RINGS AND FIELDS
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INTRODUCTION
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.
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Rings
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 ∈ R2. Addition is associativei.e. (a+b)+c=a+(b+c) for all a, b, c ∈ R 3. Addition is commutativei. e. a+b=b+a for all a,b ∈ R
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4. Existence of additive identityi.e. there exists an additive identity in R
denoted by 0 in R such that 0+a=a=a+0 for all a ∈ R5.Existence of additive inversei.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. multiplicationi.e. if a,b ∈ R, then a.b ∈ R
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7. Multiplication is associativei.e. a.(b.c)=(a.b).c for all a,b,c ∈ R8.Multiplication is associativei.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
TYPES OF RINGS
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.
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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.
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Example of Ring
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
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3. Existenceof additive identity: for 0 ∈ Z , 0+a=a=a+0 for all a ∈ ZTherefore 0 is the additive identity.4. Existence of additive inverse: for each a
∈ Z, there exists -a ∈ Z such that a+(-a)=0=(-a)+aWhere 0 is the identity element.5. Commutative property: a+b=b+a for all a,b ∈ Z
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Properties under multiplication:6. Closure property w.r.t. multiplication: as
product of two integers is also an integerTherefore a.b ∈ Z for all a,b ∈ Z7.Multiplication is associative: a.(b.c)=(a.b).c for all a,b,c ∈ Z8. Multiplication is distributive w.r.t.
addition:For all a,b,c ∈ Z, a.(b+c)=a.b+a.cAnd (b+c).a=b.a+c.aHence Z is a ring w.r.t. addition and
multiplication of integers.10
RING WITHOUT OR WITH ZERO DIVISOR , INTEGER DOMAIN , DIVISION RING
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Definition
A ring (R, +, .) is said to be without zero divisor if for all a, b ∈ R
a.b=0 => either a=0 or b=0On 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.
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EXAMPLES
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.
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INTEGER DOMAIN
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.
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Division ring or skew field
Definition: a ring is said to be a division ring or skew field if its non-zero elements forms a group under multiplication.
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Field
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
2. Associative law: a+(b+c)=(a+b)+c for all a,b,c∈ F3. Existence of identity: for all a∈ F, there
exists an element 0 ∈ F such that a+0=a=0+a4. Existence of inverse: for each a ∈ F,
there exists an element b ∈ F such thata+b=0=b+a
Element b is called inverse of a and is denpted by –a.
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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∈ F7. Commutative law: a.b=b.a for all a,b ∈ F8. Associative law: multiplication is
associativei.e., (a.b).c=a.(b.c) for all a,b,c ∈ F9. Existence of multiplication identity: for
each a ∈ F, there exists 1 ∈ F such that
a.1=1.a=a
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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
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THEOREM: A commutative ring is an integer domain if and only if
cancellation law holds in the ring
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 thatab=ac
Þ ab-ac=0Þ a(b-c)=0Since R is an integral domain, so R is
without zero divisors. 20
Therefore either a=0 or b-c=0Since a≠0 so b-c=0 => b=cHence, 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.
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If a=0, then we have nothing to prove.If a≠0 then ab=0=a.0Þ b=0Þ R is without zero divisors.Hence R is an integral domain.
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SUBRINGS
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Definition
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.
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Examples of subrings
(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.
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Theorem: 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
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
Þ 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 ∈ SConditions are sufficient:Suppose S is a non-empty subset of R such
that for all a,b ∈ S, a-b ∈ S and a.b ∈ SNow a ∈ S, a ∈ S => a-a ∈ S => 0 ∈ S
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Therefore additive identity exists.0 ∈ S, a ∈ S => 0-a= -a ∈ Si.e., each element of S possesses additive
inverse.a∈ S, -b ∈ S, => a-(-b)=a+b ∈ SThus 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
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Therefore S is closed w.r.t. multiplicationAs 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.
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Theorem: The intersection of two subrings is a subring.
Proof: let S1 and S2 be two subrings of R.
As additive identity 0 is common element of S1 and S2.
Therefore S1 ∩ S2 ≠
To show that S1 ∩ S2 is a subring, it is sufficient to prove that
(i) a,b ∈ S1 ∩ S2 => a-b ∈ S1 ∩ S2
(ii) a,b ∈ S1 ∩ S2 => a.b ∈ S1 ∩ S2
let a,b be any two elements of S1 ∩ S2 30
a,b ∈ S1 and a,b ∈ S2
Since S1 and S2 are subrings
Therefore a-b ∈ S1 and a-b ∈ S2
a.b ∈ S1 and a.b ∈ S2
Þ a-b ∈ S1 ∩ S2 and a.b ∈ S1 ∩ S2
Hence S1 ∩ S2is a subring of R.
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Example: prove that the set of integers is a subring of ring of
rational numbers.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 ∈ ZHence Z is a subring of Q.
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Assignment1. 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
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Assignment
(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.
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Test
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
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Test
3. . Prove that The intersection of two subrings is subring.
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