Lemma 15.23.15. Let $R$ be a Noetherian domain. Let $M$ be a finite $R$-module. The following are equivalent

1. $M$ is reflexive,

2. for every prime $\mathfrak p$ of $R$ one of the following happens

1. $M_\mathfrak p$ is a reflexive $R_\mathfrak p$-module, or

2. $\text{depth}(M_\mathfrak p) \geq 2$.

Proof. If (1) is true, then $M_\mathfrak p$ is a reflexive module for all primes of $\mathfrak p$ by Lemma 15.23.4. Thus (1) $\Rightarrow$ (2). Assume (2). Set $N = \mathop{\mathrm{Hom}}\nolimits _ R(\mathop{\mathrm{Hom}}\nolimits _ R(M, R), R)$ so that

$N_\mathfrak p = \mathop{\mathrm{Hom}}\nolimits _{R_\mathfrak p}( \mathop{\mathrm{Hom}}\nolimits _{R_\mathfrak p}(M_\mathfrak p, R_\mathfrak p), R_\mathfrak p)$

for every prime $\mathfrak p$ of $R$. See Algebra, Lemma 10.10.2. We apply Lemma 15.23.14 to the map $j : M \to N$. This is allowed because $M$ is finite and $N$ is torsion free by Lemma 15.22.12. In case (2)(a) the map $M_\mathfrak p \to N_\mathfrak p$ is an isomorphism and in case (2)(b) we have $\text{depth}(M_\mathfrak p) \geq 2$. $\square$

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