## Page 34

34 ALGEBRA II UNIT 1

2

ISOMORPHISM OF GROUPS

Unit Structure

2.0 Objectives

2.1 Homomorphism And Isomorphism

2.2 Cayley’s Theorem

2.3 Automorphisms and Inner Automorphisms

2.4 External Direct Product of groups.

2.5 Normal sub groups and Quotient (Factor groups)

2.6 Isomorphism Theorems

2.7 Classification of groups of order b7 upto isomorphism

2.8 Fundamental Theorem Of Finite Abelian Groups

2.9 Summary

2.10 Unit and Exercises

2.0 Objectives

After going through this unit you shall come to know about

x Relation between the groups and the existence of isomorphisms between

groups

x Special kind of subgroups called normal subgroups and its impotance in the

construction of factor groups

x Classification of groups upto order 7 using various results of isomorphism

theorem

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## Page 35

35Chapter 2: Isomorphism of Groups

2.1 Homomorphism and Isomorphism

Definition: 1] Let G, G be groups. A map f : G Gl is called a homomorphism

of group G to group G if f(ab) = f(a) f(b) for each a, b G (i.e. f preserves group

operation)

Definition: 2] Let G, G be groups. A map f : G Gl is called an is isomorphism

of group G to group G if i) f (ab) = f(a) f(b) for for each a,b G (ie f is a group

homomorphism) and ii) f is bijective.

This is denoted by G x G

Properties of group homomorphisms and isomorphisms.

Let G, G be groups and f : G l Gbe a group homomorphism and a G

Then

(i) f(e) = eawhere e, ea are identity elements of G, G respectively.

(ii) f (a1) = (f (a))1

(iii) f(ak) = (f(a))k for each k ]

Proof : (i) ea= f(e) = f(e.e) = f(e) . f(e)

= By right cancellation law, f(e) = ea

(ii) f (aa1) = f(e) = ea

11 1

11() ( ) () ( () )

() ( ( ) )fa fa e fa fa

fa fa

=

=

(iii) We prove inductively, f (an) = (f(a))n nN

The result is true for n = 1

Suppose the result is true for k, k N

f (a)( ( ) )kkfa

Then, 11 1()( )( ) ( ) ( ( ) ) ( ) ( ( ) )kk k k kfa fa a fa fa fa fa fa

,( ) ( ( ) )nnInductively f a f a

01 0

11 1() ( ) ( )) ( ( ) ) ( ) ( ( ( ) ) )() ( ) ( ( ) ) , ,

,{ , , },

nm m m mfa fa fa fa b yi i fanNfaf e e f a t h e r e s u l t i s t r u e f o r

for n n n m m N

]

(( ) ) (( ) )mnfaf a

Thus, the result is true for each n ]. munotes.in

## Page 36

36 ALGEBRA II

Properties : Let G, G be groups and f : G Gl be an onto group homomorphism

(i) If G is abelian, so is G

(ii) If G is cyclic, so is G.

(iii) O(f(a)) () .oa If f is an is isomorphism, then o(f(a)) = o(a).

Proof : (i) Let ,,ab Gaa As f is onto, for a,b, G such that

() , ()faa f bb aa

Then, () () ( ) ( ) () () ab f a fb f a b fb a fb f a b a aa aa

Thus, G is abelian.

(ii) Let G = < a > = {: }nan ]

() ( { : } ) {( ) : } { (( ) : } ( )

.nn nGfGfanfanfanfa

Thus G is cyclic

]]]

(iii) Suppose o(a) = n, then nae

(( ) ) ( )

(( ) )( )nnfaf a e

fa o a

a =

=

If f : G Gl is an isomorphism, then

(( ) ) (( ) ) ( ) (( ) ) ( )mm mofa m fa e fa fa e fe aa º º

() /maef i s o n e o n e n m º º '

() / .By iii m n m n =

Result : If G, G, are groups, and f : G Gl i s a n i s o m o r p h i s m , t h e n

1:fGGl is an isomorphism.

Examples of homomorphism and isomorphism .

(1) For any group G, I : G l G (I = identity) and the trivial may f : G lG defined

by f(x) = e are group homomorphism.

(2) The map :RR l defined by 3()xx is a group homomorphism.

(3) Consider the map f : ]n l] defined by f(m) = m is a group homomorphism. munotes.in

## Page 37

37Chapter 2: Isomorphism of Groups

Theorem 1: Any finite cyclic group of order n is isomorphic to Z ,nthe group of

integer residue classes modulo under addition.

Proof : Let G be a cyclic group of order n generated by `a’.

Then, G = {e, a, … a1n} and for m ,mraa]

01

:, m o d

() 1 ( m o d )

,, 1

,.

\mr

nn

r

rs

rsrn

Ga a b y w h e r e m r n

we define a r for o r n r r n

is well defined for a a r,s n

ae o r s n

nr s a n d r s

R

bb

l w

b b

b b

º b

º º]

rs rs º

is a group homomorphism, since

mod

() ()rsand r s t n

trsrs a a

w

ºw

.

{, 1 , 1 } , ( ) .

.r

nnis clearly onto

on a n d f o r r r a

is a group isomorphism

=

=]]

() () , , 1

,1rs

rsaa r s n

rs rs rsn n

aa

is one one

R

b b

ºº bb

º

='

Corollary: Any two finite cyclic groups of same order are isomorphic

(Consider 1)rrab b y a b o r n l l bb

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## Page 38

38 ALGEBRA II

Theorem 2: An infinite cyclic group is isomorphic to the group of integers under

addition.

Proof : Let G be an infinite cyclic group generated by `a’, {}nGa n z

Consider the map :( ) .rGd e f i n e d b y a r f o r r l ]]

For ,( , )rsaa Gr s ]

(, ) ( ) () ()rs r s r saa a r s a a

is a group homomorphism=

() ()rs r saa r s a a º º

is one one=

,( )rrFor r a G such that a r ] . is onto=

Thus, is a group isomorphism

Corollary : Any two infinite cyclic groups are isomorphic

(If < a >, < b > are infinite cyclic groups consider ,) .rrab r l ]

2.2 Cayley’s Theorem

Cayley’s Theorem : Every group is isomorphic to a group of permutations

Proof : Let G be a group

For a G, we define f a : G l G by f a (x) = ax (f a is multiplication by a on left)

We show f a is bijective map, ie f a is a permutations on G.

fa (x) = f a (y) for x, y Ga x a yx yº º

= fa is one one

For y G, 1xayG such that

1

.

{: }a

a

afx a x a y y

fi s o n t o

let G f a G

=

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## Page 39

39Chapter 2: Isomorphism of Groups

Then G i s a g r o u p u n d e r c o m p o s i t i o n o f m a p s f o r a , b f o r

,, ( ) ( ) ( ) ( ) ( )ab a b a a b ab G f of x f f x f b x ab x a b x f x

,ab ab a bfof G f of f =

ef is the identity element of G.

()efx e x x V ' ()eaxGfx ax

() () () ()ae a e a eafof x f x f x f of x

1

af is the inverse of af

111 1

aa a a e a a aafof f f f f of

'

Gis a group and :GG l defined by

a af is an group isomorphism.

' For

,,ab a b ab G a b f f of a o b

() ()ab a b ab f f f e f e a e b e a b º º º º

is clearly onto.

= G is isomorphic to ,Ga group of permutations.

Note : The group G consummated above is called the left regular representation

of G

2.3 Automorphisms and Inner Automorphisms

Definition: An isomorphism of a group G onto itself is called an automorphism

of G

Examples

1) Identity map is an automorphism of any group to itself

2) f : G lG defined by f(a + ib) = a – ib is automorphism of the group (G, +).

Let Aut G = {f : f : G lG, f automorphism of G }

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## Page 40

40 ALGEBRA II

Definition : Let G be a group and a .G T h e m a p :aiG G l d e f i n e d b y

1()aix a x a f o r xG is called the inner automorphism of G induced by a.

Let Inn (G) =1{: ( ) , }aaiG G ix a x a aG l

Theorem: Aut G is a group and Inn G < Aut G

Proof : If f, g aut G, f D g Aut G, D is associative

i Aut G, For f Aut G. then map f1 Aut G,

= Aut G is a group

11 1.

,, ,

.ed

ab

ab ab a biiI n n GFor i i Inn G where a b G

ii ii i I n n G

Inn G Ant G

=

2.4 External Direct Product of Groups

Definition : Let, 12,, . . . .n GG G be finite collection of groups. The external direct

product of 1,...n GG denoted by 12 ...n GG G is defined by

12 1 {( , .... ) / 1 }nn i i GG G g g gG i n b b where

12 1 2 1(, , ) ( , . . . . , ) ( , . . . . . . . . . , )nn n n gg g g g g g g g g aa a a a <

It can be easily verified that 12 .... .n GG G i s a g r o u p

11 2 1 1

111 1 1 1

111(, . . . ) , ( , . . . . , ) , (" . . " ) . . . . . .

(( , ..... ) ( .. )) ( ",..... ") ( , ...., ) ( ,".. ")

(( ", ... ( ") "nn n n

nn n n n n

nn nFor g g g g g g g G Ggg g g g g g g g g g gggg g g g

aa a

aa

a

11 1 1

11 1(( " ) ( " ) ) ( , . . )

(, . . . . , ) ( ( ) (" . . . . . " ) ) .nn n n

nn ngg g g gg B y a s s o c i a t i v i t y i n G G

gg g g g g

aa

aa

Thus, associatively holds in 1 n GG munotes.in

## Page 41

41Chapter 2: Isomorphism of Groups

Let 11,......., .......nn ee b e i d e n t i t y e l e m e n t s g r o u p s o f G G r e s p e c t i v e l y

11

11 1 1 1

11 1,......., .......

,( , . . . . . , ) ( , ) ( , . . . . . . , ) ( , . . . . . , )

(, . . . . . . . . . . . . . , ) ( ) (nn

nn n n n

nn nee b e i d e n t i t y e l e m e n t s g r o u p s o f G G r e s p e c t i v e l y

Then g g e e g e g e g g

eg eg e e

¸

1 )n gg

1(, . . . )n ee= is identity element of 1 .n GG

Let

1

11 1 1

11 1

1

1111 .

1.

,( ,. . )( ,. . . . , ) ( )

(, . . . . , ) ( ,ii i

ii i

nn n n

ngG f o r i n a n d G i s a g r o u p f o r i n

Let g be the inverse of g in G for i n

Then g g g g g g g g

ee g g

b b b b

bb

11 1

11 )( , . . . . . . , ) ( , ) .nn n ngg gg g

Thus, 11

11(, . . . . . . . . . . , ) ( , . . . . . . . . . . , )nn gg h a s a n i n v e r s e g g

11......... . .........nn in G G G G is a group =

This is called the external direct product of 1.........n GG

Examples :

1) 23 {(0, 0), (1, 0), (0,1), (0, 2), (1,1), (1, 2)} ZZ

This is an abelian group.

Properties of external direct product.

Let 1,n GG be finite groups and 1 n GG be their external direct

product.

(1) 11 2 1 2() ( ) ( ) . . . . ( ) .nn n OG G O G O G O G o r G G G ¸ ¸¸

(2) For 11,.., ) .... ,nn gg G G then 11(, . . , ) ) ( () , . . , ( ) ) .nn Og g l c mo g og

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## Page 42

42 ALGEBRA II

Proof : (1) is clear by property of cardinality of Cartesian product

(2) Let () 1iiOg n f o r i n b b

1

1111 1

111

1[, . . . , ]

1, 1 .

(, . . . . , ) ( ) ( , )

(, . . . , )( , . . . . . , )

(k

ii i

nr nr nnnnn

r r nnnLet cm n n

Then n for i n let n r for i n

Then g g g g g g

ee ee

Og

bb bb

=AA AAA

AA

,) \ngA

Then 1(, . . . ) \nLet O g g k ie k A

11

1

2(, ) ( , ) ( * )

(, )

1

.kk k

nn

n

k

igg g g

ee

ge f o r i n

k

= b b

A

Note, by induction, the result is true m

Theorem : Let 1,n GG be finite cyclic groups of orders 1,n nn respectively,

Then, 1....n GG is cyclic if and only if g.c.d (, ) 1 1 ,ijnn f o r ij n b b

Corollary: Let 1. k mn n Then , mn n k ! '' '

,.ij if and only if n n are relatively prime for i j v

Note : If 12,GG are abelian groups, so is 12GG

Theorem: Let m and n be positive integers of g.c.d (m,n) = 1, then U(mn) is

isomorphic to U(m) U(n).

Proof :Consider the map f : U(mn) l U(m) U(n) defined by,

f(x) = (x mod m, x mod m)

The map f is well-defined,

For if x, y U(mn), f(xy) = (xy mod m, xy mod m)

= x mod m, y mod m, x modm, y mod m)

= (x mod m, y mod m) (y modm, y mod m)

= f(x) f(y)

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## Page 43

43Chapter 2: Isomorphism of Groups

Let f(x) = f(y) º(x mod m , x mod m) = (y mod m, y mod m)

ºx mod m = y mod m x mod m = y mod m

ºx w y mod mn ( = m, n are relatively prime)

= f is one – one

f is onto by Chinese Remainder Theorem.

Therefore, f is bijective

Clearly f is a homomorphism.

Thus, U(mn) is isomorphic to U(m) U(n).

Corollary : Let 12 1,. . , . . ( , ) 1j mn n g c d n n f o r i j v

12 () () ( ) . . . ( )k ThenU m U n U n U n !

2.5 Normal Sub Groups and Quotient (Factor Groups)

Normal Subgroup

Definition: Let G be a group. A subgroup H of G is called a normal subgroup of

G if aHa1 H for each aG

1(, )or aha H for each a G each h H

This is denoted by H G

Theorem 1 : Let H be a subgroup of a group G.

Then, the following statements are equivalent.

11) (HG i e a H a H

1)

2)

3)

4) ,a

ab a baG

aHa H for each a G

aH H for each a G

HH H f o r e a c h a b G

(or aHbH = abH for each a, b G)

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## Page 44

44 ALGEBRA II

Proof :- () ( )ii iº

Suppose HG

1

11

11 1

1,

,( * )

(( * )

.Let a G then aHa H

Also a G a Ha H

Ha a a a a H a b y

aHa H

+

=

=

=

1() ( )

.ii iii

Ha aHa a a He aHº

() ( )

()abiii iv

HH H H a b H a b H i s c l o s e d u n d e r m u l t i p l i c a t i o n

º

=

()

(Hab Heab HHab HaHb Ha aH

Hab HaHb

HaHb Hab or aHbH abHH ab H

=

=

=

.

()He HH H

aHbH abHH abH

=

(iv)º(i)

11 1 1,

.For a H aHa eaHa HaHa Haa H

HG

=

Note : If G is abelian, and H < G, then H G.

Theorem : Let H G, Let /{ :}GH H a a G

Then, the operation HaHb = Hab on G/H is a well defined binary operation in G/H

and G/H is a group under this binary operation.

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## Page 45

45Chapter 2: Isomorphism of Groups

Proof : We first show that this binary operation is well defined

Let

11

12 1 2

11 1 1 1 1 1 1

22 1 2

11

13 3 2,, .

()

()Let aa h bb h h h H

we show Hab Ha b

ab a b ab b a ah a aa a h a h a h a

hh w h e r eh ab a H H G

a a

aa

aa a a a a aaa

a =

1

13 () .

.

.ab a b h h H

Hab Ha b

ie HaHb a Hbaa=

aa =

aa

+

and the operation is well defined.

For

() (),/ ,

(abc

ab c a b c a b c a b c a b c a bcHH G H

H +

+ + + + + + + + + ++

= The binary operation is associative:

/e

ae a e a e a eaHH G H

HH H H H HH

=He = H is, the identity element of G/H

For

1

11 1 1/, /aa

aa a a e a a aHG H H G H s u c h t h a t

HH H H a H H

+

1

aH= is the inverse of aH

Thus, G/H is a group.

It is called the quotient groups of G by H (or the factor group of G/H).

Note : If G is finite o(G/H) = o(G)/o(H) = [G : H].

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## Page 46

46 ALGEBRA II

Example: Let us consider ]/n]

Let us find the cosets ]/n]

For a Z, we know ,qr ] such that ,1 aq nr rn b b

., 1 .

,, 1arn an rn rn

Moreover for r s n

= = bb

bb ]]]

,rn sn rs n

rs o r rs n rs rs

º

º ºº]] ]

= The distinct cosets of Z/n ] are o + ] , 1 + ],……, (n – 1 ) + ].

Propositions 1 : Let G be a group. If G/Z(G) is cyclic, then G is abelian.

Proof : Suppose G/Z(G) = {(Z(G)x)n : n]} =

,( ) , ( ) ,( ) , , , ,nmFor a b Z G let a Z G x b Z G m n ]

12 1 2

12 1 2 2 1 2 1,, , ( )

.nm

nm n m m n n naz x bz xz z Z G

ab z x z x z z x x z z x x z x z x ba

Gi sa b e l i a n=

=

Proposition 2 : G/Z(G) x Inn G

Proof : C onsider f : G/Z(G) l Inn G defined by f(Z(G)x) = ix where

1(( ) )xia x a x

11

11 1 1 1 1 1() () () , , ()

,( ) () ( ) ( )x yZGx Gy x y G L e t x y zz ZG

For a G i a xax zya zy zyay z yay zz yay i a = =

=º

.xyiifis well defined=

=

f (Z(G)x. Z(G)y) = f (Z(G) = y) = ixy munotes.in

## Page 47

47Chapter 2: Isomorphism of Groups

11 1,( )( ) ( ) ( ) ( )

(() ()) (() ) (())

hom .xy x y

xy x y

xy x yFor a G i a xy a xy x yay x i i a

ii ifZG x ZGy i ii f ZGx f ZGy

fi s a g r o u p o m o r p h i s m

=

=

=

,, ( ( ) ) ( ( ) ) ( ) ( )xy x y For x y G f z G x f z G y i i i a i a º º

11aG

xax yay

º11 yaG yx a a y x º 1(

() ()aG yx G

zG y zGx

fi s o n e o n e =

º

º

=

Clearly for ix Z(G), f(Z(G)x) = xi and f is onto

=f is a group isomorphism.

Definition: Internal Direct Product :- Let 12,nHHH be normal

subgroups of G we say G is the internal direct product of 12HH if

(i) G = 12 1 2,{ / , 1 }nn i iHHHh hh h Hi n b b

(ii) 12 1,{ } 1 , 1 .ni HH H H ef o r i n

Note : {} )ijHH e i f ij v

In particular, if H, K are normal subgroups of G, then G is the internal direct product

of H, K if.

(i) G = HK = {/ , }hk h H k k

(ii) . Hk

Theorem : If a group G is the internal direct product of subgroups H1,..,Hn then

12 ..n GH H H

Proof: For 11 1 1,, ( )ii j j i j i j i j i j jhH hH h h h h h h h h H

11 1 1

1()

{}ij i j i j i j i

ij i i j

ij j ihh h h h h h h H

hh h H H e

hh h h

=

= munotes.in

## Page 48

48 ALGEBRA II

1

11

12 12

11 1 1

12 1 1 2 2 1

11 1 1 1

11 2 2 1 1,, 1

() ( ) ( )ii l

nn

nn n n n

nnFor h h H i n

hh h hh h

hh h h h h h h h

hh h h h h

= b b

º

11

1

12 1 1nn

nn

ni nhh e

hh

hh h h h

º

º

=

By similar argument, 1

11nnhh

Proceeding in this manner, 11iihh f o r i nb b

We define 1 :. . . . . . .n GH H b y l

11() () ( )nn hh hhi s w e l l d e f i n e d

11 1 1 1 1

11 1 1 1 1

11 1 1

11 1 1(( ,..., ) ( )) ( ,.., ) ( ,..., )

(, . . , ) ( ) (, . . . , ) ( )nn n n n n

nn n nhh h h h hh h h h h h

hh h h h h h h

= is a group homomorphism

11 11

11

11

1() ( ) () ( )

1.

