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\title{Computational Verification of the Birch and Swinnerton-Dyer
Conjecture for Individual Elliptic Curves\footnote{This material is
based upon work supported by the National Science Foundation under
Grant No. 0400386.}}
\author{Grigor Grigorov \and Andrei Jorza\and Stefan Patrikis\and William A.~Stein\and Corina Tarni\c{t}\v{a}-P\v{a}tra\c{s}cu}
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\begin{document}
\begin{abstract}
We describe theorems and computational methods for verifying the
Birch and Swinnerton-Dyer conjecture for specific elliptic curves
over~$\Q$. We apply our techniques to show that if $E$ is a non-CM
elliptic curve over~$\Q$ of conductor $\leq 1000$ and rank $\leq 1$,
then the full Birch and Swinnerton-Dyer conjecture is true for~$E$
up to odd primes that divide either a Tamagawa number of~$E$ or the
degree of some rational cyclic isogeny with domain~$E$.
\end{abstract}
\maketitle
\tableofcontents
\section{Introduction}
Let $E$ be an elliptic curve over~$\Q$. The $L$-function $L(E,s)$
of~$E$ is a holomorphic function on~$\C$ that encodes deep arithmetic
information about~$E$. This paper is about a connection between the
behavior of $L(E,s)$ at $s=1$ and the arithmetic of~$E$.
We use theorems and computation to attack the following conjecture for
many specific elliptic curves of conductor $\leq 1000$:
\begin{conjecture}[Birch and Swinnerton-Dyer]\label{conj:bsd}
The order of vanishing $\ord_{s=1} L(E,s)$ equals the rank~$r$ of $E$,
the group $\Sha(E)$ is finite, and
$$
\frac{L^{(r)}(E, 1)}{r!} = \frac{\Omega_E\cdot \Reg_E \cdot\prod_p c_p
\cdot \#\Sha(E)}{(\#E(\Q)_{\tor})^2}.
$$
\end{conjecture}
For more about Conjecture~\ref{conj:bsd}, see \cite{lang:nt3,
wiles:cmi} and the papers they reference. See also Section~\ref{sec:notation}
below for the notation used in the conjecture.
Henceforth we call it the BSD conjecture.
\begin{definition}[Analytic $\Sha$]
If $E$ has rank~$r$, let
$$
\#\Sha(E)_{\an} = \frac{L^{(r)}(E,1)\cdot (\#E(\Q)_{\tor})^2}
{r! \cdot \Omega_E \cdot \Reg_E \cdot \prod_p c_p}
$$
denote the order of $\Sha(E)$ predicted by
Conjecture~\ref{conj:bsd}. We call this the {\em analytic order} of
$\Sha(E)$.
\end{definition}
\begin{conjecture}[BSD$(E,p)$]
Let $(E,p)$ denote a pair consisting of an elliptic curve $E$ over
$\Q$ and a prime~$p$.
We also call the assertion that $\ord_{s=1}L(E,s)$ equals the rank~$r$,
that $\Sha(E)[p^\infty]$ is finite,
and
$$
\ord_p(\#\Sha(E)[p^\infty]) = \ord_p(\#\Sha(E)_{\an})
$$
the {\em BSD conjecture at~$p$}, and denote it $\BSD(E,p)$.
\end{conjecture}
The BSD conjecture is invariant under isogeny.
\begin{theorem}[Cassels]\label{thm:bsdiso}
If $E$ and $F$ are $\Q$-isogeneous and $p$ is a prime, then
$\BSD(E,p)$ is true if and only if $\BSD(F,p)$ is true.
\end{theorem}
\begin{proof}
See \cite{cassels:7, milne:duality, jorza:seniorthesis}.
\end{proof}
%In light of Theorem~\ref{thm:bsdiso}, we will frequently
%restrict our attention to optimal elliptic curves (see Definition~\ref{defn:optimal} below).
One way to give evidence for the conjecture is to compute
$\#\Sha(E)_{\an}$ and note that it is a perfect square, in accord with
the following theorem:
\begin{theorem}[Cassels]\label{thm:shasqr}
If $E$ is an elliptic curve over~$\Q$ and $p$ is a prime such that
$\Sha(E)[p^\infty]$ is finite, then $\#\Sha(E)[p^\infty]$ is a
perfect square.
\end{theorem}
\begin{proof}
See \cite{cassels:arithmeticiii, poonen-stoll}.
\end{proof}
We use the notation of \cite{cremona:onlinetables} to refer to
specific elliptic curves over~$\Q$.
\begin{conjecture}[Birch and Swinnerton-Dyer $\leq$ 1000]\label{conj:bsd1000}
For {\em all} optimal curves of conductor $\leq 1000$ we have $\Sha(E)=0$,
except for the following four rank~$0$ elliptic curves,
where $\Sha(E)$ has the indicated order:
{\rm
\begin{center}
\begin{tabular}{|c|c|c|c|c|}\hline
{\bf Curve} & 571A & 681B & 960D & 960N \\\hline
$\#\Sha(E)_{\an}$ & 4 & 9 & 4 & 4\\\hline
\end{tabular}
\end{center}
}
\end{conjecture}
\begin{theorem}[Cremona]\label{thm:cremona1000}
Conjecture~\ref{conj:bsd} is true for all elliptic curves of
conductor~$\leq 1000$ if and only if Conjecture~\ref{conj:bsd1000}
is true.
\end{theorem}
\begin{proof}
In the book \cite{cremona:algs}, Cremona computed $\#\Sha(E)_{\an}$
for every curve of conductor $\leq 1000$. By
Theorem~\ref{thm:bsdiso} it suffices to consider only the optimal
ones, and the four listed are the only ones with nontrivial
$\#\Sha(E)_{\an}$.
\end{proof}
In view of Theorem~\ref{thm:cremona1000}, the main goal of this paper is to
obtain results in support of Conjecture~\ref{conj:bsd1000}. The
results of Section~\ref{sec:results} below together imply the theorem
we claimed in the abstract:
\begin{theorem}\label{thm:main}
Suppose that $E$ is a non-CM elliptic curve of rank $\leq 1$,
conductor $\leq 1000$ and that $p$ is a prime. If $p$ is odd,
assume further that the mod~$p$ representation $\rhobar_{E,p}$ is
irreducible and~$p$ does not divide any Tamagawa number of~$E$. Then
$\BSD(E,p)$ is true.
\end{theorem}
\begin{proof}
Combine Theorem~\ref{thm:bsdr1}, Theorem~\ref{thm:bsd2}, and Theorem~\ref{thm:bsdr0}.
\end{proof}
For example, if $E$ is the elliptic curve 37A, then according to
\cite{cremona:algs}, all $\rhobar_{E,p}$ are irreducible and the
Tamagawa numbers of $E$ are $1$. Thus Theorem~\ref{thm:main} asserts
that the full BSD conjecture for~$E$ is true.
There are $18$ optimal curves of conductor $\leq 1000$ of rank $2$
(and none of rank $>2$). For these~$E$ of rank~$2$, nobody has proved
that $\Sha(E)$ is finite in even a single case. We exclude CM elliptic
curves from most of our computations. The methods for dealing with
the BSD conjecture for CM elliptic curves are different than for
general curves, and will be the subject of another paper. The same is
true for $\BSD(E,p)$ when $\rhobar_{E,p}$ is reducible.
\subsection{Acknowledgement}
We thank Michael Stoll for suggesting this project at an American
Institute of Mathematics meeting and for initial feedback and ideas,
and Stephen Donnelly and Michael Stoll for key ideas about
Section~\ref{sec:relaxkoly}. We thank John Cremona for many
discussions and his immensely useful computer software. Finally we
thank Benedict Gross and Noam Elkies for helpful feedback and
encouragement throughout the project.
We thank the Harvard College Research Program for funding P\u{a}tra\c{s}cu's
work on this paper, and the Herchel Smith Harvard Undergraduate
Research Fellowship for supporting Patrikis. Jorza and Stein were
supported by the National Science Foundation under Grant No.~0400386.
\subsection{Notation and Background}\label{sec:notation}
If~$G$ is an abelian group, let $G_{\tor}$ denote the torsion subgroup
and $G_{/\tor}$ denote the quotient $G/G_{\tor}$.
For an integer~$m$, let $G[m]$ be the kernel of
multiplication by~$m$ on~$G$.
For a commutative ring $R$, we let $R^*$ denote the group
of units in~$R$.
\subsubsection{Galois Cohomology of Elliptic Curves}
For a number field~$K$, let $G_K=\Gal(\Qbar/K)$.
Let~$E$ be an elliptic curve defined over a number field~$K$, and
consider the first Galois cohomology group $\H^1(K,E) = \H^1(G_K,
E(\Kbar))$, and the local Galois cohomology groups
$\H^1(K_v,E) = \H^1(\Gal(\Kbar_v/K_v),E(\Kbar_v))$, for
each place~$v$ of~$K$.
\begin{definition}[Shafarevich-Tate group]
The {\em Shafarevich-Tate group}
$$
\Sha(E/K)= \Ker\Bigl(\H^1(K, E) \rightarrow \bigoplus_v \H^1(K_v, E)\Bigr),
$$
of~$E$ measures the failure of global cohomology classes to be
determined by their localizations at all places.
\end{definition}
If $E$ is an elliptic curve over a field~$F$ and the field~$F$ is
clear from context, we write $\Sha(E)=\Sha(E/F)$. For example, if $E$
is an elliptic curve over~$\Q$, then $\Sha(E)$ means $\Sha(E/\Q)$.
\begin{definition}[Selmer group]
For each positive integer~$m$, the {\em $m$-Selmer group} is
$$\Sel^{(m)}(E/K)= \Ker\Bigl(\H^1(K, E[m])\rightarrow
\bigoplus_v \H^1(K_v, E)\Bigr).
$$
\end{definition}
The Selmer group relates the Mordell-Weil and Shafarevich-Tate
groups of~$E$ via the exact sequence
$$
0 \to E(K)/m E(K) \to \Sel^{(m)}(E/K) \to \Sha(E/K)[m] \to 0,
$$
where $\Sha(E/K)[m]$ denotes the $m$-torsion subgroup of $\Sha(E/K)$.
Note that $\Sha(E/K)$ is a torsion group since $\H^1(K,E)$ is
torsion.
\subsubsection{Elliptic Curves over $\Q$}
See \cite[pp.~360--361]{silverman:aec} for the definition of $L(E,s)$
and \cite{wiles:fermat, breuil-conrad-diamond-taylor} for why
$L(E,s)$ is entire.
Let $E$ be an elliptic curve over~$\Q$. We use the notation of
\cite{cremona:onlinetables} to refer to certain elliptic curves.
Thus, e.g., 37B3 refers to the third elliptic curve in the second
isogeny class of elliptic curves of conductor 37, i.e., the curve $y^2
+ y = x^3 + x^2 - 3x +1$. The ordering of isogeny classes and curves
in isogeny classes is as specified in \cite{cremona:algs}. If the
last number is omitted, it is assumed to be~$1$, so 37B refers to the
first curve in the second isogeny class of curves of conductor~$37$.
Let $\Reg_E$ be the absolute value of the discriminant of the
canonical height pairing on $E(\Q)_{/\tor}$. Let $c_p =
[E(\Q_p):E_0(\Q_p)]$ be the Tamagawa number of~$E$ at $p$, where
$E_0(\Q_p)$ is the subgroup of points that reduce to a nonsingular
point modulo~$p$. Let $\Omega_E = \int_{E(\R)}|\omega|$, where
$$\omega=\frac{dx}{2y+a_1x+a_3}$$ is the invariant differential
attached to a minimal Weierstrass model for~$E$.
For any prime~$p$, let $\rhobar_{E,p}:G_\Q \to \Aut(E[p])$ denote the
mod~$p$ representation and $\rho_{E,p}:G_\Q \to \Aut(T_p E)$ the
representation on the $p$-adic Tate module $T_p E$ of~$E$.
It follows from \cite{breuil-conrad-diamond-taylor}
that every elliptic curve $E$ over~$\Q$ is a factor of the modular curve
$X_0(N)$, where $N$ is the conductor of~$E$.
\begin{definition}[Optimal]\label{defn:optimal}
An elliptic curve~$E$ over~$\Q$ is {\em optimal} if for every
elliptic curve~$F$ and surjective morphisms $X_0(N) \to F \to E$, we
have~$E\isom F$. (Optimal curves are also called ``strong Weil curves''
in the literature.)
\end{definition}
We say $E$ is a {\em complex multiplication} (CM)
curve, if $\End(E/\Qbar)\neq \Z$.
\section{Elliptic Curve Algorithms}\label{sec:algs}
\subsection{Images of Galois Representations}
Let $E$ be an elliptic curve over~$\Q$. Many theorems that provide
explicit bounds on $\#\Sha(E)[p^\infty]$ have as a hypothesis that
$\rhobar_{E,p}$ or $\rho_{E,p}$ be either surjective or irreducible.
In this section we explain how to prove
that $\rhobar_{E,p}$ or $\rho_{E,p}$ is surjective or irreducible,
in particular cases.
\subsubsection{Irreducibility}\label{sec:irred}
Regarding irreducibility, note that $\rhobar_{E,p}$ is irreducible if and
only if there is no isogeny $E\to F$ over~$\Q$ of degree~$p$. The
degrees of all such isogenies for curves of conductor $\leq 1000$ are
recorded in \cite{cremona:algs}, which were computed
using Cremona's program {\tt allisog}. This program uses results of
Mazur \cite{mazur:rational} along with computations involving
modular curves of genus~$0$.
