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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Infinite set

Infinite set :-

A set is said to be an infinite set whose elements cannot be listed if it has an unlimited (i.e. uncountable) by the natural number 1, 2, 3, 4, ………… infinite , for any natural number n is called a infinite set. 

A set which is not finite is called an infinite set.
example of infinite sets:-
set of natural numbers , N={1,2,3,4,.........................}
set of whole numbers  ,W= {0,1,2,3,.........................}
set of real numbers    , R etc

Since we cannot ever finish counting an infinite set, we need a different approach to

thinking about “how many elements” such a set contains. In the early 1900s, Georg

Cantor presented an idea that has clarified thinking about this issue and had enormous

impact on opening new pursuits in logic.

Two (finite or) infinite sets A and B are said to have the same cardinality (equipotent) if

there is a one-to-one, onto function (i.e., a bijection) f: A Æ B.

 Since it is impossible to ascribe an ordinary number to the size of an infinite set, we

don’t try. Instead, the comparative idea above is used. If we can “line the elements up” in

exactly corresponding pairs, we say the two sets have the same cardinality (i.e., size).

So, for instance, the set of natural numbers N = {0, 1, 2, … } has the same cardinality as

the collection of all strings over the alphabet.

 ifnfinite sets that have the same cardinality as N = {0, 1, 2, … } are called countably

infinite. A set that has a larger cardinality than this is called uncountably infinite.

Assertion: there are uncountably infinite sets.

Proof by contradiction:

Let [0,1] denote the interval of all real numbers x, 0≤x≤1 — this set is uncountably

infinite. First of all, note that there is an injection f: N Æ [0,1] defined by f(n) = n/(n+1).

This maps N to a proper subset of [0,1], so [0,1] has cardinality at least as large as N.

To show that the cardinality of [0,1] is actually larger than N, we proceed by

contradiction. Suppose that [0,1] is countably infinite. Suppose that g :N Æ [0,1] is a

bijection, and therefore {g(n) | nŒN} is the set of all real number in the interval [0,1]. We

show that there must be at least one real number missing in this enumeration providing

a contradiction to the existence of g, and hence there is no bijection. Specifically, we

construct the decimal expansion of a number a that is missing from {g(n) | nŒN}. Take

the decimal expansion of a to be .d1d2d3 …, where dn = 5 if the nth digit of g(n) is 6, and

dn = 6 otherwise. Then for each n, the nth digit of a differs from the nth digit of g(n).

Therefore a differs from every number in the enumeration in at least one digit and so

aœ{g(n) | nŒN}, a contradiction since this set was assumed to be all such numbers.

Ref from homepage.

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