Let $A/k$ be a simple abelian variety.

Does there exist a non-simple abelian variety $B/k$ and a finite homomorphism $f:B\to A$ over $k$?

I don't need $f:B\to A$ to be etale.

  • $\begingroup$ I believe the question should truly be "Is any simple abelian variety covered by a non-simple abelian variety?" $\endgroup$ Aug 1, 2012 at 14:48
  • $\begingroup$ I think I see what you mean. Thanks for correcting my English. $\endgroup$
    – Harry
    Aug 1, 2012 at 14:56
  • $\begingroup$ It isn't necessarily a question of English, more of mathematics. $\endgroup$ Aug 1, 2012 at 15:43

1 Answer 1


No. If $End_A$ and $End_B$ are the endomorphism rings then an isogeny $B\to A$ will give an isomorphism $End_A\otimes\mathbb Q\to End_B\otimes\mathbb Q$. But the first of these algebras is a division algebra and the second is not.

  • 2
    $\begingroup$ In case this wasn't clear: a finite homomorphism is necessarily etale since if it etale over one point it is etale over every point, and finite morphisms are etale on a nonempty open subset, thus it is an isogeny. $\endgroup$
    – Will Sawin
    Aug 1, 2012 at 14:10
  • 2
    $\begingroup$ To elaborate: If $f:B\to A$ is finite and separable (e.g. if $char\, k=0$) then Riemann-Hurwitz implies that the ramification divisor $R= K_A-f^*K_B=0$. Therefore $f$ is etale. $\endgroup$ Aug 1, 2012 at 14:53

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