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Oct 24, 2023 at 2:14 comment added Vik78 Interesting-- thank you for the details
Oct 22, 2023 at 14:09 comment added Uriya First @Vik78 Actually, this is a bit more subtle. "Abstract" Morita equivalence and Brauer equivalence are not the same even in the affine case, and even when the base ring $R$ is a field! As in Caldararu's example, this can happen when the base ring $R$ has an automorphism acting nontrivially on the Brauer group. In the affine case, the solution is to replace Morita equivalence with the stronger Morita-equivalence-over-$R$, which is indeed equivalent to Brauer equivalence for Azumaya algebras. This finer condition is unfortunately missing in Caldararu's Thm.1.3.15.
Oct 20, 2023 at 2:26 comment added Vik78 It’s worth noting that Morita equivalence (in the sense of two Azumaya algebras having equivalent categories of right modules) is the same as the equivalence relation described in the original post when working over an affine scheme, but in general the former is coarser than the latter. See example 1.3.16 here:people.math.wisc.edu/~caldararu/publications/…
Oct 27, 2019 at 11:38 vote accept CommunityBot
Oct 27, 2019 at 11:18 history answered Uriya First CC BY-SA 4.0