Corollary 1

Let $P$ be an extremal divisor of $A$ and $p$ the corresponding divisorial ideal. Then $p$ is prime, the ring of $v_P$ is $A$, and the residue field of $v_P$ is identified with the field of fractions of $A/p$.

Let $S = A - p$. By Proposition 5, $v_P$ is zero on $S$ and $> 0$ on $p$. Hence $p$ is the intersection of $A$ and the ideal of $v_P$ and is therefore prime. On the other hand, for every extremal divisor $Q \neq P$, $Q \not\subset P$ and hence the divisorial ideal $q$ corresponding to $Q$ is not contained in $p$; thus $q \cap S \neq \varnothing$ and hence, by Proposition 5, $v_Q$ is not zero on $S$. This being so, the corollary follows from Proposition 6 and Chapter II, § 3, no. 1, Proposition 3.