Let R R R be a ring. Recall that Tor
is the left-derived functor of the tensor product (right exactness of tensor
), constructed using a projective resolution
(see also derived functor
).
Theorem (long exact sequence in Tor) Let
0 ⟶ A ′ → u A → v A ′ ′ ⟶ 0
0 \longrightarrow A' \xrightarrow{u} A \xrightarrow{v} A'' \longrightarrow 0
0 ⟶ A ′ u A v A ′′ ⟶ 0 be a short exact sequence
of right R R R -modules, and let B B B be a left R R R -module. Then there are natural connecting homomorphisms
δ n : Tor n R ( A ′ ′ , B ) ⟶ Tor n − 1 R ( A ′ , B ) ( n ≥ 1 ) ,
\delta_n:\operatorname{Tor}_n^R(A'',B)\longrightarrow \operatorname{Tor}_{n-1}^R(A',B)
\quad (n\ge 1),
δ n : Tor n R ( A ′′ , B ) ⟶ Tor n − 1 R ( A ′ , B ) ( n ≥ 1 ) , (see connecting homomorphism
) such that the following sequence is exact:
⋯ → Tor 2 R ( A ′ ′ , B ) → δ 2 Tor 1 R ( A ′ , B ) → Tor 1 R ( A , B ) → Tor 1 R ( A ′ ′ , B ) → δ 1 A ′ ⊗ R B → A ⊗ R B → A ′ ′ ⊗ R B → 0.
\cdots \to \operatorname{Tor}_2^R(A'',B)\xrightarrow{\delta_2}\operatorname{Tor}_1^R(A',B)\to
\operatorname{Tor}_1^R(A,B)\to \operatorname{Tor}_1^R(A'',B)
\xrightarrow{\delta_1} A'\otimes_R B \to A\otimes_R B \to A''\otimes_R B \to 0.
⋯ → Tor 2 R ( A ′′ , B ) δ 2 Tor 1 R ( A ′ , B ) → Tor 1 R ( A , B ) → Tor 1 R ( A ′′ , B ) δ 1 A ′ ⊗ R B → A ⊗ R B → A ′′ ⊗ R B → 0. Equivalently, for every n ≥ 1 n\ge 1 n ≥ 1 one has exactness at the three-term window
Tor n R ( A ′ , B ) ⟶ Tor n R ( A , B ) ⟶ Tor n R ( A ′ ′ , B ) → δ n Tor n − 1 R ( A ′ , B ) .
\operatorname{Tor}_n^R(A',B)\longrightarrow \operatorname{Tor}_n^R(A,B)\longrightarrow \operatorname{Tor}_n^R(A'',B)\xrightarrow{\delta_n}\operatorname{Tor}_{n-1}^R(A',B).
Tor n R ( A ′ , B ) ⟶ Tor n R ( A , B ) ⟶ Tor n R ( A ′′ , B ) δ n Tor n − 1 R ( A ′ , B ) . This is a special case of the general long exact sequence for derived functors
.
Examples Computing Tor 1 Z ( Z / n , Z / m ) \operatorname{Tor}_1^{\mathbb Z}(\mathbb Z/n,\mathbb Z/m) Tor 1 Z ( Z / n , Z / m ) . Use the short exact sequence
0 → Z → ⋅ n Z → Z / n → 0.
0\to \mathbb Z \xrightarrow{\cdot n}\mathbb Z \to \mathbb Z/n \to 0.
0 → Z ⋅ n Z → Z / n → 0. Tensor with Z / m \mathbb Z/m Z / m . Since Z ⊗ Z / m ≅ Z / m \mathbb Z\otimes \mathbb Z/m\cong \mathbb Z/m Z ⊗ Z / m ≅ Z / m , the relevant part of the long exact sequence becomes
0 → Tor 1 Z ( Z / n , Z / m ) → Z / m → ⋅ n Z / m → ( Z / n ) ⊗ Z / m → 0.
0 \to \operatorname{Tor}_1^{\mathbb Z}(\mathbb Z/n,\mathbb Z/m) \to \mathbb Z/m \xrightarrow{\cdot n} \mathbb Z/m \to (\mathbb Z/n)\otimes \mathbb Z/m \to 0.
0 → Tor 1 Z ( Z / n , Z / m ) → Z / m ⋅ n Z / m → ( Z / n ) ⊗ Z / m → 0. Hence
Tor 1 Z ( Z / n , Z / m ) ≅ ker ( ⋅ n : Z / m → Z / m ) ≅ Z / gcd ( n , m ) .
\operatorname{Tor}_1^{\mathbb Z}(\mathbb Z/n,\mathbb Z/m)\cong \ker(\cdot n:\mathbb Z/m\to \mathbb Z/m)\cong \mathbb Z/\gcd(n,m).
Tor 1 Z ( Z / n , Z / m ) ≅ ker ( ⋅ n : Z / m → Z / m ) ≅ Z / g cd( n , m ) . (Also Tor i Z ( Z / n , − ) = 0 \operatorname{Tor}_i^{\mathbb Z}(\mathbb Z/n,-)=0 Tor i Z ( Z / n , − ) = 0 for i ≥ 2 i\ge 2 i ≥ 2 because Z / n \mathbb Z/n Z / n has a length-1 projective resolution.)
Over a field, higher Tor vanishes. If R = k R=k R = k is a field and V , W V,W V , W are k k k -vector spaces, then V V V is free (hence projective), so Tor i k ( V , W ) = 0 \operatorname{Tor}_i^k(V,W)=0 Tor i k ( V , W ) = 0 for all i ≥ 1 i\ge 1 i ≥ 1 . The long exact sequence above reduces to exactness of
0 → V ′ ⊗ k W → V ⊗ k W → V ′ ′ ⊗ k W → 0 ,
0\to V'\otimes_k W \to V\otimes_k W \to V''\otimes_k W \to 0,
0 → V ′ ⊗ k W → V ⊗ k W → V ′′ ⊗ k W → 0 , reflecting that − ⊗ k W -\otimes_k W − ⊗ k W is exact.
Dual numbers: Tor 1 k [ ε ] / ( ε 2 ) ( k , k ) ≅ k \operatorname{Tor}_1^{k[\varepsilon]/(\varepsilon^2)}(k,k)\cong k Tor 1 k [ ε ] / ( ε 2 ) ( k , k ) ≅ k . Let R = k [ ε ] / ( ε 2 ) R=k[\varepsilon]/(\varepsilon^2) R = k [ ε ] / ( ε 2 ) and k = R / ( ε ) k=R/(\varepsilon) k = R / ( ε ) . The sequence
0 → R → ⋅ ε R → k → 0
0\to R\xrightarrow{\cdot \varepsilon}R\to k\to 0
0 → R ⋅ ε R → k → 0 is a projective resolution of k k k of length 1 1 1 . Tensoring with k k k makes ⋅ ε \cdot\varepsilon ⋅ ε become 0 0 0 (since ε \varepsilon ε acts as 0 0 0 on k k k ), so
Tor 1 R ( k , k ) ≅ ker ( 0 : k → k ) ≅ k .
\operatorname{Tor}_1^R(k,k)\cong \ker(0:k\to k)\cong k.
Tor 1 R ( k , k ) ≅ ker ( 0 : k → k ) ≅ k .