 CoCalc Public Fileswww / papers / index / stein-index.tex
Author: William A. Stein
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7%  AUTHOR:  William A. Stein                                        %
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54% Math operators
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73% Theorem styles
74\theoremstyle{plain}
75\newtheorem{theorem}{Theorem}[section]
76\newtheorem{proposition}[theorem]{Proposition}
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78\theoremstyle{definition}
79\newtheorem{definition}[theorem]{Definition}
80\theoremstyle{remark}
81\newtheorem{remark}[theorem]{Remark}
82
83\title{\sc{}There are genus one curves over~$\Q$\\of every odd index}
84\author{William A. Stein\footnote{Supported by an NSF Mathematical
85Sciences Postdoctoral Research Fellowship.}\\
86\small email: {\sf was@math.harvard.edu}}
87
88\begin{document}
89\maketitle
90\begin{abstract}
91The index of a genus one curve~$X$ over a field~$K$ is the smallest
92degree of an extension~$L$ of~$K$ such that $X(L)$ is nonempty.
93Let~$K$ be a number field.  We prove that for every integer~$r$ not
94divisible by~$8$, there is a genus one curve~$X$ over~$K$ of
95index~$r$.  Our proof involves an analysis of Kolyvagin's Euler system
96of Heegner points combined with explicit computations on the modular
97curve $X_0(17)$.
98\end{abstract}
99
100\section{Introduction}
101How complicated are curves of genus one?  One possible measure of the
102complexity of a curve is the smallest degree of an extension of the
103base field in which the curve has a point.  Consider a curve~$X$ of
104positive genus~$g$ over a number field~$K$.  The canonical divisor
105class on~$X$ contains a $K$-rational effective divisor of degree
106$2g-2$, so the greatest common divisor of the degrees of the extension
107fields in which~$X$ has a rational point divides $2g-2$.  When $g=1$
108this is no condition at all!
109
110In the 1950s, S.~Lang and J.~Tate asked in \cite{lang-tate} whether,
111given a positive integer~$r$, there exists a genus one curve~$X$ such
112that~$r$ is the smallest of all degrees of extensions of~$K$ over
113which~$X$ has a point.  Using Kolyvagin's Euler system of Heegner
114points, we answer their question in the affirmative, under the
115hypothesis that~$r$ is odd.  The curves we produce are torsors for the
116elliptic curve $X_0(17)$, though our methods apply to a more
117general class of genus one curves.
118The following theorem is proved in Section~\ref{sec:conclusion}.
119\begin{theorem}\label{thm:overk}
120Let~$K$ be a number field and let~$r$ be an integer not divisible
121by~$8$.  Then there are infinitely many genus one curves
122over~$K$ of index~$r$.
123\end{theorem}
124
125In Section~\ref{sec:indexes} we recall standard facts about indexes
126of genus one curves.  Section~\ref{sec:kolyvagin} contains a brief
127discussion of Heegner points, and summarizes the relevant results
128about Kolyvagin's Euler system from \cite{rubin:kolyvagin}.
129In Section~\ref{sec:nonvanishing}, which forms the heart of our
130paper, we prove a nonvanishing result for Kolyvagin's cohomology
131classes.  Finally, in Section~\ref{sec:genusoneexamples}, we prove
132Theorem~\ref{thm:overk} by combining a general result about Galois
133representations with explicit computations on $X_0(17)$.
134
135{\bf Acknowledgement:} The author would like to thank H.~Lenstra for
136introducing him to this problem, K.~Buzzard for teaching him about
137Kolyvagin's Euler system, K.~Rubin and M.~Flach for extensive
138comments, D.\thinspace{}Y.~Logachev, C.~O'Neil, and K.~Ribet for
139inspiring conversations, and N.~Elkies and G.~Grigorov for useful
141
142
143\section{Indexes of genus one curves}\label{sec:indexes}
144Let~$E$ be an elliptic curve over an arbitrary field~$k$.  The Galois
145cohomology group $H^1(k,E)=H^1(\Gal(\ksep/k),E(\ksep))$ classifies the
146isomorphism classes of torsors (principal homogeneous spaces)
147for~$E$ over~$k$.
