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Let $X$ be a normal projective complex variety. A theorem of Fujita-Zariski says that if $L$ is a Cartier divisor on $X$ such that the base locus $Bs(|L|)$ is a finite set then $L$ is semiample. It seems to me that by using this theorem it is possible to prove that the stable base locus $\mathbb{B}(L):=\bigcap_{m \in \mathbb{N}} Bs(|mL|)$ of a Cartier divisor on a variety as above cannot contain isolated points. Do you know a reference for this (or a counterexample)?

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Yes, you are right, and the result is almost immediate using the FZ theorem. A proof can be found in the recent paper Restricted volumes and base loci of linear series. by Ein, Lazarsfeld,Mustaţă, Nakamaye and Popa. For the sake of completeness I sketch the argument here.

Just to clarify: The Fujita–Zariski Theorem says that if a line bundle $L$ is ample on its base locus, then it is is semiample. If the base-locus is finite then this condition is automatically satisfied. Here semiampleness means that a multiple $L^{\otimes n}$ is globally generated.

Since $X$ is normal, I'll switch from a line bundle $L$ to a divisor $D$ (just for the sake of notation). Suppose that $x$ is an isolated point in $\mathbb{B}(D)$. We have $Bs(mD)\supset Bs((m+1)D)\supset \cdots $ so by Noetherianness we can choose a large $m$ so that that the stable base locus $\mathbb{B}(D)$ equals $Bs(|mD|)$. Let $X'$ be the blow-up of $X$ with center $Z=Bs(mD)\setminus \{x\}$. The total transform of $mD$ is can be decomposed as $E+M$ where $E=\pi^{-1}(Z)$ and $M$ is a divisor with a base locus at $f^{-1}(x)$ (with some multiplicity). Now, since the base locus of $M$ is finite, Fujita–Zariski implies that $pM$ is base-point free for some large $p$. But then $x\not\in Bs(mpD)$, which contradicts $x\in \bigcap_{m\ge 1} Bs(mD)$.

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Thanks a lot. This is more or less what I had in mind. –  Gianni Bello Jan 20 '11 at 9:46
    
Actually in the paper by ELMNP the variety is assumed to be smooth. However it seems to me that this is not used in the proof. Right? –  Gianni Bello Jan 20 '11 at 10:57
    
I think the proof goes through without the smoothness assumption yes, although normality is of course essential. –  J.C. Ottem Jan 20 '11 at 11:25
    
i've read the paper and there's a remark that states there's a way to prove it with the multiplier ideal. can someone tell me how it works? thank you –  oxydo Mar 18 '11 at 16:26

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