Lemma 31.12.15. Let X be a regular scheme of dimension \leq 2. Let \mathcal{F} be a coherent \mathcal{O}_ X-module. The following are equivalent
\mathcal{F} is reflexive,
\mathcal{F} is finite locally free.
Lemma 31.12.15. Let X be a regular scheme of dimension \leq 2. Let \mathcal{F} be a coherent \mathcal{O}_ X-module. The following are equivalent
\mathcal{F} is reflexive,
\mathcal{F} is finite locally free.
Proof. It is clear that a finite locally free module is reflexive. For the converse, we will show that if \mathcal{F} is reflexive, then \mathcal{F}_ x is a free \mathcal{O}_{X, x}-module for all x \in X. This is enough by Algebra, Lemma 10.78.2 and the fact that \mathcal{F} is coherent. If \dim (\mathcal{O}_{X, x}) = 0, then \mathcal{O}_{X, x} is a field and the statement is clear. If \dim (\mathcal{O}_{X, x}) = 1, then \mathcal{O}_{X, x} is a discrete valuation ring (Algebra, Lemma 10.119.7) and \mathcal{F}_ x is torsion free. Hence \mathcal{F}_ x is free by More on Algebra, Lemma 15.22.11. If \dim (\mathcal{O}_{X, x}) = 2, then \mathcal{O}_{X, x} is a regular local ring of dimension 2. By More on Algebra, Lemma 15.23.18 we see that \mathcal{F}_ x has depth \geq 2. Hence \mathcal{F} is free by Algebra, Lemma 10.106.6. \square
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