Proposition 16

Let A be a Zariski ring such that its completion $\hat{A}$ is a Krull domain. Then A is a Krull domain and the canonical homomorphism $\tilde{i}$ from $C(A)$ to $C(\hat{A})$ (which is defined since A is a flat A-module; cf. Chapter III, § 3, no. 4, Theorem 3) is injective.

As A is an integral domain and $A \subset \hat{A}$, A is an integral domain. Let L be the field of fractions of $\hat{A}$ and K $\subset$ L that of A. As $A = \hat{A} \cap K$ (Chapter III, § 3, no. 5, Corollary 4 to Proposition 9), A is a Krull domain (no. 3, Example 4). The fact that $\tilde{i} : C(A) \to C(\hat{A})$ is injective follows from Proposition 15 (b) and the fact that, if $b \hat{A}$ is principal, b is principal (Chapter III, § 3, no. 5, Corollary 3 to Proposition 9).

Now let A be a Krull domain and S a multiplicative subset of A not containing 0. The group $D(A)$ (resp. $D(S^{-1}A)$) is the free commutative group with basis the set of $\operatorname{div}(p)$ (resp. $\operatorname{div}(S^{-1}p)$), where p runs through the set of prime ideals of A of height 1 (resp. the set of prime ideals of A of height 1 such that $p \cap S = \varnothing$) (no. 4, Proposition 6) and, if $p \cap S = \varnothing$, then $i(\operatorname{div}(p)) = \operatorname{div}(S^{-1}p)$. Thus $D(S^{-1}A)$ is identified with the directfactor of $D(A)$ generated by the elements $\operatorname{div}(p)$ such that $p \cap S = \varnothing$ and admits as complement the free commutative subgroup of $D(A)$ with basis the set of $\operatorname{div}(p)$ such that $p \cap S \neq \varnothing$; we shall denote this complement by G. As $i : D(A) \to D(S^{-1}A)$ is surjective, so is $i : C(A) \to C(S^{-1}A)$; and:

for if an element of $D(S^{-1}A)$ is equal to $\operatorname{div}_{S^{-1}A}(x/s)$, where $x \in A$ and $s \in S$, it is the image under $i$ of the principal divisor $\operatorname{div}_A(x)$ (Proposition 14).

Suppose now that S is generated by a family of elements $(p_i)_{i \in I}$ of A such that the principal ideals $Ap_i$ are all prime. Then, if $p$ is a prime ideal of A of height 1 such that $p \cap S \neq \varnothing$, $p$ contains a product of powers of the $p_i$, and therefore one of the $p_i$, say $p_\alpha$; as $Ap_\alpha$ is non-zero and prime and $p$ is of height 1, it follows that $p = Ap_\alpha$. In the above notation, therefore $G \subset F(A)$ and (5) shows that the kernel of $\bar{i}$ is zero. We have therefore shown the following result: