Lemma 15.23.6. Let $R$ be a Noetherian domain. Let $M$ be a finite $R$-module. The following are equivalent

$M$ is reflexive,

there exists a short exact sequence $0 \to M \to F \to N \to 0$ with $F$ finite free and $N$ torsion free.

Lemma 15.23.6. Let $R$ be a Noetherian domain. Let $M$ be a finite $R$-module. The following are equivalent

$M$ is reflexive,

there exists a short exact sequence $0 \to M \to F \to N \to 0$ with $F$ finite free and $N$ torsion free.

**Proof.**
Observe that a finite free module is reflexive. By Lemma 15.23.5 we see that (2) implies (1). Assume $M$ is reflexive. Choose a presentation $R^{\oplus m} \to R^{\oplus n} \to \mathop{\mathrm{Hom}}\nolimits _ R(M, R) \to 0$. Dualizing we get an exact sequence

\[ 0 \to \mathop{\mathrm{Hom}}\nolimits _ R(\mathop{\mathrm{Hom}}\nolimits _ R(M, R), R) \to R^{\oplus n} \to N \to 0 \]

with $N = \mathop{\mathrm{Im}}(R^{\oplus n} \to R^{\oplus m})$ a torsion free module. As $M = \mathop{\mathrm{Hom}}\nolimits _ R(\mathop{\mathrm{Hom}}\nolimits _ R(M, R), R)$ we get an exact sequence as in (2). $\square$

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