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Let $G$ be group and let $M$ be a $G$-module. The module $M$ is said to be of type $FP_n$ if there exists an exact seqeunce $$ P_n \rightarrow P_{n-1} \rightarrow \ldots \rightarrow P_1 \rightarrow P_0 \rightarrow M \rightarrow 0$$ where all the $P_i$ are finitely generated projective $G$-modules.

It is known that a $G$-module $M$ is of type $FP_n$ iff for every exact colimit, the natural map $$\mathrm{colim}\ \mathrm{Ext}^k_{\mathbb{Z}[G]}(M,N_{\ast}) \rightarrow \mathrm{Ext}^k_{\mathbb{Z}[G]}(M,\mathrm{colim}\ N_{\ast})$$ is an isomorphims for $k<n$ and a monomorphism for k=n.

Using this result one can show that a direct summand of a module of type $FP_n$ is also of type $FP_n$. My question is: is there a more explicit proof of this fact, which does not use the result mentioned above?

user68316
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1 Answers1

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I realise this is an old question, but as no one else has answered:

Suppose $M = A \oplus B$ and $M$ has type $FP_n$. By induction, assume that $A$ and $B$ have type $FP_{n-1}$ (the case $n=0$, finite generation, is clear). Take a finite type resolution $P$ of $A$ of length $n-1$ and a finite type resolution $Q$ of $M$ of length $n$. $f: A \to M$ extends to a map $g: P \to Q$, and the mapping cone of $g$ is a finite type resolution of $B$ of length $n$.

This avoids actually calculating homology, but proving that the mapping cone is a resolution of $B$ is still non-trivial. So whether this answers your question depends on what you want it for.

Ged Corob Cook
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