\documentclass[11pt]{article} \hoffset=-0.05\textwidth \textwidth=1.1\textwidth \newcommand{\bndone}{87} \newcommand{\bndzero}{19} \bibliographystyle{amsalpha} \usepackage[all]{xy} \include{macros} \newcommand{\extra}[1]{} \title{Verifying the Full Birch and Swinnerton-Dyer Conjecture for Specific Elliptic Curves of Analytic Rank $0$ or $1$} \author{Stephen Donnelly and Andrei Jorza and Stefan Patrikis and William A. Stein and Michael Stoll\footnote{Some of these authors might not even know they are authors, so don't blame them for any mistakes below. Blame William Stein.}} \begin{document} \section{Introduction} For an elliptic curve $E$ defined over a number field $K$ the following diagram defines the Selmer and the Shafarevich-Tate groups as $Sel(E/K)_p=\ker f,\Sha(E/K)=\ker g$: \[\xymatrix{ 0\ar[r] & (E(K)/pE(K)) \ar[r] & H^1(G(\bar{K}/K), E[p]) \ar[r]\ar[d]_{Res}\ar[rd]^f & H^1(G(\bar{K}/K), E)[p] \ar[d]^{g_p} &\\ & & H^1(G(\bar{K_\lambda}/K_\lambda), E[p]) & \prod_{v} H^1(G(\bar{K_v}/K_v), E)[p] &\\ }\] Recall that $H^1(K, E)= \varinjlim H^1(L/K, E)$, where the direct limit is taken over all finite Galois extensions $L/K$; since each one of these groups is torsion, (killed by $[L:K]$,) so is $H^1(K, E)$. Therefore, $\Sha(E/K)$ is a torsion group, a fact we will later to use to conclude that $\Sha(E/K)=0$ for curves having $\Sha(E/K)[p]=0$ for all primes $p$. In what follows, we will obtain results of the form $\Sha(E/K)[p]=0$ for almost all $p$, where $[K:\Q]=2$. The canonical map $\Sha(E/\Q) \rightarrow \Sha(E/K)$ has kernel contained in $H^1(K/\Q, E)$, a finite abelian $2$-group, so $\Sha(E/K)[p]=0 \Rightarrow \Sha(E/\Q)[p]=0$ as long as $p \neq 2$. The difficulty in studying the Shafarevich-Tate group is the major obstacle to progress on the Birch and Swinnerton-Dyer conjecture: \begin{conjecture}[Birch-Swinnerton-Dyer] Let $E$ be an elliptic curve defined over $\Q$. Then the order of vanishing at $s=1$ of $L(E/\Q, s)$equals the rank of $E(\Q)$. More precisely, letting $r= \hbox{rk}_\Z E(\Q)$, and phrasing the conjecture in terms of $\# \Sha(E/\Q)$, \[\# \Sha(E/\Q)= \frac{L^r(E/\Q,1) |E(\Q)_{tors}|^2}{r! R(E/\Q) \prod_v c_v(E)}\] $R(E/\Q)$ denotes the elliptic regulator, the determinant of the canonical height pairing matrix on a set of free generators. The $c_v=[E(\Q_v):E_0(\Q_v)]$ for finite places are the Tamagawa numbers, measuring bad reduction at $v$ (in particular, they are $1$ when $E$ has good reduction at $v$), and $c_{\infty}= \int_{E(\R)}|\omega|$, where $\omega$ is the invariant differential $\frac{dx}{2y+a_1x+a_3}$ attached to a global minimal Weierstrass equation. \end{conjecture} Note that a theorem of Cassels (see \cite{MR31:3420}) asserts that, assuming the Shafarevich-Tate group is finite, the full Birch and Swinnerton-Dyer conjecture is invariant under isogeny. For elliptic curves of analytic rank at most 1, Kolyvagin showed that $\Sha(E/\Q)$ is finite; for these curves it is enough to check the full BSD conjecture for one curve in the isogeny class. \section{Derivatives of $L$-functions} \subsection{Gross-Zagier} One of the factors in the BSD formula is the derivative of the $L$-function. For $K=\Q(\sqrt{-D})$ a quadratic extension, we define $L(E/K,s)=L(E/\Q,s)L(E^D/\Q,s)$, where $E^D$ is the elliptic curve twisted by the quadratic character associated with $D$. If the newform of $E$ is $f$, then the newform of $E^D$ is $f\otimes \left(\frac{\cdot}{D}\right)$. On the level of Weierstrass equations, if $E$ has equation $y^2=x^3+ax+b$, then $E^D$ has equation $y^2=x^3+D^2ax+D^3b$. We say that $K$ satisfies the Heegner hypothesis for the elliptic curve $E/\Q$ of conductor $N$ if all prime factors of $N$ split in $K$. This allows the construction of a Heegner point $y_K$ on $E/K$. The following limit formula was proved by Gross and Zagier (\cite{gross-zagier}) \begin{theorem}[Gross-Zagier]If $(D,2N)=1$ then \[\hat{h}(y_K)=\frac{u^2\sqrt{D}}{c\int_{E(\mathbb{C})}\omega\wedge \overline{i\omega}}L'(E/K,1)=\alpha L'(E/K,1),\](where $\omega$ is the invariant differential associated with the elliptic curve and $u$ is half the number of units of $\mathcal{O}_K$). \end{theorem} The following decomposition theorem relates the behavior of the $L$-function for $E/K$ and $E/\Q$: \begin{proposition} \[E(K)\otimes\Z\left[\frac{1}{2}\right]=E(\Q)\otimes\Z\left[\frac{1}{2}\right]\oplus E^D(\Q)\otimes\Z\left[\frac{1}{2}\right].\] \end{proposition} \proof Denote complex conjugation by $\tau$. We will decompose $E(K)$ into its eigenspaces under the action of $\tau$. Note that tensoring with $\Z\left[\frac{1}{2}\right]$ kills 2-torsion. For $P\in E(K)$ we have the following decomposition into +1 and -1 eigenspaces of $\tau$: \[P= \frac{1+\tau}{2}P + \frac{1-\tau}{2}P.