ls \documentclass[11pt]{article} \hoffset=-0.05\textwidth \textwidth=1.1\textwidth \newcommand{\bndone}{87} \newcommand{\bndzero}{19} \bibliographystyle{amsalpha} \include{macros} %\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} \section{TODO} \begin{enumerate} \item Define $\Sha$ -- mention that it is torsion since $\H^1(\Q,E)$ is torsion. \item Write up argument that $E(K)$ is a direct sum of $E(\Q)$ and $E^K(\Q)$ up to powers of $2$ (i.e., up to tensoring with $\Z[1/2]$). \item Summarize that Kolyvagin proves triviality of Sha for 23 curves in ``On the MW and Sha for Weil Elliptic Curves''. Verify that this proves BSD for these curves (by looking them up in Cremona). \item Cremona's remark that says one should do a project like this. Where is it in his book? \item Formalize how to use the Gross-Zagier theorem to reduce computing the index $I_K = [E(K):\Z y_K]$ to computing only quantities associated to elliptic curves over~$\Q$. Distinction between the two cases when $E/\Q$ has rank $0$ and when it has rank $1$. \item Stoll can show triviality of $\Sha$ when there is a $p$-isogeny or $p=2$. \item Mention Cremona-Mazur table of nontrivial Sha, and that we don't expect any nontrivial Sha for levels up to $500$. \item Write something about isogeny invariance. \item Karl Rubin and other people have papers that prove full BSD up to power of 2 or 3 for CM elliptic curves of rank 0 or 1, maybe. I haven't read any of these papers in detail. Look at them, and list the ones of conductor up to 500 that they deal with. Deal with power of 2 as well. (E.g., MathSciNet...) \item Write up in tex with more details Donnelly's argument, justify details more with precise reference, and attribute everything to Donnelly. \item To what extent do we need to compute Heegner points? \begin{enumerate} \item Pete Green's package. \item My package. \item Cremona probably has PARI code. \item ??? \end{enumerate} \item Domain of applicability of Kato's theorems on BSD. Just copy Theorem 0.3 from Grigor's kato3.dvi with appropriate reference. Will need Proposition 1.1 in Section 1.1.1 of Ribet-Stein. \item Ultimate goal: Give a provably correct complete {\em algorithm} that takes as input an elliptic curve $E$ over $\Q$ of analytic rank at most $1$ and outputs either ``yes'' the BSD conjecture is true for $E$, or ``no'' it is not true for $E$ and here is why. This should be an algorithm in the classical sense that it terminates on any valid input. Connection with old paper of Manin. \end{enumerate} \section{Introduction} \begin{theorem}\label{thm:main} Suppose~$E$ is an elliptic curve over~$\Q$ of rank one with conductor at most~$bndone$ or rank zero and conductor at most~$\bndzero$. Then the full Birch and Swinnerton-Dyer conjecture is true for~$E$. \end{theorem} The rest of this paper is devoted to proving Theorem~\ref{thm:main}. By work of Cremona, Wiles et al., and Tate it suffices to prove that $\#\Sha(E)=1$ except for $\nf{681B}$ and $\nf{571A}$ where we must show that $\Sha(E)$ has order $9$ and $4$, respectively. \section{BSD Theorems of Kolyvagin, Kato, et al.} \subsection{Mazur-Rubin: Kolyvagin Systems} Let $E$ be an elliptic curve. Suppose $p>3$ and $\rho_{E,p}:G_\Q \ra \Aut(E[p^\infty])$ is surjective. \section{Problems} \begin{enumerate} \item {\bf Surjective:} Give an algorithm that, given a non-CM elliptic curve $E/\Q$, determines the primes~$p$ such that $\rho_{E,p}$ is not surjective. Should be reasonable using Serre's paper, etc. \item {\bf Tamagawa Number Problem:} Consider the following rank~$0$ example: \begin{verbatim} 114C 5 4 [5] [20,1,1] 2C \end{verbatim} Our methods only proved that $\#\Sha(E)\mid 25$, but in fact we have to show that $\#\Sha(E)=1$. Using Kato, and assuming that $r_E=1$, an