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MODULE THEORY 2

M.SC. COURSE , KSU SA 2008-2009 OCTOBER

THE TITLE OF THE COURSE IS:

 

 

MODULE THEORY

 

INTRODUCTION

 

Module Theory form the core of commutative algebra, which is essential in many important fields of mathematics, including:

 ALGEBRIC GEOMETRY.

 HOMOLOGICAL ALGEBRA AND ALGEBRAIC TOPOLOGY.

 REPRESENTATIONTHEORY OF GROUPS.

 It is also a generalization of theory of abelian groups ( An abelian

   group is a module over ). [ note be the ring of integers]  

 It is also a generalization of vector spaces.(A vector space is in fact a

   module over a field).

 

 Modules can be quite a bit more complicated than vector spaces; for instance, not all modules have a basis, and even those that do, free modules , behave differently from vector spaces in some respects.

 

Part I of this course is mainly concerned with carrying various results of  “ group theory to module theory “.

 

 Part II of  mainly concerns with :

  (i) classification ( up to isomorphism) of all free modules .

  (ii) classification ( up to isomorphism) of finitely generated modules over a principle ideal domain.

 

BOOKS

 

A. Algebra    by Thomas W.Hungerford, Springer- verlag.

B. Any others: (i) Topics in Algebra, by : I.N Herstein.

                       (ii) Rings, Modules and linear Algebra. By:

                             B. Hartley and T.O.Hawkes.

                       (iii) Rings and categories of modules by:

                              Anderson, Frank Wylie and Fuller, Kent R

                              (1992) Springer Verlag.

 

                       (iv) Lectures on modules and Rings, Graduate

                             texts in mathematics by: Lam, Tsit-Yuen(1999)

                          Springer- Verlag,

                       (v) Abelian groups that are directed summands of

                           every containing ablian group, Bulletin of

                          American Mathematics Society 46:800-806.

                     (vi) Direct product of modules,Chase,Stephen

                           U.(1960). Transactions of the American

                          Mathematical Society 97:457-473. And so on.

 

Prerequisites

 

Module is a graduate – level concept that would be first encountered in an abstract algebra course covering rings and fields.

 

 Field: A ring in which every non Zero element has a multiplicative

             inverse. The real numbers and the complex numbers are both fields.

 

 Ring: An abelian group together with a rule for multiplying its

   elements.

 Vector space: A set that is closed under finite vector addition and

    scalar multiplication. The basic example is dimensional Euclidean

     Space.

 

 Contents

(1) Modules and Submodules.

(2) Isomorphism theorems of modules.

(3) Direct product and direct sum of modules

(4) Exact sequences.

(5) Types of modules:

     (i) Free Modules

     (ii) Projective Modules.

     (iii) Injective Modules.

     (iv) Torsion Modules.

(6) Direct decomposition of finitely generated modules over a

     principal ideal domain.

(7) Application to group theory.

( 1—1 )

Modules

Basic Definitions And Examples

 

Definition

Let be a ring  , specifically, a left - module  consists of :

 (II) An action ( a mapping ) :  such that 

 

Called module  ( scaler )  multiplication such that for all and for all      we have :

Distributive laws: (1)

                                  (2) 

Associative law:   (3)

If  has an identity element  ( i.e. the ring  is unital )   and

                             (4)        ( unitary law)

Then  is said to be a unitary   module

 

If   is a division Ring then a unitary  (nunital )   module is called a left vector space.

 

 Remarks

 

(1) We simply write  as a left module or   in order to

      indicate the ring which is involved.

(2) A right   module or  is defined similarly, only the ring acts

     on the right, i.e. we have a module (scalar)  multiplication of the form

    and the above axioms are written with scalars on

    the right of .

(3)  A bimodule is a module which is both a left module and a right

      module.

(4)  If is commutative, then left  modules are the same as

      right  modules and are simply called  modules.

