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Let $X$ be a scheme, let $\mathcal{A}$ be a sheaf of locally free algebras on $X$. We say $\mathcal{A}$ is an azumaya algebra, if the natural map $$\mathcal{A}\otimes_{\mathcal{O}_X}\mathcal{A}^{opp}\to\mathcal{E}nd_{\mathcal{O}_X}(\mathcal{A}), $$ $$a\otimes b\mapsto (x\mapsto axb)$$ is an isomorphism.

Two azumaya algebras $\mathcal{A},\mathcal{B}$ are called Morita equivalent, if there exists locally free sheaves $\mathcal{F},\mathcal{G}$, such that $$\mathcal{A}\otimes\mathcal{E}nd_{\mathcal{O}_X}(\mathcal{F})\cong\mathcal{B}\otimes\mathcal{E}nd_{\mathcal{O}_X}(\mathcal{G}).$$

Let $\mathcal{A}$ be an Azumaya algebra, which is Morita equivalent to $\mathcal{B}:=\mathcal{O}_X$, does $\mathcal{A}$ necessarily have the form $\mathcal{E}nd_{\mathcal{O}_X}(\mathcal{H})$ for some locally free sheaf $\mathcal{H}$ on $X$?(I think this is true by the comment of Eoin). Is there a way to express this $\mathcal{H}$?

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    $\begingroup$ I think so. I'm looking at Milne's Etale cohomology book, Chapter IV, Section 2. For X a scheme you have an exact sequence $1\rightarrow G_{m,X}\rightarrow GL_{n,X}\rightarrow PGL_{n,X}\rightarrow 1$ (of etale sheaves on X). The group of azumaya algebras includes into $H^2_{et}(X,G_m)$ by Theorem 2.5 of loc. cit., so if one is trivial then it comes from an element of $H^1_{et}(X,GL_n)$. I think this correspondence sends a locally free sheaf $\mathcal{F}$ of rank $n$ to $\mathcal{E}nd(\mathcal{F})$. $\endgroup$
    – Eoin
    Commented Oct 27, 2019 at 5:33
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    $\begingroup$ @Eoin Thanks! Furthermore, would there be a way to express this $\mathcal{H}$ in terms of $\mathcal{F},\mathcal{G}$? (The result seems to claim we can always pick some line bundle $\mathcal{L}$, so that we can write $\mathcal{G}\otimes\mathcal{L}$ as $\mathcal{F}\otimes\mathcal{H}$? I am a bit confused how to show a module has a specific "tensor summand") $\endgroup$
    – user39380
    Commented Oct 27, 2019 at 6:55
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    $\begingroup$ The usual way to recover $k^n$ from $M_n(k)$ is to take a minimal left ideal. Does that work? $\endgroup$ Commented Oct 27, 2019 at 9:17
  • $\begingroup$ @MartinBright Thanks for the nice suggestion! Let me check it.. $\endgroup$
    – user39380
    Commented Oct 27, 2019 at 11:16

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The answer is yes and one can take $$\mathcal{H}:=\mathcal{Hom}_{\mathcal{End}(\mathcal G)}(\mathcal F\otimes \mathcal A, \mathcal G).$$ Here, $\mathcal F\otimes \mathcal A$ is viewed as a left $\mathcal{End}(\mathcal G)$-module via the isomorphism $\mathcal{End}(\mathcal G)\cong \mathcal{End}(\mathcal F)\otimes \mathcal A$.

More precisely, the right action of $\mathcal{A}$ on $\mathcal F\otimes\mathcal A$ determines a left $\mathcal{A}$-module structure on $\mathcal H=\mathcal{Hom}_{\mathcal{End}(\mathcal G)}(\mathcal F\otimes \mathcal A, \mathcal G)$, which in trun determines a morphism of $\mathcal O_X$-algebras $\psi:\mathcal A\to \mathcal{End}(\mathcal H)$. This morphism is an isomorphism.

Since both the source and target of $\psi$ are locally free $\mathcal O_X$-modules, it is enough to check that $\psi$ is an isomorphism after specializing to geometric points. Then the claim follows by noting that the source and target are central simple algebras of the same dimension.


I do feel that something should be said about the use of Morita theory lying in the background, so let me elaborate on this. I will consider consider the affine case $X=\mathrm{Spec} R$ for simplicity, and write $A=\Gamma(X,\mathcal{A})$, etc.

Our goal is to construct an $(A,R)$-progenerator $H$, i.e., an $(A,R)$-bimodule $H$ such that $H_R$ is projective (i.e. locally free), finitely generated and the natural map $A\to\mathrm{End}_R(H_R)$ is an isomorphism. (In this case, Morita theory tells us that ${}_AH$ is f.g. projective and $R=\mathrm{End}_A({}_AH)$ if $A$-endomorphism are written on the right.)

Write $F'=F\otimes A$ and view it as a right $A$-module. Then, since $F$ is f.g. projective, there is an $R$-algebra isomorphism $$ \mathrm{End}_A(F')\cong \mathrm{End}_R(F)\otimes A\cong \mathrm{End}_R(G). $$ Thus, both $F'$ and $G$ can be regarded as left $\mathrm{End}_R(G)$-modules.

Since $G$ is f.g. projective over $R$ and $F'$ is f.g. projective over $A$, we see that $G$ is an $(\mathrm{End}(G),R)$-progenerator and $G$ is an $(\mathrm{End}_A(F'),A)$-progenerator, which we view as a $(\mathrm{End}_R(G),A)$-progenerator. Now, Morita theory tells us that $H=\mathrm{Hom}_{\mathrm{End}_R(G)}(F',G)$, which is naturally an $(A,R)$-bimodule, is an $(A,R)$-progenerator, which is exactly what we want.

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  • $\begingroup$ 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/… $\endgroup$
    – Vik78
    Commented Oct 20, 2023 at 2:26
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    $\begingroup$ @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. $\endgroup$ Commented Oct 22, 2023 at 14:09
  • $\begingroup$ Interesting-- thank you for the details $\endgroup$
    – Vik78
    Commented Oct 24, 2023 at 2:14

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