Suppose $C \to Y$ is a closed immersion of schemes (not necessarily of codimension 1), and suppose that every prime divisor $C\subset Z \subset Y$ is equivalent to a prime divisor which does not contain $C$. Is it true that there exists induced map $Cl Y \to Cl C$?
My idea was something like this: 1. Define a map $R: L \subset Div Y \to Div C$, where $L$ is the subgroup spanned by prime divisors which does not contain $Z$. The map $R$ takes a prime divisor $Z \in L$, decompose $Z \cap C$ in irreducible components $Z_1, \ldots, Z_n$, computes somehow multiplicities $n_i$, and yields finally $\sum n_i Z_i$. Note that $Z_i$ have codimension 1 and are reduced, thus $Z_i$ is indeed a prime divisor. 2. By assumption $Cl(Y)= Div(Y)/ prDiv(Y) \simeq L / L \cap prDiv(Y)$, where $prDiv(Y)$ is the subgroup of principal divisors. This is just an algebraic restatement of the hypothesis. 3. If $f \in K(Y)$, then it comes from a $f \in O_Y(U)$ for some U not containing $C$. This gives a $\varphi(f) \in O_C(U\cap C)\subset K(C)$, and hopefully $div(\varphi(f))=R(div(f))$. In this way we would have a well defined map on divisor class groups.
Of course, we assume that $Y,C$ are such that divisor class group is defined, i.e. noetherian integral separated regular in codimension one. Thank you!