Compute $\mathrm{Tor}_{n}^{\mathbb{Z}_{8}}(\mathbb{Z}_{4},\mathbb{Z}_{4})$ Let $\mathbb{Z}_{4}$ as a $\mathbb{Z}_{8}$-module. How can I prove 
$\mathrm{Tor}_{n}^{\mathbb{Z}_{8}}(\mathbb{Z}_{4},\mathbb{Z}_{4})=\mathbb{Z}_{4}$?
Please see this.
 A: $\def\ZZ{\mathbb Z}$The first thing to do is to construct a resolution of $\ZZ_4$.
There is an obvious map $p:\ZZ_8\to\ZZ_4$ mapping (the class of) $1$ to $1$, and the kernel is generated by $4$. Clearly, the kernel is isomorphic to $\ZZ_2$ as a $\ZZ_8$-module, so we have a short exact sequence of $\ZZ_8$-modules $$0\to\ZZ_2\xrightarrow{q}\ZZ_8\xrightarrow{p}\ZZ_4\to0$$ with $q$ mapping $1$ to $4$. 
In a similar way, we find a short exact sequence $$0\to\ZZ_4\xrightarrow{r}\ZZ_8\xrightarrow{s}\ZZ_2\to0$$ with $s$ mapping $1$ to $1$ and $r$ mapping $1$ to $2$.
Now consider the diagram $$\cdots \ZZ_8\xrightarrow{r\circ p}\ZZ_8\xrightarrow{q\circ s}\ZZ_8\xrightarrow{r\circ p}\ZZ_8\xrightarrow{q\circ s}\ZZ_8\xrightarrow{p}\ZZ_4$$


*

*Prove that this is a complex, and that it is exact. Since the modules (apart from $\ZZ_4$) are free $\ZZ_8$-modules, this is then a projective resolution of $\ZZ_4$.

*Now apply the functor $\ZZ_4\otimes_{\ZZ_8}(\mathord-)$ to the complex, and compute its homology.

*Rejoice.