.ni n n i n

ii ni nhh h h hh h h

hh f o r i n h h h h

is one one

is clearly onto

º

º b b º

=

= is a group isomorphism.

2.6 Isomorphism Theorems

We recall that a homomorphism from a group G to group G is a map :fGGl

such that f(ab) = f(a) f(b) for each a, b G.

Let G, G be groups and f : G Gl be a group homomorphism we define kernel

of f (denoted by ker f) asker f {x G/f x =e } c is the identity element of G.

Proposition 1: Let G, G be groups and f : G Gl be a group homomorphism.

Then, Kerf G. munotes.in

## Page 49

49Chapter 2: Isomorphism of Groups

Proof :- We know, ()fe e e K e r f K e r f a= º v

11 1 1 1 1 1

1,, ( ) ( ) ( ) ( ) ( ( ) )

.For a b Kerffabfafbfafbe e e

ab Ker f and Ker f G

=

11 1,k e r , ( )( ) ( ) ( )( ) ( ( ) )For a G h f f aha f a f h f a f a e f a e aa

1ker , ker

keraha f for a G h f

fG

=

=

Proposition 2: Let G, G be groups and f : G Gl be a group homomorphism.

Then, f is one – one if and only if ker f = {e}.

Proof :- Suppose f is one-one

Then x kerf

() () () k e r { }fxe f x f ee x e f e aa º º º =

Conversely, suppose ker f = {e}

11

11 1,, , ( ) ( ) ( ) ( ) ( ) ()

() k e r

.Then for x y G f x f y f x f y e f x f y e

fx y e x y f x y e x y

fi s o n e o n e

aa º º

a º º º º

=

Note: If G, G are groups and f : G Gl is a group homomorphism, then

Imf = {f(x) : x G} is a subgorup of G.

Proposition: Let G be a group and H ,GThen p: G lG/H is surjective group

homomorphism, where p(x) = Hx. For x G and ker p = H

Proof: For x, y G, (p(xy) = () ()xy x yHH H p x p y

=p is a group homomorphism.

p is clearly onto. (For /, ( ) ( )xHGHp x H x x G s u c h t h a t

kerxxpH H ( identity element of G/H is H)

kerxH

pH

=

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## Page 50

50 ALGEBRA II

Note:- Thus, if H G, where G is a group, then H is a kernel of a group

homomorphism from G to a suitable group.

We next prove the Isomorphism Theorems.

First Isomorphism Theorem : Let f : G Gl be a homomomorphism of groups.

If f is onto, G/kerf Gx (or in general G/kerf x Im f).

Proof : Consider the mapping f: G/ker f lG defined by,

(ker ) ( ) ( ( ) ( )) kerffa f a o rf K a f a w h e r e f .

This map is well defined

For Ka = Kb, a, b G

11 1(k e r) ( ) ( )( ) ab K f f ab e f a f b e aa º º º

1() ( () ) () ()faf b e f a f b a º º f(Ka.Kb) = f(Kab) = f(ab) = f(a) f(b) = f(Ka)f(Kb)

=f is a group homomorphism. f(Ka) = f(Kb) º f(a) = f(b)

=f is one – one.

If f is onto, for aGc a G such that f(a) = aa

() ( )

.fKa f a a

fi s o n t o

a =

=

Note :- 1. If f is not onto, we consider f : G/ker f l Imf and G/kerf x Imf.

2. The above theorem, is also called. The Fundamental Theorem Of

Homomorphism Of Groups.

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## Page 51

51Chapter 2: Isomorphism of Groups

Second Isomorphism Theorem: Let H and K be subgroups of a group G and

K G. Then, HHKHKK

Proof : We note H K < G and K < HK,

:/ ( )KG KH K

we define a map f H HK K by f h hK for h H

º

l

Then, for 12 1 2 1 2 1 2 1 2,, ( ) ( ) ( ) ( ) ( ) ( )hh H fh h h h h Kh K fhfh .

= f is a group homomorphism.

f is onto for any element in HK / K is of the form hkk(hk)

And hk = f(h)

Ker f = {h H : hK = K} hK h = H ..

= By First Isomorphism Theorem of groups HHKHKK

Third Isomorphism Theorem of groups:

Let G be a group and H, K be normal subgroups of G If K H, then

(G/K) / (H/K) x G/H

Proof : Consider the mapping f: G/K lG/H given by f(Kx) = Hx

This map is well defined, for

11

() ( ) ( ) ( )xyK Ȁxy k xy H Hx Hy

fK x K y fK x y H x y ǾxHy f Kx f Ky

º º º

= f is a group homomorphism.

f is clearly onto.

' Any element of G/H is of the form Hx, x G

()fKx Hx =

ker { / ( ) } / , } /for f Kx x G f Kx H Kx x G Hx H H K ^ =

= By First Isomorphism Theorem, G/K/H/K x G/H munotes.in

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52 ALGEBRA II

Theorem: Let 12 1 1 2 2,, ,GGb e g r o u p s a n d H GH G Then

12 1 2HHG G a n d 12 1 2

12 1 2GG G G

HH H H

x

Proof : Consider the map 12

12

12:GGfG GHHl

12 1 1 2 2(. ) ( , ) defined by f g g H g H g

Then, for 12 1 2 1 2(, ) , ( )gg g g G G aa

12 1 2 1 1 2 2 1 1 1 2 2 2(( , ) ( , )) ( , ) ( , )fgg g g f g g g g H g g H g g aa a a a a

1

11 1 1 2 2 2 2 11 2 2 11 2 2(, ) ( , ) ( )Hg H g H g H g Hg H g Hg H g aa a a

12 1 2

12

11 2 2

12(, )( , )

hom .

(, )fgg f g gfis a group omorphism

GGFor H g H gHH

aa

=

11 2 2 1 2(, ) ( , )Hg H g f g g

fi s o n t o

=

12 1 2 12 1 12 2 1 2

11 2 2 1 2 1 2(, ) k e r ( , )(, ) ( , )( )

,( , )gg f f g g H H H g H g H H

gH g H g g

+ +

= By First Isomorphism Theorem,

12 1 2

12 1 2GG G G

HH H H

x

Correspondence Theorem: Let G, G be groups and f : G l G b e a

homomorphism of group G onto G.

Then

i] H < G ()fHG º

ii] H < G 11() (){ :( ) }fHG fH x G f x Hº munotes.in

## Page 53

53Chapter 2: Isomorphism of Groups

iii] () HG f H G º

iv] 1() HG f H G º

v] 1ker ( ( )). HG a n d H f H f f H º

vi] The map H l f(H) is a 1-1 corespondence between the family of subgroups

of G containing ker f and the family of subgroups of G. Furthermore normal

subgroups of G correspond to normal subgroups of G.

Proof :

i] ,( ) ( ) ( ) eH f e e f H a n d f H a = v

,( ) , ( ) , ( ) , . For a b f H a f a b f b for some a b H aa a a

11

11 1 1

1,( ) ( )

() ( ) ( ) ( ) ( ( ) ) ( )

() ()ab H and f ab f H

fa b fa fb fa fb ab

ab fHa n d fH G

=

aa

aa=

ii] 1()Let H G and H f H

1

1,( ) ( ) ( )

() .Then e G and e f e f e H e f H

fH

aa º º

=v

Let 1,( ) ( ) , ( ) ,ab f H t h e nf a f b H a n dH G

111 1

11 1() ( () ) . . () ( ) , ( ) () ( )

() ()faf b H i e f a f b H f a b f a f b H

ab f H f H G

= =

iii] Let , HG

11 1

11,( ) .

() , () , ,

() () ( () ) () () ( 1 )

() ( ) ( )Let a G h f H

Then a f a h f h where a G h H

ah a f a f h f a f a f h f

fa h a fh a h a H

aa

aa

aa

'

() .fHG= munotes.in

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54 ALGEBRA II

iv] Let 1,( ) ( ) HG L e t a G hf H i e f hH

11 1

11 1() ( ) ( ) ( ) ( ) ( ) ( ( ) )

() ()faha f a f h f a f a f h f a H

aha f H f H G

=

= =

v] Let H < G and H kerf.

Then, 1(( ) ) .ffH H

1

1

11

11

11

1

1(( ) ) ( ) ( )

() () () ()

() ( ) ( )

ker ,

,( ( ) )

(( ) ) .Let x f f H f x f Hfxf h f o r s o m e h H f x f h e

fxfh e a n d fx h e

xh f H xh h h H

xh hHa n d f f H H

ff HH

=

a = =

aa =

= =

=

=

vi] Let H be a subgroup of G.

Then, 1()fH is a subgroup of G containing ker f

1(( ) ) .

() .ff H H

The map H f H is onto=

=l

12 1 2

11

11 2 2

12 1 2() ( )

(( ) ) (( ) )

,k e r .For H H G f H f H

Then f f H H f f H H

HH f H H

The map is one oneº

=

=

Corollary: Let G be a group and N G.

Given any subgroup H of G/N, there is a unique subgroup H of G containing N

such that /, , / /HHN F u t h e r H G i f f HN GN a

Proof : Consider the homomorphism f : G lG/N defined by f(x) = Nx. f is an

onto homomorphism and ker f = N = By correspondence Theorem, there is a

unique subgroup H of G containing N such that f(H) = H/N = Hc

Further H // GH N G N munotes.in

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55Chapter 2: Isomorphism of Groups

Proposition : 2nnAS f o r n p then !.2nnR$

Proof : Consider :{ 1 , 1 } ( )nSd e f i n e d b y s i g n l T T

12 12 1 2 1 2

12() () ( ) ( )( ) ( )

,nThen sign Sign Sign

for S

T T T T T T T T

TT

= is a group homomorphism.

((12)) 1, ( ) 1,

ker { / ( ) 1 }

.,

/{ 1 , 1 }

() / ()2nn

nn

nn

n nis onto for I

SAASa n d b y F i r s t I s o m o r p h i s m T h e o r e m o f G r o u p s

SA

OS oA

T T

=

x

=

() !()22n

nOS nOA =

Note : We have seen that by Lagrange’s Theorem if G is a finite group and

H < G, then O(H)\O(G).

However, converse of Lagrange’s Theorem is not true is general.

We give an example below.

Example : A4 has no subgroup of order 6. (O(A4) = 12).

Proof : Suppose A4 has a subgroup H of order 6.

Then 4[: ] () / ( ) 1 2 / 62n AH O A o H

4

4.

3HAAhas eight cycles

=

Let T be a 3-cycle in A4.

If T/H, then H T vH.

If 22H Ht h e n H T T T munotes.in

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56 ALGEBRA II

21

22 1... ,

,( ) .

,3 .

()Hi e Hw h i c h i s n o t t r u eIfH t h e n a H W h i c h i s n o t t r u e

Ht h u s a l l c y c l e s a r e i n H

OH

=TT T

T T

=T

=p

A contradiction.

4A= has no subgroup of order 6.

2.7 Classification of Groups of Order b 7 Upto Isomorphism.

We note that group of order 1 is a the trivial group which is unique upto

isomorphism. Group of order 2, 3, 5, 7 are cyclic (as prime order) and are unique

upto isomorphism.

Let G be a group of order 4. Then, any element of G has order which is divisor of

4 i.e. 1, 2 or 4.

If G has an element of order 4, then G is cyclic.

Suppose G has no element of order 4.

Then any non-trivial element in G has order 2. Let G = {e, a, b, c} which e, a, b, c

and distinct elements 222abce

We find ab ev

,, ,ab e ab aa b a which is not true

ab a ab ae b e which is not true

ab b ab eb a e which is not true

ab c similarly ba e ba a ba b

ba c ab

º ºº ºº º= v v v

= ' ¬ ®

Similarly, we can show ac = b = ca

bc = a = cb.

=Composition table of G is

e a b c

e e a b c

a a e c a

b b c e a

c c b a e munotes.in

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57Chapter 2: Isomorphism of Groups

Thus, any non-cyclic group G of order 4 is of the above type (and is called the

Klein’s four group 4V)

Let G be a group of order 6. Then, order of any element of G divides 6.

? Order of any element in G is 1, 2, 3 or 6. G is a group of even order and so has

an element of order 2.

Suppose every non-trivial element of G has order 2. Then, 2,xex G

1,, ( )For x y G xy xy?

11yx x y? i.e. yx xy

G? is abelian

If ,, ,ab Ga eb ezz then ^`,,, He a b a b is a finite subset of G having closure

property.

e a b ab

e e a b ab

a a e ab b

b b ab e a

ab ab b a e

HG? , By Lagrange’s Theorem () / ( )oH oG i.e. 4/6.

A contradiction.

G?has an element of order 3 or 6.

Case 1: If G has an element of order 6 then G is cyclic.

Case 2 : G has no element of order 6. Let aG have order 3 and bG

have order 2. Then 22,, ( ( ) ( ) 3 ,( ) 2 ) be a ao a o a o bz . If ab ba , then

6 oa b which contradicts that G has no element of order 6.

12 2,, ba ab ba e a a ba a b e ba a b a?z z z z z z ''

, ba b a e ba abzzz' . munotes.in

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58 ALGEBRA II

We note 22,, ,, ,eaa ba bab are distinct

22 2 2, ab e b a ab a a b e ab a b e zz z z z z'''

2ab a b a e zz'

^`22,, ,, , Ge a a b a b a b?

2ba a b ?

Thus, 32 2,, Ga b abe b a a b

3 GS?|

? There are two non isomophic groups of order 6, one is cyclic, and the other is

isomorphic to S 3

Note : An abelian group of order 6 is cyclic.

2.8 Fundamental Theorem of Finite Abelian Groups

Let G be a finite abelian group of order n. Then

12 k nn n G|]] ] where

1iinnfor 11ikdd and the above decomposition is unique.

This is also expressed as.

Let G be a finite abelian group of order n > 1 and let 12

12 ...k

k npp pD DD where

1,...k pp are distinct primes, then

12 ....KGG G G| where () 1i

iioG p i kD d d and

1...

ii ip i p i iGr DD| ]] where 12 ... 1ii i rDD Dtt tt and

1...ii r iDD D .

We note that if ,...ikpp are primes dividing o(G), then in /ipn for . iikbb

Recall that .. ( , ) 1mnm nZZ Z i f a n d o n l y i f g c d m nx

Example : Let us list all abelian groups (upto isomorphism) of order 180.

22180 2 3 5. ¸¸ munotes.in

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59Chapter 2: Isomorphism of Groups

Abelian groups.

180 4 9 5

90 2 9 5 2 2

60 6 5 3 2 3 2

]] ] ]

]] ] ] ] ]

]] ] ] ] ] ]

Let us consider classification of abelian groups of order 234,,ppp u p t o

isomorphism.

Partitions Abelian Groups

1] 2p 2 2p]

2 = 1 + 1 ]pp]

2] 3p 3 3p]

3 = 2 + 1 3pp]]

3] 4p 4 4p]

4 = 3 + 1 3pp]]

4 = 2 + 2 22pp]]

4 = 2 + 1 + 1 2ppp]] ]

4 = 1+1+1+1 pppp]]]]

Example: Expressing abelian groups in the form 12...pp p nGG G

And 12nn n kZZZ where 1icnn for group of order 360.

360 = 3223 5 qq

30 6 2

2233

51] 2 3 5

235

23

211ppp A b e l i a n g r o u p s

]] ]

]]]]

]

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60 ALGEBRA II

2

60 6 2

222 3 322] 2 3

35

2

21 1 1pp p

]] ]

]] ] ] ]

5]

22

180 2

42953] 2 3 5

23 5

21 1pp p

]]

]]]]

32

3604] 2 3 5

23 5

11 1pp p

]

Thus, there are 4 non-isomorphic abelian groups of order 360.

An important consequence of the Fundamental Theorem of finite abelian groups.

Theorem: If G is a finite abelian groups of order n, and m is a positive divisor of

n, then G has a subgroup of order m. (This is converse of Lagrange’s Theorem)

Note : The above partition may not be unique.

2.9 Summary

1) Let G, G be groups. A map f : G Gl is called a homomorphism of group G

to group G if f(ab) = f(a) f(b) for each a, b G

(i.e. f preserves group operation)

2) Let G, G be groups. A map f : G Gl is called an is isomorphism of group G

to group G i f i ) f ( a b ) = f ( a ) f ( b ) f o r f o r e a c h a , b G (ie f is a group

homomorphism) and ii) f is bijective.

3) Any finite cyclic group of order n is isomorphic to Z ,nthe group of integer

residue classes modulo under addition.

4) Cayley’s Theorem : Every group is isomorphic to a group of permutations

5) An isomorphism of a group G onto itself is called an automorphism of G

6) Let G be a group and a .G T h e m a p :aiG G l d e f i n e d b y

1()aix a x a f o r xG is called the inner automorphism of G induced by a. munotes.in

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61Chapter 2: Isomorphism of Groups

7) For 11,.., ) .... ,nn gg G G then 11(, . . , ) ) ( () , . . , ( ) ) .nn Og g l c mo g og

8) Let G be a group. A subgroup H of G is called a normal subgroup of G if aHa

1 H for each aG

1(, )or aha H for each a G each h H

9) First Isomorphism Theorem : Let f : G Gl be a homomomorphism of

groups. If f is onto, G/kerf Gx (or in general G/kerf x Im f).

10) Second Isomorphism Theorem: Let H and K be subgroups of a group G and

K G. Then, HHKHKK

11) Third Isomorphism Theorem of groups:

Let G be a group and H, K be normal subgroups of G If K H, then

(G/K) / (H/K) x G/H

12) There exist unique groups of order 1, 2, 3, 5 as they are prime and 2 groups of

order 4 and 6 upto isomorphism.

13) Fundamental theorem of finite abelian groups: Let G be a finite abelian

group of order n. Then

12 k nn n G|]] ] w h e r e 1iinnfor

11ikdd and the above decomposition is unique.

2.10 Unit and Exercises

1) Show that Z(G) is a normal subgroup of a group G

2) Show that a subgroup of index 2 is a normal subgroup of any group G

(Let H < G, [G : H] = 2, Let a v H, then H, Ha are distinct right cosets of H of G.

H, aH are distinct left cosets of H in G. G = H U Ha = H UaH.

(Ha = aH = G H)

= For x G, x Hº Hx = xH = H

x v H º Hx = xH = G (H). munotes.in

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62 ALGEBRA II

3) What is the order of 5 + <6> in the quotient group 18/6? .]

4) Let G=222{1 , , , } 1 ij k w h e r e i jk oooo

,, ij ji k jk kj i ki ik j

Construct the composition table of G.Show that every subgroup of G is normal.

5) Show that H = {I, (12)} is a not a normal subgroup of S3 but {I, (123), (132)}

is a normal subgroup of S3.

6) If H and K are normal subgroups of a group then H KG

7) If H is a subgroup of a group G and K G, then H K < G.

11 2 2

11 1 1

11 2 2 112 2 1 2

11

12 2 2(, , , , ,

() ,

)ii eH K F o r a bH K a h k b h kh H k K

hk h k hkk h hk h k

hh h k h H K

.

