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Let $X$ be a smooth projective variety such that $-K_X$ is ample. Let $f:X\dashrightarrow Y$ be a small $\mathbb{Q}$-factorial transformation. I would like to know if is true or not that:

  • $-K_Y$ is big,
  • there exists an effective $\mathbb{Q}$-divisor $D\subset Y$ such that $-(K_Y+D)$ is ample.
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Yes to both. 1) is because $-K_Y$ and $-K_X$ have the same section rings (by smallness of $f$); 2) is the so-called Kodaira lemma (big is Q-linearly equivalent to ample+effective). –  Artie Prendergast-Smith Jun 17 at 11:45

1 Answer 1

up vote 1 down vote accepted

Let $f:X\dashrightarrow Y$ be a small map. Let $D$ be a big divisor on $D$. Then for $m >>0$ the map $\phi_{mD}:Y\dashrightarrow Z$, induced by $mD$ is birational. Now, if $\Delta:=f^{*}D$, then the map induced by $m\Delta$ is $\phi_{mD}\circ f$, which is birational as well. Therefore $\Delta$ is big. We conclude that $D$ is big if and only if $\Delta$ is big. In particular $-K_Y$ is big if and only if $-K_X$ is big.

In you case, $-K_X$ ample $\Rightarrow$ $-K_X$ big $\Rightarrow$ $-K_Y$ big.

By Corollary 2.2.7. in Lazarsfeld, "Positivity in Algebraic Geometry I", a divisor $D$ is big if and only if there exists an ample divisor $A$ and a positive integer $m$ such that $mD = A+N$, where $N$ is effective. Therefore $-(-D+\frac{1}{m}N) = \frac{1}{m}A$. In you case $D = -K_Y$ and you get that $-(K_Y+\frac{1}{m}N) = \frac{1}{m}A$ is ample.

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