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
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