Visibility in an Abelian Variety of Dimension $20$

Using the methods described in [AS02], we find that $S_2(\Gamma_0(389))$ contains exactly five Galois-conjugacy classes of newforms, and these are defined over extensions of  $\mathbf{Q}$ of degrees $1$, $2$, $3$, $6$, and $20$. Thus $J=J_0(389)$ decomposes, up to isogeny, as a product $A_1\times A_2 \times A_3\times A_6 \times A_{20}$ of abelian varieties, where $d=\dim A_d$ and $A_d$ is the quotient corresponding to the appropriate Galois-conjugacy class of newforms.

Next we consider the arithmetic of each $A_d$. Using [AS02], we find that

$$L(A_1,1)=L(A_2,1)=L(A_3,1)=L(A_6,1)=0,$$

and

$$\frac{L(A_{20},1)}{\Omega_{A_{20}}} = \frac{5^2\cdot 2^?}{97},$$

where $2^?$ is a power of $2$. Using [AS02], we find that $\char93 A_{20}(\mathbf{Q}) = 97$ and the Tamagawa number of $A_{20}$ at $389$ is also $97$. The BSD Conjecture then predicts that $\char93 {\mbox{{\fontencoding{OT2}\fontfamily{wncyr}\fontseries{m}\fontshape{n}\selectfont Sh}}}(A_{20}) = 5^2\cdot 2^?$. The following proposition provides support for this conjecture.

Proposition 4.1   There is an inclusion

$$(\mathbf{Z}/5\mathbf{Z})^2 \cong A_1(\mathbf{Q})/5 A_1(\mathbf{Q}... ...\fontfamily{wncyr}\fontseries{m}\fontshape{n}\selectfont Sh}}}(A_{20}^{\vee})).$$

Proof. Let $A=A_{20}^{\vee}$, $B=A_{1}^{\vee}=A_1$ and $J=A+B\subset J_0(389)$. Using algorithms in [AS02], we find that $A\cap B \cong (\mathbf{Z}/4)^2\times (\mathbf{Z}/5\mathbf{Z})^2$, so $B[5] \subset A$. Since $5$ does not divide the numerator of $(389-1)/12$, it does not divide the Tamagawa numbers or the orders of the torsion subgroups of $A$, $B$, $J$, and $J/B$ (we also verified this using a modular symbols computations), so Theorem 3.1 implies that there is an injective map

$$A_1(\mathbf{Q})/5 A_1(\mathbf{Q}) \hookrightarrow \Vis_{J}({\mbo... ...}\fontfamily{wncyr}\fontseries{m}\fontshape{n}\selectfont Sh}}}(A_{20}^{\vee}).$$

To finish, note that Cremona [Cre97] has verified that $A_1(\mathbf{Q}) \approx \mathbf{Z}\times \mathbf{Z}$. $\qedsymbol$