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Let $L$ be a pseudo-effective divisor, we may define its numerical fixed part $N_{\sigma}(L)$. How to prove it is a divisor? I know there is a proof in Nakayama's book, but I can't find this book.

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$N_{\sigma}(L)$ is a divisor by definition so it is not clear what you think needs to be proved. –  ulrich Oct 4 '11 at 14:36
How to prove there are only a finite number of prime divisors $\Gamma$ such that $\sigma_{\Gamma}(L)>0$ –  Zhengyu Hu Oct 4 '11 at 15:37

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up vote 3 down vote accepted

By definition, you first define $\sigma_\Gamma(L)$ for big divisors and then you take the limit.

In other words, if $L$ is big, then clearly $\sigma_\Gamma(L)$ is non zero for only finitely many divisors. Indeed, $L=A+B$ with $A$ ample $\mathbb Q$-divisor, and $B\ge 0$. Thus $\sigma_\Gamma(L)>0$ implies $\Gamma$ is contained in the support of $B$.

If $L$ is pseudo-effective, then you define $$\sigma_\Gamma(L)=\lim_{\epsilon\to 0^+} \sigma_\Gamma(L+\epsilon A)$$ where $A$ is ample (it is easy to check that the definition does not depend on the choice of $A$).

It follows that if $\sigma_\Gamma(D)=\alpha>0$ then $D-\alpha\Gamma$ is pseudo-effective. Finite generation of $N^1(X)$ implies that there can only be finitely many $\Gamma$ with such a property.

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How do you know that they remain a finite number when you pass to the limit? –  Gianni Bello Oct 4 '11 at 18:56
I added two lines of explanations... let me know if it is clear. –  mrw Oct 4 '11 at 19:00
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. –  Gianni Bello Oct 4 '11 at 19:16
That is correct. –  mrw Oct 4 '11 at 19:25
This is correct. The finite generation of Neron-Severi group gives it. –  Zhengyu Hu Oct 5 '11 at 2:04

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