Recall: a mapping f : A B C (where A, B, C are R-modules) is called R-bilinear if f is R-linear in each coordinate, i.e.,

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1 23 Hom and We will do homological algebra over a fixed commutative ring R. There are several good reasons to take a commutative ring: Left R-modules are the same as right R-modules. [In general a right R-module is the same as a left R op -module.] If A, B are R-modules then Hom R (A, B) is also an R-module. Hom R (A, B) = {f Hom(A, B) f(rx) = rf(x) r R, x A} If s R then sf(rx) = srf(x) = rsf(x) so sf Hom R (A, B). Composition of functions is also R-bilinear: Hom R (B, C) Hom R (A, B) Hom R (A, C) (1) Recall: a mapping f : A B C (where A, B, C are R-modules) is called R-bilinear if f is R-linear in each coordinate, i.e., f(rx + sy, b) = rf(x, b) + sf(y, b) (2) f(a, rx + sy) = rf(a, x) + sf(a, y) (3) The first condition (2) is equivalent to saying that, for each fixed b B, the function ˆf(b) = f(, b) : A C is R-linear, i.e., ˆf(b) HomR (A, C). The second condition (3) is equivalent to saying that the function ˆf : B Hom R (A, C) is R-linear. Therefore we have a bijection: BiLin R (A B, C) = Hom R (B, Hom R (A, C)) (4) where BiLin R (A B, C) denotes the set of R-bilinear maps f : A B C. This set has an obvious structure of an R-module since the sum f + g of two R-bilinear maps is R-bilinear and rf is R-bilinear for any r R. It is straightforward to see that the bijection (4) is an isomorphism of R-modules. Tensor product The tensor product A R B of two R-modules can be defined in two ways. The first is by a universal construction. Definition A R B is defined to be any R-module satisfying the following conditions. 1. There is an R-bilinear mapping f : A B A R B 1

2 2. Given any R-bilinear mapping g : A B C,!h Hom R (A R B, C) so that g = h f. As with any universal property this defines A R B uniquely up to isomorphism. The second condition in the definition can be written as follows. BiLin R (A B, C) = Hom R (A R B, C). Combining this with (4) we get the following. Theorem Hom R (A R B, C) = Hom R (B, Hom R (A, C)). We would like to say that tensor product (A R ) and Hom R (A, ) are adjoint additive functors R-M od R-M od. However, tensor product is (so far) only defined up to isomorphism. We need a specific model for A R B in order to have a functor. Let f : A B A R B be the universal R-bilinear mapping of Def Let a b = f(a, b). In order for f to be bilinear we need these symbols to satisfy the following. (rx + sy) b = r(x b) + s(y b) a (rx + sy) = r(a x) + s(a y) Thus an explicit model for A R B is given by taking the free R-module generated by A B (with (a, b) written as a b) module the two relations above. [Thus elements of A R B are subsets of R[A B] which in turn is a subset of the set R A B of functions A B R, namely, those which are zero almost everywhere.] Definition An R-category is a category where the Hom sets are R- modules and composition is R-bilinear. E.g., R-Mod is an R-category. A functor F : C D between R-categories is called R-linear if the induced maps: Hom C (A, B) Hom D (F A, F B) are R-linear for all A, B in C. Lemma Hom R (A, ) and A R are R-linear endofunctors 1 on R- Mod. 1 An endofunctor is a functor from a category to itself. 2

