# Help to compute Tor [closed]

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_addictedMar 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

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 $n$.
You need the rightmost memeber of your free resolution to be $\mathbb{Z}_2$ – Thomas Andrews Apr 9 '11 at 20:09
@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