# Imaginary quadratic fields $F$ with $X_{0}(15)(F)$ finite.

Tim Evink  Institute of Algebra and Number Theory, Ulm University,
Helmholtzstr. 18, 89081 Ulm, Germany<EMAIL_ADDRESS>
###### Abstract.

Caraiani and Newton have proven that if $F$ is an imaginary quadratic number
field such that $X_{0}(15)$ has rank $0$ over $F$, then every elliptic curve
over $F$ is modular. This paper is concerned with the quadratic fields
$F=\mathbb{Q}(\sqrt{-p})$ for a prime number $p$. We give explicit conditions
on $p$ under which the rank is $0$, and prove that these conditions are
satisfied for $87,5\%$ of the primes for which the rank is expected to be even
based on the parity conjecture. We also show these conditions are satisfied if
and only if rank $0$ follows from a $4$-descent over $\mathbb{Q}$ on the
quadratic twist $X_{0}(15)_{-p}$. To prove this, we perform two consecutive
$2$-descents and prove this gives rank bounds equivalent to those obtained
from a $4$-descent using visualisation techniques for $\Sh[2]$. In fact we
prove a more general connection between higher descents for elliptic curves
which seems interesting in its own right.

##  1\. Introduction

Consider the modular curve $X_{0}(15)$, which is an elliptic curve over
$\mathbb{Q}$ of rank $0$. Caraiani and Newton [7] have proven that if $F$ is
an imaginary quadratic number field such that $X_{0}(15)$ has rank $0$ over
$F$, then every elliptic curve over $F$ is modular. For positive square-free
integers $d$, the global root number of $X_{0}(15)$ over
$\mathbb{Q}(\sqrt{-d})$ turns out to be $1$ precisely when

(1) $d\equiv 0,1,2,3,4,5,8,12\bmod 15.$

Thus, based on the parity conjecture we expect the rank of $X_{0}(15)$ over
$\mathbb{Q}(\sqrt{-d})$ to be even for precisely those $d$ satisfying (1).
Forthcoming work of Smith will show that we have rank $0$ for a proportion of
100% of such $d$, see the introduction of [7]. However, this density result
might give a misleading picture for the amount of occurrences of positive rank
up to some bound: out of the $3158$ positive square-free integers $d\leq
10^{4}$ satisfying (1), we have $514$ instances of positive rank111In all
these instances of positive rank, the rank is in fact $2$. The smallest
positive square-free $d$ such that
$\mathrm{rank}(X_{0}(15)/\mathbb{Q}(\sqrt{-d}))=4$ is $d=20445$. These
conclusions were found by calling the Magma [2] function Rank on the quadratic
twist $X_{0}(15)_{-d}$..
In this paper we focus on the case where $d=p$ is a prime number, and give
explicit conditions on $p$ for which rank $0$ occurs. The main result
concerning occurrences of rank $0$ is formulated using Rédei symbols $[a,b,c]$
as developed in [21], a tri-linear number theoretic symbol taking values in
$\\{\pm 1\\}$ that encodes splitting behaviour of rational primes in certain
$D_{4}$ or $C_{4}$-extensions of $\mathbb{Q}$.

######  Theorem 1.1.

Let $p$ be a prime satisfying one of the following conditions.

  1. (i)

$p\equiv 1,2,3,4,5,8\bmod 15$ and $p\not\equiv 1\bmod 8$,

  2. (ii)

$p\equiv 1,4\bmod 15$, $p\equiv 1\bmod 8$ and $[-1,10,p]=-1$,

  3. (iii)

$p\equiv 2,8\bmod 15$, $p\equiv 1\bmod 8$ and $[-1,2,p]=1$.

Then $X_{0}(15)$ has rank $0$ over $\mathbb{Q}(\sqrt{-p})$. Moreover, the
density of these primes amongst the primes $p$ of the form $p\equiv
1,2,3,4,5,8\bmod 15$ is $87,5\%$.

The stated Rédei symbols in Theorem 1.1 can be computed using the following.

  1. (i)

If $p\equiv 1,9\bmod 40$, then $[-1,10,p]=1\,\Leftrightarrow\,p$ splits
completely in $\mathbb{Q}(\sqrt{3+\sqrt{-1}})$.

  2. (ii)

If $p\equiv 1\bmod 8$, then $[-1,2,p]=1\,\Leftrightarrow\,p$ splits completely
in $\mathbb{Q}(\sqrt{1+\sqrt{-1}})$.

See [21, §7]. Combining Theorem 1.1 with [7, Thm. 1.1] we immediately obtain
the following corollary.

######  Corollary 1.2.

Let $p$ be a prime satisfying one of conditions (i)-(iii) of Theorem 1.1. Then
every elliptic curve defined over $\mathbb{Q}(\sqrt{-p})$ is modular.

Let us illustrate the manner in which Theorem 1.1 is proven. For a prime
number $p$ we will see that

$\mathrm{rank}(X_{0}(15)/\mathbb{Q}(\sqrt{-p}))=\mathrm{rank}(X_{0}(15)_{-p}/\mathbb{Q}),$

which makes the quadratic twist $X_{0}(15)_{-p}$ a natural object of study to
determine the rank of $X_{0}(15)$ over $\mathbb{Q}(\sqrt{-p})$. When $p\equiv
1,2,3,4,5,8\bmod 15$, a basic $2$-Selmer group computation of $X_{0}(15)_{-p}$
over $\mathbb{Q}$ will show that the rank is $0$ when $p\not\equiv 1\bmod 8$,
and at most $2$ when $p\equiv 1\bmod 8$. In the latter case we improve on the
upper bound of $2$ by computing $2$-Selmer groups over suitable quadratic
number fields in a similar fashion to [12] and [13], leading to a proof of
Theorem 1.1.
We also analyse how this improvement relates to a natural way of improving on
a $2$-descent, namely higher descents. We will show that Theorem 1.1 implies
rank $0$ precisely when this follows from a $4$-descent on $X_{0}(15)_{-p}$
over $\mathbb{Q}$. In proving this connection, the following result is
instrumental.

######  Theorem 1.3.

Let $E$ be an elliptic curve defined over a number field $K$. Let
$L=K(\sqrt{d})$ be a quadratic extension such that the following conditions
are satisfied.

  1. (1)

We have $\Sh(E_{d}/K)[2]=0$.

  2. (2)

The corestriction map $\Sh(E/L)[2]\to\Sh(E/K)[2]$ is injective.

Then the restriction map $\Sh(E/K)\to\Sh(E/L)$ induces an isomorphism

(2) $\Sh(E/K)[2^{\infty}]/\Sh(E/K)[2]\xrightarrow{\sim}\Sh(E/L)[2^{\infty}].$

Theorem 1.3 is proven in the setting of visualisation of $\Sh[2]$ as in [4,
§4]. In loc. cit. an explicit necessary condition for $2$-divisibility in
$\Sh[2]$ is given, which can be used to improve on a $2$-descent. The novelty
of Theorem 1.3 lies in the fact that it gives conditions under which the
necessary condition is also sufficient, thus connecting a $2^{n+1}$-descent of
$E/K$ to a $2^{n}$-descent of $E/L$. For example, elements of order $2$ in
$\Sh(E/L)$ give rise to elements of order $4$ in $\Sh(E/K)$, a conclusion that
doesn’t follow from the results in [4]. The main idea is to generalise [5,
Lemma 7.1] to the setting of a $2$-isogeny between Abelian surfaces as in [4,
§4], which allows control over the image of $\Sh(E/K)\to\Sh(A/K)$ as in loc.
cit.
Using Theorem 1.3 we show that the rank results of Theorem 1.1 are equivalent
to those obtained from performing a $4$-descent on the twist $X_{0}(15)_{-p}$
over $\mathbb{Q}$. Specifically, using the Chinese remainder theorem to
combine the congruence conditions mod $15$ and mod $8$ from Theorem 1.1 $(ii)$
and $(iii)$, we prove the following results.

######  Theorem 1.4.

Let $p\equiv 1,49\bmod 120$ be a prime and let $E=X_{0}(15)$.

  1. (i)

If $[-1,10,p]=-1$, then $\mathrm{rank}(E_{-p}/\mathbb{Q})=0$ and
$\Sh(E_{-p}/\mathbb{Q})[4]\cong(\mathbb{Z}/2\mathbb{Z})^{2}$.

  2. (ii)

If $[-1,10,p]=1$, then $\Sh(E_{-p}/\mathbb{Q})[4]\xrightarrow{\cdot
2}\Sh(E_{-p}/\mathbb{Q})[2]$ is surjective.

  3. (iii)

If $[-1,10,p]=1$ and $\mathrm{rank}(E_{-p}/\mathbb{Q})=0$, then
$\Sh(E_{-p}/\mathbb{Q})[4]\cong(\mathbb{Z}/4\mathbb{Z})^{2}$.

######  Theorem 1.5.

Let $p\equiv 17,113\bmod 120$ be a prime and let $E=X_{0}(15)_{-p}$.

  1. (i)

If $[-1,2,p]=1$, then $\mathrm{rank}(E_{-p}/\mathbb{Q})=0$ and
$\Sh(E_{-p}/\mathbb{Q})[4]\cong(\mathbb{Z}/2\mathbb{Z})^{2}$.

  2. (ii)

If $[-1,2,p]=-1$, then $\Sh(E_{-p}/\mathbb{Q})[4]\xrightarrow{\cdot
2}\Sh(E_{-p}/\mathbb{Q})[2]$ is surjective.

  3. (iii)

If $[-1,2,p]=-1$ and $\mathrm{rank}(E_{-p}/\mathbb{Q})=0$, then
$\Sh(E_{-p}/\mathbb{Q})[4]\cong(\mathbb{Z}/4\mathbb{Z})^{2}$.

The structure of the paper is as follows. In section 2 we discuss the explicit
method of $2$-descent we will use. Section 3 covers all the $2$-Selmer group
computations and provides a proof of Theorem 1.1. In section 4 we cover the
visualisation arguments that enable us to prove Theorem 1.3, which combined
with the $2$-Selmer group computations of section 3 allows for a proof of
Theorems 1.4 and 1.5. Lastly, section 5 has some closing remarks about higher
descents, a discussion of the applicability of Theorem 1.3 in general, and the
rank of $X_{0}(15)$ over other families of imaginary quadratic fields like
$\mathbb{Q}(\sqrt{-2p})$.

##  2\. 2-descent

Let $K$ be a number field and $A$ and Abelian variety over $K$. The Mordell-
Weil theorem states that $A(K)$ is a finitely generated Abelian group, and the
method of $2$-descent provides an upper bound on its rank $A(K)$. Especially
in the case of elliptic curves, this is a well-known procedure, see for
example [20, Ch. X]. The main reference of this section is [18].

First some notation. The Galois cohomology groups $H^{*}(G_{K},M)$, for
$G_{K}$ the absolute Galois group of $K$ will be denoted by $H^{*}(K,M)$, and
when $M=G(\overline{K})$ for a commutative group scheme $G$ over $K$ we will
further abbreviate the notation to $H^{*}(K,G)$. For example if $E$ is an
elliptic curve over $K$ we will write $H^{1}(K,E)$ and $H^{1}(K,E[2])$ for
$H^{1}(G_{K},E(\overline{K}))$ and $H^{1}(G_{K},E(\overline{K})[2])$,
respectively.

The short exact sequence $0\to A[2]\to A\xrightarrow{2}A\to 0$ gives through
Galois cohomology the long exact sequence

(3) $0\to A(K)[2]\to
A(K)\xrightarrow{2}A(K)\xrightarrow{\delta}H^{1}(K,A[2])\to H^{1}(K,A)\to
H^{1}(K,A)\to\dotsb$

For a completion $K_{\mathfrak{p}}$, we can identify $G_{K_{\mathfrak{p}}}$
with a decomposition group $D_{\mathfrak{p}}\subset G_{K}$ over
$\mathfrak{p}$, hence we have restriction maps $\mathrm{res}_{\mathfrak{p}}$
on cohomology222These do not depend on the choice of decomposition group as
different choices differ by a conjugation map, which is trivial on
cohomology., which form compatible maps with the long exact sequence (3) and
its local version involving
$A(K_{\mathfrak{p}})\xrightarrow{\delta_{\mathfrak{p}}}H^{1}(K_{\mathfrak{p}},A[2])$.
We define

$\begin{split}S^{2}(A/K)&:=\ker\left(H^{1}(K,A[2])\to\prod_{\mathfrak{p}}H^{1}(K_{\mathfrak{p}},A)\right),\\\
\Sh(A/K)&:=\ker\left(H^{1}(K,A)\to\prod_{\mathfrak{p}}H^{1}(K_{\mathfrak{p}},A)\right).\end{split}$

where $S^{2}(A/K)$ is the $2$-Selmer group of $A$, and $\Sh(A/K)$ the Tate-
Shafarevich group of $A$, which fit in the short exact sequence

(4) $0\to A(K)/2A(K)\to S^{2}(A/K)\to\Sh(A/K)[2]\to 0.$

Let $S$ be the set consisting of the primes of $K$ that are either infinite,
lying over $2$, or are primes of bad reduction. Denote by $H^{1}(K,A[2];S)$
the subgroup of $H^{1}(K,A[2])$ consisting of the cocycles unramified outside
$S$. Then $H^{1}(K,A[2];S)$ is finite and contains $S^{2}(A/K)$. In fact

(5) $S^{2}(A/K)=\\{\xi\in
H^{1}(K,A[2];S):\mathrm{res}_{\mathfrak{p}}(\xi)\in\mathrm{im}(\delta_{\mathfrak{p}})\text{
for all }\mathfrak{p}\in S\\},$

see [16, Ch. I, Cor. 6.6]. In particular, since $A(K)/2A(K)$ has
$\mathbb{F}_{2}$-dimension $\mathrm{rank}(A/K)+A(K)[2]$, we obtain from (4)
the identity

(6)
$\mathrm{rank}(A/K)+\dim_{\mathbb{F}_{2}}\Sh(A/K)[2]=\dim_{\mathbb{F}_{2}}S^{2}(A/K)-\dim_{\mathbb{F}_{2}}A(K)[2].$

The quantity $\dim_{\mathbb{F}_{2}}S^{2}(A/K)-\dim_{\mathbb{F}_{2}}A(K)[2]$,
sometimes called the $2$-Selmer rank, provides an upper bound on
$\mathrm{rank}(A/K)$, and the obstruction to its sharpness is measured by
$\dim_{\mathbb{F}_{2}}\Sh(A/K)[2]$.

Now assume that $A=E$ is an elliptic curve given by the equation $y^{2}=f(x)$,
for $f(x)\in K[x]$ monic of degree $3$. In order to compute the $2$-Selmer
group $S^{2}(E/K)$, we make the cohomological setup more explicit. Consider
the étale $K$-algebra $M_{K}=K[t]/f(t)$. The norm map $M_{K}\to K$ induces a
map $M_{K}^{*}/M_{K}^{*2}\to K^{*}/K^{*2}$, of which we denote the kernel by
$M_{K}^{\prime}$.

