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@RP_ Thanks for your input. This is essentially the idea I think in the above linked article--using the definition of smoothness you mentioned is how you get 'charts' on the $p$-adic manifold $X(F)$, but I just don't want to include this in the article I'm writing so just want a reference and don't want to have to reproduce a proof. Thanks again!
@WSL Yeah, if you'd be willing to write a fairly precise answer that would be much appreciated. Of course, I'd be very happy to save you the hassle if you just knew a reference. Thanks!
@SashaP Thanks for your suggestion. For my own sake I'd prefer to have a proof that works for general $X$, just because I do occasionally also use it in that case (even though, as I said, my main interest is groups). Still, if you want to write this as an answer, it might be useful to some readers and I'll accept it if someone else doesn't provide the full answer.
@WSL I guess this is essentially the proof in the linked article [1] but, again, it's obfuscated by the particular context it's discussing the result in. Do you know a reference for this proof?
@AriyanJavanpeykar Does the statement itself not imply the claim since the Gauss point is contained in the unit circle and the usual fact about fundamental groups :"surjective iff remains connected upon pullback"?
Just to point out something obvious, but maybe could be useful to someone trying to construct a counter example: I'm pretty sure the cover needs to not come from the special fiber of a normal integral model. In particular, any of the strange covers of $\mathbb{A}^1_{\overline{F}_p}$ won't help.
And, I suppose if you can't obviously phrase it in terms of group cohomology it's not even obvious what the above really means? I guess I was taking for granted (my bad) that the theory of 'twisting by cocycles' is something that one can make sense of over an arbitrary base, but I see that a priori that's no the case. Is that the point? Regardless, it seems as though the result is true if $X_\mathbb{Q}$ is $\mathbb{P}^1$.
@WillSawin Sure, there's always the issue that the kernel being trivial is not the same as the map being injective, but essentially it suffices to check that the cohomology on $S$ of $\mathrm{GL}_n$ by twists of cocycles in $\mathrm{PGL}_n$ has trivial cohomology. This means checking that inner forms of $\mathrm{GL}_n$ have trivial cohomology over $S$ as well. Over a field I know that taking inner forms does not change the cohomology, so the claim (that I guess I didn't realize) is that this property fails integrally? Or, maybe to the point, that you can't phrase it in terms of group coh?
We have a natural map exact sequence $H^1(S,\mathrm{GL}_n)\to H^1(S,\mathrm{PGL}_n)\to \mathrm{Br}(S)$ and so if $S$ has no non-trivial vector bundles then an injection $H^1(S,\mathrm{PGL}_n)\to \mathrm{Br}(S)$. Since $\mathrm{Br}(S)\to\mathrm{Br}(k(S))$ is injective (again with my affineness and regularity assumptions), does it now follow that the same is true for Brauer-Severi schemes? So, why can't we apply this when $S$ is the spectrum of $\mathbb{Z}[\frac{1}{n}]$ to obtain the result. Sorry if I'm being dumb.
@WillSawin Sorry, let me be more precise. Let $S$ be a sufficiently nice scheme that I don't have to worry about the distinction between the cohomological and Azumaya Brauer group (e.g. $S$ is affine and regular more than suffices, I believe)[this is not really a necessary assumption, it's just for my psychological well-being]. We have a bijection between the isomorphism classes of Brauer-Severi schemes on $S$ of dimension $n$ and the set $H^1(S,\mathrm{PGL}_n)$ (where this etale [or flat] cohomology).
@WillSawin Maybe I'm being silly here, but isn't the Brauer group classifying Brauer-Severi schemes up to isomorphism, not up to equivalence, right? So if you know that $X$ and $Y$ are Brauer-Severi schemes over $\mathrm{Spec}(\mathbb{Z}[\frac{1}{n}])$ as I tried (perhaps incorreclty) to argue in my post, then doesn't the injectivity of the localization map for Brauer groups on regular local schemes imply that generically isomorphic Brauer-Severi schemes are isomorphic? Thanks!
