CoCalc Shared Fileswww / papers / nonsquaresha / junkOpen in CoCalc with one click!
Author: William A. Stein
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\edit{{\bf Idea!} If I can give conditions under which $\Sha(J/\Q)[p]=\{0\}$
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(equivalently, $\Sha(E/K)[p]=\{0\}$) then I will know
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{\em automatically} that if there is something in
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$\Sha(A/\Q)[p]$, then that something is {\em visible}, hence
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$E(\Q)/p E(\Q)\neq 0$!!!! That is, a ``strong form of finiteness
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of $\Sha$ of~$E$'', Conjecture~\ref{conj:nonvanishing}, and
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the rank-$0$ BSD lower bound on $\#\Sha$ together imply the
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following statement: $L(E,1)=0$ implies $E(\Q)$ is infinite.
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The ``strong finiteness'' is that there exists prime(s) $p$
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such that $\Sha(E/K)[p]=\{0\}$, where $K/\Q$ is appropriate
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cyclic of degree~$p$.}
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\comment{\section{Old Stuff}
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\noindent{\bf Outline of proof:}
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\begin{enumerate}
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\item There should be a twist $E'$ of~$E$ by a quadratic character
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unramified at~$p$ such that $E'(\Q)$ has rank exactly~$1$. This
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should follow from Waldspurger, Kolyvagin, Gross-Zagier, and the other
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standard results. Without loss, replace~$E$ by~$E'$ (need to show
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still that Tamagawa numbers don't change too much under a quadratic
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twist; should be easy?)
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\item (*) Use Conjecture~\ref{thm:nonvanishing} and surjective of
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$\rho_{E,p}$ to find an~$\ell$ such that $a_\ell\not\equiv \ell+1
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\pmod{p}$ (or $a_\ell\not\equiv 2\pmod{p}$)
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and $L(E,\chi_{p,\ell},1)\neq 0$.
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\item
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Let $K$ be the degree~$p$ and totally ramified at~$\ell$ extension
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of~$\Q$ corresponding to~$\chi_{p,\ell}$,
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let $J=\Res_{K/\Q}(E_K)$, and let~$B$ be the kernel of $\Tr : J \ra E$.
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\item The map $E\hookrightarrow J \ra E$ is multiplication by $[K:\Q]=p$,
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so $E[p]\subset B$.
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\item Use the irreducible assumption on $\rho_{E,p}$ to prove that
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$B(\Q)$ has no $p$-torsion.
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\item The bad reduction of~$B$ is $\ell\cdot N$. For each prime~$q$
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dividing~$N$, the extension $K/\Q$ is unramified at~$q$ and
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$B_K \isom E^{\oplus (p-1)}$, so, because the formation of N\'eron models commutes
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with unramified base change, $c_{B,q}$ divides $\overline{c}_{E,q}^{(p-1)}$.
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In particular, the rigid hypothesis on~$p$ implies that
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$p\nmid c_{B,q}$.
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\item The only thing left to prove is that $p\nmid c_{B,\ell}$.
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This is Conjecture~\ref{conj:nonvanishing}.
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\item We conclude that $E(\Q)/p E(\Q)\hookrightarrow \Sha(A)$.
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Since $E(\Q)$ has rank~$1$, this concludes the proof.
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\end{enumerate}
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}
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\comment{
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\subsection{Quadratic Twists}
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\begin{proposition}\label{prop:quadtwist}
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If~$E$ is an elliptic curve over~$\Q$, then
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there is a (quadratic) twist $E'$ of~$E$ that has rank
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greater than~$0$.
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\end{proposition}
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\begin{proof}
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By the work of many people (see the introduction of
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Ono-Skinner \cite{}), there is a twist $E'$ of $E$ such
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that $L(E',1)\neq0$. By, e.g., Murty-Murty \cite{}, there is
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a quadratic twist $E''$ of $E'$ such that
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$\ord_{s=1} L(E'',s) = 1$, then by Kolyvagin Theorem (see, e.g.,
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\cite{}), $E''$ has rank~$1$.
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\end{proof}
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\subsection{Twists of Prime Order}
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}
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