$M_{K}^{\prime}:=\ker(M_{K}^{*}/M_{K}^{*2}\to K^{*}/K^{*2}),$

Then [18, Thm. 1.1 & 1.2] establishes an isomorphism $H^{1}(K,E[2])\cong
M_{K}^{\prime}$, under which the connecting homomorphism $E(K)\to
H^{1}(K,E[2])$ translates to the map $E(K)\to M_{K}^{\prime}$ determined by

(7) $(a,b)\mapsto a-t,\quad b\neq 0.$

This isomorphism $H^{1}(K,E[2])\cong M_{K}^{\prime}$ also holds upon replacing
$K$ with a completion $K_{\mathfrak{p}}$, and upon applying both isomorphisms,
the restriction map $H^{1}(K,E[2])\to H^{1}(K_{\mathfrak{p}},E[2])$
corresponds to the natural map $M_{K}^{\prime}\to
M_{K_{\mathfrak{p}}}^{\prime}$ induced by $K\to K_{\mathfrak{p}}$.
We will also rewrite this embedding according to how $M_{K}$ splits as a
product of fields $K_{i}$, thus representing elements of $H^{1}(K,E[2])$ as
tuples $(x_{i})$, with $x_{i}\in K_{i}^{*}/K_{i}^{*2}$. Under this
correspondence, elements in $H^{1}(K,E[2];S)$ correspond to those tuples
$(x_{i})$ for which $K_{i}(\sqrt{x_{i}})$ is unramified outside $S$ for all
$i$.

Now suppose that $E(\overline{K})[2]=E(K)[2]$, that is $f(x)$ splits
completely in $K[x]$, say as $f(x)=(x-e_{1})(x-e_{2})(x-e_{3})$. Then
$M_{K}\cong K\times K\times K$ via $t\mapsto(e_{1},e_{2},e_{3})$, which
induces an isomorphism
$M_{K}^{\prime}\cong\ker\left(\bigoplus_{i=1}^{3}K^{*}/K^{*2}\xrightarrow{N}K^{*}/K^{*2}\right)$.
Here $N(x_{1},x_{2},x_{3})=x_{1}x_{2}x_{3}$. Also let $d\in K^{*}$ and
consider the quadratic twist $E_{d}/K$ with equation
$dy^{2}=(x-e_{1})(x-e_{2})(x-e_{3})$. If $(x^{\prime},y^{\prime})=(dx,d^{2}y)$
we obtain the Weierstrass equation
$(y^{\prime})^{2}=(x^{\prime}-de_{1})(x^{\prime}-de_{3})(x^{\prime}-de_{3})$,
of which the corresponding étale algebra is isomorphic to $M_{K}$, so we can
translate the embedding of $E_{d}$ using the model
$dy^{2}=(x-e_{1})(x-e_{2})(x-e_{3})$. It is given as follows.

(8)
$\displaystyle\begin{split}E_{d}(K)/2E_{d}(K)&\xrightarrow{\delta}\ker\left(\bigoplus_{i=1}^{3}K^{*}/K^{*2}\to
K^{*}/K^{*2}\right),\\\ (a,b)&\mapsto d(a-e_{1},a-e_{2},a-e_{3}),\quad\text{
if }b\neq 0\\\ (e_{1},0)&\mapsto(*,d(e_{1}-e_{2}),d(e_{1}-e_{3})),\\\
(e_{2},0)&\mapsto(d(e_{2}-e_{1}),*,d(e_{2}-e_{3})),\\\
(e_{3},0)&\mapsto(d(e_{3}-e_{1}),d(e_{3}-e_{2}),*).\end{split}$

The $*$-values are determined by other coordinates, for example the first
coordinate of the image of $(e_{1},0)$ is $(e_{1}-e_{2})(e_{1}-e_{3})$.

######  Remark 2.1.

If $d\in K^{*}$ is a square, the explicit Kummer-embedding of $E_{d}$ from (8)
is identical to the one of $E$. In particular, the abstract isomorphism of
Selmer groups of $E$ and $E_{d}$ induced by the obvious isomorphism $E\cong
E_{d}$ becomes an actual equality in the explicit setup.

We then have the following commutative diagram

(9)
${E_{d}(K)/2E_{d}(K)}$${\ker\left(\displaystyle\bigoplus_{i=1}^{3}K^{*}/K^{*2}\to
K^{*}/K^{*2}\right)}$${E_{d}(K_{\mathfrak{p}})/2E_{d}(K_{\mathfrak{p}})}$${\ker\left(\displaystyle\bigoplus_{i=1}^{3}K_{\mathfrak{p}}^{*}/K_{\mathfrak{p}}^{*2}\to
K_{\mathfrak{p}}^{*}/K_{\mathfrak{p}}^{*2}\right),}$$\scriptstyle{\delta}$$\scriptstyle{\iota_{\mathfrak{p}}}$$\scriptstyle{\delta_{\mathfrak{p}}}$

And we interpret the $2$-Selmer group $S^{2}(E/K)$ as a subgroup of the top
right kernel in (9). For a finite set $S$ of primes of $K$, let $K(S)=\\{x\in
K^{*}/K^{*2}:v_{\mathfrak{p}}(x)\equiv 0\bmod 2\text{ for all finite
}\mathfrak{p}\notin S\\}$. Then (5) translates to

$S^{2}(E/K)=\left\\{x\in\ker\left(\bigoplus_{i=1}^{3}K(S)\to
K(S)\right):\iota_{\mathfrak{p}}(x)\in\mathrm{im}(\delta_{\mathfrak{p}})\text{
for all }\mathfrak{p}\in S\right\\},$

where $S$ is the set consisting of finite primes of $K$ where $E$ has bad
reduction, the finite primes over $2$ and the real primes of $K$, and
$K(S)=\\{x\in K^{*}/K^{*2}:v_{\mathfrak{p}}(x)\equiv 0\bmod 2\text{ for all
finite }\mathfrak{p}\notin S\\}$. For
$x=(e_{1},e_{2},e_{3})\in\ker\left(\bigoplus_{i=1}^{3}K^{*}/K^{*2}\to
K^{*}/K^{*2}\right)$ we have
$\iota_{\mathfrak{p}}(x)\in\mathrm{im}(\delta_{\mathfrak{p}})$ if and only if
$v_{\mathfrak{p}}(e_{i})\equiv 0\bmod 2$ for all $i$. Thus if we let $S$ be
the set consisting of finite primes of $K$ where $E$ has bad reduction, the
finite primes over $2$ and the real primes of $K$, we have

$S^{2}(E/K)=\left\\{x\in\ker\left(\bigoplus_{i=1}^{3}K(S)\to
K(S)\right):\iota_{\mathfrak{p}}(x)\in\mathrm{im}(\delta_{\mathfrak{p}})\text{
for all }\mathfrak{p}\in S\right\\}.$

Note that local images at complex primes are trivial so we can omit those in
$S$. The set $K(S)$ is finite: it fits in the short exact sequence

$0\to R_{S}^{*}/R_{S}^{*2}\to K(S)\to\mathrm{Cl}(R_{S})[2]\to 0,$

where $R_{S}\subset K$ is the ring of $S$-integers. We can thus compute
$S^{2}(E/K)$ provided that we have computed the local images
$\mathrm{im}(\delta_{\mathfrak{p}})$ for $\mathfrak{p}\in S$, an
$\mathbb{F}_{2}$-basis for $K(S)$, and for each basis element its image under
the natural maps $K^{*}/K^{*2}\to K_{\mathfrak{p}}^{*}/K_{\mathfrak{p}}^{*2}$
for $\mathfrak{p}\in S$. To compute the local images we use the following
Proposition.

######  Proposition 2.2.

Let $F$ be a finite extension of $\mathbb{Q}_{p}$ for some prime $p$ of
$\mathbb{Q}$. Let $E$ be an elliptic curve over $F$. Then

$\mathrm{dim}_{\mathbb{F}_{2}}(E(F)/2E(F))=\begin{cases}\mathrm{dim}_{\mathbb{F}_{2}}E(F)[2]+[F:\mathbb{Q}_{2}]&\text{if
}p=2,\\\ \mathrm{dim}_{\mathbb{F}_{2}}E(F)[2]&\text{if }2<p<\infty,\\\
\mathrm{dim}_{\mathbb{F}_{2}}E(F)[2]-1&\text{if }F=\mathbb{R},\\\ 0&\text{if
}F=\mathbb{C}\end{cases}$

###### Proof.

See [22, Lem. 4.4 & Lem. 4.8]. ∎

##  3\. $X_{0}(15)$ and 2-Selmer group computations

In this section we consider the rank of $X_{0}(15)$ over imaginary quadratic
fields. After discussing the parity of the rank in relation to global root
numbers, we discuss the torsion subgroup and some local images of an arbitrary
quadratic twist $X_{0}(15)_{d}$. These local images are then used for various
$2$-Selmer group computations, resulting in a proof of Theorem 1.1.

######  Conjecture 3.1 (Parity conjecture).

For an elliptic curve $E$ defined over a number field $K$ we have

$w(E/K)=(-1)^{\mathrm{rank}(E/K)}.$

Here $w(E/K)\in\\{\pm 1\\}$ is the global root number of $E/K$, which is the
sign of the conjectural functional equation of the $L$-function of $E/K$. The
parity conjecture is weaker than the Birch and Swinnerton-Dyer conjecture. We
refer to [10] for an overview on how the parity conjecture relates to various
other conjectures, like finiteness of .

Let us fix the notation $E=X_{0}(15)$ for the remainder of this section. A
model for $E$ is given by $v^{2}+uv+v=u^{3}+u^{2}-10u-10$, see [14, Elliptic
curve 15.a5]. The coordinate change $(x,y)=(4u,8v-4u-4)$ brings the equation
in the more desirable form for the purpose of $2$-descent, yielding the
equation

$E:y^{2}=(x+13)(x+4)(x-12).$

######  Proposition 3.2.

Let $d$ be positive square-free integer. Then

$w(E/\mathbb{Q}(\sqrt{-d}))=1\quad\Leftrightarrow\quad d\equiv
0,1,2,3,4,5,8,12\bmod 15.$

###### Proof.

Let $K=\mathbb{Q}(\sqrt{-d})$. As $E/\mathbb{Q}$ is semistable, [11, Thm. 3]
implies that

$w(E/K)=(-1)^{1+s},$

where $s$ is the number of finite primes of $K$ at which $E$ has split
multiplicative reduction. Over the rationals, $E$ has non-split multiplicative
reduction at $3$, split multiplicative reduction at $5$, and good reduction
elsewhere. We see that a prime of $K$ over $3$ has split multiplicative
reduction precisely when the residue field degree is $2$, i.e. when $3$ is
inert in $K$, in which case there is one such prime. As any prime of $K$ over
$5$ has split multiplicative reduction, we have

$\displaystyle\\#$ $\displaystyle\text{split multiplicative primes of }K\text{
over }3=\begin{cases}1,&\text{ if }\left(\tfrac{-d}{3}\right)=-1\\\
0,&\text{otherwise}\end{cases}$ $\displaystyle\\#$ $\displaystyle\text{split
multiplicative primes of }K\text{ over }5=\begin{cases}2,&\text{ if
}\left(\tfrac{-d}{5}\right)=1\\\ 1,&\text{otherwise}\end{cases}$

The result follows by going over the cases for $d\bmod 15$. ∎

The parity conjecture together with Proposition 3.2 thus immediately yields
the following.

######  Corollary 3.3.

Let $d$ be a positive square-free integer and assume the parity conjecture.
Then

$\mathrm{rank}(E/\mathbb{Q}(\sqrt{-d}))\equiv 0\bmod
2\quad\Leftrightarrow\quad d\equiv 0,1,2,3,4,5,8,12\bmod 15.$

In particular, assuming the parity conjecture, we see that $E$ has non-zero
rank over $\mathbb{Q}(\sqrt{-d})$ when $d\equiv 6,7,9,10,11,13,14\bmod 15$, so
[7, Thm. 1.1] cannot be used to prove modularity of elliptic curves over
$\mathbb{Q}(\sqrt{-d})$ for those $d$. For this reason we restrict to the
congruence conditions of Corollary 3.3 whenever we attempt to prove rank $0$
for a specific $d$.

Let $d\in\mathbb{Q}^{*}$ be a non-square. Then we have the well-known relation

$\mathrm{rank}(E/\mathbb{Q})+\mathrm{rank}(E_{d}/\mathbb{Q})=\mathrm{rank}(E/\mathbb{Q}(\sqrt{d})),$

corresponding to a decomposition of $E(\mathbb{Q}(\sqrt{d}))\otimes\mathbb{Q}$
into eigenspaces for the action of
$\mathrm{Gal}(\mathbb{Q}(\sqrt{d})/\mathbb{Q})$, which we will use freely
throughout the rest of the paper. In section 4 we will also encounter a proof,
see Proposition 4.2. In Proposition 3.5 we will see that
$\mathrm{rank}(E/\mathbb{Q})=0$, so
$\mathrm{rank}(E_{d}/\mathbb{Q})=\mathrm{rank}(E/\mathbb{Q}(\sqrt{d}))$. It
follows that we can obtain upper bounds on
$\mathrm{rank}(E/\mathbb{Q}(\sqrt{d}))$ by doing a $2$-Selmer group
computation of $E_{d}/\mathbb{Q}$. We start by computing the torsion subgroup
and the local images over $2,3,5$ and $\infty$ for any such twist. Then we
focus on the main object of interest, the twist $E_{-p}$ for a prime $p$.

For a number field $K$ and $d\in K^{*}$, set $e_{1}=-13$, $e_{2}=-4$ and
$e_{3}=12$ as in the explicit $2$-descent case of section 2 corresponding to
rational $2$-torsion. Then the Kummer-embedding for $E_{d}$ as in (9) is given
as follows.

(10)
$\displaystyle\begin{split}E_{d}(K)/2E_{d}(K)&\xrightarrow{\delta}\ker\left(\bigoplus_{i=1}^{3}K^{*}/K^{*2}\to
K^{*}/K^{*2}\right),\\\ (a,b)&\mapsto d(a+13,a+4,a-12),\quad\text{ if }b\neq
0\\\ (-13,0)&\mapsto(1,-d,-d),\\\ (-4,0)&\mapsto(d,-1,-d),\\\
(12,0)&\mapsto(d,d,1).\end{split}$

######  Proposition 3.4.

Let $d\in\mathbb{Z}$ be square-free. Then

$E_{d}(\mathbb{Q})_{\mathrm{tors}}\cong\begin{dcases*}\mathbb{Z}/4\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}&if
$d\in\\{1,-1\\}$\\\
\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}&otherwise\end{dcases*}$

The points of order $4$ are $(-8,\pm 20)$ and $(32,\pm 180)$ when $d=1$, and
$(2,\pm 30)$ and $(-28,120)$ when $d=-1$.

