TITLE: What is the Tor functor?
QUESTION [84 upvotes]: I'm doing the exercises in "Introduction to commutive algebra" by Atiyah&MacDonald. In chapter two, exercises 24-26 assume knowledge of the Tor functor.
I have tried Googling the term, but I don't find any readable sources. Wikipedia's explanation use the the term "take the homology", which I don't understand (yet).
Are there any good explanations of what the Tor functor is available online not assuming any knowledge about homology?
The first exercise:
"If $M$ is an $A$-module, TFAE:

1) $M$ is flat
2) $\operatorname{Tor}_n^A (M,N)=0$ for all $n>0$ and all $A$-modules $N$.
3) $\operatorname{Tor}_1^A (M,N)=0$ for all $A$-modules $N$."

Thanks in advance.

REPLY [44 votes]: You will be a lot more motivated to learn about Tor once you observe closely how horribly tensor product behaves. 
Let us look at the simplest example possible. Consider the ring $R=\mathbb C[x,y]$ and the ideal $I=(x,y)$. These are about the most well-understood objects, right? What is the tensor product $I\otimes_RI$? This is quite nasty, it has torsions: the element $u = x\otimes y - y\otimes x$ is non-zero, but $xu=yu=0$!
Tor gives you a black box to understand this kind of things. Take the short exact sequence $0 \to I \to R \to R/I \to 0$ and tensor with $I$ we get:
$$0 \to \text{Tor}_1(R/I,I) \to I\otimes I \to I \to I/I^2 \to 0$$ 
from which you can extract:
$$0 \to \text{Tor}_1(R/I,I) \to  I\otimes I \to I^2 \to 0$$
But $\text{Tor}_1(R/I,I) = \text{Tor}_2(R/I, R/I) = \mathbb C$ by standard homological algebra.  So now everything fits nicely: the map from $I\otimes I \to I^2$ takes $f\otimes g$ to $fg$, and the kernel is generated by the element $u$, which is killed by $I$, so it is isomorphic to $R/I \cong \mathbb C$. 
To summarize: tensor product, despite being a fundamental operation, is actually quite bad, and Tor helps you to understand it.