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\title{\sc{}There are genus one curves over~$\Q$\\of every odd index}
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\author{William A. Stein\footnote{Supported by an NSF Mathematical
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Sciences Postdoctoral Research Fellowship.}\\
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\small email: {\sf was@math.harvard.edu}}
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\begin{document}
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\maketitle
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\begin{abstract}
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The index of a genus one curve~$X$ over a field~$K$ is the smallest
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degree of an extension~$L$ of~$K$ such that $X(L)$ is nonempty.
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Let~$K$ be a number field. We prove that for every integer~$r$ not
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divisible by~$8$, there is a genus one curve~$X$ over~$K$ of
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index~$r$. Our proof involves an analysis of Kolyvagin's Euler system
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of Heegner points combined with explicit computations on the modular
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curve $X_0(17)$.
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\end{abstract}
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\section{Introduction}
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How complicated are curves of genus one? One possible measure of the
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complexity of a curve is the smallest degree of an extension of the
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base field in which the curve has a point. Consider a curve~$X$ of
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positive genus~$g$ over a number field~$K$. The canonical divisor
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class on~$X$ contains a $K$-rational effective divisor of degree
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$2g-2$, so the greatest common divisor of the degrees of the extension
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fields in which~$X$ has a rational point divides $2g-2$. When $g=1$
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this is no condition at all!
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In the 1950s, S.~Lang and J.~Tate asked in \cite{lang-tate} whether,
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given a positive integer~$r$, there exists a genus one curve~$X$ such
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that~$r$ is the smallest of all degrees of extensions of~$K$ over
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which~$X$ has a point. Using Kolyvagin's Euler system of Heegner
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points, we answer their question in the affirmative, under the
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hypothesis that~$r$ is odd. The curves we produce are torsors for the
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elliptic curve $X_0(17)$, though our methods apply to a more
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general class of genus one curves.
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The following theorem is proved in Section~\ref{sec:conclusion}.
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\begin{theorem}\label{thm:overk}
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Let~$K$ be a number field and let~$r$ be an integer not divisible
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by~$8$. Then there are infinitely many genus one curves
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over~$K$ of index~$r$.
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\end{theorem}
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In Section~\ref{sec:indexes} we recall standard facts about indexes
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of genus one curves. Section~\ref{sec:kolyvagin} contains a brief
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discussion of Heegner points, and summarizes the relevant results
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about Kolyvagin's Euler system from \cite{rubin:kolyvagin}.
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In Section~\ref{sec:nonvanishing}, which forms the heart of our
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paper, we prove a nonvanishing result for Kolyvagin's cohomology
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classes. Finally, in Section~\ref{sec:genusoneexamples}, we prove
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Theorem~\ref{thm:overk} by combining a general result about Galois
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representations with explicit computations on $X_0(17)$.
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{\bf Acknowledgement:} The author would like to thank H.~Lenstra for
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introducing him to this problem, K.~Buzzard for teaching him about
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Kolyvagin's Euler system, K.~Rubin and M.~Flach for extensive
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comments, D.\thinspace{}Y.~Logachev, C.~O'Neil, and K.~Ribet for
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inspiring conversations, and N.~Elkies and G.~Grigorov for useful
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comments.
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\section{Indexes of genus one curves}\label{sec:indexes}
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Let~$E$ be an elliptic curve over an arbitrary field~$k$. The Galois
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cohomology group $H^1(k,E)=H^1(\Gal(\ksep/k),E(\ksep))$ classifies the
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isomorphism classes of torsors (principal homogeneous spaces)
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for~$E$ over~$k$.
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\begin{definition}[Index of cohomology class]
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The \defn{index} of $c\in H^1(k,E)$, denoted
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$\ind(c)$, is the greatest common divisor of the degrees of the
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separable extensions~$K$ of~$k$ for which $\res_{K}(c)=0$.
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\end{definition}
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The torsor~$X$ corresponding to~$c$ is a genus one curve over~$k$ equipped
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with an action of~$E$. Furthermore, $X(K)\neq \emptyset$ exactly when
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$\res_K(c)=0$, so
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$$\ind(c)=\gcd\{[K : k] : X(K)\neq \emptyset\}.$$
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Thus $\ind(c)$ generates the image of the degree map
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$\deg:\Div_k(X)\ra\Z$.
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We now define $\ind(X)$ so that $\ind(X) = \ind(c)$.
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\begin{definition}[Index of curve]
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The \defn{index} of an algebraic
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curve over~$k$ is the cardinality of the cokernel of
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the degree map.
