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 ϕ: M −→= N ⊕K such that the following diagram commutes:

f g 0 / N / M / K / 0

= ϕ =∼ = 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 = {bi}i∈I be a basis of P ⊕ Q. Since g is an epimorphism for every i ∈ I we can ﬁnd mi ∈ 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 = h|P .

(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 p∈P This gives a short exact sequence M f 0 −→ Ker(f)−→ R −→ P −→ 0 p∈P By assumption on P this sequence splits. so we obtain M P ⊕ Ker(f) ∼= R p∈P 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 epimorphism 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 ∈ 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, 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) = {(r1, . . . , rn, 0) | ri ∈ R} = R We have an epimorphism:

f|M : M → Im(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 ) ⊕ Ker(f|M ) | {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 ] ∈ 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 ∈ M such that the addition is associative, commutative and 0 + x = x for all x ∈ 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 homomorphism 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 ∈ 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] + [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).

Deﬁne αM : M → Gr(M), x 7→ [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 1) f¯ is a well deﬁned group homomorphism. ¯ 2) fαM = f 3) f¯ is the only homomorphism Gr(M) → G satisfying conditions 1)-2).

Uniqueness of Gr(M) follows from the universal property.

45.5 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 [m, n] = [∞, ∞] since m + ∞ = ∞ + n.

3) If G is an abelian group then Gr(G) ∼= G.

fg 45.6 Deﬁnition. If R is a ring then K0(R) := Gr(ProjR )

fg 45.7 Notation. For [P ], [Q] ∈ ProjR denote

[P ] − [Q] :=[P,Q] ∈ K0(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 K0(R) iﬀ there exists n ≥ 0 such that P ⊕ Rn ∼= Q ⊕ Rn.

180 Proof.

n ∼ n (⇒) If P ⊕ R = Q ⊕ R then in K0(R) we have [P ] + [Rn] = [P ⊕ Rn] = [Q ⊕ Rn] = [Q] + [Rn] and so [P ] = [Q]

(⇒) If [P ] = [Q] in K0(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 ∼= Rn for some n ≥ 0.

We obtain P ⊕ Rn ∼= P ⊕ S ⊕ T ∼= Q ⊕ S ⊕ T ∼= Q ⊕ Rn

45.9 Deﬁnition. Let R be a ring with identity. We say that R-modules M, N are stably isomorphic if M ⊕ Rn ∼= N ⊕ Rn for some n ≥ 0.

45.10 Deﬁnition. Let R be a ring with identity. We say an R-module M is stably free if M ⊕ Rn ∼= Rm for some m, n ≥ 0.

45.11 Note. Let R be a ring with identity. We have a homomorphism ϕ: Z → K0(R) given by ( [Rn] for n ≥ 0 ϕ(n) := −[R−n] for n < 0

181 45.12 Proposition. Let R be a ring with identity, and let ϕ: Z → K0(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 Rn ⊕Rm ∼= 0⊕Rm 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 Rn ∼= Rm for some n > m. This gives Rn−m ⊕ Rm ∼= 0 ⊕ Rm, 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 ] = [Rn] for some n ≥ 0 or [P ] = −[Rn] for some n ≥ 0.

If [P ] = [Rn] then by Proposition 45.8 we have P ⊕ Rm ∼= Rn ⊕ Rm and so P is a stably free module.

If [P ] = −[Rn] then [0] = [P ] + [Rn] = [P ⊕ Rn] Again by Proposition 45.8 this gives 0 ⊕ Rm ∼= P ⊕ Rn ⊕ Rm, and again we obtain that P is stably free.

(⇐) The group K0(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 ⊕ Rn ∼= Rm for some n, m ≥ 0. This gives [P ] + [Rn] = [P ⊕ Rn] = [Rm] Therefore [P ] = [Rm] − [Rn] = ϕ(m) − ϕ(n) = ϕ(m − n).

182 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), 264-277.

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 = idB}

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 : E0 → 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 : E0 → 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) ∼= Γ(p ⊕ q) |{z} | {z } free free

183 Fact 3. It is possible to ﬁnd vector bundles as above. Take e.g. p: TS2 → S2 to be the tangent bundle of the 2-dimensional sphere, and q : S2 × R1 → S2 to be the 1-dimensional trivial bundle over S2.

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