A Connection with the BSD Conjecture

There is heuristic reason why equalities like the one that we proved with a computer computation in the previous section should frequently be true. First, the Birch and Swinnerton-Dyer conjecture predicts that

$$\char93 {\mbox{{\fontencoding{OT2}\fontfamily{wncyr}\fontseries{m... ...lectfont Sh}}}(E/\mathbb{Q})=\frac{L'(E,1)}{\Reg(E/\mathbb{Q}) \cdot \Omega_E},$$

and in fact one knows that ${\mbox{{\fontencoding{OT2}\fontfamily{wncyr}\fontseries{m}\fontshape{n}\selectfont Sh}}}(E/\mathbb{Q})[3]=\{1\}$. Second, because $\chi$ is congruent to the identity character modulo $3$,

$$L(E,\chi,1)/\Omega_E \equiv L(E,1)=0\pmod{3},$$

suitably interpreted, and there is no reason why this congruence should hold modulo $3^2$. Thus, in the situation of Theorem 6.1, the author expects that usually ${\mbox{{\fontencoding{OT2}\fontfamily{wncyr}\fontseries{m}\fontshape{n}\selectfont Sh}}}(A/\mathbb{Q})[p]\cong E(\mathbb{Q})/p E(\mathbb{Q})$. More generally, one suspects that usually

$$\Sel^{(p)}(A/\mathbb{Q}) \cong \Sel^{(p)}(E/\mathbb{Q}).$$

Conjecture 7.3   Let the notation be as in Conjecture 4.1. Then there exists a rigid prime $p$ and a prime  $\ell\nmid N_E$ such that $L(E,\chi_{p,\ell},1)\neq 0$, $a_\ell(E) \not\equiv 2\pmod{p}$, and ${\mbox{{\fontencoding{OT2}\fontfamily{wncyr}\fontseries{m}\fontshape{n}\selectfont Sh}}}(E/K)[p]=\{0\}$, where $K$ corresponds to $\chi_{p,\ell}$.

Proposition 7.4   Assume Conjecture 7.3 and the following weak consequence of the Birch and Swinnerton-Dyer conjecture: if $A$ is a twisted power of an elliptic curve of analytic rank 0 over  $\mathbb{Q}$ and $p\mid L(A,1)/\Omega_A$, then $p\mid \char93 {\mbox{{\fontencoding{OT2}\fontfamily{wncyr}\fontseries{m}\fontshape{n}\selectfont Sh}}}(A/\mathbb{Q})\cdot \prod c_q$. If $E$ is an elliptic curve over  $\mathbb{Q}$ and $L(E,1)=0$, then $E(\mathbb{Q})$ is infinite.