Timeline for Examples of schemes $S$ with $H^{3}(S,\mathbb G_{m})=0$
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Mar 22, 2018 at 20:10 | history | edited | user19475 | CC BY-SA 3.0 |
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| Mar 22, 2018 at 19:53 | comment | added | user19475 | 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 $0$. | |
| Mar 22, 2018 at 19:50 | comment | added | Cristian D. Gonzalez-Aviles | Or at least that's how I see now. Please correct me if I'm wrong | |
| Mar 22, 2018 at 19:47 | comment | added | Cristian D. Gonzalez-Aviles | 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. | |
| Mar 22, 2018 at 19:30 | history | answered | user19475 | CC BY-SA 3.0 |