Constructing a projective resolution I am working on a problem that is asking me to compute Tor groups. I am trying to learn this material on my own, so I haven't had any formal education in this area.
Specifically, I am given the ring $R=\mathbb{Z}[X]/(X^n)$, an $R$-module $M$, and asked to calculate the groups $Tor_i^R(M, (x^m))$ where $x=X+(X^n)$ is the image of $X$ in $R$ and $0 \leq m \leq n-1$.
I know that in order to calculate these groups, I first need to construct a projective resolution for $(x^m)$ as an $R$-module, but I'm not sure where to begin. I am familiar with the proof that every module has a projective (in fact, free) resolution, but how do I actually construct the projective (free) modules and the maps between them? That is, how do I know what the canonical free modules look like concretely in this example?
Any help would be greatly appreciated.
 A: Let $A$ be your ring, and $I$ your ideal, where $m$ is as in your post. Begin by noting there is an epimorphism $A\to I$ obtained by multiplication by $x^m$. Its kernel is the ideal $(x^{n-m})$, and you can cover this by $A$ again by multiplication by $x^{n-m}$. You thus obtain a periodic resolution
$$ \cdots\longrightarrow A\longrightarrow \cdots\stackrel{x^{n-m}}\longrightarrow A\stackrel{x^m}\longrightarrow I\to 0$$
from which you can compute Tor groups easily now, in particular you need only compute the first two Tor groups. Concretely, let $M$ be an $A$-module to the left. Tensoring the resolution above with $M$ one obtains 
$$ \cdots\longrightarrow M\longrightarrow \cdots\stackrel{x^m}\longrightarrow M\stackrel{x^{m-n}}\longrightarrow M\to 0$$
where the maps are also multiplication by the indicated term. This gives that
$$\operatorname{Tor}_i^A(I,M) = 
\begin{cases}
 \dfrac{\ker\{x^m : M \to M\} }{ x^{n-m}   M  }
  & \text{ if $i\neq 0$ is even}\\ 
\dfrac{\ker\{x^{n-m} : M \to M\} }{ x^mM}
  & \text{ if $i$ is odd}\\
 \dfrac{M }{x^{n-m} M}
  & \text{ if $i=0$ }
\end{cases}$$