###### Proof.

The twist $E_{d}$ has good reduction over $\mathbb{Q}_{7}(\sqrt{7})$, so we
have injections

$E_{d}(\mathbb{Q})[m]\xhookrightarrow{}E_{d}(\mathbb{Q}(\sqrt{7}))[m]\xhookrightarrow{}\tilde{E_{d}}(\mathbb{F}_{7})$

for any positive integer $m$ coprime to $7$. Here the reduction
$\tilde{E_{d}}$ is with respect to $\mathbb{Q}_{7}(\sqrt{7})$. There are two
options for this reduction, depending on the class of $d$ in $F^{*}/F^{*2}$
for $F=\mathbb{Q}_{7}(\sqrt{7})$. In both cases we have
$\\#\tilde{E_{d}}(\mathbb{F}_{7})=8$, which implies that
$\\#E_{d}(\mathbb{Q})[2^{\infty}]\leq 8$ and $E_{d}(\mathbb{Q})[p]=0$ for all
primes $p\neq 2,7$. Similarly, $E_{d}$ has good reduction over
$\mathbb{Q}_{11}(\sqrt{11})$ and both possibilities for the reduction have
exactly $16$ rational points, so we can also conclude that
$E_{d}(\mathbb{Q})[7]=0$. It follows that
$\\#E_{d}(\mathbb{Q})_{\mathrm{tors}}\leq 8$.
As
$E_{d}(\mathbb{Q})[2]\cong\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$
the torsion subgroup $E(\mathbb{Q})_{\mathrm{tors}}$ is isomorphic to either
$\mathbb{Z}/4\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$ or
$\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$. The latter case occurs
precisely when $E_{d}(\mathbb{Q})[2]\to E_{d}(\mathbb{Q})/2E_{d}(\mathbb{Q})$
is injective, which because of (10) is equivalent to $(1,-d,-d)$ and $(d,d,1)$
being linearly independent in
$\bigoplus_{i=1}^{3}\mathbb{Q}^{*}/\mathbb{Q}^{*2}$, which is the case
precisely when $d\neq\pm 1$.
Finally, when $d=1$ we see that $(-8,\pm 20)$ and $(32,\pm 180)$ are indeed
points on the curve, which have order $4$ as their doubles equal $(12,0)$.
Similarly the doubles of the rational points $(2,\pm 30)$ and $(-28,120)$ on
$E_{-1}$ equal $(-13,0)$. ∎

If one wishes to crack a nut with a sledgehammer, one can also appeal to
Mazur’s theorem on torsion subgroups over $\mathbb{Q}$ as an alternative to
the reduction arguments in the proof of Proposition 3.4. Then (10) is also
used to exclude the $\mathbb{Z}/8\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$ case.

Let us now prepare for various $2$-Selmer group computations by providing all
possible images of the map

$E_{d}(\mathbb{Q}_{p})/2E_{d}(\mathbb{Q}_{p})\xrightarrow{\delta_{p}}\ker\left(\bigoplus_{i=1}^{3}\mathbb{Q}_{p}^{*}/\mathbb{Q}_{p}^{*2}\to\mathbb{Q}_{p}^{*}/\mathbb{Q}_{p}^{*2}\right)$

for $p=2,3,5,\infty$. From (10) we see that the local image
$\mathrm{im}(\delta_{p})$ depends only on the class of $d$ in
$\mathbb{Q}_{p}^{*}/\mathbb{Q}_{p}^{*2}$. We have
$\mathbb{Q}_{2}^{*}/\mathbb{Q}_{2}^{*2}=\langle 2,-1,3\rangle$,
$\mathbb{Q}_{3}^{*}/\mathbb{Q}_{3}^{*2}=\langle 3,-1\rangle$,
$\mathbb{Q}_{5}^{*}/\mathbb{Q}_{5}^{*2}=\langle 5,2\rangle$ and
$\mathbb{R}^{*}/\mathbb{R}^{*2}=\langle-1\rangle$. The local images over $2$
then have the following $\mathbb{F}_{2}$-bases.

(11) $\begin{gathered}\begin{array}[]{|c|c|c|c|c|}\hline\cr
d\in\mathbb{Q}_{2}^{*}/\mathbb{Q}_{2}^{*2}&1&-1&3&-3\\\
\hline\cr\hbox{\multirowsetup$\mathrm{im}(\delta_{2})$}&(1,-1,-1)&(-1,-1,1)&(1,-3,-3)&(1,3,3)\\\
&(1,3,3)&(1,-3,-3)&(3,3,1)&(-3,-3,1)\\\
&(-3,1,-3)&(1,2,2)&(1,6,6)&(1,-1,-1)\\\ \hline\cr
d\in\mathbb{Q}_{2}^{*}/\mathbb{Q}_{2}^{*2}&2&-2&6&-6\\\
\hline\cr\hbox{\multirowsetup$\mathrm{im}(\delta_{2})$}&(1,-2,-2)&(1,2,2)&(1,-3,-3)&(1,-1,-1)\\\
&(2,2,1)&(-2,-2,1)&(1,-2,-2)&(1,6,6)\\\
&(1,-1,-1)&(1,3,3)&(6,6,1)&(-6,-6,1)\\\ \hline\cr\end{array}\end{gathered}$

The other images are as follows.

(12) $\begin{gathered}\begin{array}[]{|c|c|c|c|c|}\hline\cr
d\in\mathbb{Q}_{3}^{*}/\mathbb{Q}_{3}^{*2}&1&-1&3&-3\\\
\hline\cr\hbox{\multirowsetup$\mathrm{im}(\delta_{3})$}&(1,-1,-1)&(-1,-1,1)&(1,-3,-3)&(1,3,3)\\\
&(-1,-1,1)&(-3,-3,1)&(3,3,1)&(-3,-3,1)\\\ \hline\cr\end{array}\\\
\begin{array}[]{|c|c|c|c|c|}\hline\cr
d\in\mathbb{Q}_{5}^{*}/\mathbb{Q}_{5}^{*2}&1&2&5&10\\\
\hline\cr\hbox{\multirowsetup$\mathrm{im}(\delta_{5})$}&(5,1,5)&(1,2,2)&(1,5,5)&(1,10,10)\\\
&(2,1,2)&(2,2,1)&(5,5,1)&(10,10,1)\\\ \hline\cr\end{array}\\\
\begin{array}[]{|c|c|c|c|c|}\hline\cr
d\in\mathbb{R}^{*}/\mathbb{R}^{*2}&1&-1\\\
\hline\cr\text{}\mathrm{im}(\delta_{\infty})&(1,-1,-1)&(-1,-1,1)\\\
\hline\cr\end{array}\end{gathered}$

For the justification of these local images, observe that Proposition 2.2
gives us the $\mathbb{F}_{2}$-dimensions, and one sees quickly that the given
elements are linearly independent in all cases, so it remains to show that
they are actually in the respective local images. The $2$-dimensional image of
$E_{d}(\mathbb{Q})_{\mathrm{tors}}$ yields points in the local images as well,
which depending on $p$ and $d$ have to be supplemented by additional local
points. It turns out these can be found by taking for $x$ an integer $-6\leq
x\leq 9$. For example when $p=2$ and $d=1$, we have the elements $(1,-1,-1)$
and $(-3,1,-3)$ from the global torsion (the global $4$-torsion point
$(32,180)$ has image $(5,1,5)$), and for $x=8$ we have in
$\mathbb{Q}_{2}^{*}/\mathbb{Q}_{2}^{*2}$

$(x+13)(x+4)(x-12)=21\cdot 12\cdot-4=1,$

so we obtain a point $(8,?)\in E(\mathbb{Q}_{2})$, which has image
$(-3,3,-1)$. The other cases are left to the reader.

######  Proposition 3.5.

For $d\in\\{1,-1,-2,-3,-5\\}$ we have

$\mathrm{rank}(E_{d}/\mathbb{Q})=0,\quad\text{ and
}\quad\Sh(E_{d}/\mathbb{Q})[2]=0.$

###### Proof.

From (6) we have

$\mathrm{rank}(E_{d}/\mathbb{Q})+\dim_{\mathbb{F}_{2}}\Sh(E_{d}/\mathbb{Q})[2]=\mathrm{dim}_{\mathbb{F}_{2}}S^{2}(E_{d}/\mathbb{Q})-2,$

so the result follows by proving that
$\mathrm{dim}_{\mathbb{F}_{2}}S^{2}(E_{d}/\mathbb{Q})=2$. The twist $E_{d}$
has good reduction at primes $>5$, and bad (in fact multiplicative) reduction
at $3$ and $5$, so we take $S=\\{2,3,5,\infty\\}$. Then
$\mathbb{Q}(S)=\langle-1,2,3,5\rangle\subset\mathbb{Q}^{*}/\mathbb{Q}^{*2}$.
Recall that $\delta$ and the local versions $\delta_{p}$ for $p\in S$ fit in
the commutative diagram

${E_{d}(\mathbb{Q})/2E_{d}(\mathbb{Q})}$${\ker\left(\displaystyle\bigoplus_{i=1}^{3}\mathbb{Q}^{*}/\mathbb{Q}^{*2}\to\mathbb{Q}^{*}/\mathbb{Q}^{*2}\right)}$${E_{d}(\mathbb{Q}_{p})/2E_{d}(\mathbb{Q}_{p})}$${\ker\left(\displaystyle\bigoplus_{i=1}^{3}\mathbb{Q}_{p}^{*}/\mathbb{Q}_{p}^{*2}\to\mathbb{Q}_{p}^{*}/\mathbb{Q}_{p}^{*2}\right),}$$\scriptstyle{\delta}$$\scriptstyle{\iota_{p}}$$\scriptstyle{\delta_{p}}$

and that

$S^{2}(E_{d}/\mathbb{Q})=\left\\{x\in\ker\left(\bigoplus_{i=1}^{3}\mathbb{Q}(S)\to\mathbb{Q}^{*}/\mathbb{Q}^{*2}\right):\iota_{p}(x)\in\mathrm{im}(\delta_{p}),\,\forall
p\in S\right\\}.$

Consider the case $d=1$ and let $x=(x_{1},x_{2},x_{3})\in
S^{2}(E/\mathbb{Q})$. Then from (11) we see that
$\iota_{2}(x)\in\mathrm{im}(\delta_{2})$ implies that
$x_{i}\in\langle-1,3,5\rangle$ for all $i$. Similarly, considering the
$3$-adic image $\mathrm{im}(\delta_{3})$ implies
$x_{i}\in\langle-1,2,5\rangle$ for all $i$, hence for each $i$ we have
$x_{i}\in\langle-1,3,5\rangle\cap\langle-1,2,5\rangle=\langle-1,5\rangle$. As
$5=-3$ in $\mathbb{Q}_{2}^{*}/\mathbb{Q}_{2}^{*2}$, the $2$-adic image implies
$x_{1}\in\langle 5\rangle$, and from the $5$-adic image we deduce
$x_{2}\in\langle-1\rangle$. This leaves $4$ options in total for $x$, proving
that indeed $\mathrm{dim}_{\mathbb{F}_{2}}S^{2}(E/\mathbb{Q})=2$.
Now suppose that $d=-1$. Then in $S^{2}(E_{-1}/\mathbb{Q})$ we have the
$2$-dimensional subspace generated by $(-1,-1,1)$ and $(15,6,10)$, coming from
$E_{-1}(\mathbb{Q}_{\mathrm{tors}})$. This subspace injects onto the subspace
of the $2$-adic image spanned by $(-1,-1,1)$ and $(1,-6,-6)$, hence we have
$S^{2}(E_{-1}/\mathbb{Q})=H\oplus\langle(-1,-1,1),(15,6,10)\rangle$ for

$H=\\{x\in S^{2}(E_{-1}/\mathbb{Q}):\iota_{2}(x)\in\langle(1,2,2)\rangle\\}.$

Let $x=(x_{1},x_{2},x_{3})\in H$. Then $x_{3}\in\langle 10\rangle$ by the
$3$-adic and real image, which combined with the $2$-adic image forces
$x_{3}=1$. Then $x$ has trivial $2$-adic image, which combined with the
$5$-adic image implies $x_{2}=1$, hence $H=0$ and $S^{2}(E_{-1}/\mathbb{Q})$
is $2$-dimensional as desired.
The remaining cases are similar and are left to the reader. ∎

We now focus on when the rank of $E$ over $\mathbb{Q}(\sqrt{-p})$ can be $0$,
where $p$ is some prime. In accordance with Corollary 3.3 we restrict
attention to the case $p\equiv 1,2,3,4,5,8\bmod 15$ (no prime can be $\equiv
0,12\bmod 15$), for otherwise we expect an odd, hence non-zero rank. From the
$d=1$ case of Proposition 3.5 we have

(13)
$\mathrm{rank}(E/\mathbb{Q}(\sqrt{-p}))=\mathrm{rank}(E_{-p}/\mathbb{Q}).$

Proposition 3.5 thus implies that $\mathrm{rank}(E/\mathbb{Q}(\sqrt{-p}))=0$
for $p\leq 5$, so assume $p>5$. This leaves the possible congruences $p\equiv
1,2,4,8\bmod 15$.

######  Proposition 3.6.

Let $p\equiv 1,2,4,8\bmod 15$ be a prime $>5$. Then the image of
$E_{-p}(\mathbb{Q})[2]$ gives two linearly independent elements $(-p,-p,1)$
and $(1,p,p)$ in $S^{2}(E_{-p}/\mathbb{Q})$. The dimension and additional
generators for $S^{2}(E_{-p}/\mathbb{Q})$ depend on $p\bmod 8$ and $p\bmod 15$
as follows.

$\begin{array}[]{c|c|c|c}p\bmod 8&p\bmod
15&\mathrm{dim}_{\mathbb{F}_{2}}S^{2}(E_{-p}/\mathbb{Q})&\text{additional
generators}\\\ \hline\cr 1&1,4&4&(15,6,10),(-1,-1,1)\\\
1&2,8&4&(1,2,2),(-1,-1,1)\\\ \text{not }1&\text{any}&2&\text{none}\end{array}$

###### Proof.

We take $S=\\{2,3,5,p,\infty\\}$. Then
$\mathbb{Q}(S)=\langle-1,2,3,5,p\rangle$. Besides the local images over
$2,3,5$ and $\infty$ from (11) and (12) we also need the image at $p$. We have
$\mathbb{Q}_{p}^{*}/\mathbb{Q}_{p}^{*2}=\langle r,p\rangle$ with
$r\in\mathbb{Z}$ a non-square mod $p$, which we will take to be $-1$ when
$p\equiv 3\bmod 4$. The image comes from the $2$-torsion and is given as
follows depending on $p\bmod 4$.

