\documentclass[11pt]{article} \hoffset=-0.05\textwidth \textwidth=1.1\textwidth \newcommand{\bndone}{87} \newcommand{\bndzero}{19} \bibliographystyle{amsalpha} \include{macros} %\include{header} %\title{The Birch and Swinnerton-Dyer Conjecture for %Elliptic Curves Over $\Q$ Of Rank One and Conductor at Most $\bndone$ and %of Rank Zero and Conductor at Most $\bndzero$ is True} \title{Verifying the Full Birch and Swinnerton-Dyer Conjecture for Specific Elliptic Curves of Analytic Rank $0$ or $1$} \author{Stephen Donnelly 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} \maketitle \begin{abstract} Suppose~$E$ is an elliptic curve over~$\Q$ with conductor at most~$\bndone$ and rank one or conductor at most~$\bndzero$ and rank zero. Then the full Birch and Swinnerton-Dyer conjecture is true for~$E$. We prove this by ... \end{abstract} \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{}[Find $p$-isogenies] Decide for which primes $p$ the curve $E$ has a rational % $p$-isogeny. Let $t$ be the product of these numbers and $2$. \item{}[Choose $K$] Choose the first quadratic imaginary field $K$ that satisfies the Heegner hypothesis, is such that $E/K$ has analytic rank 1, and whose discriminant is divisible by at least two primes. Let $D$ be the discriminant of $K$. Note that the condition that $D$ be divisible by two primes and that $(D,N_E)=1$, implies that $\Q(E[p])$ is linear disjoint from~$K$ for all primes~$p$. In fact, this is a necessary and sufficient condition, since if $D$ is divisible by only one prime~$p$, then because of the Weil pairing $\Q(E[p])$ will automatically contain $K$. [[Note: we'll need a density argument here to know that such a $K$ exists.]] \begin{proof} Having chosen the discriminant $D$ this way, for any prime $p$, there is a prime factor of $D$ coprime with $N_Ep$. Therefore $K$ is not a subfield of $\Q(E[p])$ and since $[K:\Q]=2$ we actually have linear disjointness. This in turn implies that $G(\Q(E[p])/\Q)\equiv G(K(E[p])/K)$. \end{proof} \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. \begin{proof} Let $p$ be a prime so that for any elliptic curve $E'$ which is $\Q$-rational $p$-isogenous to $E$ we have $E'(K)[p]=0$. \begin{proposition}$H^i(K(E[p])/K,E[p])=0$ for all $i>0$. \end{proposition} Since $G(\Q(E[p])/\Q)\equiv G(K(E[p])/K)$ there is an isomorphism $H^i(G(\Q(E[p])/\Q),E[p])\equiv H^i(G(K(E[p])/K),E[p])$. Therefore it is enough to show this on the level of $\Q$. First observe that $G(\Q(E[p])/\Q)=H$ is precisely the image $G$ of the Galois representation $\rho_p: G(\bar{\Q}/\Q) \rightarrow \hbox{Aut}(E[p]) \cong GL_2(\F_p).$ 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 $GL_2(\F_p)$. If $G$ contains $SL_2(\F_p)$ then $G$ contains $-I_2\in SL_2(\F_p)$ and so a nontrivial scalar. Lemma $4.1$ [state it here or something] implies that $\hbox{det}: G \rightarrow \F_p^*$ is surjective, so $G= GL_2(\F_p)$ and contains all of the nontrivial scalars, which we denote by $Z$. Applying the inflation-restriction exact sequence to the subgroup $Z$ of $G$ we get that $H^1(G/Z,E[p]^Z)\to H^1(G, E[p])\to H^1(Z,E[p])^{G/Z}\to\ldots$. Since $\# Z-p-1\nmid p$ and $Z$ leaves no subgroup of $E[p]$ invariant the central group is trivial. By induction this holds in all dimensions $>0$. Therefore $G$ is contained in a Borel subgroup. This implies that the representation $\rho_p$ is reducible since $E[p]$ is reducible under the action of a Borel subgroup. This implies the existence of a $\Q$-rational $p$-isogeny. 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 $E/<\left( \begin{array}{c} 1 \\ 0 \end{array} \right)> \cong E^\prime$). 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 $W \subset H$, for otherwise $H$ has order prime to $p$, and the cohomology clearly vanishes. We begin by explicitly computing the third cohomology group using the fact that $W$ is cyclic (generated by $w= \left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right)$, for instance). Recal 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 \[H^i(W, E[p])= \left\{ \begin{array}{ll} \hbox{ker}(1-w)/\hbox{im}(\hbox{Norm}(w))= <\left( \begin{array}{c} 1 \\ 0 \end{array} \right)> & \hbox{if $i$ is even}; \\ \hbox{ker(Norm}(w))/\hbox{im}(1-w)= \F_p^2/<\left( \begin{array}{c} 1 \\ 0 \end{array} \right)> & \hbox{if $i$ is odd} \end{array} \right\}.\] Since $\chi$ and $\psi$ are nontrivial by assumption, the $H/W$-invariants for both of these groups are trivial. Thus, $H^i(W, E[p])^{H/W}=0$ for $i>0$. Let us then consider the Hochschild-Serre spectral sequence $$H^i(H/W, H^j(W, \F_p^2)) \Rightarrow H^{i+j}(H, \F_p^2).