Let $E:y^2=x^3+17x$.
Let $K=\Bbb{Q}(\sqrt{D})$($D$ is a negative integer). If $D\neq -17$, I want to prove $E(K)_{tor}\cong \Bbb{Z}/2\Bbb{Z}$ admitting the fact that $E(\Bbb{Q})_{tor}\cong \Bbb{Z}/2\Bbb{Z}$.
If I once could prove $E(K)_{tor}=E(K)[2^∞]$, I can prove $E(K)[2^∞]=E(K)[2]\cong \Bbb{Z}/2\Bbb{Z}$ by explicit calculation.
But how can I prove $E(K)_{tor}=E(K)[2^∞]$ ?
I first came up with trace map. Define $E(K)\to E(\Bbb{Q})$ by $P\to P+P^{\sigma}$ where $\sigma$ is generator of $Gal(K/\Bbb{Q})$.
Thank you for your help.
Let $q \equiv 3 \pmod{4}$ be a prime. Since $E$ has CM by $\mathbb{Q}(i)$, $a_q(E)=0$, and therefore $E(\mathbb{F}_{q^2})=(q+1)^2$.
Since $E$ has good reduction at $q$, prime-to-$q$ torsion in $E$ reduces injectively mod $q$, so that the prime-to-$q$ part of $E(K)_{tors}$ is a subgroup of $E(\mathbb{F}_{q^2})$, for any quadratic field $K$.
In particular, the cardinality of the prime-to-$3$ part of $E(K)_{tors}$ divides $(3+1)^2=16$, so any torsion in a quadratic number field is either $2$-torsion or $3$-torsion.
In particular, the cardinality of $E(K)_{tors}$ divides $(7+1)^2=64$, so $E(K)_{tors}=E(K)[2^{\infty}]$.
We can also refine this result: take $q=5$. $E$ still has good reduction at $q$ and $a_q(E)=0$, so $E(\mathbb{F}_{25})$ has cardinality $(q+1)^2=36$. In particular, $E(K)[2^{\infty}]$ is a subgroup of $G=E(\mathbb{F}_{25})[2^{\infty}]$, which has order $4$.
Now, $E(\mathbb{Q}(\sqrt{-17}))[2^{\infty}] \cong (\mathbb{Z}/2\mathbb{Z})^{\oplus 2}$, so $G=(\mathbb{Z}/2\mathbb{Z})^{\oplus 2}$, and any torsion point defined over a quadratic number field $K$ is a $2$-torsion point.
In particular, if $K \neq \mathbb{Q}(\sqrt{-17})$, then $E(K)_{tors}=E(\mathbb{Q})_{tors}=\mathbb{Z}/2\mathbb{Z}$.