Lemma 31.15.13. Let $X$ be an integral regular scheme of dimension $2$. Let $i : D \to X$ be the immersion of an effective Cartier divisor. Let $\mathcal{F} \to \mathcal{F}' \to i_*\mathcal{G} \to 0$ be an exact sequence of coherent $\mathcal{O}_ X$-modules. Assume

1. $\mathcal{F}, \mathcal{F}'$ are locally free of rank $r$ on a nonempty open of $X$,

2. $D$ is an integral scheme,

3. $\mathcal{G}$ is a finite locally free $\mathcal{O}_ D$-module of rank $s$.

Then $\mathcal{L} = (\wedge ^ r\mathcal{F})^{**}$ and $\mathcal{L}' = (\wedge ^ r \mathcal{F}')^{**}$ are invertible $\mathcal{O}_ X$-modules and $\mathcal{L}' \cong \mathcal{L}(k D)$ for some $k \in \{ 0, \ldots , \min (s, r)\}$.

Proof. The first statement follows from Lemma 31.12.15 as assumption (1) implies that $\mathcal{L}$ and $\mathcal{L}'$ have rank $1$. Taking $\wedge ^ r$ and double duals are functors, hence we obtain a canonical map $\sigma : \mathcal{L} \to \mathcal{L}'$ which is an isomorphism over the nonempty open of (1), hence nonzero. To finish the proof, it suffices to see that $\sigma$ viewed as a global section of $\mathcal{L}' \otimes \mathcal{L}^{\otimes -1}$ does not vanish at any codimension point of $X$, except at the generic point of $D$ and there with vanishing order at most $\min (s, r)$.

Translated into algebra, we arrive at the following problem: Let $(A, \mathfrak m, \kappa )$ be a discrete valuation ring with fraction field $K$. Let $M \to M' \to N \to 0$ be an exact sequence of finite $A$-modules with $\dim _ K(M \otimes K) = \dim _ K(M' \otimes K) = r$ and with $N \cong \kappa ^{\oplus s}$. Show that the induced map $L = \wedge ^ r(M)^{**} \to L' = \wedge ^ r(M')^{**}$ vanishes to order at most $\min (s, r)$. We will use the structure theorem for modules over $A$, see More on Algebra, Lemma 15.124.3 or 15.124.9. Dividing out a finite $A$-module by a torsion submodule does not change the double dual. Thus we may replace $M$ by $M/M_{tors}$ and $M'$ by $M'/\mathop{\mathrm{Im}}(M_{tors} \to M')$ and assume that $M$ is torsion free. Then $M \to M'$ is injective and $M'_{tors} \to N$ is injective. Hence we may replace $M'$ by $M'/M'_{tors}$ and $N$ by $N/M'_{tors}$. Thus we reduce to the case where $M$ and $M'$ are free of rank $r$ and $N \cong \kappa ^{\oplus s}$. In this case $\sigma$ is the determinant of $M \to M'$ and vanishes to order $s$ for example by Algebra, Lemma 10.121.7. $\square$

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