148\begin{definition}[Index of cohomology class]
149The \defn{index} of $c\in H^1(k,E)$, denoted
150$\ind(c)$, is the greatest common divisor of the degrees of the
151separable extensions~$K$ of~$k$ for which $\res_{K}(c)=0$.
152\end{definition}
153The torsor~$X$ corresponding to~$c$ is a genus one curve over~$k$ equipped
154with an action of~$E$.  Furthermore, $X(K)\neq \emptyset$ exactly when
155$\res_K(c)=0$, so
156$$\ind(c)=\gcd\{[K : k] : X(K)\neq \emptyset\}.$$
157Thus $\ind(c)$ generates the image of the degree map
158$\deg:\Div_k(X)\ra\Z$.
159We now define $\ind(X)$ so that $\ind(X) = \ind(c)$.
160\begin{definition}[Index of curve]
161The \defn{index} of an algebraic
162curve over~$k$ is the cardinality of the cokernel of
163the degree map.
164\end{definition}
165
166Any canonical divisor is an element of $\Div_k(X)$ of degree $2g-2$,
167where~$g$ is the genus of~$X$, so $\ind(X)$ divides $2g-2$.  As
168mentioned in the introduction, when $g=1$ this is no condition; in
169fact, E.~Artin conjectured, and Lang and Tate proved
170in~\cite[pg. 670]{lang-tate}, that for every integer~$r$ there is some
171genus one curve~$X$ over some field~$L$ such that $\ind(X)=r$.  The
172construction of \cite{lang-tate} requires the existence of an
173$L$-rational point of order~$r$ on the elliptic curve $E=\Jac(X)$.
174The torsion subgroups of elliptic curves are uniformly bounded'', so
175for~$K$ a fixed number field and for almost all~$r$, the results of
176\cite{lang-tate} do not imply the existence of genus one curves
177over~$K$ of index~$r$.
178
179Let~$E$ be an elliptic curve over a number field~$K$, and let~$r$ be a
180positive integer.  Is there an element of $H^1(K,E)$ of index~$r$?  In
181\cite{shafarevich:exp}, Shafarevich proved that $H^1(K,E)$ contains
183\cite[\S27]{cassels:diophantinecurve} where Cassels sketches an
184alternative approach to proving Shafarevich's theorem).  However, this does
185not answer the question of Artin, because the order need not equal the
186index as Cassels remarked in \cite{cassels:arithmeticV}, where he
187found an elliptic curve~$E$ and a class $c\in{}H^1(\Q,E)$ such
188that~$c$ has order~$2$ and index~$4$.
189
191
192We pause to state some basic facts about the order and index, which we
193will use later.  Fix an elliptic curve~$E$ over a number field~$K$,
194and let~$c$ and~$c'$ be elements of $H^1(K,E)$.
195
196\begin{proposition}\label{prop:lang-tate}
197$\ord(c)\mid \ind(c)$, and they have the same prime factors.
198\end{proposition}
199\begin{proof}
200See \cite[\S2, Prop.~5]{lang-tate}.
201\end{proof}
202
203\begin{lemma}\label{lem:indsplit}
204There is an extension~$L$ of~$K$ such that $[L:K]=\ind(c)$
205and $\res_L(c)=0$.
206\end{lemma}
207\begin{proof}
208See the paragraph before the corollary in
209\cite[\S2]{lang-tate}.
210\end{proof}
211
212
213
214\begin{proposition}\label{prop:indmul}
215Suppose $c'$ has order coprime to~$c$.
216Then $\ind(c+c')=\ind(c)\cdot\ind(c')$.
217\end{proposition}
218\begin{proof}
219If~$M$ is a field that splits $c+c'$, then~$M$ also splits
220$\ord(c')(c+c')=\ord(c')c$, so~$M$ splits~$c$.  Likewise,~$M$
221splits~$c'$, so $\ind(c)\cdot \ind(c')\mid \ind(c+c')$.  For the other
222divisibility, note that by Lemma~\ref{lem:indsplit}, there are
223extensions~$L$ and~$L'$ such that $[L:K]=\ind(c)$, $[L':K]=\ind(c')$,
224and $\res_L(c)=\res_{L'}(c')=0$.  Then the compositum $L.L'$ splits
225$c+c'$ and $[L.L':K]=\ind(c)\cdot \ind(c')$.  Thus $\ind(c+c')$
226divides $\ind(c)\cdot \ind(c')$.
