The Possible Orders of Shafarevich-Tate Groups

On page 306-307 of [10], Tate discusses results about the structure of the group ${\mbox{{\fontencoding{OT2}\fontfamily{wncyr}\fontseries{m}\fontshape{n}\selectfont Sh}}}(A/K)$, where $A$ is an abelian variety over a number field $K$. He asserts that if $A$ is a Jacobian then $\char93 {\mbox{{\fontencoding{OT2}\fontfamily{wncyr}\fontseries{m}\fontshape{n}\selectfont Sh}}}(A/K)$ is a perfect square. Poonen and Stoll subsequently pointed out in [8] that Tate's assertion is not quite correct. In fact, Poonen and Stoll prove that when $A$ is a Jacobian, $\char93 {\mbox{{\fontencoding{OT2}\fontfamily{wncyr}\fontseries{m}\fontshape{n}\selectfont Sh}}}(A/K)$ is either a square or twice a square, and they give examples in which $\char93 {\mbox{{\fontencoding{OT2}\fontfamily{wncyr}\fontseries{m}\fontshape{n}\selectfont Sh}}}(A/K)$ is twice a square. Tate does not discuss the case when $A$ is not a Jacobian, except to mention results that imply that $\char93 {\mbox{{\fontencoding{OT2}\fontfamily{wncyr}\fontseries{m}\fontshape{n}\selectfont Sh}}}(A/K)$ is square away from $2$ and primes that don't divide the degree of some polarization of $A$.

Now suppose $A$ is an arbitrary abelian variety over a number field $K$. In this case, it has remained an unresolved problem during the last $35$ years to decide whether or not $\char93 {\mbox{{\fontencoding{OT2}\fontfamily{wncyr}\fontseries{m}\fontshape{n}\selectfont Sh}}}(A/K)$ is a square or twice a square. Let $E$ be an elliptic curve over  $\mathbb{Q}$ of rank $1$. Then the construction of the present paper gives, for suitable primes $p$, an injection

$$\mathbb{Z}/p\mathbb{Z}\approx E(\mathbb{Q})/p E(\mathbb{Q}) \hook... ...2}\fontfamily{wncyr}\fontseries{m}\fontshape{n}\selectfont Sh}}}(A/\mathbb{Q}),$$

where $A$ is an abelian variety over  $\mathbb{Q}$ which is a twist of $E^{\times p-1}$. Thus ${\mbox{{\fontencoding{OT2}\fontfamily{wncyr}\fontseries{m}\fontshape{n}\selectfont Sh}}}(A/\mathbb{Q})[p]$ has a ``natural'' subgroup of order $p$; moreover, no other natural subgroup of order $p$ presents itself. Is $\char93 {\mbox{{\fontencoding{OT2}\fontfamily{wncyr}\fontseries{m}\fontshape{n}\selectfont Sh}}}(A/\mathbb{Q})[p]\approx \mathbb{Z}/p\mathbb{Z}$? If the answer is yes for even a single $p>2$, then the question of whether or not $\char93 {\mbox{{\fontencoding{OT2}\fontfamily{wncyr}\fontseries{m}\fontshape{n}\selectfont Sh}}}(A/\mathbb{Q})$ must be a square or twice a square is finally resolved. We make the following contribution toward settling this problem.

Proposition 7.2   Let $E$ be the unique elliptic curve over  $\mathbb{Q}$ of conductor $43$. Let $K=\mathbb{Q}(\mu_7)^+$ be the real subfield of $\mathbb{Q}(\mu_7)$, let $R=\Res_{K/\mathbb{Q}}E_K$, and let $A = \ker\left(R \rightarrow E\right).$ Then $3\mid \char93 {\mbox{{\fontencoding{OT2}\fontfamily{wncyr}\fontseries{m}\fontshape{n}\selectfont Sh}}}(A/\mathbb{Q})$ and the Birch and Swinnerton-Dyer conjecture implies that $\char93 {\mbox{{\fontencoding{OT2}\fontfamily{wncyr}\fontseries{m}\fontshape{n}\selectfont Sh}}}(A/\mathbb{Q})[3]=3$ and $\char93 {\mbox{{\fontencoding{OT2}\fontfamily{wncyr}\fontseries{m}\fontshape{n}\selectfont Sh}}}(R/\mathbb{Q})[3]=1$.

