Timeline for $G_0(X) \cong G_0(X_{red})$ where X is a noetherian scheme
Current License: CC BY-SA 4.0
8 events
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| Sep 18, 2019 at 20:37 | comment | added | crystalline | No, you cannot deduce $G_0(X) = G_0(X_{red})$ just from the affine case. There is a Mayer-Vietoris sequence but it’s going to involve higher groups $G_i$ which you have to take into account. But you can apply devissage to M(X) directly and that works. | |
| Sep 18, 2019 at 9:14 | comment | added | user139827 | Okay thank you. | |
| Sep 18, 2019 at 9:06 | comment | added | François Brunault | You need to consider the categories for the schemes not just for the affine cover. It depends how comfortable you are, but if not it's better to write the full proof. | |
| Sep 18, 2019 at 8:40 | comment | added | user139827 | So for reducing it to affine case is it sufficient to say every affine open set of noetherian scheme is noetherian? | |
| Sep 18, 2019 at 8:10 | comment | added | François Brunault | This is an application of dévissage, see e.g. Weibel's K-book, Chapter 2. You can reduce to affine schemes but if you're not familiar with such reduction arguments, it's better you check in details that the abelian categories have same $K_0$ (of course at the end you use the result for affine schemes). This is also true for higher $G_n$, see Quillen's dévissage theorem in Higher algebraic K-theory I. | |
| Sep 18, 2019 at 7:01 | history | edited | András Bátkai | CC BY-SA 4.0 |
added higher order tag
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| Sep 18, 2019 at 4:05 | review | First posts | |||
| Sep 18, 2019 at 7:01 | |||||
| Sep 18, 2019 at 4:03 | history | asked | user139827 | CC BY-SA 4.0 |