\documentclass[12pt]{article}
\include{macros}
\title{Visualizing Mordell-Weil Groups of Elliptic Curves 
       Using Shafarevich-Tate Groups}
\date{February 8, 2002 at {\sc University of Arizona}}
\usepackage{xypic}
\usepackage{fancybox}
\hoffset=-0.05\textwidth
\textwidth=1.1\textwidth
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\newcounter{myroman}
\newcommand{\rome}[1]{\setcounter{myroman}{#1}\roman{myroman}}


\begin{document}
\maketitle
\section{Introduction}
Today I will tell you about a construction of elements of
Shafarevich-Tate groups of abelian varieties~$A$ over~$\Q$.
$$\hspace{-1in}
\xymatrix@C=-7pc@R=6pc{
&{\shadowbox{\begin{minipage}[l]{2.6in}
A Construction of Elements of $\Sha(A)$
\end{minipage}}}\ar[dl]
\ar[dr]\\
{\shadowbox{\begin{minipage}[l]{3.4in}
\begin{center}\underline{Birch and Swinnerton-Dyer Conjecture}
\end{center}
\begin{itemize}
\item If $L(A,1)\neq 0$, then
$$ \#\Sha(A) \stackrel{?}{=} \frac{L(A,1)}{\Omega_A}\cdot 
       \frac{\#A(\Q)_{\tor}\cdot\#A(\Q)^{\vee}_{\tor}}
                 {\prod_{p\mid N} c_{A,p}}$$
Find $A$ in nature with conjecturally nontrivial $\Sha(A)$,
and prove that $\Sha(A)$ is as big as expected.
\item 
Construct $A$ such that $\Sha(A)$ is nontrivial,
then check that the BSD conjecture is not obviously
false for~$A$.
\item 
Find a method for connecting the rank conjecture
about elliptic curves to the rank~$0$ formula
for abelian varieties.
\end{itemize}
\end{minipage}
}}
& &
{\shadowbox{\begin{minipage}[l]{3.4in}
\begin{center}
\underline{What are the possibilities for $\#\Sha(A)$?}
\end{center}
{\bf Question (Poonen, 1999 at AWS).}\\
Stoll and Poonen proved that if $A$ is a Jacobian, then $\#\Sha(A)$ is a square
or twice a square.  If~$A$ is not a Jacobian, is $\#\Sha(A)$ always a square 
or twice a square?\vspace{1ex}\\
{\bf Conjecture (Me, today).}\\
Let $G$ be any finite abelian group (of odd order).  
Then there is an abelian variety $A$ such that
$\Sha(A) \ncisom G\cross H$, where
$\gcd(\#G, \#H)=1$. 
\end{minipage}}
}}
$$

\newpage\section{A Construction of Elements of $\Sha(A)$}
\begin{theorem}\label{thm}
Let~$E$ be an elliptic curve over~$\Q$,
and suppose $\chi:(\Z/\ell\Z)^*\ra \C^*$
is a Dirichlet character of prime modulus~$\ell\nmid N_E$ and
order~$n$ such that 
\begin{itemize}
\item $L(E,\chi^a,1)\neq 0$ for $a=1,\ldots, n-1$,
\item $\ds\gcd\left(n,\,\,2N_E\prod_{p\mid N_E}\#\Phi_E(\Fpbar)\right	) = 1$, and
\item $a_\ell \not\con \ell+1 \pmod{p}$ for all $p\mid n$.
\end{itemize}
Let~$K$ be the degree~$n$ abelian extension of $\Q$ corresponding
to~$\chi$.
Then there exists a $K$-twist~$A$ of $E^{\oplus (n-1)}$ of rank~$0$ 
such that $L(A,s) = \prod_{a=1}^{n-1} L(E,\chi^a,s)$
and $$E(\Q)/n E(\Q)\subset \Sha(A/\Q).$$
\end{theorem}

