\simeq
E_{1}(K_{v})/N_{L_{w}/K_{v}}\left(E_{1}(L_{w})\right).$

Then [Bra14, Proposition 3.5] shows that

$E_{1}(K_{v})=N_{L_{w}/K_{v}}\left(E_{1}(L_{w})\right)\Leftrightarrow
t_{v}=1.$

∎

######  Lemma 3.10.

Let $E$ be an elliptic curve and $v$ be a prime of $K$. Assume that there is a
finite Galois extension $K^{\prime}/K$ with
$p\nmid[K^{\prime}\mathrel{\mathop{\mathchar 58\relax}}K]$ such that
$\gamma_{v^{\prime}}$ is injective for some prime $v^{\prime}|v$ of
$K^{\prime}$. Then, $\gamma_{v}$ itself is injective.

###### Proof.

Assume without loss of generality that $v$ is inert or ramified in $L$. Let
$w^{\prime}$ be the prime of $L^{\prime}$ that lies above $v^{\prime}$.
Setting $L^{\prime}\mathrel{\mathop{\mathchar 58\relax}}=K^{\prime}\cdot L$,
let $w$ be a prime of $L$ that lies below $w^{\prime}$, as depicted

$K$$L$$L^{\prime}$$K^{\prime}$$G$$G$$v.$$w$$w^{\prime}$$v^{\prime}$

Since $K^{\prime}/K$ is a prime to $p$-extension,
$\operatorname{Gal}(L^{\prime}_{w^{\prime}}/K^{\prime}_{v^{\prime}})$ is
naturally isomorphic to $\operatorname{Gal}(L_{w}/K_{v})$. By hypothesis, the
kernel of $\gamma_{v^{\prime}}$ is trivial, hence,

$H^{1}(\operatorname{Gal}(L_{w^{\prime}}^{\prime}/K_{v^{\prime}}^{\prime}),E)[p]=0.$

Consider the inflation-restriction sequence for the field extensions
$L^{\prime}_{w^{\prime}}/K^{\prime}_{v^{\prime}}/K_{v}$

$0\rightarrow H^{1}(K^{\prime}_{v^{\prime}}/K_{v},E)[p]\rightarrow
H^{1}(L^{\prime}_{w^{\prime}}/K_{v},E)[p]\rightarrow
H^{1}(L^{\prime}_{w^{\prime}}/K^{\prime}_{v^{\prime}},E)[p].$

Since $K^{\prime}/K$ is a prime to $p$-extension, it follows that

$H^{1}(K^{\prime}_{v^{\prime}}/K_{v},E)[p]=0,$

and we deduce that

$H^{1}(\operatorname{Gal}(L_{w^{\prime}}^{\prime}/K_{v}),E)[p]=0.$

Once again, from the inflation-restriction sequence for the fields
$L^{\prime}_{w^{\prime}}/L_{w}/K_{v}$, we deduce that

$H^{1}(\operatorname{Gal}(L_{w}/K_{v}),E)[p]=0$

and hence, $\gamma_{v}$ is injective. ∎

######  Lemma 3.11.

Let $E$ be an elliptic curve and $v|p$ be a prime of $K$ at which $E$ has
potentially good reduction. Assume that there is a finite Galois extension
$K^{\prime}/K$ such that $p\nmid[K^{\prime}\mathrel{\mathop{\mathchar
58\relax}}K]$ and for a prime $v^{\prime}|v$ of $K^{\prime}$, either of the
following conditions hold

  1. (1)

$E$ has good ordinary reduction at $v^{\prime}$ and
$\widetilde{E}(k_{v^{\prime}})[p]=0$, or,

  2. (2)

$E$ has good supersingular reduction at $v^{\prime}$ and $t_{v^{\prime}}\leq
1$.

Then, $\gamma_{v}$ is injective.

###### Proof.

The result follows from Lemma 3.7, Proposition 3.9 and Lemma 3.10. ∎

###  3.2. Rank stability in prime cyclic extensions

In this subsection, we study applications of our local computations to rank
stability questions in $\mathbb{Z}/p\mathbb{Z}$-extensions of $K$.

######  Theorem 3.12.

Let $K$ be a number field and $E$ be an elliptic curve over $K$. Let $p$ be an
odd prime number and assume that $\operatorname{Sel}_{p}(E/K)=0$. Let $L/K$ be
a degree $p$ Galois extension and $S$ be the set of primes $v$ of $K$ such
that either one of the following hold

  * •

$v|p$,

  * •

$E$ has bad reduction at $v$,

  * •

$v$ is ramified in $L$.

Write $S=\Sigma_{1}\cup\Sigma_{2}\cup\Sigma_{p}$, where $\Sigma_{1}$ is the
set of primes of $K$ at which $E$ has bad reduction and
$\Sigma_{2}\mathrel{\mathop{\mathchar
58\relax}}=S\setminus(\Sigma_{1}\cup\Sigma_{p})$. Assume that the following
conditions are satisfied.

  1. (1)

Each prime $v\in\Sigma_{1}$ splits completely in $L$.

  2. (2)

For each prime $v\in\Sigma_{2}$, we have that $\widetilde{E}(k_{v})[p]=0$.

  3. (3)

For each prime $v\in\Sigma_{p}$ which is not completely split in $L$, there is
a prime to $p$ Galois extension $K^{\prime}/K$ and a prime $v^{\prime}|v$ of
$K^{\prime}$ such that either

     * •

$E$ has good ordinary reduction at $v^{\prime}$ and
$\widetilde{E}(k_{v^{\prime}})[p]=0$, or,

     * •

$E$ has good supersingular reduction at $v^{\prime}$ and $t_{v^{\prime}}\leq
1$.

Then, the $p$-Selmer group $\operatorname{Sel}_{p}(E/L)=0$. In particular,
$\operatorname{rank}E(L)=0$ and $\Sh(E/L)[p]=0$.

###### Proof.

The Proposition 3.2 asserts that if $\gamma_{v}$ is injective for all primes
$v\in S$, then, $\operatorname{Sel}_{p}(E/L)=0$. For each prime
$v\in\Sigma_{1}$, it follows from the Lemma 3.3 that the map $\gamma_{v}$ is
injective. For each prime $v\in\Sigma_{2}$ such that $v\nmid p$, it follows
from Lemma 3.4 that $\gamma_{v}$ is injective. On the other hand, for $v|p$,
it follows from Lemma 3.11 that $\gamma_{v}$ is injective. ∎

Assuming that $K$ contains $\mu_{p}$, we shall consider Kummer extensions
$L=K(\alpha^{\frac{1}{p}})$ for $\alpha\in K^{*}/(K^{*})^{p}$. The following
is a well known criterion for the ramification and splitting behavior of
primes in such Kummer extensions. Given a prime $v\in\Sigma_{K}$, let
$v(\alpha)$ denote $v_{K_{v}}(\alpha)$ by abuse of notation.

