length of minimal projective resolution and projective dimension

Question

Let $K$ be a field, $A$ be a $K$-algebra, and $M$ be a finite generated $A$-module. How do you show that the length of a minimal projective resolution of $M$ is the projective dimension of $M$? (This statement is from Elements of the representation theory of associative algebras. Volume 1 page 426, below the definition of injective dimension.)

5.7. Definition. An exact sequence $$\dots \rightarrow P_m \xrightarrow{h_m} P_{m-1} \rightarrow \dots \rightarrow P_1 \xrightarrow{h_1} P_0 \xrightarrow{h_0} M \rightarrow 0 $$ is called a minimal projective resolution of $M$ if $h_j : P_j \rightarrow \operatorname{Im} h_j$ is a projective cover for all $j \geq 1$ and $h_0$ is a projective cover.

My thought was that given an arbitrary projective resolution $$ 0 \rightarrow Q_n \xrightarrow{q_n} Q_{n-1} \rightarrow \dots \rightarrow Q_1 \xrightarrow{q_1} Q_0 \xrightarrow{q_0} M \rightarrow 0 $$ and a fixed minimal projective resolution $$ 0 \rightarrow P_m \xrightarrow{p_m} P_{m-1} \rightarrow \dots \rightarrow P_1 \xrightarrow{p_1} P_0 \xrightarrow{p_0} M \rightarrow 0, $$ if there exists an epimorphism $g_k: Q_k \rightarrow P_k$ for $k = 0,\dots ,m$, then $Q_k \neq 0$ for each $k$ therefore $m \leq n$, as desired. But I don't know how to prove the existence of $g_k$ when $k \geq 2$.

Answer 1

you only need to prove the following $Let 0\rightarrow Ker q_0\rightarrow Q_0\xrightarrow {q_0}M \rightarrow 0$ be short exact sequence.

let $P_0 \xrightarrow{p_0} M$ be projective cover.then there exists an epimorphism $\alpha :Q_0 \rightarrow P_0$ such that $p_0\alpha =q_0$. ($\alpha $ is epimorphism since $P_0$ is projective cover.)Hence $Ker q_0\cong Ker p_0\coprod P$ for a projective module $P$.So you can get the result.

Answer 2

This is my understanding of Sky's answer and I write it down in case anyone need an explanation.

We only need to show that if $\operatorname{Ker} q_n \cong \operatorname{Ker} p_n \oplus P^{(n)}$ for some projective module $P^{(n)}$, then $ \operatorname{Ker} q_{n+1} \cong \operatorname{Ker} p_{n+1} \oplus P^{(n+1)}$ for some projective module $P^{(n+1)}$.

Because $Q_{n+1} \xrightarrow{q_{n+1}} \operatorname{Ker}q_n$ is an epimorphism and $$P_{n+1} \oplus P^{(n)} \xrightarrow{p_{n+1}\oplus 1} \operatorname{Ker} p_n \oplus P^{(n)} \cong \operatorname{Ker}q_n$$ is a projective cover, by Lemma 17.17 from Rings and Categories of Modules, we have $Q_{n+1} \cong P_{n+1} \oplus P^{(n)} \oplus P^{(n+1)}$ for some projective module $P^{(n+1)}$ which is contained in $\operatorname{Ker}q_{n+1}$. An easy calculation shows that $\operatorname{Ker}q_{n+1} \cong \operatorname{Ker}(p_{n+1}\oplus 1) \oplus P^{(n+1)}$ and $\operatorname{Ker}(p_{n+1}\oplus 1) \cong \operatorname{Ker}p_{n+1}$.