43 Projective modules

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1 43 Projective modules 43.1 Note. If F is a free R-module and P F is a submodule then P need not be free even if P is a direct summand of F. Take e.g. R = Z/6Z. Notice that Z/2Z and Z/3Z are Z/6Z-modules and we have an isomorphism of Z/6Z-modules: Z/6Z = Z/2Z Z/3Z Thus Z/2Z and Z/3Z are non-free modules isomorphic to direct summands of the free module Z/6Z Definition. An R-module P is a projective module if there exists an R- module Q such that P Q is a free R-module Examples. 1) If R is a ring with identity then every free R-module is projective. 2) Z/2Z and Z/3Z are non-free projective Z/6Z-modules Definition. Let f i f i+1... M i Mi+1 Mi+2... be a sequence of R-modules and R-module homomorphisms. This sequence is exact if Im(f i ) = Ker(f i+1 ) for all i Definition. A short exact sequence is an exact sequence of R-modules the form 0 N f M g K 0 (where 0 is the trivial R-module). 171

2 43.6 Note. 1) A sequence 0 N f M f is a monomorphism g is an epimorphism Im(f) = Ker(g). g K 0 is short exact iff 2) If M is a submodule of M then we have a short exact sequence 0 M M M/M 1 Morever, up to an isomorphism, every short exact sequence is of this form: 0 N f M g K 0 = = 0 Ker(g) M M/ Ker(g) 0 = 43.7 Definition. A short exact sequence 0 N f M g K 0 is split exact if there is an isomorphism ϕ: M = N K such that the following diagram commutes: 0 N = f M g K ϕ = 0 N N K K 0 = Proposition. Let R be a ring and let 0 N f M g K 0 be a short exact sequence of R-modules. The following conditions are equivalent. 1) The sequence is split exact. 172

3 2) There exists a homomorphism h: K M such that gh = id K. 3) There exists a homomorphism k : M N such that kf = id N Proof. Exercise Theorem. Let R be a ring with identity and let P be an R-module. The following conditions are equivalent. 1) P is a projective module. 2) For any homomorphism f : P N and an epimorphism g : M N there is a homomorphism h: P M such that the following diagram commutes: M P h f g N 3) Every short exact sequence 0 N f M g P 0 splits. Proof. (1) (2) Let Q be a module such that P Q is a free module, and let B = {b i } i I be a basis of P Q. Since g is an epimorphism for every i I we can find m i M such that g(m i ) = f(b i ). Define h: P Q M by ( ) h r i b i := i i r i m i 173

4 Check: since B is a basis of P Q the map h is a well defined R-module homomorphism and g h = f. Then we can take h = h P. (2) (3) We have a diagram P M g Since g is an epimorphism there is h: P M such that gh = id P. Therefore by (43.8) the sequence 0 N f M g P 0 splits. id P P (3) (1) We have the canonical epimorphism of R-modules: This gives a short exact sequence f : p P R P 0 Ker(f) p P R f P 0 By assumption on P this sequence splits. so we obtain and thus P is a projective module. P Ker(f) = p P R Corollary. If R is a ring with identity, P is a projective R-module and f : M P is an epimorphism of R-modules then M = P Ker(f). Proof. We have a short exact sequence which splits by Theorem Ker(f) M f P 0 174

5 44 Projective modules over PIDs 44.1 Theorem. If R is a PID, F is a free R-module of a finite rank, and M F is a submodule then M is a free module and rank M rank F Corollary. If R is a PID then every finitely generated projective R-module is free. Proof. If P is a finitely generated projective R-module then we have an epimorphism f : R n P for some n > 0. By Corollary we have an isomorphism P Ker(f) = R n Therefore we can identify P with a submodule of R n, and thus by Theorem 44.1 P is a free module Note. Theorem 44.1 is true also for infinitely generated free modules over PIDs. As a consequence Corollary 44.2 is true for all (non necessarily finitelly generated) projective modules over PIDs. Proof of Theorem 44.1 (compare with the proof of Theorem 13.6). We can assume that F = R n. We want to show: if M R n then M is a free R-module and rank M n. Induction with respect to n: If n = 1 then M R, so M is an ideal of R. Since R is a PID we have M = a for some a R. If a = 0 then M = {0} is a free module of rank 0. Otherwise we have an isomorphism of R-modules f : R = M, 175 f(r) = ra