 

THEOREM

 Let be an module with additive identity element  over a ring  with additive (zero element ) identity , then 

The following hold:

(1)                                    (2)  

(3)          (4)

 

PROOF

(1)   

L.H.S.            is an module , distributive law is satisfied”

                   “ by adding   to both sides we get “

      

 (2)  

L.H.S.              is an  module , distributive law is satisfied”

                         “ by adding   to both sides we get “

     

     

           

      

3)  

 

We have to prove that  are additive inverses of  , i.e. we want to prove that :

  

    

 Now  

                   is an  module , distributive law is satisfied”

                               

                                 

                                       

                                  

   

Also          is an  module , distributive law is satisfied”

                                               

                                              

                                              

Thus   are inverses of     and from the uniqueness of the additive inverse we get 

Or equivalently :

   

Also          

             

Thus   are inverses of  and from the uniqueness of the additive inverse we get 

 

(4)

 (i)  If

               

 

(ii) 

          

_______________________________________________________________________

 

Examples Of Modules

 

1.  Every additive abelian group  ( finite or infinite ) is a unitary    module .

 ( Every abelian group is a module over the ring of integers   ( ) ).

 

Answer

Let   be an additive abelian group , we define a map ( the scalar multiplicative)

:

 

 

Then:

(1) Let 

(i)  If

      

(ii)       

     

 

Note

An abelian group and module  are  equivalent  concepts. That means every abelian group is a module .

 

(iii) 

       

 

(2)    

(3) 

(4) 

 Thus  is a   module , in fact a unitary   module.

 

Note

every abelian group is a module . Conversely , if is any  module , a fortiori   is an abelian group, so modules are the same as abelian groups.

 

2. Let be any ring. Then  is a left  module, where the action of a ring element on a module element is just the usual multiplication in the ring .

 (similarly, is a right module over itself) 

 

 

 

Proof

 

Since  is a ring ,then is an abelian group.( from the definition of the ring )

Now, we define the action :   

 So, let  

Then  (I)  1)                2) 

    ( Two distributive laws are satisfied, since  is a ring )

           (3)         ( Associative law is satisfied, since  is a ring )

Then   is a left  module.

 

Note 

 (1)    then  is a left as well as a right  module.

( 2) If   is not commutative it has a left and right module structure over itself and these structures may be different.

 

3) Let  =  be a field. Then every vector space over is an - module and vice versa. i.e. if  is an unitary  - module , then is a vector space over .

 

Proof

 

Since is an - module then is an abelian group , i.e. (1) addition is commutative.

(2) addition is associative . (3) has additive identity element.

(4)

Now we define the action:       

Then , let

(5)        “ since  is an - module”

(6)           “ since  is an - module 

(7)                “ since  is an - module 

 

Also, since  is a field then is a ring with unity , so

(8)

Furthermore:

Since  is an abelian group , then

(9)  is closed under addition.

Also, since  is an - module then

(10) is closed under the multiplication.

 Thus From (1) to (10) then  is a vector space over .

 

(4) Let be a ring with identity 1 and let and let .

Then  is an module by componentwise addition and multiplication by elements of .

i.e.  and

    

 

Proof

 

Since is a ring then is a ring , so is an abelian group.

Now, we define the action :

Now let

Then (1)

                                                                           

                                                                        

                                                                           

 

(2)

                                                                 

                                              

                                         

Thus the two distributive laws are satisfied.

(3)    

                                              

                                               

                                                 

 Thus associative law is satisfied. 

(4) if  

     Then 

                                                          “ because  is a ring "

 

Thus from  (1) --- (4) we get  is a unitary   module. 

 

(5) Let    be a ring , let     be a left ideal of   , then  is a left   module. 

 

   Proof

 

   Let     is a subring of   is an abelian group.

  We define the action (mapping)   

 

Now,  let

Then  (I)   (1) 

                 (2) 

       So , two distributive laws are satisfied , “ since is a ring”.

         (II)   (3)

      So, the associative law is satisfied, “ since is a ring”.

    (4) If  has an identity  then

    Then   is a unitary  module.

 

Note

 If    ,  then  is a right   module.

 

(6)  In particular the set  is an module for any ring .

      Or equivalently the trivial group  is a trivial  module.

 

Proop

 

For any ring we know that   is an abelian group.

Now, we define the action: 

Then   

(1) 

(2) 

(3) 

(4) 

Thus  is a unitary module.

 

 

Note

Up to this stage

 

Terminology for   any ring                                    Terminology for   any field

 

is an module.                                                   is a vector space over .

 is an element of   .                                           m is a vector in .

a  is a ring element.                                                  a  is a scalar. 

______________________________________________________________________

 

 

  

 

   

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(1-2)   Submodule

 

Definition

Let  be a ring and let  be an module. An submodule of  is a subgroup  of   which is closed under the action of ring elements , i.e.

Or equivalently:

Submodules of   are just subsets of  which are themselves modules under the restricted operations.

