Timeline for cohomology of torsion sheaves and nilpotent sheaves
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Dec 4, 2012 at 20:54 | history | edited | Ryan Budney |
fix tags
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| Dec 4, 2012 at 12:07 | history | edited | Naga Venkata | CC BY-SA 3.0 |
edited body
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| Dec 3, 2012 at 22:06 | comment | added | Damian Rössler | @Will Savin: I agree but he write "X a non-reduced curve". | |
| Dec 3, 2012 at 20:06 | comment | added | Will Sawin | Depending on the definition one uses, one may not take that to be a curve. | |
| Dec 3, 2012 at 19:29 | comment | added | Sándor Kovács | @Naga: I also forgot to add this: What is $I_C$ in your specific example/question? Is it just $I_X$? | |
| Dec 3, 2012 at 19:28 | comment | added | Sándor Kovács | @Damian: yes, of course, you're right, I actually wanted to say that, but then got lost in my own neverending comment. | |
| Dec 3, 2012 at 19:03 | answer | added | Mohan | timeline score: 2 | |
| Dec 3, 2012 at 18:33 | comment | added | Damian Rössler | @Eric Wofsey & Sandor Kovacs: even if $F$ is torsion it might have support on the whole curve. Consider the ${\cal O}_X$-module $(\epsilon)$ on the curve $C[\epsilon]=C\times_k{\rm Spec}\, k[\epsilon]/\epsilon^2$. This has support $=C$. | |
| Dec 3, 2012 at 18:14 | comment | added | Sándor Kovács | @Naga: you should think about this question a little more. The notion of torsion does not include the notion of nilpotent for the simple reason that $\mathcal F$ may not have a multiplication. Perhaps you want to look at $\mathcal O_X$-algebras? Torsion and nilpotent are still two different notions, but perhaps you can figure out what it is that you are asking. Furthermore, in your particular example you are asking if the $0^{\mathrm{th}}$ or $1^{\mathrm{st}}$ cohomology of a sheaf on a curve vanishes. It is unlikely that anything like that could happen. If it is torsion... (cont'd) | |
| Dec 3, 2012 at 16:26 | comment | added | Eric Wofsey | If $X$ is a curve and $\mathcal F$ is torsion and coherent, then its support is 0-dimensional, so the higher cohomology automatically vanishes. If $\mathcal F$ is torsion and quasicoherent, it is a filtered colimit of torsion coherent sheaves so its higher cohomology still vanishes. | |
| Dec 3, 2012 at 15:27 | history | edited | Naga Venkata | CC BY-SA 3.0 |
added 28 characters in body
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| Dec 3, 2012 at 15:11 | history | asked | Naga Venkata | CC BY-SA 3.0 |