%%%%%%%%%%%%%%%%%%%%%%%
%% Tom Fisher
%% A counterexample to a conjecture of Selmer
%% For SwDyer 75th birthday book, CUP
%% latex2e
%%%%%%%%%%%%%%%%%%%%%%%

\documentclass[12pt,twoside]{article}
\usepackage{amsmath,amsfonts,amssymb}
\textheight215truemm \textwidth 138truemm \pagestyle{myheadings} \date{}

\setcounter{page}{119}
 \markboth{\qquad A counterexample to a conjecture of Selmer
\hfill}{\hfill Tom Fisher \qquad}

\usepackage{theorem}

\newtheorem{prop}{Proposition}[section]
\newtheorem{thm}[prop]{Theorem}
\newtheorem{lem}[prop]{Lemma}
\newtheorem{cor}[prop]{Corollary}
\newtheorem{conj}[prop]{Conjecture}

\makeatletter
\begingroup
\gdef\th@upshape{\normalfont
  \def\@begintheorem##1##2{%
        \item[\hskip\labelsep \theorem@headerfont ##1\ ##2]}%
\def\@opargbegintheorem##1##2##3{%
   \item[\hskip\labelsep \theorem@headerfont ##1\ ##2\ (##3)]}}
\endgroup
\makeatother

\theoremstyle{upshape}
\newtheorem{rem}[prop]{Remark}
\newtheorem{ex}[prop]{Example}

\newcommand{\qed}{\ifhmode\unskip\nobreak\fi\quad\ensuremath{\square}}
\newenvironment{proof}{\paragraph{Proof}}{\qed\par\medskip}
\newenvironment{pfof}[1]{\paragraph{Proof of #1}}{\qed\par\medskip}

% \newcommand{\proof}[1][Proof.]{\smallskip\noindent{\em #1}}
% \def\endproof{\hfill\ensuremath{\square}\par\medskip}

\newcommand{\Sha}{\mbox{\wncyr Sh}}
\newfont{\wncyr}{wncyr10 at 12pt}

 \def\Gal{{\operatorname{Gal}}}
 \def\im{\operatorname{im}}
 \def\isom{\simeq}
 \def\la{{\lambda}}
 \def\o{\mathfrak o}
 \def\ord{\operatorname{ord}}
 % \def\P{{\bf{P}}} 
 \def\phihat{{\widehat{\phi}}}
 \def\p{\mathfrak p}
 \def\Aut{{\operatorname{Aut}}}
 \def\Norm{{\operatorname{Norm}}}
 \def\Q{{\bf{Q}}} 
 \def\ra{{\, \longrightarrow \,}}
 \def\rank{\operatorname{rank}}
 \def\tors{{\operatorname{tors}}}
 \def\Z{{\bf{Z}}} 

 % \date{6 November 2002}
\title{A counterexample to a conjecture of Selmer}
\author{Tom Fisher}
 \date{}

\begin{document}

\maketitle
\addcontentsline{toc}{chapter}{Tom Fisher, A counterexample to a
conjecture of Selmer}
% \tableofcontents

\begin{abstract}
We present a counterexample to a conjecture cited by
Cassels~\cite{CaI} and attributed to Selmer. The issues 
raised have been given new significance by the 
recent work of Heath-Brown~\cite{HB} and Swinnerton-Dyer~\cite{SwD}
on the arithmetic of diagonal cubic surfaces.
\end{abstract}

% \renewcommand{\Large}{\large}
% \renewcommand{\Huge}{\huge}

\section{Introduction}
\label{intro}

Let $E$ be an elliptic curve over a number field $k$,
with complex multiplication by $\Z[\omega]$ where 
$\omega$ is a primitive cube root of unity. Let
$K = k(\omega)$, so that $[K:k]=1$ or 2 according
as $\omega \in k$ or $\omega \not\in k$. 
In his work on cubic surfaces, Heath-Brown \cite{HB} 
makes implicit use of the following statement.
\begin{thm}
\label{thm0}
If $[K:k]=2$ and the Tate--Shafarevich 
group $\Sha(E/k)$ is finite, then the order
of $\Sha(E/K)[\sqrt{-3}]$ is a perfect square.
\end{thm}

We explain how this result follows from
the work of Cassels \cite{CaIV}, and give an example
to show that the condition $[K:k]=2$ is necessary.

For the application to cubic surfaces, we only need a
special case of the theorem, namely that 
$\Sha(E/K)[\sqrt{-3}]$ cannot have order 3. This result,
still conditional on the finiteness of the Tate--Shafarevich group,
has already appeared in \cite{BF} and \cite{SwD}.
In fact Swinnerton-Dyer \cite{SwD} vastly
generalises Heath-Brown's results. In the case $[K:k]=2$
he proves the Hasse principle for diagonal cubic
3-folds over $k$, conditional only on the finiteness
of the Tate--Shafarevich group for elliptic curves over $k$.
The condition $[K:k]=2$ is unnatural, and 
conjecturally should not appear. However, the counter\-example 
presented in this article suggests that, 
if we are to follow the methods of Heath-Brown and Swinnerton-Dyer,
then this condition on $k$ is unavoidable. 

In \S\ref{dgm} we recall how it is possible to pass between the fields
$k$ and $K$. Then in \S\ref{compsel} we give a modern 
treatment of the descent by 3-isogeny studied by Selmer~\cite{S1}
and Cassels~\cite{CaI}. In \S\S\ref{selconj}--\ref{ctp} we recall
how the conjectures of Selmer may be deduced from properties 
of the Cassels--Tate pairing. This culminates in a proof 
of Theorem~\ref{thm0}. Finally in \S\ref{myex} we present our new 
example.
 