¸

8) If H, K < G, where G is a group and H G then HKK

Miscellaneous problem on Group Theorem

1] Compute order of each element in the following groups

i) 3D- Dihedral group of order 6

ii) 4D- Dihedral group of order 8.

iii) u(30) iv) 4S v) 8

2] Let X = (1 2 3 4 .. 11 12) 12.S for which integers I, 12,iii i s xbba 12

cycle ?.

3] If X = (1 2) (3 4) (5 6) (7 8) (9 10). Is there a n cycle

(1 0 )kns u c h t h a t xTp T for some positive integer k?

4] Show that 1{/ }01nHn ¬®] is a cyclic subgroup of GL2(\)

5] Let G be an abelian group prove or disprove

{: }nHx G xe f o r s o m e n ` is a subgroup of G. munotes.in

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63Chapter 2: Isomorphism of Groups

6] Let G be a finite group of order 2,np G cannot have a subgroup H of order

n-1.

7] Let 12 1 2 2 1,.GGb e g r o u p s P r o v e G G G G x

8] Show that the following are subgroups of 12GG where 12,GG are groups

then,

12 2 2 1 2 1 1

11 21] {( , ) : } 2] {( , ) : }

3]{( , ) : }eg g G ge g G

gg g Gw h e r e G G G

9] i] Prove that Z 2] is not isomorphic to ].

ii] Prove that 2is not isomorptic c to ]

10] Find the subgroup of nS generated by {(1 2), (1 2) (3 4)}

11] Prove that 4S = <(1 2 3 4), (1 2 4 3)>

12] Find a group which contains a, b such that o(a) = , o(b) = 5 and o(ab) = 2.

13] Let G be a group and , HG If o(H) = 2, Prove that H C Z(G).

14] Show that 4S has a unique subgroup of order 12.

15] From the given pairs of subgroups, find isomorphic pairs, justify your answer

1 ] 42 2andN]] 2 ] 39 2 7 and ]] ]

3 ] and]] ] 4 ] 35 1 5 and ]] ]

5 ] (8) (10)andNN 6 ] (8) (12)andNN

7 ] ] and 2 ] (under additional)

16] Find a non-cyclic group of order 4 in ]4 10]

17] Find all generators the following cyclic groups of

1] 30 20 2] / 30 3] (13) N]]] ]

18] Find a cyclic group of order 4 and a non-cyclic group of order 4 in 4S. munotes.in

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64 ALGEBRA II

19] Let 2{: , } { 2 : , }abGa b G a n d H a b a b Gba ¬ ® s h o w t h a t

GHx (operation being addition in both groups)

20) Prove that every subgroup of nD of odd order is cyclic.

21] Find orders of all elements in nAand nD

22] Let G be a finite group and NGif [G : H] and o(n) are relatively prime,

then for any (),oNxGx e i m p l i e s x N

23] Let G be a group and N be a cyclic subgroup of G such that NG%.

If H < N, show that H < G

24] Let =

{ {: , , ,} {: }00 1ab abGa b d a d o a n d N bd ¬ ¬ v ® ®\\

show that NG and G/N is abelian.

25] Prove or disprove: If G is a group and K < H < G.

KHa n dH Gt h e nK G

24] Show that i) () { }nZDe i f n i s o d d

ii) 2() { , } . ( ( ) ) .m

mZDe a w h e r e a e o a m

25] Let 31

{0 1 : , , }

001ab

Hc a b c ¬ ®]

Is H abelian? Is 33() ? HS L] Justify your answer.

26] Prove that there is no homomorphism from 82 44 onto ]] ]]

27] Determine all group homomorphism from.

1 ] 20 8onto]] 2] 422to ]]]

3] 42 .St o ] munotes.in

## Page 65

65Chapter 2: Isomorphism of Groups

28] Suppose :( 3 0 ) ( 3 0 )N l N is a group homomorphism and ker

{1, 1 1} . If (7) 7, find all elements of (30)N that map to 7.

29] Suppose : zG l] (G is a group) is a group homomorphism such that

(3, 2) a and (2,1) , b D e t e r m i n e (4, 4) (Assume operation in G is

addition).

30] Show that a group of order 65 is cyclic.

31] Let H, K be distinct subgroups of a group G of index 2 Prove that HKG

and [: ]GH K In /GH K Cyclic? Justify your answer.

32] Let G be an abelian group of order 8. Prove or disprove. G has a cyclic

subgroup of order 4.

33] Classify upto isomorphism abelian groups of under 108.

34] If G is an abelian group of order 120 and G has exactly 8 elements of order 2.

Determine the isomorphism class of G.

35] Find the number of abelian groups (upto isomorphism) of order [1] 15 [2] 42

[3] 16 [4] 48.

36] Let G be an abelian group of order 16. and ,,ab G such that o(a) = o(6) = 4

and 22abv then determine the isomorphism class of G.

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## Page 66

66 ALGEBRA II UNIT 2

3

SYLOW THEOREMS

Unit Structure

3.0 Objectives

3.1 Centralizers , Normalizers and Stabilizers

3.2 Groups Actions

3.3 Orbits & Stabilizer

3.4 Sylow Theorems

3.5 Classification of Group of Order p] Where p is a Prime

3.6 Classification of groups of order 15 upto isomorphism

3.7 Summary

3.8 Unit And Exercises

3.0 Objectives

After going through this unit you shall come to know about

x The concept of group action and stabilizers

x The notion of the order and its relation with the order of the groups

x Classification of groups upto order 15 using various results of Sylov

theorems

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## Page 67

67Chapter 3: Sylow Theorems

3.1 Centralizers and Normalizers Stabilizer

We recall that centre of a group G is defined as

/ ZG x G x g g x v \^ gG

ZGG, Moreover, if

/GZ G is cyclic, G is abelian.

Let G be a group and A be a non-empty subset of G.

Definition : The centralizer of a ()GCA i n t h e g r o u p G i s d e f i n e d a s

\^1() /GCA gG g a g aaA .

We note 1 gag a ga ag .

()GCA is the set of all elements in G which commute with every element of A.

Proposition : ()GCA G

Proof : () 'GeC A e a a e a v aA

Let ,( )GxyCA then , xa ax ya ay v aA .

= For 11 1 1;, ( ) aA ya y a y a y a

111 11 1 ay a a y a

11 1 1 1 11 xya x y a x a y a a x a y a x y a x y

1()= GxyC A .

We note that if \^ \^

, n

G Aa aCa v `n .

When \^\^

, GAaC a is denoted by

GCa .

Definition : The normalizer of GA in G, is defined as

\^1() / GNA g G g A g A .

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## Page 68

68 ALGEBRA II

Proposition : ()GNA G

Proof : 1 eAe A

()G eN A=

Let ,( )GxyA/ then 11xa x = A, ya y = A .

=

11 1 1 1 1 1 1 1xy A y x x y y A y y x x y y A yy x x A xA

1()

()G

GxyN A

NA G=

Clearly, () ()aGCA N A

Again, if \^Aa, we denote ()GNA by ()GNa and we shall drop a when there

is no ambiguity.

i) () ()GCa Z a

ii) \^()GCa a iff () aZ G.

Examples :

1) If 3 GS and (123)a. Then \^( ) , (123), (132)Ca I

I f \^,( 1 2 3 ) ,( 1 3 2 ) AI .Then () , ()GGCA A NA G .

2) If G is a group and H < G then H, ()GHNH .

= For 11,xHx H x H x H .

3) IF G is a group and H < G then

()aHCH iff H is abelian

()G HCH v ,( )G hH hCH

v1, hH a h a h v . aH

ah ha v

, aH v i s a b e l i a n hH H

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## Page 69

69Chapter 3: Sylow Theorems

3.2 Group Actions

Let G be a group and S be a non-empty set. Then map B:GS Sq l denoted by

(,)gS g Sl ¸is an action of G on S if it satisfies the following conditions.

1) e.S = S for each sS

2)

12 1 2.. ..gg s g gs for each sS and each 12,gg G

S i s c a l l e d a G - s e t .

Example :

1) Group acting on itself by left multiplication g.a = a a for each , gG aG

i ) .ea e a a for each aG.

i i )

12 12 12 1 2 1 2.. .. .. .ggaggaggaggagga f o r e a c h

aG, for each 12,gg G .

2) Group acting on itself by conjugation 1.gag a g for gGand aG

1 ) 1eae a e a

2 )

11 1

12 12 2 1 2 2 1.. .gg a gg a g g g a g g

11

12 2 1gg ag g

1

12 12 12 . gg a gg gg a

3) A group G acts on ދ(G) (The set of all subsets of G) by conjugation.

1.gSg S g for () SP G. As in (2), it is a group action.

4) Let n be a positive integer. Then n GS by

.iiTT for \^1, ....in . For \^1, ....in

\^/inGS i i TT .

5) Let G be a group acting on a non empty set S.

T h e n , f o r , gG the map :gS Sldefined by

ga ga ¸is a permutation

of S and the map :n GS l d e f i n e d b y

gg T i s a m o n o m o r p h i sm .

(Cayley’s Theorem).

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70 ALGEBRA II

3.3 Orbits and Stabilizers

If G is a group acting on a set S and sS (is fixed) then, the stabilizer of s in G is

the set \^/SGgGgss .

SGG

.seS S

eG

=

Let ,sxyG then ,xss yss .

11 1

11.. ( . ) . .

.. .ysy y s y y Se ss

xy s x y s x s s

1

ssxyG G G= =

Orbit of sS is defined as \^./ sgs S and is denoted by orbit (S)

Proposition : Let G be a group and S be a subset of G. Then the number of

conjugates of a subset S is :( )a GN S and the number aSis :( )GGC a.

We first prove a lemma (The proof may be read quickly)

Lemma : Let G be a group acting on a set S. Then, the relation on S defined by

for ,ab S abiff . ag bfor some gGis an equivalence relation. For each

aS, the number of elements in the equivalence containing a is <>:aGG, the

index of the stabilizer of a.

Proof : We first prove is an equivalence relation. For aS, e - a = a.

aa= ( is reflexive)

,,ab Sa b g G º such that . ag b

11 1.. .gag g b g g b e b b º

baº ( is reflexive)

12 ,, b ,ab Sa b c g g G
º such that 12,. ag b bg c¸

12 1 2 1 2.( . ) . , aggcggcggG º acº ( is transitive)

<>\^/ ag a g G ¸ (orbit of a). munotes.in

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71Chapter 3: Sylow Theorems

Now if <> bg aa ¸ , then agG is the left cost of aG in G and .a bg a g G l

.

= We have a map from <>al left costs of aG in G defined by .a ga g G.

The map is well defined for ..ga ba

11 1

1.. ( . ) ( ) . .

aa ahg a h h a hh a e a a

hgG h G g G

º

º º

Similarly 1

aa ehG gG h g G º

1.. . hg a a h a g aº º

and the map is one - one.

The map is clearly onto, for ,. .aa gG g aG g a g G º .

=The use of elements in the equivalence classes of a = no. of left costs of aG in

G.

= Elements in orbit of <>:a aa G .

Proof of Proposition: \^1/( )sGGg G g S g S N S .

No. of conjugates of <>:( )G SG N S when

\^ () () ()GG Sa N S N a C aº .

=No. of conjugates of <>:a aa G

Class equation : Let G be a finite group and let 1....rIIbe representatives of the

distinct conjugacy classes of G which are not contained in ()ZG. Then

<>

1() : ( )r

Gi

iGZ G G C G

.

Proof : For gG, the conjugacy class of g is \^g iff () gZ G.

1xgx g v xg x g g x v

()xGg Z G

Let \^1 () . . .mZGz z . Let 1,...rKK be conjugacy classes of G not contained in

centre, and ....nrIIbe representatives of conjugacy classes. Then, the conjugacy

classes are \^ \ ^\^11... , ...mr ez z K , .

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## Page 72

72 ALGEBRA II

This partitions G

()11 11( ) ( )

gZGomr r

ii

ii iGk Z G G C g

=

Note :

:( )i GC gO G

Theorem : A group G of order npwhere p is a prime and 1nphas non-trivial

centre.

Proof : B y c l a s s e q u a t i o n .

()1:() ()

gZ Gir

i

iGG C g Z G

As () , ( ) ,iigZ G C g G v

Now , :() / ()ii GCg G Cg so each term in

()1:()

gZ Gir

i

iGCg

is of the form

kp.

Now

()1:() ()

gZ Gir

i

iGG C g Z G

1p\, p \ : ( )n

r

i

iGp

GG C g

= p\( )

() 1G

ZG= ;

= v

G= has non-trivial centre.

Note : A group of order np, p prime, 1np is called a p group. A subgroup whose

order is power of prime is called a p - subgroup.

Corollary: A group of order 2p is abelian.

Proof : By class equation, () 1ZG v2()ZGp o r p

If 2()ZGp then ()ZG=G and G is abelian.

()ZGp then ()GZ G is cyclic

G= is abelian. munotes.in

## Page 73

73Chapter 3: Sylow Theorems

3.4 Sylow Theorems

Definition : Let G be a finite group and let p be a prime divisor of G. If pk divides

G and pk+1 does not divide order Gthen any subgroup of G of order pk is called

Sylow p - sub group of G.

Sylow’s First Theorem :

Theorem : Let G be a finite group and p be a prime. If /kpG t h e n G h a s a

subgroup of orderkp.

Proof : We prove the result by induction on G. If G=1, the theorem is trivially

true. We now assume that the theorem is true for all groups of order less than G.

If G has a proper subgroup H such that kp d i v i d e s H, then by induction

hypothesis H has a subgroup of order kp, which is a subgroup of G.

So, we now assume that kp does not divide the order of any proper subgroup of

G we next consider the class equation,

()() :()

aZ GGZ G G C A

.

/, ( )kkpG pC G v () aZ G

p/ :( )GGv = () aZ G

= From class equation, it follows that /( )pZ G . From the structure theorem of

finite abelian groups Z(G) has an element x of order p, ()xZG xG =

Consider the factor group 1/, p /kGx Gx .

= By induction hypothesis, /Gx has a subgroup of order 1kp. This subgroup

is of the form /Hx where HG and xH.

1,k H px px .

kHp= and thus G has a subgroup of order kp. munotes.in

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74 ALGEBRA II

Corollary 1:

Cauchy’s Theorem : Let G be a finite group and p be a prime that divides the

order of G. Then, G has an element of order p.

Proof : By Sylow’s Theorem, G has a subgroup H of order p.

, Hp H = is cyclic and Hx.

() .Ox p=

Corollary 2 : Let G be a finite group, ,1 ( , ) 1kGp m k p m p then G has a p-

sylow subgroup.

Definition : Conjugate Subgroups :Let H, K be subgroups of a group G. We say

that H, K are conjugates in G if there exists gGsuch that H = g k g.

Sylow’s Second Theorem :

Theorem : If H is a subgroup of a finite group G and H is a power of a prime P,

then H is contained in a sylow - p subgroup of G.

Proof : L e t K b e a S y l o w p - s u b g r o u p o f G . L e t \^ 12,, . . .n C ,, , , be the

set of all conjugates of K in G.

Since conjugation is an automorphism, each member of C is also a Sylow p-

subgroup of G. Let S c denote the group of permutations of C.

For gG, define :gCC l by

1

gi iKg K g .

Then,

11

gi g j i jKK g g g K g º ,

11 1 1

ij i j gg K gg gg K gg º º, , gG= is one one.

If 1,jj iKC h K h , for some hG

11 1 11 1

11 1 1 1()ii

iigg h h gg g g h h g g

gg h g h g g g

, ,

, , munotes.in

## Page 75

75Chapter 3: Sylow Theorems

Where 11 1()iiKg h K g h C .

() ,jg i i KC =, , g=is onto.

Thus, :gCC l is bijective and gcS .

We define a map :c TG S l by ()gTg .

For 11 11, ( ) ( )( ) ( ) ( )ig h i i i in T g h K K g h K g h g h K hg b b

11() ( ) ( ) ()ig h i i gh K h g K TgTh T K

ghTg h TgTh = .

T= is a group homomomorphism.

We next consider T(H).

H is a power of p. ()TH= is also a power of p.

Now

() \( )TH iOrb K T H

()TH iOrb K= = power of p (or 1)

Now

() 1rTH iOb k for some ,1ii n b b

gi iKK v 1

ii gH g K g K v gH

i gN Kv

iigHH N K H K

&ii Hi sap g r o u pa x N k x H x k = º '

Now :() CG N K .

:: ( ) ( ) : , :GK GNK NK K GK i s n o t d i v i s i b l e b y p ( K i s s y l o w - p

subgroup.

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76 ALGEBRA II

Now, C is sum of no. of elements in the orbits.

:GK is not divisible by p.

:()GNKº is not divisible by p. Cº is not divisible by p.

()TH i Co r b K

= There is at least one orbit having only one element.

i H= , for some I by

Sylow’s Third Theorem :

Theorem 3 : t G be a finite group and p be a prime dividing G. Then, the number

of sylow p subgroups of G is equal to 1 module p and divides G.

Furthermore, any two Sylow-p subgroups of G are conjugates.

Proof : L e t K b e a n y S y l o w - p - s u b g r o u p s o f G a n d \^ 2,, . . . .n CK K K K be

the set of all conjugates of K in G.

We show 1nw mod p.

For gG, define :gCC l by

1.gi iKg K g

11 1 1 1 1

gi g j i j i j

ijKK g K g g K g g g K g g g g K g g

KK º º

º g=

is one-one.

For 1,ii iKC K h K h for some hG.

11 1 1 1 1 1()ii igg h h gg g g h g h g =, , ,

1

jgg , where

111

jjKg h K g h C

()gjK g= is onto.

:gCC = l is bijective. gcS = , where cS is the group of permutations of set C.

Let :c TG S l be defined by ()gTg .

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77Chapter 3: Sylow Theorems

Then, for ,.gh G

()gh Tg h

For

1 111,gh i i i in K g h K g h g h K hg b b

11 1

ig i g h i g h i gh K h g h K h K K D

() ( ) ( )gh g h

Tg h TgTh=

=

and T is a group homomorphism.

Consider() () 0 /()TK i TK r b K TK .

But K is a power of p. ()KTK

()TK= is a also a power of p.

() 1TK iorb K= when 1i, and () pTK iorb K otherwise

()

11( 1 )1 m o dn

Tk i

iCo r b K p n m p

=

We next show every sylow p subgroup is a member of C. Suppose H is a sylow p-

subgroup of G which is not a member of C.

We consider T(H), then sum of the orbits size blender action of T(H) is sum of

terms each divisible by p () 11TH iorb k v ' for any i.

mod nC p= w

A contradiction.

HC=

Now <>:( ) nG N K

nG=

Corollary 1 : A unique sylow p - subgroup is normal. munotes.in

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78 ALGEBRA II

Proof : If H is the only Sylow p - subgroup of a group G, then for each

1,gGg H g is also a sylow - p- subgroup of G.

1gHg H v= gGG = )

Corollary 2 : A sylow - p - subgroup of a group which is normal is unique.

Proof : Let H be a sylow p - subgroup of a group G.

Then, any sylow-p-subgroup of G is conjugate to H, and is of the form 1gHg for

some gG.

G)1gHg H= .

H= is a unique sylow p - subgroup of G.

3.5 Classification of Group of Order p] Where P is A Prime

Theorem: Let p G] where p is an odd prime. Then 2pp Go r G Dx x]

Proof : By Cauchy’s Theorem, G has subgroups of order 2 and p.

G= has an element of order 2 and an element of order p.

