Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
    
I believe the question should truly be "Is any simple abelian variety covered by a non-simple abelian variety?" –  David Corwin Aug 1 '12 at 14:48
    
I think I see what you mean. Thanks for correcting my English. –  Harry Aug 1 '12 at 14:56
    
It isn't necessarily a question of English, more of mathematics. –  David Corwin Aug 1 '12 at 15:43
add comment

1 Answer

up vote 7 down vote accepted

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.

share|improve this answer
2  
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. –  Will Sawin Aug 1 '12 at 14:10
2  
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. –  Donu Arapura Aug 1 '12 at 14:53
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.