\section{Decomposition into Galois eigenspaces}
\label{dgm}

Let $E$ be an elliptic curve over $k$ with
complex multiplication by $\Z[\omega]$. 
The isogeny $[\sqrt{-3}]\colon E \to E$ 
is defined over $K=k(\omega)$. But the kernel $E[\sqrt{-3}]$
is defined over $k$. It follows that there is a 3-isogeny
$\phi\colon E \to \widetilde{E}$ defined over $k$ with $E[\sqrt{-3}]=E[\phi]$.
Here $\widetilde{E}$ is a second elliptic curve defined over $k$, which
we immediately recognise as the $-3$-twist of $E$. The dual 
isogeny $\phihat\colon \widetilde{E} \to E$ satisfies $\phi \circ \phihat =[3]$ 
and $\phihat \circ \phi= [3]$. Our notation
for the Selmer groups and Tate--Shafarevich 
groups follows Silverman \cite[Chapter X]{Si}.

\begin{lem}
\label{lemdecomp}
If $[K:k]=2$ then the exact sequence
\begin{equation}
\label{exseqK}
0 \ra E(K)/\sqrt{-3}E(K) \ra S^{(\sqrt{-3})}(E/K) \ra
\Sha(E/K)[\sqrt{-3}] \ra 0 
\end{equation}
is the direct sum of the exact sequences
\begin{equation}
\label{exseqE}
0 \ra \widetilde{E}(k)/\phi E(k) \ra S^{(\phi)}(E/k) \ra
\Sha(E/k)[\phi] \ra 0 
\end{equation}
and
\begin{equation}
\label{exseqEE}
0 \ra E(k)/\phihat \widetilde{E}(k) \ra 
S^{(\phihat)}(\widetilde{E}/k) \ra \Sha(\widetilde{E}/k)[\phihat] \ra 0. 
\end{equation}
\end{lem}

 \begin{proof}
Since arguments of this type have already appeared in 
\cite{BF}, \cite{N}, \cite{SwD} and presumably countless
other places in the literature, we will not dwell on the proof. 
Suffice it to say that we decompose (\ref{exseqK}) into eigenspaces
for the action of $\Gal(K/k)$, and then use the inflation-restriction
exact sequence to identify these eigenspaces as 
(\ref{exseqE}) and (\ref{exseqEE}). The observation that 
$[K:k]=2$ is prime to $\deg \phi=3$ is crucial throughout 
the proof.
 \end{proof}

\begin{rem}
Each term of the exact sequence (\ref{exseqK}) is a $\Z/3\Z$-vector
space with an action of $\Gal(K/k)$. Thus 
each term is a direct sum of the Galois modules $\Z/3\Z$ and $\mu_3$.
If we replace $E$ by $\widetilde{E}$ in (\ref{exseqK}) 
we obtain the same exact sequence of abelian groups, but as 
Galois modules the summands $\Z/3\Z$ and $\mu_3$ are interchanged.
% This gives a convenient way of thinking of the decomposition in
% Lemma~\ref{lemdecomp}.
\end{rem}

\section{Computation of Selmer groups}
\label{compsel}

Let $k$ be a number field. Let $T[a_0,a_1,a_2]$ be the 
diagonal plane cubic 
\begin{equation}
\label{diagcubic}
a_0 x_0^3 + a_1 x_1^3 + a_2 x_2^3 = 0 
\end{equation}
where $a_0, a_1, a_2 \in k^*/k^{*3}$. 
Let $E_A$ be the elliptic curve $T[A,1,1]$ with
identity element $0=(0:1:-1)$. It is well known \cite{St} that $E_A$
has Weierstrass equation $y^2 = x^3 - 432 A^2.$
An alternative proof of the following lemma may be found
in \cite[\S18]{CaL}.

\begin{lem}
\label{lem1}
The diagonal plane cubic $T[a_0,a_1,a_2]$ is a smooth curve of 
genus $1$ with Jacobian $E_A$ where $A=a_0 a_1 a_2$.
\end{lem}

 \begin{proof}
There is an isomorphism $T[a_0,a_1,a_2] \isom E_A$ 
defined over $k(\sqrt[3]{\alpha})$ where $\alpha=a_1 a_2^2$, given by
$$ \psi\colon (x_0:x_1:x_2) \mapsto 
(a_2 x_0: \alpha^{2/3} x_1: \alpha^{1/3} a_2 x_2). $$
The cocycle $\sigma(\psi) \psi^{-1}$
takes values in the subgroup $\mu_3 \subset \Aut(E_A)$
generated by $x_i \mapsto \omega^i x_i$. But since $\mu_3$ acts on $E_A$
without fixed points, this action belongs to the translation
subgroup of $\Aut(E_A)$. It follows that $T[a_0,a_1,a_2]$ is a torsor
under $E_A$ and that $E_A$ is the Jacobian of $T[a_0,a_1,a_2]$.
 \end{proof}

Temporarily working over $K=k(\omega)$ we note that $E_A$ 
has complex multiplication by $\Z[\omega]$ where 
$\omega\colon (x_0:x_1:x_2) \mapsto (\omega x_0:x_1:x_2)$ and 
that $E_A[1-\omega]=E_A[\sqrt{-3}]$ is generated by $(0:\omega:-\omega^2)$.
So as in \S2 there is a map $\phi$ which gives an exact sequence 
of Galois modules 
$$ 0 \ra \mu_3 \ra E_A \stackrel{\phi}{\ra} \widetilde{E}_A \ra 0 $$
where $\widetilde{E}_A$ is the $-3$-twist of $E_A$.
Taking Galois cohomology we obtain an exact sequence
\begin{equation}
\label{gcoh}
0 \ra \widetilde{E}_A(k)/\phi E_A(k) \stackrel{\delta}{\ra} k^*/k^{*3}
\ra H^1(k,E_A)[\phi] \ra 0.
\end{equation}
The group $H^1(k,E_A)$ parametrises the torsors under
$E_A$. We write $C_{A,\alpha}$ for the torsor under $E_A$ described
by $\alpha \in k^*/k^{*3}$. The proof of Lemma~\ref{lem1} shows that 
\begin{equation}
\label{torseq}
\begin{array}{rcl}
T[a_0,a_1,a_2] \isom C_{A,\alpha} & & 
\text{ for } A=\prod a_\nu \text{ and }
\alpha = \prod a_\nu^\nu 
\end{array}
\end{equation}
where the products are over $\nu \in \Z/3\Z$. 
Since $T[a_0,a_1,a_2] \isom T[a_1,a_2,a_0]$ it is clear
that $A \in \im \delta$. If $\widetilde{E}_A$ has Weierstrass equation
$Y^2Z=-4A X^3+Z^3$ then the 3-covering map $T[a_0,a_1,a_2] \to 
\widetilde{E}_A$ is given by 
$$ (x_0:x_1:x_2) \mapsto (x_0x_1x_2:a_1x_1^3-a_2x_2^3:a_0x_0^3). $$

