Timeline for Justification of the term "invertible sheaf"
Current License: CC BY-SA 2.5
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 22, 2010 at 12:39 | comment | added | Martin Brandenburg | Alright, this makes sense. | |
| Nov 22, 2010 at 1:12 | comment | added | Tom Goodwillie | On the neighborhood where it is defined, the map $M'\otimes N\to M\otimes N$ is such that its image has in it a section that, all alone, generates that sheaf of modules. | |
| Nov 21, 2010 at 23:47 | comment | added | Martin Brandenburg | Hm, I think that the proof has a flaw. If you define $M'$ as above, then it is not clear at all why $M' \to M$ is an epi after tensoring with $N$ on some neighborhood; this is only clear at $x$. | |
| Jul 27, 2010 at 14:43 | comment | added | Tom Goodwillie | Because of the nature of a tensor product: on some neighborhood of a given point there exist sections $m_i$ and $n_i$ such that the sum of $m_i\otimes n_i$ is the section of $M\otimes N$ that corresponds to the generator $1$ under the given isomorphism with $\cal O_X$. Take the $m_i$ to generate $M'$ (either as a submodule of the restriction of $M$ to the neighborhood, or freely if you prefer). | |
| Jul 27, 2010 at 13:06 | comment | added | Andrea Ferretti | Maybe it is a dumb question, but could explain better your last sentence? Why should such a finitely generated $M'$ exist locally? | |
| Jul 27, 2010 at 12:58 | vote | accept | Martin Brandenburg | ||
| Jul 27, 2010 at 11:34 | history | answered | Tom Goodwillie | CC BY-SA 2.5 |