The Stacks project

Proof. We prove this by induction on $d$. If $d = 0$, then $M$ is projective. In this case there is a splitting $F_0 = \mathop{\mathrm{Ker}}(F_0 \to M) \oplus M$, and hence $\mathop{\mathrm{Ker}}(F_0 \to M)$ is projective. This finishes the proof if $e = 0$, and if $e > 0$, then replacing $M$ by $\mathop{\mathrm{Ker}}(F_0 \to M)$ we decrease $e$.

Next assume $d > 0$. Let $0 \to P_ d \to P_{d-1} \to \ldots \to P_0 \to M \to 0$ be a minimal length finite resolution with $P_ i$ projective. According to Schanuel's Lemma 10.109.1 we have $P_0 \oplus \mathop{\mathrm{Ker}}(F_0 \to M) \cong F_0 \oplus \mathop{\mathrm{Ker}}(P_0 \to M)$. This proves the case $d = 1$, $e = 0$, because then the right hand side is $F_0 \oplus P_1$ which is projective. Hence now we may assume $e > 0$. The module $F_0 \oplus \mathop{\mathrm{Ker}}(P_0 \to M)$ has the finite projective resolution

of length $d - 1$. By induction applied to the exact sequence

of length $e - 1$ we conclude $\mathop{\mathrm{Ker}}(F_ e \to F_{e - 1})$ is projective (if $e \geq 2$) or that $\mathop{\mathrm{Ker}}(F_1 \oplus P_0 \to F_0 \oplus P_0)$ is projective. This implies the lemma. $\square$