3 Proof. The adjoint of the R-bilinear mapping (1) gives an R-linear mapping: Hom R (B, C) Hom R (Hom R (A, B), Hom R (A, C)), i.e., Hom R (A, ) is R-linear. For tensor product suppose f : A A, g : B B are R-linear. Then we get an R-linear map f g : A R B A R B by (f g)(a b) = f(a) g(b). To show that this is well-defined we need to show that f(a) g(b) is R-bilinear in (a, b) but this is easy: f(rx + xy) g(b) = (rf(x) + sf(y)) g(b) = r(f(x) b) + s(f(y) b) and similarly for b. What our lemma states is that f(a) g(b) is R-bilinear in the letter g. But this is also trivial: f(a) (rg + sh)(b) = f(a) (rg(b) + sh(b)) = r(f(a) g(b)) + s(f(a) h(b)). Theorem Hom R (A, ) and A R are adjoint R-linear endofunctors on R-Mod. Corollary A R Bα = (A R B α ). Proof. All left adjoint R-linear functors have this property. Before proving this we need to go over the definition of direct sum in any additive category. If B α is a collection of objects in a preadditive category C then B α is defined to be the coproduct B α = B α. In other words, there are inclusion morphisms j α : B α B α for all α so that for any collection of morphisms f α : B α C (for any C Ob(C)), there is a unique morphism g : B α C so that f α = g j α for all α. In other words, Hom C (B α, C) = Hom C ( B α, C) (5) α for all C Ob(C). Now suppose that F : D C and G : C D are adjoint R-linear functors and suppose that the direct sum B α exists in D. Then we want to show that F ( B α ) is the direct sum of the objects F B α in C. By adjunction we have: Hom C (F ( B α ), C) = Hom D ( B α, GC). By definition of direct sum in D we have: Hom D ( B α, GC) = Hom D (B α, GC). Taking adjoints again we get: HomD (B α, GC) = Hom C (F B α, C) for all C Ob(C). F ( B α ) = F B α. By definition of direct sum in C this implies that 3

4 Corollary Tensor product is right exact, i.e., given an exact sequence of R-modules: we get another exact sequence: A f B g C 0 X R A id X f X R B id X g X R C 0. Since tensor product is obviously symmetric, i.e., we have a natural isomorphism: A R B = B R A it follows that R B = B R is also right exact. Flat and injective modules An R-module X is called flat if X R is an exact functor, i.e., if for any exact sequence of R-modules: we get an exact sequence: 0 A B C 0 0 X R A X R B X R C 0 Note that since X R is right exact it is only the injectivity of the mapping X R A X R B which is in question. Proposition There is a natural isomorphism R R A = A. Proof. Natural R-linear maps f : R R A A and g : A R R A are given by f(r a) 2 = ra and g(a) = 1 a. The compositions are easily seen to be the identity: g(f(r a)) = g(ra) = 1 ra = r(1 a) = r a and fg(a) = f(1 a) = 1a = a. Corollary R is a flat R-module. Theorem HomZ(R, Q) is an injective R-module. More generally, HomZ(F, D) is injective for any divisible group D and any flat R-module F. Before we prove this we need to go over some definitions. First, Hom(M, G) = HomZ(M, G) is an R-module for any R-module M and additive group G. The action of R is given by (rf)(x) = f(rx). This uses the commutativity of R: r(sf)(x) = sf(rx) = f(srx) = f(rsx) = (rs)f(x). Note that (rs)f(x) means (rsf)(x) since rs(f(x)) is not defined. 2 General elements of a tensor product are linear combinations of simple tensors, e.g., ai b i is an arbitrary element of A R B. The elements a b are generators. 4

5 Definition An R-module I is called injective if for any R-linear map f : A I defined on a submodule A B extends to an R-linear map f : B I. 0 A j B f f I Another way to say this is that I is injective iff Hom R (, I) is an exact (contravariant) functor, i.e., iff for every exact sequence 0 A j B C 0 in R-Mod we get an exact sequence: 0 Hom R (A, I) j Hom R (B, I) Hom R (C, I) 0 To see this take any element f Hom R (A, I). Since I is injective, this extends to B, i.e., f Hom R (B, I) which maps to f, i.e., j is surjective. The exactness at the other two points is routine and holds for any module in place of I. We also need the following version of the adjunction isomorphism: Hom(A R B, C) = Hom R (A, Hom(B, C)) (6) Proof of Theorem We want to prove that Hom(F, D) is an injective R-module, i.e., that the functor Hom R (, Hom(F, D)) is exact. By (6) we have: Hom R (, Hom(F, D)) = Hom(F R ( ), D). Given an exact sequence E = (0 A B C 0), F R E = (0 F R A F R B F R C 0) is also exact since F is flat. Since D is divisible and therefore Z-injective, Hom(, D) is an exact functor and therefore takes the exact sequence F R E to an exact sequence Hom(F R E, D) = Hom R (E, Hom(F, D)) Therefore, Hom(F, D) is injective. The adjunction (6) will be proved in the next section. 5