The Selmer group attached to $\phi$ is 
 $$
 S^{(\phi)}(E_A/k) = \bigl\{ \alpha \in k^*/k^{*3} \bigm | C_{A,\alpha}
(k_\p) \ne \emptyset \text{ for all primes } \p \bigr\}.
 $$
Since $\deg \phi =3$ is odd we have ignored the infinite places.
We write $\delta_\p$ for the local connecting map obtained when 
we apply~(\ref{gcoh}) to the local field $k_\p$.
Then the condition $C_{A,\alpha} (k_\p) \ne \emptyset$
may also be written $\alpha \in \im \delta_\p$.
Using~(\ref{torseq}) to give equations for $C_{A,\alpha}$ it is easy to 
prove 
\begin{lem}
\label{lem2}
Let $k$ be a number field, and $\p$ a prime not dividing $3$.
Let $\o_\p$ denote the ring of integers of $k_\p$. Then 
$$
 \im \delta_\p= \begin{cases}
 \o_\p^* / \o_\p^{*3} & \text{ if } \ord_\p(A) \equiv 0 \pmod{3} \\
\langle A \rangle & \text{ if } \ord_\p(A) \not\equiv 0 \pmod{3}.
\end{cases}
 $$
\end{lem}
If $\p$ divides 3 the situation is more complicated,
although we still have
\begin{equation}
\label{stillhave}
\im \delta_\p \subset \o_\p^* / \o_\p^{*3} \quad \text{ if } 
\ord_\p(A) \equiv 0 \pmod{3}. 
\end{equation}
% Working over $\Q$, ????? was able to complete the picture.
% \begin{lem}
% Let $A \in \Z$ be nonzero and cube-free. Then
% $$ \im \delta_3= \left\{ \begin{array}{cl} 
% \Q_3^*/\Q_3^{*3} & \text{ if } A \equiv \pm 3 \pmod{9} \\
% \langle A \rangle & \text{ otherwise } \\
% \end{array} \right. $$
% \end{lem}
%  \begin{proof}
% We note that $\alpha \in \Z_3^*$ is a cube if and only if 
% $\alpha \equiv \pm 1 \pmod{9}$. The lemma now follows from~(\ref{})
% and the following observations concerning the solubility
% of $T=T[a_0,a_1,a_2]$ over $\Q_3$. \\
% (i) If $\ord_3(a_0,a_1,a_2)=(2,0,0)$ then $T(\Q_3) \ne\emptyset$
% if and only if $a_1 a_2^2 \in \Z_3^3$. \\
% (ii) If $\ord_3(a_0,a_1,a_2)=(1,0,0)$ then $T(\Q_3) \ne\emptyset$. \\ 
% (iii) The curve $T[1,4,7]$ is insoluble over $\Q_3$.
%  \end{proof}

If $\omega \in k_\p$ Tate local duality tells us that
$\im \delta_\p$ is a maximal isotropic subspace
with respect to the Hilbert norm residue symbol
\begin{equation}
\label{hnrs}
k_\p^*/k_\p^{*3} \times k_\p^*/k_\p^{*3} \to \mu_3. 
\end{equation}

The next lemma treats the case $k=\Q(\omega)$. This
field has ring of integers $\Z[\omega]$ and class number 1.
The unique prime above 3 is $\pi=\omega-\omega^2$.
\begin{lem}
\label{lem4}
Let $A \in \Z[\omega]$ be nonzero and cube-free. Then
$$
 \im \delta_\pi= \begin{cases}
\langle \, A,(1+A)/(1-A) \, \rangle & \text{ if } \ord_\pi(A) \ne 0 \\
\langle \, A,1-\pi^3 \, \rangle & \text{ if } \ord_\pi(A)=0 
\text{ and } A \not\equiv \pm 1 \, (\pi^3) \\
\langle \, \omega(1+3a) ,1-\pi^3 \, \rangle & \text{ if } A=\pm(1+a\pi^3)
\text{ for some } a \in \Z[\omega]. \end{cases}
 $$
\end{lem}

 \paragraph{Proof}
We recall \cite[Exercise 2.13]{CF} that $k_\pi^*/k_\pi^{*3}$ 
has basis $\pi, \omega, 1-\pi^2, 1-\pi^3$ and that these
elements define a filtration compatible with the pairing~(\ref{hnrs}).
By Tate local duality it follows that $\im \delta_\pi$ 
has order 9. So to prove the lemma it suffices to prove 
the inclusions $\supset$.
As always $A \in \im \delta_\pi$, 
whereas (\ref{stillhave}) and 
%and if $\ord_\pi(A)=0$ then 
%all $\alpha \in \im \delta_\pi$ must be units, whence 
% \subset \langle \omega, 1-\pi^2, 1-\pi^3 \rangle$,
%\Z[w]_\pi^*/\Z[w]_\pi^{*3}$, 
Tate local duality 
% $\im \delta_\pi \supset \langle 1-\pi^3 \rangle$.
tell us that $1-\pi^3 \in \im \delta_\pi$.
There is at most one more element to find.
 \begin{enumerate}
 \renewcommand{\labelenumi}{(\roman{enumi})}
 \item Suppose $\ord_\pi(A) \not = 0$. If $\alpha$ satisfies 
$\alpha-\alpha^{-1}=A$ then $T[A,\alpha,\alpha^{-1}]$ is soluble.
Splitting into the cases $\ord_\pi(A) = 1$ or 2 we find 
 $$
 4A/(1-A^2) \ \equiv\ A \pmod{\pi^4}.
 $$
Thus $\alpha=(1+A)/(1-A)$ provides a solution mod $\pi^4$.
% (ii) Suppose $\ord_\pi(A)=0$. If $\ord_\pi(\alpha)\not\equiv 0 \pmod{3}$
% then it is easy to show $C_{A,\alpha}(K_\pi)=\emptyset$. By 
% Tate local duality $1-\pi^3 \in \im \delta_\pi$.
 \item Suppose $A=1+a\pi^3$ for some $a\in \Z[\omega]$. If $\alpha$ 
satisfies $A+\alpha+\alpha^{-1}=0$ then 
$T[A,\alpha,\alpha^{-1}]$ is soluble. In view of the identity
$$ (1+\pi^3 a)+ \omega (1+3a) + \omega^2(1-3a) = 0 $$
we see that $\alpha=\omega(1+3a)$ provides a solution mod $\pi^4$. \qed
 \end{enumerate}

