Theorem 3

Let $\mathbf{A}$ be a Krull domain and $\mathfrak{p}$ an integral ideal of $\mathbf{A}$. For $\mathfrak{p}$ to be the divisorial ideal corresponding to an extremal divisor, it is necessary and sufficient that $\mathfrak{p}$ be a prime ideal of height 1.

If $\mathfrak{p}$ is the divisorial ideal corresponding to an extremal divisor, we know (no. 4, Corollary 1 to Proposition 6) that $\mathfrak{p}$ is prime and that $\mathbf{A}_\mathfrak{p}$ is a discrete valuation ring; as $\mathbf{A}_\mathfrak{p}$ has no prime ideals other than $(0)$ and $\mathfrak{p}\mathbf{A}_\mathfrak{p}$, $(0)$ and $\mathfrak{p}$ are the only prime ideals of $\mathbf{A}$ contained in $\mathfrak{p}$ (Chapter II, § 3, no. 1, Proposition 3); hence $\mathfrak{p}$ is of height 1. Conversely, we shall show first that every prime ideal $\mathfrak{p} \neq (0)$ of $\mathbf{A}$ contains a divisorial prime ideal $\mathfrak{q}$ corresponding to an extremal divisor: for, as $\mathbf{A}_\mathfrak{p} \neq K$, $\mathbf{A}_\mathfrak{p}$ is the intersection of a non-empty family $(\mathbf{A}_t)$ of essential valuation rings (no. 4, Proposition 6); each $\mathbf{A}_t$ is of the form $\mathbf{A}_{\mathfrak{q}_t}$ (no. 4, Corollary 1 to Proposition 6) and from $\mathbf{A}_\mathfrak{p} \subset \mathbf{A}_{\mathfrak{q}_t}$ we deduce that $\mathfrak{q}_t \subset \mathfrak{p}$. Thus, if $\mathfrak{p}$ is of height 1, then $\mathfrak{p} = \mathfrak{q}$, which shows that $\mathfrak{p}$ is the divisorial ideal corresponding to an extremal divisor.