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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   ( ) ).

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.

(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

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 :

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

(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)

Let   then

L.H.S

Or equivalently:

Let  then

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.

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)