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Consider $\mathbb{Z}_{2}$ as a $\mathbb{Z}_{4}$-module. How to compute $\mathrm{Tor}_{n}^{\mathbb{Z}_{4}}(\mathbb{Z}_{2},\mathbb{Z}_{2})$?

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closed as off-topic by user26857, Shailesh, Jonas, Antonios-Alexandros Robotis, choco_addicted Mar 15 at 1:22

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Indeed, this is one situation where "just apply the definition" works... – Mariano Suárez-Alvarez Apr 9 '11 at 20:07
up vote 7 down vote accepted

You need a free (or projective) resolution of $\mathbb{Z}_2$. One is $$\dots\to\mathbb{Z}_4\to\mathbb{Z}_4\to\mathbb{Z}_4\to\mathbb{Z}_4$$ where every arrow is multiplication by $2$. Now you tensor this complex (over $\mathbb{Z}_4$) with $\mathbb{Z}_2$ and you get $$\dots\to\mathbb{Z}_2\to\mathbb{Z}_2\to\mathbb{Z}_2\to\mathbb{Z}_2$$ where the arrows are now zero. Your Tor's are the cohomology of this complex. As a result, they are $\mathbb{Z}_2$ for every $n$.

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You need the rightmost memeber of your free resolution to be $\mathbb{Z}_2$ – Thomas Andrews Apr 9 '11 at 20:09
    
@user8268: thanks, I just didn't know where to start, thanks again. – user6495 Apr 9 '11 at 20:10
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@Thomas: (it may depend on conventions, but) there is no $\mathbb{Z}_2$ in the resolution; in principle it continues to the right, with zeroes. The resolution should be a non-positively graded complex (of free $\mathbb{Z}_4$-modules) with cohomology equal to $\mathbb{Z}_2$ (only in deg. $0$ - for other degrees the cohomology should vanish). (but as I said, it might depend on conventions, so I'm just explaining what I meant) – user8268 Apr 9 '11 at 20:21
    
Ah, I suppose that makes sense, but that requires that you define a project resolution in terms of cohomology, when you can instead define it "independent" of cohomology using just exact sequences. – Thomas Andrews Apr 9 '11 at 21:09
    
Hi my friend.Could you describe it more?I couldn't understand your answer. – Iman Bahmani Jul 25 '13 at 19:50

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