$$ For $i>0$, since $|H/W|$ is prime to $p$, and $H^j(W, \F_p^2)$ is a $p$-group ($\forall j$), the group $H^i(H/W, H^j(W, \F_p^2))$ is trivial. But when $i=0$ we have just computed that $H^i(H/W, H^j(W, \F_p^2))= H^j(W, \F_p^2)^{H/W}=0$, so the entire spectral sequence is trivial, and we conclude that $H^{n}(H, E[p])=0$ for all $n \geq 0$. The condition that $E(K)[p]=0$ implies that another cohomology group is trivial. \begin{proposition} Let $E$ be an elliptic curve with $E(K)[p]=0$, where $p>3$ or, if $p=3$, $K \neq \Q(\mu_3)$. Then $H^i(F/K, E(F)[p]) =0$ for all $i \geq 1$ and $F/K$ abelian extension. \end{proposition} \proof We may write the abelian group $G(F/K)$ 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 $E(F)[p]$ invariant under $P^\prime$ is trivial. Let $G = G(F/K)/H$, where $H$ is the subgroup of $G(F/K)$ that acts trivially on $E(F)[p]$. If $(\#G, p)=1$, $P\subseteq H$, so $P^\prime$ surjects onto $G$. As there is no nontrivial element of $E(F)[p]$ invariant under all of $G(F/K)$ (by the assumption on $E(K)[p]$, the same then holds for $P^\prime$. If $p|\#G$, we cannot have $E(F)[p]= \Z/p\Z$: 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, $E(F)[p]$ is the full $p$-torsion subgroup of $E$, and we can identify $G$ with a subgroup of $GL_2(\Z/p\Z)$ acting on $E(F)[p]= (\Z/p\Z)^2$. 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 $E[p] \subset E(F)[p]$, the representation $G(\bar{\Q}/K) \rightarrow \hbox{Aut}(E[p])$ factors through $G(F/K)$ (recall that the image of the representation is $G(K(E[p]/K)$). We argued before that this image contained the scalars corresponding to squares in $\F_p^*$, so $P^\prime$ contains at least $\frac{p-1}{2}$ (which is $>1$ for $p>3$) elements that leave no subgroup of $E(F)[p]$ invariant. Now, the result will follow from an application of the inflation-restriction exact sequence: \[0\rightarrow H^1(P, E(F)[p]^{P^\prime}) \rightarrow H^1(F/K, E(F)[p]) \rightarrow H^1(P^{\prime}, E(F)[p])\] The first group vanishes since $E(F)[p]^{P^{\prime}}=0$, and the third group vanishes since the order of $P^{\prime}$ is prime to $p$, and thus to the order of $E(F)[p]$. We conclude that the middle group is trivial, as desired. We can therefore extend the sequence to the second cohomology groups, deduce the same triviality result, and by induction conclude that $H^m(F/K, E(F)[p])=0$ for all $m \geq 1$. If $p=3$, $E(F)[3]=E[3] \Rightarrow \mu_3 \subset F \Rightarrow K=\Q(\mu_3)$. The last implication holds because $G(F/\Q)$ is abelian since $G(F/K)$ 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 $E(F)[3]=0$, in which case the cohomology is clearly trivial. \endproof Gross's account of Kolyvagin's proof of triviality of $p$ torsion for the Shafarevich-Tate group relies on two cohomological statements. One of them is precisely the proposition that $H^i(K(E[p])/K,E[p])=0$ (by the Hochshild-Serre spectral sequence he obtains the isomorphism $H^1(K, E[p]) \cong H^1(K(E[p]), E[p])^{G(K(E[p])/K)}$). The other one is that $H^i(K_n/K,E(K_n)[p])=0$ which follows from the previous proposition since $K_n/K$ is abelian (he uses this to construct Kolyvagin's cohomology classes and to show 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 $\mathcal{G}_n= G(K_n/K)$). S. Donnelly has suggested replacing Kolyvagin's original hypothesis with these weaker and more easily computable ones. \end{proof} Kolyvagin's work implies that $\Sha(E/K)[p]=0$ for all $p\nmid B I_K$, where $I_K=[E(K):\Z y_K]$. The remainder of the algorithm will deal with these bad primes. \item{}[Root number] Compute the root number of $E$ using the algorithm in [section blah of Cohen's Algorithms for ...]. \item{}[Compute Mordell-Weil] \begin{itemize} \item If the root number is $-1$, compute $E(\Q)$ (using ..., [possible because of [] implies that the algebraic rank equals the analytic rank since the analytic rank is at most $1$]), 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{itemize} \begin{proof} The fact that this yields an effective computation of the Mordell-Weil group comes from the following proposition: \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 This part of the algorithm uses the root number to determine a generator of $E(K)/E(K)_\textrm{tors}$. Since we are dealing with elliptic curves of rank 0 or 1 over $K$, the generator of the free group over $\Q$ comes from the generator of either $E(\Q)$ or $E^D(\Q)$. The root number will determine which one. \end{proof} \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: $$ h_K(y_K) = \alpha\cdot L'(E/K,1) = \begin{cases} \alpha \cdot L^{\prime}(E_D/\Q, 1)\cdot L(E/\Q, 1) & \text{if $E$ has rank $0$} \\ \alpha \cdot L(E_D/\Q, 1)\cdot L^{\prime}(E/\Q, 1) & \text{if $E$ has rank $1$,} \end{cases} $$ 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{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}}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} [[$K$ exists by Murty-Murty or Bump-Friedberg-Hoffstein....]] \end{proof} \end{enumerate} \bibliography{biblio} \end{document}