Mathspics: 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.