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Feb 2, 2012 at 20:28 vote accept CommunityBot moved from User.Id=19475 by developer User.Id=69903
Feb 1, 2012 at 19:20 answer added user19475 timeline score: 0
Dec 2, 2011 at 12:44 comment added user19475 Does anyone have a vague idea how to approach 3?
Dec 1, 2011 at 11:04 comment added naf The paper of Murre you refer to is indeed the one I meant. The result of Abhyankar I was referring to is Theorem 1.2 in VI.1 of Kollar's book "Rational curves...". I now see that Abhyankar's theorem actually predates Murre's and it suffices for what I said.
Dec 1, 2011 at 10:30 history edited user19475 CC BY-SA 3.0
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Dec 1, 2011 at 10:12 comment added user19475 Dear ulrich, thank you for your response. By the theorem of Murre, do you mean the article "On a connectedness theorem ..."? And which paper of Abhyankar do you mean?
Nov 25, 2011 at 15:27 comment added naf 2. is also true if the order of $X \in H^1(S\backslash Z,A)$ is prime to the characteristic. If this holds then $X$ comes from an element of $H^1(S \backslash Z,A[n])$ for some $n$ prime to the characteristic . As you suggested, we can use purity, actually just the classical Zariski-Nagata purity theorem, to extend this to an element of $H^1(S,A[n])$ and then we get the desired element of $H^1(S,A)$.
Nov 25, 2011 at 13:03 comment added naf 1. is true because $S$ is smooth. The trivialisation over $S \backslash Z$ gives a rational map from $S$ to the principal homogenous space and any such map (with $S$ a regular scheme) extends to a morphism. The reason for this is that abelian varieties do not contain rational curves but all positive dimensional fibres of a birational proper morphism $S' \to S$ are covered by rational curves (by a theorem of Murre or Abhyankar (depending on the level of generality)).
Nov 25, 2011 at 12:43 history edited user19475 CC BY-SA 3.0
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Nov 25, 2011 at 12:32 history edited user19475 CC BY-SA 3.0
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Nov 23, 2011 at 14:12 history asked user19475 CC BY-SA 3.0