Commutative Algebra chapter IV

   Chapter-IV

                                    Projective  Modules.

Definition:4.1

                   A Sequence of R-modules and R-homomorphism  of the type M M₁ M₂→……. 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 Z₂→0 is an exact sequence where i is the inclusion map and þ is the projection of Z onto       Z₂= 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(M₁N)→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_ie_i) = Ʃa_iy_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 β₁ = β|P : P→ M has the property that, g* (β₁) = 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.