\section{Selmer's conjectures}
\label{selconj}

In this section we take $k=\Q$, so that $K=\Q(\omega)$.
We consider the elliptic curves $E_A$ and $\widetilde{E}_A$ over $\Q$
where $A \ge 2$ is a cube-free integer. 
% Let $E_A$ be the elliptic
% curve over $\Q$ 
% $$ x^3 + y^3 = A z^3 \qquad 0 = (1:-1:0). $$
% We write $\widetilde{E}_A$ for the $-3$-twist of $E_A$. 
\begin{lem}
If $A \ge 3$ then the torsion subgroups are
% $E_A(\Q)_\tors = 0$ and $\widetilde{E}_A(\Q)_\tors \isom \Z/3\Z$.
$$ E_A(\Q)_\tors = 0 \quad \text{ and } \quad \widetilde{E}_A(\Q)_\tors
\isom \Z/3\Z. $$
\end{lem}

 \begin{proof}
See \cite[\S6]{St} or \cite[Chapter 1, Problem 7]{K}.
 \end{proof}

Lemma~\ref{lemdecomp} gives a decomposition into
$\Gal(K/\Q)$-eigenspaces
\begin{equation}
\label{Sdecomp}
S^{(\sqrt{-3})}(E_A/K) \isom
S^{(\phi)}(E_A/\Q) \oplus S^{(\phihat)}(\widetilde{E}_A/\Q).
\end{equation}
The following examples were found by Selmer \cite{S1}, \cite{S2}.
\begin{ex}
\label{ex60}
Let $A=60$. Lemmas~\ref{lem2} and~\ref{lem4} tell us that
$$ S^{(\sqrt{-3})}(E_{60}/K) \isom \langle 2,3,5 \rangle \subset K^*/K^{*3}. $$
Then~(\ref{Sdecomp}) gives
$S^{(\phi)}(E_{60}/\Q) \isom (\Z/3\Z)^3$ and 
% \quad \text{ and } \quad 
$S^{(\phihat)}(\widetilde{E}_{60}/\Q) = 0$. 
But a 2-descent \cite[\S15]{CaL}, \cite{mwrank} shows that 
$E_{60}(\Q)$ has rank 0. We deduce
$$\Sha(E_{60}/\Q)[3] \isom (\Z/3\Z)^2.$$ 
\end{ex}

\begin{ex}
\label{ex473}
Let $A=473$. Lemmas~\ref{lem2} and~\ref{lem4} tell us that
$$ S^{(\sqrt{-3})}(E_{473}/K) \isom \langle 11, 
1-6\omega, 1-6\omega^2 \rangle \subset K^*/K^{*3}. $$
Then~(\ref{Sdecomp}) gives
$S^{(\phi)}(E_{473}/\Q) \isom (\Z/3\Z)^2$ and 
% \quad \text{ and } \quad
$S^{(\phihat)}(\widetilde{E}_{473}/\Q) \isom \Z/3\Z$. 
But a 2-descent \cite{S2}, \cite{mwrank} shows that 
$E_{473}(\Q)$ has rank 0. We deduce
$$\Sha(E_{473}/\Q)[\phi] \isom \Z/3\Z \quad \text{ and } \quad 
\Sha(\widetilde{E}_{473}/\Q)[\phihat] \isom \Z/3\Z. $$ 
\end{ex}

\begin{rem}
According to the formulae and tables of Stephens \cite{St}, 
the above examples have $L(E_A,1) \ne 0$. 
So the claims $\rank E_A(\Q)=0$ could equally
be deduced from the work of Coates and Wiles \cite{CW}.
\end{rem}

Example~\ref{ex60} tells us that each of the curves
\begin{equation}
\label{selcubics}
% T[3,4,5], \quad T[1,3,20], \quad T[1,4,15], \quad T[1,5,12] $$
% \begin{array}{rclcrcl} \medskip
% 3 x_0^3+ 4 x_1^3+ 5 x_2^3 & = & 0 & \quad &
% x_0^3+ 3 x_1^3+ 20 x_2^3 & = & 0 \\
% x_0^3+ 4 x_1^3+ 15 x_2^3 & = & 0 & \quad &
% x _0^3+ 5 x_1^3+ 12 x_2^3 & = & 0 
% \end{array} 
\begin{array}{r@{\quad}rcl} 
T[3,4,5]: & 3 x_0^3+ 4 x_1^3+ 5 x_2^3 & = & 0 \\ 
T[1,3,20]: & x_0^3+ 3 x_1^3+ 20 x_2^3 & = & 0 \\
T[1,4,15]: & x_0^3+ 4 x_1^3+ 15 x_2^3 & = & 0 \\
T[1,5,12]: & x _0^3+ 5 x_1^3+ 12 x_2^3 & = & 0 
\end{array}
\end{equation}
is a counterexample to the Hasse Principle for smooth curves of genus
1 defined over $\Q$. Selmer proves this 
without the need for a 2-descent. Instead he shows that 
the equations~(\ref{selcubics}) are insoluble over $\Q$ by writing 
them as norm equations. % from a pure cubic field. 
As Cassels explains \cite[\S11]{CaI} this is equivalent 
to performing a second descent, {\em i.e.}\ computing the middle group in 
\begin{equation}
\label{mfilt1}
\widetilde{E}_A(\Q)/\phi E_A(\Q) \subset \phihat 
S^{(3)}(\widetilde{E}_A/\Q) \subset S^{(\phi)}(E_A/\Q). 
\end{equation}
In fact Selmer's calculations suffice to show that
% \footnote{As usual $A(p)$ denotes the $p$-primary part of
% the abelian group $A$.}
$\Sha(E_{60}/\Q)(3) \isom (\Z/3\Z)^2$. In other words
$\Sha(E_{60}/\Q)$ does not contain an element of order 9.
More recent work of Rubin \cite{M} improves 
this to $\Sha(E_{60}/\Q) \isom (\Z/3\Z)^2$.

