Reflexive surface modules and codimension-one lattice extension

At a point $x\in X$ with $\dim {\mathcal{O}}_{X,x}=2$ the local ring $A={\mathcal{O}}_{X,x}$ is a regular local ring of dimension two, hence a domain with depth two; the localizations of $A$ are regular by [F3], and those of dimension at most one are fields or discrete valuation rings in which finite torsion-free modules are free.

For the second assertion, let $A$ be the given normal Noetherian domain, $K$ its fraction field, and $v\in M^{\ast \ast }{\otimes }_{A}K$. By [F10], $A$ satisfies $(S_2)$, so [F7] applies. For every $\phi \in M^{\ast }={\mathrm{Hom}}_A(M,A)$, evaluation of $v$ on $\phi$ belongs to each $A_{\mathfrak{p}}$ of height one, because $v\in (M^{\ast \ast }{)}_{\mathfrak{p}}$. It therefore lies in $A$ by [F7]. Thus $\phi \mapsto v(\phi )$ is an $A$-linear map $M^{\ast }\to A$, that is, an element of $M^{\ast \ast }$ with generic value $v$. This also treats a field, using the empty-intersection convention.

Let $N$ be a finite $A$-module with a finite presentation $A^b\to A^a\to N\to 0$ for $A={\mathcal{O}}_{X,x}$. Dualizing gives an exact sequence $0\to N^{\ast }\to A^a\to A^b$ whose image $I$ is torsion-free. If $I=0$, then $N^{\ast }=A^a$ is free, and the asserted depth bound holds (with the zero module treated separately). Otherwise $I$ is a nonzero finite module over the domain $A$, hence has depth at least one by the depth-zero criterion for torsion-free modules over a domain; applying the long exact sequence of Ext groups from the residue field to $0\to N^{\ast }\to A^a\to I\to 0$ gives $\mathrm{depth}{N}^{\ast }\ge 2$.

If $N^{\ast \ast }=0$, it is already free. Otherwise applying the same computation to the finite module $N^{\ast }$ shows $\mathrm{depth}{N}^{\ast \ast }\ge 2$. Since $A$ is regular, every finite module over it has finite projective dimension by the Auslander--Buchsbaum--Serre criterion, so the Auslander--Buchsbaum formula gives $\mathrm{pd}{N}^{\ast \ast }=2-\mathrm{depth}{N}^{\ast \ast }=0$; a finite module of projective dimension zero over a local ring is free, so $N^{\ast \ast }$ is free.

A coherent reflexive module equals its double dual, so on an affine cover of $X$ the module of sections of $M$ is isomorphic to its double dual and is free at every point of local dimension two by step 3.1, free at points of local dimension one because their local rings are discrete valuation rings and the module is torsion-free, and free at points of local dimension zero because their local rings are fields. Over a DVR, a finite torsion-free module is free: in a nonzero relation between a minimal generating family divide the coefficients by their common lowest uniformizer power and cancel that power by torsion-freeness; one coefficient is a unit, contradicting minimality. Hence $M$ is locally free.

Work locally on an integral component of the regular surface and put $N=M\mathrm{/}\mathrm{tors}(M)$. A map from a torsion module to the domain $A$ vanishes, so $N^{\ast }=M^{\ast }$ and $N^{\ast \ast }=M^{\ast \ast }$. At every height-one point $N$ is finite torsion-free over a DVR and hence free; there $N\to N^{\ast \ast }$ is an isomorphism. The surjection ${\wedge }^{r}M\to {\wedge }^{r}N$ has torsion kernel, since it becomes an isomorphism over $K$, so its double dual is an isomorphism. The map ${\wedge }^{r}N\to {\wedge }^r{N}^{\ast \ast }$ is also generically an isomorphism and an isomorphism at height one. Applying step 1.2 to these two finite modules identifies their double duals inside their common generic exterior power. By steps 3.1 and 4.1, $N^{\ast \ast }=M^{\ast \ast }$ is locally free; hence ${\wedge }^r{M}^{\ast \ast }$ is already an invertible sheaf. The canonical identifications glue, giving $({\wedge }^{r}M{)}^{\ast \ast }\cong \det (M^{\ast \ast })$. For $r=0$, both sides are ${\mathcal{O}}_X$.

The Axiom of Choice and the Axiom of Dependent Choice are inherited from the cited commutative-algebra and Ext suppliers; no further choice is used.