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.

  • 6
    $\begingroup$ 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
  • 3
    $\begingroup$ 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

1 Answer 1


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 $\phi^*D\cdot C=D\cdot\phi_*C\geq0$.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.