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. $\square$

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.)

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).