The Stacks project

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

I think this is probably true over any ring . Indeed, if is a surjection, then naturality of the evaluation map shows that the diagram 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

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

There are also:

  • 1 comment(s) on Section 15.23: Reflexive modules

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0B36. Beware of the difference between the letter 'O' and the digit '0'.



View Lemma 15.23.3 as pdf