Let R R R ring
and let
⋯ → d n − 1 C n → d n C n + 1 → d n + 1 ⋯
\cdots \xrightarrow{d^{n-1}} C^n \xrightarrow{d^{n}} C^{n+1} \xrightarrow{d^{n+1}} \cdots
⋯ d n − 1 C n d n C n + 1 d n + 1 ⋯ be a cochain complex
of R-modules
, i.e. d n + 1 ∘ d n = 0 d^{n+1}\circ d^{n}=0 d n + 1 ∘ d n = 0 n n n
The n n n n n n
Z n ( C ∙ ) : = ker ( d n ) ⊆ C n , B n ( C ∙ ) : = im ( d n − 1 ) ⊆ C n .
Z^n(C^\bullet) := \ker(d^{n})\subseteq C^n,
\qquad
B^n(C^\bullet) := \operatorname{im}(d^{n-1})\subseteq C^n.
Z n ( C ∙ ) := ker ( d n ) ⊆ C n , B n ( C ∙ ) := im ( d n − 1 ) ⊆ C n . Since d n ∘ d n − 1 = 0 d^{n}\circ d^{n-1}=0 d n ∘ d n − 1 = 0 B n ( C ∙ ) ⊆ Z n ( C ∙ ) B^n(C^\bullet)\subseteq Z^n(C^\bullet) B n ( C ∙ ) ⊆ Z n ( C ∙ ) n n n
H n ( C ∙ ) : = Z n ( C ∙ ) / B n ( C ∙ ) .
H^n(C^\bullet) := Z^n(C^\bullet) / B^n(C^\bullet).
H n ( C ∙ ) := Z n ( C ∙ ) / B n ( C ∙ ) . A cochain complex is exact (as a sequence of modules) iff all its cohomology modules vanish; see exact complex
.
Examples Example 1: Two-term cochain complex over Z \mathbb Z Z Let C 0 = Z C^0=\mathbb Z C 0 = Z C 1 = Z C^1=\mathbb Z C 1 = Z d 0 = × n d^0=\times n d 0 = × n C i = 0 C^i=0 C i = 0
H 0 ( C ∙ ) = ker ( × n ) = 0 , H 1 ( C ∙ ) = coker ( × n ) ≅ Z / n Z .
H^0(C^\bullet)=\ker(\times n)=0,\qquad
H^1(C^\bullet)=\operatorname{coker}(\times n)\cong \mathbb Z/n\mathbb Z.
H 0 ( C ∙ ) = ker ( × n ) = 0 , H 1 ( C ∙ ) = coker ( × n ) ≅ Z / n Z . Example 2: E x t \mathrm{Ext} Ext Let R = Z R=\mathbb Z R = Z projective resolution
of Z / n Z \mathbb Z/n\mathbb Z Z / n Z
0 → Z → × n Z → Z / n Z → 0.
0\to \mathbb Z \xrightarrow{\times n} \mathbb Z \to \mathbb Z/n\mathbb Z \to 0.
0 → Z × n Z → Z / n Z → 0. Apply Hom
( − , Z ) (-,\mathbb Z) ( − , Z )
0 → Hom ( Z , Z ) → × n Hom ( Z , Z ) → 0 ,
0 \to \operatorname{Hom}(\mathbb Z,\mathbb Z)
\xrightarrow{\times n}
\operatorname{Hom}(\mathbb Z,\mathbb Z)\to 0,
0 → Hom ( Z , Z ) × n Hom ( Z , Z ) → 0 , which identifies with
0 → Z → × n Z → 0.
0\to \mathbb Z \xrightarrow{\times n} \mathbb Z \to 0.
0 → Z × n Z → 0. Thus
E x t Z 1 ( Z / n Z , Z ) ≅ H 1 ( Hom ( P ∙ , Z ) ) ≅ Z / n Z .
\mathrm{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.
Ext Z 1 ( Z / n Z , Z ) ≅ H 1 ( Hom ( P ∙ , Z )) ≅ Z / n Z . (See Ext
.)
Example 3: Vanishing over a field If k k k V , W V,W V , W k k k k k k projective
and injective
. Hence
E x t k n ( V , W ) = 0 for all n > 0
\mathrm{Ext}^n_k(V,W)=0\quad \text{for all }n>0
Ext k n ( V , W ) = 0 for all n > 0 (see Ext
),
because one may take a length-0 projective (or injective) resolution and the resulting cochain complex has zero cohomology in positive degrees.