# Modules over an Azumaya algebra and modules over the associated Brauer-Severi variety

Assume $\mathcal{A}$ is an Azumaya algebra of rank $r^2$ on a smooth projective scheme $Y$ over $\mathbb{C}$. Let $f: X\rightarrow Y$ be the Brauer-Severi variety associated to $\mathcal{A}$.

I read here in a comment that the category of modules over $\mathcal{A}$ is equivalent to the categroy of modules on $X$ which restrict to every fiber as a multiplicity of $O(1)$. I did not find a reference in the literature of this fact.

Here is what I think of the situation:

Let us start with a left $\mathcal{A}$-module $M$, then $f^{*}M$ is a left $f^{*}\mathcal{A}$-module. But $f^{*}\mathcal{A}=\mathcal{E}nd_X(I)^{op}\cong \mathcal{E}nd_X(I^{\vee})$ for a locally free $O_X$-module $I$ which restricts to $O(-1)^r$ over every closed point. So using Morita equivalence $f^{*}M=I^{\vee}\otimes N$ for a module $N$ on $X$. So $M$ gives me the module $N$ on $X$, but I don't see any reason why we should know how $N$ restricts to the fibers of $f$. Or am I missing something?

What is the exact equivalence of categories in this situation? I am also happy if we could only say something about locally free $\mathcal{A}$-modules and locally free $O_X$-modules. References are also very welcome.

• Just to point out, the sheaf $I$ that you mention is the unique locally free sheaf of rank $r$ sitting in the short exact sequence $$0\to \Omega_{X/Y}\to I\to \mathcal{O}_X \to 0,$$ coming from the canonical generator of $R^1f_*\Omega_{X/Y}$. If you make any base change to $Y'$ such that $\mathcal{A}' = \text{End}(E)$, then $X'$ is $\mathbb{P}_Y(E)$, and $I'$ is $(f')^*E^\vee\otimes \mathcal{O}_{\mathbb{P}(E)}(-1)$. Thus the pullback of $\text{End}(E)$ equals the pullback of $\text{End}(I)$. – Jason Starr Sep 4 '15 at 12:18
• ... Thus, by descent and Skolem-Noether, there is a canonical isomorphism of $f^*\mathcal{A}$ with $\text{End}(I)$ (or its opposite ring, as you write, I am not quite sure). For restriction to fibers, surely sasha meant "geometric fibers". Thus, base change again to $Y'$ and check there. – Jason Starr Sep 4 '15 at 12:22
• @Jason : My problem is not with the sheaf $I$ (all about $I$ is in Quillen's article Higher algebriac K-theory I), but with sheaves of $End(I)$ modules. For examples, assume $M$ has rank one as an $A$-module and is locally free, then the corresponding $O_X$-module, is locally free of rank $r$ on $X$. Does every rank $r$ bundle appaer this way? Does every rank $r$ bundle on $X$ give an $A$-module of rank one after tensoring with the dual of $I$ and pushing down to $S$? I don't see how to find properties of the bundles on $X$, maybe all possible bundles can appear? – Bernie Sep 4 '15 at 15:07
• I am afraid that I do not understand what you are asking. The functors, $M \mapsto I^\vee\otimes_{f^*\mathcal{A}} f^*M$ and $N\mapsto f_*\textit{Hom}_{\mathcal{O}_X}(I^\vee,N)$ establish an equivalence between the category of $\mathcal{A}$-modules that are locally free as $\mathcal{O}_Y$-modules and the category of locally free $\mathcal{O}_X$-modules whose restriction to all geometric fiber is isomorphic to a direct summand of $(I^\vee)^{\oplus n}$ for some choice of $n$. Once you have the functors, checking that they are inverse can be done after base change to $Y'$, where it is easy. – Jason Starr Sep 4 '15 at 18:16
• If you begin with a sheaf $N$ that is (etale locally) isomorphic to $(I^\vee)\otimes_{\mathcal{O}_X}f^*L$, then $G(N)$ is $f_*(\textit{Hom}_{\mathcal{O}_X}(I^\vee,I^\vee)\otimes_{\mathcal{O}_X} f^*L)$, which by the projection formula is $\mathcal{A}\otimes_{\mathcal{O}_Y} L$. Thus $F(G(N))$ is $I^\vee\otimes_{f^*\mathcal{A}} f^*(\mathcal{A}\otimes_{\mathcal{O}_Y} L) \cong I^\vee\otimes_{\mathcal{O}_X}f^*L$. – Jason Starr Sep 7 '15 at 15:07