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\end{definition}
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Any canonical divisor is an element of $\Div_k(X)$ of degree $2g-2$,
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where~$g$ is the genus of~$X$, so $\ind(X)$ divides $2g-2$. As
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mentioned in the introduction, when $g=1$ this is no condition; in
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fact, E.~Artin conjectured, and Lang and Tate proved
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in~\cite[pg. 670]{lang-tate}, that for every integer~$r$ there is some
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genus one curve~$X$ over some field~$L$ such that $\ind(X)=r$. The
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construction of \cite{lang-tate} requires the existence of an
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$L$-rational point of order~$r$ on the elliptic curve $E=\Jac(X)$.
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The torsion subgroups of elliptic curves are ``uniformly bounded'', so
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for~$K$ a fixed number field and for almost all~$r$, the results of
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\cite{lang-tate} do not imply the existence of genus one curves
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over~$K$ of index~$r$.
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Let~$E$ be an elliptic curve over a number field~$K$, and let~$r$ be a
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positive integer. Is there an element of $H^1(K,E)$ of index~$r$? In
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\cite{shafarevich:exp}, Shafarevich proved that $H^1(K,E)$ contains
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infinitely many elements of every {\em order} (see also
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\cite[\S27]{cassels:diophantinecurve} where Cassels sketches an
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alternative approach to proving Shafarevich's theorem). However, this does
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not answer the question of Artin, because the order need not equal the
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index as Cassels remarked in \cite{cassels:arithmeticV}, where he
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found an elliptic curve~$E$ and a class $c\in{}H^1(\Q,E)$ such
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that~$c$ has order~$2$ and index~$4$.
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\subsection{Elementary facts about the index}
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We pause to state some basic facts about the order and index, which we
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will use later. Fix an elliptic curve~$E$ over a number field~$K$,
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and let~$c$ and~$c'$ be elements of $H^1(K,E)$.
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\begin{proposition}\label{prop:lang-tate}
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$\ord(c)\mid \ind(c)$, and they have the same prime factors.
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\end{proposition}
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\begin{proof}
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See \cite[\S2, Prop.~5]{lang-tate}.
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\end{proof}
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\begin{lemma}\label{lem:indsplit}
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There is an extension~$L$ of~$K$ such that $[L:K]=\ind(c)$
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and $\res_L(c)=0$.
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\end{lemma}
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\begin{proof}
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See the paragraph before the corollary in
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\cite[\S2]{lang-tate}.
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\end{proof}
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\begin{proposition}\label{prop:indmul}
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Suppose $c'$ has order coprime to~$c$.
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Then $\ind(c+c')=\ind(c)\cdot\ind(c')$.
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\end{proposition}
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\begin{proof}
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If~$M$ is a field that splits $c+c'$, then~$M$ also splits
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$\ord(c')(c+c')=\ord(c')c$, so~$M$ splits~$c$. Likewise,~$M$
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splits~$c'$, so $\ind(c)\cdot \ind(c')\mid \ind(c+c')$. For the other
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divisibility, note that by Lemma~\ref{lem:indsplit}, there are
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extensions~$L$ and~$L'$ such that $[L:K]=\ind(c)$, $[L':K]=\ind(c')$,
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and $\res_L(c)=\res_{L'}(c')=0$. Then the compositum $L.L'$ splits
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$c+c'$ and $[L.L':K]=\ind(c)\cdot \ind(c')$. Thus $\ind(c+c')$
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divides $\ind(c)\cdot \ind(c')$.
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\end{proof}
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\begin{remark}
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In \cite{lichtenbaum:duality}, Lichtenbaum proved that
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$\ind(c)\mid \ord(c)^2$ for any $c\in H^1(K,E)$, and
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Cassels proved in \cite{cassels:arithmeticIV} that if $c\in\Sha(E/K)$,
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then $\ord(c)=\ind(c)$.
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\end{remark}
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If~$E$ is an elliptic curve over~$\Q$ such that
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$\#\Sha(E/\Q)=\#E(\Q)_{\tor}=1$,
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then the results mentioned above do not rule out the possibility that every
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element of $H^1(\Q,E)$ has index a perfect square. We prove, under
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the assumption that~$L(E,1)\neq 0$, that there is an
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integer~$B$ such that $H^1(\Q,E)$ contains infinitely many elements of
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index~$n$, for every integer~$n$ that is prime to~$B$ (see
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Theorem~\ref{thm:index}). For example, in
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Section~\ref{sec:genusoneexamples} we prove that one can take $B=2$ for
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the elliptic curve $X_0(17)$.
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\section{Kolyvagin's Euler system}\label{sec:kolyvagin}
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In this section, we recall the definition of Heegner points and
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several basic results about the system of cohomology classes Kolyvagin
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attaches to these points. We also state the main theorem of this paper.
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\subsection{Kolyvagin classes}
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Let~$E$ be an elliptic curve over~$\Q$ of conductor~$N$, and denote
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by $X_0(N)$ the modular curve that classifies cyclic isogenies
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of degree~$N$.
