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\title{Shafarevich-Tate Groups of Higher Rank\\Elliptic Curves}
\author{William Stein}
\date{May 15, 2011}
\begin{document}
\maketitle
\begin{abstract}
These are the notes for a talk I gave at the conference on BSD in
Cambridge, UK in May 2011. The talk is about some approaches to
trying to prove something about finiteness of Shafarevich-Tate
groups of elliptic curves over $\Q$ of rank at least $2$.
\end{abstract}
\section{Finiteness of $\Sha(E/\Q)$}
Consider elliptic curves $E$ and primes $p$. Let
\begin{equation}\label{eqn:X}
X=\{(E,p) : \rank(E)\geq 2\text{ and }\Sha(E/\Q)(p)\text{ is finite}\}.
\end{equation}
Since Shafarevich and Tate conjectured that $\Sha(E/\Q)$ is finite for
all $E$, we have the following much weaker but still open conjecture:
\begin{conjecture}
$X$ is infinite.
\end{conjecture}
\begin{theorem}[M.~Bhargava, Wei Ho, --]
Assume one of the following conditions holds:
\begin{enumerate}
\item $\#\Sha(E/\Q)[2]$ is a perfect square for all $E$,
\item $\#\Sha(E/\Q)[3]$ is a perfect square for all $E$,
\item $\rank(E) \equiv r_{\an}(E) \pmod{2}$ for all $E$.
\end{enumerate}
Then $X$ is infinite.
\end{theorem}
\begin{proof}
Use the method of Bhargava applied to a certain family of curves
with either 1 or 2 marked points. The details are substantial.
\end{proof}
Next suppose we fix a particular $E$ and consider the set
$$
X_E = \{p : \Sha(E/\Q)(p) \text{ is finite}\},
$$
and the subset
$$
X_E^0 = \{p : \Sha(E/\Q)(p) = 0\}.
$$
If we wish to show that either of these sets is infinite, there are at
least two directions in which we can go: (1) $p$-adic $L$-series and
Iwasawa theory, and (2) Heegner point methods. First we consider
$p$-adic methods.
\subsection{$p$-adic Methods}
%The use of $p$-adic methods goes back to Mazur et al.'s introduction
%of $p$-adic analogues of the Birch and Swinnerton-Dyer conjecture.
Much theoretical and computational work on $p$-adic heights and
Iwasawa theory has made explicit application of $p$-adic analogues of
the Birch and Swinnerton-Dyer conjecture much more effective in
particular cases. For example:
\begin{theorem}[Coates-Sujatha-Liang]
Let $E$ be the CM elliptic curve $y^2=x^3 - 82x$, which has rank $r=3$.
Then
$$
X_E^0 \supset \{ p : p < 30000\text{ is prime}, \quad p\equiv 1\pmod{4}, \quad p \neq 41\}.
$$
\end{theorem}
The following ``theorem'' is an enormous computation that is currently only 99.9999...\% done.
\begin{theorem}[Stein-Wuthrich \cite{stein-wuthrich}]
For every non-CM elliptic curve $E$ with $N_E\leq 30000$ and $r_E\geq 2$, we have
$$
X_E^0 \supset \{ p : 5 \leq p \leq 1000,\quad p\text{ is good ordinary, and } \rho_{E,p}\text{ is surjective}\}.
$$
\end{theorem}
Let $E$ be the rank 2 elliptic curve 389a, which is the unique curve
of conductor 389. Using optimized code in Sage (and Psage), the
calculation of the theorem takes a total of 4 minutes: 2 minutes to
compute $p$-adic $L$-functions, and $2$ minutes more to compute
$p$-adic regulators. The above gives:
\begin{proposition}\label{prop:389a_padic}
$$
X_E^0 \supset \{ 5 \leq p < 1000 : p \neq 107, 599\}.
$$
\end{proposition}
I personally see {\em no hope} that I can extend this $p$-adic
strategy to prove something about infinitely many $p$, or even about
infinitely many pairs $(E,p)$.
At least, from the above we know that the set $X$ of Equation~\eqref{eqn:X} has
cardinality at least one million.
\subsection{Heegner Points}
The second approach is to use Heegner points, which is a strategy
pioneered by Kolyvagin in the 1980s. The following theorem, which is
stated in the abstract of one of his first papers on Euler systems,
is notable because it does not rely on any other deep theorems
such as Gross-Zagier, modularity, nonvanishing of twists, etc.; those
deep results are needed only to verify the hypothesis of the argument
are satisfied by any curve with analytic rank at most $1$.
\begin{theorem}[Kolyvagin]
Let $E$ be the elliptic curve $X_0(17)$. Then
$\Sha(E)$ is finite.
