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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?

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up vote 12 down vote accepted

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.

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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 '10 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. –  user1594 Apr 7 '10 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 '10 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 '10 at 18:29
    
that makes lots of sense. thanks a lot everyone! –  Tuan Apr 7 '10 at 22:43

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