isogenies between abelian varieties that induce isomorphisms? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T18:13:37Z http://mathoverflow.net/feeds/question/41830 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/41830/isogenies-between-abelian-varieties-that-induce-isomorphisms isogenies between abelian varieties that induce isomorphisms? Tuan 2010-10-11T20:07:24Z 2010-10-11T23:05:23Z <p>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.</p> http://mathoverflow.net/questions/41830/isogenies-between-abelian-varieties-that-induce-isomorphisms/41841#41841 Answer by Pete L. Clark for isogenies between abelian varieties that induce isomorphisms? Pete L. Clark 2010-10-11T23:05:23Z 2010-10-11T23:05:23Z <p>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. </p> <p>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</p> <p><code>$A/K \stackrel{\sim}{\rightarrow} A$</code>.</p> <p>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 <code>$A/(\operatorname{ker}(\varphi)) \rightarrow B$</code>. Thus all examples arise from a surjective endomorphism of $A$ as above, well-defined up to isomorphisms on the source and target.</p>