\subsubsection{Surjectivity}\label{sec:surj}
We discuss surjectivity of $\rho_{E,p}$ in the rest of this section.
\begin{theorem}[Mazur]
If~$E$ is semistable and
$p\geq 11$, then $\rhobar_{E,p}$ is surjective.
\end{theorem}
\begin{proof}
See \cite[Thm.~4]{mazur:rational}.
\end{proof}
\begin{example}
Mazur's theorem implies that the representations $\rhobar_{E,p}$
attached to the semistable elliptic curve $E=X_0(11)$ are surjective
for $p\geq 11$. Note that $\rhobar_{E,5}$ is reducible.
\end{example}
\begin{theorem}[Cojocaru, Kani, and Serre]\label{thm:cojocaru-kani}
If~$E$ is a non-CM elliptic curve of conductor~$N$, and
$$
p\geq 1+ \frac {4\sqrt{6}}{3}\cdot N\cdot
\prod_{\text{prime }\ell|N}\left(1+\frac{1}{\ell}
\right)^{1/2},
$$
then $\rhobar_{E,p}$ is surjective.
\end{theorem}
\begin{proof}
See Theorem~2 of \cite{cojocaru-kani:surj}, whose proof
relies on the results of \cite{serre:propgal}.
\end{proof}
\begin{example}\label{ex:cojocaru-kani}
When $N=11$, the bound of Theorem~\ref{thm:cojocaru-kani} is $\sim
38.52$. When $N=997$, the bound is $\sim 3258.8$. For $N=40000$, the
bound is $\sim 143109.35$.
\end{example}
%Kolyvagin's method yields triviality of the $p$-primary component of
%$\Sha(E)$ for all primes that do not divide $I_K=[E(K):\Z y_K]$ and
%for which $\rho_{E,p}$ is surjective, and in theory one could bound
%$\Sha(E)$ by computing $\Sel^{(p)}(E/\Q)$ directly.
%In general such computations are not feasible as soon as~$p$ is at all
%large (e.g., $\geq 7$), since they require class group computations
%with number fields of large degree.
\begin{proposition}\label{prop:surj}
Let $E$ be a non-CM elliptic curve over~$\Q$ of conductor~$N$ and let
$p\geq 5$ be a prime. For each prime $\ell\nmid p\cdot N$ with
$a_\ell \not\con 0\pmod{p}$, let
$$
s(\ell) = \kr{a_\ell^2 - 4\ell}{p} \in \{0, -1, +1\},
$$
where the symbol $\kr{\cdot}{\cdot}$ is the Legendre symbol.
If $-1$ and $+1$ both occur as values of $s(\ell)$, then
$\rhobar_{E,p}$ is surjective. If $s(\ell) \in \{0,1\}$ for
all~$\ell$, then $\Im(\rhobar_{E,p})$ is contained in a Borel
subgroup (i.e., reducible), and if $s(\ell) \in
\{0,-1\}$ for all $\ell$, then $\Im(\rhobar_{E,p})$ is a
nonsplit torus.
\end{proposition}
\begin{proof}
This is an application of \cite[\S4]{serre:propgal}, where we use
the quadratic formula to convert the condition that certain
polynomials modulo~$p$ be reducible or irreducible into a quadratic
residue symbol.
\end{proof}
For computational applications we apply Proposition~\ref{prop:surj} as
follows. We choose a bound $B$ and compute values $s(\ell)$; if both
$-1$ and $+1$ occur as values of $s(\ell)$, we stop computing $s(\ell)$
and conclude that $\rhobar_{E,p}$ is surjective. If for
$\ell\leq B$ we find that $s(\ell) \in \{0,1\}$, we suspect that
$\Im(\rhobar_{E,p})$ is Borel, and attempt to show this (see
Section~\ref{sec:irred}). If for $\ell\leq B$, we have $s(\ell) \in
\{0,-1\}$, we suspect that $\Im(\rhobar_{E,p})$ is contained in a nonsplit
torus, and try to show this by computing and analyzing the
$p$-division polynomial of~$E$. If this approach is inconclusive, we
can alway increase~$B$ and eventually the process terminates. In practice
we often apply some theorem under the hypothesis that $\rhobar_{E,p}$
is surjective, which is something that in practice we verify
for a particular~$p$ using Proposition~\ref{prop:surj}.
Example~\ref{ex:cojocaru-kani} suggests that the bound of
Theorem~\ref{thm:cojocaru-kani} is probably far larger than necessary.
Nonetheless, it is small enough that in a reasonable amount of time we
can determine whether $\rhobar_{E,p}$ is surjective, using the above
process, for all~$p$ up to the bound. In this way we determine the
exact image of Galois.
\begin{remark}
% There is a proper subgroup $H$ of $\GL_2(\F_3)$ for which
% $(\Tr(g),\det(g))$ takes all possible values. For example the group
%$$H=\left\{ \pm \mtwo{1}{0}{0}{1}, \pm \mtwo{1}{0}{0}{-1},\pm
% \mtwo{0}{1}{-1}{0}, \pm \mtwo{0}{1}{1}{0}\right\}$$ of order $8$.
% <--- commented out, since it is simply WRONG. That group doesn't
% gives all possible values.
%This means that looking only at $a_\ell$ cannot give a criterion for
%surjectivity of $\rhobar_{E,p}$. On the other hand for a point
We can also determine surjectivity of the mod~$2$ and mod~$3$
representations directly using the $3$-division polynomial of~$E$.
For $p\leq 3$ one can show that $\rhobar_{E,p}$ is surjective if and
only if the $p$-division polynomial (of degree~$n$) has Galois group
$S_n$.
% Proof:
% # Algorithm: Let f be the 3-division polynomial, which is
% # a polynomial of degree 4. Then I claim that this
% # polynomial has Galois group S_4 if and only if the
% # representation rhobar_{E,3} is surjective. If the group
% # is S_4, then S_4 is a quotient of the image of
% # rhobar_{E,3}. Since S_4 has order 24 and GL_2(F_3)
% # has order 48, the only possibility we have to consider
% # is that the image of rhobar is isomorphic to S_4.
% # But this is not the case because S_4 is not a subgroup
% # of GL_2(F_3). If it were, it would be normal, since
% # it would have index 2. But there is a *unique* normal
% # subgroup of GL_2(F_3) of index 2, namely SL_2(F_3),
% # and SL_2(F_3) is not isomorphic to S_4 (S_4 has a normal
% # subgroup of index 2 and SL_2(F_3) does not.)
% # (What's a simple way to see that SL_2(F_3) is the
% # unique index-2 normal subgroup? I didn't see an obvious
% # reason, so just used the NormalSubgroups command in MAGMA
% # and it output exactly one of index 2.)
% # Here's Noam Elkies proof for the other direction:
% #> Let E be an elliptic curve over Q. Is the mod-3
% #> representation E[3] surjective if and only if the
% #> (degree 4) division polynomial has Galois group S_4? I
% #> can see why the group being S_4 implies the
% #> representation is surjective, but the converse is not
% #> clear to me.
% # I would have thought that this is the easier part: to
% # say that E[3] is surjective is to say the 3-torsion
% # field Q(E[3]) has Galois group GL_2(Z/3) over Q. Let
% # E[3]+ be the subfield fixed by the element -1 of
% # GL_2(Z/3). Then E[3] has Galois group PGL_2(Z/3), which
% # is identified with S_4 by its action on the four
% # 3-element subgroups of E[3]. Each such subgroup is in
% # turn determined by the x-coordinate shared by its two
% # nonzero points. So, if E[3] is surjective then any
% # permutation of those x-coordinates is realized by some
% # element of Gal(E[3]+/Q). Thus the Galois group of the
% # division polynomial (whose roots are those
% # x-coordinates) maps surjectively to S_4, which means it
% # equals S_4.
% For a point $P=(x,y)$, $3P=0$ implies $2P=-P$ and that happens if
% and only if $x(2P)=x(P)$, so you can write the explicit polynomial
% that $x(P)$ must satisfy if $P$ is a $3$-torsion point and try to
% check explicitely whether it is irreducible
\end{remark}
\begin{theorem}[Serre]\label{thm:serre_surj}
If $p\geq 5$ is a prime of good reduction, then $\rho_{E,p}$ is
surjective if and only if $\rhobar_{E,p}$ is surjective.
\end{theorem}
\begin{proof}
% This follows from \cite[\S3.4, Lem.~3, pg.~IV-23]{serre:ladic}.
This is proved in greater generality
as \cite[Thm.~$4'$, pg.~300]{serre:propgal}.
\end{proof}
\begin{remark}
This result does not extend to $p=3$ (see \cite[Ex.~3,~pg
IV-28]{serre:ladic}). In fact, there are infinitely many elliptic
curves with $\rhobar_{E,p}$ surjective, but $\rho_{E,p}$ not
surjective (see forthcoming work of Noam Elkies).
\end{remark}
\subsection{Special Values of $L$-Functions}\label{sec:Lprec}
Let $E$ be an elliptic curve over~$\Q$ of conductor~$N$, and
let $f = \sum a_n q^n$ be the corresponding cusp form.
The following lemma will be useful in determining how many terms of
the $L$-series of $E$ are needed to compute the $L$-series to a given
precision. (We could give a strong bound, but for our application
this will be enough, and is simplest to apply in practice.)
\begin{lemma}\label{lem:fourier}
For any positive integer~$n$, we have $|a_n|\leq n$.
\end{lemma}
\begin{proof}
For $p$ prime we know that $a_p=\alpha+\beta$, where $\alpha$ and
$\beta$ are the roots of $x^2-a_px+p=0$. Note that
$|\alpha|=|\beta|=\sqrt{p}$.
Since $a_n$ is multiplicative, it is enough to show $|a_n|\leq n$ for
prime powers $p^r$.
Let $r>1$.
Then $a_{p^r}=a_pa_{p^{r-1}}-pa_{p^{r-2}}$, and by induction,
$$a_{p^r}=\frac{\alpha^{r+1}-\beta^{r+1}}{\alpha-\beta}.$$
Then \[|a_{p^r}|\leq
\frac{2p^{(r+1)/2}}{|\alpha-\beta|}=
\frac{2p^{(r+1)/2}}{\left|\sqrt{4p-a_p^2}\right|}.\]
Note that the sign is changed since we only deal with absolute
values.
We need to show that this is $\leq p^r$. This happens if
\[\frac{2}{\sqrt{4p-a_p^2}}\leq p^{(r-1)/2}.\]
Since $a_p^2<4p$ the difference is at least 1 so it is enough to
show that $2\leq p^{(r-1)/2}$. This is true as long as $p>3$. For
$p=2$ and $p=3$ note that $a_p$ is an integer with $|a_p|<2\sqrt{p}$. For
$p=2$ this integer is at most 2 and so $4p-a_p^2\geq 4$. Similarly
for $p=3$ this is at most 3 and so $4p-a_p^2\geq 4$. Therefore it is
enough to show that $1\leq p^{(r-1)/2}$, which is true for all $r>1$.
\end{proof}
Suppose $E$ has even analytic rank.
By \cite[\S2.13]{cremona:algs} or \cite[Prop.~7.5.8]{cohen:course_ant}, we have
\begin{equation}\label{eqn:lseries_sum}
L(E,1)=2\sum_{n=1}^{\infty}\frac{a_n}{n}e^{-2\pi n/\sqrt{N}},
\end{equation}
where $a_n$ are the Fourier coefficients of the normalized eigenform
associated with~$E$. Using the bound $|a_n|\leq n$ of
Lemma~\ref{lem:fourier}, we see that if we truncate the series
(\ref{eqn:lseries_sum}) at the $k$th term, the error is at most
$$
\eps = 2\sum_{n=k}^{\infty} e^{-2\pi n/\sqrt{N}} =\frac{2e^{-2\pi
k/\sqrt{N}}}{1-e^{-2\pi/\sqrt{N}}}, $$ and the quantity on the right
can easily be evaluated.
Next suppose $E$ has odd analytic rank.
In \cite[\S2.13]{cremona:algs} or \cite[Prop.~7.5.9]{cohen:course_ant} we find that
$$
L'(E,1)=2\sum_{n=1}^{\infty}
\frac{a_n}{n}G_1(2\pi n/\sqrt{N}).
$$
We have
$$
G_1(x)=\int_1^\infty e^{-xy}\frac{dy}{y} = \int_x^\infty e^{-y} \frac{dy}{y}
\le e^{-x},
$$
and we obtain the same error bound as
for $L(E,1)$. (In fact, $G_1(x) \leq e^{-x}/x$ but we will not need this
stronger bound.)
\subsection{Mordell-Weil Groups}\label{sec:mwgroups}
If~$E$ is an elliptic curve over~$\Q$ of analytic rank~$\leq 1$, there
are algorithms to compute $E(\Q)$ that are guaranteed to succeed.
This is because $\#\Sha(E)$ is finite, by
\cite{kolyvagin:mordellweil}. Independent implementations of these
algorithms are available as part of {\tt mwrank} \cite{mwrank} and
MAGMA \cite{magma}. We did most of our computations of $E(\Q)$ using
{\tt mwrank}, but use MAGMA in a few cases, since it implements
$3$-descents, $4$-descents and Heegner points methods
(thanks to work of Tom Womack, Mark Watkins, and others).
%\edit{Double
% checking everything with MAGMA would sound very good here, no?}
\subsection{Other Algorithms}
We use many other elliptic curves algorithms, for example, for
computing root numbers and the coefficients $a_n$ of the modular form
associated to~$E$. For the most part, we used the PARI (see
\cite{pari}) C-library via SAGE (see \cite{sage}). For descriptions
of these general elliptic curves algorithms, see \cite{cohen:course_ant,
cremona:algs}.
