Commutative Algebra chapter III


  Chapter-III
Free module

Definition:3.1
         The annihilator of an R-module M is defined as
                                Ann (M)  ={ a Є R /aM =0}.
Note:3.2
i)               Ann (M) is an ideal of R.
ii)             If M is cyclic and is generated by x, Ann(M)  denoted by Ann(x).
Definition:3.3
           M is called a faithful R - module if Ann (M) =(0)
Definition:3.4
            M  is called a finitely generated R –module if
M =M1 +M2+…….+Mn where each Mi is cyclic.
              If Mi =R xi; ,then {x1,x2,……., xn} is called as generating  set for  M
Example:3.5
            The module of polynomials over R of degree atmost n is generated by 1,X,X2,……Xn
1,1+X, X2,…..Xn is also a generating set for the same module .
 The generating set is not unique.
Definition:3.6
       M is called a direct  sum of submodules M1,M2,…..,Mn ,
if every xЄ M can be uniquely expressed as      
                  x =x1+x2+………..+xn, xi Є Mi ,1≤ i≤n.
The direct sum is denoted by M= M1 M2 ……. Mn.
Theorem :3.7
    An R-module M =M1 M2 ……. Mn  iff
i)                 M = M1 + M2 + ………. +Mn and
ii)               M i (M1 +M2 +…….+ M i -1 + M i+1 +……. +Mn) =0
for all i , 1≤ i ≤ n.

Proof :
             Suppose M = M1 M2 …… Mn.
 Then clearly (i) is true .
To prove (ii)
          Suppose x Є Mi  (M1 + M2+……+ M i-1 +M i + 1………Mn)
x Є M i and x Є M1 +M2 +……..+M i-1+M i+1+…… + Mn
 x Є Mi and x =y1 +y2 +…….y i -1 +  y i+1 +…..yn, yj Є Mj, j  i Since x= 0 + 0 +…….. +0 +x +0 +…..+ 0 with x is in the  ith place,
We have by uniqueness , x= 0.
         M i  (M1 +M2 +…….+M i-1 +M i+1 +……..+Mn) =0.
         Thus  (ii) is true.
          Conversely assume the conditions (i)  and (ii)
         By(i) each xЄ M can be expressed as x =x1 + x2 +……….. +xn , xi Є Mi.
          Suppose x =y1 +y2 +…+yn, y i Є M i
 Then  0 =(x1 –y 1) + (x2 – y2) +…….+(x i – yi) +……+ (xn –y n)
          so that  xi – yi Є Mi
 (xi– yi ) = - [(x1 –y1 ) + ……+(x i-1  - y i-1) + (x i+1 – y i+1) +……
                                                                                                               …..+ ( xn – yn )]
xi - yi Є M1 +M2 +……+ Mi -1 +M i+1 + …..+Mn.
 xi – yi Є M i  (M1 +M2+….+Mi-1 + Mi+1 +………Mn)
       By (ii) Mi  (M1 +M2 +……+Mi-1 +M i+1 +……Mn) = 0,
        x i – y i =0 ,
          x i =yi , 1≤ i≤ n.
        Hence  every xЄ M can be uniquely written as
        x = x1 + x2 +…….+xn , xi Є Mi, 1≤ i≤ n.
M = M1 M2 ……. Mn.
Definition:3.8
      If M= M1 M2, x Є M then x can be uniquely expressed as
       x =x1 +x2,  x1 Є M1, x2 Є M2.
      The mappings 1 : M à M1, M à M2   defined by
 1(x) =x1,  2(x) =x2 are called projections.
Remark :
       The definition of direct sum can be extended to any collection of modules.
        For, An R-module M is a direct sum of a  collection of   submodules {Mα} α Є I  if each  x Є M can be expressed uniquely as
x=  + +…..+ ,  Є , α12…..αnЄI
We denote this by M =  
Definition :3.9
  A cyclic R-module M = R x is called free if  Ann (x) =0.
Definition :3.10
         An R –module M is called free  if it can be expressed as a direct sum M =   where each M  α is  a free cyclic R-module.
   If M α =Rx α, then the collection {X α } is  called a basis of the free module M.

