If we have a nonconstant map of nonsingular curves $\varphi:X\rightarrow Y$, then Hartshorne defines a map $\varphi^\* Div(Y)\rightarrow Div(X)$ using the fact that codimension one irreducibles are just points, and looking at $\mathcal{O}\_{Y,f(p)} \rightarrow \mathcal{O}\_{X,p}$. My question is if we don't have a nice map of curves, what conditions can we put on the morphism so that we may pull divisors back? Clearly it's not true in general, since we can take a constant map and then topologically the inverse image doesn't even have the right codim.


Thinking about this in terms of Cartier divisors (and assuming the schemes are integral), it seems like we just need a way to transport functions in $K(Y)$ to functions in $K(X)$. If $\varphi$ is dominant, then we'll get such a map. Is this sufficient? Also is there something we can say when $\varphi$ is not dominant? Something like we have a way to map divisors with support on $\overline{\varphi(X)}$ to divisors on $X$?



When do divisors pull back?
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.


