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In his book Higher-Dimensional Algerbraic Geometry, Debarre claimed that the pull back of a nef and big divisor under a generically finite morphism is still nef and big, but he only state the result and no proof. Can somebody tell me why or show me a reference? Thanks.

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This is a nice exercise. Some hints: Consider nefness and bigness separately. For preservation of bigness, look at the Stein factorization of your morphism, and use the characterizations of bigness given in Volume 1 of Lazarsfeld's book (Section 2.2, if I recall correctly). –  Yusuf Mustopa Aug 12 '12 at 5:39
    
I agree with Yusuf. Please note, the "generically finite morphism" here also must be dominant (when restricted to every irreducible component of the domain). –  Jason Starr Aug 12 '12 at 14:46
    
Thanks to Yusuf and Jason. –  MZWang Aug 25 '12 at 9:37

1 Answer 1

One can consider the following characterization of neg and big divisors:

Let $D$ a divisor on an irreducible projective variety $X$. Then $D$ is nef and big if and only if there exist an effective divisor $E$ and a rational number $0 <\epsilon\ll 1$ such that $D-\epsilon E$ is ample.

proof: Let $D$ be a nef and big divisor. Since $D$ is big there exist an ample divisor $A$, an effective divisor $E$, and a positive integer $k$ such that $kD\equiv A+E$. If $h>k$ we can write $hD\equiv (h-k)D+A+E$. The divisor $D^{'} = (h-k)D+A$ is a sum of a nef and an ample divisor. Therefore $D^{'}$ is ample. If $\epsilon = \frac{1}{h}$ we get that $$D-\epsilon E\equiv \epsilon D^{'}$$ is ample.

You have a generically finite morphism $\phi:X\rightarrow Y$, and a nef and big divisor $D$ on $Y$. Consider the Stein factorization $\phi = f\circ b$, where $b$ has connected fibers and $f$ is finite. In you case, since $\phi$ is generically finite, $b$ is birational and $f$ is finite. Since $D$ is nef and big there exist an effective divisor $E$ and a rational number $0 <\epsilon\ll 1$ such that $D-\epsilon E$ is ample. Since $f$ is finite $f^{*}(D-\epsilon E) = f^{*}D-\epsilon f^{*}E$ is ample. Now, $f^{*}E$ is effective and then $f^{*}D$ is nef and big. Finally $\phi^*D = b^*f^*D$ is nef and big bacause $b$ is birational.

Onother approach:

In particular $D$ is big $\Leftrightarrow$ $Vol(D)>0$ $\Leftrightarrow$ $Vol(\phi^{*}D) = d\cdot Vol(D)>0$ $\Leftrightarrow$ $\phi^{*}D$ is big.

  • If $\phi:X\rightarrow Y$ is a proper morphism, $D$ nef $Rightarrow$ $\phi^{*}D$ nef. If $C$ is an integral curve on $X$, by the projection formula we have $\phi^*D\cdot C = D\cdot \phi_*C \geq 0$.
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