,,

:2Ga b b p

Gb b G

= =

1

1,1kaba b

ab a b for some k k p

=

=

21 1 1 1 2 2

2() ( ) ( )

2xk k k

kbb a b aa b a a a b a aa b a

ab b

=

212(1 )( 1 ( 1 ) ( 1 ) 1kbe p k p k k k p = º º º

11 1 1ko r k P ko r k p º º

Case 1: If k = 1,

1

222

paba b ab ba ab p and G is cyclic of order p

GZ

= =

=x munotes.in

## Page 79

79Chapter 3: Sylow Theorems

Case 2: if k = p - 1 11 paba b

1

1ppaba b

ba ab

Gi saD i h e d r a lg r o u p=

Groups of order pq. where p,q are distinct prime .

Theorem : If G is a group of order p.q where p.q are primer, p < q and p does not

divide q-1, then G is cyclic and isomorphic to pqZ.

Proof : Let H be a sylow p subgroup of G and K be a sylow q subgroup of G.

The no of sylow p subgroups of G is of the form 1 + kp and divides pq.

But p x q-1

= k = 0 and H is the unique sylow.

p – subgroup of a . Ha=

Similarly, no of q – sylow subgroups of G are 1 + kq and divides pq.

k =

= K is the unique sylow subgroups of G. =. KG

H and K are cyclic subgroups of G

Let H = , k=

we show xy = yx.

11 11 1 1() , ( ) .xyxyyxyHxyxyk

11{}xyx y h k e H K H HNK K

xy yx= =

=

xyp q G

But xy G=

,pqxy G

Gi sc y c l i c G Z=

== x

3.6 Classification 0f Groups 0f Order 15 Upto Isomorphism

We have already seen that groups of order 2, 3, 5, 7, 11, 13 are cyclic (prime order)

and are unique upto isomorphism.

A group of order 1 is trivial and is unique upto isomorphism. munotes.in

## Page 81

81Chapter 3: Sylow Theorems

3) Class equation : Let G be a finite group and let 1....rIIbe representatives of

the distinct conjugacy classes of G which are not contained in ()ZG. Then

<>

1() : ( )r

Gi

iGZ G G C G

.

4) A group G of order npwhere p is a prime and 1nphas non-trivial centre.

5) Let G be a finite group and let p be a prime divisor of G. If pk divides G and

pk+1 does not divide order Gthen any subgroup of G of order pk is called Sylow

p - sub group of G.

6) Sylow’s First Theorem: Let G be a finite group and p be a prime. If /kpG

then G has a subgroup of orderkp.

7) Cauchy’s Theorem : Let G be a finite group and p be a prime that divides the

order of G. Then, G has an element of order p.

8) Let G be a group. A subgroup H of G is called a normal subgroup of G if aHa

1 H for each aG

1(, )or aha H for each a G each h H

9) Sylow’s Second Theorem : If H is a subgroup of a finite group G and H is a

power of a prime P, then H is contained in a sylow - p subgroup of G.

10) Sylow’s Third Theorem: If G be a finite group and p be a prime dividing G

. Then, the number of sylow p subgroups of G is equal to 1 module p and

divides G.

Furthermore, any two Sylow-p subgroups of G are conjugates.

11) A unique sylow p - subgroup is normal.

12) A sylow - p - subgroup of a group which is normal is unique.

13) Let p G] where p is an odd prime. Then 2pp Go r G Dx x]

14) If G is a group of order p.q where p.q are primer, p < q and p does not divide

q-1, then G is cyclic and isomorphic to pqZ.

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82 ALGEBRA II

3.8 Unit And Exercises

1) For the given group G and the H subgroup of G. Show that ()GCH H and

()aNH G .

i ) \^22

4,1 , , , GD H b a a b where 42 3ae b b a a b

i i ) \^ 3,1 , GS H

.

2) Find all groups (up to isomorphism) of order 99.

Answer : Let G be a group of order 99. Let H be a sylow 3 subgroup of G and

K be a sylow 11 subgroup of G Then, no of sylow 11 subgroups of G is

congruent to 1 mod 11 and divides 99 = K is unique sylow subgroup of G.

.

,{ }KG

Similarly H G H K e

=

It follows that elements from H and K commute and therefore GH Kq

33 3 9 9Gi sa b e l i a n

GZ Z o r GZ=

x x

3) Determine all groups of order 66 upto isomorphism.

Answer: Let G be a group of order 66. Let H be a sylow 3 subgroup of G and

K be a sylow 11 subgroup 8 G.

Then, 1 is the only positive divisor of 66 which is congruent to 1 mod 11,

Therefore , . KG

() 3 3 ( 3 3 3 1 1 3 1 1 1 ) HK G and O HK and

HK is cyclic

= q q

=

Let HK = < x >, HK has index 2 in G . HK G=

Let 2. yG a n d y

1yxy x

Let 113 2 ,iiyxy x for some i for to Now x x munotes.in

## Page 83

83Chapter 3: Sylow Theorems

211,( )iiii i iyx x y x y x y y xy x x

2,ixe

233 \ 1

11/ 1 11/ 1i

io ri =

=

,1 , 1 1 1 , 1 1 3 3 1 .ii i = o r i = o

1, 10, 23 32.io r =

= T h e r e a r e a t m o s t 4 g r o u p s o f o r d e r 6 6 . w e o b s e r v e t h a t

66 33 11 3 3 4,,ZDD Z a n d D Z a r e o f o r d e r 6 6 , a n d n o t w o o f t h e m a r e

isomorphic.

4) Show that the only group of order 255 is 255.Z

Proof:- Let G be a group of order 255.

2 5 5 = 3 . 5 . 1 7 .

Let H be a 17 sylow sub group of G Then, number of 17 sylow subgroups of G is

congruent to 1 mod 17 and divides 255.

The 17 sylow subgroup is unique.(Divisors of 255 are 1,3,5,15,51,85)

17 ()

()HG HZ

NH G

=x

=

17 () /() () ( 1 7 ) 1 6NH CH d i v i d e s A u t H A u t U = ]

Since () /() /() / 1 6NH CH G CH we have () /() 1NH CH

() GC H=

= every element of G commutes with every element of H.

()

17 ( ) , ( ) 255HZ G

ZG ZG

=

=

/( ) 1 5 , 5 , 3

/( )GZ G

GZ G i s c y c l i c

=

=or 1 munotes.in

## Page 84

84 ALGEBRA II

Gi sa b e l i a n

Gi sc y c l i c=

=

5) Show that a group of order pqr, where p,q,r are distant primer is cydic.

6) If (G) = 36 and G is non-abelian, probe that either G has more than one 2 sylow

subgroup or more than one 3 sylow subgroup.

7) Show that a group of order 56 has a proper non-triral normal subgroup.

8) Let G be a group of order 60. If the sylow 3 subgroup is normal, show that the

5 sylow subgroup is also normal.

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## Page 85

85Chapter 4: Introduction to Ring UNIT 3

4

INTRODUCTION TO RING

Unit Structure

4.0 Objective

4.1 Introduction

4.2 Ring

4.3 Characteristic Of Ring

4.4 Subring

4.5 Summary

4.6 Unit and Exercise

4.0 Objective

The objective of the this unit is to introduce the concepts of

x Ring integral domain and fields with example.

x The characteristic of ring.

x How to check a given subset of ring is subring?

4.1 Introduction

Basically there are two operations ‘ +’ and ‘ .’ which we can apply on scalars &

vectors. We know how to add and multiply two scalars. In vector space, we saw

two operation vector addition and scalar multiplication. In group theory we have

seen a set with one operation whether addition or multiplication satisfying certain

property. So for we had not seen any set with both operation addition as well as

multiplication. Ring is one such algebraic structure which has both these operation.

Loosely speaking ring means closed structure that is why (may be) an algebraic

structure which is closed with addition and vector multiplication is called as ring.

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86 ALGEBRA II

4.2 Ring

A ring R is set together with two binary operations + (addition) and ‘ ¸‘

(Multiplication) satisfying

i) (R, +) is an abelian group.

ii) `.’ is associative : i.e.

ab c a bc¸¸ ¸¸ for all a, b, c, R.

iii) The distributive laws hold in R.

i.e. for all

,,abc R a b c ac bc <<< and

ab c a b a c << <

Before we see example of ring let understand versions terms involved in ring with

the help of following remarks.

Remarks:

1) If R is ring, then (R, +) is abelian group. Multiplication (''¸) n e e d n o t b e

commutative. When ''¸ is commutative we say R is commutative Ring.

2) As (R,+) is abelian group, o n e l e m e n t s a y ` 0 ’ , s u c h t h a t

aooaa v . aR T h i s e l e m e n t ` 0 ’ i s c a l l e d a s z e r o e l e m e n t o r

additive identity of R. A ring may, may not have multiplicative identity. But

when it has, i.e. if ring R has an element 1 such that 11aa a¸ ¸ for all a

R then such element `1’ is called as unity in R & R is said to be ring with

unity. Note that 01 .v Throughout the course we assume R is ring with unity

otherwise

3) For each a in ,Rb R such that 0 ab such element is called additive

inverse of ‘ a’ and is unique.

4) A non zero element ‘a’ of R is said to be unit if there exists b in R such that

aºb = 1 . b is said to multiplicative inverse of a. Multiplicative inverse of

element may or may not exists.

5) An element aR is said to be zero divisor if there exists bov such that

ab o¸ . Note that zero divisor cannot be unit.

(as if a is unit and zero divisor then ,1bc s u c h t h a t ab ba ¸ ¸ and ac

1 bac bc c c º¸¸¸º¸ º a contradiction.

6) A commutative ring (with unity) is said to Integral domain (ID) if it has no non

zero zero divisor. (i.e. in R the only zero divisor is zero element 0). munotes.in

## Page 87

87Chapter 4: Introduction to Ring

One more way is there to define ID is that, whenever ab in ID a º

or b = 0. N o t t h a t i f R i s n o t ID a b o v e r e l a t i o n m a y n o t b e s a t i s f i e d a s i n

62. 3 0 ] but 23 0 .vv

In case of ID cancellation law hold.

as if .. ( ) ( ) ab a c ab ac a b c a or b c since ID º º º

if a o b cvº ab ac b c =º if aov

7) In the definition of units we have seen that every element of ring R need not be

unit.

A commutative ring in which every non zero element is unit is called as field.

A commutative ring R (with unity) is said to be field if for all a v, in R, there

exists b in R such that a.b = 1.

They are some standard examples of ring which are easy to show. We will list

then without proving.

1)

Set Commutative

Ring Zero

element Unity Zero

Divisor ID Units Field

,,¸] Yes 0 1 No Yes 1o No.

,, ¸_ Yes 0 1 No Yes all non Zero

element Yes

,,¸\ Yes 0 1 No Yes All non Zero

element Yes

,,< Yes 0 1 No Yes all non zero

element Yes

()nM\ No. Null

Matrix Identify

Matrix Yes No ()

..n AM

st A

v\

No. <><><>,

&xxx_

\

] Yes Zero

Polynomial 1 No Yes Non Zero

Constant

Poly. No

2) IF R is a ring than the set {( , ): , } RR a ba bRq with + and ¸ define as

(,) (, ) ( , ) & (,)(, ) ( , )ab ad a cb d ab cd a cb d ¸ is not integral

domain as if a,b R, ,( , 0 ) . ( 0 , ) ( 0 , 0 ) ab t h e n a b vv Hence not ID.

(a,0) is zero element of this ring and if 1 is unity in R. then (1,1) is unity in R

x R. If R is commutative, R x R is also commutative. munotes.in

## Page 88

88 ALGEBRA II

3) Let R = ] [i] = {: , , 1 }ab i a b i ] is also ring with 0 as zero element

and 1 as unity. This is commutative ring with no non zero divisor and hence

integral domain. Note that only 1o and iohas multiplicative inverse (as ( o

1) (o1) = 1, i (-i) = 1). So it is not field.

This ring is known as ring of Gaussian integers.

These are some important example of rings. Many examples can be found in

reference books. We end this section by given two more important class of ring.

4) Consider {0,1, .. 1}n n ] the set of residue classes modulo n. define + and

¸as abab and .ab a b Then n] is ring.

Some important results aboutn].

i) Every non zero element of n] is either zero divisor or unit.

Proof: - Let n a ]such that 0av suppose a is not zero divisor.

We will show that a is unit.

Claim: (a,n) = 1 . ((a, n) means G.C.D. of a and n)

Suppose (a, n) = d, d > 1 . Then d/a and d/n.

Let ,0nnbba n d bd v]

and

0.na aab a ndd d ¬ ¬ ¬ ® ® ®

º a is zero divisor which is not the case.

Hence (a, n) = 1 .

Therefore there exists x, y ] such that

1 ax yn

11 ax yn ax= º

yn o' x=is multiplicative inverse of a.

=a is unit.

Hence if a is not zero divisor then a is unit.

(ii) If n is prime then n] is field.

' n is prime, therefore

,1an, for all a, 1 < a < n .

Hence for any a in n] multiplicative inverse of a exists. munotes.in

## Page 89

89Chapter 4: Introduction to Ring

5) Polynomial Ring : The polynomial plays very important role in Algebra.

Finding root of polynomial is one of the central problems of Algebra. Set of all

polynomial over \with respect to operation polynomial addition and

multiplication from a ring. Let we recognize it’s important and discuss it

separately.

Definition: Let R be a commutative ring. The set of formal symbols

1

11 [] { . . : : ,nn

nn o iRxa x a x a x a a R n

is non negative integer} is

called the ring of polynomials over R in indeterminate x.

Remark:

i) Let 1

11 () . .nn

nn ofxa x ax a x a

and

1

1 () . .mm

mm o gx b x b x bx b

Then

1

11 1 1 () () ( ) ) . . ( )sssss s o ofxg x ab x a bx a b x ab

where s = max {m, n}

and 1

11 () .() . .mn mn

mn mn ofxg x c x c x c x c

were i

ii k k

koca b

Definition: Let 1

11 () . .nn

nn ofxa x a x a x a

If 0nv then n is called degree of f(x), and na is called leading coefficient of f(x).

If 1na, then polynomial is called monic polynomial. The degree of polynomial

is denoted by deg

fx.The polynomial

0 fxhas no degree and is called zero

polynomial. The polynomial

0fxa is called constant polynomial.

If R is commutative ring with 0 as zero element and 1 as unity then <>Rx is also

commutative ring with zero element as zero polynomial 0 and unity as constant

polynomial

1 fx. If R is integral domain then <>Rx is also. But <>Rx is not

field even though R is field as the only units in <>Rx are constant polynomial.

In this next unit we have one separate chapter for polynomial ring where we discuss

one important concept of ring irreducible polynomial.

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## Page 90

90 ALGEBRA II

Let we discuss some theorems giving relationship between integral domain and

fields.

Theorem 1: Every field is integral domain.

Proof : Let R be a field and let ,ab R such that

01 ab l

Let assume . a vWe will show b = 0 .

Since . a v and R is field. xR = such that 1 ax xa¸ ¸ .

Multiply both side of (1) by x we get

0xab x¸¸ ¸

()xa b =

bb º º

And as R field, therefore R is commutative ring.

Therefore R is integral domain.

Note : Every field is integral domain. But every integral domain need not be field.

For example ] which integral domain but not field.

A ring R is said to be finite ring if R is finite Set.

Theorem 2: Every finite integral domain is field.

Proof : Let R be finite integral domain.

Therefore R is commutative ring.

Let a R be such that a v

then 23,,, . . . .aa a R But as R is finite all this indices may not be distinct.

That is there exist ijv such that ijaa

let i > j then ..ijjjaa a a

1ija º

Also note that i – j > 0

Therefore 1 ijaR such that 1.1ij ijaa a

Therefore 1 ija is multiplicative inverse of a. a v

Therefore all non zero element of R has multiplicative inverse.

Therefore R is a field . munotes.in

## Page 91

91Chapter 4: Introduction to Ring

4.3 Charctristic of Ring

Let R be a finite ring. Then (R, +) is a finite abelian group. If O(R) = m then for all

a R, ma = 0 . Hence for finite ring ( R, +) there exist m` such that 0 ma¸

for all a R.

Note that such number is not unique. If m is one such number then, all multiple of

m satisfies same condition. We are interested smallest positive integer n such that

0 na¸ , such number is called characteristic of ring.

Definition: Let R be a ring. A least positive integer n such that 0 na¸ for all a

R is called characteristic of R & is de noted by char R. If no such number exists

then char 0R.

For example char ] char _ char 0\, char .nn]

Characteristic of infinite ring is obviously zero. But characteristic of finite ring is

also not easy to find. The theorem given below helps to find characteristic of ring.

Theorem 3: The characteristic of ring R with unit 1 is n if and only if n is least

positive integer such that 10n¸ .

Proof : Let assume that char R = n

= n is least positive integer such that n.a = o for all a R.

= 10n¸

Conversely let n is least positive integer such that 10n¸ .

To prove that char R = n.

Let a\ be arbitrary.

Consider

11 0 0 na n a n a a¸ ¸ ¸ ¸¸ ¸ and n is least such positive integer.

= char R = n .

Theorem 4: The characteristic of integral domain is (hence field) is either zero or

prime.

Proof: Let R be an integral domain.

If char R = 0 Then nothing to prove.

Hence assume char R = n, n v 0

To prove n is prime number,

Let assume n = x.y where 1 < x, y < n

' char R = n

= for each ,0 aR n a ¸

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92 ALGEBRA II

i.e. n.a2 = 0

i.e. (xy)a2 = 0

= (xa) (ya) = 0

= x.a = o or y.a = o ( since R is integral domain)

which is not possible as x, y < n and char R = n.

Hence no such x, y exists

= n is prime

Corollary 1: Characteristic of a finite field is prime.

Proof: Let F be a field.

(F, +) is a finite group.

Hence order of F exist & Let o(F) = m

10m= ¸ .

Let n be least positive integer such that 10n¸ as F is field and hence integral

domain. Therefore by above theorem n is prime.

4.4 Subring

Subrings are non empty subset of a ring which itself is ring with respect to the

operation of ring.

So if we want to prove a non empty subset S o f r i n g R i s s u b r i n g , o n e h a s t o

prove all the property of ring, which is a lengthy procedure. The following theorem

gives a easy method to determine whether a given subset of Ring R is subring or

not.

Subring Test:

Theorem 5: Let s be a non empty subset of ring R. Then S is subring of R if for any

a, b S,

a-b S and ab S¸ .

Proof: Let S be subring of R.

Then S be itself a ring.

Therefore for any a, b S, a-b S and a.b S munotes.in

## Page 93

93Chapter 4: Introduction to Ring

Conversely let assume that S be non empty subset of R, such that a – b and a. b

S whenever

a, b S.

To prove S is subring.

' for a, b S a-b S

=(S, +) is subgroup of (R, +)

= (S, +) is abelian group.

For ,,ab S a b S ¸

S is closed under multiplication .

Also as multiplication ( ''¸) distribute over addition in R and therefore in S also.

Therefore S is a ring.