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By \cite{breuil-conrad-diamond-taylor}, there is a surjective
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map $\pi:X_0(N)\ra E$.
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(Note that for the proof of Theorem~\ref{thm:overk} we do not
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need any modularity theorems, because we take $E=X_0(17)$.)
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Let~$K$ be a quadratic imaginary extension of~$\Q$ in which
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all primes dividing~$N$ split, and let~$D_K$ be the discriminant
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and~$\cO$ the ring of integers of~$K$.
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Since all primes dividing~$N$ split, there is
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an ideal $\ga\subset \cO$ such that $\cO/\ga$ is cyclic of order~$N$.
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Let~$H$ be the Hilbert class field of~$K$, and
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$x_H\in X_0(N)(H)$ be the Heegner points corresponding to
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$(\C/\cO,\ga^{-1}/\cO)$.
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Set $y_H=\pi(x_H)\in E(H)$,
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$y_K=\tr_{H/K}(y_H)\in E(K)$,
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and $y=y_K-y_K^{\tau}\in E(K)^{-}$, where~$\tau$ denotes
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complex conjugation.
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Assume that $L(E,1)\neq 0$, so by \cite{bump-friedberg-hoffstein, murty-murty}
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there are infinitely many ways in which to choose~$K$ as above
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so that~$y$ has infinite order.
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Under this nonvanishing hypothesis on $L(E,1)$, Kolyvagin proves in
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\cite{kolyvagin:structureofsha} that the groups $E(\Q)$
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and $\Sha(E/\Q)$ are both finite.
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In the course of his proof, Kolyvagin considers more general Heegner
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points $y_{\ell}\in E(\Qbar)$, for appropriate primes~$\ell$, and from
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these constructs cohomology classes $c_{\ell,p^n}\in H^1(\Q,E)[p^n]$
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that are used to bound the orders of certain Selmer groups associated
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to~$E$. We will study Kolyvagin's classes further and prove that for
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each prime~$p$ not in an explicit finite set and each positive
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integer~$n$, there are infinitely many primes~$\ell$ such that
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$$\ord(c_{\ell,p^n}) = \ind(c_{\ell,p^n}) = p^n.$$
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We thus obtain the following theorem, which will
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be proved in Section~\ref{sec:thm-index}.
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\begin{theorem}\label{thm:index}
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Let~$E$ be an elliptic curve over~$\Q$ such that
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$L(E,1)\neq 0$. Then there is an integer~$B$ such that, for
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all integers~$r$ coprime to~$B$, there are infinitely
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many $c \in H^1(\Q,E)$ such that $\ord(c)=\ind(c)=r$.
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\end{theorem}
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\begin{remark}
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Cathy O'Neil \cite{oneil} has investigated the obstruction to
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$\ord(c)=\ind(c)$. We show that when~$E$ has analytic
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rank~$0$, this obstruction vanishes for infinitely many~$c$.
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\end{remark}
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\subsection{Basic properties of Kolyvagin's Euler system}
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In \cite{rubin:kolyvagin}, Rubin gives a concise account of
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Kolyvagin's proof of finiteness of $\Sha(E/\Q)[p^\infty]$, under the
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simplifying assumption that~$p$ is odd. Though Kolyvagin's argument
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works even when $p=2$, for simplicity, we rely exclusively on Rubin's
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paper.
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Let~$K$ be a quadratic imaginary field as above, chosen in such a way
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that the associated Heegner point $y_K$ has infinite order. Fix
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embeddings of~$\Qbar$ into~$\C$ and into each $p$-adic field $\Qpbar$.
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Let $\tau$ denote complex conjugation, and for any
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$\Z[\tau]$-module~$A$, let $A^+$ and $A^-$ denote the kernel of
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$\tau-1$ and $\tau+1$, respectively. For the remainder of this section,
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we assume that~$p$ is an odd prime, and if $K=\Q(\sqrt{-3})$
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that $p\geq 5$. If~$\ell$ is a prime that is inert in~$K$,
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let~$K_\ell$ denote the completion of~$K$ at the unique prime lying
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over~$\ell$. If $L$ is a finite Galois extension of~$\Q$, let
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$\Frob_\ell(L/\Q)$ denote the conjugacy class of some Frobenius element
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of a prime lying over~$\ell$. For each prime $\ell\nmid N$, let
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$a_\ell=\ell+1-\#E(\F_\ell)$ be the $\ell$th Fourier coefficient of
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the newform attached to~$E$.
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\begin{definition}\label{def:mv}
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For each place~$v$ of~$\Q$, let
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$$m_v=\#H^1(\Qv^{\unr}/\Qv,E(\Qv^{\unr})).$$
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By \cite[I.3.8]{milne:duality}, each $m_v$ is finite
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and $m_v=1$ for all but finitely many~$v$, so
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$$m(p)=\sup\{\ord_p(m_v): \mbox{all places } v\mbox{ of }\Q \}$$
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is well defined, and $m(p)=0$ for almost all~$p$.
