exact sequences for modules with finite projective dimension

Proposition. Let $R$ be a ring and $M$ be a (left) $R$-module, such that . If

$$0\to K\to P_n\to \mathrm{\cdots }\to P_0\to M\to 0$$

is an exact sequence of $R$-modules, such that each $P_i$ is projective, then $K$ is projective.

Proof. Since , then there exists exact sequence of $R$-modules

$$0\to P_n^{\prime }\to \mathrm{\cdots }\to P_0^{\prime }\to M\to 0,$$

Note that sequences

$$P_n\to \mathrm{\cdots }\to P_0\to M\to 0;$$

$$P_n^{\prime }\to \mathrm{\cdots }\to P_0\to M\to 0,$$

are projective resolutions of $M$. Let $\delta :P_n\to P_{n-1}$ and $\beta :P_n^{\prime }\to P_{n-1}^{\prime }$ be maps take from these resolutions. Then generalized Schanuel’s lemma implies that $\ker \delta$ and $\ker \beta$ are projectively equivalent. But $\ker \delta \simeq K$ and $\ker \beta =0$. This means, that there are projective modules $P,Q$ such that

$$K\oplus P\simeq Q.$$

Therefore $K$ is a direct summand of a free module (since $Q$ is), which completes the proof. $\mathrm{\square }$