# Module has finitely generated projective resolution

Let $M$ be a finitely generated module (over a local noetherian ring $(R,\mathfrak m))$ such that the projective dimension of $M$ is finite $(pd\ M=n<\infty)$. We know that

i) There is a free resolution $F_\bullet$ of $M$ in which the free $R$-modules $F_i$ are finitely generated,

ii) Every projective resolution of $M$ can be truncated to a projective resolution that have length $n$.

From these, can one show that there exists a projective resolution $$0\to P_n\to P_{n-1}\to\cdots\to P_0\to M$$ of $M$ for which $P_i$ are finitely generated for all $i\in\{1,\ldots, n\}$?

Thanks.

• You only need that the ring be noetherian and the module f.g. and of finite projective dimensión. Nov 16, 2013 at 1:47
• A f.g projective module be also free module on local ring. I found the answer. Thanks.
– Q.TL
Nov 16, 2013 at 2:11
• If you have solved the question, you should answer it and accept your own answer to take the question off the unanswered questions list. Nov 16, 2013 at 2:19
• The point of my comment, though, was that what you wanted does not need that the ring be local. Nov 16, 2013 at 3:08
• @ Bruno: I just wish to see that the claim of the question be true. For the detail answer, I can't see now.
– Q.TL
Nov 16, 2013 at 3:26

Let $$\cdots\to F_m\to F_{m-1}\to\cdots\to F_0\to M\to0$$ be a free resolution of $M$ with $F_i$ of finite rank for all $i\ge 0$. If one stops at step $n=pd\ M$ we have $$0\to K_{n-1}\to F_{n-1}\to\cdots\to F_0\to M\to0$$ and $K_{n-1}$ must be projective. Furthermore, since $F_n\to K_{n-1}\to0$ we have that $K_{n-1}$ is finitely generated, so it is free of finite rank.
• One can choose $K_{n-1}=\text{Im}(F_n\to F_{n-1})$ but I can't see why $K_{n-1}$ be an direct summand of $F_{n-1}$. And so why do we have $K_{n-1}$ be projective module?
• @Q.TL Since $pd\ M=n$, $K_{n-1}$ is certainly projective and this standard fact can be found in any basic textbook in Homological Algebra or can be shown, for example, by using the generalized Schanuel's Lemma.