Tor and Ext, An Introduction

Introduction

If R is a commutative ring, tor and ext are binary operators on R modules. They don't come up very often, so we haven't bothered to create infix notation for them. We simply write tor(A,B) and ext(A,B).

These functions are based on homology, hence the placement of this topic below homology. If you are not familiar with homology groups and homotopy equivalent chain complexes, take a step back. In addition, you need a good understanding of the tensor product.

Before the operators can be applied to a module, such as M, M must be "prepared". Write M as the quotient of a free module F₀ mod the kernel F₁. Then write the following exact sequence.

0 → F₁ → F₀ → M → 0

What if a different set of generators spans M, leading to a different sequence 0 → G₁ → G₀ → M → 0? Although the representation of M is quite different, the two presentations are homotopy equivalent, as we shall demonstrate below. That's all we need for tor and ext to work properly.

For any x in M, let u be a basis element in F₀ with image x in M, and let v be a basis element in G₀ with the same image x. (Let u = v = 0 when x = 0.) Let a map on F₀ take u, and all other basis elements leading to x, to v, and let an analogous map on G₀ take v and its associates to u. When such a map is applied to anything in F₁, the result is 0 in G₀/G₁, and pulls back uniquely to something in G₁, giving a map from F₁ to G₁ that necessarily commutes with the boundary operator. And M maps onto itself via the identity map. The result is a chain map from one exact sequence into another.

The composition of these two chain maps, from F₀ to G₀ to F₀, takes all basis elements leading to x onto u. If b is the boundary operator, i.e. the inclusion of F₁ into F₀ and the implied map onto M, we need a homotopy homomorphism d with bd+db equal to the identity map. Starting with x ∈ M, d(x) can be anything in the preimage of M. Sure enough, b(d(x))+d(0) = x. For convenience, let d(x) = u, where u was the specific basis element leading to x that we selected earlier. Of course, selecting a particular u for each x in M assumes the axiom of choice. At a minimum, the basis of F₀ can be well ordered.

How does d map F₀ back into F₁? Let d(u) = 0, and for any other basis element e leading to x, let d(e) = e-u. (If x = 0, u = 0, and d(e) = e.) It is enough to evaluate bd+db on the basis of F₀. Test this out for u and for e, and get the identity map. Thus the homotopy equation is satisfied.

This argument is symbolic and runs in both directions, so the two chain complexes are homotopy equivalent. With this in hand, we're ready to define tor and ext.