Field theory

Feild ;-

An algebric expression (F,+,.) where F is a non-empty set with two operations “addition and dot” is called ring if it is satisfies following axioms..

D1. (F, + ) be an abelian group.

1)      closure for addition:-  if a,b ∈F => a+b∈F for all a,b∈F

2)      associate for addition:-if a,b,c ∈F then (a+b)+c=a+(b+c) for all a,b,c∈F

3)      identity for addition:-  if for all a∈F there are exist 0∈F such that a+0=a for all a∈F

4)      inverse for addition:- if for all a∈F then there are exist –a∈F such that a+(-a) ∈F for all a∈F

5)      commutative low    :-  for all a,b∈F such that a+b=b+a for all a,b∈F

D2.  for  (R,.).

      1)   closure for multifi. :-   for all a,b∈F => a.b∈F  for all a,b∈F

      2)  associate for dot    :-   for all a,b,c∈F st    (a.b).c=a.(b.c)  for all a,b,c∈F

      3) Commutative for dot:- for all a,b∈F => a.b =b.a

      4) unit element :- 1 ∈F st a.1 = a =1.a for all a∈F

      5) Inverse for dot:- for all a(≠0) ∈F there are exist  a^-1 ∈F st a a^-1 = 1

D3.  distribution law :- for all a,b,c∈F

       1) left distribution:-             a.(b+c)=a.b+a.c for all a,b,c∈F

       2) Right distribution :-       (b+c).a=b.a+c.a for all a,b,c∈F.

So (F,+,.) is a Field…………..

Subfield:-  A subset F' of F is called subfield if it is satisfies all conditions of a field.i.e. if F' shall be a Feild itself then it is called subfield.
Prime field :- A field ( F,+,.) is called prime field if it have not any proper subfield .
 Theoram:-Prove that Every field always be a integral domain.
proof :- :- let (F,+,.) be a field. Since we know that every field be a commutative ring with a unit element.

To show:- (F,+,.) will a integral domain.

                  For this we will only to show that every field does not contains a zero divisor.

                 Let a,b∈F be orbitary element of F .and let a≠0 such that ab≠0

                Since a≠0 => there are exist a^-1 ∈F .

                So, ab=0 => a^-1 (ab) = a^-1 .0    [a.0=0]

                                  =>( a^-1 a) b = 0      [associate law]

                                  =>1.b = 0                              [a^-1 a = 1]

                                  => b=0                  [1.b=b]

                 Similarly, suppose  ab = 0 and b≠0

                b≠0 => b^-1 ∈F.

therefore ab = 0 => (ab) b^-1 = b^-1 .0

                                => a(b b^-1 ) = 0

                                => a.1 = 0

                                => a =0

Therefore in a field ab=0 => a=0 or b=0

Therefore there are no zero divisor in a field and hence every field is a integral domain.

                                                                                                Hence proved………………

{note : - but its conversaly is not true. We can show it like below}

Proof :- let (F,+,.) be a field and let it contains distinct unit and zero elements i.e. 1≠0 .

                Let a be a non-zero element of F then

a^-1 = 0 => a a^-1 = a.0                [left multiple with a]

                => 1  = 0 [a a^-1 =1 and a.0 =0]

                => a.1 = a.0             [left multiplel with a]

                => 1 = 0  [ cencellation law]

This is a contradiction therefore integral domain is not a field.

                                                                Hence proved………

Integral Domain

Integral Domain à An algebric expression (D,+,.) where D is a non-empty set with two operations “addition and dot” is called ring if it is satisfies following axioms..

D1. (R, + ) be an abelian group.

1)      closure for addition:-  if a,b ∈D => a+b∈D for all a,b∈D

2)      associate for addition:-if a,b,c ∈D then (a+b)+c=a+(b+c) for all a,b,c∈D

3)      identity for addition:-  if for all a∈D there are exist 0∈D such that a+0=a for all a∈D

4)      inverse for addition:- if for all a∈D then there are exist –a∈D such that a+(-a) ∈D for all a∈D

5)      commutative low    :-  for all a,b∈D such that a+b=b+a for all a,b∈D

D2.  (R,.) be a semi group.