Selmer also gave practical methods for 
computing the two right hand groups in 
\begin{equation}
\label{mfilt2}
E_A(\Q)/\phihat \widetilde{E}_A(\Q) \subset \phi 
S^{(3)}(E_A/\Q) \subset S^{(\phihat)}(\widetilde{E}_A/\Q). 
\end{equation}
Following Stephens \cite{St} we write $g_1+1$, $\la_1'+1$,
$\la_1+1$ for the dimensions of the $\Z/3\Z$-vector 
spaces~(\ref{mfilt1}) and $g_2$, $\la_2'$, $\la_2$ for the dimensions 
of the $\Z/3\Z$-vector spaces~(\ref{mfilt2}).
Trivially we have $0 \le g_1 \le \la_1' \le \la_1$, 
$0 \le g_2 \le \la_2' \le \la_2$ and $\rank E_A(\Q)=g_1+g_2$. 
Based on a large amount of numerical evidence, Selmer \cite{S3} 
made the following
 
\begin{conj} 
\label{Selmerconj}
Let $A \ge 2$ be a cube-free integer. 
Let $E_A$ be the elliptic curve $x^3+y^3=Az^3$ defined 
over $\Q$. Then

\begin{description}
 \item[Weak form] The second descent excludes an even number
of generators, i.e.\ $\la_1 \equiv \la_1' \pmod{2}$ and 
 $\la_2 \equiv \la_2' \pmod{2}$.

 \item[Strong form] The number of generators of infinite order for 
$E_A(\Q)$ is an even number less than what is 
indicated by the first descent, 
i.e.\ 
 \[
 \la_1 +\la_2 \equiv g_1+ g_2 \pmod{2}.
 \]
\end{description}
\end{conj}

For $A=473$, Selmer found
$\la_1=\la_1'=\la_2=\la_2'=1$ yet $g_1=g_2=0$. 
He was thus aware of the need to combine the contributions
from $\phi$ and $\phihat$ in the strong form of his conjecture.

\begin{rem}
In Heath-Brown's notation~\cite{HB} we have $$r(A)=\rank E_A(\Q)=g_1+g_2
\quad \text{ and } \quad s(A)= \la_1+ \la_2.$$ 
By~(\ref{Sdecomp}) the order of
$S^{(\sqrt{-3})}(E_A/K)$ is $3^{s(A)+1}$ and in fact it is this
relation that Heath-Brown uses to define $s(A)$. Naturally he
writes the strong form of Selmer's conjecture as $r(A) \equiv s(A) \pmod{2}$.
\end{rem}

Now let $k$ be any number field. 
Conjecture~\ref{Selmerconj} is equivalent to the case $k=\Q$ of
the following
\begin{conj}
\label{gensel}
Suppose that $A \in k^*$ is not a perfect cube and let $E_A$ be the
elliptic curve $x^3+y^3=A z^3$ defined over $k$. Then

\begin{description}
 \item[Weak form] The index of the subgroup $\phihat
(\Sha(\widetilde{E}_A/k)[3]) \subset \Sha(E_A/k)[\phi]$ is a perfect
square. The same is true for $\phi (\Sha(E_A/k)[3]) \subset
\Sha(\widetilde{E}_A/k)[\phihat]$.

 \item[Strong form] The order of $\Sha(E_A/k)[\phi]$ times that of
$\Sha(\widetilde{E}_A/k)[\phihat]$ is a perfect square. 
\end{description}
\end{conj}

In the next section we recall how Conjecture~\ref{gensel} follows from 
the work of Cassels, the strong form being conditional on 
the finiteness of $\Sha(E_A/k)$. 

\section{The Cassels--Tate pairing}
\label{ctp}

Let $E$ be an elliptic curve over a number field $k$.
For $\phi\colon E \to E'$ an isogeny of elliptic curves over $k$ we
shall write $\phihat\colon E' \to E$ for the dual isogeny.
Cassels \cite{CaIV} defines an alternating bilinear pairing
\begin{equation}
\label{ctpairing}
\langle~,~\rangle \colon \Sha (E/k) \times \Sha (E/k) \to \Q/\Z
\end{equation}
with the following nondegeneracy property.
\begin{thm}
\label{ctisog}
Let $\phi\colon E \to E'$ be an isogeny of elliptic curves
over $k$. Then $x \in \Sha(E/k)$ belongs to the image of 
$\phihat\colon \Sha(E'/k) \to \Sha(E/k)$
if and only if $\langle x, y \rangle = 0$ 
for all $y \in \Sha(E/k)[\phi]$.
\end{thm}

 \begin{proof} This was proved by Cassels \cite{CaIV} in the 
case $\phi=[m]$ for $m$ a rational integer. The general case follows
by his methods and is explained in \cite{F}.
 \end{proof}

The pairing was later generalised to abelian varieties by 
Tate, and so is known as the Cassels--Tate pairing. The most
striking applications in the case of elliptic curves come 
from the following easy lemma.
\begin{lem}
\label{easylem}
If a finite abelian group admits a nondegenerate
alternating bilinear pairing, then its order must be a perfect
square.
\end{lem}

The weak form of Conjecture~\ref{gensel} is a special case of
\begin{cor}
\label{cor1}
Let $\phi\colon E \to E'$ be an $m$-isogeny of elliptic curves over $k$.
% with $\deg \phi=m$.
Then the subgroup $\phihat (\Sha(E'/k)[m]) \subset \Sha(E/k)[\phi]$ 
has index a perfect square.
\end{cor}

 \begin{proof} According to Theorem~\ref{ctisog} the 
pairing~(\ref{ctpairing}) restricted to $\Sha(E/k)[\phi]$
has kernel $\phihat (\Sha(E'/k)[m])$. We are done by Lemma~\ref{easylem}.
 \end{proof}

