The examples I have are: $S$ is equal to the spectrum of a global field; or a proper non-empty open subscheme of the spectrum of the ring of integers $\mathcal O_{K}$ of a number field $K$ (proper means $S$ is not all of ${\rm Spec}\,\mathcal O_{K}$); or $S$ is a non-empty open subscheme of a smooth, complete and irreducible curve over a finite field. Can anyone supply other examples, please?


There's some information on the groups $H^q(S,\mathbb{G}_m)$ in Grothendieck's Le groupe de Brauer, II in Dix Exposés. Proposition 1.4 says that if $S$ is regular then these groups are torsion for $q \ge 2$. Corollaire 3.2 says that, under suitable finiteness hypotheses, the Kummer sequence gives isomorphisms $H^q(S,\mathbb{G}_m)[\ell^\infty] \cong H^q(S,\mu_{\ell^\infty})$ for $q \ge 3$ and $\ell$ invertible on $S$. These finiteness hypotheses are satisfied, for example, if $S$ is either proper or smooth over a field that is either separably closed or finite.

Using this you can find plenty of examples with $H^3(S,\mathbb{G}_m)=0$, such as any projective space over an algebraically closed field of characteristic zero: the standard calculation of the cohomology of projective space gives $H^3(\mathbb{P}^m,\mu_n)=0$ for all $n$.

  • $\begingroup$ Dear Martin, many thanks. Perhaps I should've been more precise. What I really need are examples of schemes $S$ with interesting Brauer group such that the canonical map $H^{3}(S,\mathbb G_{m})\to H^{3}(S',\mathbb G_{m})$ is injective, where $S'\to S$ is a quadratic Galois covering. $\endgroup$ – Cristian D. Gonzalez-Aviles Mar 22 '18 at 17:40

Let $\pi: S' \to S$ be a smooth projective relative curve with $S$ a regular variety. The $\mathrm{R}^q\pi_*\mathbf{G}_m = 0$ for $q > 1$, so there is an exact sequence coming from the Leray spectral sequence $$\mathrm{H}^2(S', \mathbf{G}_m) \to \mathrm{H}^1(S,\mathrm{R}^1\pi_*\mathbf{G}_m) \to \mathrm{H}^3(S,\mathbf{G}_m) \stackrel{\pi^*}{\to} \mathrm{H}^3(S', \mathbf{G}_m) \to \mathrm{H}^2(S,\mathrm{R}^1\pi_*\mathbf{G}_m) \to \mathrm{H}^4(S, \mathbf{G}_m).$$

  • $\begingroup$ Dear TKe, thanks. It seems that you have forgotten that $R^{0}\pi_{*}\mathbb G_{m}=\pi_{*}\mathbb G_{m, S'}$ (rather than $\mathbb G_{m, S'}$). So, when $\pi$ is finite, then the map you call $\pi^{*}$ is just the canonical isomorphism coming from the degeneration of the spectral sequence. This is not the same as the map I have in mind. My map is induced by the canonical closed immersion $\mathbb G_{m, S}\to \pi_{*}\mathbb G_{m, S'}$ composed with the isomorphism mentioned above. $\endgroup$ – Cristian D. Gonzalez-Aviles Mar 22 '18 at 19:47
  • $\begingroup$ Or at least that's how I see now. Please correct me if I'm wrong $\endgroup$ – Cristian D. Gonzalez-Aviles Mar 22 '18 at 19:50
  • $\begingroup$ I see what you mean. If $\pi: S' \to S$ has geometrically connected fibres, then $\pi_*\mathbf{G}_{m,S'} = \mathbf{G}_{m,S}$. So it works for relative curves, but not for relative dimension $0$. $\endgroup$ – TKe Mar 22 '18 at 19:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.