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2.  We have studied groups, which is an algebraic structure equipped with one binary operation. Now we shall study rings which is an algebraic structure equipped with two binary operations. 2

3. DEFINITION: A non-empty set R equipped with two binary operations called addition and multiplication denoted by (+) and (.) is said to form a ring if the following properties are satisfied: 1. R is closed w.r.t. addition i.e. a,b ∈ R, then a+b ∈ R 2. Addition is associative i.e. (a+b)+c=a+(b+c) for all a, b, c ∈ R 3. Addition is commutative i. e. a+b=b+a for all a,b ∈ R 3

4. 4. Existence of additive identity i.e. there exists an additive identity in R denoted by 0 in R such that 0+a=a=a+0 for all a ∈ R 5.Existence of additive inverse i.e. to each element a in R, there exists an element –a in R such that -a+a=0=a+(-a) 6. R is closed w.r.t. multiplication i.e. if a,b ∈ R, then a.b ∈ R 4

5. 7. Multiplication is associative i.e. a.(b.c)=(a.b).c for all a,b,c ∈ R 8.Multiplication is associative i.e. for all a,b,c in R a.(b+c)=a.b+a.c [left distribution law] And (b+c).a=b.a+c.a [right distribution law] REMARK: any algebraic structure (R, +,. ) is called a ring if (R,+) is an abelian group and the properties 6, 7, 8 given above also satisfied. 5

6. 1. Commutative ring: a ring in which a.b=b.a for all a,b ∈ R is called commutative ring. 2. Ring with unity: if in a ring, there exists an element denoted by 1 such that 1.a=a=a.1 for all a ∈ R, then R is called ring with unit element. The element 1 ∈ R is called the unit element of the ring. 6

7. 3. Null ring or zero ring: The set R consisting of a single element 0 with two binary operations denoted by 0+0=0 and 0.0=0 is a ring and is called null ring or zero ring. 7

8.  Prove that the set Z of all integers is a ring w.r.t. the addition and multiplication of integers as the two ring compositions. Solution: Properties under addition; 1. Closure property: As sum of two integers is also an integer, therefore Z is closed w.r.t. addition of integers. 2. Associativity: As addition of integers is an associative composition therefore a+(b+c)=(a+b)+c for all a,b,c ∈ Z 8

9. 3. Existenceof additive identity: for 0 ∈ Z, 0+a=a=a+0 for all a ∈ Z Therefore 0 is the additive identity. 4. Existence of additive inverse: for each a ∈ Z, there exists -a ∈ Z such that a+(-a)=0=(-a)+a Where 0 is the identity element. 5. Commutative property: a+b=b+a for all a,b ∈ Z 9

10.  Properties under multiplication: 6. Closure property w.r.t. multiplication: as product of two integers is also an integer Therefore a.b ∈ Z for all a,b ∈ Z 7.Multiplication is associative: a.(b.c)=(a.b).c for all a,b,c ∈ Z 8. Multiplication is distributive w.r.t. addition: For all a,b,c ∈ Z, a.(b+c)=a.b+a.c And (b+c).a=b.a+c.a Hence Z is a ring w.r.t. addition and multiplication of integers. 10

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12.  A ring (R, +,.) is said to be without zero divisor if for all a, b ∈ R a.b=0 => either a=0 or b=0 On the other hand, if in a ring R there exists non- zero elements a and b such that a.b=0, then R is said to be a ring with zero divisors. 12

13.  Sets Z, Q, R and C are without zero divisors if for all a,b ∈ R  The ring ({0, 1, 2, 3, 4, 5}, + 6, *6 ) is a ring with zero divisors. 13

14. Definition: a commutative ring without zero divisors, having atleast two elements is called an integer domain. For example, algebraic structures (Z, +,.), (Q, +,.), (R, +,.),(C, +,.) are all integral domains. 14

15.  Definition: a ring is said to be a division ring or skew field if its non-zero elements forms a group under multiplication. 15

16.  A field is a commutative division ring.  Another definition: let F be a non empty set with atleast two elements and equipped with two binary operations defined by (+), (.) respectively. Then the algebraic structure (F, +,.) is a field if the following properties are satisfied: axioms of addition: 1. Closure property: a+b ∈ F for all a,b ∈ F 16

17. 2. Associative law: a+(b+c)=(a+b)+c for all a,b,c ∈ F 3. Existence of identity: for all a ∈ F, there exists an element 0 ∈ F such that a+0=a=0+a 4. Existence of inverse: for each a ∈ F, there exists an element b ∈ F such that a+b=0=b+a Element b is called inverse of a and is denpted by –a. 17

18. Commutative law: a+b=b+a for all a, b ∈ F Axioms of multiplication: 6. Closure law: for all a,b ∈ F, the element of a.b ∈ F 7. Commutative law: a.b=b.a for all a,b ∈ F 8. Associative law: multiplication is associative i.e., (a.b).c=a.(b.c) for all a,b,c ∈ F 9. Existence of multiplication identity: for each a ∈ F, there exists 1 ∈ F such that a.1=1.a=a 18

