Lemma 1

If $A$ is a Krull domain and $(v_t)_t$, is a family of normed valuations on $K$ satisfying (AK$_I$), (AK$_{II}$) and (AK$_{III}$), the mapping $a \mapsto (v_t(a))_{t \in I}$ is a decreasing injective mapping of the set of divisorial integer ideals of $A$ (ordered by $\subset$) to the set of positive elements & the ordered group the direct sum $\mathbf{Z}^{(I)}$.

This being so, every non-empty set of positive elements of $\mathbf{Z}^{(I)}$ has a minimal element (Algebra, Chapter VI, § 1, no. 13, Theorem 2). Hence $A$ certainly satisfies property (b) of the statement.

Conversely, let $A$ be an integral domain satisfying properties (a) and (b) of the statement. Since $A$ is completely integrally closed, $D(A)$ is an ordered group (no. 2, Theorem 1). This group is a lattice (no. 1, Proposition 2). By condition (b) of the statement, every non-empty family of positive elements of $D(A)$ has a minimal element. Let $P(A)$ be the set of extremal elements of $D(A)$. Then (Algebra, Chapter VI, § 1, no. 13, Theorem 2) $P(A)$ is a basis of the $\mathbf{Z}$-module $D(A)$ and the positive elements of $D(A)$ are the linear combinations with positive integer coefficients of the elements of $P(A)$.

Thus, for $x \in K^*$, rational integers $v_P(x)$ are defined (for $P \in P(A)$) by writing:

We also write $v_P(0) = +\infty$.

From the relations

and

for $x, y$ and $x + y$ in $K^*$, we deduce that the $v_P$ are discrete valuations on $K$. In order that $x \in A$, it is necessary and sufficient that $\text{div}(x) \geq 0$, that is that $v_P(x) \geq 0$ for all $P \in P(A)$. Thus the $v_P$ satisfy conditions (AK$_1$) and (AK$_{11}$) and obviously also (AK$_{III}$).