Exact sequences with two FL-modules

Question

Let $R$ be a ring. An $R$-module $M$ is called FL (FP) if it has a finite resolution consisiting of finitely generated free (projective) modules.

Given an exact sequence of $R$-modules, $0\to M_1\to M_2\to M_3\to 0$, if two of the modules $M_1$, $M_2$ or $M_3$ are FP, then the third is FP as well. This follows, for example, from Proposition 1.4 and Proposition 4.1b of

Bieri, Robert Homological dimension of discrete groups. Second edition. Queen Mary College Mathematics Notes. Queen Mary College, Department of Pure Mathematics, London, 1981.

My question is

If two of the modules $M_1$, $M_2$ or $M_3$ are FL, is it true that the third one is also FL?

Answer 1

Yes, this is true. See Bourbaki, N. Éléments de mathématique. Algèbre. Chapitre 10. Algèbre homologique. (French) [Elements of mathematics. Algebra. Chapter 10. Homological algebra], Theorem 3.9.1

Let $K_0(R)$ be the Grothendieck group of $R$ and $\widetilde K_0(R)=K_0(R)/\langle [R]\rangle$. Then $[P]=0$ in $\widetilde K_0(R)$ if and only if $P$ is stably free.

For any $FP$ $R$-module $M$ having the following resolution consisting of finitely generated projective $R$-modules $$0\to P_k\to \ldots \to P_1\to P_0\to M\to 0,$$ we denote $$\chi_u^R(M)=\sum_{i=0}^k (-1)^i [P_i]\in \widetilde K_0(R).$$ Observe that $\chi_u^R(M)=0$ if and only if $M$ is $FL$.

Thus, in order to prove the claim it is enough to show that if $0\to M_1\to M_2\to M_3\to 0$ is an exact sequence of $FP$ $R$-modules, then $\chi_u(M_2)=\chi_u(M_1)+\chi_u(M_3)$.

This can easily be done by induction on $pd_R(M_1)+pd_R(M_2)+ pd_R(M_3)$, where $pd_R(M)$ is the projective dimension of $M$.