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\end{definition}
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Let~$n$ be a positive integer.
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\begin{proposition}\label{prop:rubin2}
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Let~$p$ be a prime that does not divide the class
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number of~$K$ and for which $m(p)=0$.
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Suppose $\ell\nmid p D_K N$ and $\Frob_\ell(K(E[p^n])/\Q)=[\tau]$.
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Then there is an element $c_{\ell,p^n}\in H^1(\Q,E)[p^n]$ such that
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%$\res_v(c_{\ell,p^n}) = 0$ for
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%all places $v\neq \ell$ of $\Q$, and
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the order of $\res_\ell(c_{\ell,p^n})$ in $H^1(\Ql,E)[p^n]$ is equal
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to the order of the image of~$y$ in $E(K_\ell)/p^n E(K_\ell)$,
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and the index of $c_{\ell,p^n}$ divides~$p^n$.
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\end{proposition}
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\begin{proof}
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The existence of $c_{\ell,p^n}$ and statement about its order is
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proved in \cite[Prop.~5]{rubin:kolyvagin}, where $c_{\ell,p^n}$ is
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constructed from Heegner points on $X_0(N)$. For the index bound,
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note that in the proof of \cite[Prop.~5]{rubin:kolyvagin}, when
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$p\nmid [H:K]$, Rubin constructs a class $c'\in
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H^1(K'/K,E(K'))[p^r]^+$, where $r=n+m(p)$ and~$K'$ is the unique
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extension of~$K$ of degree~$p^r$ in a certain class field of~$K$.
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Since~$p$ is odd, the restriction map $\res : H^1(\Q,E)[p^r]\ra
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H^1(K,E)[p^r] ^+$ is an isomorphism. Rubin takes
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$c_{\ell,p^n}=\res^{-1}(c')$. Since $c_{\ell,p^n}$ splits
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over the degree $2p^r$ extension~$K'$ of~$\Q$, the index of
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$c_{\ell,p^n}$ divides $2p^r$. But $c_{\ell,p^n}$ has odd order and,
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by Proposition~\ref{prop:lang-tate}, $\ind(c_{\ell,p^n})$ has the same
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prime factors as $\ord(c_{\ell,p^n})$, so $\ind(c_{\ell,p^n})$
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divides~$p^r$.
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\end{proof}
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\begin{remark}
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The author does not know whether or not
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the proposition is true
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if~$p$ is allowed to divide the class number of~$K$.
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\end{remark}
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\section{Nonvanishing of cohomology classes}\label{sec:nonvanishing}
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In this section, we prove a nonvanishing result
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about the cohomology classes $c_{\ell,p^n}$ of
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Proposition~\ref{prop:rubin2}, then use it to deduce
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Theorem~\ref{thm:index}.
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\subsection{Local nonvanishing}
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Let~$E$ be as above.
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For any point $x\in E(K)$,
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let $K([p^n]^{-1}x)$ denote the field obtained by adjoining
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the coordinates of all $p^n$th roots of~$x$ to~$K$.
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Without imposing further hypothesis, this field need
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not be Galois over~$\Q$.
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\begin{lemma}\label{lem:isgalois}
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If $x\in E(K)^+\union E(K)^{-}$, then
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$K([p^n]^{-1}x)$ is Galois over~$\Q$.
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\end{lemma}
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\begin{proof} Since $G_\Q$ acts on~$x$ by $\pm 1$, the
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subgroup~$\Z x$ is $G_\Q$-invariant. Since
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$[p^n] \colon{} E\ra E$ is a $\Q$-rational isogeny
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the inverse image $[p^n]^{-1}\Z x$ is also $G_\Q$-invariant,
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so $K([p^n]^{-1}x)=K([p^n]^{-1}\Z x)$ is Galois over~$\Q$.
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\end{proof}
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\begin{definition}
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An odd prime~$p$ is {\it firm} for~$E$ if
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$m(p)=0$, there are no nontrivial
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$\Q$-rational cyclic subgroups of $E[p^\infty]$,
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and $H^1(K(E[p^n])/K, E[p^n])=0$
403
for all $n\geq 1$.
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\end{definition}
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\begin{remark}
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The set of primes that are not firm is finite, by
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Serre's theorem \cite{serre:propgal} and the theory
409
of complex multiplication.
410
\end{remark}
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412
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Let~$p$ be an odd prime that is firm for~$E$.
414
The following proposition produces infinitely many
415
primes~$\ell$ such that we have control over the
416
orders of the image in $E(K_\ell)/p^n E(K_\ell)$
417
of a global point.
