**Chapter-IV**

**Projective Modules.**

**Definition:4.1**

A Sequence of R-modules and R-homomorphism of the type M M M M M

_{1}_{2}→……._{n}_{n+1 }is called an**exact sequence**if Image

_{ί}= Ker_{ί+1,}0≤ί≤n-1**Examples: 4.2**

1. The sequence o→2Z Z

_{2}→0 is an exact sequence where i is the inclusion map and þ is the projection of Z onto Z_{2}= Z/2Z2. For any two R-modules M and N, the sequence O→M M N N→O is exact where ί is the inclusion given by ί(x)=(x,0) and is the projection given by (x,y)=y,xЄM,yЄN

**Definition**:

**4.3**

An exact sequence O→M M M

^{″}→o of R-modules**splits**, if there exists an R-homomorphism t: M"→ M such that gt=I_{M}^{″}, the identity map on M^{″}.**Example:4.4**

Example (ii) above is a split exact sequence with the splitting map t:N→M

**Remark:4.5**

o→M M M is injective, g is surjective and

^{′}^{″}→o is exact ifg induces an isomorphism of coker ( )=M/f(M

^{′}) onto M″.**Proof:**

Assume that the sequence is exact.

Obviously injective and g is surjective.

Since M M″ and g is surjective we have M/kerg M″

But kerg=Im = (M

^{,}) Thus M/ (M M″.

^{′}) conversely suppose M/f(M M

^{′})″^{′}.Also M/kerg M″ kerg=f(M

^{′})=Imf.**Theorem: 4.6**

If o→M

^{′}

**Proof:**

Suppose o→M M M″→o is a split exact sequence.

^{′}Let t:M″→M be a splitting so that gt= I

_{M}″ This implies that t is injective.

For i t(x)=o,x M″,then

x=gt(x) ( gt=I

_{M}_{″}) =g(0)

x=0

I M, then =tg( )+{ -tg( )}

Now g{ -tg( }=g( )-g(tg ))

=g( )-g( )

=o.

To prove t(M″) kerg={0}

Let y t (M″) kerg

y=t(z)=t(o)=o

y=o

M M″ M′ since and t are injective maps.

**Corollary:4.7**

Let o→M M M″→o be an exact sequence of R-modules

^{′}which splits.Then there exists an R-homomorphism s:M→M’

such that sf=I

_{M’}.**Proof:**Let o→M

^{′}

Let t: M″→M be a splitting so that by above theorem,

we have M= (M t(M″)

^{′})Take (M

_{1}to be the projection of M onto^{′}),and let s=f^{-1 }_{1}Now sf:M (x)=( (x)

^{′}→M^{′}is such that s_{ }^{-1}_{1}) s (x)= (x))=x M

^{-1}(_{M}

^{′}

**Definition4.8**

Let M and N be R-modules.The set S of all R-homomorphisms of M in N has a natural R-module structure for the operations,

( +g)(x)= (x)+g(x) x M, s.

_{,}g (a )(x)=a (x), a R,x M, S.

This module is denoted by Hom

_{R}(M,N).**Remark:**

Let M be a fixed R-modules.Any homomorphism →N′ of

R-modules induces a homomorphism

f )= α,α Hom

^{*}:Hom_{R}(M_{1}N)→Hom_{R}(M_{’}N^{′}) given by^{*}(_{R}(M,N). Then (gf)*=g Hom

^{*}f^{*},g_{R}(N,N″) and I

^{*}_{N}=Id,the identity map of Hom_{R}(M,N) Similarly,for any fixed module N and a homomorphism g:M→M

^{′}, there exists an R-module homomorphismg* : Hom Hom

_{R }(M^{′}, N) à Hom_{R}(M, N) given by g*(β) = βg, β_{R }(M^{′}, N) Then (gh)* = h*g*, h Hom

_{R}(M^{′}, M^{″}) and I

_{M}^{*}= Id, the identity map of Hom_{R }(M, N)**Theorem:4.9**

For any R-module M, and an exact sequence oà N N N

^{′ }^{″}→0 ----(1) The induced sequence 0→ Hom Hom Hom

_{R}(M, N^{′})_{R }(M,N)_{R }(M,N″)is exact.

**Proof:**

Clearly * is injective, for if

_{R}(M,N

^{′}) then f α = 0.

This implies α = 0 as is injective.

Now g (x) = g( (x))

g (x) = 0 since Im = kerg.

