Timeline for vanishing of local cohomology $H^2_{(x,y)}\left(\frac{\Bbb Z[x,y]}{(5x+4y)}\right)=0$
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| Dec 12, 2012 at 17:26 | vote | accept | Angel | ||
| Dec 8, 2012 at 17:55 | answer | added | user26857 | timeline score: 2 | |
| Dec 6, 2012 at 11:31 | answer | added | Damian Rössler | timeline score: 1 | |
| Dec 6, 2012 at 0:00 | history | edited | Angel | CC BY-SA 3.0 |
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| Dec 5, 2012 at 23:29 | history | edited | Angel | CC BY-SA 3.0 |
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| Dec 5, 2012 at 16:04 | comment | added | Angel | please,see my idea. | |
| Dec 5, 2012 at 16:03 | history | edited | Angel | CC BY-SA 3.0 |
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| Dec 5, 2012 at 15:57 | history | edited | Angel | CC BY-SA 3.0 |
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| Dec 5, 2012 at 15:52 | history | edited | Angel | CC BY-SA 3.0 |
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| Dec 5, 2012 at 15:44 | comment | added | Damian Rössler | (continued) vanishes. | |
| Dec 5, 2012 at 15:44 | comment | added | Damian Rössler | I am sorry, I miscalculated: 6.1.2 implies only that $H^2_{(x,y)}({\bf Q}[x,y]/(5x+4y))$. | |
| Dec 5, 2012 at 13:47 | comment | added | Angel | How my example be solved by 6.1.2? | |
| Dec 5, 2012 at 12:39 | comment | added | Damian Rössler | In your example, the vanishing takes place because of a general vanishing result of Grothendieck. See 6.1.2 in "Local cohomology..." by Brodmann and Sharp (Cambdridge Univ. Press). Like Ralph, I don't understand how your example is a special case. | |
| Dec 5, 2012 at 10:39 | comment | added | Angel | Yes.$M=S/(xu+yv)$. | |
| Dec 5, 2012 at 10:37 | history | edited | Angel | CC BY-SA 3.0 |
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| Dec 5, 2012 at 10:35 | comment | added | Ralph | Do you mean $M=S/(xu+yv)$ ? Also I don't understand, why your example fits into this pattern. | |
| Dec 5, 2012 at 10:26 | history | edited | Angel | CC BY-SA 3.0 |
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| Dec 5, 2012 at 8:59 | history | asked | Angel | CC BY-SA 3.0 |