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Feb 15, 2011 at 20:21 vote accept Hailong Dao
Oct 10, 2010 at 20:55 history edited Francesco Polizzi CC BY-SA 2.5
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Oct 10, 2010 at 16:14 comment added Francesco Polizzi ...But then $B$ and $B'$ are contained in fibres of a morphism $f \colon X \to Y$. By adjunction formula, the fibres of $f$ must be Abelian subvarieties of $X$, so this is not possible when $X$ is simple.
Oct 10, 2010 at 16:14 comment added Francesco Polizzi Dear Hailong, I do not know any bounds for surfaces, but maybe other people could give more precise answers. About Prop 1, since Abelian varieties have the group of translations, if you take two cycles $B$ and $B'$ of complementary dimension, you can always move $B$ in its rational equivalence class in such a way that it intersects $B'$ properly. The only exception is when $B$ is a translate of $B'$, which does not intersect $B'$...
Oct 10, 2010 at 14:49 comment added Hailong Dao And +1. Also, can you explain your last sentence a bit, why two effective divisors always intersect implies Prop 1?
Oct 10, 2010 at 14:44 comment added Hailong Dao Dear Francesco, thanks, this is very interesting. So I gather from it that there is no known bounds even when X is a surface?
Oct 10, 2010 at 14:26 history answered Francesco Polizzi CC BY-SA 2.5