418
It will be used as input to Proposition~\ref{prop:rubin2} to
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produce cohomology classes of known index.
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The proof, which involves an application of the
421
Chebotar\"ev density theorem, follows a strategy similar
422
to that used in the proof of Kolyvagin's theorem on
423
page 135 of~\cite{rubin:kolyvagin}.
424
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\begin{proposition}~\label{prop:ordind}
426
Let $p$ be a prime that is firm for~$E$, and let $x\in E(K)^{\pm}$.
427
Then there is a set of primes~$\ell$ of positive Dirichlet density
428
such that $\Frob_\ell(K(E[p^n])/\Q)=[\tau]$ and
429
the orders of the images of~$x$ in $E(K)/p^n E(K)$
430
and in $E(K_\ell)/p^n E(K_\ell)$ are the same.
431
\end{proposition}
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\begin{proof}
433
Let $p^a$ be the order of the image of~$x$ in $E(K)/p^n E(K)$. If
434
$a=0$, then there is nothing to prove, so assume that $a>0$.
435
If~$\ell$ is a prime such that the orders of the images of $p^{a-1}x$ in
436
$E(K)/p^n E(K)$ and $E(K_\ell)/p^n E(K_\ell)$
437
both equal~$p$,
438
then the images of~$x$ in
439
$E(K)/p^n E(K)$ and
440
$E(K_\ell)/p^n E(K_\ell)$ both have
441
order $p^a$.
442
It thus suffices to prove the proposition in the case when
443
the order of the image of~$x$ in $E(K)/p^n E(K)$ is~$p$.
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Let $L=K(E[p^n])$, suppose~$\ell$ is a prime such that
446
$\Frob_\ell(L/\Q)=[\tau]$, and let~$\lambda$ be one of the prime
447
ideals of~$L$ that lies over~$\ell$. We have a diagram
448
$$\xymatrix{
449
{E(K)/p^n E(K)\,\,}\[email protected]{^(->}[r]\ar[d]
450
& {H^1(K,E[p^n])\,\,}\[email protected]{^(->}[r]\ar[d] & {\Hom(G_L,E[p^n])}\ar[d]\\
451
{E(K_\ell)/p^n E(K_\ell)}\ar[r]
452
& {H^1(K_\ell,E[p^n])}\ar[r] & {\Hom(G_{L_\lambda},E[p^n]),}
453
}$$
454
Let $\varphi : G_L \ra E[p^n]$ be the element of
455
$\Hom(G_L,E[p^n])$ that~$x$ maps to. The top row
456
is injective, because~$p$ is firm, so it suffices
457
to show that the image $\varphi_\ell$ of~$\varphi$ in
458
$\Hom(G_{L_\lambda},E[p^n])$ is nonzero.
459
460
Let~$M$ be the fixed field of the kernel of~$\varphi$.
461
Since~$M$ is the compositum of
462
the two Galois extensions $K([p^n]^{-1}x)$ and $\Q(E[p^n])$
463
of~$\Q$, it is also Galois (see Lemma~\ref{lem:isgalois}).
464
Because $\Frob_{\ell}(M/\Q)|_L = [\tau]$, there is an element
465
$\sigma\in \Gal(M/L)$ such that
466
$\Frob_{\ell}(M/\Q) = [\sigma\tau]$.
467
The order of $\sigma\tau$
468
equals the degree of $M_{\lambda'}$ over $\Q_\ell$,
469
where $\lambda'$ is a prime of~$M$ lying over~$\ell$.
470
If $\varphi_\ell=0$, then $M_{\lambda'}=L_\lambda=K_\ell$,
471
so $\sigma\tau$ would have order~$2$.
472
473
The image of~$\varphi$ is a nonzero subgroup~$H$ of $E[p^n]$,
474
which is defined over~$\Q$ since $x\in E(K)^{\pm}$.
475
If every $\sigma\in \Gal(M/L)$ has the property that
476
$\sigma\tau$ has order~$2$,
477
then $H\subset E[p^n]^{-}$. This contradicts our assumption
478
that~$p$ is firm, since $H$ is a nontrivial cyclic subgroup
479
of $E[p^\infty]$. Thus there exists $\sigma \in \Gal(M/L)$
480
such that $\sigma\tau$ has order different than~$2$.
481
For this~$\sigma$ and for any
482
prime~$\ell$ such that
483
$\Frob_\ell(M/\Q) = [\sigma\tau]$, we see that
484
$\varphi_\ell\neq 0$.
485
The Chebotar\"ev density theorem provides a positive
486
density of such~$\ell$.
487
\end{proof}
488
489
\subsection{Proof of Theorem~\ref{thm:index}}
490
\label{sec:thm-index}
491
\begin{proof}[Proof of Theorem~\ref{thm:index}]
492
Let~$E$ be an elliptic curve over~$\Q$ such that $L(E,1)\neq 0$.