\] But if $P=(x,y)\in E(K)$ satisfies $\tau P=P$, then $P\in E(\Q)$; if $\tau P=-P= (x, -y)$, then $x\in \Q$ and $y\in \sqrt{D}\Q$. In particular, $(Dx, D\sqrt{D}y)\in E_D(\Q)$, and conversely we can obtain any such point in the $-1$ eigenspace of $E(K)$ from a point of $E_D(\Q)$. We may rewrite the decomposition into eigenspaces as the statement of the proposition. \endproof Assume that $L'(E^D/\Q,1)\neq 1$. For elliptic curves of rank 0 or 1 (the only ones for which we will check the full BSD conjecture) this implies that $E/K$ has rank 1 and exactly one of the curves $E/\Q,E^D/\Q$ has rank 1. Therefore the Heegner point $y_K$ will have infinite order. The parity of the root number (the sign of the functional equation of the $L$-function) is the same as the parity of the rank of $E/\Q$. \begin{enumerate} \item If the root number is $-1$ then a generator of $E(K)\otimes\Z\left[\frac{1}{2}\right]$ comes from a generator of $E(\Q)\otimes\Z\left[\frac{1}{2}\right]$ and \[L'(E/K,1)=L'(E/\Q,1)L(E^D/\Q,1).\] \item If the root number is $+1$ then a generator of $E(K)\otimes\Z\left[\frac{1}{2}\right]$ comes from a generator of $E^D(\Q)\otimes\Z\left[\frac{1}{2}\right]$ and \[L'(E/K,1)=L(E/\Q,1)L'(E^D/\Q,1).\] \end{enumerate} Note that this method is faster than computing the rank directly. \subsection{Index of Heegner points} The Birch and Swinnerton-Dyer conjecture may be rephrased \cite{mccallum}, using the Gross-Zagier formula as \begin{conjecture}[Birch-Swinnerton-Dyer]\label{bsd-heegner}For $E$ an elliptic curve of rank 1 over $K$, a quadratic extension of $\Q$, such that the Heegner points $y_K$ has infinite order, we have \[|\Sha(E/K)|=\left(\frac{[E(K):\Z y_K]}{c\prod c_p}\right)^2.\] \end{conjecture} We will use Kolyvagin's approach to computing torsion of $\Sha(E/\Q)$. For this we need to compute $[E(K):\Z y_K]$ efficiently. Note that $[E(K):\Z y_K]^2=h(y_K)/h(z)$, where $z$ is a generator of $E(K)$. We saw that we may compute $z$ up to a power of 2, which implies that we may compute $h(z)=2h_\Q(z)$ up to a power of 2. All that is left is a computation of $L$-functions and of $\alpha = \frac{u^2 \sqrt{|D|}}{c\int_{E(\mathbb{C})}\omega\wedge \overline{i\omega}} $. We may compute \begin{eqnarray*}\int_{E(\mathbb{C})}\omega\wedge \overline{i\omega}&=& \int_{E(\mathbb{C})}\frac{dx}{y}\wedge \overline{i\frac{dx}{y}}= i\int_{\mathbb{C}/\Lambda}\wp'(z)\frac{dz}{\wp'(z)}\wedge \overline{\wp'(z)\frac{dz}{\wp'(z)}}\\ &=& i\int_{\mathbb{C}/\Lambda}dz\wedge d\bar{z}=\int (dx+idy)\wedge (idx+dy)=2\int dx\wedge dy=2a, \end{eqnarray*} where $a$ is the volume of the lattice $\mathbb{C}/\Lambda$. \subsection{Efficient Computations} In the course of our computations of $[E(K):\Z y_K]=\sqrt{h(y_K)/h(z)}$ we need to compute the heights with sufficient precision to obtain the square of $[E(K):\Z y_K]$ after rounding. Assume that the rank of $E$ is 0, which is the case for most of the curves in our databases. Then $[E(K):\Z y_K]^2=h(y_K)/h(z)=\alpha L(E/\Q,1)L'(E^D/\Q,1)/h(z)$. \begin{enumerate} \item To find $\alpha$ all we need to do is find the area of the period lattice, which can be done with fast convergence (\cite{cremona-algorithms}, section 3.7). \item To find a generator $z$ of $E/\Q$ we use Cremona's mwrank (\cite{cremona-mwrank}). \item The computation of the height of the generator of $E/\Q$ is detailed in \cite{cremona-algorithms} (section 3.4). Moreover, the truncation error is exponentially small; there is an explicit bound on truncation that ensures an error of at most $10^{-k}/2$. \item Using \cite{cremona-algorithms} section 2.13 we get that \[L(E/\Q,1)=2\sum\frac{a_n}{n}e^{-2\pi n/\sqrt{N}},\](where $a_n$ are the Fourier coefficients of the normalized newform associated with $E$). We have the trivial bound $a_n\leq n$ and so the truncation error is $2\sum e^{-2\pi n/\sqrt{N}}=2/(1-e^{-2\pi/\sqrt{N}}) e^{-2\pi k/\sqrt{N}}$. \item If $F$ is an elliptic curve of rank 1 and conductor $N$, then a similar formula in the same section yields that \[L'(F/\Q,1)=2\sum \frac{a_n}{n}G_1(2\pi n/\sqrt{N}),\]where $G_1(x)=\int_1^\infty e^{-xy}dy/y\leq e^{-x}/x$. Therefore we may get the same truncation bound as before. We need this for the curve $F=E^D$ \end{enumerate} A similar analysis holds when the rank of the elliptic curve is 1. \extra{In \cite{mazur-swinnerton-dyer:arithmetic} Mazur and Swinnerton-Dyer introduce an analytic and an algebraic $p$-adic $L$-function attached to $E$ and they conjecture that the two are equal. The analytic $L$-function satisfies a Gross-Zagier formula proved by Perrin-Riou (\cite{perrin-riou}): \begin{theorem} Assume that $p$ splits in $K$. For any $\rho\in G(K_\infty/K)$ (where $K_\infty$ is the unique $\Z_p^2$-extension of $K$) we have \[\frac{d}{ds}L_p(f,1)(\rho^s)|_{s=0}=\left(1-\frac{1}{\alpha_p(f)}\right)^2\frac{\hat{h}(y_{1,K})}{hu^2},\]where $f$ is the newform attached to $E$, $h$ is the class number of $K$ and $\alpha_p(f)$ is the unit root of $X^2-a(p)X+p=0$. \end{theorem} The algebraic $p$-adic $L$-function is defined as the characteristic series of the Pontryagin dual of the $p^\infty$ Selmer group. Rubin proved the Mazur and Swinnerton-Dyer conjecture for elliptic curves with complex multiplication by $K$ and primes that split in $K$ so that $E$ has ordinary good reduction at $p$ (\cite{rubin:main-conjectures}). } \section{Kolyvagin's Method and Consequences} \subsection{Kolyvagin's approach to $\Sha_\textrm{tors}$} \extra{Assume that $E$ does not have complex multiplication.