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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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    $\begingroup$ 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). $\endgroup$ Aug 12, 2012 at 5:39
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    $\begingroup$ I agree with Yusuf. Please note, the "generically finite morphism" here also must be dominant (when restricted to every irreducible component of the domain). $\endgroup$ Aug 12, 2012 at 14:46

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One can consider the following characterization of nef 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 $0<\epsilon\ll1$ 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 $D^{\prime}=(h-k)D+A$ is a sum of a nef and an ample divisor. Therefore $D^{\prime}$ is ample. If $\displaystyle\epsilon=\frac{1}{h}$ we get that $$D-\epsilon E\equiv\epsilon D^{\prime}$$ 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 $0<\epsilon\ll1$ 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 $\phi^*D\cdot C=D\cdot\phi_*C\geq0$.
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