493
Let~$K$ be one of the infinitely many imaginary quadratic fields
494
such that the associated Heegner point~$y$ has infinite order.
495
Let $B_K$ be an integer that is divisible by~$2$ and
496
\vspace{-1ex}
497
\begin{itemize}
498
\item[---] the primes~$p$ such that $y\in pE(K)$,
499
\vspace{-1.5ex}
500
\item[---] the primes~$p$ that are not firm,
501
\vspace{-1.5ex}
502
\item[---] the order $\#E(K)_{\tor}$, and
503
\vspace{-1.5ex}
504
\item[---] the class number of $K$.
505
\vspace{-1ex}
506
\end{itemize}
507
If $K=\Q(\sqrt{-3})$, assume in addition that~$3$ divides~$B_K$.
508
509
Fix a prime $p\nmid B_K$.
510
Since $E(K)$ has rank~$1$ (see, e.g., \cite[Thm 1.3]{gross:kolyvagin})
511
and $p\nmid \#E(K)_{\tor}$, the image of~$y$ in $E(K)/p^n E(K)$
512
has order $p^n$. By Proposition~\ref{prop:ordind} there are
513
infinitely many primes~$\ell$ such that
514
$\Frob_\ell(K(E[p^n])/\Q)=[\tau]$ and the image of~$y$
515
in $E(K_\ell)/p^n E(K_\ell)$ has order~$p^n$.
516
For these~$\ell$, Proposition~\ref{prop:ordind} produces infinitely many
517
cohomology classes $c_{\ell,p^n}$ having order and index both
518
equal to~$p^n$. (Note that if $\ell\neq \ell'$ then
519
$c_{\ell,p^n}\neq c_{\ell',p^n}$.)
520
521
Let~$B$ be the greatest common divisor of the set of integers
522
$B_K$, as~$K$ varies over all quadratic imaginary extensions such that
523
the associated Heegner point has infinite order.
524
For each prime power $p^n$ that does not divide $B$,
525
we have produced infinitely many $c\in H^1(\Q,E)$
526
having order and index both equal to $p^n$.
527
If the orders of~$c$ and~$c'$
528
are coprime, then $\ord(c+c') = \ord(c)\cdot \ord(c')$ and,
529
by Proposition~\ref{prop:indmul},
530
$\ind(c+c')=\ind(c)\cdot \ind(c')$.
531
This proves the theorem.
532
\end{proof}
533
534
\section{Computing the bound $B_K$}
535
\label{sec:genusoneexamples}
536
In this section we compute, in some cases, the the bound $B_K$ that
537
appears in Section~\ref{sec:thm-index}. First we prove a general
538
theorem about semistable elliptic curves. Next we compute the index
539
of a Heegner point, and finally in Section~\ref{sec:conclusion}
540
we prove Theorem~\ref{thm:overk}.
541
542
\subsection{Galois representations attached to isolated curves}
543
The following proposition sometimes permits us to compute
544
the integer $B_K$, which appears in Section~\ref{sec:thm-index}.
545
%Note that we do not require this general of a proposition in
546
%order to prove Theorem~\ref{thm:overk}.
547
548
\begin{proposition}\label{prop:explicitindex}
549
Let~$E$ be a semistable elliptic curve over~$\Q$ of conductor~$N$,
550
let~$p$ be an odd prime, and let $K$ be a quadratic imaginary field
551
such that $\gcd(D_K, p N)=1$. Assume that $p\nmid \ord_\ell(j(E))$,
552
for each prime~$\ell$, and that~$E$ admits no isogenies of degree~$p$.
553
Then $p \nmid \#E(K)_{\tor}$ and~$p$ is firm for~$E$.
554
\end{proposition}
555
556
Before giving the proof, we summarize its main ingredients. First, we
557
observe that the assertion that $m(p)=0$ (see Definition~\ref{def:mv})
558
uses a standard result that relates unramified Galois cohomology to
559
component groups. Next, we use the semistability and isogeny
560
hypotheses to deduce that $\rho_{E,p}$ is surjective.
561
Then we use standard group cohomology to deduce that~$p$ is firm.
562
563
\begin{proof}
564
Let~$\ell$ be a prime.
565
By \cite[I.3.8]{milne:duality},
566
$H^1(\Q_\ell^{\unr}/\Q_\ell,E(\Q_\ell^{\unr})) \isom
567
H^1(\Fbar_\ell/\F_\ell, \Phi_{E,\ell}(\Fbar_\ell)),$
568
where $\Phi_{E,\ell}$ is the component group of~$E$ at~$\ell$.
569
If $\ell\nmid N$, there is nothing further to prove, so assume $\ell\mid N$.
