If you want to pull back a Cartier divisor $D$, you can do that provided the image of $f$ is not contained in the support of $D$: just pull back the local equations for $D$.
If this does not happen, on an integral scheme, you can just pass to the associated line bundle $mathcal{O}_X(D)$ and pull back that, obtaining $f^{*} mathcal{O}_X(D)$; of course you lose some information because a line bundle determines a Cartier divisor only up to linear equivalence.
Fulton invented a nice way to avoid this distinction. Define a pseudodivisor on $X$ to be a triple $(Z, L, s)$ where $Z$ is a closed subset of $X$, $L$ a line bundle and $s$ a nowhere vanishing section on $X setminus Z$, hence a trivialization on that open set. Then you can simply define the pullback of this triple as $(f^{-1}(Z), f^{*} L, f^{*} s)$, so you can always pull back pseudo divisors, whatever $f$ is.
The relation with Cartier divisors is the following: to a Cartier divisor $D$ you can associate a pseudodivisor $(|D|, mathcal{O}_X(D), s)$, where $s$ is the section of $mathcal{O}_X(D)$ which gives a local equation for $D$.
This correspondence is not bijective. First, a pseudodivisor $(Z, L, s)$ determines a Cartier divisor if $Z subsetneq X$; note that in this case enlarging $Z$ will not change the associated Cartier divisor, so to obtain a bijective correspondence with Cartier divisors you have to factor out pseudodivisors by an equivalence relation, which I leave to you to formulate. But if $Z = X$, you only obtain a line bundle on $X$, and you have no way to get back a Cartier divisor.
If you want to know more about this, read the second chapter of Fulton's intersection theory.
No comments:
Post a Comment