Lemma 15.23.5. Let $R$ be a Noetherian domain. Let $0 \to M \to M' \to M''$ an exact sequence of finite $R$-modules. If $M'$ is reflexive and $M''$ is torsion free, then $M$ is reflexive.

**Proof.**
We will use without further mention that $\mathop{\mathrm{Hom}}\nolimits _ R(N, N')$ is a finite $R$-module for any finite $R$-modules $N$ and $N'$, see Algebra, Lemma 10.71.9. We take duals to get a sequence

Dualizing again we obtain a commutative diagram

We do not know the top row is exact. But, by assumption the middle vertical arrow is an isomorphism and the right vertical arrow is injective (Lemma 15.23.2). We claim $j$ is injective. Assuming the claim a diagram chase shows that the left vertical arrow is an isomorphism, i.e., $M$ is reflexive.

Proof of the claim. Consider the exact sequence $\mathop{\mathrm{Hom}}\nolimits _ R(M',R)\to \mathop{\mathrm{Hom}}\nolimits _ R(M,R)\to Q \to 0$ defining $Q$. One applies Algebra, Lemma 10.10.2 to obtain

But $M \otimes _ R K \to M' \otimes _ R K$ is an injective map of vector spaces, hence split injective, so $Q \otimes _ R K = 0$, that is, $Q$ is torsion. Then one gets the exact sequence

and $\mathop{\mathrm{Hom}}\nolimits _ R(Q,R)=0$ because $Q$ is torsion. $\square$

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