Cochain complex - Imported Knowlpedia Content

Let $R$ be a ring and let $\{C^n\}_{n\in\mathbb Z}$ be R-modules. A cochain complex $(C^\bullet, d)$ is a collection of $R$ -linear maps

d^n : C^n \longrightarrow C^{n+1}\qquad (n\in\mathbb Z)

such that

d^{n+1}\circ d^n = 0\quad\text{for all }n.

Its cohomology modules are

H^n(C^\bullet)=\ker(d^n)/\operatorname{im}(d^{n-1}),

Remarks

Examples

Cochain complex concentrated in degree 0. For an $R$ -module $M$ , the diagram $\cdots \to 0 \to M \to 0 \to \cdots$ with $M$ in degree $0$ is a cochain complex. Then $H^0\cong M$ and $H^n=0$ for $n\neq 0$ .

Hom of a chain complex is a cochain complex. If $(C_\bullet,d)$ is a chain complex and $M$ is an $R$ -module, define $\operatorname{Hom}_R(C_\bullet,M)^n := \operatorname{Hom}_R(C_n,M),$ and set $\delta^n(\varphi) := \varphi\circ d_{n+1} \in \operatorname{Hom}_R(C_{n+1},M).$ Then $\delta^{n+1}\circ \delta^n=0$ because $d\circ d=0$ , so this is a cochain complex. This construction underlies the computation of Ext from a projective resolution.

“Multiplication by $x$ ” as a cochain complex. For $x\in R$ , the 2-term cochain complex $0 \to R \xrightarrow{\,\cdot x\,} R \to 0$ (degrees $0\to 1$ ) has $H^0 \cong \operatorname{Ann}_R(x),\qquad H^1 \cong R/xR.$