Cohomology module

The nth cohomology H^n(C) = ker(d^n)/im(d^{n-1}) of a cochain complex of modules.

Cohomology module

Let $R$ be a ring and let

\cdots \xrightarrow{d^{n-1}} C^n \xrightarrow{d^{n}} C^{n+1} \xrightarrow{d^{n+1}} \cdots

be a cochain complex of R-modules , i.e. $d^{n+1}\circ d^{n}=0$ for all $n$ .

Definition

The $n$ th cocycles and $n$ th coboundaries are

Z^n(C^\bullet) := \ker(d^{n})\subseteq C^n, \qquad B^n(C^\bullet) := \operatorname{im}(d^{n-1})\subseteq C^n.

Since $d^{n}\circ d^{n-1}=0$ , one has $B^n(C^\bullet)\subseteq Z^n(C^\bullet)$ . The $n$ th cohomology module is

H^n(C^\bullet) := Z^n(C^\bullet) / B^n(C^\bullet).

A cochain complex is exact (as a sequence of modules) iff all its cohomology modules vanish; see exact complex .

Let $C^0=\mathbb Z$ , $C^1=\mathbb Z$ , $d^0=\times n$ , and $C^i=0$ otherwise. Then

H^0(C^\bullet)=\ker(\times n)=0,\qquad H^1(C^\bullet)=\operatorname{coker}(\times n)\cong \mathbb Z/n\mathbb Z.

Let $R=\mathbb Z$ . A projective resolution of $\mathbb Z/n\mathbb Z$ is

0\to \mathbb Z \xrightarrow{\times n} \mathbb Z \to \mathbb Z/n\mathbb Z \to 0.

Apply Hom $(-,\mathbb Z)$ to get a cochain complex

0 \to \operatorname{Hom}(\mathbb Z,\mathbb Z) \xrightarrow{\times n} \operatorname{Hom}(\mathbb Z,\mathbb Z)\to 0,

which identifies with

0\to \mathbb Z \xrightarrow{\times n} \mathbb Z \to 0.

Thus

Ext^1 _{\mathbb Z}(\mathbb Z/n\mathbb Z,\mathbb Z) \;\cong\; H^1(\operatorname{Hom}(P_\bullet,\mathbb Z)) \;\cong\; \mathbb Z/n\mathbb Z.

If $k$ is a field and $V,W$ are $k$ -vector spaces, then every $k$ -module is projective and injective . Hence

Ext^n _k(V,W)=0\quad \text{for all }n>0,

because one may take a length-0 projective (or injective) resolution and the resulting cochain complex has zero cohomology in positive degrees.