$\mathrm{im}(\delta_{p}):\begin{array}[]{|c|c|}\hline\cr p\equiv 1\bmod
4&p\equiv 3\bmod 4\\\ \hline\cr(p,p,1)&(-p,-p,1)\\\ (1,p,p)&(1,p,p)\\\
\hline\cr\end{array}$

Let $H=\\{x\in S^{2}(E/K):\iota_{p}(x)=1\\}$. As $\iota_{p}$ maps the
$\delta$-image of $E_{-p}(\mathbb{Q})[2]$ onto the $p$-adic image, we have

$S^{2}(E/K)=H\oplus\delta(E_{-p}(\mathbb{Q})[2]).$

Case 1: $p\equiv 1,4\bmod 15$.
In this case, the images of the generators for $\mathbb{Q}(S)$ under the maps
$\mathbb{Q}^{*}/\mathbb{Q}^{*2}\to\mathbb{Q}_{q}^{*}/\mathbb{Q}_{q}^{*2}$ for
$q\in S$, which we will sometimes write as $\mathrm{im}_{q}(x)$ for
$x\in\mathbb{Q}(S)$, are given as follows

$\begin{array}[]{c|ccccc}&2&3&5&p&\infty\\\ \hline\cr-1&-1&-1&1&&-1\\\
2&2&-1&2&&1\\\ 3&3&3&2&&1\\\ 5&-3&-1&5&1&1\\\ p&&1&1&p&1\end{array}$

Here the empty entries depend on $p\equiv 1\bmod 8$. Let
$x=(e_{1},e_{2},e_{3})\in H$. Then $e_{i}\in\langle-1,2,3,5\rangle$ for all
$i$ as $\iota_{p}(x)=1$. Since $\mathrm{im}_{3}(e_{3})$ and
$\mathrm{im}_{\infty}(e_{3})$ are both trivial, we get $e_{3}\in\langle
10\rangle$. As $\mathrm{im}_{5}(e_{2})=1$ we have
$e_{2}\in\langle-1,6\rangle$.
If $p\not\equiv 1\bmod 8$, we see that $\mathrm{im}_{2}(e_{3})=-6$ cannot
occur, hence $e_{3}=1$, which when $p\equiv 3,5\bmod 8$ implies $e_{2}=1$ by
considering the $2$-adic image. When $p\equiv 7\mod 8$ considering the
$2$-adic image yields $e_{2}\in\langle-1\rangle$, but $e_{2}=-1$ cannot happen
as it doesn’t have trivial $p$-adic image. We conclude that $H=0$ when
$p\not\equiv 1\bmod 8$.
If $p\equiv 1\bmod 8$, then one checks that $(15,6,10)\in H$, so we obtain a
complement for $\langle(15,6,10)\rangle$ in $H$ by setting $e_{3}=1$. Then the
$2$-adic image of $x$ is contained in $\langle(-1,-1,1)\rangle$, so
$e_{2}\in\langle-1\rangle$. This leaves two options, and as indeed
$(-1,-1,1)\in H$, it follows that $H$ is $2$-dimensional and generated by
$(15,6,10)$ and $(-1,-1,1)$.

Case 2: $p\equiv 2,8\bmod 15$.
In this case we take $\mathbb{Q}_{p}^{*}/\mathbb{Q}_{p}^{*2}=\langle
3,p\rangle$, and the table of local images of the generators for
$\mathbb{Q}(S)$ is then as follows.

$\begin{array}[]{c|ccccc}&2&3&5&p&\infty\\\ \hline\cr-1&-1&-1&1&&-1\\\
2&2&-1&2&&1\\\ 3&3&3&2&&1\\\ 5&-3&-1&5&3&1\\\ p&&-1&2&p&1\end{array}$

Let $x=(e_{1},e_{2},e_{3})\in H$. Then the $q$-adic images for $q=3,5,p$ imply
$e_{i}\in\langle-1,2\rangle$ for all $i$. By the real image also
$e_{3}\in\langle 2\rangle$. When $p\equiv 3,5\bmod 8$, the $2$-adic image
implies $e_{3}=1$, and then also $e_{2}=1$ so $H=0$. When $p\equiv 7\bmod 8$
we get $e_{3}=1$ and $e_{2}\in\langle-1\rangle$ from the $2$-adic image. But
$-1$ has non-trivial $p$-adic image, hence $e_{2}=1$. It follows that $H=0$
when $p\not\equiv 1\bmod 8$.
When $p\equiv 1\bmod 8$ we see that $(1,2,2)\in H$ so we have a complement for
$\langle(1,2,2)\rangle$ in $H$ by setting $e_{3}=1$. Then
$e_{2}\in\langle-1\rangle$ by the $2$-adic image, and since $(-1,-1,1)\in H$,
it follows that $H$ is $2$-dimensional and generated by $(1,2,2)$ and
$(-1,-1,1)$. ∎

It follows that the rank of $E$ over $\mathbb{Q}(\sqrt{-p})$ is at most $2$
when $p\equiv 1\bmod 8$, and equal to $0$ when $p\not\equiv 1\bmod 8$. In
particular, combined with Proposition 3.5 this settles the case $(i)$ of
Theorem 1.1.
We will now focus on the goal of improving the rank bound when $p\equiv 1\bmod
8$. To this end, observe that for a square-free integer $d\neq-p$, the twists
$E_{-p}$ and $E_{d}$ are quadratic twists of each other by $-dp$, hence

(14)
$\mathrm{rank}(E_{-p}/\mathbb{Q})+\mathrm{rank}(E_{d}/\mathbb{Q})=\mathrm{rank}\left(E_{d}/\mathbb{Q}(\sqrt{-dp})\right).$

By choosing $d$ appropriately, a $2$-Selmer group computation of $E_{d}$ over
$\mathbb{Q}(\sqrt{-dp})$ can improve the rank bound from Proposition 3.6 for
some primes $p\equiv 1\bmod 8$. In fact, choosing $d=-1$ will settle case
$(ii)$ of Theorem 1.1, while $d=-17$ settles case $(iii)$. In section 4 these
choices for $d$ are clarified.

The remaining $2$-Selmer group computations will make use of Rédei symbols as
in [21]. Let us for convenience extend the definition of minimal ramification
as follows. In the notation of [21, Def. 7.6] we say simply that $L$ is
minimally ramified if $K\subset L$ is minimally ramified over $E$, which
causes no confusion as $K$ and $E$ are determined by $L$. Additionally, we say
that both $\mathbb{Q}(\sqrt{a},\sqrt{\beta})$ and
$\mathbb{Q}(\sqrt{b},\sqrt{\beta})$ are minimally ramified if their normal
closure over $\mathbb{Q}$, which is $L$, is minimally ramified.

######  Proposition 3.7.

Let $p\equiv 1,49\bmod 120$ be a prime. Then

$\mathrm{dim}_{\mathbb{F}_{2}}S^{2}(E_{-1}/\mathbb{Q}(\sqrt{p}))=\begin{cases*}4&if\,
$[-1,10,p]=1$\\\ 2&if\, $[-1,10,p]=-1$\end{cases*}$

###### Proof.

Let $K=\mathbb{Q}(\sqrt{p})$. Let $S$ consist of the primes of $K$ lying over
$30$ together with the real primes. Then $|S|=8$, and the completions at the
primes in $S$ are given by $\mathbb{Q}_{2},\mathbb{Q}_{3},\mathbb{Q}_{5}$ and
$\mathbb{R}$, hence we can take the local images from the $d=-1$ cases in (11)
and (12). They are given as follows.

$\begin{array}[]{|c|}\hline\cr\mathrm{im}(\delta_{2})\\\ \hline\cr(-1,-1,1)\\\
(1,-6,-6)\\\ (1,2,2)\\\
\hline\cr\end{array}\,\,\begin{array}[]{|c|}\hline\cr\mathrm{im}(\delta_{3})\\\
\hline\cr(-1,-1,1)\\\ (3,3,1)\\\
\hline\cr\end{array}\,\,\begin{array}[]{|c|}\hline\cr\mathrm{im}(\delta_{5})\\\
\hline\cr(2,1,2)\\\ (5,1,5)\\\
\hline\cr\end{array}\,\,\begin{array}[]{|c|}\hline\cr\mathrm{im}(\delta_{\infty})\\\
\hline\cr(-1,-1,1)\\\ \hline\cr\end{array}$

We have the short exact sequence

$0\to R_{S}^{*}/R_{S}^{*2}\to K(S)\to\mathrm{Cl}(R_{S})[2]\to 0,$

where $R_{S}=\\{x\in K:v_{\mathfrak{p}}(x)\geq 0\text{ for all finite
}\mathfrak{p}\in S\\}$ is the ring of $S$-integers. Since $\mathrm{Cl}(R_{S})$
is a quotient of $\mathrm{Cl}_{K}$ and $\mathrm{Cl}_{K}[2]=0$, we have
$\mathrm{Cl}(R_{S})[2]=0$. The finitely generated abelian group $R_{S}^{*}$
has rank $7$ and cyclic torsion subgroup of even order, so it follows that
$R_{S}^{*}/R_{S}^{*2}$, and hence also $K(S)$, has $\mathbb{F}_{2}$-dimension
$8$.
To find a convenient basis for $K(S)$, observe that for $d\in\\{-1,2,3,5\\}$
we have $(p,d)_{q}=1$ for all primes $q$, hence there exists $x_{d}\in K$ with
$N(x_{d})\in d\mathbb{Q}^{*2}$ such that $K(\sqrt{x_{d}})/K$ is minimally
ramified. Also denote by $y_{d}$ the conjugate of $x_{d}$. Let

$H:=\langle x_{-1},y_{-1},x_{2},y_{2},x_{3},y_{3},x_{5},y_{5}\rangle\subset
K^{*}/K^{*2}.$

The minimal ramification of the extensions $K(\sqrt{x_{d}})/K$ implies
$H\subset K(S)$. In fact equality holds because the norm-induced map
$H\to\mathbb{Q}^{*}/\mathbb{Q}^{*2}$ has kernel and image of dimension $4$, so
$H$ and $K(S)$ have the same $\mathbb{F}_{2}$-dimension.

For each $d\in\\{-1,2,3,5\\}$, we may replace both $x_{d}$ and $y_{d}$ with
$tx_{d}$ and $ty_{d}$, respectively, for some $t$ in the twisting subgroup
$T_{p,d}\subset\mathbb{Q}^{*}/\mathbb{Q}^{*2}$ as in [21, Eq. (43) and Lemma
7.7]. This doesn’t change the fact that they form a basis for $K(S)$. These
subgroups are given as follows

$\begin{array}[]{c|c}d&T_{p,d}\\\ \hline\cr-1&\langle-1,p\rangle\\\ 2&\langle
2,p\rangle\\\ 3&\langle-1,-3,p\rangle\\\ 5&\langle 5,p\rangle\end{array}$

We now make various choices, using up part of the elbow room provided by the
twisting subgroups.
Let $\mathfrak{p}_{\infty}:K\to\mathbb{R}$ be the infinite prime such that
$\mathfrak{p}_{\infty}(x_{-1})>0$ and let $\mathfrak{q}_{\infty}$ be the other
infinite prime. Let $\mathfrak{p}_{2}$ be the prime of $K$ over $2$ that is
unramified in $K(\sqrt{x_{-1}})$ and let $\mathfrak{q}_{2}$ be the conjugate
of $\mathfrak{p}_{2}$. Twist $x_{2}$ by $2$ if necessary such that
$\mathfrak{p}_{2}$ is unramified in $K(\sqrt{x_{2}})$. For $r\in\\{3,5\\}$,
let $\mathfrak{p}_{r}$ be the prime of $K$ over $r$ that splits in
$K(\sqrt{x_{2}})$ and let $\mathfrak{q}_{r}$ be the other prime over $r$.
Twist $x_{r}$ by $r^{*}$ if necessary such that $\mathfrak{p}_{r}$ is
unramified in $K(\sqrt{x_{r}})$. Finally, twist $x_{3}$ by $-1$ if necessary
such that $\mathfrak{p}_{2}$ is unramified in $K(\sqrt{x_{3}})$. For
$d\in\\{3,5\\}$, we get from
$\prod_{\mathfrak{p}}(x_{2},y_{d})_{\mathfrak{p}}=1$ that
$\mathrm{im}_{\mathfrak{p}_{2}}(x_{d})=1$. We now have the following table,
where the entries in a column corresponding to a prime $\mathfrak{p}\in S$ are
the images of the generators of $K(S)$ under the map $K(S)\to
K_{\mathfrak{p}}^{*}/K_{\mathfrak{p}}^{*2}$. We will often write
$\mathrm{im}_{\mathfrak{p}}(x)$ for the image of $x\in K(S)$ under this map.

(15)
$\begin{array}[]{c|cccccccc}&\mathfrak{p}_{2}&\mathfrak{q}_{2}&\mathfrak{p}_{3}&\mathfrak{q}_{3}&\mathfrak{p}_{5}&\mathfrak{q}_{5}&\mathfrak{p}_{\infty}&\mathfrak{q}_{\infty}\\\
\hline\cr
x_{-1}&\pagecolor[gray]{0.8}\hfil&&\pagecolor[gray]{0.8}\hfil&&\pagecolor[gray]{0.8}\hfil&&1&-1\\\
y_{-1}&&&&&&&-1&1\\\
x_{2}&\pagecolor[gray]{0.8}\hfil&&1&-1&1&2&\pagecolor[gray]{0.8}\hfil&\\\
y_{2}&&&-1&1&2&1&&\\\
x_{3}&1&3&\pagecolor[gray]{0.8}\hfil&&\pagecolor[gray]{0.8}\hfil&&\pagecolor[gray]{0.8}\hfil&\\\
y_{3}&3&1&&&&&&\\\
x_{5}&1&-3&\pagecolor[gray]{0.8}\hfil&&\pagecolor[gray]{0.8}\hfil&&\pagecolor[gray]{0.8}\hfil&\\\
y_{5}&-3&1&&&&&&\\\ \end{array}$

For conjugate primes $\mathfrak{p}$ and $\mathfrak{q}$ of $K$ we have
$\mathrm{im}_{\mathfrak{p}}(x_{d})=\mathrm{im}_{\mathfrak{q}}(y_{d})$, and
also
$\mathrm{im}_{\mathfrak{p}}(x_{d})=d\cdot\mathrm{im}_{\mathfrak{p}}(y_{d})$.
It follows that any $2\times 2$ block of the table corresponding to two
conjugate primes and $x_{d}$ and $y_{d}$ for some $d\in\\{-1,2,3,5\\}$ is
determined by any one of the four entries. Each gray-colored entry has only
two options because $\mathfrak{p}_{i}$ is unramified in $K(\sqrt{x_{d}})$ for
$i\in\\{2,3,5,\infty\\}$ and $d\in\\{-1,2,3,5\\}$, hence there are at most
$2^{11}$ possibilities for the table. This upper bound can be reduced to
$2^{7}$ due to various reciprocity relations333Tests with Magma indicate that
all of these $2^{7}$ options actually occur, meaning that we have found all
possible relations. Proving this is not necessary to complete the computation
however.:

  1. (a)

$\mathrm{im}_{\mathfrak{p}_{2}}(x_{-1})=1$ $\Leftrightarrow$
$\mathrm{im}_{\mathfrak{p}_{\infty}}(x_{2})=1$ as
$\prod_{\mathfrak{p}}(x_{-1},y_{2})_{\mathfrak{p}}=1$.

