Theorem 2

Let $A$ be an integral domain. For $A$ to be a Krull domain, it is necessary and sufficient that the twofollowing conditions be satisfied: (a) $A$ is completely integrally closed; (b) every non-empty family of divisorial integral ideals of $A$ admits a maximal element (with respect to the relation $\subset$). Moreover, if $P(A)$ is the set of extremal elements of $D(A)$, $P(A)$ is then a basis of the $\mathbf{Z}$-module $D(A)$ and the positive elements of $D(A)$ are the linear combinations of the elements of $P(A)$ with coefficients $\geq 0$.

Let $A$ be a Krull domain. It is completely integrally closed (Chapter VI, § 4, no. 5, Corollary to Proposition 9). Let $(v_i)_{i \in I}$ be a family of valuations on the field of fractions $K$ of $A$ satisfying (AK$_I$), (AK$_{II}$) and (AK$_{III}$). The $v_i$ may be assumed to be normed (Chapter VI, § 3, no. 6, Definition 3). For all $a \in I(A)$, we shall write:

then $v_t(a) \in \mathbf{Z}$, for, if $a$ is a non-zero element of $a$, the relation $Ax \supset Aa$ implies that $v_t(x) \leq v_t(a)$ (by (AK$_{III}$)), which shows that the family of $v_t(x)$ ($a \subset Ax$) is bounded above. We establish the following properties:

(1) Let $a$ be a divisorial fractional ideal; in order that $y \in a$, it is necessary and sufficient that $v_t(y) \geq v_t(a)$ for all $t \in I$.

As $a$ is divisorial, the relation $y \in a$ is equivalent to the relation "“$a \subset Ax$ implies $y \in Ax$”. Now, by (AK$_{II}$), the relation $y \in Ax$ is equivalent to “$v_t(y) \geq v_t(x)$ for all $t \in I$”. Whence (1).

(2) Let $a$ and $b$ be two divisorial fractional ideals of $A$; in order that $a \subset b$, it is necessary and sufficient that $v_t(a) \geq v_t(b)$ for all $t \in I$.

This follows immediately from property (1).

(3) If $x \in K^*$, then $v_t(Ax) = v_t(x)$.

If $Ay \supset Ax$, then $v_t(y) \leq v_t(x)$ by (AK$_{II}$) and the minimum value of $v_t(y)$ is taken at $y = x$.

(4) For all $a \in I(A)$, the indices $t \in I$ such that $v_t(a) \neq 0$ are finite in number.

There exist $x, y$ in $K^*$ such that $Ax \subset a \subset Ay$. By properties (2) and (3), $v_t(x) \geq v_t(a) \geq v_t(y)$ for all $t \in I$. It then sufficesto apply (AK$_{III}$).

We have therefore shown the following lemma: