Return to Revisions

If you remove "perhaps at the cost of extending the initial set to a larger 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 $x$.