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

It is not clear to me that is injective, which seems to be required for the diagram chase. This is because in the exact sequence one applies Lemma 0583 to obtain . But is split injective so , that is, is torsion. Then one gets the exact sequence and because is torsion.

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