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Mar 14 at 3:52 comment added Ralle Moreover, $V$ is always a generalized Sklyanin algebra as defined in Section 7 of the paper Natalia Iyudu, Stanislav Shkarin, Three dimensional Sklyanin algebras and Gröbner bases, Journal of Algebra, Volume 470, 2017, Pages 379-419.
Mar 13 at 9:48 comment added YCor So over an algebraic closure (and over a cubic extension) it is always a Sklyanin algebra. In general, it's a twisted form of a Sklyanin algebra.
Mar 13 at 5:30 comment added Peter Wu One can rescale $x,y,z$, so whether $V$ can be transformed into a Sklyanin algebra depend on it's orbit in $\mathbb{F}_p^{\times}/\mathbb{F}_p^{\times 3}$
Mar 13 at 4:01 history answered Ralle CC BY-SA 4.0