Lemma 82.12.1. In Situation 82.2.1 let $X/B$ be good. Assume $X$ is integral.
If $Z \subset X$ is an integral closed subspace, then the following are equivalent:
$Z$ is a prime divisor,
$|Z|$ has codimension $1$ in $|X|$, and
$\dim _\delta (Z) = \dim _\delta (X) - 1$.
If $Z$ is an irreducible component of an effective Cartier divisor on $X$, then $\dim _\delta (Z) = \dim _\delta (X) - 1$.
Proof. Part (1) follows from the definition of a prime divisor (Spaces over Fields, Definition 72.6.2), Decent Spaces, Lemma 68.20.2, and the definition of a dimension function (Topology, Definition 5.20.1).
Let $D \subset X$ be an effective Cartier divisor. Let $Z \subset D$ be an irreducible component and let $\xi \in |Z|$ be the generic point. Choose an étale neighbourhood $(U, u) \to (X, \xi )$ where $U = \mathop{\mathrm{Spec}}(A)$ and $D \times _ X U$ is cut out by a nonzerodivisor $f \in A$, see Divisors on Spaces, Lemma 71.6.2. Then $u$ is a generic point of $V(f)$ by Decent Spaces, Lemma 68.20.1. Hence $\mathcal{O}_{U, u}$ has dimension $1$ by Krull's Hauptidealsatz (Algebra, Lemma 10.60.11). Thus $\xi $ is a codimension $1$ point on $X$ and $Z$ is a prime divisor as desired. $\square$
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