570
Since~$E$ is semistable,
571
$\#\Phi_{E,\ell}(\Fbar_\ell) = -\ord_\ell(j)$. By hypothesis,
572
$p\nmid \ord_\ell(j)$. Thus $m(p)=0$.
573
574
Since~$E$ admits no isogenies of degree~$p$, the Galois representation
575
$\rho_{E,p}:G_\Q\ra\GL(2,E[p])$ is irreducible, and there are no
576
nontrivial $\Q$-rational cyclic subgroups of $E[p^\infty]$. Since~$E$
577
is semistable, work of Serre~\cite[Prop.~21]{serre:propgal} and
578
\cite[\S3.1]{serre:travauxwiles} implies that $\rho_{E,p}$ is
579
surjective. Thus $p\nmid \#E(K)_{\tor}$ because a point in $E(\Qbar)$
580
of order~$p$ must generate an extension of~$\Q$ of degree at least
581
$p^2-1\geq 3$.
582
583
The field~$K$ and $\Q(E[p])$ are linearly disjoint, since
584
$\gcd(D_K,pN)=1$, so
585
$$H^1(K(E[p])/K,E[p])\isom H^1(\Q(E[p])/\Q,E[p])\ncisom
586
H^1(\GL(2,\Fp), \Fp^2).$$
587
588
%The following argument shows that
589
%$H^1(\GL(2,\Fp), \Fp^2)=\{0\}$.
590
%(This is true even when $p=2$: use inf-res applied to the subgroup
591
%$A_3$ of $S_3=\GL(2,\F_2)$.)
592
%Since~$p$ is odd, there is a scalar
593
%$g=\abcd{\alpha}{0}{0}{\alpha}\in \GL(2,\Fp)$ such that $\alpha\neq 1$.
594
%Note that~$g$ is in the center of $\GL(2,\Fp)$,
595
%so~$g$ induces the morphism of
596
%pairs that is the identity on $\GL(2,\Fp)$ and
597
%multiplication by~$\alpha$ on $\Fp^2$.
598
%The morphism of pairs induced by an inner
599
%automorphism is the identity map
600
%$$H^1(\GL(2,\Fp), \Fp^2)\ra{}H^1(\GL(2,\Fp), \Fp^2).$$
601
%This map is also multiplication by~$\alpha$, so
602
%$H^1(\GL(2,\Fp), \Fp^2)$ has exponent dividing~$\alpha-1$,
603
%which is not divisible by~$p$.
604
%The exponent of $H^1(\GL(2,\Fp), \Fp^2)$ must divide~$p$,
605
%so $H^1(\GL(2,\Fp), \Fp^2)=0$.
606
607
%We verify that there is a nonidentity scalar $\alpha\in
608
%\Gal(K(E[p^n])/K)\subset\GL(2,\Z/p^n\Z)$.
609
The group~$H=H^1(K(E[p^n])/K,E[p^n])$ has exponent a power of~$p$.
610
If an element~$\alpha$ in $\Gal(K(E[p^n])/K) \subset \GL_2(\Z/p^n\Z)$ is scalar,
611
then every element of~$H$ has order dividing $\alpha-1$. This is
612
because the scalar is central, so the morphism of pairs it induces is
613
both the identity and multiplication by $\alpha$. It is necessary
614
only to choose $\alpha$ such that $\gcd(\alpha-1,p)=1$. Since~$p$ is
615
odd, $-1$ is a nonidentity element of $\Aut(E[p])=\Gal(K(E[p])/K)$.
616
Every automorphism lifts, so~$-1$ lifts to some~$g$ in
617
$\Gal(K(E[p^n])/K) \subset \Aut(E[p^n])$. Then $g^{p^{n-1}} = -1$ in
618
$\Aut(E[p^n])$, so $-1 \in \Gal(K(E[p^n])/K)$ and every element of~$H$
619
has order dividing~$2$.
620
(To show that $g^{p^{n-1}}=-1$, we use that
621
$\ord_p\binom{p^n}{k} = n + \ord_p\left(\frac{1}{k}\right)$.)
622
\end{proof}
623
624
\subsection{The number $B_K$ for $X_0(17)$}
625
In this section, we show that for $E=X_0(17)$ and $K=\Q(\sqrt{-2})$,
626
we have $B_K=2$. This is accomplished
627
by showing that the index $[E(K):\Z y]$ is a power of~$2$.
628
The elliptic curve $E=X_0(17)$ given by the Weierstrass equation
629
$$y^2 + xy + y = x^3 - x^2 - x -14$$
630
satisfies the hypothesis
631
of Proposition~\ref{prop:explicitindex} for each odd prime~$p$.