      1)   closure for multifi. :-   for all a,b∈D => a.b∈D  for all a,b∈D

      2)  associate for dot    :-   for all a,b,c∈D st    (a.b).c=a.(b.c)  for all a,b,c∈D

      3) Commutative for dot:- for all a,b∈D => a.b =b.a

      4) unit element :- 1 ∈ D st a.1 = a =1.a for all a∈D

D3.  distribution law :- for all a,b,c∈D

       1) left distribution:-             a.(b+c)=a.b+a.c for all a,b,c∈D

       2) Right distribution :-       (b+c).a=b.a+c.a for all a,b,c∈D

D4. D is without zero divisor :- i.e. a.b = 0 => a=0 or b=0 or both zero.

So (D,+,.) is a Integral Domain…………..

Integral Domain:- A ring (R,+,.) is called integral domain if it satisfies three more exioms

(1)   Commutative.

(2)   It have a unit element.

(3)   Without zero divisor.

But it have atleast two element.

Examples :- set of rational numbers is a integral domain which have a zero element and unit element.

2) set of real number is a integral domain .

Theoram:- Ring with modulo p of integers be a integral domain iff p be a prime number.

Proof :- suppose  (Ip,+p , .p) be a ring of integers with modulo p.

To show:- p be a prime number.

Assume contrary , p is not a prime number and let p =m . n where 1<m<p ,1<n<p.

Therefore         [n] .p [m] = [n.m] 

     =[p] = [0] (mod p)

                        i.e.     [n] .p [m] =[0] (mod p)

            but       [m] …[0] and [n]≠ [0].

So, (Ip,+p , .p) is not a integral domain.

Therefore  p is a prime number.

 Conversaly :-  let p be a prime number and [m],[n]∈Ip such that [m].p[n]=[0]
 =>  r.s = 0 (modp)
 =>    r.s is divided by p
   =>either r divided by p or s divided s divided by p, since p is prime number.
 =>Either r=0(mod p) or s = 0(mod P)
    =>[r] = [0] or [s] = [0]
Therefore (Ip,+p , .p) is a integral domain.

Ordered Integral Domain:-  A Integral domain (D,+,.) is called ordered integral domain if a subset D+ of D such that

1)      D+ ,will be closure corresponding to D with respect to ‘+’ and ‘.’ i.e. if a,b∈D+ => a+b∈D+ and a.b∈D+.

2)      For all a∈D+ its true for one and only one, a=0 , a∈D+,  -a∈D+ .

Theoram:- prove that set of complex numbers is not a ordered integral domain.

Proof:- let C be a set of complex numbers.we know that (C,+,.) be a integral domain. suppose C+ be a set of all positive complex numbers of C. where 0+i0 is a zero element of C+.

since i≠0 therefore for trichotomy law it will either i∈C+ or -i∈C+.

now C+ is always closure for '.' and '+'. 

let i∈C+ => i.i ∈C+=> -1∈C+

so i∈C+ , -1∈C+ => i(-1)∈C+ => -i∈C+

so i∈C+ => -i∈C+ this is contradiction of trichonotomy law. 

similarly -i∈C+ => i∈C+  ,this is again contradiction .

therefore (C+,+,.) is not a ordered integral domain.

                                                                    Hence proved.....

Subring

 Subring-

Definition-  suppose S be a non-empty set of (R, +, .) and if S be stable in R for addition and multification then it is called subring if

                                a ∈S , b∈S => a+b∈S

                             a ∈S , b∈S => ab ∈S for all a,b∈S.

                in other word, A non-empty set S is called subring if it is satisfies all axioms of a Ring.

Theoram:- suppose (R,+,.) be a Ring and S be a non-empty set of R.then (S,+,.)  will be subring of (R,+,.) iff

a) a ∈S , b∈S => a-b∈S

b) a ∈S ,b∈S => a.b ∈S for all a,b∈S.

Proof:- suppose (S,+,.) be a subring of (R,+,.) i.e. S is a ring itself.

To show:-  a) a ∈S , b∈S => a-b∈S

                  b) a ∈S ,b∈S => a.b ∈S for all a,b∈S

 since          a ∈S , b∈S    => a∈S, -b∈S      [inverse law]

                        =>a+(-b) ∈S              [closure low]

                        =>a-b∈S     for all a,b∈S

Now ,                 a ∈S ,b∈S => a.b ∈S for all a,b∈S.  [ closure for dot]

                                                                                    Proved first part.

Conversaly :-

                        Let S is a non empty set of R which is satisfies two axioms.

                        a) a ∈S , b∈S => a-b∈S     ………(1)

                         b) a ∈S ,b∈S => a.b ∈S for all a,b∈S…..(2)

to show:-          S is a subring of (R,+,.).