Hence S is subring of R

Eg:- 1) Let R = 2()M]

i) 1 :,aaSa bbb£² ¯¦¦¦¦¡°¤»¡°¦¦¢±¦¦¥¼]

Let ,,aa xxAB S a n d l e tA Bbb yy ¯ ¯¡°¡° ¡°¡°¢±¢±

then A – B = 1axazSybyb ¯¡° ¡°¢±

and AB = 1ax ay ax aySbx by bx by ¯¡° ¡°¢±

1S= is subring of R

ii) 2 :,aa bSb aab a £² ¯ ¦¦¦¦¡°¤»¡°¦¦¢±¦¦¥¼]

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94 ALGEBRA II

Let 2aa b xx yAand B Sab a z y x ¯ ¯ ¡° ¡° ¡° ¡°¢± ¢±

2ax abxyABSabx y ax ¯ ¡° ¡° ¢±

2.() () () ().() () () ()ax a b x y a x y a b xABSab x a xy abxy a x ¯ ¡°¡° ¢±

2S= is subring

2) Let R]]]

{( , , ) / }Sa b c R a b c

Take a = (1, 2, 3), b = (4, 2, 6) then a, b S

a.b = (1, 2, 3) ¸(4, 2, 6) = (4, 4, 18)

but 4 + 4 v 18 =(4, 4, 18) S

Hence S is not subring.

4.5 Summary 1) A ring R is set together with two binary operations + (addition) and ‘ ¸‘

(Multiplication) satisfying

i) (R, +) is an abelian group.

ii) `.’ is associative : i.e.

ab c a bc¸¸ ¸¸ for all a, b, c, R.

iii) The distributive laws hold in R.

i.e. for all

,,abc R a b c ac bc <<< and

ab c a b a c << <

2) Every field is integral domain.

3) Every finite integral domain is field.

4) Smallest positive integer n s u c h t h a t 0 na¸ , such number is called

characteristic of ring.

5) The characteristic of ring R with unit 1 is n if and only if n is least positive

integer such that 10n¸ . munotes.in

## Page 95

95Chapter 4: Introduction to Ring

4.6 Unit and Exercises

1) Show that q]]under component wise addition and multiplication is a

ring. Is it an integral domain? Justify.

2) Let

,R ¸ be a ring with multiplicative identity 1.R Show that

1,,R:

is a ring where, 1Rabab and ababa b :

3) Show that the ring

( i ) \^ :,, da b d a b ¯ ¡°¢±]] d is an integer is integral domain.

( i i ) {: , } , da b d a b ¯ ¡°¢±__ d is an integer is field.

4) Let R = {0,2,4,6,8} under addition and multiplication modulo 10. Prove that

R is field.

5) Let F be a field of order 2n. Prove that char F = 2 .

6) Let D be an integral domain of characteristic P. For any ,xyDshow that

( i ) ()pppxyx y (ii) ()nnnpppxyxy for all n N

( i i ) F i n d e l e m e n t x a n d y i n a r i n g o f c h a r a c t e r i s t i c 4 s u c h t h a t

444() .xyx yv

7) Let F be a field and let K be a subset of F with at least two elements. Prove

that K is subfield. (i.e. prove that for any

1,, 0 i n , a n dab F b K a b a b K v ¸ )

(This is known as subfield Test)

8) An element a R is said to be nilpotent element if there exist n` such

that 0na. Prove that a R is nilpotent then 1 – a is unit in R.

(Hint : 11nnao a=

9) An element a R is said to be idempotent if 2. aa Prove that

(i) The set of idempotent of a commutative ring is closed under

multiplication.

(ii) The only idempotent in an integral domain is 0 and 1.. munotes.in

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96 ALGEBRA II

10) Determine all zero- divisors, units & idempotent elements in (i) 18]

(ii) 36q]] (iii) q]_.

11) Let d be a positive integer. Prove that [] { , }da b d a b __ is a

field.

12) In the following examples, show that S is a subring of the given Ring R.

( i ) 2 :( )abSa c b d Rcd£² ¯¦¦¦¦¡° ¤»¡°¦¦¢±¦¦¥¼\ 0

(ii) 22:, ( )abSa b Q R Mba£² ¯¦¦¦¦¡° ¤»¡°¦¦¢±¦¦¥¼_

13) Determine which of the following are subrings of

,, ¸_. Justify your

answer.

( i ) :, , (,) 1 ,aSa b a b b i s o d db£²¦¦¦¦ ¤»¦¦¦¦¥¼]

( i i ) :, , (,) 1 , 0 ,aSa b a b b b i s e v e nb£²¦¦¦¦ v¤»¦¦¦¦¥¼]

(iii) :, , 0 , (,) 1 ,aSa b b a b a i s o d db£²¦¦¦¦ v¤»¦¦¦¦¥¼]

( i v ) \^2: Sx x _

( v ) \^:, Sx x x o _

14) Let ,,zwHz wwz £² ¯¦¦¦¦¡°¤»¡°¦¦¢±¦¦¥¼\ S h o w H is a non commutative subring of

2M \ i n w h i c h e v e r y n o n z e r o e l e m e n t h a s a n i n v e r s e w i t h r e s p e c t t o

multiplication. munotes.in

## Page 97

97Chapter 5: Ideal and Quotient Ring UNIT 3

5

IDEAL AND QUOTIENT RING

Unit Structure

5.0 Objective

5.1 Ideal

5.2. Quotient Ring

5.3 Types of Ideal

5.4 Unit Exercises

5.0 OBJECTIVE:

x The objective of this chapter is to introduce the concept of

x Ideal and its importance.

x Theorem used to determine whether given set is ideal or not.

x Quotient ring

x Types of ideal i.e. Prime and maximal ideal.

x Their relation with Quotient ring.

5.1 Ideals

Our aim is to define something similar to quotient space (in case of vector space)

and quotient group (in case of group). Note that if W is any subspace of vector

space we can very well defined quotient space which itself a vector space. To define

quotient group of group G w e n e e d s p e c i a l t y p e o f s u b g r o u p c a l l e d a s n o r m a l

subgroup. In group theory we have seen that if H is normal subgroup of G then G/H

is defined.

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98 ALGEBRA II

Quotient ring is also similar to quotient space (or group). It is denoted as R/I where

R is ring and I must be similar to normal subgroup (not just subring).

This leads to concepts of ideal.

Definition (Ideals): Let R be a ring a subring I of R is said to be ideal of R if for

, , a n d a xxR aI x a *

T h a t i s i d e a l s a r e t h o s e s u b r i n g o f R which absorbs the element of R.

I f I is proper subset of R then I is said to be proper ideal of R.

A ring having no proper ideals is called as simple ring.

Ideal Test: A non empty subset I of ring R is an ideal of R if

(i) a-b I whenever a, b I

(ii) (ii) ,xaax I¸¸ whenever a I and x R.

Example :

1) For any , nn ]]is ideal of ] as n]are only subgroup of ].

2) Let <>Rx \the set of all polynomial with real coefficient.

Let A be subset of all polynomials with constant term zero. Then A is ideal of R.

As let f(x), g(x) A

Therefore (0) 0 (0) (fg since constant term of f & g is zero)

() () () ( 0 ) 0 therefore if h z f x g x then h

Also for any <>()kx x\

(0) (0) (0) 0 0kf k ¸ ¸ , there fore () ()kxf x I

Therefore I is ideal of \.

Note that I 1 {( ) : ( ) . . . . }n

nfxR f xa x a x

1

1 {( ) : ( ) ( . . . ) }n

nfxR f x x a x a

{( ) : ( ) .( ) }fxR f x x g x

x

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## Page 99

99Chapter 5: Ideal and Quotient Ring

3) Let R be ring of the real valued functions of real variable. The subset of S of all

differentiable functions is a subring of R but not an ideal of R.

xR, x i s n o t d i f f e r e n t i a b l e , x S , s i n c e x i s d i f f e r e n t i a b l e b u t xxS¸ as xx¸ is not differentiable hence S is not an ideal.

4) Let R be a ring and a R. Then, \^: aR ar r R is ideal of R, known as

principal ideal of R generated by a.

Principal ideal plays very important role in ring theory. An integral domain in

which every ideal is principal ideal is known as principal Ideal Domanin which

we will see in coming chapters.

Theorem 1: L e t I, J be any ideal of ring R.

Then I + J = {x + y : x, I, y I}, IJ =

1:,k

nn n n

nxyx I y J

£²¦¦¦¦ ¤»¦¦¦¦¥¼

(i.e., IJ is sum of finite product of element of I and J), and IJ is ideal of R.

Proof:

(i) Let a, b I + J

11 2 2 ax y a n d bx y=

then 12 1 2() ( ab x x y y I J

Also for any 11 1 1 ,. ( ) rR a r x yr x r y rI J ¸

11 ,) as I J are ideals therefore x r I and y r J

Similarly r.a I + J

Hence I + J is ideal.

(ii) Let a, b I, J

11nm

ii j j

ijax y a n d b p q

where ij i jxpI a n dyqJ

then a – b =

11nm

ii j j

ijxyp q

1

11()nm

ij j

ljxyp q I J

Let r R

ar = 11

11()nn

ii

Lixyr x ry I J

¬ ®

(, )Since I is ideal therefore xr I

Similarly ra IJ.

Therefore IJ is ideal munotes.in

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100 ALGEBRA II

Theorem 2: A Field is simple ring. That is the only ideal in a Field is {0} and F

itself.

Let A be any non zero ideal of field F.

Let a A, a v 0. F is field therefore b F

Such that a.b = 1

But A is ideal also.

?1.ab A (Since I is ideal, r.a I for r F)

? for any x R, F x.1 = x A

?FA

. FAº

5.2. Quotient Ring

Let R be a ring and I be its ideal define R/I = {x + I : x R}

Define `+’ and `.’ as

(x+I)+(y+I) = (x + y) + I a n d (x + I)(y +I) = x y + I

We claim that R / I is a ring with respect to above operation, called quotient ring.

(i) First we prove that ‘+’ and ‘ ’ defined as above is well defined.

Let assume that ''xIxI a n d yIy I

'' therefore x x I and y y I º (By definition of cosets)

''xxyyI º ''() ( )xyxyI º

'1 ' ''( ) ( ) ( ) ( )xyIxyIx IyIx IyI º º

? + is well defined.

also as '',xxyyI

'',, therefore let x x ayybfor some a b I

? ''xxa a n d yyb munotes.in

## Page 101

101Chapter 5: Ideal and Quotient Ring

?''() ()xyxayb '' ' 'xyxbyaa b

? '' ''xyIxyxbyaa bI ''xyI (Since I is ideal therefore

'',xbyaa b I hence''.xbyaa bII

?''() () ( ) (xIyI x Iy ,

? ‘¸’ is well defined.

As (, )R is abelian group

(/ ,)RI= is also abelian group.

and

xIyI zI x yIzI x y zI

Similarly

xIy I z I x y z I

? ‘’ is associative.

() ( ) ( )

()xI yI zI xy I zIxyzI x zyzIx z IyzI

xI zI yI zI

¸

¸

Therefore ‘ ’ is distribute over ‘+’.

Hence (R / I, +, ) is ring.

Remark :

1) If I is subring only, R/I is not defined.

2) If 0 is zero element of R then 0 + I = I is zero element of R/I.

3) If R is commutative ring then quotient ring R/I is also commutative.

4) If R is ring with unity 1, then R/I is also ring with unity 1 + I.

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102 ALGEBRA II

Examples :

1) \^/4 4 , 1 4 ,2 4 ,3 4 ]] ] ] ] ]

2) \^ 2/ 6 6 , 2 6 , 4 6 ]] ]] ]] ]

3) Let R = ][i], the ring of Gaussian integer and I = <2-i>

t h e n R/I = {a+bi+<2-1> : a + bi R}

= 2 – i <2 – i> = I

= 2 – i + I = I in R/I 20ii n R = ,

24 1 /ii n R i n RI , = =

50 / 1 35 1 in R I Let i R = =

u s i n g 2 = i , a n d 5 = 0 w e g e t

13+ 51i + I = 3 + 5 q 2 + (5 q10+1) (2) + I

= 3 + 2 + I = 5 + I ( S i n c e 5 = 0 i n R / I ) = I

Thus using 2 = i, 5 = 0 in R/I any element a + bi + I of R/I is reduced to one

of the element I, 1 + I, 2 + I, 3 + I or 4 + I .

Also note that all this elements are distinct as additive order of 1 + I is 5

Hence R/I = {I, 1 + I, 2 + I, 3 + I, 4 + I}

4) Let 2() :,,,abRM a n d S a b c dcd £² ¯¦¦¦¦¡° ¤»¡°¦¦¢±¦¦¥¼]] then S is ideal of R.

Let we find R/S:

T a k e o n e e x a m p l e . C o n s i d e r o n e l e m e n t

34

17AS ¯¡°¡°¢±

34

17ASS ¯¡° ¡°¢± 12 4

11 6S ¯¡°¡°¢±

10 2 4 10 2 4,11 0 6 11 0 6SS s i n c e S ¬ ¯ ¯ ¯ ¯ ¡° ¡ ° ¡° ¡ ° ¡° ¡ ° ¡° ¡ ° ® ¢± ¢ ± ¢± ¢ ± munotes.in

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103Chapter 5: Ideal and Quotient Ring

This means that it lmSnp ¯¡°¡°¢± is element of R/S then the only choice of l, m ,n , p

are 0 and 1 Hence there total 421 6 element in R/S, therefore |R/S|=16.

5) Let <>x\ d e n o t e t h e r i n g o f p o l y n o m i a l s w i t h r e a l c o e f f i c i e n t a n d l e t

21 Ix

22[] / 1 { () 1 : () [] } RR x x f x x f x R x

Let us see what quotient ring R is;

By Division algorithm, for any

<>fxx\ <> ,( ) ( )pxa n d rxi n x \

Such that2() ()( 1 ) () () 0fx p x x r xw h e r e r x ¸

2deg ( ) deg ( 1) . . ( ) 0 ( ) , , .rx x i erx o r rx a x ba b R

2() 1 ()fxI p x x r xI=

2

22 2() ( 1 ) ()

() , 1 , () ( 1 ) () ( 1 )px x I rx I

rx I S i n c e x I t h e r e f o r e px x I px x II

2[] / 1 { ., } Rx x a x b a b , = \\

5.3 Types of Ideal

Prime Ideal: Let R be a commutative ring. An ideal P of R is said to be prime ideal

if whenever a, b P º either a P or b P.

Maximal Ideal: Let R be a commutative ring An ideal M of R is said to be maximal

ideal if (i) M v R (ii) MTR then either M = I or I = R.

Before we see example of prime and maximal ideal we just prove two theorems.

These theorem use to characterise ideal with the help of quotient ring.

Theorem 3: (P is prime /RPºis integral domain)

Let R be a commutative ring. An ideal P of R is prime ideal if and only if R/P is

integral domain. munotes.in

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104 ALGEBRA II

Proof: Let R be commutative ring and P is prime ideal of R.

To prove R/P is Integral domain.

Let / aaP a n d bb p R P

Such that ./ab P i n R P (P is zero element in R/P)

.. ( ) ( ) / ..ie a P b P Pi nR Pie a b P P ab Pº

But P is prime ideal

Therefore ab P a Pº or bPaPP o r bPP º

Hence if .ab P a P o r b Pº

Conversely let assume that R/P is integral domain To prove P is prime ideal.

Let ,ab R such that ab Pº ab + P = P º (a +P) (b + P) = P

But R/P is integral domain.

() . ()aPbP P aPP o r bPP ap o r bp º º

.ab P a Po rb P= º

= P is Prime ideal.

Theorem 4: M is Maximal ideal R/M is Field.

Let R be a commutative ring with unity. An ideal M of R is maximal ideal of R if

and only if R/M is field.

Proof: Let R be a commutative ring with unity and M is Maximal ideal of R.

To prove /RM is field. Let / aMR M such that aM M v aMº

Consider the ideal {:: } aR a r r R of R.

= M is ideal, aR is ideal.

Then N = M + aR is also ideal such that MR 1

But M is maximal ideal and M z N.

Therefore NR, hence 11RN º

,xMr R = such that 1xar munotes.in

## Page 105

105Chapter 5: Ideal and Quotient Ring

() 1 () 1xa r M M xM a rM 0= =

() () 1 ( )aMrM M x MxM M= = '

Hence a + M has multiplicative inverse.

= Every non zero element of R/M has multiplicative inverse.

= R/M is field.

Conversely, let assume that R/M is field.

To prove M is Maximal ideal.

Let assume that N is any ideal of R such that MR 1

then N/M is ideal in R/M

but R/M is field & hence the only ideals in R/M is zero ideal which M or R/M itself.

/

//If N M M N M

if N M R M N Rº

º

If MR 1 either M = N or N = R

= M is maximal ideal of R/M

Remark :

1) Since every field is integral domain therefore if M is maximal ideal º R/M is

field º R/M is integral domain º M is prime ideal

Hence every maximal ideal is prime ideal. But every prime ideal need not be

maximal ideal which we will see in the examples let see some examples of prime

and maximal ideals.

Examples:

1) Let R = ][x] and I = = set of all polynomial with constant term zero.

Claim : I is prime ideal.

Let

fxh x x¸

() []. . () () .()

(0) (0) 0 0 0there exist g x x s t f x g x x g x

fh g=

= ¸ ]

(0) 0 (0) 0fo r hº (']is integral domain)

() ()fxx o r h xx º

[]xis prime ideal in x = ]

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106 ALGEBRA II

2) Let R = ][x] and M = = set of all polynomial with even constant term.

Note that M = = {x.f(x) + 2g(x) : f(x), g(x) ] [x]}

Claim : M is maximal ideal in R

Clear MRv

Let N be any other ideal of R such that MR 1

we will prove that M = N or N = R

Let assume that MNv

Therefore we prove that N = R

,( ) ( )MNf x N s u c h t h a t f x Mv= '

= constant term of f(x) i.e. f(o) is not even

1

11 () . .nn

nn oLet f x a x a x a x a

where 0ais odd.

1 .. 2 1( )n

no oax a x b a i s o d d ' = g(x) + 1

where 1 () . . 2n

no gx a x ax b M N

() , ()gx N f x N=

1( )( )

() , 1 ()fxg xNfor any k x R k x N

=

= ¸

RN B u t NR

NR

=

=

º M is maximal ideal of R

(i) Note that ,2IxM x = is not maximal ideal.

Hence prime ideal need not be maximal ideal.

(ii) By same argument we can show for any prime is maximal ideal

of ][x]

(iii) Since there are infinitely many prime, hence there are infinitely many

maximal ideal in ][x]

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## Page 107

107Chapter 5: Ideal and Quotient Ring

3) Let p is prime, then p] is prime ideal of ].

As . Let x y p]xy p k f o r s o m e k º¸ ]

/. / /px y px o r py ==12,.xpk or y p k x p or y p º º ]]

. pi s p r i m e i d e a l =]

Also note that ]/p] is ]p which ring of residue modulo p. And as p is prime it

is field. Hence p] is maximal ideal also. Thus in ], prime and maximal ideal are

some.

4) Let {( , 0): } Ra n d I a a ]] ] consider R/I,

Let (, ) (, )ab R t h e nab I (( , 0) (0, )) ( , ) ( , )ab I a I a b I

(0, ) ( ( , ) , ( , ) )bI S i n c e a I a I I '

/{ ( 0 , ): }RI b I b = ]

Let ((x, y) + I) . ((p, q) + I) = I ( I is zero element in I)

(, ) (, ) . .xp yq I I xp yq I y q º º º

0( )

(2, ) ( , )yo r q i s I n t e g r a l d o m a i n

Io r p I

º

º ']

(, ) ( , )xII o r p II º

Hence R/I is Integral domain

= I is prime ideal

' (1, 1) is unity in ]]

and R/I = {(0, b) + I : b ]} does not contain unity hence it is not field.

