
Lemma 10.108.3. Let $R$ be a ring. Suppose that $M$ is an $R$-module of projective dimension $d$. Suppose that $F_ e \to F_{e-1} \to \ldots \to F_0 \to M \to 0$ is exact with $F_ i$ projective and $e \geq d - 1$. Then the kernel of $F_ e \to F_{e-1}$ is projective (or the kernel of $F_0 \to M$ is projective in case $e = 0$).

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 10.108.1 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

$0 \to P_ d \to P_{d-1} \to \ldots \to P_2 \to P_1 \oplus F_0 \to \mathop{\mathrm{Ker}}(P_0 \to M) \oplus F_0 \to 0$

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

$F_ e \to F_{e-1} \to \ldots \to F_2 \to P_0 \oplus F_1 \to P_0 \oplus \mathop{\mathrm{Ker}}(F_0 \to M) \to 0$

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. $\square$

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