# Free presentations of $\mathbb{Z}G$-modules

Dear All, I have a doubt about a specific definition, but I cannot find any help on the web or on the books that I have. Talking about $\mathbb{Z}G$-modules, what does one intend saying "take a free resolution $K\to F\to A$ of the module $A$"? First of all, it is clear that every module is the quotient of a free module but, can we say that every module is the quotient of a free module over another free module (i.e. is $K$ free as well?)? Also, it's not clear to me if $K\to F\to A$ has to be intended like a short exact sequence (but in this case I would have written $0\to K\to F\to A\to 0$) or not. The paper where I've found that is the Bieri and Eckmann's one "Groups with homological duality generalizing Poincaré duality", Invent. Math., 20, 103-124, (1973).

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On page 109 of Bieri–Eckmann (1973), the follow appears:

For any $G$-module $A$ and $G$-free presentation $K \rightarrowtail F \twoheadrightarrow A$ the exact coefficient sequence yields $$H^k(G;A) \cong H^{k+1}(G;K)$$

The context is that $H^k(G;F) = H^{k+1}(G;F) = 0$.

This follows whenever $0 \to K \to F \to A \to 0$ is exact and $F$ is $G$-free (so that the context holds). In particular, there is no need for $K$ to be free, and the sequence is short exact.

You can tell this because (1) the $\rightarrowtail$ has a weird tail meaning monomorphism and (2) the projective dimension of a module is not bounded by 2 (I believe take $G$ to be any polycyclic group of Hirsch Length 3, such as $G=\mathbb{Z}^3$).

Be careful that a free presentation $K \to F \twoheadrightarrow A$ may mean both $K$ and $F$ are free, $F \twoheadrightarrow A$ is surjective, and the sequence is exact. In other words, $K$ is not the kernel, but rather a free group projecting onto the kernel. This is not the meaning in this paper, but I believe is commonplace when talking about the relations module.

Bieri, Robert; Eckmann, Beno. "Groups with homological duality generalizing Poincaré duality." Invent. Math. 20 (1973), 103–124. MR340449 DOI:10.1007/BF01404060

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Thanks a lot: very clear explanation! –  fatoddsun Jun 2 '11 at 16:42