  2. (b)

$\mathrm{im}_{\mathfrak{p}_{3}}(x_{-1})=1$ $\Leftrightarrow$
$\mathrm{im}_{\mathfrak{p}_{\infty}}(x_{3})=1$ as
$\prod_{\mathfrak{p}}(x_{-1},y_{3})_{\mathfrak{p}}=1$.

  3. (c)

$\mathrm{im}_{\mathfrak{p}_{5}}(x_{-1})=1$ $\Leftrightarrow$
$\mathrm{im}_{\mathfrak{p}_{\infty}}(x_{5})=1$ as
$\prod_{\mathfrak{p}}(x_{-1},y_{5})_{\mathfrak{p}}=1$.

  4. (d)

$\mathrm{im}_{\mathfrak{p}_{5}}(x_{3})=1$ $\Leftrightarrow$
$\mathrm{im}_{\mathfrak{p}_{3}}(x_{5})=1$ as
$\prod_{\mathfrak{p}}(x_{3},y_{5})_{\mathfrak{p}}=1$.

We now take four cases, depending on $\mathrm{im}_{\mathfrak{p}_{2}}(x_{-1})$
and $\mathrm{im}_{\mathfrak{p}_{5}}(x_{-1})$.

Case 1: $\mathrm{im}_{\mathfrak{p}_{2}}(x_{-1})=1$ and
$\mathrm{im}_{\mathfrak{p}_{5}}(x_{-1})=1$.
In this case the table of local images take the following form.

(16)
$\begin{array}[]{c|cccccccc}&\mathfrak{p}_{2}&\mathfrak{q}_{2}&\mathfrak{p}_{3}&\mathfrak{q}_{3}&\mathfrak{p}_{5}&\mathfrak{q}_{5}&\mathfrak{p}_{\infty}&\mathfrak{q}_{\infty}\\\
\hline\cr x_{-1}&1&-1&&&1&1&1&-1\\\ y_{-1}&-1&1&&&1&1&-1&1\\\
x_{2}&&&1&-1&1&2&1&1\\\ y_{2}&&&-1&1&2&1&1&1\\\ x_{3}&1&3&&&&&&\\\
y_{3}&3&1&&&&&&\\\ x_{5}&1&-3&&&&&1&1\\\ y_{5}&-3&1&&&&&1&1\\\ \end{array}$

Let $e=(e_{1},e_{2},e_{3})\in S^{2}(E_{-1}/K)$. As
$\iota_{\mathfrak{p}_{2}}(e)$ and $\iota_{\mathfrak{q}_{2}}(e)$ are inside
$\mathrm{im}(\delta_{2})$, of which the first coordinate is contained in
$\langle-1,3\rangle\subset\mathbb{Q}_{2}^{*}/\mathbb{Q}_{2}^{*2}$, we see
$e_{1}\in\langle x_{-1},y_{-1},x_{3},y_{3},x_{5},y_{5}\rangle$. Using that in
fact the first coordinate of elements in $\mathrm{im}(\delta_{2})$ are in
$\langle-1\rangle$ further reduces this to $e_{1}\in\langle
x_{-1},y_{-1},x_{3}x_{5},y_{3}y_{5}\rangle$, leaving $2^{4}$ options. All of
these options in fact occur: the conjugate elements $(x_{-1},x_{-1},1)$ and
$(y_{-1},y_{-1},1)$ are in $S^{2}(E_{-1}/K)$. If
$\mathrm{im}_{\mathfrak{p}_{3}}(x_{5})=1$ we also find
$(x_{3}x_{5},x_{2}x_{3},x_{2}x_{5})$ in $S^{2}(E_{-1}/K)$ with its conjugate,
while if $\mathrm{im}_{\mathfrak{p}_{3}}(x_{5})=-1$ we have
$(x_{3}x_{5},y_{2}x_{3},y_{2}x_{5})$ and its conjugate in $S^{2}(E_{-1}/K)$.
Regardless of $\mathrm{im}_{\mathfrak{p}_{3}}(x_{5})$, we see that a
complement $H$ inside $S^{2}(E_{-1}/K)$ to the subspace generated by these
four elements is obtained by setting $e_{1}=1$. The $3$-adic and $5$-adic
images then imply $e_{2}=e_{3}\in\langle x_{-1},y_{-1},x_{2},y_{2}\rangle$,
upon which the $5$-adic and real images imply that $e_{2}=e_{3}=1$. It follows
that $H=0$ and $S^{2}(E_{-1}/K)$ is $4$-dimensional.

Case 2: $\mathrm{im}_{\mathfrak{p}_{2}}(x_{-1})=-3$ and
$\mathrm{im}_{\mathfrak{p}_{5}}(x_{-1})=1$.
The table now takes the following form.

(17)
$\begin{array}[]{c|cccccccc}&\mathfrak{p}_{2}&\mathfrak{q}_{2}&\mathfrak{p}_{3}&\mathfrak{q}_{3}&\mathfrak{p}_{5}&\mathfrak{q}_{5}&\mathfrak{p}_{\infty}&\mathfrak{q}_{\infty}\\\
\hline\cr x_{-1}&-3&3&&&1&1&1&-1\\\ y_{-1}&3&-3&&&1&1&-1&1\\\
x_{2}&&&1&-1&1&2&-1&-1\\\ y_{2}&&&-1&1&2&1&-1&-1\\\ x_{3}&1&3&&&&&&\\\
y_{3}&3&1&&&&&&\\\ x_{5}&1&-3&&&&&1&1\\\ y_{5}&-3&1&&&&&1&1\\\ \end{array}$

Let $e=(e_{1},e_{2},e_{3})\in S^{2}(E_{-1}/K)$. The $3$-adic image imply
$e_{3}\in\langle x_{-1},y_{-1},x_{2},y_{2},x_{5},y_{5}\rangle$, upon which the
real image implies $e_{3}\in\langle
x_{-1}y_{-1},x_{2}y_{2},x_{5},y_{5}\rangle$. As the $3$’rd coordinate of
elements in $\mathrm{im}(\delta_{2})$ is contained in $\langle 2,-3\rangle$,
we further conclude that $e_{3}\in\langle x_{2}y_{2},x_{5},y_{5}\rangle$.
Considering both the $\mathfrak{p}_{3}$ and $\mathfrak{q}_{3}$-adic image in
either case of $\mathrm{im}_{\mathfrak{p}_{3}}(x_{5})$ implies
$e_{3}\in\langle x_{2}y_{2}x_{5}y_{5}\rangle=\langle 10\rangle$. In
$S^{2}(E_{-1}/K)$ we find the element $(15,6,10)$ (which is the image of
$(-28,120)\in E_{-1}(K)[4]$), hence $H=\\{(e_{1},e_{2},e_{3})\in
S^{2}(E_{-1}/K):e_{3}=1\\}$ is a complement to $\langle(15,6,10)\rangle$
inside $S^{2}(E_{-1}/K)$. For $e=(e_{1},e_{2},e_{3})\in H$ we find
$e_{1}=e_{2}\in\langle x_{-1},y_{-1},x_{3},y_{3}\rangle$ by considering the
$2$-adic and $5$-adic images. The $2$-adic image the further implies
$e_{1}=e_{2}\in\langle x_{-1}y_{-1},x_{-1}x_{3}y_{3}\rangle$, upon which the
$5$-adic image yields $e_{1}=e_{2}\in\langle
x_{-1}y_{-1}\rangle=\langle-1\rangle$. Since indeed $(-1,-1,1)\in
S^{2}(E_{-1}/K)$, we conclude that $H=\langle(-1,-1,1)\rangle$ and that
$S^{2}(E_{-1}/K)$ has $\mathbb{F}_{2}$-dimension $2$.

The other two cases are covered similarly and are left to the reader. The
results are as follows: in $S^{2}(E_{-1}/K)$ we always have a $2$-dimensional
subspace coming from $E_{-1}(K)[4]$, which is generated by $(-1,-1,1)$ and
$(15,6,10)$. A basis for $S^{2}(E_{-1}/K)$ is obtained by adding the following
additional generators.

$\begin{array}[]{c|c|c|c|c}\mathrm{im}_{\mathfrak{p}_{2}}(x_{-1})&\mathrm{im}_{\mathfrak{p}_{5}}(x_{-1})&\mathrm{im}_{\mathfrak{p}_{3}}(x_{5})&\mathrm{dim}_{\mathbb{F}_{2}}S^{2}(E_{-1}/\mathbb{Q}(\sqrt{p}))&\text{additional
generators}\\\ \hline\cr
1&1&1&4&(x_{-1},x_{-1},1),(x_{3}x_{5},x_{2}x_{3},x_{2}x_{5})\\\
1&1&-1&4&(x_{-1},x_{-1},1),(x_{3}x_{5},y_{2}x_{3},y_{2}x_{5})\\\
1&2&1&2&\textit{none}\\\ 1&2&-1&2&\textit{none}\\\ -3&1&1&2&\textit{none}\\\
-3&1&-1&2&\textit{none}\\\
-3&2&1&4&(3x_{-1},3x_{-1},1),(x_{3}x_{5},x_{2}x_{3},x_{2}x_{5})\\\
-3&2&-1&4&(3x_{-1},3x_{-1},1),(x_{3}x_{5},y_{2}x_{3},y_{2}x_{5})\\\
\end{array}$

For $d\in\\{2,5\\}$ we have that $K(\sqrt{x_{d}})$ is minimally ramified,
hence we have $\mathrm{im}_{\mathfrak{p}_{d}}(x_{-1})=1$ if and only if
$[p,-1,d]=1$. Thus we see that $S^{2}(E_{-1}/K)$ has dimension $2$ if
$[p,-1,2][p,-1,5]=-1$, and dimension $4$ if $[p,-1,2][p,-1,5]=1$. Using tri-
linearity and reciprocity of Rédei symbols, we have
$[p,-1,2][p,-1,5]=[p,-1,10]=[-1,10,p]$, thus completing the proof. ∎

######  Proposition 3.8.

Let $p,q\equiv 17,113\bmod 120$ be distinct primes. Then

$\mathrm{dim}_{\mathbb{F}_{2}}S^{2}(E_{-q}/\mathbb{Q}(\sqrt{pq}))=\begin{cases*}6&if\,
$[-1,2,pq]=1$\\\ 4&if\, $[-1,2,pq]=-1$\end{cases*}$

###### Proof.

Let $K=\mathbb{Q}(\sqrt{pq})$. The primes of $K$ where $E_{-q}$ has bad
reduction are those over $2$, $3$ and $5$. Note that the good reduction at the
prime over $q$ is due to the completion being a ramified quadratic extension
of $\mathbb{Q}_{q}$. The completions at the primes of $K$ over $2,3,5$ and
$\infty$ are $\mathbb{Q}_{2},\mathbb{Q}_{3},\mathbb{Q}_{5}$ and $\mathbb{R}$,
respectively. The local images for these completions are as follows.

$\begin{array}[]{|c|}\hline\cr\mathrm{im}(\delta_{2})\\\ \hline\cr(-1,-1,1)\\\
(1,-6,-6)\\\ (1,2,2)\\\
\hline\cr\end{array}\,\,\begin{array}[]{|c|}\hline\cr\mathrm{im}(\delta_{3})\\\
\hline\cr(1,-1,-1)\\\ (-1,-1,1)\\\
\hline\cr\end{array}\,\,\begin{array}[]{|c|}\hline\cr\mathrm{im}(\delta_{5})\\\
\hline\cr(2,2,1)\\\ (1,2,2)\\\
\hline\cr\end{array}\,\,\begin{array}[]{|c|}\hline\cr\mathrm{im}(\delta_{\infty})\\\
\hline\cr(-1,-1,1)\\\ \hline\cr\end{array}$

From the $3$-adic and $5$-adic images, we see that in fact

$S^{2}(E_{-q}/K)\subset\ker\left(\bigoplus_{i=1}^{3}K(T)\to K(T)\right),$

where $T$ consists just of the primes of $K$ over $2$ together with the real
primes.
For $d\in\\{-1,2\\}$ we have $(pq,d)_{r}=1$ for all primes $r$ so, there
exists $x_{d}\in K$ with $N(x_{d})\in d\mathbb{Q}^{*2}$ such that
$K(\sqrt{x_{d}})/K$ is minimally ramified. Let $y_{d}$ be the conjugate of
$x_{d}$ in $K$. Consider

$H:=\langle x_{-1},y_{-1},x_{2},y_{2},q\rangle\subset K^{*}/K^{*2}.$

Then $H$ has $\mathbb{F}_{2}$-dimension $5$ as the norm induced map
$H\to\mathbb{Q}^{*}/\mathbb{Q}^{*2}$ has $2$-dimensional image and a
$3$-dimensional kernel. The minimal ramification of the extensions
$K(\sqrt{x_{d}})$ for $d=-1,2$, and the fact that $q$ is ramified in $K$,
implies that $H\subset K(T)$. This inclusion is an equality, as the short
exact sequence

$0\to R_{T}^{*}/R_{T}^{*2}\to K(T)\to\mathrm{Cl}(R_{T})[2]\to 0,$

implies $\dim_{\mathbb{F}_{2}}K(T)\leq 5$ as $R_{T}^{*}/R_{T}^{*2}$ has
dimension $4$ and $\mathrm{Cl}(R_{T})[2]$ has at most dimension $1$ as as
$\mathrm{Cl}_{K}[2]\cong\mathbb{Z}/2\mathbb{Z}$ and $\mathrm{Cl}(R_{T})$ is a
quotient of $\mathrm{Cl}_{K}$.
The twisting subgroups are now as follows.