632
Since the $j$-invariant of~$E$ is $3^3\cdot 11^3/17^4$,
633
every odd prime~$p$ is firm for~$E$ and
634
$\#E(K)_{\tor}$ is a power of~$2$.
635
636
The conductor $17$ of~$E$ splits in~$K$, and the
637
quadratic twist $E'$ of~$E$ by~$K$ is the curve
638
$y^2 = x^3 - 44x + 7120$, which is labeled {\bf 1088K4}
639
in \cite{cremona:onlinetables}.
640
Using MAGMA (or {\tt mwrank}), one
641
finds that $E'(\Q)\isom \Z{}P \times\Z/2$,
642
where $P=(-3,85)\in E'(\Q)$ has infinite order.
643
Since the rank of~$E'$ is~$1$, we set~$K=\Q(\sqrt{-2})$
644
in Section~\ref{sec:thm-index}.
645
Then~$B_K$ is divisible only by~$2$ and the index
646
$[E(K):\Z y]$. This index can only change
647
by a power of~$2$ if~$y$ is replaced by $y_K$, so we instead
648
consider the index $[E(K):\Z y_K]$.
649
The cokernel of the natural map $E(\Q)\oplus E'(\Q)\ra E(K)$
650
is a $2$-group and $E(\Q)\isom \Z/4\Z$, so
651
$[E(K):\Z y_K]$ is a power of~$2$ times $h(y_K)/h(P)$, where~$h$
652
is the N\'eron-Tate canonical height on~$E_K$.
653
By the Gross-Zagier formula (see~\cite[Thm.~6.3]{gross-zagier}),
654
\begin{equation}\label{eqn:index-grosszagier}
655
h(y_K) =
656
\frac{u^2 |D|^{\frac{1}{2}}}{||\omega_f||} L'_{E'}(1) L_E(1),
657
\end{equation}
658
where $D=-8$ is the discriminant of~$K$, $u=1$ is half the number of
659
units, and $||\omega_f||$ is the Peterson norm of the newform~$f$
660
corresponding to~$E$.
661
Generators for the period lattice of~$E$ are
662
$\omega_1 \sim 1.547079$ and $\omega_2 \sim 0.773539 + 1.372869i$;
663
taking the determinant gives $||\omega_f|| \sim 2.123938$.
664
Furthermore, again from \cite{cremona:onlinetables}, we find that
665
$L_E(1) \sim 0.386769$
666
and
667
$L'_{E'}(1) \sim 2.525026$,
668
so $h(y_K) \sim 1.300533$.
669
Using a computer, we find that $h(P) \sim 1.300533$
670
as well, so $[E(K):\Z y_K]$ is a power of two.
671
672
\subsection{Elements of index~$2$ and~$4$}
673
The torsion subgroup of $E=X_0(17)$ is isomorphic to $\Z/4\Z$, so
674
\cite[pg.~670]{lang-tate} implies that there are infinitely
675
many elements of $H^1(\Q,E)$ having order and index equal to~$2$,
676
and also infinitely many having order and index equal to~$4$.
677
678
679
\subsection{Proof of Theorem~\ref{thm:overk}}\label{sec:conclusion}
680
To prove Theorem~\ref{thm:overk}, we combine the above
681
computations with Theorem~\ref{thm:index}, and an observation
682
about the local properties of Kolyvagin's classes $c_{\ell,p^n}$.
683
684
\begin{proof}[Proof of Theorem~\ref{thm:overk}]
685
Let~$E=X_0(17)$ as above, and let~$K$ be an arbitrary number field.
686
Let $p^n$ be either an odd prime power, or $2$, or $4$.
687
The computations of the previous section combined with
688
Theorem~\ref{thm:index} prove that there are
689
infinitely many elements $c_{\ell,p^n}$ of $H^1(\Q,E)$ whose index and
690
order both equal $p^n$.
691
Let~$A$ be the subgroup of $H^1(\Q,E)$ generated by these classes.
692
The kernel~$B$ of $\res_K:A\ra H^1(K,E)$ is finite, so the set
693
$\mathcal{S}$ of primes~$\ell$ such that $\res_\ell(c)\neq 0$
694
for some $c\in B$ is finite. By Proposition~\ref{prop:rubin2},
695
we have $\res_v(c_{\ell,p^n})=0$ for all places $v\neq \ell$,
696
so the subgroup $A'$ of~$A$ generated by all $c_{\ell,p^n}$ with
697
$\ell\not \in \mathcal{S}$ has trivial
698
intersection with~$B$. Thus $\res_K(A')$ consists of infinitely
699
many classes in $H^1(K,E)$ having order and index both equal
700
to $p^n$, and the theorem now follows from Proposition~\ref{prop:lang-tate}.
701
\end{proof}
702
703
\providecommand{\bysame}{\leavevmode\hbox to3em{\hrulefill}\thinspace}
704
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