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+…….+M_n where each M_i is cyclic.

              If M_i =R x_i; ,then {x₁,x₂,……., x_n} 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,…..,M_n ,

if every xЄ M can be uniquely expressed as      

                  x =x₁+x₂+………..+x_n, x_i Є M_i ,1≤ i≤n.

The direct sum is denoted by M= M₁ M₂ ……. M_n.

Theorem :3.7

    An R-module M =M₁ M₂ ……. M_n  iff

i)                 M = M₁ + M₂ + ………. +M_n and

ii)               M _i (M₁ +M₂ +…….+ M _{i -1} + M _i+1 +……. +M_n) =0

for all i , 1≤ i ≤ n.

Proof :

             Suppose M = M₁ M₂ …… M_n.

 Then clearly (i) is true .

To prove (ii)

          Suppose x Є M_i  (M₁ + M₂+……+ M _i-1 +M _{i + 1}………M_n)

x Є M _i and x Є M₁ +M₂ +……..+M _i-1+M _i+1+…… + M_n

 x Є M_i and x =y₁ +y₂ +…….y _{i -1} +  y _i+1 +…..y_n, y_j Є M_j, j  i Since x= 0 + 0 +…….. +0 +x +0 +…..+ 0 with x is in the  i^th place,

We have by uniqueness , x= 0.

         M _i  (M₁ +M₂ +…….+M _i-1 +M _i+1 +……..+M_n) =0.

         Thus  (ii) is true.

          Conversely assume the conditions (i)  and (ii)

         By(i) each xЄ M can be expressed as x =x₁ + x₂ +……….. +x_n , x_i Є M_i.

          Suppose x =y₁ +y₂ +…+y_n, y _i Є M _i

 Then  0 =(x₁ –y ₁) + (x₂ – y₂) +…….+(x _i – y_i) +……+ (x_n –y _n)

          so that  x_i – y_i Є M_i

 (x_i– y_i ) = - [(x₁ –y₁ ) + ……+(x _i-1  - y _i-1) + (x _i+1 – y _i+1) +……

                                                                                                               …..+ ( x_n – y_n )]

x_i - y_i Є M₁ +M₂ +……+ M_{i -1} +M _i+1 + …..+M_n.

 x_i – y_i Є M _i  (M₁ +M₂+….+M_i-1 + M_i+1 +………M_n)

       By (ii) M_i  (M₁ +M₂ +……+M_i-1 +M _i+1 +……M_n) = 0,

        x _i – y _i =0 ,

          x _i =y_i , 1≤ i≤ n.

        Hence  every xЄ M can be uniquely written as

        x = x₁ + x₂ +…….+x_n , x_i Є M_i, 1≤ i≤ n.

M = M₁ M₂ ……. M_n.

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 {M_α} _{α Є I}  if each  x Є M can be expressed uniquely as

x=  + +…..+ ,  Є , α₁,α₂…..α_nЄI

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,……a_n) / a_i Є R } is a free R-module with basis

e₁ =(1,0,0,…..0) , e₂ =(0,1,0,…..), e_n = (0,0,……0,1).

(ii)          Z_n , the group of integers module n is not free Z-module as each

xЄ Z_n 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        α_ίϵI

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 x_1, x_{2,….. .}x _n

         Let yϵN.

         Since  is onto, x M such that (x) = y.

         Since x_1, x_2……, x _n generate M,  

         x= for some a_1, a_2……, a_n ϵR

         y= (x) =  ( )

                        = f(x_i)

         (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, x_2,…… x _n generate M.

         Define : Rⁿ → M by  (a₁, a_2,…,. an) =a₁x₁+a₂x₂+.., +a_n x _n

Now,  ((a_1, a _2,…, a_n) + (b₁, b₂,…, b _n)) =  ((a₁+b_1,…., a_n +b _n))

                                                       = (a₁+b₁) x₁+ (a₂+b₂) x₂+….+ (a_n +b _n) x _n

                                                                                      = (a₁x₁+a₂x₂+…+a _n x _n) + (b₁x₁+b₂x₂+b_nx_n)

                                                      =  (a_1, a_2…, a_n) +  (b_1, b_2…, b _n)

 (a (a_1, a_2…, a_n) =  (aa₁+aa₂+…+aa _n)

                            = (aa₁) x₁+ (aa₂) x₂+…+ (aa _n) x _n

                                         =a (a₁x₁+a₂x₂+…+a_n x _n)

                                         =a  (a_1, a_2…, a _n)

 is an R-module homomorphism.

 Clearly  is onto since x_1, x_{2, …….}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.