                        from (1)

a)       a ∈S , a∈S => a-a∈S

                           => 0   i.e. identity exist in R.

Now 0 ∈S , b∈S => 0-b∈S          [from 1]

                         => -b∈S.  i.e. inverse of every element exist.

                        Now a ∈S , b∈S    => a∈S, -b∈S      [inverse law]

                        =>a-(-b) ∈S                         [from 1]

                        =>a+b∈S     for all a,b∈S

i.e. closure for addition axist.

Now take a,b,c ∈S and S⊆R then a,b,c ∈R.

Therefore a+(b+c) = (a+b)+c for all a,b,c ∈S

i.e. associate law satisfies.

a,b∈S and S⊆R then a,b ∈R.

Therefore a+b = b+a  for all a,b ∈S

i.e. commutative law satisfies.

Therefore        (S,+) is an abelian group.

 Now                 (S,.) is an semi group.

                        Form (2)

                        a ∈S ,b∈S => a.b ∈S for all a,b∈S

                        closure for dot is satisfies.

Associate satisfies because a,b,c ∈S and S⊆R then a,b,c ∈R.

Therefore a.(b.c) = (a.b).c for all a,b,c ∈S

Therefore (S,.) is a semi group.

Since a,b,c ∈S and S⊆R then a,b,c ∈R.

 a(b+c) = a.b+a.c for all a,b,c ∈S

(b+c).a = b.a+c.a for all a,b,c ∈S

                   Therefore (S,+,.) is a subring … Hence Proved

Theoram :- prove that Intersection of two subrings is also a subring.

Proof:-      suppose (S,+,.) and (K,+,.) are two subrings. Since S ≠ i, K ≠ i=> S∩K ≠ i

To prove:- S∩K is also a subgroup.

For this we shall prove :-  a) a ∈ S∩K , b∈ S∩K => a-b∈ S∩K

b)       a ∈ S∩K ,b∈ S∩K => a.b ∈ S∩K for all a,b∈ S∩K.

now

                                a ∈ S∩K => a∈S and a∈K

                                b ∈ S∩K => b∈S and b∈K

Since S and K are subring then

                                a ∈S , b∈S => a-b∈S ,ab∈S ………………A

                        a ∈K , b∈K => a-b∈K ,ab∈K……………..B

now                  a ∈ S∩K ,b∈ S∩K =>a,b∈S and a,b∈K

                                      =>a-b∈S ,ab∈S ,and a-b∈K ,ab∈K

                                       => a-b∈S and a-b∈K, ab∈S and ab∈K

                                       =>a-b S∩K , a.b ∈ S∩K

Therefore S∩K is a subring..

Definations:-Left Ideals: suppose (R,+,.) be a ring then a non-empty set of S of R is called left ideals if

1)       S is a additive subgroup of R.

2)       For all s∈S, r∈R => sr∈S.

Right Ideals: suppose (R,+,.) be a ring then a non-empty set of S of R is called right ideals if

1) S is a additive subgroup of R.

2) For all s∈S, r∈R => rs∈S.

Ideals :- suppose (R,+,.) be a ring then a non-empty set of S of R is called ideals if

1) S is a additive subgroup of R.

2) For all s∈S, r∈R => sr∈S , rs∈S

Ring

Ringà An algebric expression (R,+,.) where R is a non-empty set with two operations “addition and dot” is called ring if it is satisfies following axioms..

R1. (R, + ) be an abelian group.

1)      closure for addition:-  if a,b ∈R => a+b∈R for all a,b∈R

2)      associate for addition:-if a,b,c ∈R then (a+b)+c=a+(b+c) for all a,b,c∈R

3)      identity for addition:-  if for all a∈R there are exist 0∈R such that a+0=a for all a∈R

4)      inverse for addition:- if for all a∈R then there are exist –a∈R such that a+(-a) ∈R for all a∈R

5)      commutative low    :-  for all a,b∈R such that a+b=b+a for all a,b∈R

R2.  (R,.) be a semi group.

      1)   closure for multifi. :-   for all a,b∈R => a.b∈R   for all a,b∈R

      2)  associate for dot    :-   for all a,b,c∈R st    (a.b).c=a.(b.c)  for all a,b,c∈R

R3.  distribution law n:- for all a,b,c∈R

       1) left distribution:-             a.(b+c)=a.b+a.c for all a,b,c∈R

       2) Right distribution :-       (b+c).a=b.a+c.a for all a,b,c∈R

So (R,+,.) is a Ring…………..