= I is not maximal ideal

5) Let us find maximal ideal of

( i ) 8] (ii) 10] ( i i i ) 12]

Let us see lattice of ideals for 8]

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## Page 108

108 ALGEBRA II

I) 8] 8]

2

4

0

Clearly 2is maximal ideal of 8].

II) 10]

2 5

0

Clearly 2 and 5 is maximal ideal of 10].

III)

C l e a r l y 2 and 3 is maximal ideal of 12].

Thus in general for any prime divisor p of n,

is maximal ideal of .n]

5.4 Unit and Exercises

1) Let S = {a + bi : a, b ], b is even}. Show that S is subring of <>i] but

not on ideal.

2) Check whether the following set I is ideal of ring R?

){ ( , ) : } ){ ( ,) : }

){ ( 2 , 2 ) : , }iI a a a R i iI a a a R

iii I a b a b R

q q

q]] ] ]] ]

]] ]

(In all three problems R is ring under com ponent wise addition and multiplication.)

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## Page 109

109Chapter 5: Ideal and Quotient Ring

(iv) 10 0 {( ) [ ] : ( ) . . 3 | }n

n If x x f x a x a x a a ] []Rx ]

(v) 10 {( ) [ ] : ( ) . .n

nIfx K x fx ax ax a ]

^10 .. 0naa a and <>Rx]

(vi) :: , ,oa abIa R Ra b doo od£² £ ²¬ ¬¦¦ ¦ ¦¦¦ ¦ ¦ ¤» ¤ »¦¦ ¦ ¦® ®¦¦ ¦ ¦¥¼ ¥ ¼\

(vii) {4 : , }, [ ]Ia b i a b Ri ]]

3) Determine the number of element in quotient ring

(i) 3 ] / 9 ] ( i i ) <>/3ii ]

4) Let 22Ii . Determine R = ][i]/<2+2i>.

What is characteristic of R? Is R is Integral domain? Is I = <2 + 2i> is prime

ideal.

5) Show that the following ideals are prime ideals in the given ring

( i ) 2

2 1[ ]xxi n x ]

(ii) I = {(3x, y) : x, y ]} in ] x ] under component wise addition and

multiplication.

6) Show that <1 – i> is maximal ideal in ][i]. (Show that ][i]/<1-i> is field.

Note that i =1 and 2 = 0 in ][i]/<1-i>)

7) Let R be the ring of continuous functions from to\\. Show that A = {f

R / f(0) = 0} is a maximal ideal of R.

8) Prove that every prime ideal is maximal ideal. munotes.in

## Page 110

110 ALGEBRA II UNIT 3

6

RING HOMOMORPHISM AND

ISOMORPHISM

Unit Structure

6.0 Objective

6.1 Ring homomorphism & isomorphism

6.2 Fundamental theorem of isomorphism

6.3 Ring of fraction, Quotient field

6.4 Summary

6.5 Unit and Exercises

6.0 Objective

After going through this unit you shall come to know about

x The concept of ring homomorphism and isomorphism.

x Fundamental Theorem of isomorphism and its application.

x Method to construct ring of fraction and Quotient field.

One way to study property of a ring is to examine its interaction with other ring by

finding some relation between them. This relation is something which must

preserve the operation of the respective ring. Such relation is called as ring

homomorphism.

6.1 Ring Homomorphism

Let R, S be any two ring. A map f:R is said to be ring homomorphism from R

to S if for any , ab R

ff fab a b and

ff fab a b i.e. f is

operation preserving mapping.

Properties of Ring Homomorphisms :

Let : lRS be a ring homomorphism. Let A be a subring of R and B is an ideal

of S.

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## Page 111

111Chapter 6: Ring Homomorphism and Isomorphism

1. For any rR a n d a n y p o s i t i v e i n t e g e r

, nn rn r a n d

n nrr .

2.

\^: Aaa A is a subring of S.

3. If A is ideal and onto S, then

A is an ideal.

4.

\^1/( ) Br Rr B is an ideal of R.

5. If R is commutative, then

R is commutative.

Proof :

1)

... nr r r r (n times)

() rr n r

2)

... nrr r r (n times)

() ( () )nrr r r

Let ,( )xyA

,ab A= such that

,xayb

xya b a b A a b A = '

A= is subring

3) :RS l is onto. = for any , yS xR such that

xy

To prove:

A is ideal of S. From (2),

A is subring.

Hence it is enough to show that for any

,,xAySxyA ¸ .

xAa A = ' such that

,, axyS is onto, bR such

that

by .

Hence

.xyab a b A (Since in ideal A ideal,

therefore ab A)

A= is ideal of S.

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112 ALGEBRA II

4) \^1() / ( )Br Rr S

Let

1,( ) ,xyB x y B = as B is ideal

xyB =

1()xyBxyB=

=

Hence 1()B is subring.

Let rR be arbitrary.

rS =

As B is ideal of S

therefore xrB x rB =

1, th e r e f o r e ( )xrB x r B=

Similarly we can show that 1.( )rx B

Hence 1()B is ideal of R.

V) Let R is commutative ring.

= for any ,, ..xyR x y y x

..

.xy yxxyy x=

=

R º is commutative ring.

Kernel of Ring homomorphism: Let :RS l be a ring homomorphism then

Kernel of , denoted as ker and is defined as

\^/0 Ker r R r .

Note that ker is ideal of R.

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113Chapter 6: Ring Homomorphism and Isomorphism

Example:

1) Let :,abRa bba£² ¯¦¦¦¦¡° ¤»¡°¦¦¢±¦¦¥¼]. Let :R l] b e d e f i n e d b y

ababba¬ ¯¡° ¡°®¢± then is ring homomorphism.

As let ab x yABba yx ¯ ¯¡° ¡°
¡° ¡°¢± ¢±

then

ax byABa x b yby ax¬ ¯ ¡° ¡° ®¢±

() ()ab xy

AB

.ab xy a xb ya y b xABba yx b xa ya xb y ¯ ¯ ¯ ¡° ¡° ¡ ° ¡° ¡° ¡ ° ¢± ¢± ¢ ±

.( ) ( )ABa x byaybx = ……………… (1)

Also

ab x yABba yx¬ ¬ ¯ ¯¡°¡° ¡°¡°® ®¢±¢±

() ab xy a xa yb xb y

() ()ax by ay bx ……… (2)

From (1) and (2)

.ABA B

Hence is homomorphism.

Let us find Ker ,

Ker 0ab abRba ba£² ¬ ¯ ¯¦¦ ¦¦ ¡° ¡° ¤» ¡° ¡° ¦¦ ® ¢± ¢±¦¦¥¼

/0 /ab abRa b Ra bba ba£² £ ² ¯ ¯¦¦ ¦ ¦¦¦ ¦ ¦¡° ¡° ¤» ¤ »¡° ¡°¦¦ ¦ ¦¢± ¢±¦¦ ¦ ¦¥¼ ¥ ¼

:aaaaa£² ¯¦¦¦¦¡° ¤»¡°¦¦¢±¦¦¥¼\ 11 11/11 11aa R£² £ ² ¯ ¯¦¦ ¦ ¦¦¦ ¦ ¦¡° ¡° ¤» ¤ »¡° ¡°¦¦ ¦ ¦¢± ¢±¦¦ ¦ ¦¥¼ ¥ ¼ munotes.in

## Page 114

114 ALGEBRA II

2) Define

:m o dn kk n l
]] . Then i s a n h o m o m o r p h i s m

known as natural homomorphism.

3) Let R be commutative ring of prime characteristic P.

Define :RR l as

pxx is ring homomorphism

as

12 112 2 1() () . . .

ppppp pp p p p

CC Cxyx y x x y x y x y y

=

Each of

111p

C ipb b is multiple of p and char R = p 10ppi iCx y= for all

i, 11ipb b

..pp

p ppxyx y x yxyx y x y x y=

Therefore is ring homomorphism, known as Frobenius homomorphism.

4) Let R be a ring and A be its any ideal.

Define a map :/fRR A l a s ()frr A , then f is ring homomorphism

known as a natural homomorphism.

Ring Isomorphism: A ring homomorphism :RS l which is one-one and onto

is known as ring isomorphism.

If :RS l is ring isomorphism we say that ring R and S is isomorphic to each

other and this is denoted by RSx.

Isomorphic rings are exactly similar in terms of property of element and their

behavior. In Other words isomorphic rings are nothing but different way of looking

of same ring.

Theorem 1. First Isomorphism Theorem of Ring:

Statement: Let :RS l be a onto ring homomorphism then /k e r ( )RR x .

Proof : Let Ker = W.

To prove

/RW R x

C l e a r l y WK e r is ideal of R.

Define a map

:/fRW R l as () ( ) ,frW r rR . munotes.in

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115Chapter 6: Ring Homomorphism and Isomorphism

Claim :

1) is well defined one - one map.

Let assume that

s, , rW W r sR rs W K e r

0 rs ( B y

Definition of Ker )

0 rs (By definition of homomorphism)

() rs f r w f r W

= f is well defined one-one map.

2) f is onto.

Let () yR

Therefore there exist xR such that

xr .

Therefore there exist /xWR W such that

fxW xy .

Therefore f is onto.

3) f is homomorphism.

Consider

()fxWyWfxyW

xy (By definition of f)

xy (is homomorphism)

() ()fxW f yW

()fxWyWfxyW

xy

xy

() . ()fxW f yW

'f is homomorphism

Hence f is isomorphism.

/RWR = x

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116 ALGEBRA II

Theorem 2. Second Isomorphism Theorem for Rings :

Statement: L e t A b e a s u b r i n g a n d l e t B b e a n i d e a l o f R. Then

\^:,ABa b a A b B i s a s u b r i n g o f ,RA B i s a n i d e a l o f A a n d

() / /ABBA AB x

Proof : Let we give sketch of the proof.

To prove this we use first isomorphism theorem of rings.

Let xAB

,,

()xab aA b BxBabBa bB aB b B=

= '

Hence define a map

:/AABB l such that ()aa B .

You prove this map is onto homomorphism.

Then apply first isomorphism theorem, according to which

/k e r /AABB x .

Now prove that ker AB .

Theorem 3. Third Isomorphism Theorem of Ring:

Statement : Let I and J be ideals of R with IJ. Then /JI is an ideal of /RI

and (/ ) / (/ ) /RI JI RJ x.

Proof : This proof is also similar to proof of second isomorphism theorem.

Here define :/ /RI RJ l as ()rI rJ .

Show that is onto homomorphism.

Then by first isomorphism theorem //RI K e r RJ x .

Show that / Ker J I .

The first isomorphism theorem has lot of application. One of the application is to

prove that a field contains p] or _. We prove this by following steps.

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## Page 117

117Chapter 6: Ring Homomorphism and Isomorphism

Theorem 4 : Let R b e a r i n g w i t h u n i t y e . T h e m a p p i n g : R l] g i v e n b y

. nn e is a ring homomorphism.

Proof : Let ,, , 0mn mn ]

() . . . ( ) m n m n e e e e m n times

.. ..

mt i m e s nt i m e see e e e

me ne m n

Let assume ,0mn

Then Let , mp nq where ,0pq.

() () ( ) () ( )

() ( ) ( )

() ( ) ( ) ( )mn mn e pq e pq e

peq e

pe qe m e n e m n=

Similarly if 0mp 0n we can prove () ( ) ( )mn m n .

Also () () () ( . ) () ( )mn mn e mn e e me ne

.mn

= i s r i n g h o m o m o r p h i s m .

Corollary 1: Let R is the ring with unity and the characteristic of R is 0n, then

R c o n t a i n s a s u b r i n g i s o m o r p h i c t o n]. If the characteristic of R i s 0 , t h e n R

contains a subring isomorphic to ].

Proof: Let \^: Sk e k ]

Claim : S is subring of R.

Let ,ab S

,nm= ] such that an e and bm e

() ab n em e nm es=

.( ) ( )( ) ( ) .ab n e m e n e m e n m e e n m e S

Hence S is subring of R.

Define : S l] such that

nn e .

Then is onto and above theorem is homomorphism.

Hence by first isomorphism theorem /.Ker Sx]

If Char R = n munotes.in

## Page 118

118 ALGEBRA II

Then Ker

\^/0 mm ] \^/0 mm e ]

\^

:/ mn m c h a r R n ]' n

/

nnS

S= x

= x]

]

Hence R contains subring isomorphic ton].

If char R = 0 ºThere exist no integer n such ne=0 .

\^/0 Ker m m ] \^ \ ^/0 0 ( S i n c e 0 ) mm e c h a r R ]

/k e r S

S= x

= x]

]

Hence R contains a subring isomorphic to ].

Corollary 2: A field contains p]or_.

Proof: Characteristics of a field is either zero or prime. Let R be a field.

Case I: Char R = p, p is prime.

Then by previous corollary R contains a subring isomorphic to p], p] is field,

as p is prime

Hence R containsp].

Case II: Char R = 0

Then subring S of R is isomorphic to ].

Let \^1.: , , 0 Ta b a b S b v

The T is subfield of R and isomorphic to _.

Hence a field contains p] or _.

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## Page 119

119Chapter 6: Ring Homomorphism and Isomorphism

6.3 Ring Of Fractions

The aim of this section is to prove that a commutative ring R is always a subring of

a larger ring _ in which every non zero zero divisor of R is unit in _. In case of

integral domain such ring is field and called as field of fraction or Quotient field.

Construction of quotient field _ from ring R is exactly as construction of _ from

].

Theorem 6 : Let D be an integral domain. Then there exists a field F (called the

field of fraction of D) that contains a subring isomorphic to D.

Proof: Let S be the set of all formal symbols of the form a/b, where ,ab D and

0bv. Define an equivalence relation w on S by //ab cd wifad bc. Let F

be the set of equivalence classes of S u n d e r t h e r e l a t i o n w a n d d e n o t e t h e

equivalence class that contains /xy by<>/xy. We define addition and

multiplication on F b y <> <>

// /ab cd a d b c b d ¯ ¢± a n d <><><>/./ /ab cd a cb d .

We will prove F is field with respect to the operation and ¸ define above.

To prove this first of all the operation and ¸ is well defined.

Let assume that 1

1aa

bb and1

1cc

d d.

To prove 11

11aca c

bd bd

1

11

1

1

11

1aaab a bbb

cccd c ddd º

º '

' To prove 11

11aca c

bd bd i.e. to prove 11 11

11ad bc a d c d

bd bd

i.e. to prove that

11 11 11() ad bc b d a d c d bd .

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## Page 120

120 ALGEBRA II

i.e. to prove that 11 11 11 11adb d bcb d a d bd c d bd

In LHS put 11ab a b and 11cd c d we get RHS.

Hence + is well defined.

Similar To prove 11

11..ac a c

bd bd w h i c h i s o b v i o u s a s 11

11 and aa c c

bd bd

11

11ac a c

bd bdº

It is trivial to prove that (F, +) is abelian group with zero element as 01ªº¬¼ and

additive inverse of abªº¬¼ as abªº¬¼.

Multiplication is obviously distributive over addition and associative.

The unity element of F is 11ªº¬¼and multiplicative inverse of non zero element

abªº¬¼ is baªº¬¼ for , 0abv.

Hence F is field. Finally let we define a map :DFo as 1xx ªº ¬¼then

is ring isomorphism from D to Das,

>@>@>@/1 /1 /1 /1 /1xyx y x y x y ªº¬¼ xy

>@>@>@ /1 /1 /1xyx y x y x y

^` : 0Ker x D x

>@^`^`^`:/ 10 : 0 0 xDx xD x

?By first isomorphism theorem,

/

()DK e r D

DD|

?|

Hence F contains a subfield isomorphic to D.

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## Page 121

121Chapter 6: Ring Homomorphism and Isomorphism

Examples :

1) Show that the ring 2ªº¬¼] a n d H a r e i s o m o r p h i c w h e r e

2:,abHa bba½ªº ®¾«»¬¼¯¿] under addition and multiplication of 22u matrixes.

Solution : :2Hªºo¬¼] as 22ababbaªº «»¬¼

Claim :

1) is well defined, one-one map

L e t 22ab x y

ba yxªº ªº «» «»¬¼ ¬¼

& axby ?

22 ab xy

22 2 2 ab xy ab xy ?

? is well defined one-one map.

2) is onto

For any 2,2 2xyHx yyx ¯ ¯ ¡° ¡°¢± ¡°¢±] such that

22xyxyyx ¯¡° ¡°¢±. Hence is onto.

3) is homomorphism.

22 ( ) 2 ab xy ax byªº ¬¼

22 () 2

2

22ab x y ax by

ba yx by a

ab xyªº ªº ªº «» «» «» ¬¼ ¬¼ ¬¼

22 2 ( ) 2 ab xy a x b y a yb xªº ¬¼ munotes.in

## Page 122

122 ALGEBRA II

222

2ax by ay bx

ay bx ax byªº «»¬¼

22 2 2 2222ab x y a xb ya yb xab xyba yx a y b x a x b y ªº ªº ª º «» «» « » ¬¼ ¬¼ ¬ ¼

= 22 ab xy

Hence is homomorphism.

? is isomorphism.

2 H ªº?|¬¼] .

2) Let :,abRa bba½ªº ®¾«»¬¼¯¿]. Let :Ro] defined ababba§·ªº ¨¸«»¬¼©¹. Show

that is a ring homomorphism. Determine Ker . Is Ker a prime ideal? Is it a

maximal ideal? Justify.

Solution: Let abAbaªº «»¬¼ and xyBRyxªº «»¬¼

ab x y a xb yABa x b yba yx b ya x

ab xy A B¬ ¯ ¯ ¯ ¡° ¡° ¡ ° ¡° ¡° ¡ ° ® ¢± ¢± ¢ ±

ab x y a xb ya y b xABba yx a yb xa xb y¬ ¬ ¯ ¯ ¯ ¡° ¡° ¡ ° ¡° ¡° ¡ ° ® ®¢± ¢± ¢ ±

. ax byaybx a xybyxa b xyAB

=is homomorphism

Ker\^:( )0 AR A

\^ :0 : 0abAR AR abba£² ¬ ¯ ¦¦ ¦¦ ¡° ¤» ¡° ¦¦ ®¢± ¦¦¥¼ .aaaaa£² ¯¦¦¦¦¡° ¤»¡°¦¦¢±¦¦¥¼]

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## Page 123

123Chapter 6: Ring Homomorphism and Isomorphism

= First Isomorphism theorem

/RK e r x] as ] is Integral domain, /RK e r is integral domain.

=Ker is prime ideal.

Also as ] is not field. =/RK e r is not field.

=Ker is not maximal ideal.

3)Prove that

2M \ contains a subring that is isomorphic to ^.

Solution :

2 :,,,abMa b c dcd£² ¯¦¦¦¦¡° ¤»¡°¦¦¢±¦¦¥¼\\ .

Define a map

2 : M l ^\ as

abai bba ¯¡° ¡°¢±.

Claim : is homomorphism.

() ai bci d aci bd () ac bd

bd ac ¯ ¡°¡°¢±

()abcdai b ci dba d c ¯ ¯¡° ¡° ¡° ¡°¢± ¢±

ab cd a c b d a d b cai b ci dba d c a d b ca c b d

ai b ci d ¯ ¯ ¯ ¡° ¡° ¡ ° ¡° ¡° ¡ ° ¢± ¢± ¢ ±

Hence is homomorphism.

Ker

\^ :0 ai b ai b

\^ \ ^00:: 0 , 0 000abai b ai b a bba£² ¯ ¯ ¦¦¦¦ ¡° ¡ ° ¤»¡° ¡ ° ¦¦¢± ¢ ± ¦¦¥¼

= By First isomorphism theorem

2 .Ker MGx ^ \i.e.