Thus Im * kerg*

Conversely let β ker g* so that

g*(β) = gβ=0

Hence β(x) = (y) for some unique y M

This defines a map →N by setting α(x) = y where

β(x) = (y)

Clearly α is a homomorphism and

= β(x) ⩝x M

Hence β Im( *)

0→Hom Hom Hom

_{R}(M, N^{′})_{R}(M,N)_{R}(M, N″) is exact.**Remark:4.10**

By a similar argument we can also show that for any

R-module N and an exact sequence of R-modules

0→M’ M M″ → 0 the induced sequence

0 → Hom Hom Hom

_{R}(M″, N)_{R}(M,N)^{ }_{R }(M**, N) is exact.**^{′}**Definition:4.11**

A submodule N of M is called a N

**direct summand**of M if M=N**’**for some submodule N**of M.**^{′ }**Result:4.12**

Let K N M are sub-modules.

i) If N is a direct summand of M, N/K is a direct summand of M/K.

ii) If K is a direct summand of N and N is a direct summand of M, then K is a direct summand of M. Then k is a direct summand of M

iii) If K is a direct summand of M, then K is a direct summand of N. if further N/K is a direct summand of N/K, then N is a direct summand of M.

**Remark:4.13**

Certain types of modules P have the property that for any surjective homomorphism g: M → M″, the induced homomorphism g*: Hom

_{R}(P, M) → Hom_{R}(P, M″) is surjective. These are the projective modules defined as follows.**Definition:4.14**

An R-modules M is called

**projective**if it is a direct summand of a free R-module.**Theorem:4.15**

An R-module P is projective if and only if for any surjective homomorphism g: M → M″, the induced homomorphism

g*:Hom

_{R}(P, M) → Hom_{R}(P,M″) is surjective.**Proof:**

Assume first that P is free, with basis {e

_{i}}_{i}_{I}Given α Hom

_{R}(P, M^{″}), let α(e_{i}) = x_{i}Since g is surjective choose y M with g(y

_{i}_{i}) = x_{i}Then the map β : P → M I can be extended to an R-linear map β by defining β(Ʃa

_{,}defined by β(e_{i}) = y_{i}, i_{i}e_{i}) = Ʃa_{i}y_{i} [ (g*(β)(e

_{i}) = g*(yi)= g(y_{i}) = x

_{i} g*(β)(α

_{i}) = α(e_{i}) ] clearly g* (β)=α

This shows that g* is surjective.

If P is projective, there exist an R-module Q with P Q = F, free.

Let : F → P be the projective map

Given α Hom Hom

_{R}(P,M^{″}), consider α_{R}(F,M^{″})._{R}(F, M) with g* (β) = α

Then β

_{1 }= β|P : P→ M has the property that, g* (β_{1}) = αHence g* is surjective.

Conversely assume that P has the property stated in the theorem.

Express P as a quotient of a free module F.

Hence there is a surjective map g: F→ P.

g*: Hom

_{R}(P,F) → Hom_{R}(P,P) is surjective.In particular, there exists β Hom

_{R}(P, F) with g* (β) = gβ=I_{p}This shows that the exact sequence 0 kerg F P 0 split

P is isomorphic to a direct summand of F

Hence P is projective

**Example:4.16**

_{}

(i) Any free R –module is projective

(ii) If R is a principal ideal domain then any projective

R-module is free, as any submodule of a free

R-module is free.

**Theorem :4.17**

P = is projective iff is projective for each α.

**Proof**:

If p is projective, it is a direct summamd of a free module F.

Since is a direct summamd of P, it is also a direct summamd

Of F and hence projective .

Conversely assume that each is projective and consider a diagram

P

M M O

Let i

_{α}=_{α}, where i_{α}: P_{α}àP is the inclusion .Since is projective , there exists an R-linear map

_{α}=

_{α}

Define Ψ : P à M by setting

Ψ (x) = ( ) if x =

Then Ψ is R-linear and g Ψ = , showing that p is projective .

**Theorem :4.18**

An R-module p is projective if and only if every exact sequence

0 à M ′ M P à 0 splits

**Proof :**If P is projective , the identity map I

_{p}:P à P can be lifted to a map

_{p}

i.e the exact sequence 0 à M ′ M Pà 0splits.

To prove the converse, suppose the exact sequence

0 à M ′ M P à 0 splits.

Express P as the quotient of a free module F with kernal K so that the sequence 0à KàFàP à0 is exact

**Corollary:**

P is finitely generated projective if and only if P is a direct summand of a free module of finite rank.