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→……. 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

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=IM, 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= IM
 This implies that t is injective.
 For  i t(x)=o,x M,then
         x=gt(x) ( gt=IM)
           =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 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’.
Proof:  
        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

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 HomR(M,N).
Remark:
          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)
Theorem:4.9
   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.
Proof:
            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.
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 HomR (M, N) HomR (M,N) HomR (M, N) is exact.


Definition:4.11
            A submodule N of M is called a direct summand of M if M=N Nfor some submodule Nof 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*: HomR (P, M) HomR (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*:HomR (P, M) HomR (P,M) is surjective.

Proof:
              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


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

Veg Special - சைவ உணவு

உளுந்து பாயாசம்

ரொம்ப ஒல்லியா இருக்கீங்களா இதை மட்டும் குடிங்க.

மருத்துவ குணம் நிறைந்த கற்பூரவள்ளியில் பஜ்ஜி செய்யலாமா?

சத்துமிக்க அவரைக்காய் வெந்தயக்கீரை பருப்பு கூட்டு

சத்தான காலை நேர உணவு.. அரிசி மாவு புட்டு!

சாஃப்ட் இட்லி: பிசுபிசுன்னு ஒட்டாமல் இருக்கும் ரகசியம் இதுதான்!

பல நோய்களை தீர்த்து, உடலுக்கு அற்புத ஆற்றலை வழங்கும் துளசி.!

பிசி பெலே பாத் - சாம்பார் சாதம் - செய்முறை (தமிழில்)

உடலில் உள்ள கெட்ட கொழுப்புகளை கரைக்க கூடிய உணவுகள்...!

இவ்வளவு மருத்துவ குணங்களை கொண்டுள்ளதா முருங்கைக் கீரை....?

எலுமிச்சை சாறு பருகுவதால் உண்டாகும் பயன்கள்...!

Veg Grilled Sandwich in Tamil with English Subtitles.

சர்க்கரை நோய் குணமாக இது ஒன்று போதும்!