2M x^\ .

Hence

2M \ contains a subring that is isomorphic to .

4) Is 2] is isomorphic to 3] munotes.in

## Page 124

124 ALGEBRA II

Solution : No. Let :2 3 l]] is isomorphism &

2 a .

42 222 2 2 2 a = ….. (I)

Also

2 242 . 22 . 2 2 a …………. (II)

Therefore From (I) & (II).

22 aa

220 2 0 aa a aº º

0aº or 2a if 0a then

2. 2 0nn .

Hence is zero map which is not isomorphism.

0 =v 2a= but then 3a].

Hence 2] is not isomorphic to 3].

5) Is 2 ¯¡°¢±] isomorphic to 5 ¯¡°¢±] .

Solution : N o , 2 ¯¡°¢±] i s n o t i s o m o r p h i c t o 5 ¯¡°¢±] . As if

:2 5 ¯ ¯ l ¡°¡°¢±¢±]] is isomorphism

1 a .

() (. 1 ) () . ( 1 ) . ()xxx a x =G G G G G

() . ()xax ºG G

º Hence a is multiplicative identity in 5. ¯¡°¢±]

1a=

,( )for any a a a= G ]

Let assume

25 G B C

2 222 : 22 5 x = G G C

2225 2 5x = C BC

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## Page 125

125Chapter 6: Ring Homomorphism and Isomorphism

Comparing we get BC = o

ºB = o or C = o

if B = o 225 º C 22/5 ºC

But no such C in ] exists

if C = o then 22 B

again so such Bexists.

Hence no such G exists

2] =G ]> is not isomorphic to 5. ¯¡°¢±]

6.4 Summary

1) First Isomorphism Theorem of Ring: Let :RS l be a onto ring

homomorphism then /k e r ( )RR x .

2) Second Isomorphism Theorem for Rings: Let A be a subring and let B be an

ideal of R. Then \^:,ABa b a A b B is a subring of ,RA B is

an ideal of A and () / /ABBA AB x

3) Third Isomorphism Theorem of Ring: Statement : Let I and J be ideals of R

with IJ. Then /JI is an ideal of /RI and (/ ) / (/ ) /RI JI RJ x.

4) Let D be an integral domain. Then there exists a field F (called the field of

fraction of D) that contains a subring isomorphic to D.

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126 ALGEBRA II

6.5 Unit and Exercises

Exercises :

1) Let :,abRa bba£² ¯¦¦¦¦¡° ¤»¡°¦¦¢±¦¦¥¼\ under addition and multiplication of 2 x 2

matries. Prove that R is isomorphic to .

2) Show that the map 2 :[ ] ()xM Gl\\ defined as

1(. . )o n

on

oaaaa x a xoa¬ G ® is ring homomorphism. Find ker G.

3) Let :,abRa b dod£² ¯¦¦¦¦¡°¤»¡°¦¦¢±¦¦¥¼] . Show that :RGl q]] defined by

(, )abadod¬ ¯¡°G¡°®¢± is a ring homomorphism. Find Ker G.

4) Let Gbe a ring homomorphism from a commutative ring R onto a commutative

ring S and let A be un ideal of S.

(i) If A is prime in S, show that 1() { : ( ) }AxR x AG G is pprime in R.

(ii) If A is maximal in S, show that 1()AG is maximal in R.

5) Let <>i] = {a + bi : a, b ]}. Show that the field of quotients of <>i] is ring

– isomorphic to <>\^:/ ,irs i r S __ (Hint: Let F is field containing ] and i,

then for any 1 ,, , aa Fa v ] Hence F contain _ and i. Also prove that for

any ,0 [ ] , [ ]ab iab i cd i i icd i v ]_ .

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## Page 127

127Chapter 7: Euclidean Domain, Principal Ideal

Domain Unique Factorization UNIT 4

7

EUCLIDEAN DOMAIN, PRINCIPAL IDEAL

DOMAIN UNIQUE FACTORIZATION DOMAIN

Unit Structure

7.0 Objective

7.1 Introduction

7.2 Prime and irreducible element

7.3 Euclidean domain (ED)

7.4 Principal ideal domain (PID)

7.5 Unique factorization domain (UFD)

7.6 Summary

7.7 Unit and Exercises

7.0 Objective The Objective this chapter is to make you understand

x Prime and irreducible elements and difference between them.

E u c l i d e a n d o m a i n (ED)

x Principal ideal domain (PID)

x Unique factorization domain (UFD)

x Difference between E D, PID, UFD

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128 ALGEBRA II

7.1 Introduction

Primes plays central role in theory of integers. Lots of famous theorems are there

for prime. For example, there are infinite numbers of primes; there are infinite

numbers of prime of the type 4n-1 (or 4n+1). Euclid lemma which states that p is

the prime, plab then either pla or pla, etc.

There some conjectures which are simple to state but yet not proved. One of them

is twin prime theorem. Prime like 3 and 5, 11 and 13, 17 and 19 are called twin

prime. Twin prime theorem states that “there are infinitely many twin primes”. This

conjecture yet to be proved.

We want to introduce same notion of prime to general ring. That is we want find

element in general ring which has property similar to prime of integers.

Note that in case of integer we define prime as positive integers which are divisible

by 1(unity) and itself but in case of general ring this definition may not be

appropriate. As if u is unit in ring R and a

a|b then ua|b. Hence in a ring if there more than one unit then definitely there more

than two divisors.

Hence we need different approach to define prime. One such approach given by

Euclid lemma.

7.2 Prime and Irreducible Element

Definition: Let R be an integral domain. A non zero non unit element p of R is said

to be prime element in R if whenever pa b pao rpb ¸º .

Definition: A non zero non unit element P o f i n t e g r a l d o m a i n R i s s a i d t o b e

irreducible element if whenever p=ab then either a is unit or b is unit.

Two elements a and b of integral domain R is said to be associates of each other if

they differ by unit (i.e., a=ub for some unit u). The other way to say same thing is

a/b and b/a.

Theorem 1: In an integral domain prime element are irreducible.

Proof: Let R be an integral domain and let pR be prime element in R. To Prove

P is irreducible.

Let p = a b

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## Page 129

129Chapter 7: Euclidean Domain, Principal Ideal

Domain Unique Factorization

Hence To prove either a or b is unit.

Since . pp pa b º pao rpbº (Since p is prime element)

If pa then a = kp for some kR

pa b= 1 pk p b k b º º bº is unit in R

Similarly if pb then we can prove a is unit in R. Hence whenever p = ab º

either a or b is unit.

Therefore p is irreducible.

Remark:

1. In case of ], the prime and irreducible are same.

2. The above theorem say that prime and irreducible are same in an integral

domain. But converse need not be true.

That is there are integral domain in which irreducible need not be prime.

Let see one such example. Consider the ring d ¯¡°¢±] where d is square free integer.

(That is d is not divisible by square of any number).

We define a function :Nd ¯l¡°¢±]] define as

22. Na b d a b d this

function is called as norm function.

This function has some trivial property which is easy to prove (of course involve

some calculation)

1.

. NxyNx Ny for all ,xyd ¯¡°¢±] .

2. If u is unit in d ¯¡°¢±]if and only if N(u) = 1

(This is clear because the only unit in d ¯¡°¢±] are 1+ and 1+ , + i in case d = i )

3. If ,xyd ¯¡°¢±] such that x|y then

.NxNy

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130 ALGEBRA II

4. If pd ¯¡°¢±] is prime then N(p) is prime number (This can be proved from

III property and Euclid Lemma)

Now consider particular example 3 ¯¡°¢±]

Consider 13 3 ¯ ¡°¢±]

Claim : 13 is irreducible element

Let 13xy for some ,3xy ¯¡°¢±]

13NN x y=

11 3 NxNy=

Nx . Ny = 4 =

1 4 or 4 1, or 2 2 Nx = , Ny= N x = , Ny=N x = , Ny= ง

But in this case either x or y is unit in 3 ¯¡°¢±]

Hence 13become irreducible.

The other possible case is

2, 2 Nx Ny

Let 33 xab ¯ ¡°¢±] then

2232 Nx a b but this is not possible

for any value of ,.ab] Hence this case is not possible.

Thus we have shown that 1+ -3 is irreducible.

Also

1+ -3 1- -3 =4=2.2

Thus

1+ -3 2.2

but 1+ -3 2 as if 1+ -3 2 then

2 = 1+ -3 a+b -3

2 = a - 3 b + a + b - 3º a-3b = 2 º and abo

But the above equations have no integer solution 1 + -3 † 2

Hence 1+ -3 2 2 ¸but

1+ -3 † 2

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131Chapter 7: Euclidean Domain, Principal Ideal

Domain Unique Factorization

3. If p is a prime element in an integral domain R pj is prime ideal.

as let ab p¸ ab k p p ab º¸ º ¸ pao rpbº [Since p is prime element]

or ap bpº

ab p a p =¸ º or bp

p= is prime ideal.

Conversely, let p is prime ideal.

Let pa b a b p ¸º¸ apº or bp (Since p is prime ideal)

pao rpbº pºis prime element.

There are different classes of ring. Let we study them. In first year we study division

algorithm in ] and in []x\. Actually they are Euclidean algorithm. The integral

domain having Euclidean algorithm is called Euclidean domain (ED) . Similarly in

F.Y.B.S.c we study the fundamental theorem of Arithmetic (i.e. Unique

factorization theorem) applicable in ] and [] .x\

The integral domain having this property is called unique factorization domain

(UFD). The class of integral domain in which every ideal is principal ideal is called

principal ideal domain (PID). We will study these class of ring one by one.

7.3 Euclidean Domain (ED)

An integral domain R is said to be Euclidean domain (ED). If there exist a function \^:0dR l] such that

(i)

dab da ¸ for all ,ab R.

(ii) For any ,,ab b o v\ there exist ,prR such that a = bp+r with r=0 or

< . dr db

Note: Such function d is called Euclidean function.

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132 ALGEBRA II

Example:

1. ] is an Euclidean domain.

I n ], the Mod Function | | is an Euclidean Function and division algorithm

is Euclidean algorithm.

2. <>x\, the ring of polynomial over \ is Euclidean domain.

I n [] ,x\the degree function, degree of polynomial is Euclidean Function and

division algorithm of polynomials are Euclidean algorithm.

3. Any Field is by default Euclidean domain. In a field multiplication is Euclidean

Function. If ,,ab Fb ov then

1ab b a this is nothing but Euclidean

algorithm.

4. Consider the ring of Gaussian integers<>\^:, ia b i a b ]] .

Then ][i] is Euclidean domain.

Define the function <>\ ^:Ni U o l]` as

22.Nai b a b =

Then for any <>,xyi]

.NxyNxNy

To See Euclidean algorithm.

Let xai b and <>yci d i ] s u c h t h a t . ci d v C o n s i d e r t h e

Quotient .xy

Let sxity where ,.st_

Let m, n be the integers closest to s and t respectively such that 1s2mb and

12nt b

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133Chapter 7: Euclidean Domain, Principal Ideal

Domain Unique Factorization

Then s+itx

y

mms i nnt i

mi n sm i tn ¯ ¢±

Then

xmi n sm i tn y ¯ ¢±

We claim thath the division algorithm of the definition of a Euclidean domain is

satisfied si satisfied with q = m + ni and

rm i n s mi t n ¯ ¢±

Clearly, q belongs to Z[i], and since r = x - qy , so does r. Finally

() Nr N s m it q ni Ny ¸

22smt n N y

11

44Ny Ny¬ ®

H e n c e f o r a n y <><> ,, 0 q , rxyiyi v ]] S u c h t h a t qrxy w i t h

rNNy.

Hence <>i] has Euclidean algorithm. <>i=] is Euclidean domain.

These are some example of Euclidean domain. Let us understand the importance

of being Euclidean domain.

Theorem 2: In an Euclidean domain every ideal is principal ideal (i.e generated by

single element).

Proof: Let R be an Euclidean domain and let I be any non zero ideal of R.

Let d be an Euclidean function of R choose aIsuch that d(a) is minimum.

Claim : Ia.

Let bI be arbitrary. Then by Euclidean algorithm there exist ,prR. Such that

b = ap + r w i t h r = 0 o r

. dr da I f r v t h e n rba p I

since is ideal I but then d(r) < d(a) will contradiction as a is the element of I

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134 ALGEBRA II

rb a q == baº

but bI be arbitrary.

Hence Ia

IaIis principal ideal

=

=

The Integral domain in which every ideal is principal is known as Principal ideal

domain (PID). Thus above theorem says that every Euclidean domain is principal

ideal domain.

Now we see this class of ring in details.

7.4 Principal Ideal Domain

An Integral domain R is said to be principal ideal domain (PID) if every ideal of R

is Principal ideal.

Example:

1. Every field is by default Principal ideal domain.

S i n c e o n l y i d e a l i n f i e l d F is zero ideal and F itself. (Since fields are simple

ring). The zero ideal is generated by zero elements and F is generated by unity.

2. ] is PID.Since any ideal of ]are of the form ,.mm ]]

3. <>x\is PID.One can argue like, as <>x\ i s E u c l i d e a n d o m a i n , e v e r y

Euclidean domain is PID, hence <>x\ is PID. But we can prove directly that <>x\ i s P I D . P r o o f i s e x a c t l y s i m i l a r t o t h e w a y w e p r o v e “ I n E u c l i d e a n

domain every ideal is principal”. We request the students to understand the

similarity between two proofs.

Proof: Let I be any nonzero ideal of <>x\ Since Let

fxI such that f(x) is

monic and deg f(x) is minimum.

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## Page 135

135Chapter 7: Euclidean Domain, Principal Ideal

Domain Unique Factorization

Claim:-

If x

Let

gxI be arbitrary.

Let <>x\is Euclidean domain, then by division algorithm, there exist

pxand

<> rx x\ s u c h t h a t

gxp xfxr x w i t h

rx o or deg

rx d egfx .

If

rx O vthen

rxgxp xfxI (Since I is ideal)

But this is contradiction to minimality of deg

fxas

,fxIfx is monic &

deg

fx is minimum.

Hence

rx

gxp xfx =

gx f x I f x º º

If x=

Hence I is principal ideal.

= Every ideal of <>x\ is principal Hence <>x\ is PID.

4. <>x]is not PID.

Consider the ideal

\^:0 If x f LVHYHQ i.e. I is ideal of <>x] with even

constant term.

\^ 12 .. : ,n

no o Ia x a x a a a ] LVHYHQ

\^1

1 .. 2 : 2 , 2n

no o oxax a b a b x

Let assume that

,2xfx for some

<> .fxx]

<>xf xxfx g x g x x

=

= ] IRUVRPH

= 1 = deg x = deg

fxgx d e g f x d e g gx

ºdeg

1 fx d e g g x but as degree is non negative number,

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136 ALGEBRA II

Hence either

1, deg f x deg g x

or

,1 deg f x deg g x .

If deg f(x) = 1, then f(x) = ax + b , for some ,ab] b u t t h e n

22fxa x b º is not possible. Hence

. degfx1v

deg f x =

fx ºis constant polynomial.

Let

2fxhx for some

hx x ] t h e n

2f1h1

f1 2 1 ,º ' b u t

fx2 º ( A s f(x) i s c o n s t a n t p o l y n o m i a l b u t

then

x2gx which is nonsense.

Hence

,2xfx is not possible.

= I is not principal ideal.

A ring being Principal ideal domain has lots of advantage. We see them one by one.

We already seen “In an integral domain primes are irreducible”. But we also seen

the example of integral domain where irreducible are not prime. The one advantage

of being PID is that irreducible are also prime.

Theorem 3: I n P r i n c i p a l i d e a l d o m a i n i r r e d u c i b l e a r e p r i m e ,

Proof: Let R be principal ideal domain, and let r be irreducible element of R.

To prove r is prime.

Let ,, .rb c cb R

Clearly r is non zero, non unit element.

To prove rb or rc consider the ideal \^ /,Ixr byxyR Since R is PID Id = for some dR

Since rI d

ra d = for some a in R. (as r is irreducible )

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## Page 137

137Chapter 7: Euclidean Domain, Principal Ideal

Domain Unique Factorization

If d is unit then I = R and as 1,RR= B C such that 1 rbB C

, cc rb c r c r r b c r b c = B C = B ºC

rc r b c r c=B Cº

Now if a is unit then rd! ! and as . bd br r bº º

=r is prime

1. In short we say that in PID, prime and irreducible are same.

2. We had seen that in 3 ¯¡°¢±], irreducible are not prime from this and above

theorem we conclude that 3 ¯¡°¢±] is not PID.

3. One more advantage of being PID is existence of GCD . Before we see let recall

the definition.

Definition:

(i) Let R be a ring. We say that element a of R divides bRor b is divisible by

a if there exist cR such that b = ac and this we denote by ab.

(ii) Let R b e a r i n g ,.ab R We say that dR i s g r e a t e s t c o m m o n d i v i s i o r

(GCD) of a and b if (i) da and db (ii) If dR such that 'da and 'db

then 'dd . It is denoted by

,.ab

In short the greatest common divisor of a, b at is the largest among all common

divisor.

In school, even if in F.Y.B.Sc. we seen how to find GCD of two positive integer.

The Euclidean algorithm is powerful technique to find GCD. We also seen how to

GCD of two polynomial with the help of division algorithm.

Do you think GCD of any two number exists in all integral domain. To surprise

you the answer is no.

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138 ALGEBRA II

Let see this example.

Let a4 &

b=2 1+ - 3 in -3 ¯¡°¢±]

as

a = 2 2 = 1+ -3 1- -3

b = 2(1 3) then 2a n d 2aband

1+ -3 a and

1+ -3 b

=Both 2 a n d

1+ -3 a r e c o m m o n d i v i s o r o f a & b. But 2† 3 ) and

†2 1+ -3 as both are irreducible.

Like this you can see so many example of non existence of GCD, Note that -3 ¯¡°¢±]

is not PID.

So we hope you can guess the second advantage being yes. PID guarantees the

existing of GCD.

Theorem 4: Let R Principal ideal domain. Then for any ,ab R , greatest common

divisor of a and b exists.

Proof: R is PID and ,.ab R Consider the ideal \^:,Iax byxyR As R

is PID,

= there exist d R such that I=< d >.

Claim: d is gcd of ,.ab

(1) ,ba I d '

11 1 1&, , & aa dbb d a bR d ad b= º

dd I'

,mn R= such that d=am + bn .

Now let assume that 1dR such that 1da and 1.db

1 R ºB C such that 11. ad a n d bdB C

then

1 11 1 dda m b n m d n d m n d d B C B C º

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139Chapter 7: Euclidean Domain, Principal Ideal

Domain Unique Factorization

Hence d is GCD of a and b.

1. Note that the GCD d of a and b is generator of ideal generated by a and b. This

may the reason why GCD is denoted by (,)ab .

2. Every Euclidean domain is PID. Hence in Euclidean domain also GCD of any

two elements exists. In fact Euclidean algorithm is best way to calculate the

GCD.

3. We have shown that Euclidean domain º PID but PID does not mean

Euclidean domain. For example it shown that the ring

,11 9|,2Ra b a b £²¦¦ ¦¦ R R ¤»¦¦¦¦¥¼] is PID but not Euclidean domain,

but detail is beyond the scope of syllabus. So we skip that.

Now let we move to next advantage of being PID. That is PID every ascending

chain of ideal is finite.