\begin{remark}
Note that $K$ is contained in the totally real subfield
$\Q(\mu_\ell)^+$ of $\Q(\mu_\ell)$ because the order
of $\chi(-1)$ divides the odd number~$n$.
\end{remark}

\begin{proof}[Sketch of Proof]
Let $R=\Res_{K/\Q}(E_K)$ be the Weil restriction of scalars 
of $E_K$ down to~$\Q$. For any $\Q$-scheme~$S$,
we have $R(S)=E_K(S\cross_\Q K)$, and 
as $\Gal(\Qbar/\Q)$-modules
$$
  R(\Qbar) = E(\Qbar \tensor K) \isom E(\Qbar)\tensor_{\Z} \Z[\Gal(K/\Q)],
$$
where $\tau\in \Gal(\Qbar/\Q)$ acts on 
$\sum P_\sigma\tensor \sigma \in E(\Qbar)\tensor_{\Z}\Z[\Gal(K/\Q)]$ by
$$\tau\left(\sum P_\sigma\tensor \sigma\right) = 
    \sum \tau(P_\sigma)\tensor \sigma\tau_{|K}.
$$
The $L$-series of~$R$ is $\prod_{a=1}^{n} L(E,\chi^a,s)$, and $R$ has
good reduction at all $p\nmid \ell\cdot N$.

Let $\Delta : E\hra R$ be the diagonal embedding, which sends~$P$ 
to $\sum_{\sigma\in\Gal(K/\Q)} P\tensor \sigma$, 
and let $\Sigma : R\ra E$ be the summation map, which sends
$\sum P_\sigma \tensor \sigma$ to $\sum P_\sigma$.
Note that both $\Delta$ and $\Sigma$ are defined over~$\Q$
and that $\Sigma\circ \Delta = [n]$.
If $A=\ker(\Sigma)$ then
$$
  A_{\Qbar} = \ker \left(+: E_{\Qbar}^{\oplus n} \ra E_{\Qbar}\right) 
             \isom E^{\oplus (n-1)},
$$
the isomorphism being the one that sends
$(P_1,\ldots,P_{n-1})$ to $(P_1,\ldots,P_{n-1},-(\sum P_i))$.
In particular,~$A$ is a twist of $E^{\oplus (n-1)}$.
We summarize this information in the following diagram:
\begin{equation}\label{eqn:d1}
\xymatrix@=3pc{
{E[n]}\ar[d] \ar[r] & {E}\ar@{^(->}[d]^{\Delta} \ar[r]^{[n]} & {E}\ar@{=}[d]\\
A \ar@{^(->}[r] & R \ar@{->>}[r]^{\Sigma} & E.
}
\end{equation}

Now pass to $\Q$-rational points in diagram (\ref{eqn:d1}) and
rearrange things to obtain the following diagram:
$$
\xymatrix@=3pc{
0 \ar[r]& {E(\Q)}\ar[r]^{[n]}\ar[d] & {E(\Q)}\ar@{=}[d] \ar[r] & {E(\Q)/n E(\Q)}\ar[r]\ar[d]^{\iota} & 0\\
0 \ar[r]& {R(\Q)/A(\Q)}\ar[r] & {E(\Q)}\ar[r]&{\ker(H^1(\Q,A)\ra H^1(\Q,R))}\ar[r] & 0.\\
}
$$
Here we have used that $E(\Q)[n]=0$, since $E[p]$ is irreducible for
$p\mid n$, and we've included the beginning of the long exact sequence of
Galois cohomology associated to $0\ra A\ra{}R\ra{}E\ra0$.
Using the snake lemma, we see that~$\iota$ is surjective and has kernel
a subgroup of $R(\Q)/(A(\Q)+E(\Q))$.  
One can use that $a_\ell\not\con \ell+1\pmod{p}$ for any $p\mid n$ and that
$A(\Q)$ is finite (which follows from Kato's Euler system work!)  to
show that $R(\Q)/(A(\Q)+E(\Q))$ contains no $p$-torsion for $p\mid n$,
hence $\ker(\iota)=0$.  

