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Jul 19, 2020 at 10:19 comment added Mikhail Bondarko Yes, this question may be attacked by means of decomposing the diagonal; yet I am not sure that looking at smooth projectives is optimal here. On the other case, the injectivity question appears to be much easier; it suffices to have a degree 1 zero-cycle on any (and thus, on all) smooth X such that $k(X)=K$.
Jul 19, 2020 at 2:42 comment added Jinhyun Park I believe this is in general open. When K is purely transcendental over k, one can have injectivity maybe? But surjectivity does not seem immediate to me in general. Though I think there are some works in this direction, by e.g. Vishik, Karpenko, Merkurjev, Zainoulline, S. Gille, R. Fino, S. Baek, etc. and I think you may find some partial answers from some of their papers. Like A. Vishik, in Geometric methods in the algebraic theory of quadratic forms, 25–101, Lecture Notes in Math., 1835, Springer, Berlin, 2004? This long paper seems to consider the function field case.
Jul 18, 2020 at 18:06 comment added Jason Starr I assume that you aware that you are asking which projective $k$-varieties have an integral decomposition-of-the-diagonal.
Jul 18, 2020 at 8:48 history asked Mikhail Bondarko CC BY-SA 4.0