Timeline for Cartier divisor that is not a difference of two effective Cartier divisors
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Sep 15, 2022 at 2:11 | comment | added | Daebeom Choi | OK, I have one plausible approach. Let $\pi:\mathbb{P}_k^3\to X$ be the map. Then this induces a morphism of sheaves $O_X^\times\to \pi_\ast O_{\mathbb{P}_k^3}^\times$. Since $\pi$ is finite, we can calculate their stalk (at least in the etale topology). Based on this calculation, I think this is injective, and the cokernel is $i_\ast O_{\mathbb{P}_k^1}^\times$, where $i$ is the inclusion of "bad locus". Then I think the LES of (etale) cohomology associated to this SES can solve the problem... Is this the right way to do this? | |
| Sep 15, 2022 at 1:47 | comment | added | Johan | Take a few days to think about it. Fitst do some easier examples of glueings. Etc, etc. | |
| Sep 15, 2022 at 1:44 | comment | added | Daebeom Choi | How can I calculate that Picard group of the example? | |
| Sep 15, 2022 at 1:40 | comment | added | Johan | The Picard group of the example is nontrivial as it equals $k^*$. The Picard group of a variety is equal to the group of Cartier divisors up to rational equivalence. | |
| Sep 15, 2022 at 1:28 | history | edited | Daebeom Choi | CC BY-SA 4.0 |
added 23 characters in body
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| Sep 15, 2022 at 1:27 | comment | added | Daebeom Choi | Sorry. I mean "Cartier divisor up to linear equivalence". Fixed! | |
| Sep 15, 2022 at 0:33 | comment | added | Johan | OK,I think your notion of a Cartier divisor is different from Hartshorne's definition: a Cartier divisor on a variety $X$ is an element of $\Gamma(X, \mathcal{K}^*/\mathcal{O}^*)$ where $\mathcal{K}$ is the sheaf of rational functions. For a variety of positive dimension, this is always very big. | |
| Sep 14, 2022 at 21:10 | history | asked | Daebeom Choi | CC BY-SA 4.0 |