Lemma 15.70.7. Let $R$ be a ring. Let $I \subset R$ be an ideal. If $M$ is a finite, $I$-projective $R$-module, then $M^\vee = \mathop{\mathrm{Hom}}\nolimits _ R(M, R)$ is $I$-projective.

**Proof.**
Assume $M$ is finite and $I$-projective. Choose a short exact sequence $0 \to K \to R^{\oplus r} \to M \to 0$. This produces an injection $M^\vee \to R^{\oplus r} = (R^{\oplus r})^\vee $. Since the extension class in $\mathop{\mathrm{Ext}}\nolimits ^1_ R(M, K)$ corresponding to the short exact sequence is annihilated by $I$, we see that for any $a \in I$ we can find a map $M \to R^{\oplus r}$ such that the composition with the given map $R^{\oplus r} \to M$ is equal to $a : M \to M$. Taking duals we find that $a : M^\vee \to M^\vee $ factors through the map $M^\vee \to R^{\oplus r}$ given above and we conclude.
$\square$

## 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Your email address will not be published.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.

## Comments (0)