1
$\begingroup$

Let ($X$,$\Delta$) be projective klt pair and $f \colon X \rightarrow Z$ be contraction of ($K_X + \Delta$) - negative extremal ray $R$. If $X$ is $\mathbb{Q}$ -factorial and $\mathrm{dim}Z < \mathrm{dim} X $, then Z is $\mathbb{Q}$- factorial?

I have read proof of above claim in Kollar and Mori "Birational geometry of algebraic varieties" but I have some trouble.The proof is following:

Let $D$ be effective weil divisor on $Z$ and $Z_0 \subset Z$ be smooth locus of $Z$. $D'$ be closure of $f^{-1}(D \cap Z_0 )$. By $\mathbb{Q}$-factoriality of X, $mD'$ is cartier for some $m \in \mathbb{Z}$. Since $D'$ does not meet general fibre of $f$ we have $D'.R = 0$ Hence $mD'$ is pullback of some cartier divisor on $Z$ and $D$ is $\mathbb{Q}$ cartier.

My trouble is the last statement.

Why $D$ is $\mathbb{Q}$ cartier since $mD'$ is pullback of some cartier divisor on Z?

$\endgroup$

1 Answer 1

1
$\begingroup$

Let $D''$ be a Cartier divisor on $Z$ such that $f^*\mathscr O_Z(D'')\simeq \mathscr O_X(mD')$. Then by the projection formula $\mathscr O_Z(D'')\simeq f_*\mathscr O_X(mD')$ (since $f_*\mathscr O_X\simeq \mathscr O_Z$).

On the other hand, from the construction we see that $\mathscr O_Z(D'')|_{Z_0}\simeq f_*\mathscr O_X(mD')|_{Z_0}\simeq \mathscr O_Z(mD)|_{Z_0}$. Finally, because $Z$ is normal, the complement of $Z_0$ has at least codimension $2$ in $Z$. Both $\mathscr O_Z(D'')$ and $\mathscr O_Z(mD)$ are reflexive sheaves of rank $1$, and they are isomorphic on the complement of a codimension $2$ closed set, so they are isomorphic everywhere. Hence $mD\sim D''$ which is Cartier.

$\endgroup$

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.