% An elliptic curve over~$\Q$ has complex multiplication if and only
% if its $j$-invariant is in the following set:
% $$
% \{0, 54000, -12288000, 1728, 287496, -3375, 16581375, 8000,
% -32768, $$
% $$\quad -884736, -884736000,
% -147197952000, -262537412640768000\}.
% $$
\section{The Kolyvagin Bound}
In this section we describe a bound due to Kolyvagin on $\#\Sha(E)$,
and compute it for many specific elliptic curves over~$\Q$. In fact,
the bound is on $\#\Sha(E/K)$, where $K$ is a quadratic imaginary
field; this is not a problem, because the natural map $\Sha(E/\Q)\to
\Sha(E/K)$ has kernel of order a power of~$2$, so the bound is also a
bound on the odd part of $\#\Sha(E)$.
Let $E$ be an elliptic curve over~$\Q$ of conductor~$N$.
For any quadratic imaginary field $K=\Q(\sqrt{-D})$, let $E^D$ denote the
twist of~$E$ by~$D$. If $E$ is defined by $y^2=x^3+ax+b$, then
$E^D$ is defined by $y^2=x^3+D^2ax+D^3b$, and
$$L(E/K,s)=L(E,s)\cdot L(E^D,s).$$
\begin{definition}[Heegner Hypothesis]\label{defn:hh}
We say that $K$ satisfies the {\em Heegner hypothesis} for~$E$ if
$K\neq \Q(\sqrt{-1}), \Q(\sqrt{-3})$, and every prime factor of~$N$
splits as a product of two distinct primes in the ring of integers
of~$K$. (The condition $K\neq \Q(\sqrt{-1}), \Q(\sqrt{-3})$ is not
necessary for some of the results below, but we include it for
simplicity.)
\end{definition}
If~$K$ satisfies the Heegner hypothesis for~$E$, then there is a
Heegner point $y_K \in E(K)$, which is the sum of images of certain
complex multiplication (CM) points on $X_0(N)$ (see
\cite[\S{}I.3]{gross-zagier}). Properties of this point impact the
arithmetic of~$E$ over~$K$.
\subsection{Bounds on $\#\Sha(E/K)$}\label{sec:boundshakoly}
Suppose that~$K$ is an imaginary quadratic extension of~$\Q$ that
satisfies the Heegner hypothesis for~$E$. Kolyvagin proved the
following theorem in \cite{kolyvagin:euler_systems}:
\begin{theorem}[Kolyvagin]\label{thm:kolysurj}
Let $R=\End(E/\C)$ and let $F=\Frac(R)$, so if $E$ is non-CM then $F=\Q$.
If~$p$ is an odd prime unramified in $F$ such that $\Gal(F(E[p])/F)
= \Aut_R(E[p])$, i.e., $\Im(\rhobar_{E,p})$ is as large as possible,
then
$$
\ord_p (\#\Sha(E/K)) \leq 2 \cdot \ord_p ([E(K): \Z y_K]).
$$
\end{theorem}
Note that if~$E$ does not have complex multiplication, the hypotheses of
both these theorems imply that $p\nmid \#E(K)_{\tor}$
(see Lemma~\ref{lemma:stein}).
%\begin{example}
%Give an example here. \edit{todo}
%\end{example}
Cha \cite{cha:shabound, cha:vanishing} extended Kolyvagin's method to
provide better bounds on $\Sha(E/K)$ in some cases. Let~$K$ be a
number field, let $D_K$ be the discriminant of~$K$, and let~$N$ be the
conductor of~$E$.
\begin{theorem}[Cha]\label{thm:cha}
If $p\nmid D_K$, $p^2\nmid N$, and
$\rhobar_{E,p}$ is irreducible, then
$$\ord_p (\#\Sha(E/K)) \leq 2 \cdot \ord_p([E(K):\Z y_K]).$$
\end{theorem}
As we will see in the proof of Theorem~\ref{thm:bsdkato} below, there
is one curve that satisfies the hypotheses of that theorem, but for
which we cannot use Theorem~\ref{thm:kolysurj} to prove $\BSD(E,5)$.
The problem is that $\rhobar_{E,5}$ is not surjective. We can use
Cha's theorem though:
\begin{lemma}\label{lem:608B}
Let $E$ be the elliptic curve {\rm 608B}, which has rank~$0$. Then
$\BSD(E,5)$ is true for~$E$.
\end{lemma}
\begin{proof}
Since $E$ admits no $5$-isogeny (see \cite{cremona:algs}),
$\rhobar_{E,5}$ is irreducible. Also, $5^2\nmid 608$, so for any
Heegner $K$ of discriminant coprime to~$5$ we can apply
Theorem~\ref{thm:cha}. Taking $K=\Q(\sqrt{-79})$, we find that the
odd part of $[E(K):\Z y_K]$ is~$1$, so $5\nmid\#\Sha(E/K)$. It
follows that $5\nmid \#\Sha(E)$, so $\BSD(E,5)$ is true, according
to Theorem~\ref{thm:cremona1000}.
\end{proof}
Cha's assumption on the reduction of~$E$ at~$p$ and that $p\nmid D_K$
is problematic when there is a prime $p\geq 5$ of additive reduction
or one uses only one $K$. This situation does occur in several cases,
which motivated us to prove the following theorem:
\begin{theorem}\label{thm:us}
Suppose $E$ is a non-CM elliptic curve over~$\Q$.
Suppose $K$ is a quadratic imaginary field that satisfies the
Heegner hypothesis and~$p$ is an odd prime such that $p\nmid
\#E'(K)_{\tor}$ for any curve $E'$ that is $\Q$-isogenous to~$E$.
Then
$$
\ord_p (\#\Sha(E)) \leq 2 \ord_p ([E(K): \Z y_K]),
$$
unless $\disc(K)$ is divisible by exactly one prime~$\ell$, in
which case the conclusion is only valid if $p\neq \ell$.
\end{theorem}
Since the proof of Theorem~\ref{thm:us} is somewhat long and
technical, we defer the proof until Section~\ref{sec:relaxkoly}.
\begin{remark}
If in Theorem~\ref{thm:us}, $\rhobar_{E,p}$ is irreducible, then
$p\nmid \#E'(K)_{\tor}$ for all $E'$ isogenous to~$E$. This is
because the isogeny $E\to E'$ has degree coprime to~$p$, so
$E[p]\isom E'[p]$. Also, since $E[p]$ is irreducible, if $E'(K)$
were to contain a $p$-torsion point, it would have to contain all of
them, a contradiction since $\bmu_p \not\subset K$ (recall
that we exclude $\Q(\sqrt{-3})$ and $\Q(\sqrt{-4})$).
\end{remark}
% Insert example that satisfies us but not Cha or Kolyvagin.\edit{todo}
% \end{example}
% \edit{Perhaps we need to discuss relation between Cha's condition
% and our condition, when $p^2\nmid N$. E.g., Irreducible
% implies the $p\nmid \#E'(K)_{\tor}$ condition, by
% Lemma later. The converse not so clear, but maybe true.
% In any case, the best part about our condition is that we
% don't require that $p^2\nmid N$. This is very important,
% since typically non-surj non-irred representations have $p^2\mid N$.
% }
%In
%some cases, $\Sha(E)$ at the remaining primes can then be
%analyzed using $p$-descent.
%\subsubsection{Existence of Suitable Quadratic Imaginary Extensions}~\label{Kexists}
\begin{theorem}[Bump-Friedberg-Hoffstein, Murty-Murty, Waldspurger]\label{thm:bfh}
There are infinitely many quadratic imaginary extensions $K/\Q$ such
that~$K$ satisfies the Heegner hypothesis and $\ord_{s=1}L(E/K)=1$.
\end{theorem}
\begin{proof}
If $\hbox{ord}_{s=1}L(E,s) =0$, then the papers
\cite{murty-murty:mean} and
\cite{bump-friedberg-hoffstein:nonvanishing} both imply the existence
of infinitely many~$K$ such that $y_K$ has infinite order. If
$\hbox{ord}_{s=1}L(E,s) =1$, then a result of Waldspurger
(\cite{waldspurger:valeurs}) applies, as does
\cite{bump-friedberg-hoffstein:nonvanishing}.
\end{proof}
Theorem~\ref{thm:bfh} is not used in our computations, but ensures
that our procedure for bounding $\#\Sha(E)$, when $E$ has analytic
rank $\leq 1$, is an algorithm, i.e., it always terminates with a
nontrivial upper bound.
%See also \cite{darmon:ratpoints} for an overview of how these results
%and the work of Gross-Zagier and Kolyvagin fit together to settle the
%rank part of the BSD conjecture for elliptic curves of analytic rank~$0$ or~$1$.
% In our computations, however, we do not merely take any $K$ satisfying
% the Heegner hypothesis and the analytic rank hypothesis. We instead
% choose $K$ to be linearly disjoint from the fields $\Q(E[p])$ for all
% primes $p$, and a simple way to ensure this is to require the
% discriminant of $K$ (i.e., $D$) to be divisible by at least $2$ distinct
% odd primes. A conjecture of Goldfeld says that the density of
% discriminants $D$ such that $K= \Q(\sqrt{D})$ satisfies our hypotheses
% is roughly $\frac{1}{2}$ (see \cite{goldfeld:conjectures_quadratic}).
% Thus far the best proven result is due to Ono and Skinner, who showed
% (\cite{ono-skinner:nonvanish}) that, in the case
% $\ord_{s=1}L(E,s)=0$, the number of such discriminants has density at
% least on the order of magnitude of $\frac{1}{\log X}$. Unfortunately,
% this is precisely the density of the prime numbers, so a density
% argument will not help us here (Ono and Skinner's result also does not
% distinguish between discriminants that give rise to the same imaginary
% quadratic extension). Note that Ono later improved this theorem (by a
% small power of $\log X$) under the assumption that $E(\Q)[2]=0$ (see
% \cite{ono:nonvanish_twist}).
% In the worst case, the only $K$ that exist and satisfy both of our
% hypotheses have only a single odd prime divisor $p$ of their
% discriminants. But then for $\ell \neq p$, $K$ is linearly disjoint
% from $\Q(E[\ell])$, so we can run our algorithm as before, only adding
% $p$ to the list of ``bad primes'' on which we have to perform
% descents. We have used the fact that the only ramified primes in
% $\Q(E[\ell])/\Q$ are, by the criterion of N\'eron-Ogg-Shafarevich (see
% \cite{silverman:aec}), primes dividing the conductor $N$ of $E$ and
% $\ell$ itself.
% This bad prime $p$ might be large, however, making the $p$-descent
% cumbersome. In that case, it would be better in practice to produce a
% second field $K'$ satisfying our hypotheses (recall that infinitely
% many exist). Unless $K'=K$, $K'$ is linearly disjoint from $\Q(E[p])$
% since $p$ will not ramify in $K'$. There is a good chance that we
% might be able to show $\Sha(E/K)[p]=0$, but it is possible that a
% curve in $E$'s isogeny class has $p$-torsion or that $y_K$ is a
% multiple of $p$ in $E(K)$. There are universal bounds on the possible
% $p$-torsion for quadratic fields, so the problematic primes resulting
% from torsion will still be 'small,' but the ever-mysterious index of
% the Heegner point may keep us from getting information at large primes
% (in particular, the contributions from nontrivial elements of the
% Shafarevich-Tate group, or large Tamagawa numbers). In practice, the
% bad primes are usually small, but see Remark~\ref{rmk:bigind}.
\subsection{The Gross-Zagier Formula}\label{sec:gz}
We use the Gross-Zagier formula to compute the index
$[E(K):\Z y_K]$ without explicitly computing $y_K$.
The modularity theorem of \cite{breuil-conrad-diamond-taylor} asserts
that there exists a surjective morphism $\pi:X_0(N)\to E$.
Choose~$\pi$ to have minimal degree among all such morphisms.
Let $\pi^*(\omega)$ be the pullback of a minimal invariant
differential~$\omega$ on~$E$.
Then $\pi^*(\omega) = \alpha\cdot f$, for some constant~$\alpha$ and some
normalized cusp form~$f$. By
\cite[Prop.~2]{edixhoven:manin}, we know that $\alpha\in\Z$.
\begin{definition}[Manin Constant]
The {\em Manin constant} of $E$ is $c=|\alpha|$.
\end{definition}
Manin conjectured in \cite[\S5]{manin:parabolic} that $c=1$ for the
optimal curve in the $\Q$-isogeny class of~$E$.
\begin{theorem}[Gross-Zagier]\label{thm:gz}
If $K$ satisfies the Heegner hypothesis for~$E$, then
the N\'eron-Tate canonical height of $y_K$ is
$$
h(y_K)=\frac{\sqrt{D}}{c^2\cdot{}\int_{E(\C)}\omega\wedge
\overline{i\omega}}\cdot L'(E/K,1).
$$
\end{theorem}
\begin{proof}
Gross and Zagier proved the following formula in \cite{gross-zagier}
under the hypothesis that~$D$ is odd.
For the general assertion see \cite[Thm.~6.1]{zhang:gzgl2}.
\end{proof}
\subsection{Remarks on the Index}
Suppose that $E$ is an elliptic curve over~$\Q$ of conductor~$N$
and that $E$ has analytic rank~$1$ over a quadratic imaginary
field~$K$ that satisfies the Heegner hypothesis. In
\cite{mccallum:kolyvagin}, McCallum rephrases the analogue of
Conjecture~\ref{conj:bsd} for~$E$ over~$K$ using the Gross-Zagier
formula as follows:
\begin{conjecture}[Birch and Swinnerton-Dyer]\label{conj:bsd-heegner}
Suppose $K$ is a quadratic imaginary field that satisfies
the Heegner hypothesis, and that $E$ has analytic rank~$1$ over $K$.
