Twisted line bundles Brauer class This question is mainly a reference request about the order of a Brauer class on a smooth projective variety over $\mathbb{C}$. Namely, let $X$ be a smooth complex projective variety and $\alpha$ be a Brauer class on $X$.
I have read in a few papers on the ArXiv that the order of $\alpha$ divides the rank of any $\alpha$-twisted sheaf. In particular, the existence of a $\alpha$-twisted line bundle would imply that $\alpha$ is trivial.
On the other hand, let $ p : Y \longrightarrow X$ be a non-trivial Brauer-Severi variety associated to the class $\alpha \in Br(X)$. It seems that one can define a $p^*\alpha$-twisted line bundle $\mathcal{O}_{Y/X}(1)$ and it seems this twisted line bundle is not a line bundle (so that the class $p^* \alpha$ is non trivial).
I am not sure how to reconcile these two claims. Is there something obvious I am missing?
Thanks a lot!
 A: There are many functors from the category of twisted sheaves to the category of untwisted sheaves. Typically, if $E$ is a locally free $\alpha$-twisted sheaf then $\mathcal{H}om(E,-)$ is such a functor. It identifies the category of $\alpha$-twisted sheaves with the category of $\mathcal{E}nd(E)$-modules. 
In many cases, by a rank of an $\alpha$-twisted sheaf $F$ people understand the rank of the corresponding sheaf $\mathcal{H}om(E,F)$. In this sense, existence of a rank 1 twisted sheaf is equivalent to the vanishing of the Brauer class. 
Sometimes, however, one can divide this by the rank of $\mathcal{E}nd(E)$. This "divided rank" may be equal to 1 without $\alpha$ being trivial. For instance, the divided rank of $E$ itself is equal to 1.
A: OK so I think I understand what I missed from the beginning. If I take a particular $\alpha$-twisted sheaf, say $F$, then the object $F$ itself, as a twisted sheaf, depends on the Cech cocycle I use to represent $\alpha$. In particular, the Cech cocycle used to define $\mathcal{O}_{Y/X}(1)$ on a Severi Brauer variety is non-trivial.
On the other hand, the category of $\alpha$-twisted sheaves only depends (up to equivalence) on the cohomology class $\alpha \in H^2(X,\mathcal{O}^{\star}_X)_{et}$. 
Hence, though $\mathcal{O}_{Y/X}(1)$ is not a line bundle, one still has that the category of $p^* \alpha$-twisted sheaves on $Y$ is equivalent to the category of untwisted sheaves on $Y$!
