Lemma 15.70.6. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let

$0 \to K \to P \to M \to 0$

be a short exact sequence of $R$-modules. If $M$ is $I$-projective and $P$ is projective, then $K$ is $I$-projective.

Proof. The element $\text{id}_ K \in \mathop{\mathrm{Hom}}\nolimits _ R(K, K)$ maps to the class of the given extension in $\mathop{\mathrm{Ext}}\nolimits ^1_ R(M, K)$. Since by assumption this class is annihilated by any $a \in I$ we see that $a : K \to K$ factors through $K \to P$ and we conclude. $\square$

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