Let R R R ring
and M M M R-module
.
Definition A projective resolution of M M M chain complex
⋯ → d 2 P 1 → d 1 P 0 → ε M → 0
\cdots \xrightarrow{d_2} P_1 \xrightarrow{d_1} P_0 \xrightarrow{\varepsilon} M \to 0
⋯ d 2 P 1 d 1 P 0 ε M → 0 such that:
Each P i P_i P i projective R-module
. The sequence is exact (equivalently, the augmented complex is an exact complex
). Equivalently, if P ∙ P_\bullet P ∙ ⋯ → P 1 → P 0 → 0 \cdots\to P_1\to P_0\to 0 ⋯ → P 1 → P 0 → 0
H 0 ( P ∙ ) ≅ M , H i ( P ∙ ) = 0 for all i > 0 ,
H_0(P_\bullet)\cong M,\qquad H_i(P_\bullet)=0 \text{ for all } i>0,
H 0 ( P ∙ ) ≅ M , H i ( P ∙ ) = 0 for all i > 0 , where H i H_i H i homology
of the underlying chain complex.
A projective resolution is free if each P i P_i P i free
.
Existence in module categories is guaranteed by projective resolutions exist
.
What resolutions are for For any R R R N N N P ∙ ⊗ R N P_\bullet\otimes_R N P ∙ ⊗ R N Tor
:
H i ( P ∙ ⊗ R N ) ≅ H A H A H U G O S H O R T C O D E 308 s 9 H B H B ( M , N ) .
H_i(P_\bullet\otimes_R N)\cong Tor_i^R
(M,N).
H i ( P ∙ ⊗ R N ) ≅ H A H A H U GOS H ORTCO D E 308 s 9 H B H B ( M , N ) . right exact
, so one needs derived functors.) Applying Hom R ( P ∙ , N ) \operatorname{Hom}_R(P_\bullet,N) Hom R ( P ∙ , N ) cochain complex
whose cohomology computes Ext
:
H i ( Hom R ( P ∙ , N ) ) ≅ H A H A H U G O S H O R T C O D E 308 s 13 H B H B ( M , N ) .
H^i(\operatorname{Hom}_R(P_\bullet,N))\cong Ext_R^i
(M,N).
H i ( Hom R ( P ∙ , N )) ≅ H A H A H U GOS H ORTCO D E 308 s 13 H B H B ( M , N ) . Examples Example 1: A projective resolution of Z / n Z \mathbb Z/n\mathbb Z Z / n Z Over R = Z R=\mathbb Z R = 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 is exact, and Z \mathbb Z Z Z / n Z \mathbb Z/n\mathbb Z Z / n Z 1 1 1
Example 2: Computing T o r 1 Z ( Z / n , Z / m ) \mathrm{Tor}_1^{\mathbb Z}(\mathbb Z/n,\mathbb Z/m) Tor 1 Z ( Z / n , Z / m ) Tensor the resolution in Example 1 with Z / m Z \mathbb Z/m\mathbb Z Z / m Z
0 → Z / m → × n Z / m → 0.
0\to \mathbb Z/m \xrightarrow{\times n} \mathbb Z/m \to 0.
0 → Z / m × n Z / m → 0. Then
H A H A H U G O S H O R T C O D E 308 s 14 H B H B 1 Z ( Z / n , Z / m ) ≅ H 1 ≅ ker ( × n : Z / m → Z / m ) ≅ Z / gcd ( m , n ) Z .
Tor_1
_1^{\mathbb Z}(\mathbb Z/n,\mathbb Z/m)
\cong H_1
\cong \ker(\times n:\mathbb Z/m\to\mathbb Z/m)
\cong \mathbb Z/\gcd(m,n)\mathbb Z.
H A H A H U GOS H ORTCO D E 308 s 14 H B H B 1 Z ( Z / n , Z / m ) ≅ H 1 ≅ ker ( × n : Z / m → Z / m ) ≅ Z / g cd( m , n ) Z . Example 3: A resolution of k k k k [ x ] k[x] k [ x ] Let R = k [ x ] R=k[x] R = k [ x ] M = R / ( x ) ≅ k M=R/(x)\cong k M = R / ( x ) ≅ k
0 → R → × x R → k → 0
0\to R \xrightarrow{\times x} R \to k \to 0
0 → R × x R → k → 0 is a free (hence projective) resolution of k k k 1 1 1
H A H A H U G O S H O R T C O D E 308 s 15 H B H B 1 k [ x ] ( k , k ) ≅ H 1 ( ( 0 → R → x R → 0 ) ⊗ k [ x ] k ) ≅ H 1 ( 0 → k → 0 k → 0 ) ≅ k .
Tor_1
_1^{k[x]}(k,k)
\cong H_1\bigl((0\to R\xrightarrow{x}R\to 0)\otimes_{k[x]}k\bigr)
\cong H_1(0\to k\xrightarrow{0}k\to 0)\cong k.
H A H A H U GOS H ORTCO D E 308 s 15 H B H B 1 k [ x ] ( k , k ) ≅ H 1 ( ( 0 → R x R → 0 ) ⊗ k [ x ] k ) ≅ H 1 ( 0 → k 0 k → 0 ) ≅ k .