227\end{proof}
228
229
230\begin{remark}
231In \cite{lichtenbaum:duality}, Lichtenbaum proved that
232$\ind(c)\mid \ord(c)^2$ for any $c\in H^1(K,E)$, and
233Cassels proved in  \cite{cassels:arithmeticIV} that if $c\in\Sha(E/K)$,
234then $\ord(c)=\ind(c)$.
235\end{remark}
236
237If~$E$ is an elliptic curve over~$\Q$ such that
238$\#\Sha(E/\Q)=\#E(\Q)_{\tor}=1$,
239then the results mentioned above do not rule out the possibility that every
240element of $H^1(\Q,E)$ has index a perfect square.  We prove, under
241the assumption that~$L(E,1)\neq 0$, that there is an
242integer~$B$ such that $H^1(\Q,E)$ contains infinitely many elements of
243index~$n$, for every integer~$n$ that is prime to~$B$ (see
244Theorem~\ref{thm:index}).  For example, in
245Section~\ref{sec:genusoneexamples} we prove that one can take $B=2$ for
246the elliptic curve $X_0(17)$.
247
248
249\section{Kolyvagin's Euler system}\label{sec:kolyvagin}
250In this section, we recall the definition of Heegner points and
251several basic results about the system of cohomology classes Kolyvagin
252attaches to these points.  We also state the main theorem of this paper.
253
254\subsection{Kolyvagin classes}
255Let~$E$ be an elliptic curve over~$\Q$ of conductor~$N$, and denote
256by $X_0(N)$ the modular curve that classifies cyclic isogenies
257of degree~$N$.
258By \cite{breuil-conrad-diamond-taylor}, there is a surjective
259map $\pi:X_0(N)\ra E$.
260(Note that for the proof of Theorem~\ref{thm:overk} we do not
261need any modularity theorems, because we take $E=X_0(17)$.)
262Let~$K$ be a quadratic imaginary extension of~$\Q$ in which
263all primes dividing~$N$ split, and let~$D_K$ be the discriminant
264and~$\cO$ the ring of integers of~$K$.
265Since all primes dividing~$N$ split, there is
266an ideal $\ga\subset \cO$ such that $\cO/\ga$ is cyclic of order~$N$.
267Let~$H$ be the Hilbert class field of~$K$, and
268$x_H\in X_0(N)(H)$ be the Heegner points corresponding to
269$(\C/\cO,\ga^{-1}/\cO)$.
270Set $y_H=\pi(x_H)\in E(H)$,
271    $y_K=\tr_{H/K}(y_H)\in E(K)$,
272and $y=y_K-y_K^{\tau}\in E(K)^{-}$, where~$\tau$ denotes
273complex conjugation.
274Assume that $L(E,1)\neq 0$, so by \cite{bump-friedberg-hoffstein, murty-murty}
275there are infinitely many ways in which to choose~$K$ as above
276so that~$y$ has infinite order.
277Under this nonvanishing hypothesis on $L(E,1)$, Kolyvagin proves in
278\cite{kolyvagin:structureofsha} that the groups $E(\Q)$
279and $\Sha(E/\Q)$ are both finite.
280
281In the course of his proof, Kolyvagin considers more general Heegner
282points $y_{\ell}\in E(\Qbar)$, for appropriate primes~$\ell$, and from
283these constructs cohomology classes $c_{\ell,p^n}\in H^1(\Q,E)[p^n]$
284that are used to bound the orders of certain Selmer groups associated
285to~$E$.  We will study Kolyvagin's classes further and prove that for
286each prime~$p$ not in an explicit finite set and each positive
287integer~$n$, there are infinitely many primes~$\ell$ such that
288$$\ord(c_{\ell,p^n}) = \ind(c_{\ell,p^n}) = p^n.$$
289We thus obtain the following theorem, which will
290be proved in Section~\ref{sec:thm-index}.
291\begin{theorem}\label{thm:index}
292Let~$E$ be an elliptic curve over~$\Q$ such that
293$L(E,1)\neq 0$. Then there is an integer~$B$ such that, for
294all integers~$r$ coprime to~$B$, there are infinitely
295many $c \in H^1(\Q,E)$ such that $\ord(c)=\ind(c)=r$.