Then
$$
\#\Sha(E/K) = \left(\frac{[E(K):\Z y_K]} {c^2\cdot \prod_{p\mid
N}c_p}\right)^2.
$$
Here the $c_p$ are the Tamagawa numbers of $E$ over~$\Q$, $c$ is the
Manin constant of~$E$, and $\Z y_K$ is the cyclic group generated by $y_K$.
\end{conjecture}
\begin{remark}\label{rmk:bigind}
A serious issue is that Conjecture~\ref{conj:bsd-heegner} implies
that the index $I_K= [E(K):\Z y_K]$ will be divisible by the
Tamagawa numbers $c_p$. One sees using Tate curves that these
Tamagawa numbers can be arbitrarily large. In many cases when $E$
has analytic rank $0$, we could instead apply Theorem~\ref{thm:kato}
below, but when $E$ has analytic rank~$1$ a new approach is
required, e.g., computation of $p$-adic regulators and use of
results of P.~Schneider and others toward $p$-adic analogues of the
BSD conjecture. This will be the subject of a future paper.
\end{remark}
\begin{remark}\label{rmk:91b}
Conjecture~\ref{conj:bsd-heegner} has interesting implications in
certain special cases. For example, if $E$ is the curve 91B, then
$c_{7}=c_{13}=1$. Also $c=1$, as
Cremona has verified, and $\#E(\Q)_{\tor} = 3$. Thus for any~$K$,
we have $3 \mid [E(K) : \Z y_K]$. If $y_K$ has infinite order, then
Conjecture~\ref{conj:bsd-heegner} implies that $3^2\mid
\#\Sha(E/K)$. For $K=\Q(\sqrt{-103})$, the point $y_K$ is torsion,
and in this case $E(K)$ has rank~$3$ and (conjecturally)
$\Sha(E/K)[3]=0$. See Remark~\ref{rmk:heeg_r1_3} for another
example along these lines.
\end{remark}
\subsection{Mordell-Weil Groups and Quadratic Imaginary Fields}
Let $E$ be an elliptic curve over~$\Q$ and $K=\Q(\sqrt{D})$ a quadratic
imaginary field such that $E(K)$ has rank~$1$.
In this section we explain how to understand $E(K)$ in terms
of $E(\Q)$ and $E^{D}(\Q)$.
\begin{proposition}\label{prop:tens2}
Let $R=\Z[1/2]$ and $K=\Q(\sqrt{D})$.
For any squarefree integer~$D\neq 1$, we have
$$
E(K)\otimes R = (E(\Q)\otimes R)\oplus (E^D(\Q)\otimes R).
$$
\end{proposition}
\begin{proof}
Let~$\tau$ be the complex conjugation automorphism
on $E(K)\otimes R$. The characteristic polynomial of~$\tau$
is $x^2-1$, which is squarefree, so $E(K)\otimes R$ is a
direct sum of its $+1$ and $-1$ eigenspaces for~$\tau$.
The natural map $E(\Q)\hookrightarrow E(K)$
identifies $E(\Q)\otimes R$ with the $+1$ eigenspace for~$\tau$
since $E(K)^{G_\Q} = E(\Q)$;
likewise, $E^{D}(\Q)\hookrightarrow E(K)$
identifies $E^{D}(\Q)\otimes R$ with the $-1$ eigenspace
for~$\tau$.
\end{proof}
The following slightly more refined proposition will be important
for certain explicit Heegner point computations (directly after
Equation~\ref{eqn:index_height}).
\begin{proposition}\label{prop:mw_r1}
Suppose $E(K)$ has rank $1$.
Then the image of either $E(\Q)_{/\tor}$ or
$E^{D}(\Q)_{/\tor}$ has index at most~$2$
in $E(K)_{/\tor}$.
\end{proposition}
\begin{proof}
Since $E(K)$ has rank $1$, Proposition~\ref{prop:tens2} implies that
exactly one of $E(\Q)$ and $E^{D}(\Q)$ has rank $1$ and the other
has rank $0$. We may assume that $E(\Q)$ has rank~$1$ (otherwise,
swap $E$ and $E^{D}$). Let $i$ be the natural inclusion
$E(\Q)\hookrightarrow E(K)$, and let $\tau$ denote the automorphism
of $E(K)$ induced by complex conjugation. Then $P\mapsto (1+\tau)P$
induces a map~$E(K)\to E(K)^+=E(\Q)$ that, upon taking quotients by
torsion, induces a map $\psi: E(K)_{/\tor} \to E(\Q)_{/\tor}$. Let
$P_1$ be a generator for $E(\Q)_{/\tor}$ and $P_2$ a generator for
$E(K)_{/\tor}$, and write $i(P_1) = n P_2$, for some nonzero
integer~$n$. Then
$$
[2]P_1 = \psi(i(P_1)) = \psi(n P_2) = [n]\psi(P_2) = [nm] P_1
\pmod{E(\Q)_{\tor}},
$$
for some nonzero integer~$m$. Thus $2=nm$, so $n\leq 2$.
\end{proof}
If~$D$ satisfies the Heegner hypothesis, then
by computing the residue symbol $\kr{N}{D}$ and understanding
how the sign of the functional equation changes under twist,
we see that
$$
\ord_{s=1} L(E,s) \not \con \ord_{s=1} L(E^{(D)},s) \pmod{2}.
$$
Suppose $K$ satisfies the Heegner hypothesis and
$\ord_{s=1}L(E/K,s) = 1$. Then work of Kolyvagin (see \cite{kolyvagin:mordellweil, kolyvagin:subclass}) implies that $E(K)$ has rank~$1$.
The root number $\eps_E=\pm 1$ of~$E$ is the sign of the functional
equation of $L(E,s)$. If $\eps_E=+1$, then the analytic rank
$\ord_{s=1} L(E,s)$ is even, and if $\eps_E=-1$, then it is odd.
\begin{proposition}\label{prop:compL}
Let $E$ be an elliptic curve, let $D=D_K$ be a
discriminant that satisfies the Heegner hypothesis such that
$\ord_{s=1}L(E/K,s)=1$, and let $R=\Z[1/2]$. Then
\begin{enumerate}
\item If $\eps_E=+1$, then a generator of $E(K)\otimes R$
is the image of a
generator of $E^D(\Q)\otimes R$ and
$L'(E/K,1)=L(E,1)\cdot L'(E^D,1).$
\item If $\eps_E=-1$, then a generator of $E(K)\otimes{}R$
is the image of a
generator of $E(\Q)\otimes R$ and
$L'(E/K,1)=L'(E,1)\cdot L(E^D,1).$
\end{enumerate}
\end{proposition}
We will use the above proposition to relate computation of
$E(K)\otimes R$ to computation of Mordell-Weil groups of
elliptic curves defined over~$\Q$.
\subsection{Computing the Index of the Heegner Point}\label{sec:compind}
A key input to the theorems of Section~\ref{sec:boundshakoly} is computation of
the index $[E(K):\Z y_K]$.
We have
\begin{equation}\label{eqn:index_height}
[E(K)_{/{\tor}}:\Z y_K]^2=h(y_K)/h(z),
\end{equation}
where~$z$ is a generator of $E(K)_{/\tor}$.
In the Gross-Zagier formula we have $h=h_K$, the N\'eron-Tate
canonical height on $E(K)=E^D(K)$ relative to $K$. Let $h_\Q$ denote
the height on $E(\Q)$ or $E^D(\Q)$. Note that if $P\in E(\Q)$ or
$E^D(\Q)$, then
\begin{equation}\label{eqn:ht2}
h_\Q(P) = \frac{1}{[K:\Q]}\cdot h_K(P) = \frac{h_K(P)}{2}.
\end{equation}
Using Proposition~\ref{prop:mw_r1} and algorithms for computing
Mordell-Weil groups (see Section~\ref{sec:mwgroups}), we can
compute~$z$ or $2z$, so we can compute $h(z)$ or $2h(z)$. In
practice, even for curves of conductor up to $1000$, it can take a
huge amount of time to compute~$z$. This section about practical
methods to either compute the index or at least bound it.
It is not difficult to compute $h(y_K)$, without computing $y_K$
itself, using the Gross-Zagier formula (Section~\ref{sec:gz}). We
compute $L'(E/K,1)$ by computing $L$-functions of elliptic curves
defined over~$\Q$ as explained in Proposition~\ref{prop:compL}. It
remains to compute
\begin{equation}\label{eqn:alpha}
\alpha =
\frac{\sqrt{|D|}}{c^2\int_{E(\C)}\omega\wedge
\overline{i\omega}}.
\end{equation}
\subsubsection{The Manin Constant}
Manin conjectured that the Manin constant~$c$ for any optimal elliptic
curve factor~$E$ of $X_0(N)$ is~$1$, and there are bounds on the
possibilities for~$c$ (see, e.g., \cite{edixhoven:manin,
agashe-ribet-stein:manin}). There is an algorithm to verify in any
particular case that $c=1$, as explained in the proof of the following
proposition.
\begin{proposition}[Cremona]
If $E$ is an optimal elliptic curve of conductor at most $80000$,
then the Manin constant of $E$ is $1$.
\end{proposition}
\begin{proof}
For each level $N\leq 80000$ we do the following. Using the modular
symbols algorithms of \cite{cremona:algs}, we enumerate the rational
newforms $f_1,\ldots, f_d$, which correspond (via the modularity
theorem) to the optimal elliptic curves $E_1, \ldots, E_d$ of
conductor~$N$, respectively. For each $f_i$ we compute
approximations to {\bf xxx} decimal digits of the $c_4$ and $c_6$
invariants of the lattice $\Lambda_{E_i}$ attached to the optimal
curve in the isogeny class. We then guess integers $c_4'$ and
$c_6'$ that are close to the computed approximations, and verify
that the elliptic curve $E_i'$ with invariants $c_4'$, $c_6'$ is an
elliptic curve of conductor~$N$. We also compute the full isogeny
class of $E_i'$ using the program {\tt allisog}. Repeating this
procedure for each newform~$f$, we obtain $d$ distinct isogeny
classes of elliptic curves of conductor~$N$, and by modularity these
must be in bijection with the newforms $f_i$. However, at this
point we have not {\em proved} that $E_i=E_i'$ or even that $E_i'$
is an optimal quotient. However, we have provably found all
elliptic curves over~$\Q$ of conductor~$N$.
We next compute the $c_4$ and $c_6$ invariants of all curves of
conductor~$N$, and observe that the first 12 digits of the
$c$-invariants for these curves are sufficient to distinguish them.
(12 digits is enough for every curve up to conductor $80000$.) If
we had guessed $c_4'$ and $c_6'$ incorrectly above, so that
$E_i'\neq E_i$, there would be two curves of conductor~$N$ both of
whose $c$-invariants have the same initial {\bf xxx} decimal digits,
which is impossible since $12$ digits of precision are sufficient to
distinguish any two. Thus $E_i'=E_i$, and the $c_4'$, $c_6'$ we
computed are the correct invariants of the optimal quotient attached
to $f_i$.
Finally, we observe that $c_4'$ and $c_6'$ are the invariants of a
minimal Weierstrass equation, which implies that the Manin constant
of~$E_i$ is~$1$.
\end{proof}
\subsubsection{The Integral}
We have the following lemma
regarding the integral in (\ref{eqn:alpha}):
\begin{lemma}
We have
$ \int_{E(\C)}\omega\wedge \overline{i\omega} = 2\cdot \Vol(\C/\Lambda),$
where the volume $\Vol(\C/\Lambda)$ is the absolute value of the
determinant of a matrix formed from a basis for the lattice
in $\C$ obtained by integrating the N\'eron differential $\omega_E$
against all homology classes in $\H_1(E,\Z)$.
\end{lemma}
\begin{proof}
Fix the Weierstrass equation
$
y^2=4x^3+g_4 x + g_6
$
for $E$, so $x=\wp(z)$ and $y=\wp'(z)$.
First note that
$$
\omega = \frac{dx}{y} = \frac{d\wp(z)}{\wp'(z)}
= \frac{\wp'(z) dz}{\wp'(z)} = dz.
$$
Thus
\begin{align*}\label{page:intc}
\int_{E(\C)}\omega\wedge \overline{i\omega}
&=\int_{\C/\Lambda}dz\wedge \overline{i dz}\\
&=-i\int_{\C/\Lambda} (dx+idy)\wedge (dx-idy)\\
&=-i(2i)\int_{\C/\Lambda} dx\wedge dy=2\cdot \Vol(\C/\Lambda).
\end{align*}
\end{proof}
Note that
$\Vol(\C/\Lambda)$
can be computed to high precision using the Gauss arithmetic-geometric
mean, as described in \cite[\S3.7]{cremona:algs}.
%In the course of our computations of $[E(K):\Z y_K]$ below, we
%compute the heights and $L$-values with sufficient precision to obtain
%the correct square of $[E(K):\Z y_K]$ after rounding.
\subsubsection{Mordell-Weil Groups and Heights}
For the curves that we run our computation on, we use \cite{mwrank}
(via \cite{sage}), which computes a basis for $E^D(\Q)$, and not just
a basis for a subgroup of finite index.