######  Proposition 3.13.

Let $L/K$ be as above and $v\in\Sigma_{K}$ and
$i_{v}\mathrel{\mathop{\mathchar 58\relax}}K\hookrightarrow K_{v}$ denote the
natural embedding. Then the following assertions hold.

  1. (1)

Assume that $v\nmid p$. Then, $v$ is unramified in $L$ if and only if
$v(\alpha)\equiv 0\pmod{p}$.

  2. (2)

Assume that $v|p$. Then, $v$ is unramified in $L$ if and only if the
congruence

$\frac{\alpha}{x_{v}^{p}}\equiv 1\pmod{\pi_{v}^{pe_{v}}}$

has a solution for $x_{v}\in K_{v}^{\times}$, where $e_{v}$ is the
ramification index of $v$ in $K/\mathbb{Q}(\mu_{p})$.

  3. (3)

The prime $v$ splits in $L$ if and only if
$i_{v}(\alpha)\in(K_{v}^{\times})^{p}$.

###### Proof.

The above result is [Gra03, Theorem 6.3]. ∎

Let us consider the special case where $K\mathrel{\mathop{\mathchar
58\relax}}=\mathbb{Q}(\mu_{p})$.

######  Lemma 3.14.

Suppose that $K=\mathbb{Q}(\mu_{p})$ and let $v$ be the unique prime of $K$
that lies above $p$. Let $m$ be a rational integer such that
$m\notin(K^{\times})^{p}$. Setting $L\mathrel{\mathop{\mathchar
58\relax}}=K(m^{\frac{1}{p}})$, we make the following assumptions.

  1. (1)

Suppose that $m^{p-1}\equiv 1\mod{p^{2}}$. Then, $v$ splits in $L$.

  2. (2)

Suppose that $p\nmid m$, then, $t_{v}\leq 1$.

###### Proof.

The condition that $m^{p-1}\equiv 1\mod{p^{2}}$ implies that
$m\in(\mathbb{Q}_{p}^{\times})^{p}$. In particular, this implies that
$m\in(K_{v}^{\times})^{p}$ and thus $v$ splits in $L$ by Proposition 3.13.
This proves part (1). For Part (2), assume without loss of generality that
$p||(m^{p-1}-1)$. The result then follows from [Viv04, Theorem 5.6]. ∎

######  Theorem 3.15.

Let $p$ be an odd prime number and $K=\mathbb{Q}(\mu_{p})$. Assume that
$\operatorname{Sel}_{p}(E/K)=0$ and $m$ be a rational integer such that
$m\notin(K^{\times})^{p}$. Let $Q_{1}$ be the set of primes $v\nmid p$ at
which $E$ has good reduction and $\widetilde{E}(k_{v})[p]=0$. Let $Q_{2}$ be
the set of primes $v\nmid p$ at which $E$ has bad reduction. Then, assume that
the following conditions are satisfied.

  1. (1)

The only primes $v\nmid p$ for which $v(m)\not\equiv 0\mod{p}$ are contained
in $Q_{1}$.

  2. (2)

For each prime $v\in Q_{2}$, we have that $i_{v}(m)\in(K_{v}^{\times})^{p}$.

  3. (3)

For each prime $v\mid p$ at which $E$ has good ordinary reduction and
$\widetilde{E}(k_{v})[p]\neq 0$, we have that
$i_{v}(m)\in(K_{v}^{\times})^{p}$.

  4. (4)

For each prime $v\mid p$ at which $E$ has supersingular reduction, we have
that $v(m)\equiv 0\mod{p}$.

  5. (5)

For each prime $v\mid p$ of bad reduction for $E$, we have that
$i_{v}(m)\in(K_{v}^{\times})^{p}$.

Then, we have that $\operatorname{Sel}_{p}(E/K(m^{1/p}))=0$.

###### Proof.

Setting $L\mathrel{\mathop{\mathchar 58\relax}}=K(m^{1/p})$ it suffices to
show that the conditions of Theorem 3.12 are satisfied. This is a direct
consequence of Proposition 3.13 and Lemma 3.14. ∎

##  4\. Elliptic curves with rational $3$-isogeny : Stability results

We illustrate our results from the previous section to the case when $p=3$. In
this setting, we may consider families of elliptic curves $E_{/\mathbb{Q}}$
that have a rational $3$-isogeny. More specifically, we consider the family of
elliptic curves given by $E\mathrel{\mathop{\mathchar
58\relax}}=E_{a}\mathrel{\mathop{\mathchar 58\relax}}y^{2}=x^{3}+a$, where $a$
is a cubefree integer. Let $\varphi_{a}\mathrel{\mathop{\mathchar
58\relax}}E\rightarrow E^{\prime}$ be a rational $3$-isogeny, where
$E^{\prime}\mathrel{\mathop{\mathchar 58\relax}}y^{2}=x^{3}-27a$. Set
$\zeta\mathrel{\mathop{\mathchar 58\relax}}=\frac{-1+\sqrt{-3}}{2}$ and
$\varpi\mathrel{\mathop{\mathchar 58\relax}}=1-\zeta$. We have that
$\varpi=-\sqrt{-3}\zeta$, and hence, $\varpi^{6}=-27$. Thus, over
$K\mathrel{\mathop{\mathchar 58\relax}}=\mathbb{Q}(\mu_{3})$, the curves
$E^{\prime}$ and $E$ are isomorphic via
$\theta_{\varpi}\mathrel{\mathop{\mathchar
58\relax}}E^{\prime}\xrightarrow{\sim}E$, defined by

$\theta_{\varpi}(x,y)\mathrel{\mathop{\mathchar
58\relax}}=\left(\frac{x}{\varpi^{2}},\frac{y}{\varpi^{3}}\right).$

Hence, $E$ is $3$-isogenous to itself over $K$. Let
$\varphi\mathrel{\mathop{\mathchar
58\relax}}=\theta_{\varpi}\circ\varphi_{a}\mathrel{\mathop{\mathchar
58\relax}}E\rightarrow E$ be the isogeny which is explicitly given as follows

$\varphi(x,y)=\left(\frac{x^{3}+4a}{\varpi^{2}x^{2}},\frac{y(x^{3}-8a)}{\varpi^{3}x^{3}}\right).$

The kernel of $\varphi$ is the order $3$ subgroup
$\\{O,(0,\sqrt{a}),(0,-\sqrt{a})\\}\subset E(\bar{K})$, cf. [Top93, p. 3].
Denote by $E(K)[\varphi]$ the kernel of the map
$\varphi\mathrel{\mathop{\mathchar 58\relax}}E(K)\rightarrow E(K)$. Since
$\varphi$ is self-dual, we find that $E(K)[\varphi^{2}]=E(K)[3]$, and thus,
$E(K)[\varphi]$ is contained in $E(K)[3]$.