296\end{theorem}
297
298\begin{remark}
299Cathy O'Neil \cite{oneil} has investigated the obstruction to
300$\ord(c)=\ind(c)$. We show that when~$E$ has analytic
301rank~$0$, this obstruction vanishes for infinitely many~$c$.
302\end{remark}
303
304
305\subsection{Basic properties of Kolyvagin's Euler system}
306In \cite{rubin:kolyvagin}, Rubin gives a concise account of
307Kolyvagin's proof of finiteness of $\Sha(E/\Q)[p^\infty]$, under the
308simplifying assumption that~$p$ is odd.  Though Kolyvagin's argument
309works even when $p=2$, for simplicity, we rely exclusively on Rubin's
310paper.
311
312Let~$K$ be a quadratic imaginary field as above, chosen in such a way
313that the associated Heegner point $y_K$ has infinite order.  Fix
314embeddings of~$\Qbar$ into~$\C$ and into each $p$-adic field $\Qpbar$.
315Let $\tau$ denote complex conjugation, and for any
316$\Z[\tau]$-module~$A$, let $A^+$ and $A^-$ denote the kernel of
317$\tau-1$ and $\tau+1$, respectively. For the remainder of this section,
318we assume that~$p$ is an odd prime, and if $K=\Q(\sqrt{-3})$
319that $p\geq 5$.  If~$\ell$ is a prime that is inert in~$K$,
320let~$K_\ell$ denote the completion of~$K$ at the unique prime lying
321over~$\ell$.  If $L$ is a finite Galois extension of~$\Q$, let
322$\Frob_\ell(L/\Q)$ denote the conjugacy class of some Frobenius element
323of a prime lying over~$\ell$.  For each prime $\ell\nmid N$, let
324$a_\ell=\ell+1-\#E(\F_\ell)$ be the $\ell$th Fourier coefficient of
325the newform attached to~$E$.
326
327\begin{definition}\label{def:mv}
328For each place~$v$ of~$\Q$, let
329  $$m_v=\#H^1(\Qv^{\unr}/\Qv,E(\Qv^{\unr})).$$
330By \cite[I.3.8]{milne:duality}, each $m_v$ is finite
331and $m_v=1$ for all but finitely many~$v$, so
332  $$m(p)=\sup\{\ord_p(m_v): \mbox{all places } v\mbox{ of }\Q \}$$
333is well defined, and $m(p)=0$ for almost all~$p$.
334\end{definition}
335
336Let~$n$ be a positive integer.
337\begin{proposition}\label{prop:rubin2}
338Let~$p$ be a prime that does not divide the class
339number of~$K$ and for which $m(p)=0$.
340Suppose $\ell\nmid p D_K N$ and $\Frob_\ell(K(E[p^n])/\Q)=[\tau]$.
341Then there is an element $c_{\ell,p^n}\in H^1(\Q,E)[p^n]$ such that
342%$\res_v(c_{\ell,p^n}) = 0$ for
343%all places $v\neq \ell$ of $\Q$, and
344the order of $\res_\ell(c_{\ell,p^n})$ in $H^1(\Ql,E)[p^n]$ is equal
345to the order of the image of~$y$ in $E(K_\ell)/p^n E(K_\ell)$,
346and the index of $c_{\ell,p^n}$ divides~$p^n$.
347\end{proposition}
348\begin{proof}
349The existence of $c_{\ell,p^n}$ and statement about its order is
350proved in \cite[Prop.~5]{rubin:kolyvagin}, where $c_{\ell,p^n}$ is
351constructed from Heegner points on $X_0(N)$.  For the index bound,
352note that in the proof of \cite[Prop.~5]{rubin:kolyvagin}, when
353$p\nmid [H:K]$, Rubin constructs a class $c'\in 354H^1(K'/K,E(K'))[p^r]^+$, where $r=n+m(p)$ and~$K'$ is the unique
355extension of~$K$ of degree~$p^r$ in a certain class field of~$K$.