Proof. We first verify that the BSD conjecture predicts that $\char93 {\mbox{{\fontencoding{OT2}\fontfamily{wncyr}\fontseries{m}\fontshape{n}\selectfont Sh}}}(E/K)[3]=1$. Because $K/\mathbb{Q}$ is abelian,

$$L(E_K,s) = L(E,s)\cdot L(E,\chi,s)\cdot L(E,\chi^{-1},s),$$

where $\chi:(\mathbb{Z}/7\mathbb{Z})^*\rightarrow \mu_3$ is a Dirichlet character of order $3$. For each of the three real places $v$ of $K$, we have $\Omega_{E,v}=\Omega_{E/\mathbb{Q}}$. Next observe that

$$E(\mathbb{Q})\otimes \mathbb{Z}_3 \rightarrow E(K)\otimes \mathbb{Z}_3$$

is surjective, because

$$\frac{E(K)}{E(\mathbb{Q})} = \frac{R(\mathbb{Q})}{E(\mathbb{Q})} \hookrightarrow (R/E)(\mathbb{Q})_{\tor}$$

and $(R/E)(\mathbb{Q})[3]=\{0\}$ by Proposition 5.2. If $P\in E(\mathbb{Q})$ then

$$\langle P, P\rangle_\mathbb{Q}= \frac{1}{[K:\mathbb{Q}]} \langle P, P \rangle_K,$$

so $\Reg(E/K) \sim 3\cdot \Reg(E/\mathbb{Q})$, where $\sim$ denotes ``equality up to a number coprime to $3$''. By Proposition 5.2, $E(K)_{\tor}[3]=\{0\}$ and $3\nmid c_v$ for places $v$ of  $\mathbb{Q}$ (because this is true for $R$). Finally,

$$\char93 {\mbox{{\fontencoding{OT2}\fontfamily{wncyr}\fontseries{m}\fontshape{n}\selectfont Sh}}}(E/K) = \frac{L'(E_K,1)\cdot \char93 E(K)_{\tor}^2} {\Reg(E/K)\cdot\Omega_{E/\mathbb{Q}}^3\cdot \prod_{v} c_{v}}$$

$$\sim \frac{L'(E,1)}{3\cdot \Reg(E/\mathbb{Q}) \cdot \Omega_E} \cdot \Norm_{\mathbb{Q}(\mu_3)/\mathbb{Q}}\left(\frac{L(E,\chi,1)}{\Omega_E}\right)$$

$$= \frac{1}{3}\cdot 3=1.$$

We verified the last nontrivial equality with a computer using standard modular symbols techniques.

We have an exact sequence

$$0 \rightarrow \Vis({\mbox{{\fontencoding{OT2}\fontfamily{wncyr}\f... ...fontfamily{wncyr}\fontseries{m}\fontshape{n}\selectfont Sh}}}(R/\mathbb{Q})[3].$$

Since

$${\mbox{{\fontencoding{OT2}\fontfamily{wncyr}\fontseries{m}\fontsh... ...2}\fontfamily{wncyr}\fontseries{m}\fontshape{n}\selectfont Sh}}}(E/K)[3]=\{0\},$$

$$E(\mathbb{Q})/3 E(\mathbb{Q})\hookrightarrow \Vis({\mbox{{\fonten... ...ontfamily{wncyr}\fontseries{m}\fontshape{n}\selectfont Sh}}}(A/\mathbb{Q})[3]),$$

and $\Vis({\mbox{{\fontencoding{OT2}\fontfamily{wncyr}\fontseries{m}\fontshape{n}\selectfont Sh}}}(A/\mathbb{Q})[3])$ is a surjective image of $R(\mathbb{Q})\otimes \mathbb{Z}_3 = E(K)\otimes \mathbb{Z}_3$, which, as mentioned above, is a surjective image of $E(\mathbb{Q})\otimes \mathbb{Z}_3=\mathbb{Z}_3$, it follows that

$$\Vis({\mbox{{\fontencoding{OT2}\fontfamily{wncyr}\fontseries{m}\fontshape{n}\selectfont Sh}}}(A/\mathbb{Q})[3])=\mathbb{Z}/3\mathbb{Z}.$$

$\qedsymbol$