For an affine variety $V$, the Brauer group $Br(V)$ and the cohomological Brauer group $Br'(V)$ (i.e. the torsion subgroup of $H^2_{et}(V,\mathbb{G}_m)$) coincide, by results of various authors including Hoobler, Gabber etc. In fact it is true for affine schemes, but I have a simpler setup, namely I have a smooth complex variety.

However, I have been told that in fact $Br'(V) = H^2_{et}(V,\mathbb{G}_m)$ for affine $V$, and I can't seem to find a reference for this fact. This implies that all $\mathbb{G}_m$-gerbes on affine varieties have torsion class in $H^2$, a fact which I find slightly surprising, but which I can't rule out. My dream is that there are non-torsion elements in $H^2$, but I can live with their absence.

What is a reference, if true, for the result that $H^2_{et}(V,\mathbb{G}_m)$ is a torsion group?

EDIT: not that it is a hugely reliable source, but wikipedia says (without citation):

If $F$ is a coherent sheaf (or $\mathbb{G}_m$) then the étale cohomology of $F$ is the same as Serre's coherent sheaf cohomology calculated with the Zariski topology (and if $X$ is a complex variety this is the same as the sheaf cohomology calculated with the usual complex topology).

which would indicate that $H^2_{et}(V,\mathbb{G}_m) \simeq H^2(V(\mathbb{C}),\mathcal{O}^\times)$ where the cohomology on the right is sheaf cohomology for the analytic topology on $V(\mathbb{C})$.

  • $\begingroup$ It's entirely possible that whoever added that statement to Wikipedia saw the result about $H^1$ with values in $\mathbb{G}_m$, and wrongly extrapolated. $\endgroup$ Jun 11 '14 at 9:01
  • $\begingroup$ as for the wikipedia comment -- it's definitely true that cohomology of coherent sheaves is the same for the étale topology and the zariski topology on an arbitrary scheme. The statement in parentheses I don't know (for proper varieties it is GAGA; for non-proper varieties it's probably false -- at least, certainly it's false for $H^0$ of the structure sheaf.) $\endgroup$
    – hunter
    Jun 11 '14 at 10:29
  • $\begingroup$ I'm completely confident about the results with coherent sheaves and GAGA, just the statement about the multiplicative group is what I was focussing on. $\endgroup$ Jun 11 '14 at 11:18
  • $\begingroup$ I think it's relatively rare for $H^2_{et}(V,\mathbb{G}_m)\simeq H^2(V(\mathbb{C}),\mathscr{O}^\times)$ for a smooth complex variety $V$. For instance, this is impossible if $H^2(V,\mathscr{O})\neq 0$, by the exponential sequence. On the other hand, the étale cohomology group is torsion in this case. $\endgroup$ Jun 14 '14 at 3:45
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    $\begingroup$ Without the smoothness hypothesis, there are examples where $H^2_{et}(V,\mathbb{G}_m)$ can be non-torsion, even for a complex variety. I guess the earliest example is due to Grothendieck and Mumford. It is a normal surface, which I think is even affine. This is mentioned in Remark 2.8 of Milne's Chapter IV from the book on étale cohomology. By the way, that reference has a lot of very nice facts about the Brauer group. You're original question is answered as Corollary 2.6 there. $\endgroup$ Jun 14 '14 at 3:49

By the answer at https://mathoverflow.net/questions/171638, there is a theorem of Grothedieck [1] that implies that any non-singular variety $X$ has all cohomology groups $H^q_{et}(X,\mathbb{G}_m)$ torsion for $q\geq 2$. Why this is true is still a bit mysterious, even though I have hints to the proof.

[1] Le groupe de Brauer II, Proposition 1.4

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    $\begingroup$ You might look at the Leray spectral sequence for the pushforward of $\mathbb{G}_m$ from the generic point. $\endgroup$ Jun 14 '14 at 3:42
  • $\begingroup$ Yes, Grothendieck mentions this in his lead up to proposition 1.4 $\endgroup$ Jun 14 '14 at 22:12
  • $\begingroup$ Unfortunately I'm not an algebraic geometer, so I'll have to take this one on faith and come back to it when I have the leisure. $\endgroup$ Jun 14 '14 at 23:58
  • $\begingroup$ Ah, I see: H^2(X,G_m) injects into H^2(Spec K,G_m,K), and the latter is (I guess) a Galois cohomology group, hence torsion. $\endgroup$ Jun 17 '14 at 2:45
  • $\begingroup$ Exactly the point. $\endgroup$ Jun 18 '14 at 17:09

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