Characteristic 0 case

Recall the computation of the homology of the cyclic group $\bZ/n\bZ$. Let $M$ be a module over $\bZ/n\bZ$, that is a module over the group ring $k[\bZ/ n\bZ]$ for some ring $k$. to compute $\rH_i(\bZ/n\bZ; M)$ one uses a complex

$$M\xleftarrow{1-t} M \xleftarrow{N} M \xleftarrow{1-t} M \xleftarrow{N} \hdots$$

When the ring $k$ is a field of characteristic 0, there is a homotopy from $\id$ to 0,

$$M\xrightarrow{h} M \xrightarrow{h'} M \xrightarrow{h} M \xrightarrow{h'} \hdots,$$

$$\begin{align*} h &:= -\inv{n}\sum_{i=1}^{n-1} it^i, \\ h' &:= \inv{n} \id, \\ & h(1-t)+Nh'= t^n =\id. \end{align*}$$
It proves that
$$\begin{align*} \rH_0(\bZ/n\bZ; M) &= M/ 1-t, \\ \rH_n(\bZ/n\bZ; M) &= 0, \quad n\geq 1. \end{align*}$$
Now instead of considering all bicomplex $C_{\bullet\bullet}$ we can take the reduced complex $C_{\bullet}^{\la}$ which is defined as a cokernel of the map $(1-t)$ between first and zeroth column of $C_{\bullet\bullet}$

$$\xymatrix{ \vdots \ar[d] & \vdots \\ C_3/(1-t) \ar[d]_{b} &... ...ar[d]_{b} & 0 \\ C_1/(1-t) \ar[d]_{b} & 0 \\ C_0/(1-t) & 0 }$$

If $C_n=A^{\ox(n+1)}$, then $C_n^{\la}(A)=A^{\ox(n+1)}/(1-t)$ and we denote

$$\rH^{\la}_n(A):=\rH_n(C_{\bullet}^{\la})$$

As a corollary we have that if $k\supset \bQ$, then $\rH^{\la}(A)\isom \HC_n(A)$ and there exists an exact sequence

$$\hdots \to \HH_n(A) \xrightarrow{I} \rH^{\la}_n(A) \xrightar... ...S} \rH^{\la}_{n-2}(A) \xrightarrow{B} \HH_{n-1}(A) \to \hdots.$$

In the case of characteristic not equal 0 the maps are still defined, but the sequence is not exact.