Lemma 15.23.3 (0B36)—The Stacks project

Table of contents
Part 1: Preliminaries
Chapter 15: More on Algebra
Section 15.23: Reflexive modules
Lemma 15.23.3 (cite)

Lemma 15.23.3. Let $R$ be a discrete valuation ring and let $M$ be a finite $R$-module. Then the map $j : M \to \mathop{\mathrm{Hom}}\nolimits _ R(\mathop{\mathrm{Hom}}\nolimits _ R(M, R), R)$ is surjective.

Proof. Let $M_{tors} \subset M$ be the torsion submodule. Then we have $\mathop{\mathrm{Hom}}\nolimits _ R(M, R) = \mathop{\mathrm{Hom}}\nolimits _ R(M/M_{tors}, R)$ (holds over any domain). Hence we may assume that $M$ is torsion free. Then $M$ is free by Lemma 15.22.11 and the lemma is clear. $\square$

Comments (2)

Comment #3601 by Remy on October 01, 2018 at 01:35

I think this is probably true over any ring $R$ . Indeed, if $R^m \twoheadrightarrow M$ is a surjection, then naturality of the evaluation map $\operatorname{ev} \colon M \to M^{\ast\ast}$ shows that the diagram $\begin{array}{ccc}R^n & \stackrel{\operatorname{ev}}\to & R^n \\ \downarrow & & \downarrow \\ M & \stackrel{\operatorname{ev}}\to & M^{\ast\ast}\end{array}$ commutes. Since the top map is an isomorphism and the right vertical map is surjective, we conclude that the bottom map is surjective.

Comment #3602 by Remy on October 01, 2018 at 01:37

Oh I see, I don't know why the right map is surjective.

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