Cremona describes the computation of heights of points on curves defined
over~$\Q$ in detail in \cite[\S3.4]{cremona:algs}. There is an
explicit bound on the error in the height computation, which shrinks
exponentially in terms of the precision of approximating series, and
can be made arbitrarily small. For the $L$-function computations, see
Section~\ref{sec:Lprec}.
\subsubsection{Indexes of Heegner Points on Rank $1$ Curves}
Suppose $E$ is an elliptic curve over~$\Q$ of analytic rank~$1$, and
we wish to compute indexes $i_K = [E(K)_{/\tor}: \Z y_K]$ for
various~$K$. Assume that $E(\Q)$ is known, so we can compute $h(z)$
to high precision, where $z$ generates $E(\Q)/_{\tor}$. Then
computing the indexes $i_K$ is relatively easy. For each~$K$,
compute $h(y_K)$ as described above using the Gross-Zagier formula, so
$$
h(y_K) = \alpha \cdot L'(E,1) \cdot L(E^D,1).
$$
Then
$$
i_K = \sqrt{\frac{h(y_K)}{h(z)}} = \sqrt{\frac{h(y_K)}{2 h_\Q(z)}}.
$$
We emphasize that computation of the Heegner
point itself is not necessary. For the results of this index
computation
for $E$ of conductor $\leq 1000$, see Section~\ref{sec:comp_r1}.
\begin{example}
Let~$E$ be the elliptic curve 540B, which has rank~$1$, and
conductor $540=2^2 \cdot 3^3 \cdot 5$. The first~$K$ that satisfies
the Heegner hypothesis is $\Q(\sqrt{-71})$. The group $E(\Q)$ is
generated by $z=(0,1)$, and we have $h_\Q(z)\sim 0.656622630$. We
have
$$
\alpha \sim \frac{\sqrt{71}}{2 \cdot 3.832955} \sim 1.09917,
$$
so
$$
h(y_K) \sim 1.09917 \cdot 1.9340458 \cdot 5.559761726 \sim 11.819.
$$
Thus
$$
i_K = \sqrt{\frac{11.819}{2 \cdot 0.656622630}} \sim \sqrt{8.99999} \sim 3.
$$
\end{example}
\subsubsection{Indexes of Heegner Points on Rank $0$ Curves}\label{sec:methodr0}
Assume that the analytic rank of~$E$
is~$0$. In practice, computing the indexes of Heegner points in this
case is substantially more difficult than the rank~$1$
case.
For a Heegner quadratic imaginary field~$K=\Q(\sqrt{D})$, we have
$$
i_K = [E(K)_{/\tor}:\Z y_K]^2 = \frac{h(y_K)}{h(z)} =
\alpha\cdot \frac{L(E,1)\cdot L'(E^D,1)}{h(z)},
$$
so one method to find $i_K$ is to find a generator $z\in E^D(\Q)$
exactly using descent algorithms, which will terminate since we know
that $\Sha(E^D)$ is finite, by Kolyvagin's theorem. However, since
$E^D$ has potentially large conductor and rank~$1$, in practice the
Mordell-Weil group will sometimes be generated by a point of large
height, hence be extremely time consuming to find. One can use
$2$-descent, $3$-descent, $4$-descent, and Heegner points methods
(i.e., explicitly compute the coordinates of the Heegner point as
decimals and try to recognize them using continued fractions.) In
some cases these methods produce in a reasonable amount of time an
element of $E^D(\Q)$ of infinite order, and one can then saturate the
point using \cite{mwrank} to find a generator~$z$.
\begin{example}
Let $E$ be the curve 11A. The first field that satisfies the
Heegner hypothesis is $K=\Q(\sqrt{-7})$. The conductor of
$F=E^{-7}$ is $539$, and we find a generator $z\in F(\Q)$
for the Mordell-Weil group of this twist. This point has height
$h_\Q(z) \sim 0.1111361471$. We have
$$
\alpha \sim \frac{\sqrt{7}}{2 \cdot 1.8515436234} \sim 0.71447177.
$$
The height over $K$ of the Heegner point is thus
$$
h(y_K) \sim 0.71447177 \cdot 0.25384186 \cdot 1.225566874 \sim 0.2222722925.
$$
Thus by (\ref{eqn:ht2})
$$
i_K = \frac{h(z)}{h(y_K)} = \frac{2h_{\Q}(z)}{h(y_K)} \sim 1.
$$
\end{example}
There is a trick to bound the index $i_K$ without computing {\em any}
elements of $E(K)$. This is useful when the algorithms mentioned
above for computing a generator of $E^D(\Q)$ produce no information in
a reasonable amount of time.
First compute the height $h(y_K)$ using the
Gross-Zagier formula. Next compute the Cremona-Prickett-Siksek
\cite[Ch.~4]{prickett:phd} bound~$B$ for $E^D$, which is a number such
that if $P\in E^D(\Q)$, then the naive logarithmetic height of~$P$ is
off from the canonical height of~$P$ by at most~$B$. Using standard
sieving methods implemented in \cite{mwrank}, we compute all points
on~$E$ of naive logarithmic height up to some number $h_0$. If we
find any point of infinite order, we saturate, and hence compute
$E^D(\Q)$, then use the above methods. If we find no point of
infinite order, we conclude that there is no point in $E^D(\Q)$ of
canonical height $\leq h_0-B$. If $h_0-B>0$, we obtain an upper bound
on~$i_K$ as follows. If~$z$ is a generator for $E^D(\Q)$, then
$h_\Q(z) > h_0 - B$, so
using (\ref{eqn:ht2}) we have
$$
h_{\Q}(z) = \frac{1}{2} \cdot h_K(z) = \frac{h(y_K)}{2\cdot i_K^2} > h_0-B.
$$
Solving for $i_K$ gives
\begin{equation}\label{eqn:indbound}
i_K < \sqrt{\frac{h(y_K)}{2(h_0-B)}},
\end{equation}
so to bound $i_K$ we consider many $K$ (e.g., the first $30$), and
for each compute the quantity on the right side of
(\ref{eqn:indbound}) for a fixed choice of $h_0$.
We then use a $K$ that minimizes this quantity.
\begin{remark}
Another approach to finding some Heegner point, which we discussed
with Noam Elkies, is to search for small points on $E(K)$ over
various fields~$K$, until finding a $K$ that satisfies the Heegner
hypothesis and is such that $E(K)$ has rank~$1$. For example, if
$E$ is given by $y^2=x^3+ax+b$, and $x_0$ is a small integer, write
$y_0^2 \cdot D = x_0^3+ax_0 + b$, where $y_0$ and~$D$ are integers,
and~$D$ is square free. Then $(x_0,y_0)$ is a point on the
quadratic twist of~$E$ by~$D$. We did not use this approach, since
it was not necessary in order to prove Theorem~\ref{thm:main}.
\end{remark}
\begin{example}
Let $E$ be the elliptic curve 546E. Then $K=\Q(\sqrt{-311})$
satisfies the Heegner hypothesis, since the prime divisors of
$546=2\cdot 3\cdot 7\cdot 13$ split completely in~$K$. We
compute the height of the Heegner point~$y_K$. Let~$F$ be the
quadratic twist of~$E$ by $-311$. We have
$$
\alpha \sim \frac{\sqrt{311}}{2 \cdot 0.0340964942689662168001} \sim 258.60711587
$$
Thus
\begin{align*}
h(y_K) &\sim \alpha \cdot L(E,1)\cdot L'(F,1) \\
& \sim 258.60711587 \cdot 2.2783578 \cdot 12.41550 \sim 7315.20688,
\end{align*}
where in each case we compute the $L$-series using enough terms to
obtain a value correct to $\pm 10^{-5}$. Thus $7320$ is a conservative
upper bound on $h(y_K)$. The Cremona-Prickett-Siksek bound for $F$ is $B=13.0825747$.
We search for points on $F$ of naive logarithmic height $\leq 18$, and
find no points. Thus
(\ref{eqn:indbound}) implies that
$$
i_K < \sqrt{7320/(2\cdot (18 - 13.0825747))} \sim 27.28171 < 28.
$$
It follows that if $p\mid i_K$, then $p\leq 23$.
Searching up to height $21$ would (presumably) allow us to remove $23$,
but this might take much longer.
\end{example}
For the results of our computations for all $E$ of conductor $\leq
1000$, see Section~\ref{sec:comp_r0}.
\subsection{Results of Computations}
\subsubsection{Curves of Rank $1$}\label{sec:comp_r1}
First we consider curves of rank~$1$. Recall from
Conjecture~\ref{conj:bsd1000} that we expect $\Sha$ to be trivial for
all optimal rank~$1$ curves of conductor at most $1000$.
\begin{proposition}\label{prop:r1indexex}
Suppose $(E,p)$ is a pair with $E$ an optimal elliptic curve of
conductor up to $1000$ of rank~$1$.
Let~$I$ be the greatest common divisor of $[E(K)_{/\tor}:\Z y_K]$ for
the first four quadratic imaginary fields $K=\Q(\sqrt{D})$ that
satisfy the Heegner hypothesis. If~$p\mid I$, then
$$p\mid 2\cdot \#E(\Q)_{\tor}\cdot \prod_{q\mid N} c_{E,q},$$ except if
$(E,p)$ is $(540B,3)$ or $(756B,3)$.
\end{proposition}
\begin{proof}
For each rank~$1$ curve~$E$ of conductor up to $1000$ we perform
the following computation.
\begin{enumerate}
\item Let $R_E$ be the regulator of~$E$, correct to precision at
least $10^{-10}$, which we look up in the {\tt allbsd} table
of \cite{cremona:onlinetables}.
\item List the first four discriminants $D=D_0,D_1,D_2,D_3$ such that
$K=\Q(\sqrt{D})$ satisfies the Heegner hypothesis. For each $D=D_i$
do the following computation:
\begin{enumerate}
\item Compute $L'(E,1)$ to some bounded precision $\eps$, using
$2\sqrt{N} + 10$ terms. The bound $\eps$ is determined as explained
in Section~\ref{sec:Lprec}.
\item Compute $L(E^{D},1)$ to some bounded precision $\eps'$
using $2\sqrt{N} + 10$ terms.
\item Compute $\alpha = \sqrt{|D|}/(2\Vol(\C/\Lambda))$ to
precision at least $10^{-10}$ using PARI.
\item Using a simple implementation of classical interval arithmetic
(in \cite{sage}) and the bounds above, we compute an interval in which
the real number
$$
\alpha \cdot L'(E,1) \cdot L'(E^D,1) / (\Reg_E/2)
$$
must lie. If there is a unique integer in this
interval, by Theorem~\ref{thm:gz}
this must be the square of the index $[E(K):\Z y_K]^2$.
If there is no unique integer in this interval,
we increase the precision of the computation of $L'$ and $L$ and
repeat the above steps. In all cases in the range of our
computation, we find a unique
integer in the interval; as a double check on our calculations
we verify that the integer is a perfect square.
\end{enumerate}
\end{enumerate}
\end{proof}
\begin{remark}\label{rmk:heeg_r1_3}
For the curves 540B and 756B there is no $3$-torsion, but there is a
rational $3$-isogeny. In each case we verified in addition that $3$
divides the $\GCD$ of the indexes for at least the first $16$ fields
$K$ that satisfy the Heegner hypothesis. Thus as in
Remark~\ref{rmk:91b}, Conjecture~\ref{conj:bsd-heegner} asserts that
$9\mid \#\Sha(E/K)$ for the first sixteen~$K$. This illustrates
that not only Tamagawa numbers but also isogenies can be an
obstruction to applying Kolyvagin's theorem to bound $\#\Sha(E)$,
even if the irreducibility hypothesis on $\rhobar_{E,p}$ is removed.
%(One
% could likely prove nontriviality of $\Sha(E/K)$ in these cases by
% using Theorem~\ref{thm:bsdiso}.)
\end{remark}
\begin{proposition}\label{prop:nonsurjirred}
Suppose $E$ is a non-CM optimal curve of conductor $\leq
1000$ and~$p$ is an odd prime such that $\rhobar_{E,p}$ is
irreducible but not surjective.
If $E$ has rank~$0$ then $(E,p)$ is one of the following:
{\rm (245B,3), (338D,3), (352E,3), (608B,5), (675D,5), (675F,5), (704H,3), (722D,3), (726F,3), (800E,5), (800F,5), (864D,3), (864F,3), (864G,3), (864I,3)}.
If $E$ has rank~$1$, then $(E,p)$ is one of the following:
{\rm
(245A,3), (338E,3), (352F,3), (608E,5), (675B,5), (675I,5), (704L,3), (722B,3), (726A,3), (800B,5), (800I,5), (864A,3), (864B,3), (864J,3), (864L,3).}
There are no curves of rank $\geq 2$ with the above property.
\end{proposition}
\begin{proof}
Using Proposition~\ref{prop:surj} we make a list of pairs $(E,p)$
such that $\rhobar_{E,p}$ might not be surjective, and such that if
$(E,p)$ is not in this list, then $\rhobar_{E,p}$ is surjective.
Then using the program {\tt allisog}, we compute for each curve $E$,
a list of all degrees of isogenies emanating from~$E$, and remove
those pairs $(E,p)$ for which~$p$ divides the degree of one of those
isogenies. The curves listed above are the ones that remain.
\end{proof}
\begin{remark}
In Proposition~\ref{prop:nonsurjirred}, the non-surjective
irreducible $(E,p)$ come in pairs, one of rank~$0$ and one of
rank~$1$ having the same conductor. Each pair of curves are related
by a quadratic twist. This pattern is common, but does not always
occur. For example, (1184F,3) and (1184H,3) are both of rank~$0$
and have non-surjective irreducible representation, and no curve of
conductor 1184 and rank~$1$ has this property. Note that
$1184=2^5\cdot 37$ and 1184F and 1184H are quadratic twists of each
other by $-1$.