356Since~$p$ is odd, the restriction map $\res : H^1(\Q,E)[p^r]\ra 357H^1(K,E)[p^r] ^+$ is an isomorphism.  Rubin takes
358$c_{\ell,p^n}=\res^{-1}(c')$.  Since $c_{\ell,p^n}$ splits
359over the degree $2p^r$ extension~$K'$ of~$\Q$, the index of
360$c_{\ell,p^n}$ divides $2p^r$.  But $c_{\ell,p^n}$ has odd order and,
361by Proposition~\ref{prop:lang-tate}, $\ind(c_{\ell,p^n})$ has the same
362prime factors as $\ord(c_{\ell,p^n})$, so $\ind(c_{\ell,p^n})$
363divides~$p^r$.
364\end{proof}
365\begin{remark}
366The author does not know whether or not
367the proposition is true
368if~$p$ is allowed to divide the class number of~$K$.
369\end{remark}
370
371
372\section{Nonvanishing of cohomology classes}\label{sec:nonvanishing}
373In this section, we prove a nonvanishing result
374about the cohomology classes $c_{\ell,p^n}$ of
375Proposition~\ref{prop:rubin2}, then use it to deduce
376Theorem~\ref{thm:index}.
377
378
379\subsection{Local nonvanishing}
380Let~$E$ be as above.
381For any point $x\in E(K)$,
382let $K([p^n]^{-1}x)$ denote the field obtained by adjoining
383the coordinates of all $p^n$th roots of~$x$ to~$K$.
384Without imposing further hypothesis, this field need
385not be Galois over~$\Q$.
386\begin{lemma}\label{lem:isgalois}
387If $x\in E(K)^+\union E(K)^{-}$, then
388$K([p^n]^{-1}x)$ is Galois over~$\Q$.
389\end{lemma}
390\begin{proof} Since $G_\Q$ acts on~$x$ by $\pm 1$, the
391subgroup~$\Z x$ is $G_\Q$-invariant.  Since
392$[p^n] \colon{} E\ra E$ is a $\Q$-rational isogeny
393the inverse image $[p^n]^{-1}\Z x$ is also $G_\Q$-invariant,
394so $K([p^n]^{-1}x)=K([p^n]^{-1}\Z x)$ is Galois over~$\Q$.
395\end{proof}
396
397
398\begin{definition}
399An odd prime~$p$ is {\it firm} for~$E$ if
400$m(p)=0$, there are no nontrivial
401$\Q$-rational cyclic subgroups of $E[p^\infty]$,
402and $H^1(K(E[p^n])/K, E[p^n])=0$
403for all $n\geq 1$.
404\end{definition}
405
406\begin{remark}
407The set of primes that are not firm is finite, by
408Serre's theorem \cite{serre:propgal} and the theory
409of complex multiplication.
410\end{remark}
411
412
413Let~$p$ be an odd prime that is firm for~$E$.
414The following proposition produces infinitely many
415primes~$\ell$ such that we have control over the
416orders of the image in $E(K_\ell)/p^n E(K_\ell)$
417of a global point.
418It will be used as input to Proposition~\ref{prop:rubin2} to
419produce cohomology classes of known index.
420The proof, which involves an application of the
421Chebotar\"ev density theorem, follows a strategy similar
422to that used in the proof of Kolyvagin's theorem on
423page 135 of~\cite{rubin:kolyvagin}.
424
425\begin{proposition}~\label{prop:ordind}
426Let $p$ be a prime that is firm for~$E$, and let $x\in E(K)^{\pm}$.
427Then there is a set of primes~$\ell$ of positive Dirichlet density
428such that $\Frob_\ell(K(E[p^n])/\Q)=[\tau]$ and
429the orders of the images of~$x$ in $E(K)/p^n E(K)$
430and in $E(K_\ell)/p^n E(K_\ell)$ are the same.
431\end{proposition}
432\begin{proof}
433Let $p^a$ 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 $a>0$.
435If~$\ell$ is a prime such that the orders of the images of $p^{a-1}x$ in
436$E(K)/p^n E(K)$ and  $E(K_\ell)/p^n E(K_\ell)$
437both equal~$p$,
438then the images of~$x$ in
439$E(K)/p^n E(K)$ and
440$E(K_\ell)/p^n E(K_\ell)$ both have
441order $p^a$.
442It thus suffices to prove the proposition in the case when
443the order of the image of~$x$ in $E(K)/p^n E(K)$ is~$p$.