$\begin{array}[]{c|c}d&T_{pq,d}\\\ \hline\cr-1&\langle-1,p,q\rangle\\\
2&\langle 2,p,q\rangle\\\ \end{array}$

Let $\mathfrak{p}_{\infty}:K\to\mathbb{R}$ be the infinite prime such that
$\mathfrak{p}_{\infty}(x_{-1})>0$ and let $\mathfrak{q}_{\infty}$ be the other
infinite prime. Let $\mathfrak{p}_{2}$ be the prime of $K$ over $2$ that is
unramified in $K(\sqrt{x_{-1}})$ and let $\mathfrak{q}_{2}$ be the other prime
over $2$. Twist $x_{2}$ by $2$ if necessary such that $\mathfrak{p}_{2}$ is
unramified in $K(\sqrt{x_{2}})$. We then have the following table of local
images.

$\begin{array}[]{c|cccc}&\mathfrak{p}_{2}&\mathfrak{q}_{2}&\mathfrak{p}_{\infty}&\mathfrak{q}_{\infty}\\\
\hline\cr x_{-1}&\pagecolor[gray]{0.8}\hfil&&1&-1\\\ y_{-1}&&&-1&1\\\
x_{2}&\pagecolor[gray]{0.8}\hfil&&\pagecolor[gray]{0.8}\hfil&\\\ y_{2}&&&&\\\
q&1&1&1&1\end{array}$

As $K(\sqrt{x_{-1}})/K$ is minimally ramified, we find that
$\mathrm{im}_{\mathfrak{p}_{2}}(x_{-1})=1$ precisely when $[pq,-1,2]=1$.
Similarly, working with $K(\sqrt{x_{2}})$ we find that
$\mathrm{im}_{\mathfrak{p}_{\infty}}(x_{2})=1$ precisely when $[pq,2,-1]=1$.
By Rédei reciprocity we have $[pq,2,-1]=[pq,-1,2]$, giving a relation between
the two conditions, and by the reciprocity again they both also equal
$[-1,2,pq]$.

Case 1: $[-1,2,pq]=1$.
In this case the table takes the form

$\begin{array}[]{c|cccc}&\mathfrak{p}_{2}&\mathfrak{q}_{2}&\mathfrak{p}_{\infty}&\mathfrak{q}_{\infty}\\\
\hline\cr x_{-1}&1&-1&1&-1\\\ y_{-1}&-1&1&-1&1\\\ x_{2}&a&2a&1&1\\\
y_{2}&2a&a&1&1\\\ q&1&1&1&1\end{array}$

with $a\in\langle-3\rangle\subset\mathbb{Q}_{2}^{*}/\mathbb{Q}_{2}^{*2}$. Let
$(e_{1},e_{2},e_{3})\in S^{2}(E_{-q}/K)$. Then the real images imply that
$e_{3}\in\langle x_{2},y_{2},q\rangle$, while the first coordinate of the
$2$-adic images imply $e_{1}\in\langle x_{-1},y_{-1},q\rangle$. As $e_{2}$ is
determined by $e_{1}$ and $e_{3}$, this leaves $2^{6}$ options, all of which
are in fact in $S^{2}(E_{-q}/K)$. Thus $S^{2}(E_{-q}/K)$ is $6$-dimensional.

Case 2: $[-1,2,pq]=-1$.
In this case the table takes the form

$\begin{array}[]{c|cccc}&\mathfrak{p}_{2}&\mathfrak{q}_{2}&\mathfrak{p}_{\infty}&\mathfrak{q}_{\infty}\\\
\hline\cr x_{-1}&-3&3&1&-1\\\ y_{-1}&3&-3&-1&1\\\ x_{2}&a&2a&-1&-1\\\
y_{2}&2a&a&-1&-1\\\ q&1&1&1&1\end{array}$

with $a\in\langle-3\rangle\subset\mathbb{Q}_{2}^{*}/\mathbb{Q}_{2}^{*2}$. Let
$(e_{1},e_{2},e_{3})\in S^{2}(E_{-q}/K)$. Then the real images imply that
$e_{3}\in\langle 2,q\rangle$, while the first coordinate of the $2$-adic
images imply $e_{1}\in\langle-1,q\rangle$. This leaves $2^{4}$ options, all of
which are in $S^{2}(E_{-q}/K)$. Thus $S^{2}(E_{-q}/K)$ is $4$-dimensional. ∎

As final ingredient for the proof of Theorem 1.1, let us state the following
version of Chebotarëv’s density theorem. We refer to [19, Ch. I, §2.2].

######  Theorem 3.9 (Chebotarëv’s density theorem).

Let $K/\mathbb{Q}$ be a finite Galois extension and let
$C\subset\mathrm{Gal}(K/\mathbb{Q})$ be a non-empty subset closed under
conjugation. Then

$\lim_{X\to\infty}\frac{\\#\\{p\leq X:p\text{ is unramified in }K\text{ and
}\mathrm{Frob}_{p}\in C\\}}{\\#\\{p\leq X\\}}=\frac{\\#C}{[K:\mathbb{Q}]}.$

Here all instances of $p$ assume that $p$ is a prime, and $\mathrm{Frob}_{p}$
is some choice of $\mathrm{Frob}_{\mathfrak{p}}$ for $\mathfrak{p}$ a prime of
$K$ over $p$. Different choices for $\mathrm{Frob}_{p}$ are conjugate, so as
$C$ is closed under conjugation, the condition $\mathrm{Frob}_{p}\in C$ is
independent of the choice of $\mathrm{Frob}_{p}$.

###### Proof of Theorem 1.1.

Recall from (13) the identity

(18)
$\mathrm{rank}(E/\mathbb{Q}(\sqrt{-p}))=\mathrm{rank}(E_{-p}/\mathbb{Q}).$

We prove that $\mathrm{rank}(E/\mathbb{Q}(\sqrt{-p}))=0$ by showing that
$\mathrm{rank}(E_{-p}/\mathbb{Q})=0$. Suppose that $p\equiv 1,2,3,4,5,8\bmod
15$. Then $\mathrm{rank}(E_{-p}/\mathbb{Q})=0$ follows from Proposition 3.5 if
$p\leq 5$, and from Proposition 3.6 if $p>5$.
If $p\equiv 1\bmod 8$, $p\equiv 1,4\bmod 15$, and $[-1,10,p]=-1$, then
applying both Proposition 3.5 and (14) for $d=-1$ we obtain

$\mathrm{rank}(E_{-p}/\mathbb{Q})=\mathrm{rank}(E_{-1}/\mathbb{Q}(\sqrt{p})).$

The right hand side is $0$ by Proposition 3.7.
Lastly, assume that $p\equiv 1\bmod 8$, $p\equiv 2,8\bmod 15$, and
$[-1,2,p]=1$. Then $p\neq 17$ as $[-1,2,17]=-1$, and as $[-1,2,17p]=-1$, we
get from Proposition 3.8 and (14) with $d=-17$ the inequality

$\mathrm{rank}(E_{-p}/\mathbb{Q})+\mathrm{rank}(E_{-17}/\mathbb{Q})\leq 2.$

In fact we have $\mathrm{rank}(E_{-17}/\mathbb{Q})=2$: the points
$(\tfrac{-2246}{17},\tfrac{-103950}{17^{2}})$ and
$(\tfrac{52}{17},\tfrac{832}{17^{2}})$ have images $(1,2,2)$ and $(-1,-1,1)$
in $S^{2}(E_{-17}/\mathbb{Q})$, respectively. We conclude that
$\mathrm{rank}(E_{-p}/\mathbb{Q})=0$.

It remains to prove the density statement. Note that it suffices to show that

(19) $\lim_{X\to\infty}\frac{\\#\\{p\leq X:p\text{ satisfies }(i),(ii)\text{
or }(iii)\\}}{\\#\\{p\leq X\\}}=\frac{7}{16},$

because from Theorem 3.9 applied to $\mathbb{Q}(\zeta_{15})$ and
$C=\\{\overline{1},\overline{2},\overline{4},\overline{8}\\}\subset(\mathbb{Z}/15\mathbb{Z})^{*}=\mathrm{Gal}(\mathbb{Q}(\zeta_{15})/\mathbb{Q})$
we deduce

$\lim_{X\to\infty}\frac{\\#\\{p\leq X:p\equiv 1,2,3,4,5,8\bmod
15\\}}{\\#\\{p\leq X\\}}=\frac{1}{2}.$

Applying Theorem 3.9 to $\mathbb{Q}(\zeta_{120})$ we deduce

(20) $\lim_{X\to\infty}\frac{\\#\\{p\leq X:p\text{ satisfies
}(i)\\}}{\\#\\{p\leq X\\}}=\frac{4\cdot 3}{\varphi(120)}=\frac{3}{8}$

Let $E=\mathbb{Q}(\sqrt{-1},\sqrt{10})$ and $F=E(\sqrt{3+\sqrt{-1}})$. Then
$F$ is minimally ramified and can be used to compute the Rédei symbol
$[-1,10,p]$ provided it defined, which is the case if and only if $p\equiv
1,9\bmod 40$.
Consider the Galois extenion $F(\zeta_{24})/\mathbb{Q}$ of degree $32$. Let
$\sigma\in\mathrm{Gal}(F(\zeta_{24})/\mathbb{Q})$ be the generator of the
order $2$ subgroup corresponding to the intermediate field $E(\zeta_{24})$,
and let $Z_{\mathfrak{p}}$ be the decomposition field of a prime
$\mathfrak{p}$ of $F(\zeta_{24})$ over $p$. The prime $p$ splits completely in
$E(\zeta_{24})$ precisely when $p\equiv 1,49\bmod 120$, hence

$p\text{ satisfies
}(ii)\,\Leftrightarrow\,Z_{\mathfrak{p}}=E(\zeta_{24})=\,\Leftrightarrow\,\mathrm{Frob}_{p}=\sigma.$

Theorem 3.9 thus implies

(21) $\lim_{X\to\infty}\frac{\\#\\{p\leq X:p\text{ satisfies
}(ii)\\}}{\\#\\{p\leq X\\}}=\frac{1}{32}$

For condition $(iii)$, let
$E^{\prime}=\mathbb{Q}(\sqrt{-1},\sqrt{2})=\mathbb{Q}(\zeta_{8})$, and
$F^{\prime}=E^{\prime}(\sqrt{1+\sqrt{-1}})$. The standard isomorphism

$\mathrm{Gal}(F^{\prime}(\zeta_{15})/\mathbb{Q})\xrightarrow{\sim}\mathrm{Gal}(F^{\prime}/\mathbb{Q})\times\mathrm{Gal}(\mathbb{Q}(\zeta_{15})/\mathbb{Q})=\mathrm{Gal}(F^{\prime}/\mathbb{Q})\times(\mathbb{Z}/15\mathbb{Z})^{*},$

maps a Frobenius element $\mathrm{Frob}_{p}$ over $p$ to
$(\mathrm{Frob}_{p},p\bmod 15)$. Let
$C\subset\mathrm{Gal}(F^{\prime}(\zeta_{15})/\mathbb{Q})$ be the subset
corresponding under this isomorphism with
$1\times\\{\overline{2},\overline{8}\\}$. It follows that
$\mathrm{Frob}_{p}\in C$ if and only if $p\equiv 17,113\bmod 120$ and
$[-1,2,p]=1$, hence with Theorem 3.9 we obtain

(22) $\lim_{X\to\infty}\frac{\\#\\{p\leq X:p\text{ satisfies
}(iii)\\}}{\\#\\{p\leq X\\}}=\frac{2}{64}=\frac{1}{32}.$

As the conditions $(i)$, $(ii)$ and $(iii)$ are mutually exclusive, (19)
follows by taking the sum of (20), (21) and (22), thus completing the proof. ∎

######  Remark 3.10.

Any prime $q\equiv 17,113\bmod 120$ such that
$\mathrm{rank}(E_{-q}/\mathbb{Q})=2$ can be used in the proof of 1.1(iii).
Possible alternatives for the choice of $q=17$ are for example $q=233$, or
$q=617$. A priori the prime $q$ must also satisfy $[-1,2,q]=-1$, but after
having a proof of Theorem 1.1(iii) we know this follows from the rank being
non-zero.

##  4\. Relation to $4$-descent

In the previous section we encountered various instances where the rank bound
of $X_{0}(15)_{-p}/\mathbb{Q}$ obtained from a $2$-Selmer group computation
could be improved using $2$-Selmer group computations over suitable quadratic
fields. A natural question is how this improvement is related to a
$4$-descent. In this section we prove that the rank bounds are exactly as
strong as those obtained from a $4$-descent. In fact we will show a stronger
result relating higher descents. After a lemma we recall parts of [4, §4]
concerning visualisation techniques of $\Sh[2]$, and explain how these
techniques can improve on a $2$-descent. The main result of this section,
Theorem 1.3, gives a condition under which the rank bounds obtained from
visualisation of $\Sh[2]$ are in fact equivalent to those obtained from a
$4$-descent. We then use this result to show that the rank bounds obtained in
section $3$ are equivalent to those obtained from performing a $4$-descent on
$X_{0}(15)_{-p}$ over $\mathbb{Q}$.

Throughout this section, let $E$ be an elliptic curve defined over a number
field $K$.

######  Lemma 4.1.

For $L/K$ a finite extension we have the following commutative diagram.

${M_{L}^{\prime}}$${{H^{1}(L,E[2])}}$${M_{K}^{\prime}}$${{H^{1}(K,E[2])}}$$\scriptstyle{\sim}$$\scriptstyle{N}$$\scriptstyle{\mathrm{cor}}$$\scriptstyle{\sim}$

Here $N$ is induced from the relative norm map $N_{M_{L}/M_{K}}:M_{L}\to
M_{K}$ and the horizontal isomorphisms are as in section 2. Note that $M_{L}$
is a finite free $M_{K}$-algebra of rank $[L:K]$ as $M_{L}=M_{K}\otimes_{K}L$,
hence we indeed have said norm map.

###### Proof.

By transitivity of the norm, see [3, Prop. III.9.4], we have a commutative
diagram of norm maps.