} Let $E$ be an elliptic curve and $K$ be a quadratic extension staisfying the Heegner hypothesis such that $L^{\prime}(E/K, 1)\neq 0$. Then $y_K$ has infinite order. Kolyvagin (\cite{kolyvagin:euler_systems}) shows that in that case the rank of $E(K)$ is 1 and $\Sha(E/K)$ is finite. Following Gross' account of Kolyvagin's work (\cite{gross:kolyvagin}), we get the following bounds on $|\Sha(E/K)|$. Assume that $E$ does not have complex multiplication. Let $I_K= [E(K): \Z y_K]$. There exists an integer $t_{E/K}$ divisible only by primes $p$ (shown to be finite by Serre in \cite{serre:propgal}) such that the representation $G(\bar{\Q}/\Q)\rightarrow \hbox{Aut}(E[p])$ is not surjective; then $|\Sha(E/K)|\hskip 5pt| t_{E/K}I_K^2$. \extra{The existence of an imaginary quadratic extension $K$ satsifying the Heegner hypothesis and such that $L^{\prime}(E/K, 1)\neq 0$ is assumed in Kolyvagin's paper; this result was simultaneously proven (using different methods) in \cite{MR92e:11050} and \cite{MR92a:11058}. Gross notes that if $E$ is semistable then for all primes $p\geq 11$ the $p$-adic representation is surjective.} Kolyvagin uses an Euler system of Heegner points to construct certain cohomology classes. Then he applies Tate local duality to study the local cohomology classes associated with $Sel(E/K)_p$ and finally uses the Chebotarev density theorem to bound $Sel(E/K)_p$. The fundamental exact sequence \[0\to E(K)/pE(K)\to Sel(E/K)_p\to \Sha(E/K)[p]\to 0\]gives the triviality of $\Sha(E/K)[p]$ in all but finitely many cases. The main assumption on the $p$ where Kolyvagin {\it does} prove triviality of $p$-torsion of $\Sha$ (i.e., the surjectivity of the mod $p$ representation) is used in two places in the argument. \begin{enumerate} \item First, the construction of the cohomology classes requires that restriction $Res:H^1(K,E[p])\to H^1(K_n,E[p])^{G(K_n/K)}$ is an isomorphism (where $K_n$ is the ring class field of conductor $n$ over $K$). The inflation restriction sequence yields \[H^1(K_n/K,E(K_n)[p])\to H^1(K,E[p])\to H^1(K_n,E[p])^{G(K_n/K)}\to H^2(K_n/K,E(K_n)[p]).\] The result follows from the fact that $E(K_n)[p]=0$ if the representation is surjective. \item The second part where surjectivity is essential is in the definition of a nondegenerate pairing \[H^1(K,E[p])\otimes G(K(E[p]))\to E[p].\] For simplicity of notation write $L=K(E[p])$. Again, the inflation restriction sequence yields \[H^1(L/K,E(L)[p])\to H^1(K,E[p])\to H^1(L,E[p])^{G(L/K)}\to H^2(L/K,E(L)[p]).\] Since $G(L)$ acts trivially on $E[p]$, it is enough to prove that $H^i(L/K,E[p])=0$. The surjectivity of the mod $p$ representation implies that $G=G(L/K)=G(\Q(E[p])/\Q)\equiv GL_2(\F_p)$. If $Z$ is the group of scalars the Hochschild-Serre spectral sequence $H^i(G/Z,H^j(Z,E[p]))\Longrightarrow H^{i+j}(L/K,E[p])$ will give the result since $|Z|=p-1$ and $E[p]$ is a $p$-group. \end{enumerate} \subsection{Computational Difficulties} The main computational problem of this method is that there is no universal bound on $p$ so that the mod $p$ representation is surjective for all elliptic curves and primes larger than the bound. There are a few results towards this bound: \begin{enumerate} \item For $E$ semi-stable the representation is surjective if $p\geq 11$ (\cite{gross:kolyvagin}). \item For general $E$, the representation is surjective when $p\geq 1+ \frac {4\sqrt 6 N}{3} \prod_{l|N}\left(1+\frac{1}{l}\right)$ (\cite{grigor}). \end{enumerate} Kolyvagin's method yields triviality of the $p$-primary component of $\Sha(E/\Q)$ for all primes that do not divide $I_K$ and for which the mod $p$ representation is surjective. For these primes, information about $\Sha$ may be obtained by using descents, which are very hard in general. The equivalent statement of the BSD conjecture (\ref{bsd-heegner}) implies that $I_K$ will always be divisible by the Tamagawa numbers $c_p$. Therefore we may hope to reduce work by weakening the surjectivity hypothesis and by dealing with the primes that divide $I_K$ but not $\prod c_p$. \section{Weakening the Hypotheses of Kolyvagin's Method} The basic principle behind the weakening of the hypotheses of Kolyvagin's method is that Kolyvagin uses the surjectivity of the mod $p$ representation only to prove that $H^i(K(E[p])/K,E[p])=0$ and $H^i(K_n/K,E(K_n)[p])=0$. These statements are hard to check computationally but may be replaced by more restrictive conditions that can be