assumption we could check ..., we only get again that $\#\Sha(E)\mid 25$, since $L_N(E,1)/\Omega_E$ is divisible by $25$: we get one $5$ from $L(E,1)/\Omega$ because of the Tamagawa number, and another $5$ from the bad factor at $19$, which is $1/(19-(-1))$. If we use Kato as improved by Mazur and Rubin (see page 75 of ``Kolyvagin Systems''), and if their $r_E=1$, then we get that $$ \ord_5(\#\Sha(E)[5^\infty]) < \ord_5(L_N(E,1)/\Omega_E) = 2. $$ Since $\ord_5(\#\Sha(E)[5^\infty])$ is even, it follows that $5\mid \#\Sha(E)[5^\infty]$. {\bf There is a problem with this: Mazur-Rubin assume that the representation on $E[5^\infty]$ is surjective, but I only know that the representation on $E[5]$ is surjective.} The problem isn't a problem because Lemma 3 on page IV-23 of Serre's {\em Abelian $\ell$-adic representations and elliptic curves} shows that if $p\geq 5$ then $G_\Q\to \Aut(E[p])$ is surjective if and only if $G_\Q \to \Aut(E[p^\infty])$ is surjective. The programs in my directory /home/was/people/rubin/r_e seem to suggest the $5\mid r_E$. Alternative idea, which doesn't work: In this case the Heegner point reduces to a point in the identity component, but $E(K)$ reduces to something of order divisible by~$5$. Thus if we could prove the following refinement of McCallum,Kolyvagin, then we could resolve this example: \begin{conjecture} Suppose $y_K$ has infinite order. Then $E(K)$ has rank~$1$ and $\Sha(E/K)$ is finite. Further, if $p$ is an odd prime which is unramified in $F$ and such that $\rho_{E,p}$ is surjective, then $$ \ord_p(\#\Sha(E/K)) \leq 2 \ord_p([E_0(K) : \Z y_K]), $$ where $$ E_0(K) := \ker\left(E(K) \ra \bigoplus_{v} \Phi_{E,v}(\F_v)\right). $$ \end{conjecture} \item {\bf Heegner:} My Heegner point program needs to be improved. Precision; maybe use best discriminant, which might not be the smallest (in abs. value). Look at Cremona's and Peter's. OR -- just compute $L$-function and use mwrank to find what multiple it is of the generator. However, in using mwrank we need to know that we've really found a generator. \end{enumerate} \begin{verbatim} E = elliptic curve G = GCD(H) B = gcd(#(E(F_p)) : p < 1000, p odd and good) H = odd part of indexes of Heegner points corr to first five quad imag fields T = Tamagawa numbers Thus Sha(E/Q) divides G^2*B^oo*2^oo. E G B H T verify deg 11A 1 5 [1] [5] 2C5TP 1 14A 1 6 [1] [2,3] 2C3TP 1 15A 1 8 [1] [2,4] 2C 1 17A 1 4 [1] [4] 2C 1 19A 1 3 [1] [3] 2C3TP 1 20A skip 21A 1 8 [1] [4,2] 2C 1 24A skip 26A 1 3 [1] [1,3] 2C3TP 2 26B 1 7 [1] [7,1] 2C7TP 2 30A 1 12 [1] [2,3,1] 2C3TP 1 33A 1 4 [1] [2,2] 2C 35A 1 3 [1] [1,3] 2C3TP 36A 1(skip) [1] 37B 1 3 [1] [3] 2C3TP 38A 1 3 [1] [1,3] 2C3TP 38B 1 5 [1] [5,1] 2C5TP 39A 1 4 [1] [2,2] 2C 40A 1(skip) [1] 42A 1 8 [1] [8,1,1] 2C 43A 44A 1 45A 1 46A 1 2 [2,1] 2C 48A skip 49A skip 50A skip 50B skip 51A 1 3 [3,1] 2C3TP 52A skip 54A 54B 55A 1 4 [2,2] 2C 56A 56B 57B 1 4 [2,2] 2C 57C 1 5 [10,1] 2C5TP 58B 1 5 [10,1] 2C5TP 62A 1 4 [4,1] 2C 63A skip 64A skip 66A 1 6 [2,3,1] 2C3TP 66B 1 4 [4,1,1] 2C 66C ? 10 [10,5,1] 2C5TP 67A ? 1 [1] 2C 69A 1 2 [2,1] 2C 70A 1 4 [4,2,1] 2C 72A skip 73A 1 2 [3,1] [2] 2C 77B 1 3 [6,1] 2C3TP 77C 1 2 [1,2] 2C 114C 5 4 [5] [20,1,1] 2C C = Cremona's mwrank (2-descent) TP = Torsion point paper of Stoll (p-isogeny descent in presence of a rational point of order p). \end{verbatim} \section{Misc} \begin{proposition}Suppose $E$ is an elliptic curve over $\Q$ and $p\geq 5$ is a prime such that $E$ does not have a rational $p$-isogeny (equivalently, so that $E[p]$ is an irreducible $\Gal(\Qbar/\Q)$ module). If $K$ is a quadratic extension of $\Q$, then $E(K)$ has no $p$ torsion. \end{proposition} \begin{proof} Suppose $z\in E[p]$ is nonzero. Because $E[p]$ is irreducible, the Galois closure $L$ of $\Q(z)$ has Galois group isomorphic to the image $H$ of $\Gal(\Qbar/\Q)$ in $\Aut(E[p])$. Because of the Weil pairing, $H$ has order at least $p-1$, so since $p\geq 5$, we see that $L$ has degree $\geq 4$. If $z\in E(K)$, then $K=L$ since $K$ is Galois, and this contradicts that $L$ has degree $\geq 4$. \end{proof} \begin{conjecture} Suppose $p\geq 5$ and there are no $K$-rational $p$-isogenies. Then $\H^i(K(E[p])/K, E[p]) = 0$ for $i=1,2$. \end{conjecture} \newpage \section{11} \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.]] \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 $E$ 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} \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|}}{\|\omega_E\|^2} $. Here $u$ is half the number of units in the ring of integers of $K$ 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 $\omega_E$. [[andrei write down where there is a factor of $2$!!!]] 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} \begin{remark} Possible different approach, which might be especially useful when $E(\Q)$ has rank $0$: Use half integral weight forms to compute the index $[E(K):\Z y_K]$ as the coefficient of a modular form. This could be, in some sense (?), more efficient than directly computing $E^D(\Q)$. \end{remark} \begin{example} We run through the algorithm for the ``first'' elliptic curve $E=X_0(11)$: $$ y^2 + y = x^3 - x^2 - 10x - 20. $$ \begin{enumerate} \item By Cremona's table, there is only a $5$-isogeny, $t=10$. \item The Heegner hypothesis is that $11$ splits completely in $K$, and the discriminant of $K$ should be coprime to $11$. Try $K=\Q(\sqrt{-1})$. Nope. Try $K=\Q(\sqrt{-2})$; yes, this works. \item From the table, the sign is $+1$. \item The twist of $E$ by $D=-8$ has minimal model $$ y^2 = x^3 - x^2 - 41x + 199. $$ This is {\bf 704K2} in Cremona's tables. The Mordell-Weil group is generated by $(2,11)$. (Should the following MAGMA code be believed? No.) \begin{verbatim} > G, f:=MordellWeilGroup(F); > G; Abelian Group isomorphic to Z Defined on 1 generator (free) > f(G.1); (2 : 11 : 1) \end{verbatim} \item Using {\tt BG.gp}, we find that $$ L(E,1) \sim 0.2538418805947805478584144442 $$ and $$ L'(E^D,1) \sim 1.887913559019476476731614099 $$ \begin{verbatim} ? \r BG ? e=ellinit([ 0, -1, 1, -10, -20 ]); ? ellanalyticrank(e) Summing 7 a_n terms Rank is even L^(0)=0.2538418805947805478584144442 [0, 0.2538418805947805478584144442, 1.000000077760499379277370890] ? e=ellinit([ 0, -1, 0, -41, 199 ]); ? ellanalyticrank(e) Summing 53 a_n terms Rank is odd L^(1)=1.887913559019476476731614099 [1, 1.887913559019476476731614099, 0.1830190942955215687124786121] \end{verbatim} Also \begin{verbatim} ? e=ellinit([ 0, -1, 1, -10, -20 ]) ? e.omega [1.269209304279553421688794617, \ 0.6346046521397767108443973084 + 1.458816616938495229330889613*I] ? e.omega[1]*imag(e.omega[2]) 1.851543623455959317708006712 ? sqrt(8)/% 1.527604907016368150928945837 \end{verbatim} So We have $u = 1$, $D=-8$, and $\|\omega_E\|\sim 1.8515436234$, so $$ \alpha = \frac{u^2 \sqrt{|D|}}{\|\omega_E\|} \sim 1.527604907016368150928945837 $$ Finally, $$ h(y_K) \sim 0.7320764341087109483011606701 $$ \begin{verbatim} ? 1.887913559019476476731614099 * \ 0.2538418805947805478584144442 * \ 1.527604907016368150928945837 0.7320764341087109483011606701 \end{verbatim} \item \begin{verbatim} ? e=ellinit([ 0, -1, 0, -41, 199 ]); ? 2*ellheight(e,[2,11]) 0.1830190931500931069448448415 \end{verbatim} We have \begin{verbatim} ? 0.7320764341087109483011606701 / 0.1830190931500931069448448415 4.000000336076075243576248085 \end{verbatim} Thus $I_K = 2$. \item We conclude that $\Sha(E/\Q)[p]=0$ for $p\neq 2,5$. \item Leave this to Stoll... \end{enumerate} \bibliography{biblio} \end{document}