444
445Let $L=K(E[p^n])$, suppose~$\ell$ is a prime such that
446$\Frob_\ell(L/\Q)=[\tau]$, and let~$\lambda$ be one of the prime
447ideals 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}$$
454Let $\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
456is injective, because~$p$ is firm, so it suffices
457to show that the image $\varphi_\ell$ of~$\varphi$ in
458$\Hom(G_{L_\lambda},E[p^n])$ is nonzero.
459
460Let~$M$ be the fixed field of the kernel of~$\varphi$.
461Since~$M$ is the compositum of
462the two Galois extensions $K([p^n]^{-1}x)$ and $\Q(E[p^n])$
463of~$\Q$, it is also Galois (see Lemma~\ref{lem:isgalois}).
464Because $\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]$.
467The order of $\sigma\tau$
468equals the degree of $M_{\lambda'}$ over $\Q_\ell$,
469where $\lambda'$ is a prime of~$M$ lying over~$\ell$.
470If $\varphi_\ell=0$, then $M_{\lambda'}=L_\lambda=K_\ell$,
471so $\sigma\tau$ would have order~$2$.
472
473The image of~$\varphi$ is a nonzero subgroup~$H$ of $E[p^n]$,
474which is defined over~$\Q$ since $x\in E(K)^{\pm}$.
475If every $\sigma\in \Gal(M/L)$ has the property that
476$\sigma\tau$ has order~$2$,
477then $H\subset E[p^n]^{-}$.  This contradicts our assumption
478that~$p$ is firm, since $H$ is a nontrivial cyclic subgroup
479of $E[p^\infty]$.  Thus there exists $\sigma \in \Gal(M/L)$
480such that $\sigma\tau$ has order different than~$2$.
481For this~$\sigma$ and for any
482prime~$\ell$ such that
483$\Frob_\ell(M/\Q) = [\sigma\tau]$, we see that
484$\varphi_\ell\neq 0$.
485The Chebotar\"ev density theorem provides a positive
486density 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}]
492Let~$E$ be an elliptic curve over~$\Q$ such that $L(E,1)\neq 0$.
493Let~$K$ be one of the infinitely many imaginary quadratic fields
494such that the associated Heegner point~$y$ has infinite order.
495Let $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}
507If $K=\Q(\sqrt{-3})$, assume in addition that~$3$ divides~$B_K$.
508
509Fix a prime $p\nmid B_K$.
510Since $E(K)$ has rank~$1$ (see, e.g., \cite[Thm 1.3]{gross:kolyvagin})
511and $p\nmid \#E(K)_{\tor}$, the image of~$y$ in $E(K)/p^n E(K)$
512has order $p^n$.  By Proposition~\ref{prop:ordind} there are
513infinitely many primes~$\ell$ such that
514$\Frob_\ell(K(E[p^n])/\Q)=[\tau]$ and the image of~$y$
515in $E(K_\ell)/p^n E(K_\ell)$ has order~$p^n$.
516For these~$\ell$, Proposition~\ref{prop:ordind} produces infinitely many
517cohomology classes $c_{\ell,p^n}$ having order and index both
518equal to~$p^n$.  (Note that if $\ell\neq \ell'$ then
519$c_{\ell,p^n}\neq c_{\ell',p^n}$.)
520
521Let~$B$ be the greatest common divisor of the set of integers
522$B_K$, as~$K$ varies over all quadratic imaginary extensions such that
523the associated Heegner point has infinite order.
524For each prime power $p^n$ that does not divide $B$,
525we have produced infinitely many $c\in H^1(\Q,E)$
526having order and index both equal to $p^n$.
527If the orders of~$c$ and~$c'$
528are coprime, then $\ord(c+c') = \ord(c)\cdot \ord(c')$ and,
529by Proposition~\ref{prop:indmul},
530$\ind(c+c')=\ind(c)\cdot \ind(c')$.
531This proves the theorem.
532\end{proof}
533
534\section{Computing the bound $B_K$}
535\label{sec:genusoneexamples}
536In this section we compute, in some cases, the the bound $B_K$ that
537appears in Section~\ref{sec:thm-index}.  First we prove a general
538theorem about semistable elliptic curves.  Next we compute the index
539of a Heegner point, and finally in Section~\ref{sec:conclusion}
540we prove Theorem~\ref{thm:overk}.
