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.                           

No comments:

PDF FILE TO YOUR EMAIL IMMEDIATELY PURCHASE NOTES & PAPER SOLUTION. @ Rs. 50/- each (GST extra)

HINDI ENTIRE PAPER SOLUTION

MARATHI PAPER SOLUTION

SSC MATHS I PAPER SOLUTION

SSC MATHS II PAPER SOLUTION

SSC SCIENCE I PAPER SOLUTION

SSC SCIENCE II PAPER SOLUTION

SSC ENGLISH PAPER SOLUTION

SSC & HSC ENGLISH WRITING SKILL

HSC ACCOUNTS NOTES

HSC OCM NOTES

HSC ECONOMICS NOTES

HSC SECRETARIAL PRACTICE NOTES

2019 Board Paper Solution

HSC ENGLISH SET A 2019 21st February, 2019

HSC ENGLISH SET B 2019 21st February, 2019

HSC ENGLISH SET C 2019 21st February, 2019

HSC ENGLISH SET D 2019 21st February, 2019

SECRETARIAL PRACTICE (S.P) 2019 25th February, 2019

HSC XII PHYSICS 2019 25th February, 2019

CHEMISTRY XII HSC SOLUTION 27th, February, 2019

OCM PAPER SOLUTION 2019 27th, February, 2019

HSC MATHS PAPER SOLUTION COMMERCE, 2nd March, 2019

HSC MATHS PAPER SOLUTION SCIENCE 2nd, March, 2019

SSC ENGLISH STD 10 5TH MARCH, 2019.

HSC XII ACCOUNTS 2019 6th March, 2019

HSC XII BIOLOGY 2019 6TH March, 2019

HSC XII ECONOMICS 9Th March 2019

SSC Maths I March 2019 Solution 10th Standard11th, March, 2019

SSC MATHS II MARCH 2019 SOLUTION 10TH STD.13th March, 2019

SSC SCIENCE I MARCH 2019 SOLUTION 10TH STD. 15th March, 2019.

SSC SCIENCE II MARCH 2019 SOLUTION 10TH STD. 18th March, 2019.

SSC SOCIAL SCIENCE I MARCH 2019 SOLUTION20th March, 2019

SSC SOCIAL SCIENCE II MARCH 2019 SOLUTION, 22nd March, 2019

XII CBSE - BOARD - MARCH - 2019 ENGLISH - QP + SOLUTIONS, 2nd March, 2019


BUY FROM PLAY STORE

DOWNLOAD OUR APP


HOW TO PURCHASE OUR NOTES?



S.P. Important Questions For Board Exam 2020


O.C.M. Important Questions for Board Exam. 2020


Economics Important Questions for Board Exam 2020


Chemistry Important Question Bank for board exam 2020


Physics – Section I- Important Question Bank for Maharashtra Board HSC Examination


Physics – Section II – Science- Important Question Bank for Maharashtra Board HSC 2020 Examination


MUST REMEMBER THINGS on the day of Exam


Are you prepared? for English Grammar in Board Exam.


Paper Presentation In Board Exam


How to Score Good Marks in SSC Board Exams


Tips To Score More Than 90% Marks In 12th Board Exam


How to write English exams?


How to prepare for board exam when less time is left


How to memorise what you learn for board exam


No. 1 Simple Hack, you can try out, in preparing for Board Exam

How to Study for CBSE Class 10 Board Exams Subject Wise Tips?

JEE Main 2020 Registration Process – Exam Pattern & Important Dates

NEET UG 2020 Registration Process Exam Pattern & Important Dates

How can One Prepare for two Competitive Exams at the same time?

8 Proven Tips to Handle Anxiety before Exams!