19. 10. Existence of inverse: for each non-zero element a ∈ F, there exists an element b ∈ F such that a.b=b.a=1 element b is called multiplicative inverse of a and is denoted by 1/a. 11. Distributive law: multiplication is distributive w.r.t. addititon i.e., for all a,b,c ∈ F a.(b+c)=a.b+a.c (b+c).a=b.a+c.a 19

20.  Proof: let R be a commutative ring. Suppose R is an integeral domain. We have to prove that cancellation law holds in R. let a, b, c ∈ R and a≠ 0 such that ab=ac  ab-ac=0  a(b-c)=0 Since R is an integral domain, so R is without zero divisors. 20

21. Therefore either a=0 or b-c=0 Since a≠0 so b-c=0 => b=c Hence, cancellation law holds in R. Conversely, suppose that cancellation law holds in R. Now, we have to prove that R is an integral domain. As R is commutative so we only have to show that R is without zero divisors. 21

22. If a=0, then we have nothing to prove. If a≠0 then ab=0=a.0  b=0  R is without zero divisors. Hence R is an integral domain. 22

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24. Subrings: If R is a ring under two binary compositions and S is a non-empty subset of R such that S itself is a ring under the same binary operations, then S is called subring of R. 24

25. (i) Z is a subring of the ring (Q, +,. ) (ii) Q is a subring of the ring ( R, +,. ) (iii) {0} and R are always subrings of the ring R and are called improper subrings of R. Other subrings if any, are called proper subrings of R. 25

26. Proof: Conditions are necessary: Suppose ( S, +,.) is a subring of (R, +,.) Let a,b ∈ S => -b ∈ S [Since –b is additive inverse of b] 26

27.  a-b ∈ S [since S is closed w.r.t. addition] Also a.b ∈ S [since S is closed w.r.t multiplication] Hence, a,b ∈ S => a-b ∈ S and a.b ∈ S Conditions are sufficient: Suppose S is a non-empty subset of R such that for all a,b ∈ S, a-b ∈ S and a.b ∈ S Now a ∈ S, a ∈ S => a-a ∈ S => 0 ∈ S 27

28. Therefore additive identity exists. 0 ∈ S, a ∈ S => 0-a= -a ∈ S i.e., each element of S possesses additive inverse. a ∈ S, -b ∈ S, => a-(-b)=a+b ∈ S Thus S is closed w.r.t. addition. As S is a subset of R, therefore associativity an commutativity must hold in S as they hold in R. Now, a,b ∈ S => a.b ∈ S 28

29. Therefore S is closed w.r.t. multiplication As S is a subset of R, therefore associativity of multiplication and distributive laws must hold n S as they hold in R. Hence S is a subring of R. 29

30.  Proof: let S 1 and S 2 be two subrings of R. As additive identity 0 is common element of S 1 and S 2. Therefore S 1 ∩ S 2 ≠  To show that S 1 ∩ S 2 is a subring, it is sufficient to prove that (i) a,b ∈ S 1 ∩ S 2 => a-b ∈ S 1 ∩ S 2 (ii) a,b ∈ S 1 ∩ S 2 => a.b ∈ S 1 ∩ S 2 let a,b be any two elements of S 1 ∩ S 2 30

31. a,b ∈ S 1 and a,b ∈ S 2 Since S 1 and S 2 are subrings Therefore a-b ∈ S 1 and a-b ∈ S 2 a.b ∈ S 1 and a.b ∈ S 2  a-b ∈ S 1 ∩ S 2 and a.b ∈ S 1 ∩ S 2 Hence S 1 ∩ S 2 is a subring of R. 31

32. Solution: let Z be the set of integers and Q be the ring of rationals. Therefore Z is a subset of Q. 0 ∈ Z => Z≠  Let a,b be any two elements of Z. Therefore a-b ∈ Z and ab ∈ Z Hence Z is a subring of Q. 32

33. 1. Define ring and its types. 2. Prove that the set Z of all integers is a ring w.r.t. the addition and multiplication of integers as the two ring cmpositions. 3. Define field. A commutative ring is an integer domain if and only if cancellation law holds in the ring. 4. Define sub-ring. The necessary and sufficient conditions for a non-empty subset S of the ring R to be a subring of R are 33

34. (i) a,b ∈ S => a-b ∈ S (ii) a,b ∈ S => a.b ∈ S 5. Prove that The intersection of two subrings is subring. 6. Prove that the set of integers is a subring of ring of rational numbers. 34

35.  Do any two questions: 1. Prove that the set Z of all integers is a ring w.r.t. the addition and multiplication of integers as the two ring cmpositions. 2. Define sub-ring. The necessary and sufficient conditions for a non-empty subset S of the ring R to be a subring of R are (i) a,b ∈ S => a-b ∈ S (ii) a,b ∈ S => a.b ∈ S 35

36. 3.. Prove that The intersection of two subrings is subring. 36