Timeline for How to prove the existence of divisorial Zariski decomposition?
Current License: CC BY-SA 3.0
11 events
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| Oct 5, 2011 at 20:13 | comment | added | mrw | It seems quite easy to me. Assume that $\Gamma=\sum a_i \Gamma_i \equiv 0$ for some $a_i\in \mathbb N$ then $D+\Gamma^+\equiv D+\Gamma^-$, which implies the claim. | |
| Oct 5, 2011 at 15:10 | comment | added | Gianni Bello | Actually I don't see it so trivially. Of course $N^1(X)$ is finitely generated, but how do you see that the $\Gamma_i$ (I'm using the notation of my previous comments ) are not linearly dependent? This requires some effort in Nakayama's book, I cannot see this in a trivial way. | |
| Oct 5, 2011 at 2:04 | vote | accept | Zhengyu Hu | ||
| Oct 5, 2011 at 2:04 | comment | added | Zhengyu Hu | This is correct. The finite generation of Neron-Severi group gives it. | |
| Oct 4, 2011 at 19:26 | history | edited | mrw | CC BY-SA 3.0 |
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| Oct 4, 2011 at 19:25 | comment | added | mrw | That is correct. | |
| Oct 4, 2011 at 19:16 | comment | added | Gianni Bello | I reckon that to show the last assetion of your answer you need to use that if the $\Gamma_i$ are prime divisors such that $\sigma_{\Gamma_i}(D)>0$, then the numerical classes of the $\Gamma_i$ are linearly independent in the vector space $N^1(X)$. Am I wrong? This is what Nakayama does in his book. | |
| Oct 4, 2011 at 19:00 | comment | added | mrw | I added two lines of explanations... let me know if it is clear. | |
| Oct 4, 2011 at 18:56 | comment | added | Gianni Bello | How do you know that they remain a finite number when you pass to the limit? | |
| Oct 4, 2011 at 18:49 | history | edited | mrw | CC BY-SA 3.0 |
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| Oct 4, 2011 at 18:38 | history | answered | mrw | CC BY-SA 3.0 |