The Stacks project

Lemma 15.23.8. Let $R$ be a Noetherian domain. Let $M$ be a finite $R$-module. Let $N$ be a finite reflexive $R$-module. Then $\mathop{\mathrm{Hom}}\nolimits _ R(M, N)$ is reflexive.

Proof. Choose a presentation $R^{\oplus m} \to R^{\oplus n} \to M \to 0$. Then we obtain

\[ 0 \to \mathop{\mathrm{Hom}}\nolimits _ R(M, N) \to N^{\oplus n} \to N' \to 0 \]

with $N' = \mathop{\mathrm{Im}}(N^{\oplus n} \to N^{\oplus m})$ torsion free. We conclude by Lemma 15.23.5. $\square$


Comments (2)

Comment #4657 by Remy on

After the second sentence, couldn't you just conclude directly from Tag 15.23.5? (In fact, you don't need to take the image, because exactness on the right would not be needed.)

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Your email address will not be published.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 0AV3. Beware of the difference between the letter 'O' and the digit '0'.