\end{remark}
\begin{remark}
Proposition~\ref{prop:nonsurjirred} suggests that it is rare for
$\rhobar_{E,p}$ to be non-surjective yet irreducible. When this
does occur, frequently $p^2\mid N$, though not always. Continuing
the computation to conductor~$10000$ we find that $p^2\mid N$ about
half the time in which $\rhobar_{E,p}$ is non-surjective yet
irreducible. This gives a sense of the extent to which
Theorem~\ref{thm:cha} improves on Theorem~\ref{thm:kolysurj}.
\end{remark}
\begin{theorem}\label{thm:bsdr1}
Suppose $(E,p)$ is a pair consisting of a rank~$1$ non-CM elliptic
curve~$E$ of conductor $\leq 1000$ and a prime~$p$ such that
$\rho_{E,p}$ is irreducible and~$p$ does not divide any Tamagawa
number of~$E$. Then $\BSD(E,p)$ is true.
\end{theorem}
\begin{proof}
By Theorem~\ref{thm:bsd2} we may assume that $p$ is odd.
The pairs that do not satisfy the
Heegner point divisibility hypothesis in
Proposition~\ref{prop:r1indexex} are those in
$S = \{(540B,3), (756B,3)\}$. However, both of these
curves admit a rational $3$-isogeny, so are excluded by
the hypothesis of Theorem~\ref{thm:bsdr1}.
Let
\begin{align*}
T &= \{(245A,3), (338E,3), (352F,3), (608E,5), (675B,5), (675I,5), \\
& \qquad(704L,3), (722B,3), (726A,3), (800B,5), (800I,5), (864A,3), \\
& \qquad (864B,3), (864J,3), (864L,3)\}.
\end{align*}
Then Proposition~\ref{prop:nonsurjirred},
Theorem~\ref{thm:cremona1000}, and Theorem~\ref{thm:kolysurj} together
imply $\BSD(E,p)$ for all pairs as in the hypothesis of
Theorem~\ref{thm:bsdr1}, except the pairs in $S\union T$. Note that
for each $(E,p) \in T$, we have $p^2\mid N$, so Theorem~\ref{thm:cha}
does not apply either. We eliminate the pairs
$$
(245A,3), (338E,3), (352F,3), (608E,5), (704L,3), (864J,3), (864L,3)
$$
from consideration because in each case $p\mid \prod c_\ell$.
For each $(E,p) \in T$ the representation $\rhobar_{E,p}$ is
irreducible and $E$ does not have CM, so the hypothesis of
Theorem~\ref{thm:us} are satisfied. For the pairs
$$\{((245A,3), (338E,3), (352F,3), (608E,5), (704L,3), (864J,3),
(864L,3)\}$$
we have $p\mid [E(K):\Z y_K]$ for the first six Heegner~$K$, but
that is not a problem since we eliminated these pairs from consideration.
For the remaining pairs, in each case we find a~$K$ such that
$p\nmid [E(K):\Z y_K] \cdot \disc(K)$, so Theorem~\ref{thm:us}
implies that $p\nmid \#\Sha(E)$, so $\BSD(E,p)$ is true.
\end{proof}
\subsubsection{Curves of Rank $0$}\label{sec:comp_r0}
\begin{proposition}\label{prop:r0indexex}
Suppose $(E,p)$ is a pair with $E$ an optimal elliptic curve of
conductor $\leq 1000$ of rank~$0$. Let~$I$ be the greatest common
divisor of $[E(K)_{/\tor}:\Z y_K]$ as~$K$ varies over quadratic
imaginary fields that satisfy the Heegner hypothesis. If~$p\mid I$
and $\rhobar_{E,p}$ is irreducible,
then
$$
p\mid 2\cdot \#E(\Q)_{\tor}\cdot \prod_{q\mid N} c_{E,q},
$$
except possibly for the curves in the following table:
{\rm \begin{center}
\begin{tabular}{|l|l|l|}\hline
$E$ & $p\mid I$? & $D$ used \\\hline
258E & 3 & $-983$\\ \hline
378G & 3 & $-47$\\ \hline
594F & 3 & $-359$ \\ \hline
600G&3 & $-71$\\ \hline
612D&3& $-359$ \\\hline
626B&3 &$-39$ \\\hline
658A & 3 & $-31$\\\hline
676E & 5 & $-23$ \\\hline
681B & 3 & $-8$ \\\hline
735B & 3 & $-479$\\\hline
738B & 3 & $-23$ \\\hline
742F &3, 5 & $-199$ \\\hline
\end{tabular}
$\qquad$
\begin{tabular}{|l|l|l|}\hline
$E$ & $p\mid I$? & $D$ used \\\hline
777B & 3 & $-215$\\\hline
780B & 3,7 & $-191$ \\\hline
819D &3,5 & $-404$ \\\hline
850I & 3 & $-151$\\\hline
858D & 5, 7 & $-95$ \\\hline
858K & 7 & $-1031$\\\hline
900A &3 &$-71$\\\hline
906E &$p\leq 19$ & $-23$\\\hline
924A & 5 & $-1679$ \\\hline
978C &3 & $-431$\\\hline
980I &3 & $-671$\\\hline
& & \\\hline
\end{tabular}
\end{center}}
\noindent{}In this table, the first column gives an elliptic curve,
the second column gives the primes~$p$ (with $\rhobar_{E,p}$ irreducible)
that might divide the $\GCD$ of indexes, and the third column gives
the discriminant used to make this deduction.
\end{proposition}
\begin{proof}
We use the methods described in Section~\ref{sec:methodr0}, and
precision bounds as in the proof of
Proposition~\ref{prop:r1indexex}. In many cases we combined
explicit computation of a Heegner point for one prime, with the
bounding technique explained in Section~\ref{sec:methodr0}, or only
computed information using the bound.
For the curve 910E, we used four-descent via MAGMA to compute the
point $(3257919871/16641, 133897822473008/2146689)$ on the $-159$
twist $E^D$, found using \cite{mwrank} that it generates $E^D(\Q)$,
and obtained an index that is a power of $2$ and $3$. Since $3$
divides a Tamagawa number, we do not include 910E in our table.
Likewise, for 930F and $D=-119$, we used MAGMA's four-descent
commands to find a point of height $\sim 85.3$, and deduced that the
only odd prime that divides the index is $11$; since $11$ is a
Tamagawa number, we do not include 930F. Similar remarks apply for
966J with $D=-143$. We were unable to use $4$-descent to find a
generator for a twist of 906E1. (Fortunately, $906 = 2\cdot 3\cdot
151$, so Theorem~\ref{thm:bsdkato} implies $\BSD(E,p)$ except for
$p=2,3,151$, and for our purposes we will only need that $151$ does
not divide the Heegner point index.)
\end{proof}
\begin{remark}
We could likely shrink the table in Proposition~\ref{prop:r0indexex}
further using MAGMA's four descent command. However, we
will not need a smaller table for our ultimate application to
the BSD conjecture (Theorem~\ref{thm:bsdr0}).
\end{remark}
\begin{theorem}\label{thm:rank0koly}
Suppose $(E,p)$ is a pair with~$E$ a rank~$0$ non-CM curve of conductor
$\leq 1000$ and~$p$ a prime such that $\rhobar_{E,p}$ is irreducible
and~$p$ does not divide any Tamagawa number of~$E$. Then
$\BSD(E,p)$ is true except possibly if $(E,p)$ appears in
the table in the statement of Proposition~\ref{prop:r1indexex},
i.e.,~$E$ appears in column~$1$ and~$p$ appears in the column
directly to the right of~$p$.
\end{theorem}
\begin{proof}
The argument is similar to the proof of Theorem~\ref{thm:bsdr1}.
By Theorem~\ref{thm:bsd2} we may assume that $p$ is odd.
Let $S$ be the set of pairs $(E,p)$ in the table
in Proposition~\ref{prop:r1indexex}.
Let
\begin{align*}
T &= \{(245B,3), (338D,3), (352E,3), (608B,5), (675D,5), (675F,5), \\
& \qquad(704H,3), (722D,3), (726F,3), (800E,5), (800F,5), (864D,3), \\
& \qquad (864F,3), (864G,3), (864I,3)\}.
\end{align*}
%v = [('245B',3), ('338D',3), ('352E',3), ('608B',5), ('675D',5), ('675F',5), ('704H',3), ('722D',3), ('726F',3), ('800E',5), ('800F',5), ('864D',3), ('864F',3), ('864G',3), ('864I',3)]
Then Proposition~\ref{prop:nonsurjirred},
Theorem~\ref{thm:cremona1000}, and Theorem~\ref{thm:kolysurj} together
imply $\BSD(E,p)$ for all pairs as in the hypothesis of
Theorem~\ref{thm:bsdr1}, except the pairs in $S\union T$, since
the representation $\rhobar_{E,p}$ is surjective and we have
verified that $p\nmid [E(K):\Z y_K]$ for some~$K$.
We eliminate the pairs
$(722D,3)$ and $(726F,3)$
from consideration because in each case $p\mid \prod c_\ell$.
For each $(E,p) \in T$ the representation $\rhobar_{E,p}$ is
irreducible and $E$ does not have CM, so the hypotheses of
Theorem~\ref{thm:us} are satisfied. Next for each pair $(E,p)\in T$
except for $(722D,3)$ and $(726F,3)$, which we eliminated already, we
find a~$K$ such that $p\nmid [E(K):\Z y_K]$ and $\disc(K)$ is not
divisible only be~$p$. Theorem~\ref{thm:us} implies that $p\nmid
\#\Sha(E)$, hence $\BSD(E,p)$ is true.
% 245B -- i=1, D=-19
% 338D -- i=1, D=-23
% 352E -- i=1, D=-7
% 608B -- i=1, D=-15 (OK, since divisible by 2 primes)
% 675D -- , D=-11
% 704H -- i=1, D=-7
% 722D -- ???
% 726F -- ???
% 800E -- i=3, D=-31, ok
% 800F -- i=1, D=-31
% 864D -- i=1, D=-23
% 864F -- i=1, D=-23
% 864G -- i=1, D=-23
% 864I -- i=1, D=-23
\end{proof}
\subsubsection{Two Descent}\label{sec:2descent}
In this section, we explain how descent computations imply that
$\BSD(E,2)$ is true for curves of conductor $N\leq 1000$.
\begin{theorem}\label{thm:bsd2}
If $E$ is an elliptic curve with $N\leq 1000$, then
$\BSD(E,2)$ is true.
\end{theorem}
\begin{proof}
According to Theorem~\ref{thm:bsdiso}, it suffices to prove the
theorem for the set~$S$ of optimal elliptic curves with $N\leq 1000$.
By doing an explicit $2$-descent, Cremona computed
$\Sel^{(2)}(E/\Q)$ for every curve $E\in S$, as explained in
\cite{cremona:algs}. This implies that $\Sha(E)[2]$ has order the
predicted order of $\Sha(E)[2^\infty]$ for all $E\in S$. Using
MAGMA's {\tt FourDescent} command, we compute $\Sel^{(4)}(E/\Q)$ in
the three cases in which $\Sha(E)[2]\neq 0$, and find that
$\Sha(E)[4] = \Sha(E)[2]$.
%(Note: The MAGMA documentation for {\tt
% FourDescent} is misleading.)
By Theorem~\ref{thm:cremona1000}, it
follows that $\BSD(E,2)$ is true for all $E\in S$.
\end{proof}
\subsubsection{Three Descent}\label{sec:3descent}
We sharpen Theorem~\ref{thm:rank0koly} using Stoll's
$3$-descent package (see \cite{stoll:3descent}).
\begin{proposition}\label{prop:3descent}
We have $3\nmid \#\Sha(E)$ for each of the curves listed in the Table
in Proposition~\ref{prop:r0indexex} with~$3$ in the second column
and $\rhobar_{E,3}$ irreducible, except for $681B$ where
$\#\Sha(E)[3^\infty] = 9$.
\end{proposition}
\begin{proof}
We use Stoll's package \cite{stoll:3descent} to compute each of the
Selmer groups $$\Sel^{(3)}(E)\isom \Sha(E)[3],$$ and obtain the
claimed dimensions. When computing class groups in Stoll's package
one must take care to not assume any conjectures (by slightly
modifying the call to {\tt ClassGroup} in {\tt 3descent.m}).
Finally, that $\Sha(E)[3^\infty] = 9$ follows by applying
Theorem~\ref{thm:kolysurj} with $K=\Q(\sqrt{-8})$, and noting that
$\rhobar_{E,3}$ is surjective and the index is exactly divisible
by~$3$.
\end{proof}
\section{The Kato Bound}
Kato proved a theorem that bounds $\Sha(E)$ from above when
$L(E,1)\neq 0$.
\begin{theorem}[Kato]\label{thm:kato}
Let $E$ be an optimal elliptic curve over~$\Q$ of conductor~$N$, and
let~$p$ be a prime such that $p \nmid 6N$ and $\rho_{E,p}$ is
surjective. If $L(E,1) \neq 0$, then $\Sha(E)$ is
finite and
$$
\ord_p (\#\Sha(E)) \leq
\ord_p \left(\frac{L(E, 1)}{\Omega_E}\right).
$$
\end{theorem}
This theorem follows from the existence of an ``optimal'' Kato Euler
system (see \cite{kato:secret} and
\cite{mazur-rubin:kolyvagin_systems}) combined with a recent result of
Matsuno~\cite{matsuno:finitelambda} on finite submodules of Selmer
groups over $\Z_p$-extensions. For more details, look at the proof of
\cite[Cor.~8.9]{rubin:kato} where one replaces an unknown module with
the module Matsuno computes. See also \cite{grigor:phd} for further
discussion and recent results on lower bounds on $\Sha(E)$ that make
use of optimal Kato Euler systems.