######  Lemma 4.1.

With respect to notation above, the following are equivalent

  1. (1)

$a\notin(K^{\times})^{2}$,

  2. (2)

$E(K)[\varphi]=0$,

  3. (3)

$E(K)[3]=0$.

###### Proof.

Note that the point $(0,\sqrt{a})$ of $E(\bar{K})$ has order $3$ and is the
generator of $E[\varphi]$. Thus, $a$ is not a square in $K^{\times}$ if and
only if $(0,\sqrt{a})$ is not contained in $E(K)$. This shows that

$a\notin(K^{\times})^{2}\Leftrightarrow E(K)[\varphi]=0,$

i.e., conditions (1) and (2) are equivalent.

There is a natural short exact sequence

$0\rightarrow E[\varphi]\rightarrow
E[3]\xrightarrow{\varphi}E[\varphi]\rightarrow 0.$

Since $\varphi$ is defined over $K$, taking Galois invariants gives the exact
sequence

$0\rightarrow E(K)[\varphi]\rightarrow
E(K)[3]\xrightarrow{\varphi}E(K)[\varphi].$

Thus, one has $E(K)[\varphi]=0$ if and only if $E(K)[3]=0$. This proves that
conditions (2) and (3) are equivalent, completing the proof. ∎

###  4.1. Results of Mazur and Rubin on the parity of ranks

Note that $E$ is isogenous to its quadratic twist by $-3$ over $\mathbb{Q}$
and hence $\operatorname{rank}E(K)=2\operatorname{rank}E(\mathbb{Q})$. In
particular, the rank of $E(K)$ is even. It is however still natural to study
the parity of the rank of $E(K)$ as an $\mathcal{O}_{K}$-module. We briefly
discuss parity results of Mazur and Rubin [MR07] for cubic twist families.

Let $m$ be a cubefree integer. Consider the cubic twists $E_{(m)}$ and
$E_{(m^{2})}$ of $E$, defined as follows

$\begin{split}E_{(m)}\mathrel{\mathop{\mathchar
58\relax}}=E_{m^{2}a}\mathrel{\mathop{\mathchar
58\relax}}y^{2}=x^{3}+m^{2}a;\\\ E_{(m^{2})}\mathrel{\mathop{\mathchar
58\relax}}=E_{m^{4}a}\mathrel{\mathop{\mathchar
58\relax}}y^{2}=x^{3}+m^{4}a.\end{split}$

In the next subsection, it is proven that for many values of $m$, the groups
$E_{(m)}(K)$ and $E_{(m^{2})}(K)$ both have rank $0$. Setting
$L\mathrel{\mathop{\mathchar 58\relax}}=K(\sqrt[3]{m})$, we take
$\chi=\chi_{m}\mathrel{\mathop{\mathchar
58\relax}}\operatorname{Gal}(L/K)\rightarrow\mu_{3}$ to be the cubic character
of order $3$ defined as follows

$\chi(\sigma)\mathrel{\mathop{\mathchar
58\relax}}=\frac{\sigma(\sqrt[3]{m})}{\sqrt[3]{m}},$

for $\sigma\in\operatorname{Gal}(L/K)$. Since
$\operatorname{Aut}_{\bar{K}}(E)\simeq\mathcal{O}_{K}^{\times}=\mu_{6}$, we
find that the action of $\operatorname{G}_{K}$ on
$\operatorname{Aut}_{\bar{K}}(E)$ is trivial. We view $\mu_{3}$ as a subgroup
of $\operatorname{Aut}_{\bar{K}}(E)$ and thus view $\chi$ as a cohomology
class as follows

$\chi\in\operatorname{Hom}\left(\operatorname{G}_{K},\operatorname{Aut}_{\bar{K}}(E)\right)=H^{1}\left(\operatorname{G}_{K},\operatorname{Aut}_{\bar{K}}(E)\right).$

The elliptic curve $E_{(m)}$ (resp. $E_{(m^{2})}$) is the twist $E_{\chi}$
(resp. $E_{\chi^{2}}$). In other words, there is an isomorphism
$\phi\mathrel{\mathop{\mathchar 58\relax}}E_{\chi}\xrightarrow{\sim}E$ over
$\bar{K}$ such that $\phi^{\sigma}=\chi(\sigma)\phi$, cf. [Sil09, section
X.2]. For $P\in E_{\chi}(\bar{K})$, we have that

(4.1) $\sigma(\phi(P))=\chi(\sigma)\phi(\sigma(P)).$

Given a $\operatorname{G}_{K}$-module $M$, take

$M^{\chi}\mathrel{\mathop{\mathchar 58\relax}}=\\{m\in
M\mid\sigma(m)=\chi(\sigma)m\text{ for all }\sigma\in\operatorname{G}_{K}\\}.$

In view of (4.1), we have the identification $E_{\chi}(K)=E(L)^{\chi}$, and
therefore,

$E(L)\simeq E(K)\oplus E(L)^{\chi}\oplus E(L)^{\chi^{2}}\simeq E(K)\oplus
E_{\chi}(K)\oplus E_{\chi^{2}}(K).$

From (2.4), one obtains the short exact sequence

$0\rightarrow
E_{\chi}(K)\otimes\mathbb{Z}_{3}\rightarrow\mathfrak{S}_{3}(E/L)^{\chi}\rightarrow\operatorname{T}_{3}\left(\Sh(E/L)[3^{\infty}]\right)^{\chi}\rightarrow
0,$

and thus,

$\operatorname{rank}\left(\mathfrak{S}_{3}(E/L)^{\chi}\right)=\operatorname{rank}E_{\chi}(K)$

provided $\Sh(E/L)[3^{\infty}]$ is finite.

Let $S$ be the set of primes of $K$ consisting of the prime above $3$, the
primes at which $E$ has bad reduction and the primes that ramify in $L$. For
each prime $v\in S$, there is a locally defined invariant
$\delta_{v}\in\mathbb{Z}/2\mathbb{Z}$, cf. [MR07, Definition 4.5]. Let
$\mathbb{Z}_{3}[\chi]$ be the extension of $\mathbb{Z}_{3}$ generated by the
values of $\chi$. The following result follows from Theorem A of _loc. cit._

######  Theorem 4.2 (Mazur–Rubin).

With respect to notation above, we have that

$\operatorname{rank}_{\mathbb{Z}_{3}}\left(\mathfrak{S}_{3}(E/K)\right)-\operatorname{rank}_{\mathbb{Z}_{3}[\chi]}\left(\mathfrak{S}_{3}(E/L)^{\chi}\right)\equiv\sum_{v\in
S}\delta_{v}\pmod{2}.$

