Timeline for The class of the diagonal in the symmetric product of a smooth curve
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
Commonmark migration
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| Mar 31, 2016 at 5:53 | answer | added | t3suji | timeline score: 6 | |
| Mar 30, 2016 at 14:21 | history | edited | Francesco Polizzi | CC BY-SA 3.0 |
added 16 characters in body; edited title
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| Mar 30, 2016 at 13:31 | vote | accept | Francesco Polizzi | ||
| Mar 30, 2016 at 13:07 | answer | added | Jason Starr | timeline score: 8 | |
| Mar 30, 2016 at 11:49 | comment | added | Jason Starr | I will write an answer. The original reference might be Mattuck, but I need to check. | |
| Mar 30, 2016 at 11:47 | comment | added | Francesco Polizzi | @JasonStarr: thank you very much for your very useful comment. Could you please be so kind and add your comment as an answer, so that the question will not appear as unanswered? And do you know any reference for a direct proof for all curves? | |
| Mar 30, 2016 at 11:43 | comment | added | Jason Starr | The answer is yes, and you can, indeed, deduce the result for all smooth curves from the case of general curves by specialization. The point is separatedness and properness of the relative Picard scheme of a family of smooth, projective varieties. Over a smooth base, so that the total space of the family is also smooth, this is just the fact that the closure in the total space of a Cartier divisor in a general fiber is the unique flat extension, and it is a Cartier divisor (since the total space is smooth). There are also direct proofs for all curves. | |
| Mar 30, 2016 at 9:40 | history | edited | Francesco Polizzi | CC BY-SA 3.0 |
added 11 characters in body
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| Mar 30, 2016 at 9:33 | history | asked | Francesco Polizzi | CC BY-SA 3.0 |