How is $\text{Tor}_1^{\mathbb{Z}}(X,R)$ defined in a simple manner as an R-module?

Question

Wishing to understand the universal coefficient formula for homology modules, I came across this new space $\text{Tor}_1^{\mathbb{Z}}$ that I presume that it can be viewed as an $R$-module, and I do not understand how its elements look like, how the addition operation is defined, the scalar multiplication a

The formula is:

Corollary $7.57$. If $X$ is a topological space and $A$ is an abelian group, then, for all $n \geq 0$,

$$H_n(X,A) \cong H_n(X) \otimes_\mathbb{Z} A \ \oplus \ \text{Tor}_1^\mathbb{Z}(H_{n-1}(X),A).$$

My question is: Is it possible to define in a concrete way (I am not an algebrist and I do not want to use a lot of specialized concepts in order to define it) the $R$-module $\text{Tor}_1^{\mathbb{Z}}(X,R)$ for a given $R$-module $X$, with $R$ a principal ideal domain?

I found in the book Mac Lane-Homology, page 150 and in the book of Rotman-An introduction to homological algebra, page 412 the following definition which is not so clear for me:

There are other constructions of Tor. For example, $\text{Tor}_1^\mathbb{Z}(A,B)$ can be defined by generators and realtions. Consider all triples $(a,n,b)$, where $a \in A$, $b \in B$, $na=0$, and $nb=0$; then $\text{Tor}_1^\mathbb{Z}(A,B)$ is generated by all such triples subject to the relations (whenever both sides are defined)

$$(a+a',n,b) = (a,n,b) + (a',n,b)$$ $$(a,n,b+b') = (a,n,b) + (a,n,b')$$ $$(ma,n,b) = (a, mn, b) = (a, m, nb)$$

In this definition we see only an abelian group, but I want to see it as an R-module. Can someone please help me, maybe with an easy example, to understand the construction of $\text{Tor}_1^{\mathbb{Z}}(X,R)$ as an R-module?

Answer 1

First, let me mention that $\text{Tor}^\mathbb{Z}(-,-)$ is an operation that eats two $\mathbb{Z}$-modules (read: abelian groups) and outputs a new $\mathbb{Z}$-module. So when you ask for $\text{Tor}^\mathbb{Z}(A,B)$ to be an $R$-module, it's not clear what you mean. What is true is that $\text{Tor}^R$ exists for any ring $R$, and $\text{Tor}^R(-,-)$ eats two $R$-modules and spits out a new $R$-module. This is probably the level of generality that you're thinking of.

Now, $\text{Tor}$ has lots of interesting properties, relating it to other branches of math, and because of this it has many applications (see this video, for example), but it sounds like you're primarily interested in computing $\text{Tor}$ so that you can use it for the universal coefficient theorem. Let's start by giving a few important properties of $\text{Tor}$ that will use used all the time in computations:

1. $\text{Tor}^R(A,B) \cong \text{Tor}^R(B,A)$
2. $\text{Tor}^R(A, \bigoplus_\alpha B_\alpha) \cong \bigoplus_\alpha \text{Tor}^R(A, B_\alpha)$
3. $\text{Tor}^R(\bigoplus_\alpha A_\alpha, B) \cong \bigoplus_\alpha \text{Tor}^R(A_\alpha, B)$

Now let's start with homology with coefficients in an abelian group. So we've computed $H_n(X)$, the homology groups with coefficients in $\mathbb{Z}$, and we want to compute $H_n(X,A)$, the homology groups with coefficients in some abelian group $A$. The universal coefficient theorem says that

$$ H_n(X,A) \cong H_n(X) \otimes_\mathbb{Z} A \ \oplus \ \text{Tor}^\mathbb{Z}(H_{n-1}(X), A) $$

Since we're working over $\mathbb{Z}$, let me mention two more useful facts for computation in this special case:

4. $\text{Tor}^\mathbb{Z}(\mathbb{Z},B) \cong 0$ for every $B$
5. $\text{Tor}^\mathbb{Z}(\mathbb{Z}/n, B) \cong \{ b \in B \mid nb = 0 \}$

In particular, note that $5$ implies $\text{Tor}^\mathbb{Z}(\mathbb{Z}/m, \mathbb{Z}/n) \cong \mathbb{Z}/\gcd(m,n)$.

