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Let $\varphi : A \to B$ be an isogeny between 2 abelian varieties of dimension $g$. Are there known conditions for the $\ker\varphi$ so that this induces an isomorphism between $A$ and $B$? For example, if $\ker\varphi \cong (\mathbb{Z}/n\mathbb{Z})^{2g}$, then $A \cong B$, because $\varphi$ factors through the multiplication map $A \xrightarrow{\times n} A$ followed by an isomoprhism $A \to B$. I wonder if there are other cases that induce isomorphisms.

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If I understand your question correctly, then there are many other cases. Take for example an elliptic curve $E$ with complex multiplication, and take any non-zero map $\phi:E\to E$. The kernel can be cyclic but $E$ is still isomorphic to $E$ mod $\ker(\phi)$. In general this has nothing much to do with abelian varieties, it's just group theory really. You're just asking, in some slightly cryptic manner, of examples of isogenies from an abelian variety to itself. You know about multiplication by $n$ but there can be far more than that; the situation is well-understood buttoolongtofitintothisbo – Kevin Buzzard Oct 11 '10 at 20:24
x.$\mathfrak{}$ – S. Carnahan Oct 12 '10 at 2:51
Already the simplest criterion given here, seems to have non trivial applications. Here is a paper disproving the 30 year old Donagi conjecture on the Torelli problem for P=rym varieties, by Izadi and Lange. – roy smith Nov 13 '10 at 20:29
up vote 3 down vote accepted

Kevin's comment is right on the money, but here it is in more detail: I will give a general criterion for an isogeny $\varphi: A \rightarrow B$ of abelian varieties to induce an isomorphism upon passage to the kernel.

Let me work over an unnamed algebraically closed field. Suppose that $A = B$ and $\eta \in \operatorname{End}(A)$ is a surjective endomorphism of $A$. (N.B.: If $A$ is simple -- i.e., contains no proper nontrivial subvariety -- then any nonzero endomorphism is surjective. In particular this holds for all elliptic curves.) Then $\eta$ is also an isogeny: i.e., its kernel is a finite subgroup scheme, say $K$ and -- essentially, by the first isomorphism theorem for groups, as Kevin says -- it follows that there is an induced isomorphism

$A/K \stackrel{\sim}{\rightarrow} A$.

This condition is also necessary: if $\varphi: A \rightarrow B$ is an isogeny such that $B \cong A$, then composing with this isomorphism gives a surjective endomorphism of $A$ and the resulting map factors through an isomorphism $A/(\operatorname{ker}(\varphi)) \rightarrow B$. Thus all examples arise from a surjective endomorphism of $A$ as above, well-defined up to isomorphisms on the source and target.

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