Examples of schemes $S$ with $H^{3}(S,\mathbb G_{m})=0$

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 $n$.
• 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. – 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 $$\mathrm{H}^2(S', \mathbf{G}_m) \to \mathrm{H}^1(S,\mathrm{R}^1\pi_*\mathbf{G}_m) \to \mathrm{H}^3(S,\mathbf{G}_m) \stackrel{\pi^*}{\to} \mathrm{H}^3(S', \mathbf{G}_m) \to \mathrm{H}^2(S,\mathrm{R}^1\pi_*\mathbf{G}_m) \to \mathrm{H}^4(S, \mathbf{G}_m).$$
• 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. – Cristian D. Gonzalez-Aviles Mar 22 '18 at 19:47
• 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 $0$. – TKe Mar 22 '18 at 19:53