Theorem 1

Let $A$ be an integral domain. For the monoid $D(A)$ of divisors of $A$ to be a group, it is necessary and sufficient that $A$ be completely integrally closed.

Suppose that $D(A)$ is a group. Let $x \in K$; suppose that $A[x]$ is contained in a finitely generated sub-$A$-module of $K$. Then we have seen (no. 1) that $a = A[x]$ is an element of $I(A)$. Then $xa \subset a$ and hence $\operatorname{div}(x) + \operatorname{div} a \geq \operatorname{div} a$. Since $D(A)$ is an ordered group, we conclude that $\operatorname{div}(x) \geq 0$, whence $x \in A$. Thus $A$ is completely integrally closed (Chapter V, § 1, no. 4, Definition 5).

Conversely, suppose that $A$ is completely integrally closed. Let $a$ be a divisorial ideal. We shall show that $\operatorname{div} a + \operatorname{div}(A : a) = 0$, which will prove that $D(A)$ is a group. As $a(A : a) \subset A$, it suffices (no. 1) to verify that every fractional principal ideal $Ax^{-1}$ which contains $a(A : a)$ also contains $A$. Now, for $y \in K^*$, the relation $Ay \supset a$ implies $y^{-1} \in A : a$, whence $y^{-1}a \subset a(A : a) \subset Ax^{-1}$ and hence $xa \subset Ay$. As $a$ is divisorial, we deduce that $xa \subset a$, whence $x^n a \subset a$ for all $n \in \mathbf{N}$. There exist elements $x_0, x_1$ of $K^*$ such that $Ax_0 \subset a \subset Ax_1$; therefore $x^n x_0 \in Ax_1$, whence $x^n \in Ax_1 x_0^{-1}$. As $A$ is completely integrally closed, $x \in A$, that is $Ax^{-1} \supset A$, which completes the proof.

Note that, if $A$ is completely integrally closed (and even Noetherian), a divisorial ideal of $A$ is not necessarily invertible, in other words, in general $J(A) \neq D(A)$ (Exercise 2 and § 3, no. 2, Proposition 1).