Commutative Algebra chapter IV

                                    Projective  Modules.
                   A Sequence of R-modules and R-homomorphism  of the type M M1 M2→……. Mn Mn+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

 An exact sequence O→M M M→o of R-modules splits, if there exists an  R-homomorphism t: M"→ M such that gt=IM, the identity map on M.
               Example (ii) above is a split exact sequence with the splitting map t:N→M
N give by t(y)=(0,y)
           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″.
        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″       
Suppose  o→M M M″→o is a split exact sequence.
Let t:M″→M be a splitting so that gt= IM
 This implies that t is injective.
 For  i t(x)=o,x M,then
         x=gt(x) ( gt=IM)
         t is injective.
          I   M, then  =tg( )+{ -tg( )}
Now g{ -tg( }=g( )-g(tg ))
                              =g( )-g( )
  -tg( )  kerg.
=tg( )+{ -tg( )}
To prove t(M) kerg={0}
Let y t (M)  kerg
                              T(M″) and y kerg
                              y=t(z),z M″and g( )=o
        t(M) kerg={o}
                     M=t(M) kerg
            t(M)  Imf    ( kerg=Imf)
            M M″ M′ since  and t  are injective maps.
  Let oM  M M″→o be an exact sequence of R-modules
which splits.Then there exists an R-homomorphism s:MM’
such that sf=IM’.
        Let oM 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:MM  is  such that  s (x)=( -1 1) (x)
                                                      s (x)= -1( (x))=x M
                               s =IM

         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 HomR(M,N).
          Let M be a fixed R-modules.Any homomorphism N of
R-modules induces a homomorphism
f*:HomR(M1N)HomR(MN) given by *( )= α,α  HomR(M,N).
 Then (gf)*=g*f*,g HomR(N,N) and
              I*N=Id,the identity map of HomR(M,N)
           Similarly,for any fixed module N and a homomorphism g:MM, there exists an R-module homomorphism
g* : HomR (M, N) à HomR (M, N) given by g*(β) = βg, β  HomR (M, N)
       Then (gh)* = h*g*, h  HomR (M, M)
               and IM* = Id, the identity map of HomR (M, N)
   For any R-module M, and an exact sequence  oà N  N  N →0 ----(1)              
 The induced sequence 0 HomR(M, N)  HomR (M,N)  HomR (M,N″)
is exact.
            Clearly * is injective, for if
        * (α) = 0, α HomR (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→HomR (M, N)  HomR (M,N) HomR (M, N) is exact.
            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 HomR (M, N) HomR (M,N) HomR (M, N) is exact.

            A submodule N of M is called a direct summand of M if M=N Nfor some submodule Nof M.
            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.

      Certain types of modules P have the property that for any surjective homomorphism g: M M, the induced homomorphism g*: HomR (P, M) HomR (P, M) is surjective. These are the projective modules defined as follows.
      An R-modules M is called projective  if  it  is a direct summand of a free R-module.
      An R-module P is projective if and only if for any surjective homomorphism g: M M, the induced homomorphism
g*:HomR (P, M) HomR (P,M) is surjective.

              Assume first that P is free, with basis {ei}i I
Given α HomR (P, M), let α(ei) = xi
Since g is surjective choose yi M with    g(yi) = xi
Then the map β : P → M, defined by β(ei) = yi, i I can be extended to an R-linear map β by defining β(Ʃaiei) = Ʃaiyi­
               [ (g*(β)(ei) = g*(yi)= g(yi)
                              = xi
           g*(β)(α) = α(ei) ]          
                 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  HomR(P,M), consider α HomR(F,M).
 By above case, there exists β  HomR (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*: HomR (P,F) HomR (P,P) is surjective.
In particular, there exists β  HomR (P, F) with g* (β) = gβ=Ip
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

(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                                         
                                                   M M O

Let α =  i α , where iα : Pα 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.
 P is finitely generated projective if and only if P is a  direct   summand of a free module of finite rank.