Now, in some sense, this is almost everything you need to know! After all, every finitely generated abelian group looks like $\mathbb{Z}^n \oplus \bigoplus_k \mathbb{Z}/n_k$. So the above properties guarantee that we can compute $\text{Tor}^\mathbb{Z}(A,B)$ for any two finitely generated abelian groups $A$ and $B$ by splitting up the direct sums, then using $4$ and $5$ to compute the pieces. For example:

$$ \begin{align} \text{Tor}^\mathbb{Z} \left ( \mathbb{Z}^2 \oplus \mathbb{Z}/2 \oplus \mathbb{Z}/3, \ \mathbb{Z} \oplus \mathbb{Z}/4 \right ) &\overset{2}{\cong} \text{Tor}^\mathbb{Z} \left ( \mathbb{Z}^2 \oplus \mathbb{Z}/2 \oplus \mathbb{Z}/3, \ \mathbb{Z} \right ) \oplus \text{Tor}^\mathbb{Z} \left ( \mathbb{Z}^2 \oplus \mathbb{Z}/2 \oplus \mathbb{Z}/3, \ \mathbb{Z}/4 \right ) \\ &\overset{4}{\cong} \text{Tor}^\mathbb{Z} \left ( \mathbb{Z}^2 \oplus \mathbb{Z}/2 \oplus \mathbb{Z}/3, \ \mathbb{Z}/4 \right ) \\ &\overset{3}{\cong} \text{Tor}^\mathbb{Z} \left ( \mathbb{Z}, \mathbb{Z}/4 \right ) \oplus \text{Tor}^\mathbb{Z} \left ( \mathbb{Z}, \mathbb{Z}/4 \right ) \oplus \text{Tor}^\mathbb{Z} \left ( \mathbb{Z}/2, \mathbb{Z}/4 \right ) \oplus \text{Tor}^\mathbb{Z} \left ( \mathbb{Z}/3, \mathbb{Z}/4 \right ) \\ &\overset{4}{\cong} \text{Tor}^\mathbb{Z} \left ( \mathbb{Z}/2, \mathbb{Z}/4 \right ) \oplus \text{Tor}^\mathbb{Z} \left ( \mathbb{Z}/3, \mathbb{Z}/4 \right ) \\ &\overset{5}{\cong} \mathbb{Z}/\gcd(2,4) \oplus \mathbb{Z}/\gcd(3,4) \\ &\cong \mathbb{Z}/2 \end{align} $$

Hopefully it's clear how you can do a longer version of this computation to compute $\text{Tor}^\mathbb{Z}(A,B)$ for your favorite finitely generated abelian groups $A$ and $B$.

---

But what if $A$ and $B$ are $R$-modules for some PID $R$? Well remember that every finitely generated $R$-module over a PID decomposes as $R^n \oplus \bigoplus_k R/r_k$. Rules $1$, $2$, and $3$ still hold unchanged. So we can split these direct sums in the way we'd like. But what about rules $4$ and $5$? Do they have analogues in the case of general PIDs? Of course they do!

4. $\text{Tor}^R(R,B) \cong 0$ for every $R$-module $B$
5. $\text{Tor}^R(R/r, B) \cong \{ b \in B \mid rb = 0 \}$ for every $R$-module $B$

Note that, $\text{Tor}^R(R/r, B) \cong \{ b \in B \mid rb = 0 \}$ has a natural $R$-module structure, since it's a submodule of $B$.

Lastly, as in the $\mathbb{Z}$ case, $5$ tells us how to compute $\text{Tor}^R(R/r, R/s)$, and thus how to compute $\text{Tor}^R(A,B)$ for any two finitely generated $R$-modules $A$ and $B$. I'll leave this computation to you, though (or to a quick google, if you're so inclined).

(Oh, and there are other properties that I've omitted for the sake of concreteness which are also super helpful for computing $\text{Tor}^R$ in practice, even for non-finitely-generated modules. You can read about them here, for instance. The key thing you should have in the back of your head is that even though these definitions look extremely complicated, with practice they're often quite routine to compute! So if you can reduce some problem to computing a $\text{Tor}^R$ group, you've practically solved your problem! Hopefully the finitely generated examples in this answer have helped convince you of this.)

---

I hope this helps ^_^