Computing $\mathrm{Tor}_{i}^{R}(M,(x^k))$, where $R=\mathbb{Z}[x]/(x^n)$.
Let $M$ be any $R$ module. $k$ and $n$ are positive integers. I think I have a valid free resolution for $(x^k)$ given by
$$ \cdots \overset{\cdot x^{n-k}}{\longrightarrow}R\overset{\cdot x^k}{\longrightarrow}R\overset{\cdot x^{n-k}}{\longrightarrow}R\overset{\cdot x^k}{\longrightarrow} (x^k)\longrightarrow0$$
We then tensor with $M$ to and take homology. This is where I am running into trouble. With the complex$$ \cdots \overset{1\otimes \cdot x^{n-k}}{\longrightarrow}M\overset{1\otimes\cdot x^k}{\longrightarrow}M\overset{1\otimes\cdot x^{n-k}}{\longrightarrow}M\longrightarrow0$$
I know I should get that $\mathrm{Tor}_{0}^{R}(M,(x^k))=M\otimes_R(x^k)$, but I am having trouble computing that directly. I get $M/(M\otimes (x^{n-k}))$ which I am not sure how to make sense of. Similarly, describing the kernels of these maps is eluding me. Thank you for your help.
As $(x^k)\cong R/(x^{n-k})$ as $R$ modules, then the tensor product $$M\otimes_R(x^k)\cong M\otimes_R R/(x^{n-k})\cong M/(x^{n-k}M).$$ The $\text{Tor}_0^R$ as you compute it is the cokernel of $\cdot x^{n-k}$ on $M$, that is $M/(x^{n-k}M)$, so it does all check out.