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Let $K = \mathbb{Q}(\sqrt{ 2})$. Let $E_1$ be an elliptic curve over $K$. What is the structure of the set of points $E_1(K)$? Assume there is an elliptic curve $E_2$ defined over $K$ and a map $f : E_1 \rightarrow E_2$ defined over $K$ that is surjective with finite kernel. Compare $E_2(K)$ and $E_1(K)$ as precisely as possible.

My attempt:

From Mordell-Weil I know that $E_1(K) \cong \mathbb{Z}^r \bigoplus E_1(K)_{tors}$ and same for $E_2(K) \cong \mathbb{Z}^s \bigoplus E_2(K)_{tors}$. Can I say something more on the structure of $E_1(K)$??

Then we have that $f$ is surjective so my guess is that the rank of $E_2$ is less than the rank of $E_1$, i.e. $s \leq r$? I don't really know if this is right...

And the finite kernel implies that $f(E_1(K)_{tors}) \subseteq E_2(K)_{tors}$??

Any help?

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  • $\begingroup$ For any homomorphism between two groups $f\colon G \to H$ the points of finite order map to points of finite order! $\endgroup$ – Alex J Best Apr 4 at 21:16
  • $\begingroup$ @AlexJBest didn't I write it in the last line? $\endgroup$ – user289143 Apr 4 at 21:17
  • $\begingroup$ right, but its not because of the finite kernel condition, finite kernel implies something stronger. $\endgroup$ – Alex J Best Apr 4 at 21:18
  • $\begingroup$ @AlexJBest and what does it imply? I can't see it.. I can say from the first isomorphism theorem that $E_1(K) / Ker(f) \cong E_2(K)$ since the map is surjective $\endgroup$ – user289143 Apr 4 at 21:20
  • $\begingroup$ What does the structure of such points mean? Like a full classification of the possible orders and torsion groups? That would be an annals worthy paper, I'd imagine. Even doing the analogue of Mazur's Torsion theorem would be crazy (I don't think it's been done). You can look at the work of Kamienny-Mazur to get a bound on the possible sizes of the torsion subgroups. You can say some basic things like, for example, $E(N)$ can never be $(\mathbb{Z}/N\mathbb{Z})^2$, but such statements are of a much simpler character. $\endgroup$ – Alex Youcis Apr 4 at 21:24

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