Lemma 42.16.1. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be locally of finite type over $S$. Assume $X$ is integral.
If $Z \subset X$ is an integral closed subscheme, 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 (Divisors, Definition 31.26.2) and the definition of a dimension function (Topology, Definition 5.20.1). Let $\xi \in Z$ be the generic point of an irreducible component $Z$ of an effective Cartier divisor $D \subset X$. Then $\dim (\mathcal{O}_{D, \xi }) = 0$ and $\mathcal{O}_{D, \xi } = \mathcal{O}_{X, \xi }/(f)$ for some nonzerodivisor $f \in \mathcal{O}_{X, \xi }$ (Divisors, Lemma 31.15.2). Then $\dim (\mathcal{O}_{X, \xi }) = 1$ by Algebra, Lemma 10.60.13. Hence $Z$ is as in (1) by Properties, Lemma 28.10.3 and the proof is complete. $\square$
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