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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