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Jan 23, 2014 at 13:31 vote accept Daniel Loughran
Jan 23, 2014 at 10:48 answer added Chris Wuthrich timeline score: 7
Jan 23, 2014 at 9:29 comment added Daniel Loughran Thanks anon. If one of you would like to post your comments as an answer, I will accept it.
Jan 23, 2014 at 7:58 comment added abz The problem is only for torsion at primes in S. More precisely, let m be an integer not divisible by any prime in S. Then the subgroup of the S Tate-Shafarevich group killed by some power of m coincides with the similar subgroup of the usual Tate-Shafarevich group. This is a standard result (see for example Milne's Arithmetic Duality Theorems I, 6.6., but note that his S is the complement of yours).
Jan 22, 2014 at 23:23 history edited Daniel Loughran CC BY-SA 3.0
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Jan 22, 2014 at 22:30 comment added user76758 Example 7.5.1 in the paper "Finiteness theorems for algebraic groups over function fields" in Compositio Math. 148 (2012), which uses just standard methods in global Galois cohomology (duality theorems, etc.) There must be earlier references on this issue as well.
Jan 22, 2014 at 19:34 comment added Daniel Loughran I see, can you offer some references?
Jan 22, 2014 at 16:54 comment added user76758 No, once you allow non-empty $S$ it is often (maybe always?) provably infinite, in contrast with the case of linear algebraic groups, for which the "$S$-version" is provably always finite.
Jan 22, 2014 at 14:38 comment added Daniel Loughran No, my question is about what happens when $S$ is non-empty. I have edited the question to make it clearer.
Jan 22, 2014 at 14:36 history edited Daniel Loughran CC BY-SA 3.0
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Jan 22, 2014 at 13:37 comment added Jason Starr Did you just post the Shafarevich conjecture as a Math Overflow question?
Jan 22, 2014 at 13:31 history asked Daniel Loughran CC BY-SA 3.0