Weak approximation in Krull domains

Question

Suppose $R$ is a Krull domain with the field of fraction $K$. To every prime ideal $P$ of $R$ of height $1$, one can associate a $ \mathbb{Z}$-valued discrete valuation which we denote by $v_P$.

Suppose $P_1, \dots, P_k$ are pairwise distinct height one prime ideas of $R$. It is well-known that the following approximation theorem holds: for any $(n_1, \dots, n_k) \in \mathbb{Z}^k$, there exists $x \in K$ such that

1. $v_{P_i} (x)= n_i$ for all $1 \le i \le k$
2. $v_{P}(x) \ge 0$ for all prime ideals $P$ of $R$ of height one other than $P_1, \dots, P_k$.

I would like to know if one can replace (2) with $v_{P}(x) = 0$ for all all prime ideals $P$ of height one other than $P_1, \dots, P_k$, perhaps at the cost of extending the initial set to a larger set of prime ideals.

In case it helps, the case I am really interested in is when $R$ is an integral extension of $ \mathbb{Q}[X_1, \dots, X_n]$. Any reference would be highly appreciated.

Edit: To clarify the question, what I would like to know if the following refinement of the approximation lemma holds:

Given $P_1, \dots, P_k$ are above, there exists a finite set $\mathcal{P}$ of prime ideals of height $1$ only depending on $P_1, \dots, P_k$ so that for every vector $(n_i)_{1 \le i \le k} \in \mathbb{Z}^k$, there exists $x \in K$ such that in addition to satisfying (1) and (2) above, it also satisfies $v_{P}(x)=0$ for all height one prime ideals $P$ outside $\mathcal{P}$.

I interpret the comment below by @Laurent Moret-Bailly as a negative answer to this. If this is indeed the case, I would still like to see how this goes wrong.

Answer 1

If you remove "perhaps at the cost of extending the initial [finite] set to a larger [finite] set of prime ideals" from your question, then the answer is yes for a fixed Krull domain $R$ if and only if $R$ is a UFD. First, suppose that the answer is yes for $R$. Then let $x$ be an element of $K$ such that $v_P(x) = 1$ for a fixed height one prime $P$ and $v_Q(x) = 0$ for all other height one primes $Q$. It is clear, then, that $P$ is principal, generated by $x$. (Check this.) Since then all height one primes of $R$ are principal, $R$ is a UFD. Conversely, if $R$ is a UFD, then every height one prime $P$ of $R$ is principal, say, generated by $x_P$, and then $x = x_{P_1}^{n_1} x_{P_2}^{n_2} \cdots x_{P_k}^{n_k}$ satisfies the required condition for $x$.

The OP's qualified statement is true for any Krull domain $R$. Here is a proof. For each $i $ from $1$ to $k$, there exist nonzero $x_i^+, x_i^- \in K$ such that $v_{P_i}(x_i^\pm) = \pm 1$, $v_{P_j}(x_i^\pm) = 0$ for all $j \neq i$, and $v_{Q}(x_i^\pm) \geq 0$ for all $Q$ other than the $P_i$. Since for any nonzero $y \in K$ one has $v_Q(y) = 0$ for all but finitely many height one primes $Q$, one has $v_Q(x_i^{\pm}) = 0$ for all $i $ for all but finitely many height one primes $Q$, so we may collect all $Q$ such that $v_Q(x_i^{\pm}) \neq 0$ for some $i$ into a finite set $\mathcal{P}$. Then, for any $(n_1, n_2, \ldots, n_k) \in \mathbb{Z}^k$, the nonzero element $x= (x_1^\pm)^{|n_1|} (x_2^\pm)^{|n_2|} \cdots (x_k^\pm)^{|n_k|}$ of $K$, with the sign of each $\pm$ in $x_i^{\pm}$ matching the sign of $n_i$, satisfies the desired condition of the OP.