\edit{{\bf Idea!} If I can give conditions under which $\Sha(J/\Q)[p]=\{0\}$1(equivalently, $\Sha(E/K)[p]=\{0\}$) then I will know2{\em automatically} that if there is something in3$\Sha(A/\Q)[p]$, then that something is {\em visible}, hence4$E(\Q)/p E(\Q)\neq 0$!!!! That is, a ``strong form of finiteness5of $\Sha$ of~$E$'', Conjecture~\ref{conj:nonvanishing}, and6the rank-$0$ BSD lower bound on $\#\Sha$ together imply the7following statement: $L(E,1)=0$ implies $E(\Q)$ is infinite.8The ``strong finiteness'' is that there exists prime(s) $p$9such that $\Sha(E/K)[p]=\{0\}$, where $K/\Q$ is appropriate10cyclic of degree~$p$.}11121314\comment{\section{Old Stuff}15\noindent{\bf Outline of proof:}16\begin{enumerate}1718\item There should be a twist $E'$ of~$E$ by a quadratic character19unramified at~$p$ such that $E'(\Q)$ has rank exactly~$1$. This20should follow from Waldspurger, Kolyvagin, Gross-Zagier, and the other21standard results. Without loss, replace~$E$ by~$E'$ (need to show22still that Tamagawa numbers don't change too much under a quadratic23twist; should be easy?)24252627\item (*) Use Conjecture~\ref{thm:nonvanishing} and surjective of28$\rho_{E,p}$ to find an~$\ell$ such that $a_\ell\not\equiv \ell+129\pmod{p}$ (or $a_\ell\not\equiv 2\pmod{p}$)30and $L(E,\chi_{p,\ell},1)\neq 0$.3132\item33Let $K$ be the degree~$p$ and totally ramified at~$\ell$ extension34of~$\Q$ corresponding to~$\chi_{p,\ell}$,35let $J=\Res_{K/\Q}(E_K)$, and let~$B$ be the kernel of $\Tr : J \ra E$.3637\item The map $E\hookrightarrow J \ra E$ is multiplication by $[K:\Q]=p$,38so $E[p]\subset B$.3940\item Use the irreducible assumption on $\rho_{E,p}$ to prove that41$B(\Q)$ has no $p$-torsion.4243\item The bad reduction of~$B$ is $\ell\cdot N$. For each prime~$q$44dividing~$N$, the extension $K/\Q$ is unramified at~$q$ and45$B_K \isom E^{\oplus (p-1)}$, so, because the formation of N\'eron models commutes46with unramified base change, $c_{B,q}$ divides $\overline{c}_{E,q}^{(p-1)}$.47In particular, the rigid hypothesis on~$p$ implies that48$p\nmid c_{B,q}$.4950\item The only thing left to prove is that $p\nmid c_{B,\ell}$.51This is Conjecture~\ref{conj:nonvanishing}.5253\item We conclude that $E(\Q)/p E(\Q)\hookrightarrow \Sha(A)$.54Since $E(\Q)$ has rank~$1$, this concludes the proof.5556\end{enumerate}5758}5960616263646566\comment{67\subsection{Quadratic Twists}68\begin{proposition}\label{prop:quadtwist}69If~$E$ is an elliptic curve over~$\Q$, then70there is a (quadratic) twist $E'$ of~$E$ that has rank71greater than~$0$.72\end{proposition}73\begin{proof}74By the work of many people (see the introduction of75Ono-Skinner \cite{}), there is a twist $E'$ of $E$ such76that $L(E',1)\neq0$. By, e.g., Murty-Murty \cite{}, there is77a quadratic twist $E''$ of $E'$ such that78$\ord_{s=1} L(E'',s) = 1$, then by Kolyvagin Theorem (see, e.g.,79\cite{}), $E''$ has rank~$1$.80\end{proof}81\subsection{Twists of Prime Order}82}8384