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Let $S$ be a finitely generated domain with the field of fractions $F.$ Let X be a smooth, geometrically connected affine variety over $S.$ Let $A$ be an Azumaya algebra over $X.$ Assume that for all large enough primes $p,$ $A_p$ splits over $X_p$-the reduction modulo $p$ of $X.$ Does this assumption imply that $A_{\overline{F}}$ splits over $X_{\overline{F}}?$ My naive guess is that the answer should be "yes". Any suggestions or references would be greatly appreciated.

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Let $n$ be the order of $A$ in the Brauer group of $X$, then $A_{\overline F}$ splits if and only if the corresponding class $[A_{\overline F}]\in H^2_{et}(X_{\overline F}, \mu_n)$ is 0. If $X$ is smooth and proper and $n$ is invertible on $S$ then by Deligne's theorem in SGA $4\frac{1}{2}$ the pushforward $R^2p_*\mu_n$ is a local system on $S$ where $p: X\rightarrow S$ is the projection. In particular the specialization map $H^2_{et}(X_{\overline F}, \mu_n)\rightarrow H^2_{et}(X_{s}, \mu_n)$ is an isomorphism for any geometric point $s\rightarrow S$ of some characteristic $p$. The image of $[A_{\overline F}]$ under specialization is exactly $[A_s]$, thus in fact $[A_{\overline F}]=0$ if and only if $[A_s]=0$. This means that it is enough to check that $A$ splits on a single geometric fiber of characteristic $p>n$. Using excision this can be generalized to $X$ being a complement in a smooth proper scheme of a normal crossing divisor.

Now returning to your original question, by Nagata any $X$ can be compactified by a proper $S$-scheme $\overline X\supset X$, moreover by Hironaka the singularities of $\overline X_F$ can be resolved such that the complement $\overline X_F\setminus X_F$ is a normal crossing divisor. "Spreading out" one gets that for any $X\rightarrow S$, the base change of $X_{S'}$ to a Zariski open subset $S'\subset S$ is also a complement of a normal crossing divisor in a smooth proper $S'$-scheme, thus the argument above applies.

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  • $\begingroup$ Thank you, that's a great answer! $\endgroup$
    – Weiwei Z.
    Mar 30, 2021 at 15:50
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    $\begingroup$ Are you using $0 \to \text{Pic}(X_{\overline{F}}) \to H^2_{et}(X_{\overline{F}}, \mu_n) \to Br(X_{\overline{F}})[n] \to 0$ and if so why "if and only if" in the first sentence? $\endgroup$
    – Johan
    Mar 31, 2021 at 1:27
  • $\begingroup$ @Johan Ah, hm.. I was rather thinking about the boundary map $H^1_{et}(X_{\overline F},PGL_n) \rightarrow H^2_{et}(X,\mu_n)$ coming from the fiber sequence $BSL_n\rightarrow BPGL_n\rightarrow B^2\mu_n$. On the other hand I guess one could use the exact sequence in your comment equally well by lifting the class of $A$ in $Br(X_{\overline F})[n]$ to any class in $H^2_{et}(X_{\overline F},\mu_n)$ and playing the same game.. $\endgroup$
    – user42024
    Mar 31, 2021 at 9:06
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    $\begingroup$ OK, I ask again: why "if and only if"? $\endgroup$
    – Johan
    Mar 31, 2021 at 16:31
  • $\begingroup$ @Johan I finally understood that you are right and there is something essential missing from the argument (I also deleted what I wrote before since it was wrong). My proof works well for classes in $H^2_{et}(X_{\overline F},\mu_n)$, but then returning to the Azumaya algebra question, a priori all what I get is that the class in the specialization is an image of a class in $\mathrm{Pic}(X_s)/n\mathrm{Pic}(X_s)$. $\endgroup$
    – user42024
    Mar 31, 2021 at 22:26

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