Lemma 10.109.6. Let $R$ be a Noetherian ring. Let $M$ be a finite $R$-module. Let $d \geq 0$. The equivalent conditions (1) – (4) of Lemma 10.109.4 are also equivalent to

1. there exists a resolution $0 \to P_ d \to P_{d - 1} \to \ldots \to P_0 \to M \to 0$ with $P_ i$ finite projective.

Proof. Choose a resolution $\ldots \to F_2 \to F_1 \to F_0 \to M \to 0$ with $F_ i$ finite free (Lemma 10.71.1). By Lemma 10.109.4 we see that $P_ d = \mathop{\mathrm{Ker}}(F_{d - 1} \to F_{d - 2})$ is projective at least if $d \geq 2$. Then $P_ d$ is a finite $R$-module as $R$ is Noetherian and $P_ d \subset F_{d - 1}$ which is finite free. Whence $0 \to P_ d \to F_{d - 1} \to \ldots \to F_1 \to F_0 \to M \to 0$ is the desired resolution. $\square$

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