To show that the image of~$\iota$ lies in the
subgroup $\Sha(A/\Q)$ of $H^1(\Q,A)$, uses that $\gcd(n,2\cdot
N_E\cdot c)=1$, where~$c$ is the product of all Tamagawa numbers
of~$E$ and~$A$.  These last steps are fairly technical and use some
nontrivial machinery.
(That~$n$ is odd is only used to show that~$\iota$ 
maps into $\Sha(A/\Q)$.)
\end{proof}

\section{Data Collection}
Next we collect some data that both gives evidence for the Birch and
Swinnerton-Dyer conjecture and for my conjecture that if~$G$ is an
abelian group then there is an abelian variety~$A$ such that
$\Sha(A)\ncisom G\cross H$ with $\gcd(\#H,\#G)=1$.
We will always choose~$E$ below so that $N_E$ is prime,~$E$ 
is isolated in its isogeny class (hence $\rho_{E,p}$ is 
surjective for all~$p$), and $c_{E,p}=1$ for all $p\mid N$.

Let $\#\Shaan(A)^*$ denote the prime-to-$2\ell$ part of
$$
 \frac{L(A,1)}{\Omega_A}\cdot 
        \frac{\#A(\Q)_{\tor}\cdot\#A^{\vee}(\Q)_{\tor}}
            {\prod_{p\mid \ell N_E} c_{A,p}}.
$$
Remark 5.4 of Edixhoven's {\em N\'eron models and tame ramification} can
be used to show that 
$$
 \Phi_{A,\ell}(\Fbar_\ell)=E(\Fbar_\ell)[n]\ncisom (\Z/n\Z)^2,
$$
so $c_{A,\ell}=1$, since $E(\F_\ell)[p]=0$ for all $p\mid n$.
Since~$K$ is only ramified at~$\ell$
and the formation of N\'eron models commutes with unramified base change,
$c_{A,p} = c_{E,p}^{n-1}=1$ for $p\mid N_E$.
Since $A(\Q)\subset A(K) \ncisom E(K)^{\oplus (n-1)}$, 
and $E(K)_{\tor}=0$ (since all $\rho_{E,p}$ are surjective), 
we have $\#A(\Q)_{\tor} = \#A^{\vee}(\Q)_{\tor}=1$.
I think (but have not proven, yet!) that
$$
 \Omega_{A/\Q} = \left(\frac{1}{\sqrt{\ell}}\cdot \Omega_{E/\Q}\right)^{n-1}.
$$
To prove this, it would (mostly) suffice to show that
$
 \Omega_{A/K} = \Omega_{A/\Q}^n\cdot \ell^{\binom{n}{2}},
$
where $\binom{n}{2} = n(n-1)/2$.   Assume this formula for
$\Omega_{A/\Q}$, we can very quickly compute $\Shaan(A)^*$ using
modular symbols.

The elliptic curves {\bf 61A} of rank~$1$, {\bf 389A} of rank~$2$, and
{\bf 5077A} of rank~$3$ each have prime conductor, trivial torsion subgroup,
and Tamagawa number~$c_p=1$.  In the table below,
$p_d$ denotes a $d$-digit prime number (where $d$ is written in
Roman numerals), and
a $-$ means that some hypothesis of Theorem~\ref{thm} is {\em not} satisfied.
(This table took under ten minutes to compute on a Pentium III 933.)

$$\begin{array}{|c|c||c|c|c|}\hline
n & \ell & \#\Shaan^*\text{ for }{\bf 61A}& \#\Shaan^*\text{ for }{\bf 389A} & \#\Shaan^*\text{ for }{\bf 5077A}\\\hline\hline
3 & 487 & 3& 3^{4}& 3^{3}\\\hline
9 & 487 & 3^{2} \!\cdot\! 19^{2}& 3^{8}& 3^{6} \!\cdot\! 17^{2}\\\hline
27 & 487 & 3^{3} \!\cdot\! 19^{2} \!\cdot\! p_{\rome{6}}^{2}& 3^{12} \!\cdot\! 163^{2}& 3^{9} \!\cdot\! 17^{2} \!\cdot\! 433^{2} \!\cdot\! p_{\rome{6}}^{2}\\\hline
81 & 487 & 3^{4} \!\cdot\! 19^{2} \!\cdot\! p_{\rome{4}}^{2} \!\cdot\! p_{\rome{6}}^{2} \!\cdot\! p_{\rome{7}}^{2}& 3^{16} \!\cdot\! 163^{2} \!\cdot\! p_{\rome{19}}^{2}& 3^{12} \!\cdot\! 17^{2} \!\cdot\! 433^{2} \!\cdot\! p_{\rome{4}}^{2} \!\cdot\! p_{\rome{5}}^{2} \!\cdot\! p_{\rome{6}}^{2} \!\cdot\! p_{\rome{7}}^{2} \!\cdot\! p_{\rome{9}}^{2}\\\hline
5 & 251 & 5& 5^{2}& -\\\hline
25 & 251 & 5^{2} \!\cdot\! 151^{2} \!\cdot\! p_{\rome{5}}^{2}& 5^{4} \!\cdot\! 149^{2} \!\cdot\! p_{\rome{4}}^{2}& -\\\hline
125 & 251 & 5^{3} \!\cdot\! 151^{2} \!\cdot\! p_{\rome{5}}^{2} \!\cdot\! p_{\rome{18}}^{2}& 5^{6} \!\cdot\! 149^{2} \!\cdot\! p_{\rome{4}}^{2} \!\cdot\! p_{\rome{5}}^{2} \!\cdot\! p_{\rome{10}}^{2} \!\cdot\! p_{\rome{11}}^{2}& -\\\hline
7 & 197 & 7 \!\cdot\! 29^{2}& 7^{2} \!\cdot\! 13^{4}& 7^{3}\\\hline
49 & 197 & 7^{2} \!\cdot\! 29^{2} \!\cdot\! p_{\rome{10}}^{2}& 7^{4} \!\cdot\! 13^{4} \!\cdot\! p_{\rome{9}}^{2}& 7^{6} \!\cdot\! p_{\rome{4}}^{2} \!\cdot\! p_{\rome{4}}^{2} \!\cdot\! p_{\rome{5}}^{2}\\\hline
11 & 89 & 11 \!\cdot\! 67^{2}& 11^{2}& 11^{3} \!\cdot\! 67^{2}\\\hline
13 & 53 & 13& 13^{2}& -\\\hline
17 & 103 & 17 \!\cdot\! 613^{2}& 17^{2} \!\cdot\! 101^{2}& 17^{3} \!\cdot\! 67^{2}\\\hline
19 & 191 & 19 \!\cdot\! 37^{2}& 19^{2}& 19^{5} \!\cdot\! 37^{2}\\\hline
\end{array}
$$

The BSD conjecture and this table (and my ``conjecture'' about $\Omega_A$) imply that
for the integers $n$ in the first column of the table, there is an~$A$ 
such that
$$\Sha(A) \ncisom  (\Z/n\Z)\cross H$$
with $\gcd(n,\#H)=1$.  This is evidence for Conjecture 1, and also gives lots of examples
to show that $\#\Sha(A)$ is neither a square or twice a square in general.

\hd{Challenge:}  {\em Let $E$ be one of the curves considered in the table, let $r$ be its rank,
and notice that in the table $n^r\mid \#\Shaan^*$.  
The BSD conjecture predicts that this divisibility should always hold.   Prove that it
does for infinitely many $\ell$.}

\end{document}