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One fundamental conjecture on the Brauer group is that $\operatorname{Br}(X)$ is finite for $X/\operatorname{Spec} \mathbf{Z}$ proper. By class field theory (the theorem of Albert–Brauer–Hasse–Noether), this is true for $\dim{X} = 1$. For varieties $X$ over finite fields, the finiteness of $\operatorname{Br}(X)[\ell^\infty]$ is equivalent to the surjectivity of the cycle class map $\operatorname{Pic}(X) \otimes \mathbf{Z}_\ell \to \operatorname{H}^2(X,\mathrm{Z}_\ell(1))$ (the Tate conjecture in dimension $1$). The Tate conjecture in dimension $1$ over finite fields is known for smooth projective curves (trivial), Abelian varieties (Tate) and K3 surfaces (at least for characteristic $> 3$), and hence for products of such varieties.

Are there results on the finiteness (of an $\ell$-primary part) of the Brauer group of schemes flat and proper over $\operatorname{Spec} \mathbf{Z}$ of dimension $> 1$?

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    $\begingroup$ For an elliptic curve, isn't this equivalent to finiteness of Sha, and thus aren't there partial results coming from the various partial results on BSD? $\endgroup$
    – Will Sawin
    Commented Nov 7, 2017 at 15:05
  • $\begingroup$ @WillSawin: This is at least equivalent for function fields. $\endgroup$
    – user19475
    Commented Nov 7, 2017 at 15:07
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    $\begingroup$ Doesn't the same cohomological argument work? $\endgroup$
    – Will Sawin
    Commented Nov 7, 2017 at 15:17
  • $\begingroup$ @WillSawin: You are right, I have posted this as an answer. $\endgroup$
    – user19475
    Commented Nov 7, 2017 at 15:21
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    $\begingroup$ @Timo Keller What motivates your question? The question is wide open in the flat proper case over $\text{Spec}(\mathbf{Z})$ as much as in the proper case over $\text{Spec}(\mathbf{Z})$, so you must have something specific in mind. $\endgroup$
    – user95222
    Commented Nov 25, 2017 at 4:51

2 Answers 2

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Let $\mathscr{C}/X$ be a relative curve with a section, e.g. a relative elliptic curve. Then there is a short exact sequence $$0 \to \mathrm{Br}(X) \to \mathrm{Br}(\mathscr{C}) \to Ш(\mathbf{Pic}^0_{\mathscr{C}/X}/X) \to 0$$ (see https://www.timokeller.name/TateShafarevich.pdf, Theorem 4.27; for $\dim{X}=1$, this is a theorem of Grothendieck).

So let $X$ be the spectrum of the ring of integers of a number field and $\mathscr{C}/X$ a relative elliptic curve with finite Ш.

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    $\begingroup$ This is true when $X$ is a smooth projective curve over a finite field by a Thm of Grothendieck, and if $X$ is smooth projective geometrically irreducible over a finite field by fairly elementary arguments, but it's not quite true if $X$ is the spectrum of the ring of integers of an algebraic number field $K$. $\text{Br}(\mathscr{C})$ should surject onto $H^1(X_{\rm\acute{e}t}, \text{Pic}^0_{\mathscr{C}/X})$ (actually onto its image into $H^1(X_{\rm\acute{e}t}, \text{Pic}_{\mathscr{C}/X})$, which is off from the Tate-Shafarevich group by a finite group of exponent $2$ if $K$ has a real place.) $\endgroup$
    – user87684
    Commented Nov 25, 2017 at 5:04
  • $\begingroup$ Yes, I forgot the infinite places. $\endgroup$
    – user19475
    Commented Nov 25, 2017 at 6:29
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Not sure this is too interesting, but I however think that the case when $X \rightarrow \text{Spec}(\mathbb{Z})$ is proper and smooth(!) should follow from a result proven independently by Fontaine and Abrashkin that implies that the Hodge-numbers are zero for $i+j \leq 3$ and $i \neq j$.

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