Or equivalently:

Let  be an module. An submodule of is a subset  of   such that:

SM1 :    is a subgroup of the additive group of   and

SM2 : 

 

Note

SM2   says the action :  which gives the   module structure of   maps  into . It is clear that modules axioms are satisfied and so  is actually an   module.

 Or equivalently:

Let  be a ring and let  be an module. An additive subgroup  of   is called an submodule of  if it is an module under the induced scalar multiplication.

 

Lemma

A subset  of an module  is a submodule of  if and only if :

(i)     ( or equivalently   )

(ii)

(iii)

 

Proof    Try  to prove  !!!

 

Remark

If has an identity 1 and  is a unital module then so is .

 

 

 

 

 

 

Examples

 

(1)  Any   module  has the two submodules  and . (  is called the trivial submodule )

The reson : Since  are modules then are submodules  of .

(2) Let be a vector space over a field . Any subspace of  is an - Submodule of .

(3) Over a field , the only submodules of are  .

 

Proof

Since  is a field  is a ring is - module. Since every field is a simple ring

The only  ideals of  are  the only submodules of are .

 

(4)  Since a module is an abelian group , then the submodules of   are its subgroups.

 

THEOREM

Let be  an  module   and let  and  be submodules of   , then the following are submodules:

 ( a ) 

  ( b )

PROOF

( a ) 

  We have to prove that

(1)  

(2) If 

(3)

 Now (1)  Since   and  are submodules 

 

(2) If 

Now, let

(3)

Now   Let

Define the action ( mapping )

Thus from  (1) to (3) is a submodule  of  .

 

( b )

 We have to prove:

(1)

(2) If 

(3)

 

Now,  (1)

Since

Thus 

(2) If 

Now,  Let       

Then:

                     

Since,           because

Then

 

 

(3)

Now, Let

Define the action 

Now,

                          Since   

 Since

 Thus from (1) to (3)  is a submodule of   .

 

THEOREM

 Let be an  module and let  be any nonempty collection of submodules of    . Then:

(a)   is a submodule of .

(b)  is a submodule of .

 

Note: 

(a) gives the cue for defining the submodule generated by a subset of .

 

roof

(a)   is a submodule of .

We have to prove :

(1)

(2) If    

(3)

Proof

(1)

Now, Since                 since is a submodule of  

 

(2) If    

Now, Let

     

                      since is a submodule 

(3)

Let  ,  define the action:

 

 

“ because is a submodule”

Thus from (1) to (3) we get  is a submodule of .

 

(b)  is a submodule of .

 

We have to prove :

(1) 

(2) If  

(3) 

   First notice that:

 

 

(1) 

Now, since:

Also 

(2) If 

Let 

Now, 

Since 

   

   

 

(3) 

Let 

Define the action :  such that:

Now, 

                 

Since is a submodule

Thus from  (1) to (3) is a submodule of  .

Definition

Let  be a ring and let be an module and let be a subset  of  . Then

(1) The intersection of all submodules of  containing is called the submodule generated by  and is denoted by . ( or submodule spanned by )

 

(2) If   is finite and  (   generates the module ,   is said to be finitely generated.

 

Or equivalently: if    finite we write   rather than  for the submodule generated by .

 

(3) If   , then  generates the zero module .

(4) If  consists of a single element  , then the submodule generated by is called the cyclic ( sub) module generated by .

(5) If   is a family of submodules of , then the submodule generated by   is called the sum of modules  .

If the index set  I is finite. Then the sum of   is denoted by .

 

Lemma

Let  be a ring, let  be an - module and let , then

(a)  If   then 

(b) If  

Theorem

Let  be a ring, let  be an - module and let and     be a family of submodules of   and . Then

(1)  is a submodule of  .

(2) The cyclic submodule generated by is

(3) The submodule generated by  is

 

(4) If  has a unity  and   is a unity - module , then:

        

(5)   is a submodule of   .

(6)  

        is called the sum of the modules  

Proof

  

(1)  is a submodule of  .

Now, we have to prove:

(i) 

Since is a ring  

                            

                            

  (ii) If

  Now,

  Since is an module, also   , since  is a ring

     

  (iii) We have to prove that :

 Now,      Since is an - module,    , since is a ring.

 

So , from  (1) to (3) , we get .

 is a submodule of  .

