Lecture 22: The Fundamental Theorem of Homological Algebra

22. THE FUNDAMENTAL THEOREM OF HOMOLOGICAL ALGEBRA 55 TorR 0 (M, N ) = M ⊗R N since we know that M ⊗R − is right-exact. 22 The fundamental theorem of homological algebra We will now show that the R-modules TorR n (M, N ) are well-defined and functorial. This will be an application of a very general principle. Theorem 22.1 (Fundamental Theorem of Homological Algebra). Let M and N be R-modules; let 0 ← M ← E0 ← E1 ← · · · be a sequence in which each En is free; let 0 ← N ← F0 ← F1 ← · · · be an exact sequence; and let f : M → N be a homomorphism. Then we can lift f to a chain map f∗ : E∗ → F∗ , uniquely up to chain homotopy. Proof. Let’s try to construct f0 . Consider: / K0 = ker(M ) 0  g0  / L0 = ker(N ) 0 M / E0 f0 / F0 N /M  f /N /0 We know that E0 = RS for some set S. What we do is map the generators of E0 into M via M and then into F via f , and then lift them to F0 via N (which is possible because it’s surjective). Then extend to a homomorphism, to get f0 . You can restrict f0 to kernels to get g0 . Now the map d : E1 → E0 satisifes M ◦ d = 0, and so factors through a map to K0 = ker M . Similarly, d : F1 → F0 factors through a map F1 → L0 , and this map must be surjective because the sequence F1 → F0 → N is exact. We find ourselves in exactly the same situation: 0 0 / K1  g1 / L1 / E1  f1 / F1 / K0  g0 / L0 /0 56 CHAPTER 2. COMPUTATIONAL METHODS So we construct f∗ by induction. Now we need to prove the chain homotopy claim. So suppose I have f∗ , f∗0 : E∗ → F∗ , both lifting f : M → N . Then fn0 − fn (which we’ll rename `n ) is a chain map lifting 0 : M → N . We want to consruct a chain null-homotopy of `∗ ; that is, we want h : En → Fn+1 such that dh + hd = `n . At the bottom, E−1 = 0, so we want h : E0 → F1 such that dh = `0 . This factorization happens in two steps. /M E0 h F1 v / / L0 ~ `0  / F0  N 0 /N. First, N `0 = 0 implies that `0 factors through L0 = ker N . Next, F1 → L0 is surjective, by exactness, and E0 is free, so we can lift generators and extend R-linearly to get h : E0 → F1 .. The next step is organized by the diagram d E1 h F2 v / / L1 `1 ~ / E0 h  ~ /4 F1 d  `0 / F0 d This diagram doesn’t commute; dh = `0 , but the (d, h, `1 ) triangle doesn’t commute. Rather, we want to construct h : E1 → F2 such that dh = `1 − hd. Since d(`1 − hd) = `0 d − dhd = (`0 − dh)d = 0 . the map `1 − hd lifts to L1 = ker d. But then it lifts through F2 , since F2 → L1 is surjective and E1 is free. Exactly the same process continues. This proof uses a property of freeness that is shared by a broader class of modules. Definition 22.2. An R-module P is projective if any map out of P factors through any surjection: >M P  /N Every free module is projective, and this is the property of freeness that we jave been using; the Fundamental Theorem of Homological Algebra holds under the weaker assumption that each En is projective. Any direct summand in a projective is also projective. Any projective module is a direct summand of a free module. Over a PID, every projective is free, because any submodule of a free is free. But there are examples of nonfree projectives: Example 22.3. Let k be a field and let R be the product ring k × k. It acts on k in two ways, via (a, b)c = ac and via (a, b)c = bc. These are both projective R-modules that are not free. Now we will apply Theorem 22.1 to verify that our proposed construction of Tor is independent of free (or projective!) resolution, and is functorial. 22. THE FUNDAMENTAL THEOREM OF HOMOLOGICAL ALGEBRA 57 Suppose I have f : N 0 → N . Pick arbitrary free resolutions N 0 ← F∗0 and N ← F∗ , and pick any chain map f∗ : F∗0 → F∗ lifting f . We claim that the map induced in homology by 1 ⊗ f∗ : M ⊗R F∗0 → M ⊗R F∗ is independent of the choice of lift. Suppose f∗0 is another lift, and pick a chain homotopy h : f∗ ' f∗0 . Since M ⊗R − is additive, the relation 1 ⊗ h : 1 ⊗ f∗ ' 1 ⊗ f∗0 still holds. So 1 ⊗ f∗ and 1 ⊗ f∗0 induce the same map in homology. For example, suppose that F∗ and F∗0 are two projective resolutions of N . Any two lifts of the identity map are chain-homotopic, and so induce the same map H∗ (M ⊗R F∗ ) → H∗ (M ⊗R F∗0 ). So if f : F∗ → F∗0 and g : F∗0 → F∗ are chain maps lifing the identity, then f∗ ◦ g∗ induces the same self-map of H∗ (M ⊗R F∗0 ) as the identity self-map does, and so (by functoriality) is the identity. Similarly, g∗ ◦ f∗ induces the identity map on H∗ (M ⊗R F∗ ). So they induce inverse isomorphisms. Putting all this together shows that any two projective resolutions of N induce canonically 0 isomorphic modules TorR n (M, N ), and that a homomorphism f : N → N induces a well defined R R R 0 map Torn (M, N ) → Torn (M, N ) that renders Torn (M, −) a functor. My last comment about Tor is that there’s a symmetry there. Of course, M ⊗R N ∼ = N ⊗R M . ∼ This uses the fact that R is commutative. This leads right on to saying that TorR n (M, N ) = R Torn (N, M ). We’ve been computing Tor by taking a resolution of the second variable. But I could equally have taken a resolution of the first variable. This follows from Theorem 22.1. Example 22.4. I want to give an example when you do have higher Tor modules. Let k be a field, and let R = k[d]/(d2 ). This is sometimes called the “dual numbers,” or the exterior algebra over k. What is an R-module? It’s just a k-vector space M with an operator d (given by multiplication by d) that satisfies d2 = 0. Even though there’s no grading around, I can still define the “homology” of M : ker d H(M ; d) = . im d This k-algebra is augmented by an algebra map  : R → k splitting the unit; (d) = 0. This renders k an R-module. Let’s construct a free R-module resolution of this module. Here’s a picture. •o •o •o •o •o • • • • The vertical lines indicate multiplication by d. We could write this as  d d 0←k← −R← −R← − R ← ··· . Now tensor this over R with an R-module M ; so M is a vector space equipped with an operator d with d2 = 0. Each copy of R gets replaced by a copy of M , and the differential gives multiplication by d on M . So taking homology gives ( k ⊗R M = M/dM for n = 0 R Torn (k, M ) = H(M ; d) for n > 0 . So for example TorR n (k, k) = k for n ≥ 0 .