Note that the value of
$\operatorname{rank}_{\mathbb{Z}_{3}}\left(\mathfrak{S}_{3}(E/K)\right)$ is
expected to be even, and thus the above result predicts that

(4.2) $\operatorname{rank}E_{(m)}(K),\operatorname{rank}E_{(m^{2})}(K)\equiv
2\left(\sum_{v\in S}\delta_{v}\right)\pmod{4}.$

Since $\operatorname{rank}E_{(m)}(K)=2\operatorname{rank}E_{(m)}(\mathbb{Q})$,
the above relation implies that

$\operatorname{rank}E_{(m)}(\mathbb{Q}),\operatorname{rank}E_{(m^{2})}(\mathbb{Q})\equiv\left(\sum_{v\in
S}\delta_{v}\right)\pmod{2}.$

###  4.2. Main results

In this section, we prove the main results of this article. As in the previous
subsection, let $m$ be a cubefree integer. Assume now that $m$ is coprime to
$3$. Set $L_{m}\mathrel{\mathop{\mathchar 58\relax}}=K(\sqrt[3]{m})$, note
that $L_{m}/K$ is a cyclic cubic extension. Notice that the substitution

$(x,y)\mapsto(m^{2/3}x,my)$

gives us that $E$ and $E_{(m)}$ (resp. $E_{(m)}$ and $E_{(m^{2})}$) are
isomorphic over $\mathbb{Q}(\sqrt[3]{m})$. We find there are isomorphisms of
the Selmer groups

$\operatorname{Sel}_{3}(E/L_{m})\simeq\operatorname{Sel}_{3}(E_{(m)}/L_{m})\simeq\operatorname{Sel}_{3}(E_{(m^{2})}/L_{m}).$

Thus in particular, if $\operatorname{Sel}_{3}(E/L_{m})=0$, then
$\operatorname{rank}E_{(m)}(K)=0$ and $\operatorname{rank}E_{(m^{2})}(K)=0$.
Let $\mathfrak{p}\mathrel{\mathop{\mathchar 58\relax}}=\varpi\mathcal{O}_{K}$
be the unique prime of $K$ that lies above $(3)$. Note that $E$, $E_{(m)}$ and
$E_{(m^{2})}$ all have additive reduction at $\mathfrak{p}$.

From here on in, denote by $\ell$ a prime divisor of $m$, and $q$ a prime
divisor of $a$.

######  Definition 4.3.

Let $\mathcal{M}_{a}$ be the set of rational primes $\ell$ such that the
following conditions are satisfied

  * •

$\ell\nmid a$,

  * •

$\ell\equiv 1\pmod{6}$,

  * •

$\widetilde{E}(\mathbb{F}_{\ell})[3]=0$.

######  Theorem 4.4.

With respect to notation above, assume that

  1. (1)

$m\equiv 1\pmod{9}$,

  2. (2)

$\operatorname{Sel}_{3}(E/K)=0$,

  3. (3)

$m$ is only divisible by primes of $\mathcal{M}_{a}$,

  4. (4)

$m$ is a cube modulo $q$ for each prime $q|a$.

Then, $\operatorname{Sel}_{3}(E/L_{m})=0$, in particular,
$\operatorname{rank}E_{(m)}(K)=0$ and $\operatorname{rank}E_{(m^{2})}(K)=0$.

###### Proof.

We show that the conditions of Theorem 3.12 are indeed satisfied for
$L=L_{m}$.

  1. (1)

First, we consider the condition for the primes of $\Sigma_{1}$. Recall that
these are primes $v\nmid 3$ at which $E$ has bad reduction. We must show that
such $v$ must be completely split in $L$. By part (3) of Proposition 3.13, we
have that $v$ splits in $L$ if and only if $m\in(K_{v}^{\times})^{3}$. Let $q$
be the rational prime such that $v|q$. Since $v$ is a prime of bad reduction,
it follows that $q|2a$. Consider two cases. First, suppose that $q|a$. Note
that $q\nmid m$ and $q\neq 3$. Thus, $m\in(\mathbb{Z}_{q}^{\times})^{3}$ if
and only if $m$ is a cube modulo $q$. Since $m$ is a cube modulo $q$ by
assumption, we conclude that $m\in(\mathbb{Z}_{q}^{\times})^{3}$ and hence,
$m\in(K_{v}^{\times})^{3}$. Thus $v$ is completely split in $L_{m}$ for all
primes $v\in\Sigma_{1}$. Now consider the other case, namely, $q=2$. In this
case, since $m$ is odd, $m\in\mathbb{Z}_{2}^{\times}$. However, it is easy to
see that $\mathbb{Z}_{2}^{\times}=(\mathbb{Z}_{2}^{\times})^{3}$, and thus, in
particular $m\in(K_{v}^{\times})^{3}$ and thus, $v$ splits in $L$.

  2. (2)

Recall that $\Sigma_{2}$ is the set of primes $v\nmid 3$ of $K$ at which $E$
has good reduction which are ramified in $L$. The condition (2) of Theorem
3.12 requires that $\widetilde{E}(k_{v})[3]=0$ for all primes
$v\in\Sigma_{2}$. This follows from the condition that $m$ is only divisible
by primes of $\mathcal{M}_{a}$. In greater detail, let $v\in\Sigma_{2}$ and
$\ell$ be the rational prime such that $v|\ell$. Since $v$ is ramified in $L$,
it follows that $\ell|m$. Thus, $\ell\in\mathcal{M}_{a}$, and by definition,
$\widetilde{E}(\mathbb{F}_{\ell})[3]=0$. We are to show that
$\widetilde{E}(k_{v})[3]=0$. Since $\ell\in\mathcal{M}_{a}$, it follows by
definition that $\ell\equiv 1\mod{3}$, thus $\ell$ splits in $K$. Thus,
$k_{v}=\mathbb{F}_{\ell}$, and therefore, $\widetilde{E}(k_{v})[3]=0$.

  3. (3)

Since $m\equiv 1\mod{9}$, it follows that $m\in(\mathbb{Z}_{3}^{\times})^{3}$.
Thus in particular, it follows that $m\in(K_{v}^{\times})^{3}$ and thus $v$
splits in $L$ for the prime $v$ of $K$ that lies above $3$. Therefore, the
condition (3) of Theorem 3.12 is satisfied.

The result thus follows from Theorem 3.12. ∎

######  Definition 4.5.

Let $\mathcal{Q}_{a}$ be the set of primes $\ell$ such that

  1. (1)

$\ell\nmid a$,

  2. (2)

$\ell\equiv 1\pmod{18}$,

  3. (3)

$\ell$ is a cube modulo $q$ for each prime $q$ that divides $a$.

  4. (4)

$\widetilde{E}(\mathbb{F}_{\ell})[3]=0$.

######  Corollary 4.6.

Let $K=\mathbb{Q}(\mu_{3})$ and assume that $\operatorname{Sel}_{3}(E/K)=0$.
Let $m$ be an integer that is divisible only by primes of $\mathcal{Q}_{a}$.
Then, we have that $\operatorname{rank}E_{(m)}(K)=0$ and
$\operatorname{rank}E_{(m^{2})}(K)=0$.

###### Proof.

It is clear that the conditions of Theorem 4.4 are satisfied for integers $m$
that are divisible exclusively by primes in $\mathcal{Q}_{a}$. ∎

In light of Corollary 4.6, it is natural to investigate the density of the set
of primes $\mathcal{Q}_{a}$. Let $\rho_{E,3}\mathrel{\mathop{\mathchar
58\relax}}\operatorname{G}_{\mathbb{Q}}\rightarrow\operatorname{GL}_{2}(\mathbb{F}_{3})$
denote the Galois representation on $E[3]$. Set $\mathbb{Q}(E[3])$ to denote
the Galois extension of $\mathbb{Q}$ which is _cut out by_ $E[3]$. In greater
detail, $\mathbb{Q}(E[3])$ is defined as
$\bar{\mathbb{Q}}^{\operatorname{ker}\rho_{E,3}}$. It is clear that

$\operatorname{Gal}(\mathbb{Q}(E[3])/\mathbb{Q})\simeq\frac{\operatorname{G}_{\mathbb{Q}}}{\operatorname{ker}\rho_{E,3}}\simeq\operatorname{image}\rho_{E,3}.$

Given $\sigma\in\operatorname{Gal}(\mathbb{Q}(E[3])/\mathbb{Q})$, set
$\rho(\sigma)\mathrel{\mathop{\mathchar 58\relax}}=\rho_{E,3}(\sigma)$. Note
that $E$ admits a rational $3$-isogeny $\varphi\mathrel{\mathop{\mathchar
58\relax}}E\rightarrow E^{\prime}$, and therefore, there is a
$\operatorname{G}_{\mathbb{Q}}$-stable subgroup $C\subset E[3]$ that has order
$3$. This means that the representation $\rho_{E,3}$ is reducible. Choose an
$\mathbb{F}_{3}$-basis of $E[3]$ such that the image of $\rho_{E,3}$ is
contained $B$, the group of upper triangular matrices
$\left({\begin{array}[]{cc}\ast&\ast\\\ 0&\ast\\\ \end{array}}\right)$ in
$\operatorname{GL}_{2}(\mathbb{F}_{3})$. In other words, let $v_{1}$ be a
generator of $C$ and let $v_{2}\in E[3]$ be another vector not contained in
$C$. Then, with respect to the basis $\\{v_{1},v_{2}\\}$, one obtains an upper
triangular representation. Thus, there are characters
$\varphi,\psi\mathrel{\mathop{\mathchar
58\relax}}\operatorname{G}_{\mathbb{Q}}\rightarrow\mathbb{F}_{3}^{\times}=\\{\pm
1\\}$ such that

$\rho_{E,3}=\left({\begin{array}[]{cc}\varphi&\ast\\\ 0&\psi\\\
\end{array}}\right).$

As a Galois module $C\simeq\mathbb{F}_{3}(\varphi)$. Let $\omega$ denote the
mod-$3$ cyclotomic character and note that
$\det\rho_{E,3}=\varphi\psi=\omega$. Write $B=T\ltimes U$, where $T$ consists
of diagonal matrices $\left({\begin{array}[]{cc}\ast&\\\ &\ast\\\
\end{array}}\right)$ and $U$ consists of unipotent matrices
$\left({\begin{array}[]{cc}1&\ast\\\ 0&1\\\ \end{array}}\right)$.

######  Lemma 4.7.

With respect to notation above, the following conditions are equivalent

  1. (1)

$E(K)[3]\neq 0$.

  2. (2)

We have that $\\{\varphi,\psi\\}=\\{1,\omega\\}$,

  3. (3)

the image of $\rho_{E,3}$ does not contain
$-\operatorname{Id}=\left({\begin{array}[]{cc}-1&0\\\ 0&-1\\\
\end{array}}\right)$.

###### Proof.

First, we prove (1) is equivalent to (2). Since $\varphi$ and $\psi$ take
values in $\\{\pm 1\\}$, we have that $\varphi^{2}=\psi^{2}=1$. Since
$\varphi\psi=\omega$, we find that $\psi=\varphi^{-1}\omega=\varphi\omega$.
Since $\omega_{|\operatorname{G}_{K}}=1$, we find that
$\varphi_{|\operatorname{G}_{K}}=\psi_{|\operatorname{G}_{K}}$. The condition
that $E(K)[3]\neq 0$ is equivalent to the condition that
$\varphi_{|\operatorname{G}_{K}}=\psi_{|\operatorname{G}_{K}}=1$. Note that
any character $\epsilon\mathrel{\mathop{\mathchar
58\relax}}\operatorname{G}_{\mathbb{Q}}\rightarrow\mathbb{F}_{3}^{\times}$
such that $\epsilon_{|\operatorname{G}_{K}}=1$ is either $1$ or $\omega$.
Thus, the condition (1) implies that $\varphi,\psi\in\\{1,\omega\\}$. Since
$\varphi=\psi\omega$, we find that $\\{\varphi,\psi\\}=\\{1,\omega\\}$.
Conversely, if $\\{\varphi,\psi\\}=\\{1,\omega\\}$, then,
$\varphi_{|\operatorname{G}_{K}}=\psi_{|\operatorname{G}_{K}}=1$, and thus,
$E(K)[3]\neq 0$. This implies that (2) implies (1). Thus, (1) and (2) are
equivalent.

Suppose that $\\{\varphi,\psi\\}=\\{1,\omega\\}$, then,
$-\operatorname{Id}\notin\operatorname{image}\rho_{E,3}$ since either
$\varphi$ or $\psi$ is $1$. Conversely, suppose that
$\\{\varphi,\psi\\}\neq\\{1,\omega\\}$, then, $\varphi\neq 1$. Thus, there
exists $\sigma\in\operatorname{G}_{K}$ such that $\varphi(\sigma)=-1$. Since
$\varphi(\sigma)\psi(\sigma)=\omega(\sigma)=1$, we find that $\psi(\sigma)=-1$
and hence $\rho_{E,3}(\sigma)^{3}=-\operatorname{Id}$. This proves that (2)
and (3) are equivalent. ∎

Let $q$ be a prime dividing $a$. Suppose that $q\equiv 1\mod{3}$, then set
$\mathbb{L}_{q}$ to denote the unique $\mathbb{Z}/3\mathbb{Z}$ extension of
$\mathbb{Q}$ contained in $\mathbb{Q}(\mu_{q})$. On the other hand, if
$q\not\equiv 1\mod{3}$, set $\mathbb{L}_{q}\mathrel{\mathop{\mathchar
58\relax}}=\mathbb{Q}$. Denote by

$\mathbb{L}\mathrel{\mathop{\mathchar
58\relax}}=\left(\prod_{q|a}\mathbb{L}_{q}\right)\cdot\mathbb{Q}(\mu_{18})$

the compositum of all the fields $\mathbb{L}_{q}$ as $q$ ranges over the
primes dividing $a$, and the field $\mathbb{Q}(\mu_{18})$.

