Lemma 31.12.13. Let X be an integral locally Noetherian normal scheme. Let \mathcal{F} be a coherent \mathcal{O}_ X-module. The following are equivalent
\mathcal{F} is reflexive,
\mathcal{F} is torsion free and has property (S_2), and
there exists an open subscheme j : U \to X such that
every irreducible component of X \setminus U has codimension \geq 2 in X,
j^*\mathcal{F} is finite locally free, and
\mathcal{F} = j_*j^*\mathcal{F}.
Proof. Using Lemma 31.12.2 the equivalence of (1) and (2) follows from More on Algebra, Lemma 15.23.18. Let U \subset X be as in (3). By Properties, Lemma 28.12.5 we see that \text{depth}(\mathcal{O}_{X, x}) \geq 2 for x \not\in U. Since a finite locally free module is reflexive, we conclude (3) implies (1) by Lemma 31.12.12.
Assume (1). Let U \subset X be the maximal open subscheme such that j^*\mathcal{F} = \mathcal{F}|_ U is finite locally free. So (3)(b) holds. Let x \in X be a point. If \mathcal{F}_ x is a free \mathcal{O}_{X, x}-module, then x \in U, see Modules, Lemma 17.11.6. If \dim (\mathcal{O}_{X, x}) \leq 1, then \mathcal{O}_{X, x} is either a field or a discrete valuation ring (Properties, Lemma 28.12.5) and hence \mathcal{F}_ x is free (More on Algebra, Lemma 15.22.11). Thus x \not\in U \Rightarrow \dim (\mathcal{O}_{X, x}) \geq 2. Then Properties, Lemma 28.10.3 shows (3)(a) holds. By the already used Properties, Lemma 28.12.5 we also see that \text{depth}(\mathcal{O}_{X, x}) \geq 2 for x \not\in U and hence (3)(c) follows from Lemma 31.12.12. \square
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