Examples:3.11
(i)            R n ={(a1,a2,……an) / ai Є R } is a free R-module with basis
e1 =(1,0,0,…..0) , e2 =(0,1,0,…..), en = (0,0,……0,1).
(ii)          Zn , the group of integers module n is not free Z-module as each
xЄ Zn has  a  non –zero annihilator.
   Theorem:3.12
               Any two basis of a free module have the same cardinality.
 Proof:
         Let M be a free module with basis {x α} αϵI
          Choose a maximal ideal m in R and let R /m=K
          Then V=M/mM is annihilated by m hence it is a k-vector space
         Let  α= x α + mM
Claim: {  α} is a basis of V over K
             Let α ί + mϵ R/M=k,  + Mm ϵV
           Now,  Σ (α ί +m) ( +mM) =0,  ί ϵK=R/M
                       Σ α ί +mM=0
                                Σ α ί  =0
                       α ί=0        αίϵI

Let  = x + mM ϵ V=M/mM, xϵM
Since xϵM, x= Σ α ί            x+mM= Σ α ί  +mM
                          = Σ (α ί +mM)
                          = Σ(α ί +mM) ( +mM)
                         = Σ  i( +mM)
                   = Σ  i
{ α } a basis of V over K
Since any two basis of a vector space have the same Cardinality, it follows that any two basis of a free module have the same cardinality.
Corollary:3.13                                                                                                  
               If a free module F has a basis with n elements, then any other basis of F also has   n elements
Definition 3.14
          If a free module F has a basis with n element. Then n is called the rank of F.
Example: 3.15
1.    A cyclic module M =Rx with Ann (x) =0 is free of rank one.
2.    The R-module Rn is a free of rank n.
3.    The R-module R[x] is free with basis {1,x,x2,….Xn,…}and has countable rank
Definition:3.16
           Let M and N are R-modules and let : M→N The image of f is the set  imf =  (M)
       The Cokernal of  is Coker ( ) =N/im  which is a quotient module of N.
Remark:3.17    
         Homomorphic image of a finitely generated module is also finitely generated.
         For, let : M→N be an   R-module Homomorphism which is onto.
         Suppose M is generated by x1, x2,….. .x n
         Let yϵN.
         Since  is onto, x M such that (x) = y.
         Since x1, x2……, x n generate M,  
         x= for some a1, a2……, an ϵR
         y= (x) =  ( )
                        = f(xi)
         (x1),…..,  (x n) generates N.  


Theorem: 3.18
          M is a finitely generated R-module iff  M is isomorphic to a quotient of  R n for some integer n>0.
 Proof:
         Suppose M is a finitely generated R-module.
          Let x1, x2,…… x n generate M.
         Define : Rn → M by  (a1, a2,…,. an) =a1x1+a2x2+.., +an x n
Now,  ((a1, a 2,…, an) + (b1, b2,…, b n)) =  ((a1+b1,…., an +b n))
                                                       = (a1+b1) x1+ (a2+b2) x2+….+ (an +b n) x n
                                                                                      = (a1x1+a2x2+…+a n x n) + (b1x1+b2x2+bnxn)
                                                      =  (a1, a2…, an) +  (b1, b2…, b n)
 (a (a1, a2…, an) =  (aa1+aa2+…+aa n)
                            = (aa1) x1+ (aa2) x2+…+ (aa n) x n
                                         =a (a1x1+a2x2+…+an x n)
                                         =a  (a1, a2…, a n)
 is an R-module homomorphism.
 Clearly  is onto since x1, x2, …….x n generate M.
Thus  is a R-module homomorphism from Rn onto M.
          M  Rn/ker .
 Conversely suppose M Rn/ker  for some R-module homomorphism of R n onto M.
To prove M is a finitely generated R-module.
 Let e ί = (0, …,0,1, 0, 0) Є Rn
Now e ί   generate R n.
 since  is an R-module homomorphism, of R n onto M
    (e ί), 1≤ί≤n generate M.
 Hence M is a finitely generated R-module.
Remark:
       Any  R-module M can be expressed as the quotient of a free 
R- module F.