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. $\square$
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