Lemma 31.13.2. Let $S$ be a scheme. Let $D \subset S$ be a closed subscheme. The following are equivalent:

1. The subscheme $D$ is an effective Cartier divisor on $S$.

2. For every $x \in D$ there exists an affine open neighbourhood $\mathop{\mathrm{Spec}}(A) = U \subset S$ of $x$ such that $U \cap D = \mathop{\mathrm{Spec}}(A/(f))$ with $f \in A$ a nonzerodivisor.

Proof. Assume (1). For every $x \in D$ there exists an affine open neighbourhood $\mathop{\mathrm{Spec}}(A) = U \subset S$ of $x$ such that $\mathcal{I}_ D|_ U \cong \mathcal{O}_ U$. In other words, there exists a section $f \in \Gamma (U, \mathcal{I}_ D)$ which freely generates the restriction $\mathcal{I}_ D|_ U$. Hence $f \in A$, and the multiplication map $f : A \to A$ is injective. Also, since $\mathcal{I}_ D$ is quasi-coherent we see that $D \cap U = \mathop{\mathrm{Spec}}(A/(f))$.

Assume (2). Let $x \in D$. By assumption there exists an affine open neighbourhood $\mathop{\mathrm{Spec}}(A) = U \subset S$ of $x$ such that $U \cap D = \mathop{\mathrm{Spec}}(A/(f))$ with $f \in A$ a nonzerodivisor. Then $\mathcal{I}_ D|_ U \cong \mathcal{O}_ U$ since it is equal to $\widetilde{(f)} \cong \widetilde{A} \cong \mathcal{O}_ U$. Of course $\mathcal{I}_ D$ restricted to the open subscheme $S \setminus D$ is isomorphic to $\mathcal{O}_{S \setminus D}$. Hence $\mathcal{I}_ D$ is an invertible $\mathcal{O}_ S$-module. $\square$

Comments (1)

Comment #735 by Keenan Kidwell on

All the instances of $X$ should be $S$.

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