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For any scheme $X$, let $\operatorname{Br}X$ denote the (Azumaya) Brauer group of $X$, namely the Morita equivalence classes of Azumaya $\mathcal{O}_{X}$-algebras.

Is the functor $$\operatorname{Br} : \operatorname{Sch}^{\operatorname{op}} \to \operatorname{Ab}$$ sending $X \mapsto \operatorname{Br}X$ a sheaf for the Zariski topology on $\operatorname{Sch}$?

In other words, if $X = \bigcup_{i \in I} U_{i}$ is a Zariski open covering of a scheme $X$, is the sequence $$0 \to \operatorname{Br}X \to \prod_{i \in I} \operatorname{Br}U_{i} \to \prod_{i_{1},i_{2} \in I} \operatorname{Br}(U_{i_{1}} \cap U_{i_{2}}) $$ exact?

Thoughts: My naive guess is "no" since $\operatorname{Br}$ is not an etale sheaf and $\operatorname{Pic}$ is not a Zariski sheaf. Exercise 8(f) of http://www-personal.umich.edu/~bhattb/teaching/mat731fall2011/ex6.pdf shows that the functor $U \mapsto \mathrm{H}_{et}^{2}(U,\mathbb{G}_{m})$ is a Zariski sheaf if $X$ is regular Noetherian. See also the discussion on Milne's "Etale Cohomology", IV, Remark 2.10.

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  • $\begingroup$ It might be 2-descent: you need data of the form (Azumaya algebras $A_i$ over (rings corresponding to affine) $U_i$, $A_i|_{U_{ij}}$-$A_j|_{U_{ij}}$-bimodules $B_{ij}$ over $U_{ij}:=U_i\cap U_j$ together with coherent isomorphisms $B_{ii}\cong A_i$ over $U_i$ and $B_{ij}|_{U_{ijk}}\otimes_{A_j|_{U_{ijk}}}B_{jk}|_{U_{ijk}}$ $\cong$ $B_{ik}|_{U_{ijk}}$ over $U_{ijk}:=U_i\cap U_j\cap U_k$) for all (not necessarily distinct) $i$, $j$, $k$. In other words it might be a 2-stack rather than a sheaf. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Sep 25, 2015 at 6:11
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    $\begingroup$ @მამუკაჯიბლაძე - $Br$ is really the decategorification of a 2-functor (valued in something complicated and not groupoidal, as far as I've worked with it), as you say. The extent to which it is a stack is tricky: it satisfies a pseudo or lax descent for covers of separated schemes by two affines, by result of Gabber. $\endgroup$
    – David Roberts ♦
    Commented Sep 25, 2015 at 7:42
  • $\begingroup$ @DavidRoberts So is it a 2-stack (rather than just a (1-)stack) then? If one assigns to an affine open the 2-groupoid of Azumaya algebras, Morita equivalences and their natural transformations (over that open), there seems to be enough higher structure to formulate that, no? $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Sep 25, 2015 at 8:45
  • $\begingroup$ In fact I just realized there is one more level, this 2-groupoid carries a multiplication making it a group-up-to-... in 2-groupoids $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Sep 25, 2015 at 8:54
  • $\begingroup$ Yes, it should be something like a group 2-stack, but it's not entirely clear how. $\endgroup$
    – David Roberts ♦
    Commented Sep 25, 2015 at 11:46

1 Answer 1

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No, $\mathrm{Br}$ is not a Zariski sheaf: it is possible for a non-trivial Azumaya algebra on a variety to become trivial when restricted to a Zariski cover. This can happen even for a normal surface with rational singularities. Some references:

  • M. Ojanguren, "A non-trivial locally trivial algebra", J. Algebra 29, 1974
  • F. DeMeyer and T. Ford, "Nontrivial, locally trivial Azumaya algebras", Contemporary Mathematics 124, 1992
  • My answer to this question and more generally this article.
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