checked using a computer. \subsection{Irreducibility of the mod $p$ representation} In a recent thesis, Cha (\cite{cha:kolyvagin}) has in certain cases provided weaker conditions under which Kolyvagin's results hold. Let $K$ be any finite extension of $\Q$ and let $\ell$ be an odd prime. Cha assumes that \begin{enumerate} \item There is an unramified prime divisor $v$ of $\ell$ in $K/\Q$ such that $E$ has either good or multiplicative reduction at $v$. \item $E(K)[\ell]=0$. \end{enumerate} He then proves the following two theorems: \begin{enumerate} \item $H^1(K(E[\ell^i])/K, E[\ell^i])=0$ for all $i \geq 1$ unless $\ell =3$ and $G(K(E[\ell])/K) \isom G_{except}$, where $G_{except}$ is the subgroup of $GL_2(\F_\ell)$ given by $$G_{except}= \left\{ \left( \begin{array}{cc} a & b \\ 0 & 1 \end{array} \right) | a \in \F_\ell^*, b \in \F_\ell \right\}.$$ \item When $K$ is an imaginary quadratic extension of $\Q$ satisfying the Heegner hypothesis, let $y_K \in E(K)$ denote the usual Heegner point. Let $m$ be the largest integer such that $y_K \in \ell^m E(K)$ modulo $\ell$-torsion, and assume $\ell$ does not divide the discriminant of $K$ and $E$ has good or multiplicative reduction at $\ell$. Then if the Galois representation $$\rho_\ell : G(\bar{\Q}/\Q) \rightarrow Aut(E[\ell])$$ is irreducible, $$ord_\ell |\Sha(E/K)| \leq 2m.$$ \end{enumerate} Kolyvagin proved the following effective version of his theorem (see \cite{kolyvagin:euler_systems}): \begin{theorem}[Kolyvagin] Let $R$ be the ring of endomorphisms of $E$, with $F$ its field of fractions. Suppose the Heegner point $y_K$ has infinite order. Then if $\ell$ is an odd prime unramified in $F$ such that $G(F(E[\ell])/F)= Aut_R(E[\ell])$, $$\hbox{ord}_\ell |\Sha(E/K)| \leq 2 \hbox{ord}_\ell [E(K): \Z y_K].$$ \end{theorem} Assuming $E$ does not have complex multiplication, the hypotheses of both these theorems imply $E(K)[\ell]=0$, so, in particular, $\Sha(E/K)$ has no $\ell$-torsion if $\ell~\nmid~[E(K):~\Z y_K]$ (for a proof that irreducibility of the mod-$\ell$ representation over $\Q$ implies $E(K)[\ell]=0$, see Lemma~\ref{lemma:stein}). Mazur proves in \cite{mazur:rational} that for $p>37$ the representation is irreducible since $E$ does not have complex multiplication. This reduces the calculation of $|\Sha(E/\Q)|$ to dealing with its $p$-primary components for a few exceptional primes $p$($p\leq 37$ in general and $p\leq 7$ for semistable curves as well as the primes dividing the conductor and $I_K=[E(K):\Z y_K]$). \extra{These are generally dealt with by $p$-descent, however, and Cha's assumption on the reduction of $E$ at a given prime makes the result ineffective for potentially large prime divisors of the conductor of $E$. The assumption on irreducibility of the Galois representation is, from a computational perspective, a great improvement on Kolyvagin's original assumption, but we can further improve this hypothesis to one about torsion over $K$. } \subsection{Reducing the hypotheses to statements about torsion} \subsubsection{The mod $p$ representation} \extra{The following lemma (exercise $6$ in \cite{ribet-stein:serre}) will be useful in the next two propositions.} \begin{lemma}\label{weil-pairing-lemma} The determinant of the mod $p$ representation attached to $E$ is the cyclotomic character. \end{lemma} \proof The Weil pairing induces an isomorphism of $G(\bar{\Q}/\Q)$-modules $E[p] \wedge E[p] \cong \mu_{p}$. Let us fix a basis $\{e_1, e_2\}$ of $E[p]$, with respect to which $\rho_{p}(\sigma)$ has the form $\left( \begin{array}{cc} a & b \\ c & d \end{array} \right)$. Then $$\sigma(e_1 \wedge e_2)= (ae_1+ce_2) \wedge (be_1+de_2)= \hbox{det}(\rho_{p}(\sigma))e_1 \wedge e_2.$$ It follows that the above composition gives the cyclotomic character (i.e., the action of $G(\bar{\Q}/\Q)$ on $\mu_{p}$), which is clearly surjective. \endproof Let $G=G(\Q(E[p])/\Q)$ be the image of the mod $p$ representation. If $p\nmid |G|$, then $H^i(G,E[p])=0$ since $E[p]$ is a $p$-group. Therefore $p||G|$. By Proposition $15$ of \cite{serre:propgal}, $G$ either contains $SL_2(\F_p)$ or is contained in a Borel subgroup of $GL_2(\F_p)$. If $G$ contains $SL_2(\F_p)$ then \extra{$G$ contains the scalar $-1$.