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It is known that if a group $G$ is of type $F_2$ (finitely presented) and of type $FP_n(\mathbb{Z})$, then $G$ is of type $F_n$.

However, is this true also for other rings which are not $\mathbb{Z}$? Or are there conditions to have that $F_2$ and $FP_n(R)$ imply $F_n$?

I am interested in the case of $\mathbb{Q}$. Are there conditions such that if a group is of type $F_2$ and of type $FP_n(\mathbb{Q})$, then it is of type $F_n$?

Thanks in advance.

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    $\begingroup$ You could use Bestvina-Brady groups to build a group that's of type $F_2$ and type $FP_3(\mathbb{Q})$ but not type $FP_3(\mathbb{Z})$ (hence not type $F_3$) if you can build a finite flag complex that's simply connected and has non-trivial finite second integral homology. I don't know enough algebraic topology examples to think of such a thing off the top of my head though. $\endgroup$
    – Matt Zaremsky
    Commented May 25, 2022 at 19:34

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The paper https://link.springer.com/article/10.1007/BF02804017 shows there are $\mathbb Q$-acyclic $k$-dimensional simplicial complexes with a complete $k-1$-skeleton and $\binom{n-1}{k}$ $k$-simplices which have nonzero finite $k-1$-homology and $0$ in $k$-homology. If $k\geq 3$ this will give you a simply connected simplicial complex with the desired property but it is not flag so take the barycentric subdivision to get a homeomorphic flag complex. Thus you use Bestvina-Brady as @MattZaremsky suggests to get finitely presented groups that are $FP(\mathbb Q)$ and $F_{k-1}$ but not $F_k$ for $k\geq 3$.

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