\documentclass[11pt]{article} \hoffset=-0.05\textwidth \textwidth=1.1\textwidth \newcommand{\bndone}{87} \newcommand{\bndzero}{19} \bibliographystyle{amsalpha} \usepackage[all]{xy} \include{macros} \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}(BSD) 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. Thus, if we can check the conjecture for a single curve in a given isogeny class, we will have verified it for all $\Q$-isogenous curves (as we will see, Kolyvagin has proven that $\Sha(E/\Q)$ is finite for curves of analytic rank at most $1$). \section{Gross-Zagier Formulae} One of the factors in the BSD formula is the derivative of the $L$-function. Let $K=\Q(\sqrt{-D})$ be a quadratic extension of discriminant $D$. We say that $K$ satisfies the Heegner hypothesis for an 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 $E/K$. Then the following limit formula was proved by Gross and Zagier (\cite{gross-zagier}) \begin{theorem}(Gross-Zagier)\[\hat{h}(y_K)=\frac{u^2\sqrt{D}}{\int_{E(\mathbb{C})}\omega\wedge i\bar{\omega}}L'(E/K,1),\](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} Complex conjugation acts on $E(K)$ and we get a decomposition \begin{proposition} For a quadratic imaginary field $K=\Q(\sqrt{D})$, $E(K)$ decomposes, up to $2$-torsion, as a direct sum $E(\Q) \oplus E_D(\Q)$, where $E_D$ is the quadratic twist of $E$. (Recall that from a Weierstrass equation of $E/\Q$, we obtain a Weierstrass equation of $E_D/\Q$ as follows: \[\begin{array}{c} E: y^2= x^3+ax+b \\ E_D: y^2= x^3+D^2ax+D^3b \end{array})\] \end{proposition} \proof Denote complex conjugation by $\tau$. We will decompose $E(K)$ into its eigenspaces under the action of $\tau$. First, we kill $E(K)[2]$ (by tensoring with $\Z[\frac{1}{2}]$, for instance), so that, using the fact that $\tau^2=1$, we can define the projections from $E(K)$ to its $+1$ and $-1$ eigenspaces under $\tau$: for $P\in E(K)$, \[P= \frac{1+\tau}{2}P + \frac{1-\tau}{2}P.\] 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 $E_D(\Q)$. Thus, having eliminated the $2$-torsion, the decomposition of $E(K)$ into its $\tau$-eigenspaces just reads \[E(K)= E(\Q) \oplus E_D(\Q).\] \endproof There is an alternative, equally useful, formula \begin{proposition} \[\hat{h}(y_K)= \frac{u^2 \sqrt{D}}{\int_{E(\mathbb{C})}\omega\wedge i\bar{\omega}} L^{\prime}(E_D/\Q, 1)L(E/\Q, 1),\] \end{proposition} 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 and Consequences} Assume that $E$ does not have complex multiplication. Kolyvagin shows (\cite{kolyvagin:euler_systems}) that if $y_K$ has infinite order (i.e., $L^{\prime}(E/K, 1)\neq 0$), then $E(K)$ has rank $1$. Moreover, he proves $\Sha(E/K)$ is finite, with the following bounds (we follow Gross's notation and organization of Kolyvagin's results in \cite{gross:kolyvagin}). 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 $\# \Sha(E/K)| t_{E/K}I_K^2$. 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 some density theorems (Chebotarev) to bound $Sel(E/K)_p$. Then the fundamental exact sequence \[0\to E(K)/pE(K)\to Sel(E/K)_p\to \Sha(E/K)[p]\to 0\]gives 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 $p$-adic 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 to define 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 $p$-adic 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} The main computational problem of this method is that there is no universal bound on $p$ so that the $p$-adic representation is surjective for all elliptic curves and primes larger than the bound. Moreover, the surjectivity condition is hard to verify computationally. \section{Weakening the Hypotheses of Kolyvagin's Method} 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} In particular, the second theorem will imply $\Sha(E/K)$ has no $\ell$-torsion if $\ell~\nmid~[E(K):~\Z y_K]$. Mazur proves in \cite{mazur:rational} that for $p>37$ the representation is irreducible since $E$ does not have complex multiplication. This is exactly the sort of result we try to obtain, since we can reduce 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 $I_K=[E(K):\Z y_K]$). 