Let us assume that $\Sha(E/k)$ is finite.
So by Theorem~\ref{ctisog} 
% tells us that the Cassels--Tate pairing~(\ref{ctpairing}) is
% nondegenerate.
and Lemma~\ref{easylem} the order of $\Sha(E/k)$ is a perfect square.
If $\phi\colon E \to E'$ is an isogeny of elliptic curves over $k$ 
then the same conclusions will hold for $E'$. We define
\begin{equation}
\label{ctphi}
\langle~,~\rangle_\phi \colon \Sha (E/k) \times \Sha (E'/k) \to \Q/\Z \,;
\quad (x,y) \mapsto \langle \phi x, y \rangle = 
\langle x, \phihat y \rangle 
\end{equation}
where the equality on the right is \cite[Theorem 1.2]{CaVIII}.
The strong form of Conjecture~\ref{gensel} is a special case of
\begin{cor}
\label{cor2}
Let $\phi\colon E \to E'$ be an isogeny of elliptic curves over $k$.
If $\Sha(E/k)$ is finite then the order of 
$\Sha(E/k)[\phi]$ times that 
of $\Sha(E'/k)[\phihat]$ is a perfect square.
\end{cor}

 \begin{proof}
According to Theorem~\ref{ctisog} the left and 
right kernels of $\langle~,~\rangle_\phi$
are $\Sha(E/k)[\phi]$ and $\Sha(E'/k)[\phihat]$. 
We obtain a nondegenerate pairing
$$\Sha(E/k)/\Sha(E/k)[\phi] \times \Sha(E'/k)/\Sha(E'/k)[\phihat]
\to \Q/\Z. $$
We deduce that these quotients have the same order and
% $\#\Sha(E/k)/\#\Sha(E/k)[\phi] = \#\Sha(E'/k)/\#\Sha(E'/k)[\phihat]$.
are done since $\Sha(E/k)$ and $\Sha(E'/k)$ 
each have order a perfect square.
 \end{proof}

Another well known consequence is
\begin{cor}
\label{cor3}
Let $E$ be an elliptic curve over $k$ whose Tate--Shafarevich group 
is finite, and let $m$ be a rational integer.
% If the Tate--Shafarevich group $\Sha(E/k)$ is finite
Then the order of $\Sha(E/k)[m]$ is a perfect square.
\end{cor}

 \begin{proof} % We apply Lemma~\ref{easylem} to the alternating pairing
% $\langle~,~\rangle_m$. 
According to Theorem~\ref{ctisog} the kernel of $\langle~,~\rangle_m$ 
is $\Sha(E/k)[m]$. We obtain a nondegenerate alternating pairing
$$\Sha(E/k)/\Sha(E/k)[m] \times \Sha(E/k)/\Sha(E/k)[m] \to \Q/\Z. $$
We apply Lemma~\ref{easylem} to this pairing and are 
done since $\Sha(E/k)$ has order a perfect square.
 \end{proof}

\begin{rem}
We could equally deduce Corollary~\ref{cor2} from Corollaries~\ref{cor1}
and~\ref{cor3}.
\end{rem}

 \begin{pfof}{Theorem~\ref{thm0}} Let $E$ be an elliptic 
curve over $k$ with complex multiplication by $\Z[\omega]$ and suppose
that $[K:k]=2$. Lemma~\ref{lemdecomp} tells us that
$$ \Sha(E/K) [\sqrt{-3}] \isom 
\Sha(E/k)[\phi] \oplus \Sha(\widetilde{E}/k)[\phihat]. $$
Assuming $\Sha(E/k)$ is finite, Corollary~\ref{cor2} shows that
the group on the right has order a perfect square. 
So the group on the left has order a perfect square, and this
is precisely the statement of Theorem~\ref{thm0}.
 \end{pfof}

In the first of his celebrated series of papers, 
Cassels \cite{CaI} defines a pairing
$S^{(\sqrt{-3})}(E_A/K) \times S^{(\sqrt{-3})}(E_A/K) \to \mu_3$.
It is of course a special case of the pairing~(\ref{ctpairing}).
He uses it to prove the weak form of Conjecture~\ref{gensel}
in the case $[K:k]=1$. However in the introduction to the same paper
he misquotes the strong form of Selmer's conjecture.
The statement he gives is equivalent to
\begin{itemize}
\item If $[K:k]=1$ then the order of 
$\Sha(E_A/K)[\sqrt{-3}]$ is a perfect square.
\end{itemize}
It is this statement to which we have found a counterexample.
It is possible that Cassels was misled by earlier work of 
Selmer at a time when he did not appreciate the need to combine the
contributions from $\phi$ and $\phihat$ in the strong form of
his conjecture.

\begin{rem}
It is tempting to try and prove 
Theorem~\ref{thm0} also in the case $[K:k]=1$ by imitating the proof of 
Corollary~\ref{cor3}. However 
the isogeny $[\sqrt{-3}]$ has dual $[-\sqrt{-3}]$ and this 
extra sign means that the pairing $\langle~,~\rangle_{\sqrt{-3}}$ 
is symmetric rather than alternating. 
Lemma~\ref{easylem} does not apply.
\end{rem}

\section{A new example}
\label{myex}

In this section we take $K=\Q(\omega)$. 
Let $E_A$ be the elliptic curve $x^3+y^3=Az^3$. 
We aim to find $A \in K$ such that the order of 
$\Sha(E_A/K)[\sqrt{-3}]$ is not a perfect square. 
As in Example~\ref{ex473} our method is to compare
a 3-descent with a 2-descent. The form of the curves
$E_A$ makes the 3-descent easy. We use the results
of \S\ref{compsel} to compute the Selmer group
$S^{(\sqrt{-3})}(E_A/K)$. For the 2-descent we would
like to use John Cremona's program {\tt mwrank} \cite{mwrank}.
But {\tt mwrank} is written specifically for elliptic
curves over $\Q$, whereas Theorem~\ref{thm0}
tells us that there are no examples of the required form
with $A^2 \in \Q$. Fortunately we were able to use a program
of Denis Simon \cite{Sim1}, \cite{Sim2}, written using the computer
algebra package {\tt pari}
\cite{pari}, that extends Cremona's work on 2-descents
to general number fields (in practice of degrees~1 up to~5).

