Long exact sequence for Tor

Let $R$ be a ring, let

0 \longrightarrow A' \xrightarrow{u} A \xrightarrow{v} A'' \longrightarrow 0

be a short exact sequence of right $R$ -modules, and let $B$ be a left $R$ -module. Then there are natural connecting homomorphisms

\delta_n:\operatorname{Tor}_n^R(A'',B)\longrightarrow \operatorname{Tor}_{n-1}^R(A',B) \quad (n\ge 1),

(see connecting homomorphism) such that the following sequence is exact:

\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.

Equivalently, for every $n\ge 1$ one has exactness at the three-term window

\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).

Examples

Computing $\operatorname{Tor}_1^{\mathbb Z}(\mathbb Z/n,\mathbb Z/m)$ . Use the short exact sequence $0\to \mathbb Z \xrightarrow{\cdot n}\mathbb Z \to \mathbb Z/n \to 0.$ Tensor with $\mathbb Z/m$ . Since $\mathbb Z\otimes \mathbb Z/m\cong \mathbb Z/m$ , the relevant part of the long exact sequence becomes $0 \to \operatorname{Tor}_1^{\mathbb Z}(\mathbb Z/n,\mathbb Z/m) \to \mathbb Z/m \xrightarrow{\cdot n} \mathbb Z/m \to (\mathbb Z/n)\otimes \mathbb Z/m \to 0.$ Hence $\operatorname{Tor}_1^{\mathbb Z}(\mathbb Z/n,\mathbb Z/m)\cong \ker(\cdot n:\mathbb Z/m\to \mathbb Z/m)\cong \mathbb Z/\gcd(n,m).$ (Also $\operatorname{Tor}_i^{\mathbb Z}(\mathbb Z/n,-)=0$ for $i\ge 2$ because $\mathbb Z/n$ has a length-1 projective resolution.)

Over a field, higher Tor vanishes. If $R=k$ is a field and $V,W$ are $k$ -vector spaces, then $V$ is free (hence projective), so $\operatorname{Tor}_i^k(V,W)=0$ for all $i\ge 1$ . The long exact sequence above reduces to exactness of $0\to V'\otimes_k W \to V\otimes_k W \to V''\otimes_k W \to 0,$ reflecting that $-\otimes_k W$ is exact.

Dual numbers. Let $R=k[\varepsilon]/(\varepsilon^2)$ and regard $k=R/(\varepsilon)$ as an $R$ -module. The periodic free resolution $\cdots\xrightarrow{\cdot\varepsilon}R\xrightarrow{\cdot\varepsilon}R \xrightarrow{\cdot\varepsilon}R\longrightarrow k\longrightarrow0$ becomes a complex with zero differentials after tensoring with $k$ . Hence $\operatorname{Tor}_i^R(k,k)\cong k\qquad(i\ge1).$