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 $K$. %Thus, while Cha's results are of interest, in what follows we will prove somewhat stronger refinements of the hypotheses of Kolyvagin's argument. \begin{remark} One of the facts used in Kolyvagin's method is that $G(K(E[p])/K)\equiv G(\Q(E[p])/\Q)$ which is true when $K\cap \Q(E[p])=\Q$. %We now undertake to weaken the hypotheses of Gross's Proposition $2.1$, replacing $G(\Q(E[p])/\Q)= GL_2(\F_p)$ with the statement that no curve in the $\Q$-isogeny class of $E$ has any $p$-torsion over $K$. Gross only uses the hypothesis on $G(\Q(E[p])/\Q)$ at two points in the paper: section $4$, where he constructs Kolyvagin's cohomology classes and shows that restriction gives an isomorphism $H^1(K, E[p]) \cong H^1(K_n, E[p])^{\mathcal{G}_n}$, where $K_n$ denotes the ring class field of conductor $n$ over $K$ and $\mathcal{G}_n= G(K_n/K)$; and section $9$, where he requires $H^n(K(E[p])/K, E[p])=0$ for all $n > 0$, for then by the Hochshild-Serre spectral sequence he obtains the isomorphism $H^1(K, E[p]) \cong H^1(K(E[p]), E[p])^{G(K(E[p])/K)}$. It is worth noting, as in \cite{cha:kolyvagin}, that in light of the Birch and Swinnerton-Dyer conjecture, the assumption $E(K)[p]=0$ should be enough for this argument to go through, but we do not have a proof of this result. %We will need to introduce an additional condition on $K$, that $K \cap \Q(E[p])= \Q$. This ensures that $G(\Q(E[p])/\Q)= G(K(E[p])/K)$, an isomorphism that also induces an isomorphism on cohomology $H^i(\Q(E[p])/\Q, E[p])= H^i(K(E[p])/K, E[p])$. Thus, whenever we prove vanishing of the first cohomology group in what follows, we are really interested in the consequence for the second group. The field $\Q(E[p])/\Q$ is ramified precisely at $p$ and primes dividing the conductor $N$ of $E$. If we choose $K$ so that $K/\Q$ is ramified at some prime not in this list, then $K$ cannot be contained in $\Q(E[p])$. In particular, among those $K$ satisfying the Heegner hypothesis, it suffices to choose one having at least two odd prime divisors of its discriminant: these primes ramify in $K/\Q$, at most one of them can equal $p$, and the other one cannot ramify in $\Q(E[p])/\Q$ because primes dividing $N$ must split (and so not ramify) in $K$, by the Heegner hypothesis. \end{remark} The following lemma (exercise $6$ in \cite{ribet-stein:serre}) will be useful in the next two propositions. \begin{lemma} The determinant of the mod $\ell$ representation attached to $E$ is the cyclotomic character. \end{lemma} \proof The Weil pairing induces an isomorphism of $G(\bar{\Q}/\Q)$-modules $E[\ell] \bigwedge E[\ell] \cong \mu_{\ell}$. Let us fix a basis $\{e_1, e_2\}$ of $E[\ell]$, with respect to which $\rho_{\ell}(\sigma)$ has the form $\left( \begin{array}{cc} a & b \\ c & d \end{array} \right)$. Then $$\sigma(e_1 \wedge e_2)= (ae_1+ce_2) \wedge (be_1+de_2)= \hbox{det}(\rho_{\ell}(\sigma))e_1 \wedge e_2.$$ It follows that the above composition gives the cyclotomic character (i.e., the action of $G(\bar{\Q}/\Q)$ on $\mu_{\ell}$), which is clearly surjective. \endproof The next proposition shows the weakened hypothesis is sufficient to replace surjectivity at 2. \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 First observe that $G(\Q(E[p])/\Q)=G$ is precisely the image of the Galois representation $\rho_p: G(\bar{\Q}/\Q) \rightarrow \hbox{Aut}(E[p]) \cong GL_2(\F_p).$ By assumption, $G$ is the image of an irreducible representation. There are two cases to consider: if $p \nmid \#G$, the cohomology clearly vanishes because multiplication by $\#G$ kills the cohomology but is an isomorphism on the $p$-group $E[p]$. Otherwise, let $p|\#G$. By Proposition $15$ of \cite{serre:propgal}, $G$ either contains $SL_2(\F_p)$ or is contained in a Borel subgroup of $GL_2(\F_p)$. $E[p]$ is reducible under the action of a Borel subgroup, so by hypothesis $SL_2(\F_p) \subset G$. Thus, $G$ contains a nontrivial scalar (minus the identity); in fact, Lemma $4.1$ implies that $\hbox{det}: G \rightarrow \F_p^*$ is surjective, so $G= GL_2(\F_p)$ and contains all of the nontrivial scalars. Applying the inflation-restriction exact sequence