We consider all cube-free $A \in \Z[\omega]$ with $A^2 \not\in \Q$
and $\Norm(A) \le 150$. We ignore repeats of the form $\pm \sigma(A)$
for $\sigma \in \Gal(K/\Q)$. In all 123 cases a calculation 
based on Lemmas~\ref{lem2} and~\ref{lem4} shows that $S^{(\sqrt{-3})}(E_A/K)$
is isomorphic to either $\Z/3\Z$ or $(\Z/3\Z)^2$. In the 98 cases
where $S^{(\sqrt{-3})}(E_A/K) \isom \Z/3\Z$ it follows immediately
that $\rank E_A(K)=0$. In the remaining 25 cases we run Simon's
program. For 20 of these curves the program exhibits a point of
infinite order. Since $E_A(K)$ has the structure of $\Z[\omega]$-module, 
we are able to deduce that $\rank E_A(K)=2$. The remaining 5 cases are 
$$A = \pm (3+7 \omega), \, \pm (9+\omega), \, \pm (12+5 \omega), \, 
\pm (6+13 \omega), \, \pm (13+7 \omega) $$
and their Galois conjugates. In each of these cases Simon's program
reports that $\rank E_A(K)=0$. Reducing modulo some small primes 
we find $E_A(K) \isom \Z/3\Z$. Thus 
$$ \Sha(E_A/K)[\sqrt{-3}] \isom \Z/3\Z. $$

For the remainder of this article we restrict attention to the
first of these examples, namely $A=3+7 \omega$, and
give further details of the descent calculations involved.
% required for the first of these examples, namely for $A=3+7 \omega$. 
In particular we establish the counterexample of the title 
in a way that is independent of Simon's program.
% We were thus able to check a large number of values of 
% $A$, and so arrived at the example $A=3+7 \omega$ for 
% which we now give full details.
% \begin{ex}

We begin by checking the above computation of $S^{(\sqrt{-3})}(E_A/ K)$
for $A=3+7\omega$.
Since $(A)$ is prime, Lemma~\ref{lem2} tells us that
\begin{equation}
\label{computesel}
S^{(\sqrt{-3})}(E_A/ K) \subset \langle \omega, 3+7\omega \rangle. 
\end{equation}
We check the local conditions at the primes
$(\pi)$ and $(A)$ above 3 and 37 respectively. 
\begin{itemize}
\item Since $37 \equiv 1 \pmod{9}$ we know
that $\omega$ is a cube locally at $(A)$. 
\item Lemma~\ref{lem4} gives $\im \delta_\pi=\langle A, 1-\pi^3 \rangle
\subset K_\pi^*/K_\pi^{*3}$.
Since $A=\omega-\pi^3$ it is clear that $\omega$ belongs
to this subgroup. 
\end{itemize}
It follows that equality holds in~(\ref{computesel}) as required. 
% A 2-descent \cite{Sim} shows that $\rank E(K)=0$. Reducing
% modulo some small primes we find $E_A(K) \isom \Z/3\Z$. 
% Thus 
% $$ \Sha(E_A/K)[\sqrt{-3}] \isom \Z/3\Z. $$

Given the provisional nature of Simon's program
we have taken the liberty of writing out the 2-descent 
for $A=3+7 \omega$ in the style of Cassels \cite[p.~72--73]{CaL}.
The curve $E_A$ has Weierstrass form
\begin{equation}
\label{weqn}
Y^2= X^3- 2^4 3^3 (3+ 7 \omega)^2.
\end{equation}
The 2-descent takes place over 
the field $L=K(\delta)$ where $\delta^3=4(3+7\omega)$.
According to {\tt pari} \cite{pari}\footnote{These calculations 
were performed using Version 2.0.20 (beta)}, 
$L$ has class number $h=3$, and fundamental units 
$$
\begin{array}{rcl}
\eta_1&=& (-7-3\omega)+(-3-2\omega)\delta 
+(-2+\omega )\delta^2/2 \\
\eta_2&=&(-7-3\omega)+(2-\omega)\delta 
+(3+2\omega)\delta^2/2.
\end{array} 
$$
Furthermore {\tt pari} is able to certify these results, independent of
any conjecture.
We have chosen $\eta_1$ and $\eta_2$ to be $K$-conjugates. They have
minimal polynomial
$$x^3+(21+9\omega)x^2+(102-165\omega)x-1.$$
If $(X,Y)=(r/t^2,s/t^3)$ is a solution of (\ref{weqn}), with
fractions in lowest terms, then a common prime divisor of any two of 
$$ r- 3 \delta^2 t^2, \quad r- 3 \omega \delta^2 t^2, \quad
r- 3 \omega^2 \delta^2 t^2 $$
must divide $2(1-\omega)(3+7\omega)$. 
Since $2, (1-\omega), (3+7\omega)$ ramify 
completely, $r- 3 \delta^2 t^2$ must be a perfect
ideal square. Since $h$ is odd it follows that 
$S^{(2)}(E/K)$ is a subgroup
% \footnote{The notation $\langle\;\rangle$ means ``subgroup generated by''.} 
of $\langle -1,\eta_1,
\eta_2 \rangle \subset L^*/L^{*2}$. We claim that
$S^{(2)}(E/K)$ is trivial. By considering norms from $L$
to $K$, it suffices to show that the equation
% \footnote{We recall
% that $\eta_1, \eta_2, 1/(\eta_1\eta_2)$ are $K$-conjugates.}
$$r- 3 \delta^2 t^2=\eta \alpha^2 
\quad \text{ with } \eta=\eta_1, \eta_2 \text{ or } 1/(\eta_1\eta_2)$$
is insoluble for $r,t \in K$ and $\alpha \in L$. The action 
of $\Gal(L/K)$ shows that we need only consider the case $\eta=\eta_1$. 
Put $\alpha=u+v \delta + w \delta^2$. Equating coefficients
of powers of $\delta$ we obtain
$$
\begin{array}{rcl}
0 & = & (-3-2\omega)u^2+(-14-6\omega)uv+(-26-36\omega)v^2 \\
& & \quad~+(-52-72\omega)uw +(40-104\omega)vw+(-148\omega) w^2 \\[8pt]
-3t^2& = & ((-2+\omega)/2)u^2+(-6-4\omega)uv+(-7-3\omega)v^2 \\
& & \quad~+(-14-6\omega)uw +(-52-72\omega)vw+(20-52\omega)w^2. 
\end{array}
$$
On putting
$$
\begin{array}{rcl}
u & = & (-8+6\omega)e+(-6-34\omega)f+ (-20+15\omega)g \\
v & = & (-4-4\omega)e+ (12+4\omega)f+(-10-11\omega)g \\
w & = & (1-\omega)e+ (1+4\omega)f+ (2-2\omega)g 
\end{array}
$$
in the first equation, it becomes
$$0= (3+7\omega)g^2-16ef. $$
Hence there are $m, n$ such that 
$$e:f:g=m^2:(3+7\omega)n^2:4mn.$$
On substituting into the second equation, we get
$$ 
\begin{array}{rcl}
-3 t^2& = & 2(-1-4\omega)m^4+8(-4+3\omega) m^3 n+4(21+12\omega)m^2 n^2 \\
& & \qquad\qquad\qquad\qquad~+8(4-3\omega) m n^3+2(-33-40\omega)n^4. 
\end{array}
 $$
But this is impossible over the 2-adic completion of $K$.
% as may be seen by considering 2-adic valuations. 
Hence $S^{(2)}(E_A/K)$ is trivial and $\rank E_A(K)=0$ 
as claimed.

