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Consider:

  • Silverman, Ex V.5.4: Elliptic curves $E/\mathbb{F}_q$ and $E'/\mathbb{F}_q$ are isogenous if and only if $\#E(\mathbb{F}_q) = \# E'(\mathbb{F}_q)$.
  • Silverman, Proposition 4.12: any finite subgroup $\Phi \subset E$ induces an isogeny $E \rightarrow E'$, where $E'$ has group structure $E / \Phi$.
  • Silverman Theorem 2.3: Any non-constant morphism of curves is surjective.

Why do these three facts not contradict each other? In other words, if by quotienting with some nontrivial subgroup $\Phi \subset E(\mathbb{F}_q)$ I can induce a surjective isogeny $E \rightarrow E'$, why doesn't $E'$ have strictly fewer rational points than $E$? This is blowing my mind. Thank you.

BD107
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  • Did you look at isogenies of the complex torus $\mathbb{C}/\Lambda$ with $\Lambda = \mathbf{Z} + i \mathbf{Z}$ ? Those are of the form $z + \Lambda \mapsto z + \Lambda'$ with $\Lambda'$ a lattice containing $\Lambda$, which is the same as $z +\Lambda\mapsto z+H+\Lambda$ where $H$ is (a set of representatives of) $\Lambda'/\Lambda$ which is the kernel of the isogeny. Looking at the Weierstrass functions of $\Lambda,\Lambda'$ helps too, as $\wp_{\Lambda'}(z) = \sum_{a \in H} \wp_\Lambda(z+a)$ – reuns Oct 24 '18 at 19:39

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Surjective means here surjective from $E(\overline k)$ to $E'(\overline k)$ where $\overline k$ is the algebraic closure of $k=\Bbb F_q$. The image of $E(k)$ may be a proper subgroup of $E'(k)$. There may be points $Q\in E'(k)$ which are images of elements of $E(\overline k)$ all lying outside $E(k)$.

Angina Seng
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  • thank you, but what about this? https://math.stackexchange.com/questions/980399/to-show-a-morphism-of-affine-k-varieties-which-is-surjective-on-closed-points-is – BD107 Oct 24 '18 at 19:43
  • The closed points in this definition correspond to Galois orbits on $E(\overline k)$. @BD107 – Angina Seng Oct 24 '18 at 19:44
  • Ah... so even a closed point of $E$ could reside in $E(\overline k) - E(k)$? – BD107 Oct 24 '18 at 19:48