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May 16 at 17:44 comment added Firebolt2222 I think I found a nice paper (jstor.org/stable/2374333?seq=2) and a suitabale interpretation, which I will have to check myself a bit more carefully. But for other people that are interested, I will write a more elaborate answer as soon as I get a better grasp of it.
May 16 at 17:42 comment added Firebolt2222 @DanielLoughran Thank you very much for the reference! This seems very helpful. I'll have a deeper look into it. Though I have to admit that I'm not that strong in algebraic geometry.
S May 15 at 14:07 history bounty ended CommunityBot
S May 15 at 14:07 history notice removed CommunityBot
May 7 at 19:26 comment added Daniel Loughran Elements of this form appear in the paper "Surfaces defined by pairs of polynomials" by Skorobogatov and Gvirtz-Chen. it seems in general to get something non-trivial you need to be using transcendental elements (indeed this cohomology group is trivial over number fields).
May 7 at 19:25 comment added Daniel Loughran At least if you have the corresponding roots of unity in our ground field, then you can construct elements of this Galois cohomology group using the cup product $H^1(k,\mu_n) \times H^1(k,\mu_n) \times H^1(k,\mu_n) \to H^3(k,\mu_n)$, which can be related to your cohomology using Kummer theory. Whether such an element represents a non-trivial cohomology class is a tricky question.
S May 7 at 12:57 history bounty started Firebolt2222
S May 7 at 12:57 history notice added Firebolt2222 Draw attention
May 2 at 12:17 comment added Firebolt2222 @Bma Thank you I'll have a look
May 1 at 7:43 comment added Bma Arithmetic Duality Theorems by Milne may have some relevant material
Apr 30 at 14:48 history edited Firebolt2222 CC BY-SA 4.0
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S Apr 30 at 14:42 review First questions
Apr 30 at 14:55
S Apr 30 at 14:42 history asked Firebolt2222 CC BY-SA 4.0