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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?
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.
