Proposition 9

Let $v_1, \ldots, v_r$ be distinct essential valuations of a Krull domain $A$ and $n_1, \ldots, n_r$ rational integers. There exists an element $x$ of the field of fractions $K$ of $A$ such $v_i(x) = n_i$ for $1 \leq i \leq r$ and $v(x) \geq 0$ for every essential valuation $v$ of $A$ distinct from $v_1, \ldots, v_r$.

Let $p_1, \ldots, p_r$ be the divisorial ideals of $A$ corresponding to the valuations $v_1, \ldots, v_r$. There exists $y \in K$ such that $v_i(y) = n_i$ for $1 \leq i \leq r$ (Chapter VI, § 7, no. 2, Corollary 1 to Theorem 1). The essential valuations $w_1, \ldots, w_s$ of $A$ distinct from the $v_i$ for which the integer $w_j(y) = -m_j$ is $< 0$ are finite in number; let $q_1, \ldots, q_s$ be the corresponding ideals. There exists no inclusion relation between $p_1, \ldots, p_r, q_1, \ldots, q_s$ since these ideals correspond to extremal divisors and these ideals are prime (Corollary 1 to Proposition 6). Hence the integral ideal $a = q_1^{m_1} \ldots q_s^{m_s}$ is contained in none of the $p_i$ (Chapter 11, § 1, no. 1, Proposition 1) and is therefore not contained in their union (loc. cit., Proposition 2). Therefore there exists $z \in a$ such that $z \notin p_i$ for $1 \leq i \leq r$; then $v_1(z) = \cdots = v_r(z) = 0$ and $w_j(z) \geq m_j$ for $1 \leq j \leq s$; hence the element $x = yz$ solves the problem.