######  Lemma 4.8.

Let $\ell\nmid 6a$ be a prime. Then, the conditions (1)–(3) of Definition 4.5
are satisfied by $\ell$ if and only if $\ell$ splits completely in
$\mathbb{L}$.

###### Proof.

A prime $\ell$ splits in $\mathbb{Q}(\mu_{18})$ if and only if $\ell\equiv
1\mod{18}$. Note that $\ell$ splits in $\mathbb{L}_{q}$ if and only if $\ell$
is a cube modulo $q$. This proves the result. ∎

Note that the field $\mathbb{Q}(\mu_{18})$ is unramified at all primes other
than $2$ and $3$. On the other hand, $\mathbb{L}_{q}$ is totally ramified at
$q$, and unramified at all other primes. Recall that
$\mathbb{L}_{q}\mathrel{\mathop{\mathchar 58\relax}}=\mathbb{Q}$ unless
$q\equiv 1\mod{3}$. This means that the fields $\\{\mathbb{L}_{q}\\}$ and
$\mathbb{Q}(\mu_{18})$ are all mutually disjoint. Therefore, we find that

$[\mathbb{L}\mathrel{\mathop{\mathchar
58\relax}}\mathbb{Q}]=\prod_{q|a}[\mathbb{L}_{q}\mathrel{\mathop{\mathchar
58\relax}}\mathbb{Q}]\times[\mathbb{Q}(\mu_{18})\mathrel{\mathop{\mathchar
58\relax}}\mathbb{Q}]=3^{s}\times 6=2\times 3^{s+1},$

where $s$ is the number of primes $q\equiv 1\mod{3}$ which divide $a$. Set
$F\mathrel{\mathop{\mathchar 58\relax}}=\mathbb{Q}(E[3])$,
$G\mathrel{\mathop{\mathchar 58\relax}}=\operatorname{Gal}(F/\mathbb{Q})$ and
$A\subset G$ be the subset of $\sigma\in G$ such that

$\operatorname{det}\rho_{E,3}(\sigma)=1\text{ and
}\operatorname{trace}\rho_{E,3}(\sigma)\neq 2.$

######  Lemma 4.9.

With respect to notation above, let $\ell\nmid 6a$ be a prime number. Then
$\ell$ is unramified in $\mathbb{Q}(E[3])$ and the following are equivalent

  1. (1)

$\ell\equiv 1\mod{3}$ and $\widetilde{E}(\mathbb{F}_{\ell})[3]=0$,

  2. (2)

$\operatorname{Frob}_{\ell}\in A$.

###### Proof.

Since $\ell\nmid 6a$, it is a prime of good reduction and it follows from the
Neron–Ogg–Shafarevich theorem that $\ell$ is unramified in $\mathbb{Q}(E[3])$.
We then note that

$\operatorname{det}\rho_{E,\ell}(\operatorname{Frob}_{\ell})=\ell\text{ and
}\operatorname{trace}\rho_{E,3}(\operatorname{Frob}_{\ell})=\ell+1-\\#\widetilde{E}(\mathbb{F}_{\ell}).$

The result thus follows. ∎

######  Lemma 4.10.

With respect to the notation above, we have that $F\cap\mathbb{L}=K$.

###### Proof.

Set $H\mathrel{\mathop{\mathchar 58\relax}}=\operatorname{Gal}(F/K)\subset G$
and $H^{\prime}\mathrel{\mathop{\mathchar
58\relax}}=\operatorname{Gal}(F/F\cap\mathbb{L})\subseteq H$. We show that
$H^{\prime}=H$. Note that $G\simeq\rho_{E,3}(G)$ and $H\simeq\rho_{E,3}(H)$,
and hence we may view these as subgroups of $B$. Since
$\omega=\operatorname{det}\rho_{E,3}$, we find that $G$ contains a matrix with
determinant $-1$. Note that $B=\langle-\operatorname{Id}\rangle\times B_{0}$,
where $B_{0}$ consists of matrices of the form
$\left({\begin{array}[]{cc}\ast&\ast\\\ 0&1\\\ \end{array}}\right)$. Note that
$B_{0}$ contains $U$ as its unique normal subgroup of order $3$, and contains
$3$ subgroups $C_{1}$, $C_{2}$ and $C_{3}$ that are of order $2$. The
possibilities for $G$ are $B$, $B_{0}$, $C_{i}$ and
$\langle-\operatorname{Id}\rangle\times C_{i}$ as $i=1,2,3$. Note that since
$-\operatorname{Id}\in G$, the groups $U$ and
$\langle-\operatorname{Id}\rangle\times U$ are excluded from our list. All
abelian quotients of these groups are $2$-groups. Hence,
$G/H^{\prime}=\operatorname{Gal}(F\cap\mathbb{L}/\mathbb{Q})$ is an abelian
$2$-group. Since the abelian $2$-quotient of
$\operatorname{Gal}(\mathbb{L}/\mathbb{Q})$ is
$\operatorname{Gal}(K/\mathbb{Q})$, it follows that $F\cap\mathbb{L}=K$. ∎

######  Proposition 4.11.

Assume that $a$ is not a square in $K$ and is prime to $3$. Then, with respect
to notation above, the set of primes $\mathcal{Q}_{a}$ has positive density
given by

$\mathfrak{d}(\mathcal{Q}_{a})=\frac{1}{8\times 3^{s+1}},$

where $s$ is the number of prime divisors $q$ of $a$ such that $q\equiv
1\pmod{3}$.

###### Proof.

Recall that $F\mathrel{\mathop{\mathchar 58\relax}}=\mathbb{Q}(E[3])$,
$G\mathrel{\mathop{\mathchar 58\relax}}=\operatorname{Gal}(F/\mathbb{Q})$ and
$H\mathrel{\mathop{\mathchar 58\relax}}=\operatorname{Gal}(F/K)$. The
representation

$\rho=\rho_{E,3}\mathrel{\mathop{\mathchar
58\relax}}\operatorname{G}_{\mathbb{Q}}\rightarrow\operatorname{GL}_{2}(\mathbb{F}_{3}),$

induces a natural isomorphism between $G$ and the image of $\rho$. Let
$\ell\nmid 6a$ be a prime, then, $\ell$ is unramified in $F$. Denote by
$\operatorname{Frob}_{\ell}\in G$ a Frobenius element of $\ell$. Recall that
$A$ is the subset of $G$ consisting of $\sigma$ such that

$\operatorname{trace}\left(\rho(\sigma)\right)\neq 2\text{ and
}\operatorname{det}\left(\rho(\sigma)\right)=1.$

By Lemma 4.9, the following are equivalent

  1. (1)

$\ell\equiv 1\mod{3}$ and $\widetilde{E}(\mathbb{F}_{\ell})[3]=0$,

  2. (2)

$\operatorname{Frob}_{\ell}\in A$.