${M_{L}^{*}}$${M_{K}^{*}}$${L^{*}}$${K^{*}}$

Modding out by squares, we obtain an induced map $N:M_{L}^{\prime}\to
M_{K}^{\prime}$. For the commutativity, let
$\overline{M}_{K}=\overline{K}[t]/f(t)$ and similarly for $L$. Letting $G_{K}$
act trivially on $t$ we give $\overline{M}_{K}$ a $G_{K}$-module structure in
the obvious way. Then $M_{K}=\overline{M}_{K}^{G_{K}}$, and the norm map
$N_{M_{L}/M_{K}}:M_{L}\to M_{K}$ equals
$\mathrm{cor}:H^{0}(L,\overline{M}_{L})\to H^{0}(K,\overline{M}_{K})$. The
isomorphism $H^{1}(K,E[2])\to M_{K}^{\prime}$ comes from the composition

$H^{1}(K,E[2])\xrightarrow{w_{K}}H^{1}(K,\mu_{2}(\overline{M_{K}}))\xrightarrow{k_{K}}M_{K}^{*}/M_{K}^{*2},$

and similarly for $L$ (notation of the maps is as in [18], but with subscripts
to specify the base field). The desired commutativity thus follows at once
because the corestriction maps are natural and compatible with connecting
homomorphisms. ∎

Suppose that $L=K(\sqrt{d})$ is a quadratic extension of $K$, and consider the
Weil restriction $A=\mathrm{Res}_{L/K}E_{L}$. It is an Abelian surface which
by definition, see [1, §7.6], comes with bijections

(23) $\mathrm{Hom}_{L}(T_{L},E_{L})\cong\mathrm{Hom}_{K}(T,A)$

for any $K$-scheme $T$, natural in $T$. The $G_{K}$-module $A(\overline{K})$
is induced from the $G_{L}$-module $E(\overline{L})$, hence by Shapiro’s lemma
we get $H^{n}(K,A)\cong H^{n}(L,E)$ for all $n\geq 0$. For $n=0$ this
isomorphism coincides with the isomorphism $E(K)\cong E(L)$ obtained from (23)
with $T=\mathrm{Spec}(K)$. Similarly, $A(\overline{K})[2]$ is induced from
$E(\overline{L})[2]$ and we obtain isomorphic long exact sequences

(24)
${0}$${{A(K)[2]}}$${A(K)}$${A(K)}$${{H^{1}(K,A[2])}}$${{H^{1}(K,A)}}$${\cdots}$${0}$${{E(L)[2]}}$${E(L)}$${E(L)}$${{H^{1}(L,E[2])}}$${{H^{1}(L,E)}}$${\cdots}$$\scriptstyle{\sim}$$\scriptstyle{\sim}$$\scriptstyle{\sim}$$\scriptstyle{\sim}$$\scriptstyle{\sim}$

As $K_{\mathfrak{p}}\otimes
L\cong\prod_{\mathfrak{q}|\mathfrak{p}}L_{\mathfrak{q}}$ for a prime
$\mathfrak{p}$ of $K$, the isomorphism (24) also induces isomorphisms

$\begin{split}S^{2}(A/K)&\cong S^{2}(E/L),\\\
\Sh(A/K)&\cong\Sh(E/L).\end{split}$

Now suppose that $E$ and the quadratic twist $E_{d}$ are given by the
equations

$\displaystyle E$ $\displaystyle:\,\,\,y^{2}\,$ $\displaystyle=f(x)$
$\displaystyle E_{d}$ $\displaystyle:dy^{2}\,$ $\displaystyle=f(x)$

for a monic cubic polynomial $f(x)\in K[x]$. The identity on $E_{L}$ and the
isomorphism $E_{L}\to(E_{L})_{d}$ fixing the $x$-coordinate yields by (23) a
map $\phi:E\times E_{d}\to A$ defined over $K$. This is a $2$-isogeny444For
$L/K$ a finite Galois extension, $\mathrm{Res}_{L/K}E_{L}$ is isogenous to a
direct sum of twists of $E$ by the irreducible rational representations of
$\mathrm{Gal}(L/K)$, see [15, Theorem 4.5]. with kernel equal to the diagonal
embedding $\Delta$ of $E_{1}[2]\cong E_{2}[2]$, hence $\phi$ factors through
$[2]$ as $[2]=\phi^{\vee}\phi$ for a unique $\phi^{\vee}:A\to E\times E_{d}$.
As an immediate application, let us quickly give a proof of

######  Proposition 4.2.

We have $\mathrm{rank}(E/L)=\mathrm{rank}(E/K)+\mathrm{rank}(E_{d}/K)$.

###### Proof.

As the rank is an isogeny invariant we obtain the result as follows.

$\begin{split}\mathrm{rank}(E/L)&=\mathrm{rank}(A/K)\\\
&=\mathrm{rank}(E\times E_{d}/K)\\\
&=\mathrm{rank}(E/K)+\mathrm{rank}(E_{d}/K)\qed\end{split}$

In particular, since $\Delta$ is isotropic with respect to the Weil pairing
$e_{2}$ on $(E_{1}\times E_{2})[2]$, we obtain a principal polarisation of
$(E\times E_{d})/\Delta\cong A$ compatible with the isogenies $\phi$ and
$\phi^{\vee}$. Alternatively, one can use that Weil restriction preserves
principal polarisations, see [9, Prop. 2].

Denote $\phi$ composed with the inclusion $E\to E\times E_{d}$ by $p^{*}$, and
$\phi^{\vee}$ composed with the projection $E\times E_{d}\to E$ by $p_{*}$ as
in [4, §3]. Then we have the following commutative diagram

${E}$${E\times E_{d}}$${A}$${E\times
E_{d}}$${E}$$\scriptstyle{[2]}$$\scriptstyle{p^{*}}$$\scriptstyle{[2]}$$\scriptstyle{\phi}$$\scriptstyle{\phi^{\vee}}$$\scriptstyle{p_{*}}$

We see that $[2]$ on $E$ factors through $p^{*}$ and $p_{*}$. This also
induces a factorisation of $[2]$ on the long exact sequence of cohomology
associated to $[2]$ on $E(\overline{K})$, and upon applying the isomorphism of
(24) we obtain the following commutative diagram

${0}$${{E(K)[2]}}$${E(K)}$${E(K)}$${{H^{1}(K,E[2])}}$${{H^{1}(K,E)}}$${\cdots}$${0}$${{E(L)[2]}}$${E(L)}$${E(L)}$${{H^{1}(L,E[2])}}$${{H^{1}(L,E)}}$${\cdots}$${0}$${{E(K)[2]}}$${E(K)}$${E(K)}$${{H^{1}(K,E[2])}}$${{H^{1}(K,E)}}$

where the first row of vertical arrows coincide with restriction maps, and the
second row of vertical arrows with corestriction maps. Observe that for
$\xi\in\Sh(E/K)[2]$, this implies that if $\xi$ is divisible by $2$ in
$\Sh(E/K)[2]$, then $\xi$ has to be in the image of
$\Sh(E/L)[2]\xrightarrow{\mathrm{cor}}\Sh(E/K)[2]$. Using Lemma 4.1 we obtain
the following commutative diagram.

(25)
${S^{2}(E/L)}$${\Sh(E/L)[2]}$${S^{2}(E/K)}$${\Sh(E/K)[2]}$$\scriptstyle{N}$$\scriptstyle{\mathrm{cor}}$

hence checking if $\xi$ is in the image of
$\Sh(E/L)[2]\xrightarrow{\mathrm{cor}}\Sh(E/K)[2]$ can be seen by checking if
a lift of $\xi$ in $S^{2}(E/K)$ is a norm from $S^{2}(E/L)$, which can be
checked in practice provided that the two relevant $2$-Selmer groups have been
computed. This gives a necessary condition for $2$-divisibility in
$\Sh(E/K)[2]$, which is used in [4, Lemma 3.3] to improve on a $2$-descent. It
need not be sufficient however, as seen in [4, §7, Example 2].

For an abelian group $M$, let $M_{\mathrm{nd}}=M/M_{\mathrm{div}}$, where
$M_{\mathrm{div}}$ is the subgroup of $M$ consisting of the infinitely
divisible elements. The Cassels-Tate pairing on $\Sh(A/K)$ (recall that $A$ is
principally polarized) yields a non-degenerate pairing

$\langle-,-\rangle_{A}:\Sh(A/K)_{\mathrm{nd}}\times\Sh(A/K)_{\mathrm{nd}}\to\mathbb{Q}/\mathbb{Z},$

and similarly for $\Sh(E\times E_{d}/K)$. The pairings $\langle-,-\rangle_{A}$
and $\langle-,-\rangle_{E\times E_{d}}$ are compatible with respect to the
dual isogenies $\phi$ and $\phi^{\vee}$, in the sense that

$\langle\phi\xi,\xi^{\prime}\rangle_{A}=\langle\xi,\phi^{\vee}\xi^{\prime}\rangle_{E\times
E_{d}},$

see [17, Prop. 31]. This induces a non-degenerate pairing

(26) $\Sh(A/K)_{nd}[\phi^{\vee}]\times\Sh(A/K)_{nd}/\phi\Sh(E\times
E_{d}/K)_{nd}\to\mathbb{Q}/\mathbb{Z}.$

Compare also with [5, §7]. We can now give a proof of Theorem 1.3, whose
statement we recall for convenience.

######  Theorem 1.3.

Suppose that the following conditions are satisfied.

  1. (1)

We have $\Sh(E_{d}/K)[2]=0$.

  2. (2)

The map $\mathrm{cor}:\Sh(E/L)[2]\to\Sh(E/K)[2]$ is injective.

Then $\mathrm{res}:\Sh(E/K)\to\Sh(E/L)$ induces an isomorphism

$\Sh(E/K)[2^{\infty}]/\Sh(E/K)[2]\xrightarrow{\sim}\Sh(E/L)[2^{\infty}].$

###### Proof.

Using (24) we translate the statements of the theorem to maps only involving
Tate-Shafarevich groups over $K$. Specifically, condition (2) translates to
the injectivity of $\Sh(A/K)[2]\to\Sh(E/K)[2]$, and we need to show that the
map $\Sh(E/K)\to\Sh(A/K)$ induces an isomorphism

$\Sh(E/K)[2^{\infty}]/\Sh(E/K)[2]\xrightarrow{\sim}\Sh(A/K)[2^{\infty}].$

The map $p_{*}$ induces maps $\Sh(A/K)[2]\to\Sh(E/K)[2]$ and
$\Sh(A/K)[2^{\infty}]\to\Sh(E/K)[2^{\infty}]$, which we will denote by
$p_{*,2}$ and $p_{*,2^{\infty}}$, respectively. Similarly, restricting
$\Sh(p^{*})$ to the $2$-torsion and $2$-part, respectively, we obtain maps
denoted by $p^{*}_{2}$ and $p^{*}_{2^{\infty}}$.

The first step is to show that $p_{*,2^{\infty}}$ is injective. Observe that
because $p_{*,2}$ is assumed to be injective, this follows directly once we
know that $\ker(p_{*,2^{\infty}})\subset\Sh(A/K)[2]$.
As is additive and $\Sh(E_{d}/K)[2]=0$, the map $\Sh(E\times
E_{d}/K)[2^{\infty}]\to\Sh(E/K)[2^{\infty}]$ is an isomorphism. It follows
that

$\ker(p_{*,2^{\infty}})=\Sh(A/K)[\phi^{\vee}]\subset\Sh(A/K)[2],$

and hence $p_{*,2^{\infty}}$ is injective. Secondly, as
$\Sh(A/K)[\phi^{\vee}]=0$ we have

$\Sh(E\times E_{d}/K)[\phi]=\Sh(E\times E_{d}/K)[2].$

Since also $\Sh(E/K)[2^{\infty}]\xrightarrow{\sim}\Sh(E\times
E_{d}/K)[2^{\infty}]$, we deduce that $\ker(p^{*}_{2^{\infty}})=\Sh(E/K)[2]$.
Thus to complete the proof it suffices to show that $p^{*}_{2^{\infty}}$ is
surjective, which follows from showing that $\Sh(\phi)$ is surjective. As
$\Sh(A/K)[\phi^{\vee}]=0$, the non-degeneracy of the pairing (26) implies that
$\Sh(E\times E_{d}/K)_{\mathrm{nd}}\to\Sh(A/K)_{\mathrm{nd}}$ is surjective.
As $\Sh(E\times E_{d}/K)_{\mathrm{div}}\to\Sh(A/K)_{\mathrm{div}}$ is
surjective (multiplication by $2$ is clearly surjective on
$\Sh(A/K)_{\mathrm{div}}$), we deduce that also $\Sh(\phi)$ itself is
surjective, completing the proof. ∎

Recall the commutative diagram (25). If $S^{2}(E/K)$ and $S^{2}(E/L)$ have
been computed this provides a way to verify condition (2) of Theorem 1.3.
Before we apply Theorem 1.3 to prove Theorems 1.4 and 1.5, let us discuss the
conditions of Theorem 1.3. In the proof we saw that the map
$\Sh(E/K)\to\Sh(A/K)$ has kernel equal to all of $\Sh(E/K)[2]$. In the
language of visualisation, this means that all of $\Sh(E/K)[2]$ is
simultaneously visualised in the Abelian surface $A$. From [4, Cor. 3.4] we
know that a $\xi\in\Sh(E/K)[2]$ is in the kernel of $\Sh(E/K)\to\Sh(A/K)$ if
and only if $\xi$ is in the image of $E_{d}(K)\to\Sh(E/K)[2]$. Thus we see
that a necessary condition for the hypotheses of Theorem 1.3 to be valid, is
that $E_{d}(K)$ hits all of $\Sh(E/K)[2]$.

We can see that this indeed happens for the twists corresponding to the
quadratic extensions of $\mathbb{Q}$ seen in Propositions 3.7 and 3.8: in the
$p\equiv 1,49\bmod 120$ case of Proposition 3.6 we see that
$\Sh(X_{0}(15)_{-p}/\mathbb{Q})[2]$ is generated by the images of $(15,6,10)$
and $(-1,-1,1)$, which are in the image of $X_{0}(15)_{-1}(\mathbb{Q})$ as
seen in the proof of Proposition 3.5! This case is special in the sense that
the points in $X_{0}(15)_{-1}(\mathbb{Q})$ hitting
$\Sh(X_{0}(15)_{-p}/\mathbb{Q})[2]$ are all torsion. In the $p\equiv
17,113\bmod 120$ case such a twist does not exist (a consequence of
Proposition 3.4), but since the explicit generators of
$\Sh(X_{0}(15)_{-p}/\mathbb{Q})[2]$, the images of $(2,2,1)$ and $(-1,-1,1)$,
do not depend on $p$, we can take a prime $q$ distinct from $p$ satisfying the
same congruences, causing $\Sh(X_{0}(15)_{-p}/\mathbb{Q})[2]$ to get hit by
$X_{0}(15)_{-q}(\mathbb{Q})$ provided that also
$\mathrm{rank}(X_{0}(15)_{-q})=2$.
In fact this also works in the $p\equiv 1,49\bmod 120$ case: one can prove
Theorems 1.1 and 1.4 by replacing Proposition 3.7 by a variant involving
$\mathbb{Q}(\sqrt{pq})$ in the same spirit as Proposition 3.8. Two possible
choices for $q$ in this case (so $q\equiv 1,49\bmod 120$ and with
$X_{0}(15)_{-q}$ of rank $2$) are $q=241$ and $q=601$.