(2) The cyclic submodule generated by is

Now, let

 

Then we have to prove

To prove , we will prove:

(i)  is a submodule of  .

(ii)                         (iii) 

 

 

Proof

(i)  is a submodule of  .

We have to prove:

(1)

Since,

(2) We have to prove if 

Now, let,

          

Notice that :  because is an abelian group,

since it is module         

 

  is an module and module “.

                                                   

      

       (3) Let    we claim   ?

Since  

Then  

               

              

               

                 

              

 

Then from (1) to (3)    is a submodule of .           

  (ii)  

|Let      

Since   

But      is the smallest submodule containing   and it is also contained in every submodule containing m.

   

 (iii) 

    

Let

Now,  and    since  is  submodule of .

Also .

Then

So,

Thus from (ii) and (iii)

(3) The submodule generated by  is

Proof

Let 

 

 We have to prove that : , that means:

  (i)  .

  (ii)                        (iii) 

   Now, to prove , we have to prove:

(a)       

(b) If

 (c) 

 Proof

(a)       

Since  

    

   

 (b) If ?

  Let

     , 

Where,

By inserting  in both for the missing we can write:

              

Then:

          

          

          

        

       (c)    ?

     Let

         Then

                         

                       

By inserting   for the missing  we can write:

    

    

 Then from (a) to (c)   is submodule of   .

 

  (ii)  ?

 

Let  

   

But is a submodule of  and is the intersection of all submodules of containing S

To prove  

 

Let

Since      

(a) Since     is submodule   of  .

(b)  is a module,               

 

Thus from (a) and (b) and the fact   is a submodule of we have

From (1) to (3) we get

 ,   

 

(4) If  has a unity  and   is a unity - module , then:

        ?

Under the conditions is a ring with unity and  is a unitary - module.

We have to prove:

(1) (i)     and  (ii) 

Now,

Let,

(ii)   

Since is a unitary - module then 

But  is the intersection of all submodules of  containing m

Thus from (i) and (ii)

(2) We have to prove : (i)           and (ii) 

(i)

Let

Thus

(ii)

Since is a unitary module

But

But is a submodule of and  is the smallest submodule containing .

Then

Thus from (i) and (ii) we get

 

 

 

 (5)   is a submodule of   .?     Try to prove.

6)  

        is called the sum of the modules

We have to prove that

(i)              and  (ii)  

Where

  (i)   

  

    

   

  

But   is a submodule of   , so it contains the smallest submodule containing the set  

   

  

 

 

Let

Since 

But  is a submodule of

Then from (i) and (ii) we get

 =

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

EXAMPLES

 

(1) Let be a vector space over a field . Then  is finitely generated as - module

If and only if  is finite dimensional over  . Also is cyclic if and only if

(2) Let be an abelian group, then is finitely generated as a -module if and only if it is finitely generated as a group. Also is a cyclic -module if and only if it is a cyclic group.

 

(3) Let be a ring with unity. Then is a unity - module and it is cyclic because

      that means

(4) Let be a commutative  ring with unity and let , then .

Moreover is cyclic submodule of the - module if and only if   is a principal ideal of .

Remark

A finitely generated - module need not to be finitely generated as an abelian group.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

CHAPTER TWO

HOMOMORPHISM , QUOTIENT MODULES AND

ISOMORPHISM THEORMS OF MODULES

 

DEFINITION

Let  and   are  modules for some ring  . Let is a homomorphism of additive groups. Then we will say that    is an  homomorphism if

 OR EQUIVALENTLY

 Let  and   are  modules for some ring  . An module  homomorphism

Or   homomorphism  from to   is a map  such that

 (i) MH1         (i.e.  is a group momomorphism)

 (ii) MH2          

       

 

Note

(1) module  homomorphism is a homomorphism of additive abelian groups.

(2) If  is a division ring , an  module  homomorphism is called a linear transformation.

 

Remarks

 

(1) Notice that  and  are modules over the same ring  , we can not define a module homomorphism from modulento an   module , when   are different rings.

(2) If    is an module  homomorphism then:

  (a) If      is injective (  then   is called an module  momomorphism .

  (b) If    is surjective (onto) then   is called an module  epimorphism, also

        is called monomorphic image of .

 (c)  If    is injective (  and onto) then   is called an module  isomorphism,

       also and   are  called isomorphic modules and is denoted by

 (d) If   then  is called an module endomorphism.

(e) If   and  is an module isomorphism then is called an module

     automorphism.