541
542\subsection{Galois representations attached to isolated curves}
543The following proposition sometimes permits us to compute
544the 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}
549Let~$E$ be a semistable elliptic curve over~$\Q$ of conductor~$N$,
550let~$p$ be an odd prime, and let $K$ be a quadratic imaginary field
551such that $\gcd(D_K, p N)=1$.  Assume that $p\nmid \ord_\ell(j(E))$,
552for each prime~$\ell$, and that~$E$ admits no isogenies of degree~$p$.
553Then $p \nmid \#E(K)_{\tor}$ and~$p$ is firm for~$E$.
554\end{proposition}
555
556Before giving the proof, we summarize its main ingredients.  First, we
557observe that the assertion that $m(p)=0$ (see Definition~\ref{def:mv})
558uses a standard result that relates unramified Galois cohomology to
559component groups.  Next, we use the semistability and isogeny
560hypotheses to deduce that $\rho_{E,p}$ is surjective.
561Then we use standard group cohomology to deduce that~$p$ is firm.
562
563\begin{proof}
564Let~$\ell$ be a prime.
565By \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)),$
568where $\Phi_{E,\ell}$ is the component group of~$E$ at~$\ell$.
569If $\ell\nmid N$, there is nothing further to prove, so assume $\ell\mid N$.
570Since~$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
574Since~$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
576nontrivial $\Q$-rational cyclic subgroups of $E[p^\infty]$.  Since~$E$
577is semistable, work of Serre~\cite[Prop.~21]{serre:propgal} and
578\cite[\S3.1]{serre:travauxwiles} implies that $\rho_{E,p}$ is
579surjective.  Thus $p\nmid \#E(K)_{\tor}$ because a point in $E(\Qbar)$
580of order~$p$ must generate an extension of~$\Q$ of degree at least
581$p^2-1\geq 3$.
582
583The 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 586H^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).$$
601%This map is also multiplication by~$\alpha$, so
602%$H^1(\GL(2,\Fp), \Fp^2)$ has exponent dividing~$\alpha-1$,
603%which is not divisible by~$p$.
604%The exponent of $H^1(\GL(2,\Fp), \Fp^2)$ must divide~$p$,
605%so $H^1(\GL(2,\Fp), \Fp^2)=0$.
606
607%We verify that there is a nonidentity scalar $\alpha\in 608%\Gal(K(E[p^n])/K)\subset\GL(2,\Z/p^n\Z)$.
609The group~$H=H^1(K(E[p^n])/K,E[p^n])$ has exponent a power of~$p$.
610If an element~$\alpha$ in $\Gal(K(E[p^n])/K) \subset \GL_2(\Z/p^n\Z)$ is scalar,
611then every element of~$H$ has order dividing $\alpha-1$.  This is
612because the scalar is central, so the morphism of pairs it induces is
613both the identity and multiplication by $\alpha$.  It is necessary
614only to choose $\alpha$ such that $\gcd(\alpha-1,p)=1$.  Since~$p$ is
615odd, $-1$ is a nonidentity element of $\Aut(E[p])=\Gal(K(E[p])/K)$.
616Every automorphism lifts, so~$-1$ lifts to some~$g$ in
617$\Gal(K(E[p^n])/K) \subset \Aut(E[p^n])$.  Then $g^{p^{n-1}} = -1$ in
618$\Aut(E[p^n])$, so $-1 \in \Gal(K(E[p^n])/K)$ and every element of~$H$
619has order dividing~$2$.
620(To show that $g^{p^{n-1}}=-1$, we use that
621$\ord_p\binom{p^n}{k} = n + \ord_p\left(\frac{1}{k}\right)$.)
622\end{proof}
623
624\subsection{The number $B_K$ for $X_0(17)$}
625In this section, we show that for $E=X_0(17)$ and $K=\Q(\sqrt{-2})$,
626we have $B_K=2$.  This is accomplished
627by showing that the index $[E(K):\Z y]$ is a power of~$2$.
628The elliptic curve $E=X_0(17)$ given by the Weierstrass equation
629$$y^2 + xy + y = x^3 - x^2 - x -14$$
630satisfies the hypothesis
631of Proposition~\ref{prop:explicitindex} for each odd prime~$p$.