######  Proposition 4.3.

Let $E=X_{0}(15)$ and let $p\equiv 1,49\bmod 120$ be a prime. Then we have an
isomorphism

$\Sh(E_{-p}/\mathbb{Q})[2^{\infty}]/\Sh(E_{-p}/\mathbb{Q})[2]\cong\Sh(E_{-1}/\mathbb{Q}(\sqrt{p}))[2^{\infty}]$

###### Proof.

Let $K=\mathbb{Q}(\sqrt{p})$. We apply Theorem 1.3 to $E_{-p}$ and $d=p$. Also
recall Remark 2.1. Since $(E_{-p})_{p}\cong E_{-1}$ we have
$\Sh((E_{-p})_{p}/\mathbb{Q})[2]=0$ by Proposition 3.5. Let us see why
$\Sh(E_{-p}/K)[2]\xrightarrow{\mathrm{cor}}\Sh(E_{-p}/\mathbb{Q})[2]$ is
injective. When $[-1,10,p]=-1$ this follows immediately as then
$\Sh(E_{-p}/K)[2]=0$ by Proposition 3.7, so suppose that $[-1,10,p]=1$. We
have the commutative diagram

(27)
${S^{2}(E_{-p}/K)}$${\Sh(E_{-p}/K)[2]}$${S^{2}(E_{-p}/\mathbb{Q})}$${\Sh(E_{-p}/\mathbb{Q})[2]}$$\scriptstyle{N}$$\scriptstyle{\mathrm{cor}}$

Note that $\Sh(E_{-p}/K)[2]$ and $\Sh(E_{-p}/\mathbb{Q})[2]$ have the same
size: their $\mathbb{F}_{2}$-dimensions both equal
$2-\mathrm{rank}(E_{-p}/\mathbb{Q})$. Thus to prove injectivity it is
equivalent to show surjectivity, which follows from (27) and the computation
of the 2-Selmer groups: $\Sh(E_{-p}/\mathbb{Q})[2]$ is generated by the images
of $(15,6,10)$ and $(-1,-1,1)$, which are indeed norms: $(15,6,10)$ is the
norm of either $(x_{3}x_{5},x_{2}x_{3},x_{2}x_{5})$ or
$(x_{3}x_{5},y_{2}x_{3},y_{2}x_{5})$, while $(-1,-1,1)$ is the norm of either
$(x_{-1},x_{-1},1)$ or $(3x_{-1},3x_{-1},1)$. ∎

Similarly we have

######  Proposition 4.4.

Let $E=X_{0}(15)$, and let $p,q\equiv 17,113\bmod 120$ distinct primes such
that $\mathrm{rank}(E_{-q}/\mathbb{Q})=2$. Then we have an isomorphism

$\Sh(E_{-p}/\mathbb{Q})[2^{\infty}]/\Sh(E_{-p}/\mathbb{Q})[2]\cong\Sh(E_{-q}/\mathbb{Q}(\sqrt{pq}))[2^{\infty}]$

###### Proof.

Let $K=\mathbb{Q}(\sqrt{pq})$. As $\mathrm{rank}(E_{-q}/\mathbb{Q})=2$,
Proposition 3.6 implies that $\Sh(E_{-q}/\mathbb{Q})[2]=0$, thus as in the
proof of Proposition 4.3, the desired isomorphism follows from Theorem 1.3 if
we show that the right vertical map in the diagram

(28)
${S^{2}(E_{-p}/K)}$${\Sh(E_{-p}/K)[2]}$${S^{2}(E_{-p}/\mathbb{Q})}$${\Sh(E_{-p}/\mathbb{Q})[2]}$$\scriptstyle{N}$$\scriptstyle{\mathrm{cor}}$

is injective. Note that

$\begin{split}\dim_{\mathbb{F}_{2}}S^{2}(E_{-p}/K)-2&=\mathrm{rank}(E_{-q}/K)+\dim_{\mathbb{F}_{2}}\Sh(E_{-p}/K)[2]\\\
&=\mathrm{rank}(E_{-p}/\mathbb{Q})+2+\dim_{\mathbb{F}_{2}}\Sh(E_{-p}/K)[2]\end{split}$

As $\mathrm{rank}(E_{-q}/\mathbb{Q})=2$ we must have $[-1,2,q]=-1$ by Theorem
1.1, so if $[-1,2,p]=1$ we have $\dim_{\mathbb{F}_{2}}S^{2}(E_{-q}/K)=4$ by
Proposition 3.8 and hence $\dim_{\mathbb{F}_{2}}\Sh(E_{-p}/K)[2]=0$, so the
injectivity clearly follows.
When $[-1,2,p]=-1$ we have $\dim_{\mathbb{F}_{2}}S^{2}(E_{-q}/K)=6$, and we
see that $\Sh(E_{-p}/K)[2]$ and $\Sh(E_{-p}/\mathbb{Q})[2]$ both have
$\mathbb{F}_{2}$-dimension $2-\mathrm{rank}(E_{-p}/\mathbb{Q})$, so
injectivity is equivalent to surjectivity. As in the proof of Proposition 4.3,
this follows from (28) and the $2$-Selmer group computations. The images of
$(1,2,2),(-1,-1,1)\in S^{2}(E_{-p}/\mathbb{Q})$, generate
$\Sh(E_{-p}/\mathbb{Q})[2]$, and they are the norms of $(1,x_{2},x_{2})$ and
$(x_{-1},x_{-1},1)$, respectively. ∎

Combining the various results we can now prove Theorems 1.4 and 1.5.

###### Proof of Theorem 1.4.

From Proposition 3.6 we have

(29)
$\mathrm{rank}(E_{-p}/\mathbb{Q})+\dim_{\mathbb{F}_{2}}\Sh(E_{-p}/\mathbb{Q})[2]=2.$

Suppose that $[-1,10,p]=-1$. Then $\mathrm{rank}(E_{-p}/\mathbb{Q})=0$ by
Theorem 1.1, hence
$\Sh(E_{-p}/\mathbb{Q})[2]\cong(\mathbb{Z}/2\mathbb{Z})^{2}$ by (29). From
Proposition 3.7 we have $\Sh(E_{-p}/\mathbb{Q}(\sqrt{p}))[2]=0$, hence from
Proposition 4.3 we obtain

$\Sh(E_{-p}/\mathbb{Q})[2^{\infty}]=\Sh(E_{-p}/\mathbb{Q})[2],$

which proves i. Now assume $[-1,10,p]=1$. Then multiplication by $2$ induces
an injection

(30)
$\Sh(E_{-p}/\mathbb{Q})[4]/\Sh(E_{-p}/\mathbb{Q})[2]\to\Sh(E_{-p}/\mathbb{Q})[2].$

The left hand side of (30) is isomorphic to
$\Sh(E_{-p}/\mathbb{Q}(\sqrt{p}))[2]$ by Proposition 4.3, which has the same
size as $\Sh(E_{-p}/\mathbb{Q})[2]$. It follows that the map (30) is also
surjective, which proves ii.
If we additionally assume that $\mathrm{rank}(E_{-p}/\mathbb{Q})=0$, then
$\Sh(E_{-p}/\mathbb{Q})[2]\cong(\mathbb{Z}/2\mathbb{Z})^{2}$, which combined
with surjectivity of $\Sh(E_{-p}/\mathbb{Q})[4]\xrightarrow{\cdot
2}\Sh(E_{-p}/\mathbb{Q})[2]$ forces

$\Sh(E_{-p}/\mathbb{Q})[4]\cong(\mathbb{Z}/4\mathbb{Z})^{2},$

proving iii. ∎

###### Proof of Theorem 1.5.

Again we have

$\mathrm{rank}(E_{-p}/\mathbb{Q})+\dim_{\mathbb{F}_{2}}\Sh(E_{-p}/\mathbb{Q})[2]=2.$

When $p=17$ we have $[-1,2,p]=-1$, $\mathrm{rank}(E_{-p}/\mathbb{Q})=2$ and
$\Sh(E_{-p}/\mathbb{Q})[2]=0$, so the theorem clearly holds in this case. Thus
we may assume $p\neq 17$ for the rest of the proof. Let
$K=\mathbb{Q}(\sqrt{17p})$. By Proposition 3.8 and (14) we obtain

$\mathrm{rank}(E_{-p}/\mathbb{Q})+\dim_{\mathbb{F}_{2}}\Sh(E_{-17}/K)[2]=\begin{cases*}0&if\,
$[-1,2,p]=1$\\\ 2&if\, $[-1,2,p]=-1$\end{cases*}$

If $[-1,2,p]=1$ then $\mathrm{rank}(E_{-p}/\mathbb{Q})=0$, and with
Proposition 4.4 we have

$\Sh(E_{-p}/\mathbb{Q})[2^{\infty}]=\Sh(E_{-p}/\mathbb{Q})[2]\cong(\mathbb{Z}/2\mathbb{Z})^{2},$

which proves i. If $[-1,2,p]=-1$ then multiplication by $2$ gives the
injection

$\Sh(E_{-p}/\mathbb{Q})[4]/\Sh(E_{-p}/\mathbb{Q})[2]\to\Sh(E_{-p}/\mathbb{Q})[2],$

which is surjective as the left hand side is isomorphic to $\Sh(E_{-p}/K)[2]$
by Proposition 4.4, which has the same size $\Sh(E_{-p}/\mathbb{Q})[2]$. This
proves ii, and iii follows just as in the proof of Theorem 1.4. ∎

##  5\. Concluding remarks

Theorems 1.4 and 1.5 show that the upper bounds obtained on
$\mathrm{rank}(X_{0}(15)/\mathbb{Q}(\sqrt{-p}))$ in Theorem 1.1 are the same
as those obtained from performing a $4$-descent over $\mathbb{Q}$ on the twist
$X_{0}(15)_{-p}$. In fact Theorems 1.4 and 1.5 can also be proven by a
computation of the Cassels-Tate pairing on $S^{2}(X_{0}(15)_{-p}/\mathbb{Q})$,
as explained in [8]. The advantage of the proof as presented in this paper is
that Propositions 4.3 and 4.4 show that the rank bound obtained from
$2^{n}$-descent over $\mathbb{Q}$ on $X_{0}(15)_{-p}$ is equivalent to the one
obtained from a $2^{n-1}$-descent on $X_{0}(15)_{-1}$ or $X_{0}(15)_{-17}$
over a suitable quadratic field, opening up possibilities to further improve
on the rank bounds in various ways.

For example, a Cassels-Tate pairing computation on
$S^{2}(X_{0}(15)_{-1}/\mathbb{Q}(\sqrt{p}))$ for primes $p\equiv 1,49\bmod
120$ gives the same rank bounds as an $8$-descent on $X_{0}(15)_{-p}$ over
$\mathbb{Q}$. Consider for instance the prime $p=1609$. Then $p\equiv 49\bmod
120$ and $[-1,10,p]=1$, and with Magma one computes that the Cassels-Tate
pairing on $S^{2}(X_{0}(15)_{-1}/\mathbb{Q}(\sqrt{p}))$ is non-trivial, which
proves that

$\mathrm{rank}(X_{0}(15)_{-p}/\mathbb{Q})=0,\quad\text{and
}\quad\Sh(X_{0}(15)_{-p}/\mathbb{Q})[2^{\infty}]\cong(\mathbb{Z}/4\mathbb{Z})^{2}.$

Alternatively, one can work with explicit equations for the $2$-covers
corresponding to the elements of $S^{2}(X_{0}(15)_{-1}/\mathbb{Q}(\sqrt{p}))$
and perform a second descent as in [6, §9].

Let us discuss the rank of $X_{0}(15)$ over imaginary quadratic fields of the
form $\mathbb{Q}(\sqrt{-dp})$, where $d$ is positive, square-free and not
divisible by primes $>5$, and $p>5$ is a prime. A $2$-Selmer group computation
over $\mathbb{Q}$ of $X_{0}(15)_{-dp}$ gives an upper bound of either $0$ or
$2$ on the rank provided that $dp\equiv 0,1,2,3,4,5,8,12\bmod 15$. In cases
where the rank bound is $2$, depending on the class of $p\bmod 120$, one can
find $2$ independent elements in the $2$-Selmer group that are independent of
$p$, just as in in Proposition 3.6. Experimentation with Magma seems to
suggest that Theorem 1.3 is applicable and yields rank bounds equivalent to
those obtained from $4$-descent on $X_{0}(15)_{-dp}$ by finding suitable
quadratic twists satisfying the necessary condition as discussed after the
proof of Theorem 1.3 in section 4.

Theorem 1.3 is also applicable to the quadratic twists $E_{p}$ of
$E:y^{2}=x^{3}-x$, where $p\equiv 1\bmod 8$ is a prime as one encounters in
studying congruent primes. Specifically, the rank bounds obtained from
$2$-Selmer group computations over $\mathbb{Q}(\sqrt{p})$ in [13] can be shown
to be equivalent to those from a $4$-descent of $E_{p}$ over $\mathbb{Q}$ by
an application of Theorem 1.3. If one goes one step further with for example a
Cassels-Tate pairing computation over $\mathbb{Q}(\sqrt{p})$ one arrives at a
proof of a variant of [5, Theorem B] as mentioned in the last sentence of
section 10 of loc. cit. The variance lies in the fact that the number
theoretic description is different, even though they can likely be proven to
be equivalent directly through a reciprocity argument, similar to how the
$2$-descent of $E$ over $\mathbb{Q}(\sqrt{p})$ is expressed using $[p,-1,2]$,
and the $2$-descent of the isogeneous curve is expressed using $[-1,2,p]$,
which are the same because of Rédei reciprocity.

Let us conclude by discussing the restrictiveness of the hypotheses of Theorem
1.3 in general. As remarked before, if the hypotheses are fulfilled, this
implies that $\Sh(E/K)\to\Sh(A/K)$ has kernel equal to $\Sh(E/K)[2]$, forcing
surjectivity of $E_{d}(K)\to\Sh(E/K)[2]$. In particular this implies

$\mathrm{rank}(E_{d}/K)\geq\mathrm{dim}_{\mathbb{F}_{2}}\Sh(E/K)[2]-\mathrm{dim}_{\mathbb{F}_{2}}E(K)[2],$

which, as $\Sh(E/K)[2]$ increases in size, becomes increasingly restrictive.
Specialising to $K=\mathbb{Q}$, then the size of $\Sh(E/\mathbb{Q})[2]$ is
known to be unbounded as $E$ varies. Assuming that the ranks of elliptic
curves over $\mathbb{Q}$ are uniformly bounded, this implies there exist
elliptic curves over $\mathbb{Q}$ for which the hypotheses of Theorem 1.3 are
not fulfilled for any quadratic twist.

## Acknowledgements

I would like to thank Stevan Gajovic for informing me about the modularity
result of Newton and Caraiani, which initiated the process of writing this
paper. Gratitude goes to Michael Stoll, who directed me to the visualisation
techniques, which opened the door to investigate the connection of the rank
results to a $4$-descent. I also thank Nils Bruin for fruitful discussions
regarding his paper [4] during his visit to Ulm, and for helpful
clarifications of some later questions. Finally, Irene Bouw and Jeroen
Sijsling gave valuable suggestions to preliminary versions of this paper.

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