Lemma 15.23.3. Let $R$ be a discrete valuation ring and let $M$ be a finite $R$-module. Then the map $j : M \to \mathop{\mathrm{Hom}}\nolimits _ R(\mathop{\mathrm{Hom}}\nolimits _ R(M, R), R)$ is surjective.

**Proof.**
Let $M_{tors} \subset M$ be the torsion submodule. Then we have $\mathop{\mathrm{Hom}}\nolimits _ R(M, R) = \mathop{\mathrm{Hom}}\nolimits _ R(M/M_{tors}, R)$ (holds over any domain). Hence we may assume that $M$ is torsion free. Then $M$ is free by Lemma 15.22.11 and the lemma is clear.
$\square$

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