to the subgroup of $G$ generated by the scalars implies that $H^i(G, E[p])=0$ for all $i>0$, because this subgroup has order prime to $p$ ($p-1$, in fact), and it leaves no subgroup of $E[p]$ invariant. \endproof For the above result, we assumed that the representation of $G(\bar{\Q}/\Q)$ was irreducible, but we in fact improve our hypotheses further with the following proposition: \begin{proposition} In the notation of the above proposition, suppose that the image $G$ of the representation is contained in a Borel subgroup (we used our irreducibility hypothesis to avoid this situation before). Then unless $E$ or a $p$-isogenous curve (over $\Q$) $E^\prime$ has a $\Q$-rational $p$-torsion point, $H^i(\Q(E[p]/\Q, E[p])=0$ for all $i>0$. \end{proposition} \proof Choosing a basis, suppose that the action of $G$ on $E[p]= \F_p^2$ is given by $\left( \begin{array}{cc} \chi & * \\ 0 & \psi \end{array} \right)$ for characters $\chi$ and $\psi$. If $\chi$ is trivial, all matrices of the above form fix $\left( \begin{array}{c} 1 \\ 0 \end{array} \right)$. In particular, there is a point of $E[p]$ fixed by the action of $G$, a contradiction since we have assumed that $E(\Q)[p]=0$. Now suppose 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. 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 The next proposition provides a weaker hypothesis for the use of surjectivity at~1. THERE IS THIS STATEMENT HERE I DO NOT UNDERSTAND ABOUT NOT Q P ISOG IMPLIES NO K P TORS. %Note that if $E$ has no $p$-isogeny (defined over $\Q$) for $p>2$, then it has no $p$-torsion over $K$. \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 $i \geq 1$. (In the actual proof for 1 we use $L=K_n$, the ring class field of conductor $n$ which is abelian.) \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 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. \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} \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 the first quadratic imaginary field $K$ that satisfies the Heegner hypothesis, such that $E/K$ has analytic rank 1, and whose discriminant is divisible by at least two primes. \item{}[Find $p$-torsion] Decide for which primes $p$, there is a curve $E'$ that is $\Q$-isogenous to $E$ such that $E'(K)[p]\neq 0$. (We enumerate the $\Q$-isogeny class of $E$ using [standard method]. Then for each $E'$ in the $\Q$-isogeny class, compute the torsion subgroup of $E'(K)$ using [standard method], and see whether or not $p$ divides its order.) Let $B$ be the product of $2$ and these primes. \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 $h_K(y_K)$ relative to $K$ of the Heegner point associated to $K$ using the Gross-Zagier formula: $$ \hat{h}_K(y_K) = \alpha\cdot L'(E/K,1) = \begin{cases} \alpha \cdot L^{\prime}(E_D/\Q, 1)\cdot L(E/\Q, 1) & \text{if $E$ has rank $0$} \\ \alpha \cdot L(E_D/\Q, 1)\cdot L^{\prime}(E/\Q, 1) & \text{if $E$ has rank $1$,} \end{cases} $$ where $\alpha = \frac{u^2 \sqrt{|D|}}{\|\int_{E(\mathbb{C})}\omega\wedge \overline{i\omega}\|} $. Here $u$ is half the number of units in the ring of integers of $K$. We may compute \begin{eqnarray*}\int_{E(\mathbb{C})}\omega\wedge \overline{i\omega}&=& \int_{E(\mathbb{C})}\frac{dx}{y}\wedge \overline{\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}}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 $\mathbb{C}/\Lambda$. %[[ 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 $$I_K = \sqrt{h_K(y_K)/h_K(z)} = [E(K):\Z y_K],$$ up to primes that divide the $B$ from step 2. Note that $h_K(z) = 2\cdot h_\Q(z)$, and that we do {\em not} compute $E(K)$ or $\Z y_K$ directly, but instead compute the index using properties of heights. \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{General Algorithms for Elliptic Curves} \begin{enumerate} \item For root number cite the paper that Cremona said. \item For Mordell-Weil ? \item For descent there is a more than sufficient article by our own Michael Stoll online. \item Perhaps how to evaluate the $L$ series. \item In general, what are algorithms to generate the isogeny classes of $E$. \end{enumerate} \bibliography{biblio-pat} \end{document}