 

 (3) If and  are modules then we denote by  the set of all from  to .

While we denote by

the set of all on .

 

(4) Any homomorphism ( where and are both  )

is a  module homomorphism , i.e. a group homomorphism.

 

(5) A ring endomorphism   is not necessary an module endomorphism.

 

 For example

Let  and consider the map 

Where  is the complex conjugateof  .  Then   is a ring endomorphism because :

 

(i)

(ii)

but  is not a   module endomorphism because if we take  then   Since

 So  

 

An module endomorphism   is not necessary  a ring homomorphismbecause , For example let   and consider the map

then   is a module homomorphism , because

 (i) 

  (ii)

   But   is not a ring endomorphism because if we take 

Because , L.H.S 

 

 

 

 

 

 

 

 

EXAMPLES

(1) If and  are modules for some ring . Then the map defined by:

  

 is an  homomorphism and is called the zero homomorphism and is denoted by .

 

Proof

Let     :

(i) 

(ii) 

Thus from  (i) and (ii)   is an homomorphism.

 (2)  If  be an  module for some ring  . Then the map on defined

By : 

Is an

Proof

Let 

(i) 

(ii) Define an action  then

   

 Then from  (i) and (ii) ,  is an  endomorphism and is called  the identity  homomorphism.

(3)  For any abelian groups  as modules , any group homomorphism:

 is a  homomorphism.

  

Proof

The first condition for homomorphism which is the group homomorphism is given.

So, we have to prove:

(i) For          

L.H.S.  = R . H. S

                                    n - times                                   n - times

 

(ii) For we have to prove that

        ?

 From (i) 

 for any group homomorphism the image of additive inverse is the inverse of the image of the element.

 

(3) Let be an  module. For any fixed  the map

      where  is commutative is homomorphism.

Proof

(i)               “definition of

                                        

                            “definition of  

(ii)                      “definition of

                                      

                                                       

                    

                   

 

(4) Let   be a submodule of an module , define a map

        Then   is an monomorphism,   is called the inclution or injection homomorphism.

Proof

Let 

1. We have to prove  is an homomorphism

 

(ii) Define an action :  

 

 Thus from (i) and (ii)  ,    is an  homomorphism.

 2. We have to prove that    is injective  

Let  

 

Then from  (1) and (2)  is an   monomorphism.

 

Note

As a special case , where  the inclusion  homomorphism is the identity homomorphism which is denoted by  .

 

 

THEOREM

 If  and  be modules, then:

  (a)     “ the set of all   homomorphisms from  to  

      is an abelian groupn, where    is defined by :

     

   (b) If   is a commutative ring, then  is an module if we define the action:

  to be

 Proof

 

 (a) To prove is an abelian group we have to prove :

(1) is closed under addition. Or ( addition is closed ).

(2)    is associative. Or ( addition is associative )

(3) Addition is commutative .

(4) There exists a unique additive identity in   .

(5) There exists a unique additive inverse .

 Proof

(1) Let

  are homomorphism.

We have to prove 

That means  

Now, Let

 (i) 

                               

  Since  are both homomorphism

                                 

                                 

(ii)

L.H.S.

 

                             

                             

So from (i) and (ii)

(2) Addition is associative:

Let   then

L.H.S

                                     

                                  

 

Or equivalently:

                                                     

                                                   

                                                     

 

(3) Addition is commutative:

Let  then  

                    

                   

(4)  has unique additive identity.

Since the zero map  is an module homomorphism

Then:

                          

                         

 Zero map is the unique additive identity.

(5) We have to prove all elements in   have additive inverse.

Let  then

                       

                       

Now, we have to prove is an homomrphism.

 

 

 

Now, since

(i) Let

                              

                             

(ii) let   

 

                 

                

                

Thus from (1) to (5) is an abelian group.

                       

b. We have to prove  is an module.

(i)

L.H.S

                              

                              

                              

                                                        

(ii)

                               

                               

                                   

 (iii)

 

L.H.S 

                             

                              

 

Also we have to prove that   is closed under the multiplication. That means

we have to prove that :  rf is a homomorphism.

(1)  is a group homomorphism.

 

                     

                     

 

(2)  

                   

                    

                      

                      

 

Definition

If and  are modules, and is an  homomorphism of into then:

 

(a)  

(b)  

If    then               

(c)

(d)

Theorem

Let  and be two modules, and  be an module homomorphism and let  be an submodule of   and    be an submodule of  .