\end{theorem}
The proof of this theorem involves constructing Heegner points on
$X_0(17)$, and using them to define a set of elements of $\H^1(K,
E[p^n])$ for every prime power $p^n$, and finally using the elements
to bound $\Sha(E)(p)$ for all $p$.
I have not yet given up all hope that someday we will be able to
extend some argument like this to one specific elliptic curve of rank
$2$. The rest of this talk is about some details in this direction.
\section{Heegner Points on 389a}
For the rest of this article, let $E$ be the elliptic curve 389a,
which has rank $2$ with generators $P=(-1,1)$ and $Q=(0,0)$.
\begin{center}
\includegraphics[width=.6\textwidth]{389a}
\end{center}
The torsion subgroup of $E$ is trivial; moreover, the mod~$p$ representations
$E[p]$ are surjective, for all $p$.
The curve $E$ has modular degree $40$:
$$
\phi:X_0(389) \to E, \qquad\qquad \deg(\phi)=40.
$$
Also, let $K=\Q(\sqrt{-7})$, which has class number $1$ and satisfies the
Heegner hypothesis since $N=389$ is split in $K$.
We have $r_{\an}(E/K) = 3$.
Fix one of the two
choices $\cN$ of primes over $389$, so we have $\cO_K/\cN \isom
\Z/N\Z$. The Heegner point
$$
x_1 = (\C/\cO_K, \cN^{-1}/\cO_K) \in X_0(N)(K)
$$
maps to $0$ in $E$:
$$
y_1 = \phi(x_1) = 0 \in E(K).
$$
That $y_1$ is torsion follows from the Gross-Zagier formula, since $r_{\an}(E/K)>1$.
Since $y_1=0$, we are motivated to consider higher Heegner points $y_{p}$, which
are indexed by the primes $p$ that are inert in $K$, which are the primes
congruent to $3,5,6\pmod{7}$:
$$
\cI = \{\text{ inert primes }\} = \{3,5,13,17,19,31, 41, \ldots\}.
$$
For any $p\in \cI$, let $\cO_p = \Z + p\cO_K$ be the order of conductor $p$
and $\cN_p = \cN \cap \cO_p$.
We have a Heegner point
$$
x_p = (\C/\cO_p, \cN_p^{-1}/\cO_p) \in X_0(N)(K_p)
$$
where $K_p$ is the ring class field associated to $p$ (we also will sometimes write $K_1$ for the Hilbert
class field of $K$, which is just $K$ for our example); thus
$$
\Gal(K_p/K_1) \isom (\cO_K/ p \cO_K)^*/(\Z/p\Z)^*
$$
is cyclic of order $p+1$, and $K_p$ is totally ramified at $p$.
Finally, let
$$
y_p = \phi(x_p) \in E(K_p).
$$
\begin{proposition}\label{prop:ypoo}
The point $y_p$ has infinite order for {\em all} $p\in \cI$.
\end{proposition}
\begin{proof}
Using Galois theory and that $\rho_{E,p}$ is surjective, one can show that $E(K_p)_{\tor} = 0$,
so if $y_p$ has finite order, then $y_p=0$. Since $0 \in E(\Q)$ the fiber
$F = \phi^{-1}(0) \in X_0(N)(\Qbar)$ is closed under the action of $G_{\Q}$.
Since $\phi(x_p) = 0$, we have that $x_p \in F$. But by CM theory, there
is a simply transitive group action of $\Gal(K_p/K)$ on the $\Gal(K_p/K)$-orbit
of $x_p$, so $\#F \geq p+1$. We thus have
$$
p+1 \leq \#\phi^{-1}(0) \leq \deg(\phi) = 40,
$$
so $p \leq 39$.
To finish the proof we check that $y_p\neq 0$ for each inert prime
$p\leq 39$, via either of the following two approaches:
\begin{enumerate}
\item {\bf Numerical:} We use the classical complex analytic approach to numerically
computing Heegner points and find, e.g., that
$$
y_3 \sim (.63-.73i, -.47+1.39i) \neq 0.
$$
It takes only moments to compute all the other relevant $y_p$.
