Proposition 6

Let $A$ be a Krull domain, $K$ its field of fractions, $(v_t)_{t \in I}$ a family of valuations on $K$ with the properties of Definition 3 and $A$, the ring of $v_t$. Let $S$ be a multiplicative subset of $A$ not containing $0$ and $J$ the set of indices $t \in I$ such that $v_t$ is zero on $S$. Then $S^{-1}A = \bigcap_{t \in J} A_t$; in particular $S^{-1}A$ is a Krull domain.

We write $B = \bigcap_{t \in J} A_t$. Then $S^{-1} \subset B$ and $A \subset B$ and hence $S^{-1}A \subset B$. Conversely, let $x \in B$. Let $J'$ denote the finite set of indices $t$ such that $v_t(x) < 0$. If $t \in J'$, then $x \notin A_t$, hence $t \notin J$ and hence there exists $s_t \in S$ such that $v_t(s_t) > 0$. Let $n(t)$ be an integer $> 0$ such that $v_t(s_t^{n(t)}x) \geq 0$; we write $s = \prod_{t \in J'} s_t^{n(t)}$. Then $v_t(sx) \geq 0$ for all $t \in I$ and hence $sx \in A$ and $x \in S^{-1}A$. Thus $B = S^{-1}A$.