Timeline for Effective Cartier divisors corresponding to a line bundle
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Aug 11, 2021 at 7:14 | vote | accept | Aoi Koshigaya | ||
| Jul 3, 2021 at 13:46 | history | edited | Will Sawin | CC BY-SA 4.0 |
added 113 characters in body
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| Jul 3, 2021 at 13:45 | comment | added | Will Sawin | @DamianRössler Yes, this is a good point. I didn't realize this would be necessary when writing my answer (although for these inductive arguments $X$ will likely be a complete intersection and then it is unnecessary). | |
| Jul 3, 2021 at 13:43 | comment | added | Damian Rössler | Savin. I agree for the reduced case but I think it is not clear that "component" also refers to components corresponding to some associated primes. | |
| Jul 3, 2021 at 13:26 | comment | added | Will Sawin | @DamianRössler This is what the first paragraph of my answer is about. | |
| Jul 3, 2021 at 13:15 | comment | added | Damian Rössler | I see. I think you should put this in your answer because I think that this is what concerns the OP. | |
| Jul 3, 2021 at 13:05 | comment | added | Will Sawin | @DamianRössler If $X$ is reduced, you choose it to be nonzero on each irreducible component, from which it follows that it is not a zero divisor, since every element it could be a zero divisor is would be nonzero on some component. If $X$ is not reduced, you choose it to be nonzero on the induced reduced subscheme of each component and each associated prime. | |
| Jul 3, 2021 at 13:00 | comment | added | Damian Rössler | If $X$ is not integral, how to you find a section which defines a Cartier divisor? Locally, the ideal of the divisor will be defined by a single element but it is not clear that this element is not a zero divisor. | |
| Jul 3, 2021 at 12:44 | history | answered | Will Sawin | CC BY-SA 4.0 |