Lemma 15.23.5 (0EB8)—The Stacks project

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

\[ \mathop{\mathrm{Hom}}\nolimits _ R(M, R) \leftarrow \mathop{\mathrm{Hom}}\nolimits _ R(M', R) \leftarrow \mathop{\mathrm{Hom}}\nolimits _ R(M'', R) \]

Dualizing again we obtain a commutative diagram

\[ \xymatrix{ \mathop{\mathrm{Hom}}\nolimits _ R(\mathop{\mathrm{Hom}}\nolimits _ R(M, R), R) \ar[r]_ j & \mathop{\mathrm{Hom}}\nolimits _ R(\mathop{\mathrm{Hom}}\nolimits _ R(M', R), R) \ar[r] & \mathop{\mathrm{Hom}}\nolimits _ R(\mathop{\mathrm{Hom}}\nolimits _ R(M'', R), R) \\ M \ar[u] \ar[r] & M' \ar[u] \ar[r] & M'' \ar[u] } \]

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

\[ \mathop{\mathrm{Hom}}\nolimits _ K(M'\otimes _ R K,K) \to \mathop{\mathrm{Hom}}\nolimits _ K(M\otimes _ R K,K) \to Q\otimes _ R K\to 0 \]

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

\[ 0 \to \mathop{\mathrm{Hom}}\nolimits _ R(Q,R) \to \mathop{\mathrm{Hom}}\nolimits _ R(\mathop{\mathrm{Hom}}\nolimits _ R(M,R),R) \to \mathop{\mathrm{Hom}}\nolimits _ R(\mathop{\mathrm{Hom}}\nolimits _ R(M',R),R) \]

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

Comments (2)

Comment #3620 by JuanPablo on October 11, 2018 at 18:36

It is not clear to me that $\text{Hom}_R(\text{Hom}_R(M,R),R)\to\text{Hom}_R(\text{Hom}_R(M',R),R)$ is injective, which seems to be required for the diagram chase. This is because in the exact sequence $\text{Hom}_R(M',R)\to \text{Hom}_R(M,R)\to Q\to 0$ one applies Lemma 0583 to obtain $\text{Hom}_K(M'\otimes_R K,K)\to \text{Hom}_K(M\otimes_R K,K)\to Q\otimes_R K\to 0$ . But $M'\otimes_R K\to M\otimes_R K$ is split injective so $Q\otimes_R K=0$ , that is, $Q$ is torsion. Then one gets the exact sequence $0\to \text{Hom}_R(Q,R)\to \text{Hom}_R(\text{Hom}_R(M,R),R)\to\text{Hom}_R(\text{Hom}_R(M',R),R)$ and $\text{Hom}_R(Q,R)=0$ because $Q$ is torsion.

Comment #3723 by Johan on October 23, 2018 at 13:37

Welcome back. Very good, I added what you said. Thanks! See changes here.

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