Types of Ring:à
·         Ring with unity:à A ring with multification identity 1 which is define as a.1=1.a=a for all a∈R is called Ring with unity…
·         Commutative Ring:à A ring with multification identity is called commutative ring i.e. a.b=b.a for all a,b∈R ….
Examples. Set of intergers is a Ring Set of real number is a ringSet of rational is a ring…..
·         Boolean ring:à A set (R,+,.) is called Boolean Ring if its all element are idempotent. i.e. a²=a for all a∈R
·         Zero devisors:àA ring is called with zero divisor if it is difined as ab=0 but a…0 , b…0..
·         Ring without zero divisor:à A ring is called ring without zero divisor if it is defined as for all a,b∈R  ,   ab=0 => a=0 or b=0 or a=b=0 both are zero…

Theoram:à A ring R is a without zero divisor iff R is satisfies cencellation law..

·         Proof:à  firstly let R is a Ring without zero divisor i.e. a,b∈R are orbitary element  s.t. ,   ab=0 => a=0 or b=0 or a=b=0 both are zero…
T show:-                       ab=ac (a…0) => b=c
                                     ba=ca (b…0) => b=c
for this , take                ab=ac (a…0) => a(b-c) = 0
                                                       =>b-c = 0 since a…0
                                                        => b=c
                                  Similarly we can prove   ba=ca (b…0) => b=c
Conversaly:- now let R be a ring which satisfies cancellation law.
To prove:- R is a without zero divisor ring.
Assume contrary, suppose R is a ring with zero divisor then there are exist a,b∈R such that ab=0 but a…0 , b…0.

Now ,                ab= 0, a is not 0 => ab = a.0                       [a.0=0

                                           => b = 0                 [ by cancellation law]
                             This is  a contradiction because b…≠0

Similarly           ab= 0, b is not 0… => ab = 0.b                      [0.b=0]
                                       => a = 0                [ by cancellation law]
   Again this is a contradiction because a is not. therefore R is a Ring without zero divisor......
Hence proved...........
Theoram:- if R is a ring with additive identity 0 then for all a,∈R a.0=0=0.a    for all  a,∈R
Proof:-       for all a,∈R
  0+0 = 0  
 a.(0+0)=a.0     (left multification of a)
  a.0+a.0=a.0+0  (distribution and ideal)
 a.0 = 0               (cencellation law)
  similarly we can prove 0.a=0                                                  proved…..
theorem:- :- if R is a ring with additive identity 0 then for all a,b∈R   a(-b) = -(ab) = (-a)b
proof :-     for all a,b∈R
  b + (-b) = 0           (inverse law)
  a.(b + (-b) ) = 0                       ( left multification of a)
 a.b + a.(-b) =0                         (distribution)
 ab + a (-b) =0                          (inverse law)
 a (-b) = -(ab)    for all a,b∈R  
 similarly we can prove (-a) b = -(ab)   for all a,b∈R.
Theoram:- :- if R is a ring with additive identity 0 then for all a,b∈R   (-a)(-b)=ab
Proof:-     (-a)(-b) = -[(-a)b]        
                 [a (-b) = -(ab)]
                            =>-[-(ab)]       [a (-b) = -(ab)]
                            => ab  for all a,b∈R.   [ -(-a) = a ]
                                                                             proved

Theoram :- if R is a ring with additive identity 0 then for all a,b∈R  , -(a+b) = (-a) + (-b)Proof :-     now we can take
   [(-a) + (-b) ]+ (a+b) =[(-b) + (-a)] + (a+b)        (commutative)
=>(-b) + [(-a) + (a+b)]      (associate 
 =>(-b) +[ {(-a) + (a)}+b]    (associate)
 =>(-b) +[ 0+b]                 (inverse)  
=>(-b) +( b )                   (identity)
=> 0                                             (inverse)
                                           Proved…………

Theoram:- if R is a ring with additive identity 0 then for all a,b∈R  a.(b-c)= ac-bc   for all a,b,c∈R 
Proof |-  for all a,b,c∈R 
 a.(b-c)  = a.[b(+-c)]
=>a.b + a.(-c)      [distribution law]
 =>ab – ac           {a(-b)=-(ab)}
  Similarly we can prove (b-c)a = ba-ca  for all a,b,c∈R 
Theoram :- if R is Boolean ring st a² =a for all a∈R     then prove that

(1)    a+a =0

(2)    a + b = 0 => a = b for all a,b,∈R

(3)     R be a commutative ring.