It is clear that $A$ is contained in $H$. Since $a$ is assumed to not be
square in $K$, we find that $E(K)[3]=0$. By Lemma 4.7, $-\operatorname{Id}\in
A$. Given any matrix $h\in H$, we have that $\operatorname{det}(\rho(h))=1$
and $\operatorname{trace}(\rho(h))$ is either $1$ or $2$. Note that the trace
of a matrix with determinant $1$ cannot be $0$. If $h\in H$, then since
$-\operatorname{Id}\in H$, we have that $-h\in H$. Thus, we find that

(4.3) $\\#A=\frac{\\#H}{2}.$

We compute the density of $\mathcal{Q}_{a}$ by applying the Chebotarev density
theorem to the field $F\cdot\mathbb{L}$. Note that $F\cap\mathbb{L}=K$ by
Lemma 4.10. Thus, the Galois group
$\operatorname{Gal}(F\cdot\mathbb{L}/K)\simeq
H\times\operatorname{Gal}(\mathbb{L}/K)$. Let $\mathcal{S}$ be the subset
$A\times\\{1\\}$ contained in $\operatorname{Gal}(F\cdot\mathbb{L}/K)$ and
view $\mathcal{S}$ as a subset of
$\operatorname{Gal}(F\cdot\mathbb{L}/\mathbb{Q})$. Note that the set
$\mathcal{S}$ is stable with respect to the action of conjugation. Let $\ell$
be a prime which is unramified in $F\cdot\mathbb{L}$ and such that
$\operatorname{Frob}_{\ell}\in\mathcal{S}$. Then, by Lemma 4.8, the prime
$\ell$ splits completely in $\mathbb{L}$, and consequently, conditions (1)–(3)
of Definition 4.5 are satisfied. Also $\operatorname{Frob}_{\ell}\in A$, it
follows from Lemma 4.9 that (4) is also satisfied. Thus, by the Chebotarev
density theorem and (4.3), we find that

$\mathfrak{d}(\mathcal{Q}_{a})=\frac{\\#\mathcal{S}}{[F\cdot\mathbb{L}\mathrel{\mathop{\mathchar
58\relax}}\mathbb{Q}]}=\frac{\\#\mathcal{S}}{2[F\cdot\mathbb{L}\mathrel{\mathop{\mathchar
58\relax}}K]}=\frac{\\#A}{2\\#H\times[\mathbb{L}\mathrel{\mathop{\mathchar
58\relax}}\mathbb{Q}]}=\frac{1}{4[\mathbb{L}\mathrel{\mathop{\mathchar
58\relax}}\mathbb{Q}]}=\frac{1}{8\times 3^{s+1}}.$

∎

We prove the main result of this article.

###### Proof of Theorem A.

It is easy to see that $a\notin(K^{\times})^{2}$. The result is a direct
consequence of Corollary 4.6 and Proposition 4.11. ∎

######  Corollary 4.12.

Let $E$ and $K$ be as in Theorem A. Then, there is a positive density of
primes $\ell$ such that

$\operatorname{rank}E_{\ell^{2}a}(K)=0\text{ and
}\operatorname{rank}E_{\ell^{4}a}(K)=0.$

###### Proof.

The result is an immediate consequence of Theorem A, upon setting $m=\ell$ to
be a prime number in $\mathcal{Q}_{a}$. ∎

###  4.3. An illustrative example

Let us illustrate the result in Theorem A through an explicit example.
Consider the curve $E_{-1}\mathrel{\mathop{\mathchar 58\relax}}y^{2}=x^{3}-1$.
We check that $\operatorname{Sel}_{3}(E_{-1}/K)=0$. The quadratic twist of $E$
by $-3$ is given by $E_{27}$. It suffices to show that

$\operatorname{Sel}_{3}(E_{-1}/\mathbb{Q})=0\text{ and
}\operatorname{Sel}_{3}(E_{27}/\mathbb{Q})=0.$

This is computed via the following Magma code

    E:=EllipticCurve([0,0,0,0,-1]);
    ThreeSelmerGroup(E);
    Et := EllipticCurve([0,0,0,0,27]);
    ThreeSelmerGroup(Et);

    Ouput:
    Abelian Group of order 1.
    Abelian Group of order 1

Note that $\mathcal{Q}_{-1}$ (cf. Definition 4.5) consists of the primes
$\ell\equiv 1\mod{18}$ such that $\widetilde{E}_{-1}(\mathbb{F}_{\ell})[3]=0$.
Corollary 4.6 then asserts that if $m$ is divisible only by primes of
$\mathcal{Q}_{-1}$, then $E_{-m^{2}}(K)$ and $E_{-m^{4}}(K)$ both have rank
$0$. Proposition 4.11 asserts that the density of $\mathcal{Q}_{-1}$ is
$\frac{1}{24}$. We arrive at the following concrete application of Theorem 4.4
and Proposition 4.11 for the elliptic curve $E\mathrel{\mathop{\mathchar
58\relax}}=E_{-1}$.

######  Theorem 4.13.

Let $E_{/\mathbb{Q}}$ be the elliptic curve $y^{2}=x^{3}-1$. Given a cube-free
integer $m$, let $E_{(m)}$ (resp. $E_{(m^{2})}$) denote the cubic twist of $E$
defined by $y^{2}=x^{3}-m^{2}$ (resp. $y^{2}=x^{3}-m^{4}$). Let
$\mathcal{Q}\mathrel{\mathop{\mathchar 58\relax}}=\mathcal{Q}_{-1}$ be the set
of primes $\ell\equiv 1\mod{18}$ such that
$\widetilde{E}(\mathbb{F}_{\ell})[3]=0$. The set $\mathcal{Q}$ is infinite
with density $1/24$. Furthermore, if $m$ is divisible by only the primes in
$\mathcal{Q}$, then,

$\operatorname{rank}E_{(m)}(K)=0\text{ and
}\operatorname{rank}E_{(m^{2})}(K)=0.$

###### Proof.

The result follows from Theorem 4.4 and Proposition 4.11, see above for more
details. ∎

We remark that there are 15 primes in $\mathcal{Q}$ that are $<1000$, given as
follows

$\mathcal{Q}\cap[1,1000]=\\{19,127,163,199,271,307,379,487,523,631,739,811,883,919,991\\}.$

The Theorem 4.13 implies in particular that for any of the primes
$\ell\in\mathcal{Q}\cap[1,1000]$ above, the ranks of $E_{-\ell^{2}}(K)$ and
$E_{-\ell^{4}}(K)$ are both zero. This has been checked by the authors using
Magma.