6 and so M is a free module of rank 1. Next, assume that for some n every submodule of R n is a free R-module of rank n, and let M R n+1. Take the homomorphism of R-modules We have: We have an epimorphism: f : R n+1 R, f(r 1,..., r n+1 ) = r n+1 Ker(f) = {(r 1,..., r n, 0) r i R} = R n f M : M Im(f M ) Since Im(f M ) R, thus Im(f M ) is a free R-module, and so by Corollary we have M = Im(f M ) Ker(f M ) We also have: Ker(f M ) = Ker(f) M It follows that that Ker(f M ) is a submodule of Ker(f), and since Ker(f) is a free R-module of rank n by the inductive assumption we get that Ker(f M ) is a free R-module of rank n. Therefore M = Im(f M ) }{{} free rank 1 and so M is a free R-module of rank n + 1. Ker(f M ) }{{} free rank n 176

7 45 The Grothendieck group Recall. If R is a ring with IBN and F is a free, finitely generated R-module then rank F = number of elements of a basis of F Goal. Extend the notion of rank to finitely generated projective modules. Idea. 1) Rank should be additive: rank(p Q) = rank P + rank Q. 2) Rank of a module need not be an integer. Each ring determines a group K 0 (R) such that for each finitely generated projective module rank of P is an element [P ] K 0 (R). Recall. A commutative monoid is a set M together with addition M M M, (x, y) x + y and with a trivial element 0 M such that the addition is associative, commutative and 0 + x = x for all x M Example. Let Proj fg R be the set of isomorphism classes of finitely generated projective R-modules. For a projective finitely generated R-module P denote The set Proj fg R The identity element in Proj fg R [P ] = the isomorphism class of P is a commutative monoid with addition given by [P ] + [Q] := [P Q] is [0], the isomorphism class of the zero module. 177

8 45.2 Theorem. Let M be a commutative monoid. There exists an abelian group Gr(M) and a homomorphism of monoids α M : M Gr(M) that satisfies the following universal property. If G is any abelian group and f : M G is a homomorphism of monoids then there exists a unique homomorphism of groups f : Gr(M) G such that the following diagram commutes: M G α M f Gr(M) Moreover, such group Gr(M) is unique up to isomorphism. f 45.3 Note. Let CMono denote the category of commutative monoids. We have the forgetful functor U : Ab CMono Theorem 45.2 is equivalent to the statement that this functor has a left adjoint Gr: CMono Ab, M Gr(M) 45.4 Definition. Let M be a commutative monoid. The group Gr(M) is called the group completion or the Grothendieck group of the monoid M. Proof of Theorem Construction of Gr(M). Let M be a commutative monoid. Define Gr(M) := M M/ 178

9 where (x, y) (x, y ) iff x + y + t = x + y + t for some t M Check: is an equivalence relation on M M. Notation: [x, y] := the equivalence class of (x, y) (Intuitively: [x, y] = x y) Note: for any x M we have [x, x] = [0, 0] since x + 0 = 0 + x. Addition in Gr(M): [x, y] + [x, y ] = [x + x, y + y ] Check: this operation is well defined, it is associative, and it has [0, 0] as the identity element. Additive inverses in Gr(M): [x, y] = [y, x] Indeed: [x, y] + [y, x] = [x + y, y + x] = [0, 0] Construction of the homomorphism α M : M Gr(M). Define α M : M Gr(M), x [x, 0] The universal property of Gr(M). Let G be an abelian group and let f : M G be a homomorphism of commutative monoids. Define f : Gr(M) G, f([x, y]) := f(x) f(y) Check: 179

10 1) f is a well defined group homomorphism. 2) fα M = f 3) f is the only homomorphism Gr(M) G satisfying conditions 1)-2). Uniqueness of Gr(M) follows from the universal property Examples. 1) Gr(N) = Z 2) Let M = N { } with n + = for all n M. Then Gr(M) is the trivial group. Indeed, for any m, n M we have since m + = + n. [m, n] = [, ] 3) If G is an abelian group then Gr(G) = G Definition. If R is a ring then K 0 (R) := Gr(Proj fg R ) 45.7 Notation. For [P ], [Q] Proj fg R denote [P ] [Q] :=[P, Q] K 0 (R) [P ] :=[P, 0] [Q] :=[0, Q] 45.8 Proposition. Let R be a ring with identity. If P, Q are finitely generated projective R-modules then [P ] = [Q] in K 0 (R) iff there exists n 0 such that P R n = Q R n. 180

