Timeline for Checking normality of variety
Current License: CC BY-SA 2.5
12 events
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| Sep 16, 2017 at 19:25 | comment | added | Karl Schwede | Here's one way to do it by hand (there may be other options with a computer). Say $R$ is a local domain. You can form an open set $U$ by removing an interesting codimension \geq 2 set of $X = Spec R$ (outside of which $X$ is S2), and then computing $S = \Gamma(U, O_X)$. If $$S = R,$$ then $R$ is S2, otherwise not. Another option is to compute $Hom_R( \omega_R, \omega_R)$. In general that's the S2-ification of $R$. | |
| Sep 16, 2017 at 6:03 | comment | added | user100841 | @Karl Schwede how to show S2 for a local ring? S2 is local property right? actually i want to show something is deminormal..one of the property is S2 | |
| Apr 7, 2011 at 2:27 | comment | added | Karl Schwede | Moon. If your variety is (locally) a complete intersection, then it is automatically S2 (and CM). Outside of that case, it can be hard, so it depends on exactly what you know. Of course, the fact that you have an incidence variety might be helpful, those sort of things often be S2 I would suspect. For trying to prove CM, one approach is to show that $H^i(X, L^{-n}) = 0$ for $L$ ample, $n \gg 0$ and $i < \dim X$. Grothendieck duality and Serre vanishing then forces $X$ to be Cohen-Macaulay. | |
| Apr 7, 2011 at 2:16 | comment | added | Moon | Dear Karl, thank you for great answer. I have another question. To show the normality of given variety, is it the best way checking R1 and S2? In my situation, given variety $X$ is an incident variety of points and curves in projective space. By dimension estimate, R1 is not too difficult, but I can't figure out how can I prove S2 (or CM) condition. | |
| Apr 5, 2011 at 18:50 | history | edited | Karl Schwede | CC BY-SA 2.5 |
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| Apr 5, 2011 at 18:39 | history | edited | Karl Schwede | CC BY-SA 2.5 |
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| Apr 5, 2011 at 15:10 | history | edited | Karl Schwede | CC BY-SA 2.5 |
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| Apr 5, 2011 at 15:01 | history | edited | Karl Schwede | CC BY-SA 2.5 |
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| Apr 5, 2011 at 14:53 | comment | added | Karl Schwede | Yes, I was mapping $Y$ to $X$ in my head. I'll fix it now, thanks! | |
| Apr 5, 2011 at 14:52 | history | edited | Karl Schwede | CC BY-SA 2.5 |
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| Apr 5, 2011 at 14:51 | comment | added | Georges Elencwajg | Dear Karl, don't you mean that $Y$ should be semi-normal ? | |
| Apr 5, 2011 at 14:34 | history | answered | Karl Schwede | CC BY-SA 2.5 |