**Chapter-IV**

**Projective Modules.**

**Definition:4.1**

A Sequence of R-modules and R-homomorphism of the type M M_{1} M_{2}→……. M_{n} M_{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/2Z

2. 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 N give by t(y)=(0,y)

**Remark:4.5**

o→M^{′} M M^{″}→o is exact if is injective, g is surjective and

g 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.

The sequence is exact.

**Theorem: 4.6**

If o→M^{′} M M″→o is a split exact sequence, then M M′ 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

t is injective.

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

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

=g( )-g( )

=o.

-tg( ) kerg.

=tg( )+{ -tg( )}

t(M″)+kerg.

M=t(M″)+kerg

To prove t(M″) kerg={0}

Let y t (M″) kerg

T(M″) and y kerg

y=t(z),z M″and g( )=o

z=gt(z)=g(y)=0

z=o

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

y=o

t(M′) kerg={o}

M=t(M″) kerg

t(M″) Imf ( kerg=Imf)

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^{′} M M″→o be an exact sequence.

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

we have M= (M^{′}) t(M″)

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

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

s (x)= ^{-1}( (x))=x M

s =I_{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, _{,}g s.

(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^{*}:Hom_{R}(M_{1}N)→Hom_{R}(M_{’}N^{′}) given by ^{*}( )= α,α Hom_{R}(M,N).

Then (gf)*=g^{*}f^{*},g Hom_{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 homomorphism

g* : Hom_{R }(M^{′}, N) à Hom_{R} (M, N) given by g*(β) = βg, β Hom_{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_{R}(M, N^{′}) Hom_{R }(M,N) Hom_{R }(M,N″)

is exact.

**Proof:**

Clearly * is injective, for if

* (α) = 0, α Hom_{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.

g = 0

g* * = (g )* = 0 as g = 0

Thus Im * kerg*

Conversely let β ker g* so that

g*(β) = gβ=0

For any x M, gβ(x) = 0

β kerg = Im ( the sequence (1) is exact)

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

α(x) = y)

= β(x) ⩝x M

α = β

Hence β Im( *)

ker g* Im *

Im( *) = kerg* and hence the sequence

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

**Definition:4.11**

A submodule N of M is called a **direct summand** of M if M=N 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_{i} M with g(y_{i}) = x_{i}

Then the map β : P → M_{,} defined by β(e_{i}) = y_{i}, i I can be extended to an R-linear map β by defining β(Ʃa_{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_{R}(P,M^{″}), consider α Hom_{R}(F,M^{″}).

By above case, there exists β Hom_{R} (F, M) with g* (β) = α

Then β_{1 }= β|P : P→ M has the property that, g* (β_{1}) = α

Hence 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.

By our assumption the induced map,

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

By theorem4.6

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 _{α} = i _{α} , where i_{α} : P_{α} àP is the inclusion .

Since is projective , there exists an R-linear map

: à M such that g Ψ _{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 à M such that g = I _{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

By assumption , this sequence splits so that P is a direct summand of F and hence projective.

**Corollary:**

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

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