Steps for computing Tor$(\mathbb{Z}, \mathbb{Z}\times\mathbb{Z})$

Question

I'm reviewing algebraic topology, in particular the Kunneth Formula. I can't find online or in my book (by Hatcher) an explanation for how to calculate $\mbox{Tor}(G,H)$ for any two groups. My understanding is that Tor measures the failure of a ses to be exact. Could you please tell me the steps for calculating Tor in general?

I'm piecing together tidbits from my notes and other questions on Stack Exchange. Here's what I could come up with for Tor$(\mathbb{Z}, \mathbb{Z}\times\mathbb{Z})$.

First take a free resolution of the first group, $\mathbb{Z}$ such as $0 \rightarrow 0 \rightarrow \mathbb{Z} \rightarrow \mathbb{Z} \rightarrow 0$, since the integers form a free group without relations. Then tensor each group over integers and ditch the group on the left (since tensoring is right exact) to get $? \rightarrow 0 \rightarrow \mathbb{Z} \rightarrow \mathbb{Z} \rightarrow 0$. Now find what $?$ must be in order for the sequence to be exact. In this case, $?=0$ so Tor$(\mathbb{Z}, \mathbb{Z}\times\mathbb{Z})=0$.

Answer 1

Consider a commutative ring $R$ and some $R$-modules $M$ and $N.$ One can compute the $R$-modules $\operatorname{Tor}_i^R(M, N)$ in essentially two ways.

1.) Begin with a projective resolution of $N,$ e.g., $P_\bullet : \cdots \to P_2 \to P_1 \to P_0 \to N \to 0;$ then, apply the functor $M \otimes_R -$ to this to obtain a chain complex $$T_\bullet : \cdots \to M \otimes_R P_2 \to M \otimes_R P_1 \to M \otimes_R P_0 \to 0.$$ We define $\operatorname{Tor}_i^R(M, N)$ as the homology of this chain complex, i.e., $\operatorname{Tor}_i^R(M, N) = H_i(T_\bullet).$

2.) Begin with a short exact sequence of $R$-modules $0 \to K \to N \to I \to 0.$ By applying the right-exact functor $M \otimes_R -$ to this exact sequence, we obtain a long exact sequence of Tor, i.e., $$\begin{align*} \cdots \to \operatorname{Tor}_1^R(M, K) \to \operatorname{Tor}_1^R(M, N) &\to \operatorname{Tor}_1^R(M, I) \\ \\ &\to \operatorname{Tor}_0^R(M, K) \to \operatorname{Tor}_0^R(M, N) \to \operatorname{Tor}_0^R(M, I) \to 0. \end{align*}$$ One can prove that $\operatorname{Tor}_0^R(M, -) = M \otimes_R -,$ so this long exact sequence measures the failure of $M \otimes_R -$ to be left-exact. Under some mild assumptions, in this setting, one could say something about the vanishing of the higher Tor, or one could explicitly compute $\operatorname{Tor}_i^R(M, N).$

Here are just a few facts about $\operatorname{Tor}_i^R(M, N)$ that could come in handy for computations.

• We have that $\operatorname{Tor}_i^R(M, N) \cong \operatorname{Tor}_i^R(N, M),$ so always compute the easier of the two.
• If $M$ and $N$ are finitely generated, then $\operatorname{Tor}_i^R(M, N)$ is finitely generated
• Given that $M$ is a flat $R$-module, we have that $\operatorname{Tor}_i^R(M, N) = 0$ for all integers $i \geq 1.$ Conversely, if $\operatorname{Tor}_1^R(M, N) = 0$ for all $R$-modules $N,$ then $M$ is flat as an $R$-module.
• Given a commutative ring $R$ with ideals $I$ and $J,$ we have that $\operatorname{Tor}_1^R(R/I, R/J) \cong (I \cap J) / IJ.$
• Given an $R$-regular sequence $I = (x_1, \dots, x_n),$ we have that $\operatorname{Tor}_n^R(R/I, M) \cong (0 :_M I).$
• Tor commutes with direct sums, direct limits, and localization.