\subsection{Computations}
When $L(E,1)\neq 0$ the group $\Sha(E)$ is finite,
so $\ord_p(\#\Sha(E))$ is even. Thus if
$\ord_p \left(\frac{L(E, 1)}{\Omega_E}\right)$ is odd,
we conclude that
$$\ord_p (\#\Sha(E)) \leq
\ord_p \left(\frac{L(E, 1)}{\Omega_E}\right) - 1.$$
%There are {\em no} cases when $N\leq 1000$ and
%this bound is not sharp.
%\newcommand{\pair}[2]{\mbox{\rm (#1,#2)}}
\begin{lemma}\label{lem:katobad}
There are no pairs $(E,p)$ that satisfy the conditions of
Theorem~\ref{thm:kato} with $N \leq 1000$, such that
$$
\ord_p(\#\Sha(E)_{\an}) <
\ord_p \left(\frac{L(E, 1)}{\Omega_E}\right) - 1.
$$
\end{lemma}
\begin{proof}
First we make a table of ratios $L(E,1)/\Omega_E$ for all curves of
conductor $\leq 1000$. For each of these with $L(E,1)\neq 0$, we
factor the numerator of the rational number $L(E,1)/\Omega_E$. We
then observe that the displayed inequality in the statement of the
proposition does not occur.
% For each prime factor with $p^2\mid A$ and $p\nmid 6N$, we find
% that $\rho_{E,p}$ is surjective. I
\end{proof}
\begin{theorem}\label{thm:bsdkato}
Suppose $(E,p)$ is a pair such that $N\leq 1000$, $p\nmid 3N$,~$E$ is
a non-CM elliptic curve of rank~$0$, and~$\rhobar_{E,p}$ is
irreducible. Then $\BSD(E,p)$ is true.
\end{theorem}
\begin{proof}
The statement for $p=2$ follows from Theorem~\ref{thm:bsd2}.
Let $S$ be the set of pairs $(E,p)$ as in the statement of
Theorem~\ref{thm:bsdkato} for which~$E$ is optimal and
$p>2$. By Theorem~\ref{thm:cremona1000} it suffices to prove that
$p\nmid \#\Sha(E)$ for all $(E,p)\in S$. Using
Proposition~\ref{prop:surj} with $A=1000$, we compute for each
rank~$0$ non-CM elliptic curve of conductor $N\leq 1000$, all primes
$p\nmid 6N$ such that $\rho_{E,p}$ might not be surjective. This
occurs for $53$ pairs $(E,p)$, with the $E$'s all distinct. For
these $53$ pairs $(E,p)$, we find that the representation
$\rhobar_{E,p}$ is reducible (since there is an explicit $p$~isogeny
listed in \cite{cremona:algs}), except for the pair $(608B,5)$, for
which $\rhobar_{E,5}$ is irreducible.
Thus Theorem~\ref{thm:kato} implies that for each pair $(E,p)\in S$,
except $(608B,5)$, we have the bound $$\ord_p(\#\Sha(E)) \leq
\ord_p(L(E,1)/\Omega_E).$$ By Theorem~\ref{thm:shasqr},
$\ord_p(\#\Sha(E))$ is even, so $\Sha(E)[p^\infty]$ is trivial
whenever
$$\ord_p(L(E,1)/\Omega_E)\leq 1.$$
By Theorem~\ref{thm:cremona1000}, $\ord_p(\#\Sha(E)_{\an})=0$
for all $p\geq 5$.
Thus by Lemma~\ref{lem:katobad},
there are no pairs $(E,p) \in S$ with $\ord_p(L(E,1)/\Omega_E) > 1$
(since otherwise some $\ord_p(\#\Sha(E)_{\an})$ would be nontrivial).
Finally, note that we dealt with $(608B,5)$ in Lemma~\ref{lem:608B}
using Cha's theorem.
This completes the proof.
\end{proof}
\subsection{Combining Kato and Kolyvagin}\label{sec:results}
In this section we bound $\Sha(E)$ for rank~$0$ curves by
combining the Kato and Kolyvagin approaches.
\begin{theorem}\label{thm:bsdr0}
Suppose $E$ is a non-CM elliptic curve of rank~$0$ with conductor
$N\leq 1000$, that~$\rhobar_{E,p}$ is irreducible, and that~$p$ does
not divide any Tamagawa number of~$E$.
Then $\BSD(E,p)$ is true.
\end{theorem}
\begin{proof}
Let $(E,p)$ be as in the hypothesis to Theorem~\ref{thm:bsdr0}. By
Theorem~\ref{thm:bsdkato}, $\BSD(E,p)$ is true, except possibly if $p\mid
3N$. Theorem~\ref{thm:rank0koly} implies $\BSD(E,p)$, except if
$(E,p)$ appear in the Table of Proposition~\ref{prop:r0indexex}.
Inspecting the table, we see that whenever a prime $p\geq 5$ is
in the second column, then~$p$ does not divide the conductor~$N$
of~$E$. This proves $\BSD(E,p)$ for $p\geq 5$.
Let $E$ be the curve 681B. Then $\BSD(E,3)$ asserts that
$\#\Sha(E)[3^\infty]=9$. It follows from \cite{cremona-mazur} and
\cite[App.]{agashe-stein:bsd}, or from the $3$-descent of
Section~\ref{sec:3descent} that $\#\Sha(E)[3] = 9$. Also,
$\rhobar_{E,3}$ is surjective and for $D=-8$ we have
$\ord_3([E(K):\Z y_K])=1$, so $\#\Sha(E)[3^\infty] \mid 9$, which
proves $\BSD(E,3)$.
Finally Proposition~\ref{prop:3descent} implies $\BSD(E,3)$ for the
remaining curves, which proves the theorem.
\end{proof}
\section{Proof of Theorem~\ref{thm:us}}\label{sec:relaxkoly}
In this section we prove Theorem~\ref{thm:us}. Assume that~$E$
and~$K$ are as in the statement of the theorem, and assume that
$\ord_{s=1}L(E/K,1) = 1$. Then the Heegner point $y_K$ has infinite
order. Kolyvagin (\cite{kolyvagin:euler_systems}) shows that in this
case the rank of $E(K)$ is 1 and $\Sha(E/K)$ is finite.
\subsection{Gross's Account}
Gross's account of Kolyvagin's work in \cite{gross:kolyvagin}
contains a proof that if~$E$ does not have complex
multiplication, then
$$
\#\Sha(E/K) \mid t \cdot [E(K):\Z y_K]^2,
$$
where $t$ is an integer divisible only by primes~$p$ such that the
representation $\rhobar_{E,p}:\Gal(\Qbar/\Q)\rightarrow \Aut(E[p])$ is
not surjective. Gross makes no claim about the powers of primes that
divide~$t$ (though Kolyvagin does in his papers). Our
Theorem~\ref{thm:us} provides a better bound, which removes the
condition that~$E$ not have CM, and relaxes the surjectivity
hypothesis on $\rhobar_{E,p}$.
Gross uses surjectivity of $\rhobar_{E,p}$ as a hypothesis only to prove
the following two propositions. We will prove analogous propositions
below, but under weaker hypotheses, which yields our
claimed improvement to~\cite{gross:kolyvagin}.
\begin{proposition}[Gross]\label{prop:gross-surj1}
Assume that $\rhobar_{E,p}$ is surjective.
For any integer~$n$, let
$K_n$ be the ring class field of~$K$ of conductor~$n$.
The restriction map
$$
\Res:\H^1(K,E[p])\to \H^1(K_n,E[p])^{\Gal(K_n/K)}
$$
is an isomorphism.
\end{proposition}
\begin{proof}
That $\rhobar_{E,p}$ is surjective
implies that $E(K_n)[p]=0$. The
inflation-restriction-transgression sequence then implies
that $\Res$ is an isomorphism.
\end{proof}
Gross also uses surjectivity of $\rhobar_{E,p}$ when proving that
the pairing
$$
\H^1(K,E[p])\otimes \Gal(K(E[p])/K)\to E[p]
$$
is nondegenerate, as follows.
Setting $L=K(E[p])$, we have that
$$
\H^1(L/K,E(L)[p])\to \H^1(K,E[p])
\to \H^1(L,E[p])^{\Gal(L/K)}\to \H^2(L/K,E(L)[p]).
$$
To see that the pairing is nondegenerate, it suffices to know
that the groups $\H^i(L/K,E[p])$ vanish for $i=1,2$.
This is because we have
$$
\H^1(L,E[p])^{\Gal(L/K)} = \Hom(G_L, E[p])^{\Gal(L/K)}
$$
since $K(E[p]) \subset L$ and the pairing is $(c,\sigma) =
\res_L(c)(\sigma)$. Thus nondegeneracy of the pairing then follows
from the following proposition.
\begin{proposition}[Gross]\label{prop:gross-surj2}
Let $E$ be an elliptic curve over a number field~$K$ and
let $p$ be a prime.
Assume that $\rhobar_{E,p}$ is surjective. Then
$$\H^i(K(E[p])/K,E[p])=0\qquad \text{for all}\quad i\geq 1.$$
\end{proposition}
\begin{proof}
As above set $L=K(E[p])$.
The surjectivity of $\rhobar_{E,p}$ implies that
$$
G=\Gal(L/K)\isom \Gal(\Q(E[p])/\Q)\isom \GL_2(\F_p).
$$
If~$Z\subset G$ is the subgroup
corresponding to the scalars in $\GL_2(\F_p)$,
then the Hochschild-Serre spectral sequence implies that
$$
\H^i(G/Z,\H^j(Z,E(L)[p]))\Longrightarrow \H^{i+j}(L/K,E(L)[p]).
$$
Since $\#Z =p-1$, and $E(L)[p]$ is a $p$-group, and $p$ is odd,
we have $\H^j(Z,E(L)[p])=0$ for all $j\geq 1$.
Also, since $p$ is odd, and non-identity scalars have no
nonzero fixed points, $\H^0(Z,E(L)[p])=0$. Thus for
all $i, j$ we have $$\H^i(G/Z,\H^j(Z,E(L)[p]))=0,$$ which implies
that the groups $\H^{i+j}(L/K,E(L)[p])$ are all~$0$.
\end{proof}
Thus our goal is to prove analogues of
Propositions~\ref{prop:gross-surj1}--\ref{prop:gross-surj2} under
hypotheses that are more amenable to computation.
\subsection{Preliminaries}
\begin{lemma}\label{weil-pairing-lemma}
The determinant of $\rhobar_{E,p}$ is the cyclotomic character,
hence $\det(\rhobar_{E,p})$ is surjective.
\end{lemma}
\begin{proof}
For the convenience of the reader, we give a proof here.
The Weil pairing induces an isomorphism of $\Gal(\Qbar/\Q)$-modules
$E[p] \wedge E[p] \cong \mu_{p}$. Fix a basis $\{e_1, e_2\}$
of $E[p]$, with respect to which $\rho_{p}(\sigma)$ has the form
$\abcd{a}{b}{c}{d}$. Then
$$\sigma(e_1 \wedge e_2)= (ae_1+ce_2)
\wedge (be_1+de_2)= \hbox{det}(\rho_{p}(\sigma))\cdot e_1 \wedge e_2.
$$
It
follows that composition with the determinant gives the cyclotomic
character (i.e., the action of $\Gal(\Qbar/\Q)$ on $\mu_{p}$), which
is surjective since no nontrivial roots of unity lie in~$\Q$.
\end{proof}
We will choose the quadratic field $K$ to be linearly
disjoint from $\Q(E[p])$, so $\Gal(K(E[p])/K) \isom \Gal(\Q(E[p])/\Q)$.
Thus, for our application, it will suffice to show vanishing of
$\H^i(\Q(E[p])/\Q, E[p])$, for $i>0$.
Let $G\subseteq \Gal(\Q(E[p])/\Q)$ be the image of $\rhobar_{E,p}$.
If $p\nmid \#G$, then for $i>0$ we have $\H^i(G,E[p])=0$ since $E[p]$
is a $p$-group. Therefore we may assume that $p\mid \#G$. By
\cite[Prop.~15]{serre:propgal}, the image~$G$ either contains $\SL_2(\F_p)$ or
is contained in a Borel subgroup of $\GL_2(\F_p)$. \extra{A Borel
subgroup of an algebraic group is in general a maximal solvable
subgroup. In our caseit is a subgroup of $\GL_2(\F_p)$ that
preserves a fixed line in $\F_p^2$.} If $G$ contains $\SL_2(\F_p)$
then properties of the Weil pairing imply that
$$\det: G \rightarrow \F_p^*$$ is
surjective, so $G=\GL_2(\F_p)$. In this case, we already know
Propositions~\ref{prop:gross-surj1}--\ref{prop:gross-surj2}.
\begin{lemma}\label{lemma2}
Assume that $G$ is contained in a Borel subgroup of $\GL_2(\F_p)$.
Moreover, assume that there is a basis of $E[p]$ so that $G$
acts as $\abcd {\chi}{*}{0}{\psi}$ where~$\chi$ and~$\psi$ are
nontrivial characters. Then $\H^i(G,E[p])=0$.