632Since the $j$-invariant of~$E$ is $3^3\cdot 11^3/17^4$,
633every odd prime~$p$ is firm for~$E$ and
634$\#E(K)_{\tor}$ is  a power of~$2$.
635
636The conductor $17$ of~$E$ splits in~$K$, and the
637quadratic twist $E'$ of~$E$ by~$K$ is the curve
638$y^2 = x^3 - 44x + 7120$, which is labeled {\bf 1088K4}
639in \cite{cremona:onlinetables}.
640Using MAGMA (or {\tt mwrank}), one
641finds that $E'(\Q)\isom \Z{}P \times\Z/2$,
642where $P=(-3,85)\in E'(\Q)$ has infinite order.
643Since the rank of~$E'$ is~$1$, we set~$K=\Q(\sqrt{-2})$
644in Section~\ref{sec:thm-index}.
645Then~$B_K$ is divisible only by~$2$ and the index
646$[E(K):\Z y]$.   This index can only change
647by a power of~$2$ if~$y$ is replaced by $y_K$, so we instead
648consider the index $[E(K):\Z y_K]$.
649The cokernel of the natural map $E(\Q)\oplus E'(\Q)\ra E(K)$
650is 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~$h$
652is the N\'eron-Tate canonical height on~$E_K$.
653By the Gross-Zagier formula (see~\cite[Thm.~6.3]{gross-zagier}),
654\begin{equation}\label{eqn:index-grosszagier}
655h(y_K) =
656   \frac{u^2 |D|^{\frac{1}{2}}}{||\omega_f||} L'_{E'}(1) L_E(1),
657\end{equation}
658where $D=-8$ is the discriminant of~$K$, $u=1$ is half the number of
659units, and $||\omega_f||$ is the Peterson norm of the newform~$f$
660corresponding to~$E$.
661Generators for the period lattice of~$E$ are
662$\omega_1 \sim 1.547079$ and $\omega_2 \sim 0.773539 + 1.372869i$;
663taking the determinant gives $||\omega_f|| \sim 2.123938$.
664Furthermore, again from \cite{cremona:onlinetables}, we find that
665  $L_E(1) \sim 0.386769$
666and
667  $L'_{E'}(1) \sim 2.525026$,
668so  $h(y_K) \sim 1.300533$.
669Using a computer, we find that $h(P) \sim 1.300533$
670as well, so $[E(K):\Z y_K]$ is a power of two.
671
672\subsection{Elements of index~$2$ and~$4$}
673The 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
675many elements of $H^1(\Q,E)$ having order and index equal to~$2$,
676and also infinitely many having order and index equal to~$4$.
677
678
679\subsection{Proof of Theorem~\ref{thm:overk}}\label{sec:conclusion}
680To prove Theorem~\ref{thm:overk}, we combine the above
681computations with Theorem~\ref{thm:index}, and an observation
682about the local properties of Kolyvagin's classes $c_{\ell,p^n}$.
683
684\begin{proof}[Proof of Theorem~\ref{thm:overk}]
685Let~$E=X_0(17)$ as above, and let~$K$ be an arbitrary number field.
686Let $p^n$ be either an odd prime power, or $2$, or $4$.
687The computations of the previous section combined with
688Theorem~\ref{thm:index} prove that there are
689infinitely many elements $c_{\ell,p^n}$ of $H^1(\Q,E)$ whose index and
690order both equal $p^n$.
691Let~$A$ be the subgroup of $H^1(\Q,E)$ generated by these classes.
692The kernel~$B$ of $\res_K:A\ra H^1(K,E)$ is finite, so the set
693$\mathcal{S}$ of primes~$\ell$ such that $\res_\ell(c)\neq 0$
694for some $c\in B$ is finite.  By Proposition~\ref{prop:rubin2},
695we have $\res_v(c_{\ell,p^n})=0$ for all places $v\neq \ell$,
696so the subgroup $A'$ of~$A$ generated by all $c_{\ell,p^n}$ with
697$\ell\not \in \mathcal{S}$ has trivial
698intersection with~$B$.  Thus $\res_K(A')$ consists of infinitely
699many classes in $H^1(K,E)$ having order and index both equal
700to $p^n$, and the theorem now follows from Proposition~\ref{prop:lang-tate}.
701\end{proof}
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