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43 Projective Modules 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. 43.2 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. 43.3 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. 43.4 Definition. Let fi fi+1 ::: −! Mi −! Mi+1 −! Mi+2 −! ::: be a sequence of R-modules and R-module homomorphisms. This sequence is exact if Im(fi) = Ker(fi+1) for all i. 43.5 Definition. A short exact sequence is an exact sequence of R-modules the form f g 0 −! N −! M −! K −! 0 (where 0 is the trivial R-module). 171 43.6 Note. f g 1) A sequence 0 ! N −! M −! K ! 0 is short exact iff • f is a monomorphism • g is an epimorphism • Im(f) = Ker(g). 2) If M 0 is a submodule of M then we have a short exact sequence 0 −! M 0 −! M −! M=M 0 −! 1 Morever, up to an isomorphism, every short exact sequence is of this form: f g 0 / N / M / K / 0 =∼ = =∼ 0 / Ker(g) / M / M= Ker(g) / 0 43.7 Definition. A short exact sequence f g 0 −! N −! M −! K −! 0 ∼ is split exact if there is an isomorphism ': M −!= N ⊕K such that the following diagram commutes: f g 0 / N / M / K / 0 = ' =∼ = 0 / N / N ⊕ K / K / 0 f g 43.8 Proposition. Let R be a ring and let 0 ! N −! M −! K ! 0 be a short exact sequence of R-modules. The following conditions are equivalent. 1) The sequence is split exact. 172 2) There exists a homomorphism h: K ! M such that gh = idK . 3) There exists a homomorphism k : M ! N such that kf = idN Proof. Exercise. 43.9 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: P } } h } } f } } } ~} g M / N f g 3) Every short exact sequence 0 ! N −! M −! P ! 0 splits. Proof. (1) ) (2) Let Q be a module such that P ⊕ Q is a free module, and let B = fbigi2I be a basis of P ⊕ Q. Since g is an epimorphism for every i 2 I we can find mi 2 M such that g(mi) = f(bi). Define h~ : P ⊕ Q −! M by ! ~ X X h ribi := rimi i i 173 Check: since B is a basis of P ⊕ Q the map h~ is a well defined R-module ~ ~ homomorphism and gh = f. Then we can take h = hjP . (2) ) (3) We have a diagram P idP g M / P Since g is an epimorphism there is h: P ! M such that gh = idP . Therefore f g by (43.8) the sequence 0 ! N −! M −! P ! 0 splits. (3) ) (1) We have the canonical epimorphism of R-modules: M f : R ! P p2P This gives a short exact sequence M f 0 −! Ker(f)−! R −! P −! 0 p2P By assumption on P this sequence splits. so we obtain M P ⊕ Ker(f) ∼= R p2P and thus P is a projective module. 43.10 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 f 0 −! Ker(f)−!M −! P −! 0 which splits by Theorem 43.9. 174 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 . 44.2 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 epimor- phism f : Rn ! P for some n > 0. By Corollary 43.10 we have an isomorphism P ⊕ Ker(f) ∼= Rn Therefore we can identify P with a submodule of Rn, and thus by Theorem 44.1 P is a free module. 44.3 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 = Rn. We want to show: if M ⊆ Rn 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 = hai for some a 2 R. If a = 0 then M = f0g is a free module of rank 0. Otherwise we have an isomorphism of R-modules ∼ f : R −!= M; f(r) = ra 175 and so M is a free module of rank 1. Next, assume that for some n every submodule of Rn is a free R-module of rank ≤ n, and let M ⊆ Rn+1. Take the homomorphism of R-modules n+1 f : R ! R; f(r1; : : : ; rn+1) = rn+1 We have: ∼ n Ker(f) = f(r1; : : : ; rn; 0) j ri 2 Rg = R We have an epimorphism: fjM : M ! Im(fjM ) Since Im(fjM ) ⊆ R, thus Im(fjM ) is a free R-module, and so by Corollary 43.10 we have ∼ M = Im(fjM ) ⊕ Ker(fjM ) We also have: Ker(fjM ) = Ker(f) \ M It follows that that Ker(fjM ) 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(fjM ) is a free R-module of rank ≤ n. Therefore ∼ M = Im(fjM ) ⊕ Ker(fjM ) | {z } | {z } free free rank ≤ 1 rank ≤ n and so M is a free R-module of rank ≤ n + 1. 176 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 K0(R) such that for each finitely generated projective module rank of P is an element [P ] 2 K0(R). Recall. A commutative monoid is a set M together with addition M × M ! M; (x; y) 7! x + y and with a trivial element 0 2 M such that the addition is associative, commu- tative and 0 + x = x for all x 2 M. fg 45.1 Example. Let ProjR be the set of isomorphism classes of finitely generated projective R-modules. For a projective finitely generated R-module P denote [P ] = the isomorphism class of P fg The set ProjR is a commutative monoid with addition given by [P ] + [Q] := [P ⊕ Q] fg The identity element in ProjR is [0], the isomorphism class of the zero module. 177 45.2 Theorem. Let M be a commutative monoid. There exists an abelian group Gr(M) and a homomorphism of monoids αM : 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 homomor- phism of groups f¯: Gr(M) ! G such that the following diagram commutes: f M / G z= z z z αM z z f¯ z z z Gr(M) Moreover, such group Gr(M) is unique up to isomorphism. 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 7! 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 45.2. Construction of Gr(M). Let M be a commutative monoid. Define Gr(M) := M × M= ∼ 178 where (x; y) ∼ (x0; y0) iff x + y0 + t = x0 + y + t for some t 2 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 2 M we have [x; x] = [0; 0] since x + 0 = 0 + x. Addition in Gr(M): [x; y] + [x0; y0] = [x + x0; y + y0] 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 αM : M ! Gr(M).
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