\section*{Acknowledgements}
The author would like to thank Laura Basile, Alexei Skorobogatov and
Sir~Peter Swinnerton-Dyer for useful conversations.

\begin{thebibliography}{CaV}

\bibitem[BF]{BF}
C.L. Basile and T.A. Fisher,
Diagonal cubic equations in four variables with prime coefficients,
{\em Rational points on algebraic varieties}, 1--12, 
Progr. Math., {\bf 199}, Birkh\"auser, Basel, 2001

\bibitem[BBBCO]{pari}
C. Batut, K. Belabas, D. Bernardi, H. Cohen and M. Olivier, 
{\tt pari/gp}, a computer algebra package,
{\tt http://www.parigp-home.de}
 
\bibitem[CaI]{CaI}
J.W.S. Cassels,
Arithmetic on curves of genus $1$. I, On a conjecture of Selmer, 
{\em J. reine angew. Math.} {\bf{202}} (1959) 52--99

\bibitem[CaIV]{CaIV} 
J.W.S. Cassels, 
Arithmetic on curves of genus $1$. IV, Proof of the Hauptvermutung,
{\em J. reine angew. Math.} {\bf{211}} (1962) 95--112

\bibitem[CaVIII]{CaVIII}
J.W S. Cassels,
Arithmetic on curves of genus $1$. VIII, On conjectures of 
Birch and Swinnerton-Dyer,
{\em J. reine angew. Math.} {\bf{217}} (1965) 180--199

\bibitem[CaL]{CaL} 
J.W.S. Cassels,
{\em Lectures on elliptic curves}, 
LMSST {\bf{24}}, Cambridge University Press, Cambridge, 1991

\bibitem[CF]{CF} 
J.W.S. Cassels and A. Fr\"ohlich (Eds.), 
{\em Algebraic number theory},
Academic Press, London, 1967

\bibitem[CW]{CW}
J. Coates and A. Wiles,
On the conjecture of Birch and Swinnerton-Dyer,
{\em Invent. Math.} {\bf{39}} (1977) 223--251

\bibitem[Cr]{mwrank} 
J.E. Cremona, 
{\tt mwrank}, a program for performing 2-descent on elliptic
curves over $\Q$, \\
{\tt http://www.maths.nottingham.ac.uk/personal/jec/ftp/progs}

\bibitem[F]{F} 
T.A. Fisher, The Cassels--Tate pairing and the Platonic solids, {\em J.
Number Theory} {\bf98} (2003) 105--155

\bibitem[HB]{HB}
D.R. Heath-Brown, The solubility of diagonal 
cubic Diophantine equations, {\em Proc. London Math. Soc.} (3) {\bf{79}} 
(1999) 241--259

\bibitem[K]{K}
N. Koblitz, {\em Introduction to elliptic curves and modular forms},
GTM {\bf{97}}, Springer-Verlag, New York, 1993

\bibitem[M]{M}
B. Mazur,
On the passage from local to global in number theory, 
{\em Bull. Amer. Math. Soc.} {\bf{29}} (1993) 14--50

\bibitem[N]{N}
J. Nekov\'a\v r,
Class numbers of quadratic fields and Shimura's correspondence, 
{\em Math. Ann.} {\bf{287}} (1990) 577--594

\bibitem[S1]{S1}
E.S. Selmer, The diophantine equation $ax\sp 3+by\sp 3+cz\sp 3=0$,
{\em Acta Math.} {\bf{85}} (1951) 203--362

\bibitem[S2]{S2}
E.S. Selmer, The diophantine equation $ax\sp 3+by\sp 3+cz\sp 3=0$, 
completion of the tables,
{\em Acta Math.} {\bf{92}} (1954) 191--197

\bibitem[S3]{S3} 
E.S. Selmer, A conjecture concerning rational points on cubic curves,
{\em Math. Scand.} {\bf{2}} (1954) 49--54

\bibitem[Sil]{Si}
J.H. Silverman, 
{\em The arithmetic of elliptic curves},
GTM {\bf{106}}, Springer-Verlag, New York, 1986

\bibitem[Si1]{Sim1}
D. Simon, Computing the rank of elliptic curves over number fields,
{\em LMS J. Comput. Math.} {\bf{5}} (2002) 7--17

\bibitem[Si2]{Sim2}
D. Simon, {\tt ell.gp}, a program for calculating the rank of
elliptic curves over number fields,
{\tt http://www.math.unicaen.fr/$\sim$simon }

\bibitem[St]{St}
N.M. Stephens, The diophantine equation $X\sp{3}+Y\sp{3}=DZ\sp{3}$ 
and the conjectures of Birch and Swinnerton-Dyer,
{\em J. reine angew. Math.} {\bf{231}} (1968) 121--162

\bibitem[SwD]{SwD}
H.P.F. Swinnerton-Dyer, 
The solubility of diagonal cubic surfaces,
{\em Ann. Sci. \'Ecole Norm. Sup.} (4) {\bf{34}} (2001) 891--912

\end{thebibliography}

\bigskip
\noindent
Tom Fisher, \\
DPMMS, Centre for Mathematical Sciences, \\
Wilberforce Road, Cambridge CB3 0WB, UK \\
e-mail: T.A.Fisher@dpmms.cam.ac.uk \\
web: http://www.dpmms.cam.ac.uk/$\!\sim$taf1000 

\end{document}