proof- 

     since R is a ring such that a² =a for all a∈R .

 a∈R => a,a ∈R 

   =>a+a∈R 

 now,

 (a+a)² = (a+a)              [since a² =a]

  (a+a) (a+a) = (a+a)

ð       (a+a)a+(a+a)a= (a+a) [left distribution]

ð       a.a+a.a+a.a+a.a =a+a [right distribution]

ð       a² + a² + a² + a² =a+a

ð       (a+a) + (a+a) = (a+a) + 0 [ since a² =a]

ð       a+a =0

proved (1)

II proof-

    Now since  a+a =0

(1)          a+b = 0 => a+b = a+a

 =>  b=a          [ cancellation law]

  proved

III) proof-

   (a+b)² = (a+b)         [since a² =a]

  (a+b) (a+b) = (a+b)

ð       (a+b)a+(a+b)b= (a+b) [left distribution]

ð       a.a+b.a+a.b+b.b =a+b [right distribution]

ð       a² + ba + ab+ b² =a+b

ð       a+ba+ab+b =a+b

=>(a+b)+ (ab+ba) = (a+b) + 0 [ since a² =a]

=> ab+ba = 0

(2)    => ab =ba    [a + b = 0 => a = b ]

Therefore R is commutative

Hence proved…

Right Ideals: suppose (R Definations:-Left Ideals: suppose (R,+,.) be a ring then a non-empty set of S of R is called left ideals if

1)       S is a additive subgroup of R.

2)       For all s∈S, r∈R => sr∈S.

(R,+,.) be a ring then a non-empty set of S of R is called right ideals if

1) S is a additive subgroup of R.

2) For all s∈S, r∈R => rs∈S. 

Ideals :- suppose (R,+,.) be a ring then a non-empty set of S of R is called ideals if

1) S is a additive subgroup of R.

2) For all s∈S, r∈R => sr∈S , rs∈S

Improper and Proper Ideals:- suppose (R,+,.) be a ring . Ideal R and {0} are called Improper or trivals ideals and different from it are all called Proper or non-trival ideals.

Unit and zero ideals:- (R,+,.) be a ring then R and {0} are called Unit and zero ideals.

Simple ideals:- A ring (R, +, .) is called a simple ring if it does not contains any proper ideals.

Theorem:- Intersection of two ideals is also an ideal.

Proof:- suppose M and N are two ideals where M and N are both non-empty.then we can say both have 0 element ie

0∈M ,0∈N => 0∈M∩N therefore M∩N also a non-empty.

Since M is an ideal therefore it satisfies two condition ie

 M is a subgroup and s∈M , r∈R => sr,rs∈M …………….(1)

N is also an ideal therefore it also satisfies two contion ie

N is a subgroup and s∈N ,r∈R => sr,rs∈N………………..(2)

We have to prove :- M∩N is an ideal.

Since we know that intersection of two subgroup is also a subgroup. Therefore M∩N is also a subgroupFor this, take s∈M∩N , and r∈R

Now

s∈M∩N , and r∈R => s∈M and s∈N ,r∈R

=>s∈M ,r∈R and s∈N ,r∈R

  => rs,sr∈M and rs,sr∈ N       from (1) and (2)

   => rs,sr ∈ M∩N

It satisfies both conditions .

and hence M∩N is an ideal, but union of two ideals is not necessary be an ideal . let us proof this by an example. Now we take a set M ={2n : n∈Z} and N = {5n : n∈I} are ideals for Ring R. but if you try to solve , it does not satisfies first low . for example 2∈M ,5∈ N => 2,5 ∈ M∪N but 2-5 = -3 ∉MUN because -3 does not exist both ideals , so MUN is not necessary a ideal.

Note:- similarly we can prove this forcollection of  family of intersecton of all ideals.

Theoram:- if (R,+,.) be a ring and M and N are two ideals then MUN is also an ideal iff they contains each others.