} Lemma \ref{weil-pairing-lemma} implies that $\det: G \rightarrow \F_p^*$ is surjective, so $G= GL_2(\F_p)$. \begin{lemma}\label{lemma1}If $G=GL_2(\F_p)$ then $H^i(G,E[p])=0$. \end{lemma} \proof Let $Z$ be the subgroup of scalars. Clearly $E[p]^Z=0$. Consider the Hochshild-Serre spactral sequence \[H^i(G/Z,H^j(Z,E[p]))\Longrightarrow H^{i+j}(G,E[p]);\]If $j>0$ then the group $H^j(Z,E[p])=0$ because $|Z|=p-1$ and $E[p]$ is a $p$-group. If $j=0$ then $H^j(Z,E[p])=E[p]^Z=0$. Therefore $H^i(G,E[p])=0$. \endproof \begin{lemma}\label{lemma2} Assume that $G$ is contained in a Borel subgroup of $GL_2(\F_p)$. Moreover, assume that (with respect to some basis of $E[p]$), $G$ acts as $\left(\begin{matrix}\chi&*\\0&\psi\end{matrix}\right)$ such that $\chi,\psi$ are not trivial. Then $H^i(G,E[p])=0$. \end{lemma} \proof Let $W$ be the (unique) $p$-Sylow subgroup of $\left( \begin{array}{cc} * & * \\ 0 & * \end{array} \right)$ consisting of matrices of the form $\left( \begin{array}{cc} 1 & * \\ 0 & 1 \end{array} \right)$. We may assume $W \subset G$, for otherwise $G$ has order prime to $p$, and the cohomology clearly vanishes. We begin by explicitly computing $H^j(W, E[p])$ using the fact that $W$ is cyclic (generated by $w= \left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right)$, for instance). Recall that for cyclic groups we can compute cohomology using the particularly simple projective resolution $$...\rightarrow \Z[W] \rightarrow \Z[W] \rightarrow \Z \rightarrow 0$$ where the boundary maps alternate between $w-1$ and $\hbox{Norm}= \sum_{i=0}^{p-1} w^i$ (i.e., the maps are given by multiplication in the group ring $\Z[W]$). Then we immediately see that \[H^j(W, E[p])= \left\{ \begin{array}{ll} \hbox{ker}(1-w)/\hbox{im}(\hbox{Norm}(w))= \left\langle\left( \begin{array}{c} 1 \\ 0 \end{array} \right)\right\rangle & \hbox{if $j$ is even}; \\ \hbox{ker(Norm}(w))/\hbox{im}(1-w)= \F_p^2/\left\langle\left( \begin{array}{c} 1 \\ 0 \end{array} \right)\right\rangle & \hbox{if $j$ is odd} \end{array} \right\}.\] 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 $j \geq 0$. Let us then consider the Hochschild-Serre spectral sequence $$H^i(G/W, H^j(W, \F_p^2)) \Rightarrow H^{i+j}(G, \F_p^2).$$ For $i>0$, since $|G/W|$ is prime to $p$, and $H^j(W, \F_p^2)$ is a $p$-group ($\forall j$), the group $H^i(G/W, H^j(W, \F_p^2))$ is trivial. But when $i=0$ we have just computed that $H^i(G/W, H^j(W, \F_p^2))= H^j(W, \F_p^2)^{G/W}=0$, so the entire spectral sequence is trivial, and we conclude that $H^{n}(G, E[p])=0$ for all $n \geq 0$. \endproof \extra{We will apply the previous two lemmas when $G$ is the image of the mod $p$ representation.} \subsubsection{Rational Torsion Points and Rational Isogenies} The next propositions show how to reduce the hypothesis that $H^i(\Q(E[p])/\Q)=0$ to a statement about torsion and rational isogeny. In terms of the mod $p$ representation, the fact that $E$ has no $\Q$-rational $p$-isogeny corresponds to the irreducubility of the representation. \begin{proposition} Suppose $E$ has no $\Q$-rational $p$-isogeny. Then $H^i(\Q(E[p])/\Q, E[p])=0$ for all $i >0$. \end{proposition} \proof The assumption implies that the mod $p$ representation 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 Lemma \ref{lemma1}. The former case contradicts the hypothesis since $E[p]$ is reducible under the action of a Borel subgroup. \endproof 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 treats this case in general: \begin{proposition} Suppose that for any elliptic curve $E'$ $p$-isogenous to $E$ over $\Q$ we have $E'(\Q)[p]=0$. Then $H^i(\Q(E[p]/\Q, E[p])=0$ for all $i>0$. \end{proposition} \proof The proof of the previous proposition works here except for the case when $G$ is contained in a Borel subgroup. For some basis of $E[p]$, $G$ acts as $\left( \begin{array}{cc} \chi & * \\ 0 & \psi \end{array} \right)$ for characters $\chi$ and $\psi$. Lemma \ref{lemma2} proves the proposition if the characters are not trivial. Assume that $\chi$ is trivial. Then all matrices of the above form fix $\left( \begin{array}{c} 1 \\ 0 \end{array} \right)$. Therefore there is a point of $E[p]$ fixed by the action of $G$ which contradicts the fact that $E(\Q)[p]=0$. Assume that $\psi$ is trivial. Matrices of the above form preserve the line generated by $\left( \begin{array}{c} 1 \\ 0 \end{array} \right)$, so this line forms a $G(\bar{\Q}/\Q)$-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 of the complementary line generated by $\left( \begin{array}{c} 0 \\ 1 \end{array} \right)$ is a $1$-dimensional subspace of $E^{\prime}[p]$, and if $\psi=1$, $G(\bar{\Q}/\Q)$ clearly acts trivially on this subspace (we have an isomorphism of Galois modules $E/\left\langle\left( \begin{array}{c} 1 \\ 0 \end{array} \right)\right\rangle \cong E^\prime$). Thus, $E^{\prime}(\Q)[p]$ is nontrivial, contradicting our assumption. \endproof \subsubsection{$K$-rational Torsion Points} We will substitute the condition $H^i(K_n/K,E(K_n)[p])=0$ with a condition on $K$-torsion. \begin{proposition} Let $E$ be an elliptic curve with $E(K)[p]=0$, where $p>3$ or, if $p=3$, $K \neq \Q(\mu_3)$. Let $L$ be a finite abelian extension of $K$. Then $H^i(L/K, E(L)[p]) =0$ for all $i \geq 1$. \end{proposition} \proof The proof is the now standard use of Sylow groups and the Hochschild-Serre spectral sequence. Write the abelian group $G(L/K)$ as a direct sum $P \oplus P^\prime$, where $P$ is its Sylow $p$-subgroup and $(p, \#{P^\prime})=1$. We claim that the subgroup of $E(L)[p]$ invariant under $P^\prime$ is trivial. Let $G = G(L/K)/H$, where $H$ is the subgroup of $G(L/K)$ that acts trivially on $E(L)[p]$. If $(\#G, p)=1$, $P\subseteq H$, so $P^\prime$ surjects onto $G$. As there is no nontrivial element of $E(L)[p]$ invariant under all of $G(L/K)$ (by the assumption on $E(K)[p]$, the same then holds for $P^\prime$. If $p|\#G$, we cannot have $E(L)[p]= \Z/p\Z$: the latter group has automorphism group isomorphic to $(\Z/p\Z)^*$, of order $p-1$, but if $p|\#G$, $G$ would give rise to at least $p$ distinct automorphisms. Thus, $E(L)[p]$ is the full $p$-torsion subgroup of $E$, and we can identify $G$ with a subgroup of $GL_2(\Z/p\Z)$ acting on $E(L)[p]= (\Z/p\Z)^2$. We can choose a basis of $(\Z/p\Z)^2$ so that $G$ contains the subgroup $\left( \begin{array}{cc} 1 & x \\ 0 & 1 \end{array} \right)$, where $x \in \Z/p\Z$. Being abelian, $G$ must be contained in the normalizer of this subgroup, so $G \subseteq \{ \left( \begin{array}{cc} a & b \\ 0 & a \end{array} \right) | a,b \in \Z/p\Z \}$, and we claim that $G$ contains an element with $a \neq 1$. Since $E[p]= E(L)[p]$, the representation $G(\bar{\Q}/K) \rightarrow \hbox{Aut}(E[p])$ factors through $G(L/K)$ (recall that the image of the representation is $G(K(E[p])/K)$). The determinant of the mod-$p$ representation of $G(\bar{\Q}/\Q)$ is surjective (onto $\F_p^*$), and $[K:\Q]=2$, so the character $G(\bar{K}/K) \rightarrow \F_p^*$ has image of index at most $2$ in $F_p^*$. That is, it contains at least $\frac{p-1}{2}$ elements, the squares in $\F_p^*$. Thus, for $p>3$, $G$ contains an element with non-trivial determinant having the form $\left( \begin{array}{cc} a & b \\ 0 & a \end{array} \right)$ with $a \neq 1$. Now,$\left( \begin{array}{cc} a & b \\ 0 & a \end{array} \right)^p = \left( \begin{array}{cc} a & 0 \\ 0 & a \end{array} \right)$ since we're working mod $p$, and it follows that $G(L/K)$ contains an element that acts as a nontrivial scalar. In particular, since the group of scalars of $GL_2(\F_p)$ has $p-1$ elements, this nontrivial scalar must be an element of $P^\prime$. Therefore $E(L)[p]^{P^\prime}=0$. The result will now follow from the Hochschild-Serre spectral sequence $H^i(P',H^j(P,E(L)[p]^{P^\prime}))\Longrightarrow H^{i+j}(L/K,E(L)[p])$. Then we must have $H^n(L/K,E(L)[p])=0$ for $n>0$. If $p=3$, $E(L)[3]=E[3] \Rightarrow \mu_3 \subset L \Rightarrow K=\Q(\mu_3)$. The last implication holds because $G(L/\Q)$ is abelian since $G(K_n/K)$ and $G(K/\Q)$ are, so it has a unique index $2$ subgroup; both $K$ and $\Q(\mu_3)$ correspond to index $2$ subgroups by elementary Galois theory). This contradicts our assumption on $K$, so we must have $E(L)[3]=0$, in which case the cohomology is clearly trivial. \endproof \begin{corollary} \[H^i(K_n/K,E(K_n)[p])=0.\] \end{corollary} The relation between the two section is given by the following lemma: \begin{lemma}\label{lemma:stein} Let $E/\Q$ be an elliptic curve, $K$ a quadratic extension of $\Q$, and $p>3$ an odd prime such that $E(K)[p] \neq 0$. Then the mod-$p$ representation over $\Q$ is reducible. In particular, $E$ has a $\Q$-rational $p$-isogeny. \end{lemma} \proof Let $P$ be a nontrivial element of $E(K)[p]$, and let $\tau$ be a lift of complex conjugation in $G(K/\Q)$ to $G(\bar{\Q}/\Q)$. If $\tau P$ is a multiple of $P$, then the one-dimensional subspace of $E[p]$ generated by $P$ is $G(\bar{\Q}/\Q)$-stable, so the representation over $\Q$ is reducible. Else, $P$ and $\tau P$ generate all of $E[p]$. By definition, $\tau P \in E(K)$, so $E(K)[p]= E[p]$, and because of the Weil pairing, $\mu_p \subset K$. For $p>3$, this is a contradiction. \endproof \subsubsection{General Strategy} To summarize, we can now apply Kolyvagin's arguments (as given in \cite{gross:kolyvagin}) to find $\Sha(E/\Q)[p]=0$ for all odd primes $p$ such that all curves in the $\Q$-isogeny class of $E$ have no $K$-rational $p$-torsion and $p \nmid I_K$ (of course we may still discard very easily primes $>37$ in general and $>7$ in the semistable case. \extra{\begin{remark} If $E(K)[p] \neq 0$ the subsequent situation is particularly easy to deal with because $p$-descent is much more easily implemented with a known rational $p$-torsion point. \end{remark}} We will choose a quadratic extension $K$ that satisfies the Heegner hypothesis such that $L'(E/K,1)\neq 0$. \subsubsection{Existence of K} Here we collect the results that imply the existence of quadratic imaginary extensions $K/\Q$ such that $K$ satisfies the Heegner hypothesis and $\hbox{ord}_{s=1}L(E/K) =1$. Then the Heegner point $y_K$ has infinite order, so we may apply the work of Kolyvagin to conclude that $\Sha(E/K)$ is finite and $E(K)$ has rank $1$. There are two cases to consider: \begin{enumerate} \item If $\hbox{ord}_{s=1}L(E/\Q) =0$, then the papers \cite{MR92e:11050} and \cite{MR92a:11058} both imply the existence of infinitely many distinct $K$ (that is, with different fundamental discriminants) satisfying our two hypotheses. \item If $\hbox{ord}_{s=1}L(E/\Q) =1$, then a result of Waldspurger (\cite{MR87g:11061b}) does the trick, as does the above result of Bump-Friedberg-Hoffstein. \end{enumerate} See also \cite{MR2020572} for a clear overview of how all of these results-- along with the work of Gross-Zagier and Kolyvagin-- fit together to settle the rank $0$ and $1$ BSD conjecture. 