Let $A$ be a k-algebra,where k is a fixed field. We denote by $\mathfrak{D}^b(A-mod)$ the bounded derived A-module category. A complex $Z^{\bullet}=(Z^i,d^i) \in \mathfrak{D}^b(A-mod)$ such that all $Z^i$ are finitely generated projective is quasi-isomorphic to the following complex $$ \cdots \rightarrow 0 \rightarrow Coker(d^{-n-1}) \rightarrow Z^{-n+1} \rightarrow \cdots \rightarrow Z^{-2} \rightarrow Z^{-1} \rightarrow Ker(d^0) \rightarrow 0 \rightarrow \cdots $$ So we can see $H^i(Z^{\bullet})=0$ for $i <-n$ or $i >0 $.

Then how can I get that $Coker(d^{-n-1})$ is finitely presented module? And how to get the projective dimension of $Coker(d^{-n-1})$ is finite?


Well, how do we conclude that our complex $Z^\bullet$ is quasi-isomorfic to the complex

$$ \cdots \longrightarrow 0 \longrightarrow Coker(d^{-n-1}) \longrightarrow Z^{-n+1} \longrightarrow \cdots \longrightarrow Z^{m-1} \longrightarrow Ker(d^m) \longrightarrow 0 \longrightarrow \cdots \ ? $$

We use the two truncations, which are quasi-isomorphisms, due to the vanishing of cohomology with sufficiently small and sufficiently large numbers: $Z^\bullet \longleftarrow \tau_{\le m}Z^\bullet \longrightarrow \tau_{\ge -n}\tau_{\le m}Z^\bullet$.

We can use the stupid truncation $\sigma_{\le -n}$ instead to get the complex

$$ \cdots \longrightarrow Z^{-n-2} \longrightarrow Z^{-n-1} \longrightarrow Z^{-n} \longrightarrow 0 . $$

Its only cohomology lies in degree $-n$ and equals to $Coker(d^{-n-1})$, so we obtained a finite resolution of $Coker(d^{-n-1})$ by finitely generated projective modules:

$$ \cdots \longrightarrow Z^{-n-2} \longrightarrow Z^{-n-1} \longrightarrow Z^{-n} \longrightarrow Coker(d^{-n-1}) \longrightarrow 0 . $$

Therefore $Coker(d^{-n-1})$ is of finite projective dimension. $Z^{-n}$ and $Z^{-n-1}$ are finitely generated projective modules, so there are such (finitely generated projective) modules $P_0$ and $P_1$ and finitely generated free modules $F_0$ and $F_1$, that $Z^{-n} \oplus P_0 = F_0$ and $Z^{-n-1} \oplus P_0 \oplus P_1 = F_1$. So we have a resolution of $Coker(d^{-n-1})$ of the form

$$ \cdots \longrightarrow Z^{-n-2} \oplus P_1 \longrightarrow F_1 \longrightarrow F_0 \longrightarrow Coker(d^{-n-1}) \longrightarrow 0 . $$

So $Coker(d^{-n-1})$ is finitely presented.

In fact, this looks like a standard exercise from a book on homological algebra, and it is usually more useful to do such an exercise by yourself than to post a question on MO.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.