\item {\bf Algebraic:} This method is vastly more complicated and much
slower than the numerical method above, but it is purely algebraic
and generalizes to allow us to later verify nontriviality of certain
cohomology classes. The main idea is to simply reduce everything
modulo a prime $\lambda$ of $\Zbar$ over an inert prime, and use
rational quaternion algebras to make each object computable directly
modulo $\lambda$, hence avoiding any characteristic $0$
computations. Supposing, for example, that $\lambda\mid 5$, we have
the following commuting diagram of abelian groups:
$$
\xymatrix{
& {X_0(N)(K_p)}\ar[dl]\ar[dr] & \\
{\Div(X_0(N)(\F_{5^2})^{\rm ss})\tensor \F_3}\ar[dr]^{\pi} & & E(K_p)\ar[dl]\\
& E(\F_{5^2}) \tensor \F_3 &
}
$$
We compute the divisor group as the free abelian group on the set of right
ideal classes in an Eichler order of level $N$ in the rational
quaternion algebra ramified at $5$ and $\infty$. We then compute
$\overline{x}_1 = x_1 \pmod{\lambda}$ by finding (using ternary
quadratic forms as suggested by \cite{jetchev-kane:equi}) a right
ideal class $[I]$ such that the left order $R_I$ contains $\cO_K$. We
use that $T_p(\overline{x}_1) = \sum_{\sigma \in \Gal(K_p/K_1)}
\sigma(x_p)\pmod{\lambda}$ to find all conjugates of $x_p$. We compute
the map $\pi$ in the diagram only up to scaling by an automorphism by
using that it is $\T$-invariant and surjective (slight generalization
of Ihara's lemma). For more details, see \cite{stein:kolyconj2}.
\end{enumerate}
\end{proof}
\begin{remark}
For a generalize of the above proposition, see \cite[\S3]{jetchev-lauter-stein}.
\end{remark}
Though it is exciting that $y_p$ has infinite order for all inert primes $p$, we must
temper our enthusiasm with the fact that
$$
\Tr_{K_p/K_1}(y_p) = a_p(E) y_1 = 0 \in E(K),
$$
so that the $y_p$ do not provide a direct source of nonzero elements of $E(K)$.
\section{Selmer Elements for 389a: Kolyvagin's Idea}
For every positive integer $n$, the (classical) $n$-Selmer group over a field $M$ sits in the exact sequence
$$
0 \to E(M)/n E(M) \xrightarrow{\delta} \Sel^{(n)}(E/M) \to \Sha(E/M)[n] \to 0.
$$
Also, we view $\Sel^{(n)}(E/M)$ as a subgroup of the first Galois
cohomology group $\H^1(M, E[n])$.
For any prime $p\in \cI$, let $n_p = \gcd(p+1, \#E(\F_p))$. By the
Chebotarev density theorem, we can make $n_p$ divisible by any power
of any prime that we want. We have the following table of values of
$n_p$ for the first few $p$, and our elliptic curve 389a:
\begin{center}
\begin{tabular}{|l|llllllllllllll|}\hline
$p$ & 3 & 5 & 13 & 17 & 19 & 31 & 41 & 47 & 59 & 61 & 73 & 83 & 89 & 97 \\\hline
$n_p$ & 2 & 3 & 1 & 6 & 5 & 4 & 3 & 2 & 3 & 2 & 1 & 12 & 2 & 1\\\hline
\end{tabular}
\end{center}
Fix any choice of generator $\sigma_p$ of $\Gal(K_p/K_1)$, and let
\begin{equation}\label{eqn:zp}
z_p = \sum_{i=1}^{p} i \sigma_p^i(y_p) \in E(K_p).
\end{equation}
We emphasize that $z_p$ depends on the choice of generator.
From $z_p$ we obtain a class $\tau_p$ in the Selmer group as follows:
$$
\xymatrix{
{[z_p] \in (E(K_p)/n_p E(K_p))^{\Gal(K_p/\Q)}}\ar[r] & {\H^1(K_p,E[n_p])^{\Gal(K_p/\Q)}} \\
& {\H^1(\Q,E[n_p])}\ar[u]_{\isom}\\
& {\tau_p \in \Sel^{(p)}(E/\Q)}\ar[u]
}
$$
(If we replace $\sigma_p$ by a different choice, then $\tau_p$ is scaled by a unit.)
That $\tau_p$ is in the Selmer group, and not just $\H^1$ uses that $r_{\an}(E/\Q)>1$,
as explained in \cite[Prop.~6.2]{gross:kolyvagin}.
\begin{theorem}[Kolyvagin]\label{thm:koly}
If a prime $\ell$ divides $\order(\tau_p)$ for some inert prime $p$,
then
$
\Sha(E/\Q)(\ell)
$
is finite. If
$\ord_{\ell}(\order(\tau_p)) = \ord_{\ell}(n_p) \geq 1$, then
$\Sha(E/\Q)[\ell]=0$.
\end{theorem}
The following is implied by conjectures in
\cite{kolyvagin:structure_of_selmer}.
We emphasize that here we are only making a conjecture about the curve 389a.
\begin{conjecture}[Kolyvagin]\label{conj:koly}
For every prime $\ell$ there is some $p\in \cI$ with $\ell\mid \order(\tau_p)$.
\end{conjecture}
In fact, because $\Sha(E/K)_{\an}=1$, we expect more strongly that for
every prime $\ell$, there is a $p\in \cI$ such that $\ord_{\ell}(\order(\tau_p)) =
\ord_{\ell}(n_p) \geq 1$.