###  4.4. Numerical data

In the table 1 below (on the next page), we collect some numerical data for
the values of $a\in[-20,20]$ satisfying the conditions of Theorem 3.12, along
with the density of $\mathcal{Q}_{a}$. The data shows that Theorem A applies
to

$a\in\\{-17,-16,-14,-10,-9,-8,-6,-5,-1,6,7,8,13,14,20\\}.$

Thus, the result applies to $15$ out of a total of $34$ values considered for
$a$. All computations were done on Magma.

$a$ | $\operatorname{Sel}_{3}(E_{a}/\mathbb{Q})$ | $\operatorname{Sel}_{3}(E_{-27a}/\mathbb{Q})$ | $\operatorname{Sel}_{3}(E_{a}/K)$ | $s$ | $\mathfrak{d}(\mathcal{Q}_{a})$
---|---|---|---|---|---
$-20$ | $\mathbb{Z}/3\mathbb{Z}$ | $\mathbb{Z}/3\mathbb{Z}$ | $(\mathbb{Z}/3\mathbb{Z})^{2}$ | $-$ | $-$
$-19$ | $\mathbb{Z}/3\mathbb{Z}$ | $\mathbb{Z}/3\mathbb{Z}$ | $(\mathbb{Z}/3\mathbb{Z})^{2}$ | $-$ | $-$
$-18$ | $\mathbb{Z}/3\mathbb{Z}$ | $\mathbb{Z}/3\mathbb{Z}$ | $(\mathbb{Z}/3\mathbb{Z})^{2}$ | $-$ | $-$
$-17$ | $0$ | $0$ | $0$ | $0$ | $1/24$
$-16$ | $0$ | $0$ | $0$ | $0$ | $1/24$
$-15$ | $\mathbb{Z}/3\mathbb{Z}$ | $\mathbb{Z}/3\mathbb{Z}$ | $(\mathbb{Z}/3\mathbb{Z})^{2}$ | $-$ | $-$
$-14$ | $0$ | $0$ | $0$ | $1$ | $1/72$
$-13$ | $\mathbb{Z}/3\mathbb{Z}$ | $\mathbb{Z}/3\mathbb{Z}$ | $(\mathbb{Z}/3\mathbb{Z})^{2}$ | $-$ | $-$
$-11$ | $\mathbb{Z}/3\mathbb{Z}$ | $\mathbb{Z}/3\mathbb{Z}$ | $(\mathbb{Z}/3\mathbb{Z})^{2}$ | $-$ | $-$
$-10$ | $0$ | $0$ | $0$ | $0$ | $1/24$
$-9$ | $0$ | $0$ | $0$ | $0$ | $1/24$
$-8$ | $0$ | $0$ | $0$ | $0$ | $1/24$
$-7$ | $\mathbb{Z}/3\mathbb{Z}$ | $\mathbb{Z}/3\mathbb{Z}$ | $(\mathbb{Z}/3\mathbb{Z})^{2}$ | $-$ | $-$
$-6$ | $0$ | $0$ | $0$ | $0$ | $1/24$
$-5$ | $0$ | $0$ | $0$ | $0$ | $1/24$
$-4$ | $\mathbb{Z}/3\mathbb{Z}$ | $\mathbb{Z}/3\mathbb{Z}$ | $(\mathbb{Z}/3\mathbb{Z})^{2}$ | $-$ | $-$
$-2$ | $\mathbb{Z}/3\mathbb{Z}$ | $\mathbb{Z}/3\mathbb{Z}$ | $(\mathbb{Z}/3\mathbb{Z})^{2}$ | $-$ | $-$
$-1$ | $0$ | $0$ | $0$ | $0$ | $1/24$
$2$ | $\mathbb{Z}/3\mathbb{Z}$ | $\mathbb{Z}/3\mathbb{Z}$ | $(\mathbb{Z}/3\mathbb{Z})^{2}$ | $-$ | $-$
$3$ | $\mathbb{Z}/3\mathbb{Z}$ | $\mathbb{Z}/3\mathbb{Z}$ | $(\mathbb{Z}/3\mathbb{Z})^{2}$ | $-$ | $-$
$5$ | $\mathbb{Z}/3\mathbb{Z}$ | $\mathbb{Z}/3\mathbb{Z}$ | $(\mathbb{Z}/3\mathbb{Z})^{2}$ | $-$ | $-$
$6$ | $0$ | $0$ | $0$ | $0$ | $1/24$
$7$ | $0$ | $0$ | $0$ | $1$ | $1/72$
$8$ | $0$ | $0$ | $0$ | $0$ | $1/24$
$10$ | $\mathbb{Z}/3\mathbb{Z}$ | $\mathbb{Z}/3\mathbb{Z}$ | $(\mathbb{Z}/3\mathbb{Z})^{2}$ | $-$ | $-$
$11$ | $\mathbb{Z}/3\mathbb{Z}$ | $\mathbb{Z}/3\mathbb{Z}$ | $(\mathbb{Z}/3\mathbb{Z})^{2}$ | $-$ | $-$
$12$ | $\mathbb{Z}/3\mathbb{Z}$ | $\mathbb{Z}/3\mathbb{Z}$ | $(\mathbb{Z}/3\mathbb{Z})^{2}$ | $-$ | $-$
$13$ | $0$ | $0$ | $0$ | $1$ | $1/72$
$14$ | $0$ | $0$ | $0$ | $1$ | $1/72$
$15$ | $(\mathbb{Z}/3\mathbb{Z})^{2}$ | $(\mathbb{Z}/3\mathbb{Z})^{2}$ | $(\mathbb{Z}/3\mathbb{Z})^{4}$ | $-$ | $-$
$17$ | $(\mathbb{Z}/3\mathbb{Z})^{2}$ | $(\mathbb{Z}/3\mathbb{Z})^{2}$ | $(\mathbb{Z}/3\mathbb{Z})^{4}$ | $-$ | $-$
$18$ | $\mathbb{Z}/3\mathbb{Z}$ | $\mathbb{Z}/3\mathbb{Z}$ | $(\mathbb{Z}/3\mathbb{Z})^{2}$ | $-$ | $-$
$19$ | $\mathbb{Z}/3\mathbb{Z}$ | $\mathbb{Z}/3\mathbb{Z}$ | $(\mathbb{Z}/3\mathbb{Z})^{2}$ | $-$ | $-$
$20$ | $0$ | $0$ | $0$ | $0$ | $1/24$
Table 1. Numerical data for Theorem A

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