Proposition 17

Let $A$ be a Krull domain and $S$ a multiplicative subset of $A$ not containing $0$. Then the canonical homomorphism $\bar{i}$ from $C(A)$ to $C(S^{-1}A)$ is surjective. If further $S$ is generated by a family of elements $p_i$ such that the principal ideals $Ap_i$ are all prime, then $\bar{i}$ is bijective.

As a second application of formula (5), consider the following situation: let $R$ be a Krull domain; take $A$ to be the polynomial ring $A = R[X]$ (no. 9, Proposition 13) and $S$ to be the set $R - (0)$ of non-zero constant polynomials of $A$. The prime ideals $p$ of $A$ of height 1 such that $p \cap S \neq \varnothing$ are those of the form $p_0A$, where $p_0$ is a prime ideal of $R$ of height 1 (no. 9, Remark). Hence, in the notation introduced above, $G$ is identified with $D(R)$ by identifying $\mathrm{div}_A(p_0A)$ with $\mathrm{div}_R(p_0)$. On the other hand $G \cap F(A)$ is identified with $F(R)$: for if an ideal $a$, of $R$ generates a principal ideal $a_0A = f(X)A$ in $A = R[X]$, then $f(0) \in a_0A$ since $a_0A$ is a graded ideal of the ring $A$ (graded by the usual degree of polynomials) and hence $f(0) \in a$; further, for $a \in a_0A$, $a = f(X)g(X)$ where $g(X) \in R$, whence, comparing terms of degree 0, $a = f(0)g(0)$; it follows that $a$ is the principal ideal of $R$ generated by $f(0)$. Finally, denoting by $K$ the field of fractions of $R$, $S^{-1}A$ is identified with the polynomials ring $K[X]$, which is a principal ideal domain; hence $C(S^{-1}A) = (0)$. Thus, by (5), $C(A) = \mathrm{Ker}(\bar{i})$ is identified with $C(R)$ and we have proved the following result: