Timeline for Why does the de Rham double complex compute the algebraic de Rham cohomology ?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 20, 2012 at 4:18 | vote | accept | Max Flander | ||
| Aug 15, 2012 at 6:01 | comment | added | Minhyong Kim | For the algebraic De Rham complex, you take the cover to consist of affines, in which case all the intersections will be as well (for a separated scheme). After this, you just set up an obvious sheafy version of the double complex, which then gives you an acyclic resolution. You can prove that an acyclic resolution computes the same cohomology as an injective resolution the same way you do it for a single sheaf as you might find, for example, in Lang's algebra book. | |
| Aug 15, 2012 at 5:56 | comment | added | user22479 | EGA $0_{\rm{III}}$ 12.4.7. | |
| Aug 15, 2012 at 4:49 | comment | added | Mariano Suárez-Álvarez | (Presumably, the double complex in the question is constructed starting from such a good covering and not an arbitrary one) | |
| Aug 15, 2012 at 4:22 | history | answered | David E Speyer | CC BY-SA 3.0 |