11 Proof. ( ) If P R n = Q R n then in K 0 (R) we have [P ] + [R n ] = [P R n ] = [Q R n ] = [Q] + [R n ] and so [P ] = [Q] ( ) If [P ] = [Q] in K 0 (R) then P S = Q S for some finitely generated projective R-module S. Exercise: If S is a finitely generated projective R-module then there is a finitely generated projective R-module T such that S T = R n for some n 0. We obtain P R n = P S T = Q S T = Q R n 45.9 Definition. Let R be a ring with identity. We say that R-modules M, N are stably isomorphic if M R n = N R n for some n Definition. Let R be a ring with identity. We say an R-module M is stably free if M R n = R m for some m, n Note. Let R be a ring with identity. We have a homomorphism ϕ: Z K 0 (R) given by { [R n ] for n 0 ϕ(n) := [R n ] for n < 0 181

12 45.12 Proposition. Let R be a ring with identity, and let ϕ: Z K 0 (R) be the homomorphism as in (45.11). 1) ϕ is 1-1 iff R has IBN. 2) ϕ is an epimorphism iff every finitely generated projective R-module is stably free. Proof. 1) By Proposition 45.8 for n 0 we have n Ker(ϕ) iff R n R m = 0 R m for some m 0. If R has IBN this is possible only if n = 0, and so Ker(ϕ) = {0}. Conversely, assume that R does not have IBN. Then R n = R m for some n > m. This gives R n m R m = 0 R m, and so ϕ(n m) Ker(ϕ). 2) ( ) Assume that ϕ is an epimorphism. Then for every finitely generated projective R-module P we have [P ] = [R n ] for some n 0 or [P ] = [R n ] for some n 0. If [P ] = [R n ] then by Proposition 45.8 we have P R m = R n R m and so P is a stably free module. If [P ] = [R n ] then [0] = [P ] + [R n ] = [P R n ] Again by Proposition 45.8 this gives 0 R m = P R n R m, and again we obtain that P is stably free. ( ) The group K 0 (R) is generated by elements [P ] where P is a finitely generated projective R-module, so it is enough to show that for any such P we have [P ] = ϕ(k) for some k Z. Since P is stably free we have P R n = R m for some n, m 0. This gives [P ] + [R n ] = [P R n ] = [R m ] Therefore [P ] = [R m ] [R n ] = ϕ(m) ϕ(n) = ϕ(m n). 182

13 45.13 Example. Here is an example of a stable free module that is not free. For details see: R. G. Swan, Vector bundles and projective modules, Transactions AMS 105 (2) (1962), Let B be a compact, normal, topological space and let p: E B be a real vector bundle over B. Define: C(B) = {f : B R f - continuous } C(B) is a ring (with pointwise addition and multiplication). Let Γ(p) be the set of all continuous sections of p: Γ(p) = {s: B E ps = id B } Note: Γ(p) is an C(B)-module with poinwise addition and pointwise multiplication by elements of C(B). Fact 1. The module Γ(p) is free iff p is a trivial vector bundle. Fact 2. If p: E B and q : E B are real vector bundles over B then we have an isomorphism of C(B)-modules: Γ(p q) = Γ(p) Γ(q) Upshot. If p: E B, q : E B are bundles such that p is non-trivial, but both q and p q are trivial bundles then Γ(p) is stably free C(B)-module that is not free. Indeed, in such case we have: Γ(p) Γ(q) }{{} free 183 = Γ(p q) }{{} free

14 Fact 3. It is possible to find vector bundles as above. Take e.g. p: T S 2 S 2 to be the tangent bundle of the 2-dimensional sphere, and q : S 2 R 1 S 2 to be the 1-dimensional trivial bundle over S 2. Note: one can also show that a C(B)-module M is finitely generated projective module iff M = Γ(p) for some vector bundle p: E B. 184

1.1 Definition. A monoid is a set M together with a map. 1.3 Definition. A monoid is commutative if x y = y x for all x, y M.

1 Monoids and groups 1.1 Definition. A monoid is a set M together with a map M M M, (x, y) x y such that (i) (x y) z = x (y z) x, y, z M (associativity); (ii) e M such that x e = e x = x for all x M (e