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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