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 we have observed that a simple way to ensure this is to require the discriminant of $K$ ($:= D$) 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{MR81i:12014}). Thusfar the best proven result is due to Ono and Skinner, who showed (\cite{MR2000a:11077}) that, in the case $\hbox{ord}_{s=1}L(E/\Q)=0$, the number of such discriminants has density at least on the order of magnitude of $\frac{1}{\hbox{log} X}$. 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 $\hbox{log} X$) under the assumption that $E/\Q[2]=0$ (see \cite{MR2002a:11051}). 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 force us to perform descents on large primes (in particular, the contributions from nontrivial elements of the Shafarevich-Tate group). In practice, though, the bad primes are small. \section{Elliptic Curves with Complex Multiplication} The previous discussion fails in the case of elliptic curves $E$ with complex multiplication. Alternative results are obtained by Rubin in \cite{rubin:main-conjectures}. Perrin-Riou proves \cite{perrin-riou} the following corollaries to the $p$-adic Gross-Zagier type formula: \begin{proposition} \begin{enumerate} \item If $f$ is the normalized newform attached to an elliptic curve with complex multiplication then $L_p$ has analytic rank 1 if and only if $L$ has analytic rank 1. \item If the rank of $L_p$ is 1 then the $p$-part of the Birch and Swinnerton-Dyer conjecture is true up to a unit of $\Z_p$. \end{enumerate} \end{proposition} This follows from the Gross-Zagier type formula and a similar formula for the algebraic $p$-adic $L$-series obtained in \cite{perrin-riou:thesis}. Rubin proves the following theorem (\cite{rubin:main-conjectures}) \begin{theorem} Let $E$ have complex multiplication by $K$. (Since the Birch and Swinnerton-Dyer conjecture is isogeny-invariant we may assume that $End(E)=\mathcal{O}_K$.) \begin{enumerate} \item If $L(E/K,1)\neq 0$ then for all primes $\mathfrak{p}$ not dividing $|\mathcal{O}_K^\times|$ we have \[|\Sha[\mathfrak{p}^\infty|=\mathbb{N}(\mathfrak{p})^{m(\mathfrak{p})},\]where $m(\mathfrak{p})=\textrm{ord}_\mathfrak{p}\left(|E(K)|\frac{L(\bar{\psi},1)}{\Omega}\right)$, where $\psi$ is the Hecke character associated to $E$. \item If the analytic rank of $E$ is 1 then Perrin-Riou's work together with the conjecture of Mazur and Swinnerton-Dyer implies that $\Sha(E/\Q)[p]$ is as expected from the BSD conjecture whenever $p>2$ splits in $K$. \end{enumerate} \end{theorem} \section{Computations} The following algorithm will prove the full BSD for elliptic curves without complex multiplication up to conductor 20000. \begin{algorithm} {\sf Let $E$ be an elliptic curve over $\Q$ of analytic rank at most $1$. The following algorithm computes $|\Sha(E/\Q)[p]|$ for {\em all} primes $p$. \begin{enumerate} \item{}[Choose $K$] Choose two quadratic imaginary fields $K$ that satisfy the Heegner hypothesis, such that $E/K$ has analytic rank 1. \item{}[Find $p$-torsion] Decide for which primes $p$, there is a curve $E'$ that is $\Q$-isogenous to $E$ such that $E'(K)[p]\neq 0$. Let $B$ be the product of these primes and 2. \item{}[Root number] Compute the root number of $E$. \item{}[Compute Mordell-Weil] \begin{enumerate} \item If the root number is $-1$, compute $E(\Q)$ and let $z$ be a generator modulo torsion. \item If the root number is $+1$, compute $E^D(\Q)$, and let $z$ be a generator modulo torsion. \end{enumerate} \item{}[Height of Heegner point] Compute the height $h_K(y_K)$. %[[ and $\|\omega_E\|^2$ is the %volume of $E(\C)$, which is twice the volume of the %period lattice $\Z\omega_1 + \Z\omega_2$ associated %to $E$ and $\omega_E$. %The differential $\omega_E$ is the $c\cdot \omega$, where %$c$ is the Manin constant for $E$ and $\omega$ is %a N\'eron differential on $E$ (we are assuming $E$ %is modular here, which is OK.) % [[say something about %computing $c\omega$... cite Section 2.14 of Cremona's %book.]] If $\pi:X_0(N)\to E$ is the modular parametrization, %then $\pi^*(\omega) = c\cdot \omega_E$, where %$\omega_E = f(q)\frac{dq}{q} \in \H^0(X_0(N),\Omega_{X_0(N)})$ %is the normalized cuspidal eigenform corresponding to~$E$.]] \item{}[Index of Heegner point] Compute $$I_K = \sqrt{h_K(y_K)/h_K(z)} = [E(K):\Z y_K].