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Let $E$ and $F$ be two abelian varieties of dimension 1 over $\mathbb{C}$. Let $f : E \to F$ be a surjective homomorphism of abelian varieties ($f(0) = 0$). If $\ker (f) \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$, does this imply that $E$ and $F$ are isomorphic?

flag	asked Apr 7 2010 at 17:23 Tuan 343●1●8

Robin Chapman · Accepted Answer · 2010-04-07 17:28:32Z UTC

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Yes. In zero characteristic the image of an isogeny of elliptic curves is determined up to isomorphism by its kernel. Your isogeny has the same kernel as the doubling map from $E$ to itself.

answered Apr 7 2010 at 17:28

Robin Chapman
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@Robin: could you please provide a hint on how one proves this statement (that the image is determined by ker)? thanks – Qfwfq Apr 7 2010 at 17:51

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Think about the usual isomorphism theorems for groups. The map E -> F kills ker f, so factors as E -> E/(ker f) -> F. Since degrees are multiplicative the map E/(ker f) - > F has degree 1, so must be an isomorphism. – JT Apr 7 2010 at 17:54

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The same holds in positive characteristic, as long as "kernel" is interpreted scheme-theoretically. In particular, if the kernel of f is isomorphic to $(\mathbb{Z}/2\mathbb{Z})^2$ (as a constant group scheme), then $E \cong F$. – Pete L. Clark Apr 7 2010 at 18:07

To add to JT's remark, we also need that when $f_1:E\to F_1$ and $f_2:E\to F_2$ are isogenies of elliptic curves with $f_1$ separable and $\ker f_1\subseteq\ker f_2$ then there is an isogeny $g:F_1\to F_2$ with $f_2=gf_1$. Over $\mathbb{C}$ where elliptic curves are complex tori, this is quite easy to prove. Over general fields it requires more work; see Silverman's book for instance. – Robin Chapman Apr 7 2010 at 18:29

that makes lots of sense. thanks a lot everyone! – Tuan Apr 7 2010 at 22:43

isogeny of elliptic curves

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