 

Then:

(a) is an submodule of  .

(b)    is an submodule of   .

(c)  is an submodule of   .

(d)  is an submodule of  .

(e) 

(f)  is an injective (1-1) if and only if  .

 

Proof

(a) is an submodule of 

We have to prove that :

1. 

2.

3.

  

1. 

Since is an  module then   is an abelian group  

Also   “ Since   is an homomorphism “ then it is a group homomorphism  

.

2.

Let 

Consider

3.  We define the action  such that

 

 Now: let  

Now:          " since is an homomorphism.

  

Then from (1) to (3)   is an submodule  of  .

 

 

(b)    is an submodule of   .

 We have to prove is an submodule of   .

As (a) we have to prove:

 1.  

Since

2. Let :

 

Now: 

                     

Since

 

3.Define an action 

And let

Now, consider 

And since  

So erom (1) to (3)   is an submodule of  .

 

If    then               

(c)

Similarly:

 

1. Since is an submodule of

 

2. Let 

  Now ,  

Since and is an submodule

 

3. Let  , define the action

Since 

Now:

Since  

So from 1. To 3.  is an submodule of .

 

(d)  is an submodule of  .

Similarly.

 

1. Since  is an submodule of 

Since

 

 

2. Let 

Now,

 

3. We define the action:

Let

Since  is an submodule

So from 1. To 3. We get  is an submodule of  .

 

(e) 

Since  is an  homomorphism

 is a group homomorphism

(f)  is an injective (1-1) if and only if  .

Since  is an  homomorphism

 is a group homomorphism

 

 

 

Definition

A nontrivial module is called simple if the only submodules of   are .

 

Example

 is a simple .

 

 

 

QUOTIENT MODULES  

Theorem

Let be a ring , let  be an   module and let be a submodule of .  The additive abelian quotient group  where is normal, and 

 be the set of all cosets of   in .

Can be made into an module by defining an action of elements of  by

 

Or equivalently:

Theorem

Let be a ring , and  be an   module with an   submodule . Then  the set  of all cosets of   in  where  is normal , under the two binary operations

1.

2. Define an action 

Is an module.

Proof

First we have to prove: the scalar multiplication is well defined , Or equivalently

We have to prove the action of  on  is well defined , i.e, we have to show that :

If   , then 

Now , since  

Since is an submodule of

Thus the scalar multiplic

ation is well defined .

OR  Equivalently

      and this is true since is an submodule.

Second to prove  is an module we have to prove is an abelian group.

1.  is closed under the addition. Because:

Let  then

2. Addition is associative , because:

Let

                                                 

                                                  

                                                 

3. has a unique identity , because:

Let

So

4.

5. Addition is commutative, because

Let

Then

                                                                

Thus from 1. To 5. is an abelian group.

b. Let  , then

(i)

                                      

                                      

(ii)

                             

                            

                          

(iii)

                           

                          

(iv)

Thus from (a) and (b)  is an module .

 

Theorem

Let  be a finitely generated module and let be a submodule of  , then is a finitely generated module.

 

Proof

Let the set generates .

  where

Now let

Since

 

              

     

    

    is finitely generated  module.

 

Theorem

Let   be an    module for some ring    , and let    be an    submodule of    , then the natural map:

   is an    epimorphism with kernel .

 

Proof

We have to prove

1. The  natural map  is an homomorphism.

Let  , then

(i)          “ definition of

                    

                      

                    

(ii) We define an action :

     

Thus from (i) and (ii)  is an homomorphism.

2.  is onto , because:

So from 1. And 2. the natural map:

   is an   epimorphism . ( i.e module homomorphism and onto.)

3. To prove

Since

                    

                     

                      =

Thus from (1) to (3) the theorem  is valid.

 

 

THE ISOMORPHISM THEOREMS

 

Theorem

Let and   be modules , let be a submodule of and let  be the natural homomorphism. Let   be any homomorphism whose kernel contains . ( ), then   a unique homomorphism    (i.e. making the diagram  

 

                    

M                        N

                                   commute.

                 

Proof

We have to prove

1.  is well defined.

2.  is an homomorphism .

3.

4.    is unique.

First , define

1. Let  

 Then  is well defined.

 

2.  is an homomorphism .

  Let   ,  then

    (i)

                                             

                                             

                                            

(ii) Define the action

 

(i)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

         

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