$$ \item{}[Annihilate $\Sha$] Then $\Sha(E/\Q)[p] = 0$ for all primes $p\nmid B \cdot I_K$. \item{}[$p$-descent] For each prime $p\mid B \cdot I_K$, do a $p$-descent and compute $\Sha(E/\Q)[p]$. (Note that this is likely not too difficult because there is a $p$-torsion point over $K$ on a curve $F$ that is $\Q$-isogenous to $E$. Ideas: If an isogeny from $E$ to $F$ has degree divisible by $p$, then $E$ has a rational $p$-isogeny, which makes $p$-descent eiser. If an isogeny from $E$ to $F$ has degree coprime to $p$, then $\Sha(F/\Q)[p]\isom \Sha(E/\Q)[p]$, and $F$ has a $K$-rational $p$-torsion point, so $p$-descent on $F$ should be relatively easy.) To reduce the number of $p$ for which one must do a $p$-descent, use several $K$. \end{enumerate} } \end{algorithm} \begin{proof} Step 1 guarantees that $G(K(E[p])/K)\equiv G(\Q(E[p])/\Q)$. This step will always find such a $K$ due to the density theorem mentioned. Step 2 will determine the primes for which the weakened hypothesis fails and so the primes for which we must do descent. Since the root number and the rank have the same parity, the fact that the rank is at most 1 implies that the root number determines the rank of the curve $/\Q$. Therefore, by computing the Mordell-Weil group of $E$ or $E^D$ over $\Q$ (but only one of them) we can find a generator of $E(K)$. Step 6 computes the last set of primes at which we need descent. Finally, the last step takes care of the exceptional primes. \end{proof} For elliptic curves with complex multiplication by $K$, if $E$ has rank 0 we may just do descent for the primes dividing $|\mathcal{O}_K^\times|$. If $E$ has rank 1, then we need to take care of the primes that do not split in $K$. We may also apply Kolyvagin's method with our weakened (computationally viable) hypotheses to eliminate some of these primes. Then we may do descent on those. \section{Algorithms} \subsection{Computing the Mordell-Weil group} While in general an unsolved problem, finding a complete set of (free) Mordell-Weil generators when the rank of the curve is already known is a fairly simple, if sometimes time-consuming, problem. Basically, one searches for points by naive height until a point of infinite order is found (the easiest way to check whether a point on $E(\Q)Q$ has infinite order is to compute its multiples up to $12$; if none of these is zero, the point cannot have finite order by Mazur's theorem on torsion subgroups of elliptic curves over $\Q$). The first point of infinite order found (call it $P$) may not be a generator, however: it may only generate a finite-index subgroup of $E(\Q)$. But if it is a nontrivial multiple of a generator, the canonical height of the generator is at most $\frac{1}{4} \hat{h}(P)$ (the canonical height). By a result of Silverman, the naive height of the generator may then be bounded, and an exhaustive computation will then turn it up if it exists. For more details on both the theory and implementation of this method, see \cite{cremona:algs} or the updated version available on Cremona's webpage (http://www.maths.nott.ac.uk/personal/jec/book/fulltext/index.html). \subsection{p-descents} Traditionally, performing a $p$-descent on $E/\Q$ means computing the quotient group $E(\Q)/pE(\Q)$ by first computing the $p$-Selmer group and then somehow trying to get a handle on $\Sha(E/\Q)[p]$. Our task is much simpler since we already know the rank of any curve we are working with (by Kolyvagin's theorem). In particular, $\hbox{dim}_{\F_p} E(\Q)/pE(\Q) = \hbox{rk}_\Z E(\Q) + \hbox{dim}_{\F_p}E(\Q)[p]$; we know all of these quantities, so we can compute $\Sha(E/\Q)[p]$ (our ultimate goal) by simply finding the order of the $p$-Selmer group and applying the fundamental exact sequence $$0 \rightarrow E(\Q)/pE(\Q) \rightarrow Sel^p(E/\Q) \rightarrow \Sha(E/\Q)[p] \rightarrow 0.$$ $Sel^p(E/\Q)$ is effectively computable, so this poses no problem for the validity of our algorithm. In all cases this calculation can be reduced to ``standard'' computations over number fields. In particular, for $S$ a set of ``bad primes'' (traditionally $p$ and places of bad reduction for $E/\Q$), we have to determine the $p$-part of the $S$-class group and a basis of the $S$-units modulo $p^{th}$-powers. For a full discussion and improvements to the basic approach, see \cite{MR2004g:11045}. The method is fool-proof when $p=2$ or $p=3$ (given that we are working over $\Q$), but for larger primes current limitations in computational number theory may make the theoretically possible calculations infeasible. Fortunately, many of our examples are exceptional cases having $K$-rational $p$-torsion. This implies they have rational $p$-isogenies (for a proof, see below), and Schaefer and Stoll, for instance, perform a successful $13$-descent on a curve using the fact that it has a rational $13$-isogeny. \subsection{Finding Isogenies} Cremona (\cite{cremona:algorithms}, section 3.8) describes an algorithm to compute all isogenous curves for any given elliptic curve over $\Q$. \subsection{Root Number} Cremona's reference? \bibliography{biblio} \end{document}