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