\end{lemma}
\begin{proof}
Let $W=\abcd{1}{*}{0}{1}$ be the unique $p$-Sylow subgroup of
$\abcd{*}{*}{0}{*}\subset\GL_2(\F_p)$. We may assume $W \subset G$, for
otherwise $G$ has order prime to $p$, and the cohomology
vanishes.
We begin by explicitly computing $\H^j(W, E[p])$ using the fact
that~$W$ is cyclic, generated by $w=\abcd{1}{1}{0}{1}$.
Recall that for cyclic groups we can compute cohomology using the
projective resolution $$\dots \rightarrow \Z[W]
\rightarrow \Z[W] \rightarrow \Z \rightarrow 0$$
where the boundary
maps alternate between multiplication by $w-1$ and $\hbox{Norm}= \sum_{i=0}^{p-1} w^i$.
Then we see that
$$
\H^j(W, E[p])=
\begin{cases}
\Ker(1-w)/\Im(\Norm(w))=
\left\langle
\left( \begin{smallmatrix} 1 \\ 0 \end{smallmatrix} \right) \right\rangle,
& \text{if $j$ is even}\vspace{1ex},\\
\Ker(\Norm(w))/\Im(1-w)= \F_p^2/\left\langle \left(
\begin{smallmatrix} 1 \\ 0 \end{smallmatrix} \right)
\right\rangle, & \text{if $j$ is odd.}
\end{cases}
$$
Since $\chi$ and $\psi$ are nontrivial by assumption, the
$G/W$-invariants for both of these groups are trivial. Thus
$\H^j(W, E[p])^{G/W}=0$ for $j \geq 0$.
Consider the Hochschild-Serre spectral sequence
$$
\H^i(G/W, \H^j(W, E[p])) \Rightarrow \H^{i+j}(G, E[p]).
$$
For $i>0$, since $\#(G/W)$ is prime to $p$, and $\H^j(W, E[p])$ is a
$p$-group for all~$j$, the group $\H^i(G/W, \H^j(W, E[p]))$ is
trivial. But when $i=0$ we have just computed that $\H^i(G/W, \H^j(W,
E[p]))= \H^j(W, E[p])^{G/W}=0$, so the entire spectral sequence is
trivial, and we conclude that $\H^{n}(G, E[p])=0$ for all $n \geq 0$.
\end{proof}
\extra{We will apply the previous two lemmas when $G$ is the image of the mod $p$ representation.}
\subsection{Analogue of Proposition~\ref{prop:gross-surj1}}
In this section we verify that $\H^i(K_n/K,E(K_n)[p])=0$ under a
simple condition on $p$-torsion over $K$.
\begin{proposition}
Let $E$ be an elliptic curve over~$\Q$ and~$K$ be a quadratic
imaginary extension of~$\Q$. Assume that~$p$ is a prime with
$p\nmid \#E(K)_{\tor}$ and if $p=3$ assume that $K \neq
\Q(\zeta_3)$. Then for every finite abelian extension~$L$ of~$K$
we have
$$
\H^i(L/K, E(L)[p]) =0\qquad\text{for all}\quad i \geq 1.
$$
\end{proposition}
\begin{proof}
Write the abelian group $\Gal(L/K)$ as a direct sum $P \oplus
P^\prime$, where $P$ is its Sylow $p$-subgroup, so $p\nmid \#P'$.
First we show that the subgroup of $E(L)[p]$ invariant under
$P^\prime$ is trivial. Let $G = \Gal(L/K)/H$, where $H$ is the
subgroup of $\Gal(L/K)$ that acts trivially on $E(L)[p]$.
Thus $G\subset \Aut(E(L)[p])$.
\vspace{1ex}\noindent{\bf Case 1.}
If $p\nmid \#G$, then $P\subseteq H$, so $P^\prime$
surjects onto $G$. There is no nonzero element of $E(L)[p]$
invariant under $\Gal(L/K)$ by our assumption that $p\nmid \# E(K)$, so
the same holds for $P^\prime$.
\vspace{1ex}\noindent{\bf Case 2.}
If $p\mid \#G$, we cannot have $E(L)[p]= \F_p$, since $\F_p$ has
automorphism group isomorphic to $\F_p^*$, of order $p-1$, but
$G\subset \Aut(E(L)[p])$ and $\#G> p-1$.
Thus, $E(L)[p]$ is the full $p$-torsion subgroup of~$E$,
and we identify~$G$ with a subgroup of $\GL_2(\F_p)$ acting
on $E(L)[p]= \F_p^2$.
We can choose a basis of $\F_p^2$ so that $G$ contains the
subgroup generated by $\abcd{1}{1}{0}{1}$. Since~$G$ is abelian, it
must be contained in the normalizer of this subgroup, so $G
\subseteq \{ \abcd{a}{b}{0}{a} : a \in \F_p^*, b \in \F_p \}$. We
claim that~$G$ contains an element with $a \neq 1$. Since $E[p]=
E(L)[p]$, the representation $\Gal(\Qbar/K) \rightarrow
\hbox{Aut}(E[p])$ factors through $\Gal(L/K)$. The determinant of
$\rhobar_{E,p}:G_\Q\to\Aut(E[p])$ is surjective onto $\F_p^*$, and
$[K:\Q]=2$, so the character $\Gal(\Kbar/K) \rightarrow \F_p^*$ has
image of index at most~$2$ in $F_p^*$. That is, it contains at
least $(p-1)/2$ elements, the squares in $\F_p^*$. Thus, for
$p>3$, the group~$G$ contains an element with non-trivial
determinant having the form $\abcd{a}{b}{0}{a}$ with $a \neq 1$.
Now, ${\abcd {a}{b}{0}{a}}^p= \abcd{a}{0}{0}{a}$ since $a,b\in\F_p$,
so $\Gal(L/K)$ contains an element that
acts as a nontrivial scalar. Since the group of
scalars in $\GL_2(\F_p)$ has $p-1)$ elements, this nontrivial scalar
must be in $P^\prime$, so $E(L)[p]^{P^\prime}=0$.
We have shown in each case that $E(L)[p]^{P^\prime}=0$.
Because $p\nmid \#P'$ we have $\H^i(P', E(L)[p])=0$ for all $i \geq 1$,
so for each $i\geq 1$ there is an exact inflation-restriction sequence
$$
0\rightarrow \H^i(P, E(L)[p]^{P^\prime}) \rightarrow
\H^i(L/K, E(L)[p]) \rightarrow \H^i(P^{\prime}, E(L)[p]).
$$
The first group vanishes since $E(L)[p]^{P^{\prime}}=0$, and the third
group vanishes as mentioned above.
We conclude that $\H^i(L/K, E(L)[p])=0$, as claimed.
Finally we deal with the case $p=3$. The only situation in the above
argument where $p=3$ is relevant is in Case~2, when $3\mid \#G$. Our
hypothesis that $K\neq \Q(\zeta_3)$ implies that
$\det(\rho_{E,3}):\Gal(\Kbar/K)\to \F_3^*$ is surjective, since the
fixed field of the kernel of the mod~$3$ cyclotomic character is
$\Q(\zeta_3)$. If we are in Case~2, then the image of $\Gal(\Kbar/K)$
in $\GL_2(\F_3)$ is contained in $\{ \abcd{a}{b}{0}{a} : a \in \F_p^*,
b \in \F_p \}$. Since no upper triangular matrix has determinant~$2$,
this contradicts surjectivity of $\det(\rho_{E,3})$. Thus our
hypothesis that $K\neq \Q(\zeta_3)$ implies that Case~2 does not occur.
\end{proof}
\begin{corollary}
Let $E$ be an elliptic curve with $p\nmid \#E(K)_{\tor}$, where $p>3$ or, if
$p=3$, $K \neq \Q(\zeta_3)$. Let $K_n$ be the ring class field of
conductor $n$ of $K$. Then $\H^i(K_n/K, E(K_n)[p]) =0$ for all $i
\geq 1$.
\end{corollary}
\begin{lemma}\label{lemma:stein} Let $E$ be an elliptic curve
over~$\Q$, let $K$ be a quadratic imaginary extension, and let~$p\mid
\#E(K)_{\tor}$ an odd prime. If $p=3$, assume $K\neq \Q(\zeta_3)$.
Then $\rhobar_{E,p}$ is reducible.
\end{lemma}
\begin{proof}
Let $P \in E(K)[p]$ be nonzero, and let $\tau$ be a
lift of the generator of $\Gal(K/\Q)$ to $G_\Q$. If
$\tau P$ is a multiple of $P$, then the one-dimensional subspace of
$E[p]$ generated by $P$ is $G_\Q$-stable, so
$\rhobar_{E,p}:G_\Q\to \Aut(E[p])$ is is reducible.
If $\tau P$ is not a multiple of~$P$, then $P$ and $\tau P$
generate all of $E[p]$. Since $\tau P \in E(K)$, we have
$E(K)[p]= E(\Qbar)[p]$.
Because the Weil pairing in nondegenerate we have
$\mu_p \subset K$. This is a contradiction by
our hypothesis on~$K$ and $p$.
Since $p>3$, this is a contradiction.
\end{proof}
\subsection{Analogue of Proposition~\ref{prop:gross-surj2}}
%\edit{best section name? saying rational isogenies sounds like there's going to be
% some irreducibility assumption lurking}
In this section we show how vanishing of $\H^i(\Q(E[p])/\Q, E[p])$
follows from a statement about torsion and rational isogenies.
Note that~$E$ has no $\Q$-rational $p$-isogeny if and only if
$\rhobar_{E,p}$ is irreducible.
\begin{proposition}\label{prop:vanisha}
If~$p$ is an odd prime and~$E$
has no $\Q$-rational $p$-isogeny, then
$\H^i(\Q(E[p])/\Q, E[p])=0$ for all $i >0$.
\end{proposition}
\begin{proof}
Our hypothesis that $E$ has no $\Q$-rational $p$-isogeny implies
that $\rhobar_{E,p}$ is irreducible. As we already noted, the
problem reduces to the case when either~$G$ is contained in a Borel
subgroup or $G=\GL_2(\F_p)$. The latter case follows from
Proposition~\ref{prop:gross-surj2}. The former case contradicts the
hypothesis since the module $E[p]$ is reducible as a module over a
Borel subgroup.
\end{proof}
For the above result, we used the irreducibility of the representation
to deal with the case when~$G$ was contained in a Borel subgroup. The
following proposition completes the proof of the general case:
\begin{proposition}
Suppose $p$ is an odd prime and that $E(\Q)[p]=0$ and for all
elliptic curves $E'$ that are $p$-isogenous to~$E$ over~$\Q$ we have
$E'(\Q)[p]=0$. Then
$$\H^i(\Q(E[p])/\Q, E[p])=0 \qquad\text{for all}\qquad i>0.$$
\end{proposition}
\begin{proof}
If $E$ admits no $p$-isogeny, then Proposition~\ref{prop:vanisha}
implies the required vanishing. Thus $E$ admits a rational
$p$-isogeny, so $E[p]$ is reducible, and $G=\Im(\rhobar_{E,p})$ is
contained in a Borel subgroup. In particular, for some basis of
$E[p]$, the image $G$ acts as $\abcd {\chi}{*}{0}{\psi}$
%$\left( \begin{array}{cc}
%\chi & * \\
%0 & \psi \end{array} \right)$
for characters $\chi$ and $\psi$. If both $\chi$ and $\psi$ are
nontrivial, then Lemma \ref{lemma2} implies the proposition and we
are done. Thus assume that either $\chi$ or $\psi$ is trivial.
First suppose that $\chi$ is trivial. Then all matrices of the above
form fix $\left( \begin{smallmatrix} 1 \\ 0 \end{smallmatrix}
\right)$. Therefore there is a point of $E[p]$ fixed by the action
of $G$, which contradicts the assumption that $E(\Q)[p]=0$.
Next suppose that $\psi$ is trivial. Matrices of the above form
preserve the line generated by $\left( \begin{smallmatrix} 1 \\ 0
\end{smallmatrix} \right)$, so this line forms a
$\Gal(\Qbar/\Q)$-stable subspace of $E[p]$. In particular, there
exists an isogeny over $\Q$ to a curve $E^\prime$ having this line
as kernel. The image under this isogeny of the line generated by
$\left(
\begin{smallmatrix} 0 \\ 1 \end{smallmatrix} \right)$ is a
$1$-dimensional subspace of $E^{\prime}[p]$, and since $\psi=1$,
$\Gal(\Qbar/\Q)$ acts trivially on this subspace (we have an
isomorphism of Galois modules $E/\left\langle \left(
\begin{smallmatrix} 1 \\ 0 \end{smallmatrix} \right)
\right\rangle \cong E^\prime$). Thus, $E^{\prime}(\Q)[p]$ is
nontrivial, contradicting our assumption.
\end{proof}
%We finally apply Gross's arguments \cite{gross:kolyvagin} to show that
%$p\nmid \#\Sha(E)$ for all odd primes~$p$, such that $p \nmid I_K$ and
%all curves in the $\Q$-isogeny class of~$E$ have no $K$-rational
%$p$-torsion.
%\bibliographystyle{amsalpha}
%\bibliography{biblio}
\newcommand{\etalchar}[1]{$^{#1}$}
\providecommand{\bysame}{\leavevmode\hbox to3em{\hrulefill}\thinspace}
\providecommand{\MR}{\relax\ifhmode\unskip\space\fi MR }
% \MRhref is called by the amsart/book/proc definition of \MR.
\providecommand{\MRhref}[2]{%
\href{http://www.ams.org/mathscinet-getitem?mr=#1}{#2}
}
\providecommand{\href}[2]{#2}
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\end{document}