Upload
conlan
View
36
Download
0
Tags:
Embed Size (px)
DESCRIPTION
Rings and fields. 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. Rings. - PowerPoint PPT Presentation
Citation preview
RINGS AND FIELDS
1
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.
2
RingsDEFINITION: 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
3
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
4
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 RINGS1. 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
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
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
8
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
9
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
11
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.
12
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.
13
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.
14
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.
15
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.
17
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=a18
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
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.
21
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.
22
SUBRINGS
23
DefinitionSubrings: 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
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.
25
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
27
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
28
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.
29
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 subringsTherefore a-b ∈ S1 and a-b ∈ S2
a.b ∈ S1 and a.b ∈ S2Þ a-b ∈ S1 ∩ S2 and a.b ∈ S1 ∩ S2
Hence S1 ∩ S2is a subring of R.
31
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.
32
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
33
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.
34
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
35
Test3. . Prove that The intersection of two
subrings is subring.
36