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My question is similar to this one but I think it is different.

Suppose we are given an infinitely generated free abelian group, which forms a $\mathbb{Z}_{2}$-graded chain complex, such that its homology is finitely generated. Does it then follow from homological algebra, that the Euler characteristic of this homology group does not depend upon the coefficients? Or perhaps there is a counterexample?

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Yes, that's true, by the universal coefficient theorem and the fact that for any finitely generated abelian group $A$ and any prime $p$, we have $\dim_{\mathbb Q} A\otimes_{\mathbb Z}{\mathbb Q} = \text{rank}_{\mathbb Z}\ A = \text{dim}_{{\mathbb F}_p}\ A\otimes_{\mathbb Z} {\mathbb Z}/p{\mathbb Z} - \text{dim}_{{\mathbb F}_p}\text{Tor}^1_{\mathbb Z}(A,{\mathbb Z}/p{\mathbb Z})$ (by the classification of f.g. abelian groups, for example). The fact that the complex is $2$-periodic unbounded does not cause trouble.

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  • $\begingroup$ Thanks! Sorry, I got very busy and lost track of this question. Unfortunately I am not very familiar with homological algebra so terms you use in the answer are not all familiar to me (as well as the context). In particular, I only know the universal coefficient theorem in its most primitive version - would you be able to at least add a reference to the place where it is stated in this generality? $\endgroup$
    – Fyodor
    Jan 5, 2015 at 17:12

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