Proof:-let M and N are two ideals

To prove:-M ∪N is a Ideal of R.

  let  M and N are two ideal.and let they contains each others. i.e. M⊆N and N⊆M.

then  M∪N=M or N . and M and N are Ideal .therefore MUN  is a ideal of R.

conversaly:-  now let MUN is an ideal. Then we will show that they contains each other.assume contrary- MéN and NéM

now                  MéN=> there exist  a∈M then a∉N.               ……..(1)

 NéM => there exist b∈N then b∉M                        ……..(2)

From (1) and (2)

a∈M then a∉N=>a∈ M∪N

 b∈Nthen b∉M=>b∈ M∪N.

so,    a∈ M∪N , b∈ M∪N=>a-b ∈ M∪N   (since M∪N is a subgroup)

 a-b∈ MUN then a-b ∈ M or a-b∈ N

now

a ∈ M, a-b∈ M=> a-(a-b)= b∈ M but b∉M this is a contradiction.  {by 1}

Similarly a-b∈ N , b∈ N => a-b+b=a∈ N but a∉N  this a contradiction. {by 2}

Therefore M⊆N and N⊆M.

Hence proved.

 Definition :-suppose R is a commutative ring with unit element and a∈R then the ideal of it all multiples of a ie (a) = {ra : r∈R } is an ideal generate from a and its denoted by (a). then an Ideal M of R is called principal ideal if for a∈R st M=(a)..

*Result:- if M and N are two ideals of ring R then the set M+N = { x∈R ; x=a+b, a∈M ,b∈N }, is an ideal generate from MUN .

Proof:- since M and N are two ideals and we have to show M+N  is an ideal generate from MUN . for this we have to firstly show that M+N is an ideal then its an ideal generate from MUN.

Let x,y ∈M+N =>there are exist  a₁,a₂ ∈M and b₁,b₂∈N   where x=a₁+b₁ and y=a₂+b₂.

  *         Now, since x,y ∈M+N => x-y=(a₁+b₁)-(a₂+b₂)

                                                        =(a₁-a₂)+(b₁-b₂) ∈ M+N 

[because M and N are ideals therefore  a₁-a₂∈M and b₁-b₂∈N ]

So we can say that if x,y ∈M+N => x-y∈M+N for all x,y is in M+N.

·         let r∈R and x∈M+N => xr=(a₁+b₁)r = a₁r +b₁r∈M+N

=>xr∈M+N

Similarly r∈R and x∈M+N then rx∈M+N. this shows that M+N is an ideal of R.

After this we will prove that M+N is an ideal generate from MUN. for this we show that M+N=(M+N)

For this we take element a∈M so a= a+0 ∈M+N [0∈N] therefore M⊆M+N and similarly N⊆M+N…….

Since M⊆M+N , N⊆M+N => MUN⊆M+N………….(a)

We suppose that X ve an ideal of R such that MUN⊆X .

If z∈M+N then z=a+b where a∈M,b∈N.

Now , a∈MUN and b∈ MUN => a,b ∈X because MUN⊆X

                                         => a+b∈X

                                         => z∈X .

ie M+N ⊆X. then we can say M+N =(MUN) ie M+N is an ideal generate from MUN.

                                      hence proved.

*Note :- multiple of two ideals is also an ideal.

*Result :- prove that the ring of all integer be an principal ideal ring.

Proof:-

           Let (Z,+, .) be a ring of integer .let S be and Ideal of Z .

Condition 1:- if S=(0) then nothing to prove because S be a trival principal ideal generate from 0∈Z.

Condition2:- if S≠(0). Ie S is a non-trival ideal. Then we can say S contains only positive elements.then we can say there are exist n be a least positive integer such that n∈S .

then (n) ⊆ S……(1)

Now only to show that S⊆ (n). and then we can say S will be principal ideal of Z.

 Let a∈Z such that a∈S. then by the definition of division function there are exist q,r in Z such that a= qn+r where 0≤r<n.  or a-qn=r..

Since S be an ideal therefore n∈S , q∈Z=> nq∈S. [definition of ideal] .

Again a∈S ,qn∈ S => a-qn∈S [because S is a addition subgroup of Z]

ð       r∈S. since 0≤r<n where n is a least positive integer therefore n is in S and then the value of r will be zero ie r=0 .

therefore a=qn , where a is the multiple of n and then a∈(n).[definition of principal ideal]

ie

a∈S => a∈(n) so S⊆(n)……..(2)

by (1) and (2)

(n) ⊆ S , S⊆(n) => S=(n)

And hence S is a principal ideal generate form Z.

                                                            Hence proved.

Euclidean Ring:- A commutative ring is called  an Euclidean ring if for all x∈R there are exist a non-negative integer d(x)  define as

1)      d(x) = 0 iff  x=0 ,where 0∈R

2)       d(x,y) ≥ d(x) when x,y ≠ 0

3)       for all x∈R and y≠z∈R there are exist q,r∈R st x=yq+r   , 0≤d(r)<d(y).

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