Theorem 1

Let $A$ be an integral domain. The following conditions are equivalent: (a) $A$ is factorial; (b) the ordered group of non-zero fractional principal ideals of $A$ is a direct sum of groups isomorphic to $\mathbf{Z}$ (ordered by the product order); (c) condition (M) is satisfied and the intersection of two principal ideals of $A$ is a principal ideal; (d) condition (M) is satisfied and, for every extremal element $p$ of $A$, the ideal $Ap$ is prime; (e) $A$ is a Krull domain and every prime ideal of height 1 is principal.

We shall denote by K the field of fractions of A and by $\mathscr{P}^*$ (or $\mathscr{P}^*(A)$) the ordered group of non-zero fractional principal ideals of A. The proof will be carried out by proving the following implications:

(a) ⇔ (b) ⇔ (c) ⇔ (d) ⇔ (e)

We show that (a) implies (b); if A is factorial, $\mathscr{P}^*$ is isomorphic to the group of divisors of A and hence to a direct sum of groups $\mathbf{Z}$ ($§ 1$, no. 3, Theorem 2).

Note now that the relation "the intersection of two integral principal ideals of A is a principal ideal" means that every ordered pair of elements of A admits a $lcm$, that is that $\mathscr{P}^*$ is a lattice-ordered group (Algebra, Chapter VI, $§ 1$, no. 9, Proposition 8). The fact that (b) implies (c) (and even is equivalent to it) therefore follows from Algebra, Chapter VI, $§ 1$, no. 13, Theorem 2. The fact that (c) implies (d) follows from Algebra, Chapter VI, $§ 1$, no. 13, Proposition 14 (DIV).

The fact that (d) implies (b) follows from Algebra, Chapter VI, $§ 1$, no. 13, Theorem 2 applied to the group $\mathscr{S}^*$.

We show that (b) implies (e). If (b) holds, there is an isomorphism of $\mathscr{P}^*$ onto $\mathbf{Z}^{(l)}$; let $(v_i(x))_{i \in I}$ denote the element of $\mathbf{Z}^{(l)}$ corresponding to the ideal $Ax$ ($x \in K^*$). It is seen immediately that each $v_i$ is a discrete valuation on K, that A is the intersection of the rings of the $v_i$ and that, for $x \in K^*$, $v_i(x) = 0$ except for a finite number of indices $i$; hence A is a Krull domain. On the other hand, let q be a prime ideal of A of height 1; it contains a non-zero element a which is necessarily not invertible and hence also (by definition of a prime ideal) one of the extremal elements of A; as $Ap$ is prime and non zero, $q = Ap$, which proves that q is principal.

Finally we show that (e) implies (a). Let a be a divisorial ideal of A. There exist prime ideals $p_i$ of A of height 1 such that $\operatorname{div} a = \sum_i n_i \operatorname{div} p_i$ where $n_i \in \mathbf{Z}$. If (e) holds, $p_i$ is of the form $Ap_i$, whence $\operatorname{div} a = \operatorname{div} \left( \prod_i Ap_i^{n_i} \right)$ and hence $a = \prod_i Ap_i^{n_i}$ since a is divisorial.