As evidence for the conjecture, the paper \cite{jetchev-lauter-stein}
gave numerical evidence that $\tau_5$ has order $n_5=3$; however, we
emphasize that this was only evidence, since the method of that paper
was not made rigorous (in theory it could have been wrong due to
insufficient numerical precision). As more evidence, the paper
\cite{stein:kolyconj2} proves, using a generalization of the algebraic
construction of the proof of Proposition~\ref{prop:ypoo} above, that
many classes $\tau_p$ are nonzero (for a few dozen triples $(E,K,p)$,
for various $E$); that computation also led to some interesting
results about how the images of the classes $\tau_p$ are
distributed in $\Sel^{(\ell)}(E/K)$, which will appear in a future
paper of Stein-Weinstein.
We have the following table, where the three $?$ cases mean that the
indicated computation is not 100\% certain yet.
\begin{center}
\begin{tabular}{|c|c|c|c|}\hline
$p$ & $n_p$ & $\order(\tau_p)$ & $\tau_p$ (up to scalar!)\\\hline
3 & 2 & 2 & $\delta(Q)$\\
5 & 3 & 3 & $\delta(P+Q)$\\
13 & 1 & 1 & $\delta(0)$\\
17 & 6 & $1$? & $\delta(0)$\\
19 & 5 & 5 & $\delta(P)$\\
31 & 4 & $4$? & $\delta(Q)$\\
41 & 3 & 3 & $\delta(P+2Q)$\\
$\cdots$ & $\cdots$ & $\cdots$ & $\cdots$ \\
419 & 35 & $35$? & ?\\\hline
\end{tabular}
\end{center}
Applying Kolyvagin's Theorem~\ref{thm:koly}, we have
\begin{proposition}\label{prop:389a_heegner}
$\{3,5\} \subset X_E^0$.
\end{proposition}
This is much less impressive than Proposition~\ref{prop:389a_padic},
but there is (in my opinion) vastly more hope that an approach using
Heegner points could eventually generalize to show that $X_E$ is
infinite.
\section{A Higher Rank Gross-Zagier Formula}
We continue to let $E$ be the elliptic curve 389a, but allow the quadratic
imaginary field $K$ to vary. We require only that $389$ split in $K$ and
$r_{\an}(E/K)=3$, so $D=-7,-11,-19,-20,-24,-35,\ldots$ are all allowed, but
$D=-264$ is not (since $r_{\an}(E/K)=5$).
Let $\ell\in \cI$ be any inert prime and fix $\lambda\mid \ell$ a prime in $\Zbar$.
Let $p$ be an inert prime with $\# E(\F_{\ell})\mid n_p$.
Recall the point $z_p$ from \eqref{eqn:zp}, and let $\overline{z}_p$ be the image
of $z_p$ under the map $E(K_p) \to E(\F_{\ell^2})$ got by reduction modulo $\lambda$.
In fact, $z_p$ lands in $E(\F_{\ell})$, because
$z_p$ is fixed by complex conjugation.
Since $E(\F_{\ell})/ n_p E(\F_{\ell}) = E(\F_{\ell})$, the point $z_p$ does
not depend on the choice of $\lambda$ (though it does depend up to a unit scaling on
the choice of $\sigma_p$).
Let $\pi_\ell:E(\Q) \to E(\F_{\ell})$ be the reduction modulo $\ell$ homomorphism.
\begin{definition}
Let
$$
W_{\ell} = \pi_{\ell}^{-1}(\langle \overline{z}_p :\text{ all such }p\rangle) \subset E(\Q).
$$
\end{definition}
It is trivial that $W_{\ell}$ is a finite index subgroup of $E(\Q)$.
I make the following conjecture about the subgroups $W_{\ell}$, which would
imply Conjecture~\ref{conj:koly}.
\begin{conjecture}
The set of indexes $\{[E(\Q):W_{\ell}] : \ell \in \cI\}$ is bounded.
\end{conjecture}
The following conjecture is then a higher-rank analogue of the Gross-Zagier formula:
\begin{conjecture}
If $W_{\ell}$ has maximal index, then $[E(\Q):W_{\ell}] = \sqrt{\#\Sha(E/K)}$, and moreover
$$
\frac{L^{(3)}(E/K,1)}{3!} = \Omega_{E/K} \cdot \Reg(W_\ell) \cdot \Reg(E^D(\Q)),
$$
up to a power of $2$. Here $\Omega_{E/K} = 2\Vol(\C/\Lambda)/\sqrt{|D|}$.
\end{conjecture}
\begin{quote}
``It is always a good idea to try to prove true theorems.'' -- Bryan Birch
\end{quote}
\bibliography{biblio}
\end{document}