Theorem 5: Let R b e P I D a n d 12 3 n II I , "" b e s t r i c t l y i n c r e a s i n g

chain of ideals. Then this chain must be at finite length.

Proof: Let nnIU I

Claim: I is ideal of R.

12 3

,n

nnII I

Let a b I U I ,

""

,pq = such that 1 ,,ppq aIb Ib I

if let pq p q v then pqII

,qq q ab I a b I I º º 'LVLGHDO nnabIU Iº

Let r R then clearly ar and ra pI

.nnar and ra I U I =

Hence I is ideal.

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140 ALGEBRA II

But as R is PID, therefore there exist d R such that I = d

Since nndd I U I

=there exist k such that kdI

but then I = k dI nknUI Iº

Hence 12 3 n II I , ""

= chain of ideal is finite.

This is important theorem which helps us (in future) to prove that PIDs h a v e

property called unique factorization as a product of irreducible. The class of ring

having this property is called as unique factorization domain (UFD). Let we study

this in details.

7.5 Unique Factorization Domain (UFD)

An integral domain R is said to be unique factorization domain (UFD) if

(i) Every non zero non unit element R can be expressed as a product of irreducible

of R.

(ii) The factorization into irreducible is unique up to associates and the order in

which the factors appear.

Now let we prove U. PID FDº

Theorem 5: Every principal ideal domain is Unique factorization domain.

Proof: Let R b e a p r i n c i p a l i d e a l d o m a i n a n d l e t oa b e a n y n o n z e r o n o n u n i t

element of R.

We will prove that oa can be expressed as product of irreducible

Claim: oa has at least one irreducible factor.

If oa is itself irreducible, then we are done. So assume 11 oaa b where neither of

11ab are unit and 1ais non zero.

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141Chapter 7: Euclidean Domain, Principal Ideal

Domain Unique Factorization

If 1a is irreducible then 1a become irreducible factor of oa.

So assume 12 2aa b where neither of 2aand 2b are unit. It 2a is irreducible

then we done otherwise we continue some step so that sequence of element nnab

of R such that 1 .nn naa b

Thus we have 012 .... aaa

vvv

Thus we have strictly increasing sequence of ideal in R. But R PID. There for this

Chain must be finite.

That is there exist raR such that 12 ...r aa a

In particular rais irreducible factor of .oa

Thus every non zero non unit element of R has irreducible factor.

Let 01 1ap q where 1p is irreducible if 1qis also irreducible we get 0aas product

of irreducible.

Let assume 1q i s n o t i r r e d u c i b l e . c l e a r l y 1q is not unit otherwise 0abecome

irreducible which is not the case.

Hence let 12 2qp q where 2pwhere 2qis irreducible & 2q is not unit. If 2q is

also irreducible then 01 11 2 2ap qp p q , product of irreducible and we done. If

not we continue same process. Thus again getting ascending chain of ideal

01 ... aq c . Being PID this chain must be finite. Hence mqR such that

01 ... ,mm ac q CC q q being irreducible as product of irreducible.

Thus every non zero non unit element of R can be written as product uniqueness.

Let 12...k ap p p 12...r qq q where ipand jqare irreducible in R for all i and

j.

12 1 2

112

112... ...

...

...kr

k

rpp p q q p

pp p p

pq q q=

= '

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142 ALGEBRA II

And as irreducible are prime in PID.

1jpq= for some j. Without loss of generality let 11pq

11 1qp a= but both 1pand 1q are irreducible 1aºis a unit.

1p=is associate of 1q.

12 1 2 2 1 2 2 1 2 ... ... ... ... ... ...kr k r k rpp p q q q p paqq p paqq= º º

Similarly canceling 23,. . .ppand if Krwe get

12 1 1. . . . . .kk r aa a q q 1º as product of irreducible.

This is contradiction.

Hence rk and 'ips are associate of 'jqs . Hence proved.

Example :

1) A field <>,,Fx]\ as being PID, there are UFD.

One can prove directly. In case of field, there is no non zero non unit element.

Hence by default field are UFD.

I n c a s e o f <>

and x ]\ w e c a n p r o v e u n i q u e f a c t o r i z a t i o n i n p r i m e

(irreducible polynomial) as in F.Y.B.Sc.

2) A part from this one can show that it D is UFD then <>Dx also.

Thus as ] is UFD, there fore <>x] also.

T h i s t h e o r e m i s p r o v e d i n n e x t c h a p t e r .

Remark : Since PID UFDº but converse need not be true that is UFDºPID.

For example we seen that <>x] is not PID but it is UFD as ] is UFD.

Being UFD will also has lots of benefit. One such benefit is given in following

theorem.

Theorem : In unique factorization domain irreducibles are primes.

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## Page 143

143Chapter 7: Euclidean Domain, Principal Ideal

Domain Unique Factorization

Proof :

Let R be a UFD and aR be irreducible.

To show that a is prime.

As a is irreducible. a=is non zero non unit.

Let ab c for some ,bc R .

=bc =ka for some kR.

As R is UFD, therefore ,,bck can be written as product of irreducible.

Let 123 1 2 1 2 ... , ... , ...jn n bb b b bcc c c kk k k A w h e r e ,ijbcand pkare

irreducibles in R for all ,ijand p then

bc ka12 12 12... ... ... ...mn bb b cc c kk k aº A .

As factorization is unique upto associates. Hence a must be associates of some ib

or jcif a is associate of ibthen iaband hence ab if a is associate of some jc

then jacand hence ac.

or ab c ab ac= º

a= is prime.

Thus we see in case of PID and UFD primes and irreducibles are essentially same.

This is the one reason why interval domain like -3 ¯¡°¢±] is no UFD (Note that

,-3 1 + -3 ¯¡°¢±] is irreducible but not prime).

One more advantage of being UFD is that it also guarantees the existence of GCD

of any two element of UFD. This is given as an exercise.

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144 ALGEBRA II

7.6 Summary

1) Let R be an integral domain. A non zero non unit element p of R is said to be

prime element in R if whenever pa b pao rpb ¸º .

2) In an integral domain prime element are irreducible.

3) An integral domain R is said to be Euclidean domain (ED). If there exist a

function \^ :0dR U l] such that

( i )

dab da ¸ for all ,ab R.

(ii) For any ,,ab b o v\ there exist ,prR

such that a = bp+r with r=0 or

<. dr db

4) In an Euclidean domain every ideal is principal ideal (i.e generated by single

element).

5) An Integral domain R is said to be principal ideal domain (PID) if every ideal

of R is Principal ideal.

6) Let R b e P I D a n d 12 3 n II I , "" b e s t r i c t l y i n c r e a s i n g c h a i n o f

ideals. Then this chain must be at finite length.

7) An integral domain R is said to be unique factorization domain (UFD) if

(i) Every non zero non unit element R c a n b e e x p r e s s e d a s a p r o d u c t o f

irreducible of R.

(ii) The factorization into irreducible is unique up to associates and the order

in which the factors appear.

8) Every principal ideal domain is Unique factorization domain.

9) In unique factorization domain irreducibles are primes.

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145Chapter 7: Euclidean Domain, Principal Ideal

Domain Unique Factorization

7.7 Unit and Exercises

1. Define

(i) Euclidean domain

( i i ) P r i n c i p a l i d e a l d o m a i n . S h o w t h a t a E u c l i d e a n d o m a i n i s a

p r i n c i p a l i d e a l d o m a i n .

2. Show that the following rings are Euclidean domain:

(i) The ring []i], ring of Gaussian integer

(ii) The polynomial ring F[x], where F is a Field.

3. Show the polynomial ring []x\is OID.

4. Prove or disprove

I f F is PID then F[x] is also PID.

5. Show that every ascending chain of ideals

12 1 ... ...nn II I I in a PID R is finite.

6. Prove that in a PID R, an element aR is prime

i f a n d o n l y i f a is irreducible.

7. Show that a PID is a UFD.

8. Show that any prime element in integral domain is irreducible.

Is converse true? Justify your answer.

9. Explain why -5 ¯¡°¢±] is not PID.

10. Show that any two elements a & b in a PID R have a G CD which can be

expressed in the form abM Nwhere ,,, R MN

11. Show that every irreducible element in a UFD is prime.

12. Is 5 ¯¡°¢±] is UFD? Justify your answer.

13. Show that ]is PID but <>x] is not.

( H i n t : s h o w t h a t ( 2 , x ) i s n o t p r i n c i p a l i d e a l ) munotes.in

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146 ALGEBRA II

14. Let Rbe an integral domain in which every non zero, non unit

element can be expressed as a product of irreducible and every

i r r e d u c i b l e e l e m e n t i s p r i m e . S h o w t h a t R is UFD.

15. (Hint : Note that factorization is unique upto associate.

Take two factorization of an element & use the fact that irreducible

is prime.)

16. Let \ be an UFD. Show that for any ,, ,ab R gcd of a and b exist.

(Hint : let 12 1 212 1 2 ... , ..m mm kkkk ap P p bpp p A AA where all andij m A

need not be non zero then take 212 ....nn n ikk dp p p where \^max ,ii inmA

then show that d is gcd of a and b.

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## Page 147

147Chapter 8: Irreducibility in Polynomial Ring UNIT 4

8

IRREDUCIBILITY IN POLYNOMIAL RING

Unit Structure

8.0 Objective

8.1 Introduction

8.2 Definition and example of irreducible and reducible polynomial.

8.3 Gauss lemma.

8.4 Eisenstein’s Criterion.

8.5 Summary

8.6 Unit and Exercises

8.0 Objective

This chapter makes you to understand

x Irreducible, reducible polynomials

x Classification of irreducible polynomials in \[x] and []x

x Various criteria to check irreducibility.

x Gauss lemma and Eisenstein’s criteria.

8.1 Introduction

In the chapter of ring we had deals with polynomial ring in detail. In high school

students spend much time factoring polynomials and finding their roots. This is

what we going to learn in this chapter but in abstract manner. Let us understand

FIRST which polynomial we can factorize and which cannot. We start with

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148 ALGEBRA II

8.2 Definition and Example of Irreducible and Reducible

Polynomial

Definition: L e t D be an integral domain. A polynomial p(x) in D[x] is said to

be irreducible polynomial if whenever p(x) = f(x) g(x) then either f(x) or g(x) is unit

(i.e. constant polynomial) in D[x].

Definition: A n o n z e r o – n o n u n i t p o l y n o m i a l o f D[x] which is not irreducible

is said to reducible .

Example:

1. The polynomial 223xis irreducible in []x_ but reducible in \ [x].

2. The polynomial 21x is irreducible over \ but reducible over ^.

3. The polynomial 21x is irreducible over 3] but reducible over 5]

Note that \^3 0,1, 2 ]

2fx x then

01 , 12 25 m o d 3 2 .ff f

but if we take element from 5]

01 1224 1 5 0 m o d 5fff

2

5

220

Hence 2 1

1f

xx i n

x

=

=] LVURRW

LVUHGXFLEOH

4. The only irreducible polynomial in \[x] are linear polynomial or the

polynomial224xbx c b c o VXFKWKDW .

Proof: Clearly the linear polynomial ax + b, ,ab\are irreducible polynomial

because they cannot be factorize.

Similarly for polynomial 22,xbx c if b 4c o then it has complex root

hence cannot be factorize. Now we will prove polynomial of any other degree must

be reducible. Let f(x) is any polynomial of degree n, n>2.

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149Chapter 8: Irreducibility in Polynomial Ring

Case I: deg f(x) is odd.

As complex roots are always come with conjugate pair and number of roots is equal

to degree of polynomial, therefore a polynomial of odd degree must have one real

root.

Let B be real root of f(x).

fxx g x = B where degree of

gxn 1 f=is reducible.

Case II: deg f(x) is even.

In worst case let assume all roots of f(x) are complex.

Let a + ib be one root of f(x) then aíib must be other roots.

.fxx a i b 2 a i b g x =

2 2xai bx ai b g x x a b g x

2 2andxab g x = are two factors of

<>fxi n x\

fx = is reducible.

5. The only irreducible polynomial in ^ are linear polynomials.

T h e o t h e r w a y o f s a y i n g t h i s i s e v e r y p o l y n o m i a l o v e r

O n e o f i m p o r t a n t p r o b l e m i n m a t h e m a t i c s i s t o f i n d r o o t o f p o l y n o m i a l s . T h e s e

roots have different meaning and application in different context. Lots of

techniques are developed in order to find the root of polynomial.

S u p p o s e w e h a d a p p l i e d o n e t e c h n i q u e t o f i n d r o o t . B u t w e h a v e n ’ t g o t , t h e n

we apply second technique to find root but we havn’t got, then third, and so on.

And then we came to know that this polynomial has no solution.

I f i n a d v a n c e w e g e t t o k n o w t h a t t h e p o l y n o m i a l h a s n o r o o t o r i n o t h e r w o r d

it is irreducible then lots of our efforts can be saved. So we are going to see

some test to check whether given polynomial are reducible or not.

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150 ALGEBRA II

Reducibility test for Degrees 2 and 3.

Theorem 1: Let F be a field. If

<>fxF x and

degfx2 or 3 then f(x) is

reducible over F if and only if f(x) has a zero in F.

Proof : Suppose that

fxg x h x

Where

<> ,.gxh x F x

as

deg deg degfxg xh x =

as

deg orfx2 3

= one of the polynomial g(x) or h(x) must be of degree one.

Let assume deg g(x) = 1.

,0 gx a x b a = v

then clearly 11.xba F is root of g(x) and hence of f(x).

=Conversely let assume that f has root Į in F then f(x) = (x- Į) g(x)

where deg g(x) = 1 or 2 = f(x) is reducible.

The above theorem is particularly used when the underlying Field is .p] because

in this case, we can check for reducibility of f(x) by simply checking that f(a) = 0

or not for a = 0, 1, …., p-1.

Note that the polynomials of degree larger than 3 may be reducible over a field,

even though they do not have zeros in field. For example in

_[x], t h e p o l y n o m i a l 4221 xx is equal to

221, x b u t h a s n o z e r o s i n

_.

To see next tests for inedibility we need following definition.

Content of Polynomial

Definition: The content of a non zero polynomial 1

1 .. ,nn

nn oax a x a

where ia] is the greatest common divisor of 11,, . . , .nn oaa a a

Definition: A polynomial 1 .. ,n

no iax a x a a ] i s s a i d t o b e primitive

polynomial if its content is one. munotes.in

## Page 151

151Chapter 8: Irreducibility in Polynomial Ring

8.3 Gauss Lemma

Theorem 1: T h e p r o d u c t o f t w o p r i m i t i v e p o l y n o m i a l s i s P r i m i t i v e .

Proof: Let f(x) and g(x) be primitive polynomials and suppose that f(x)g(x) is not

primitive. Let p i s p r i m e d i v i s o r o f c o n t e n t o f f(x) g(x). L e t

,fxg x a n d

fxgx be the polynomials obtained from f(x), g(x) and f(x) g(x) by reducing

the coefficient modulo p

then

<> ,pfxg x x ] and

<>,pfxg x x ]LQ

( ' Each coefficient of f(x) g(x) is divisible by p)

( ' p is prime, p=] is integral domain and hence <>px] is integral domain.)

ºEither p divides coefficient of f(x) or p divides coefficient of g(x).

Which is contradiction as f(x) and g(x) is primitive.

.fxgxi s p r i m i t i v e =

Theorem 2: L e t

<>fxx]. If f(x) is reducible over _,

then it is reducible over ].

Proof: Suppose that f(x) = g(x) h(x), where g(x) and h(x) >@x_.

We may assume that f(x) is primitive because we can divide both f(x) and g(x) h(x)

by content of f(x). Let a and b be the least common multiple of the denominators

of the coefficients of g(x) and h(x) respectively.

Then ab f(x) = ag(x) bh(x) w h e r e

<>,. ag x bh x x ]

Let 12and cc be the content of ag(x) and bh(x) respectively.

11 ag x c g x= and

21 . bh x c h x where

1gx a n d h1(x) are primitive

polynomial in <>x]

12 1 1 .. ab f x c c g x h x=

Note that f(x) is of content 1, Hence content abf(x) is ab.

<>

.pfxgx i n x

fx o r g x

º

º ]

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## Page 152

152 ALGEBRA II

By Gauss lemma

11.gxhx is also primitive.

12 1 2 12

12 1 1 12.

.cc g xg xi s cc

abf x c c g x h x ab c c

=

= º FRQWHQWDW

<>

<>11 11.,

.fxg x h x w h e r e g x h x x

fx x=

=]

]

LVUHGXFLEOHRYHU

Theorem 3: Let P be a prime and suppose that

<>fxx] with deg

.fx 1

Let

fx be a polynomial in <>px] obtained from f(x) by reducing all coefficients

of f(x) modulo p. If f(x) is irreducible over p] and deg

degfxf x then f(x)

is irreducible over _.

Proof: Since by previous theorem we know that if f(x) is reducible over _ then it

is reducible over ].

Hence let assume f(x) = g(x). h(x) where g(x), h(x) <>x].

Let

,a n dfxg x h x be the polynomials obtained from f(x), g(x) and h(x) by

reducing all the coefficients modulo p. And

fx is irreducible over .p]

Since deg f(x) = deg

fx

deg deg ( ) deggx gx f x = and

deg deg deghx hx f x

But

.fxg x h x

This means that

gxa n d hx are proper factor

fx, but this is contradiction

to our assumption that

fx is irreducible over .p]

Hence our assumption that f(x) reducible over _is wrong.

()fx =is irreducible over _.

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## Page 153

153Chapter 8: Irreducibility in Polynomial Ring

Example:

Let

3221 3 2 7 fx x x x

Then over

32

2 1 fx x x ]

S i n c e

01 1ff

t h e n

2 fx ] LVLUUHGXFLEOHRYHU

and as deg f= deg f

fx = is irreducible over _.

8.4 Eisenstein’s Criterion

Theorem 4: L e t

<>1

1 ..nn

nn ofxa x ax a x

] if there is a prime p

such that 2

1 †, , . . . , b u t †nn o oPaP a P a P a then f(x) is irreducible over _.

Proof: Let assume that f(x) is reducible over _. Then by previous theorem f(x) is

reducible over ] also.

Let assume that f(x) = g(x) h(x) where

1d e g , d e g .gxh x n

Let

11 .. and .. .rk

ro ko gx b x b x b hx c x c x c

Since f(x) = g(x). h(x)

and .nr k oo oab c ab c=

Since .oo oooPap b cp b o r p cºº (Since P is prime)

but 2

0†pa, p does not divide both &.oobc Hence Let assume 0†opbb u t p c

Also .& .nr kr kpap b cp b p c ºº

†† † †nr kr kpa pb c pba n d pc ºº

†rpb' munotes.in

## Page 154

154 ALGEBRA II

Therefore there exist least integer t such that †tpb but the

11 ...tt o t o t

tab cb c b c

as p a

Now by choice of t, p must divide each term of right hand side except the first one.

11 .... . . .t

to t t o pa b c b c i e p bc Which is not possible as ,.topb pc

Hence f(x) is irreducible over _.

Example: Irreducibility of thp Cyclotomic polynomial for any prime p,

The thpCyclotomic polynomial.

12 111p

pp

pxxx x xx !! is irreducible over _.

As consider the polynomial

1

11.. 1 1111. .11pp p

pp

pxp x p xxfxx x p x pxx

Note that p divides all coefficient of f(x) except for leading coefficient 1 and also

2p does not divide constant term which p.

Hence by Eisenstein’s criteria f(x) is irreducible over Q.

But if

. Ox gxhx were a non trivial factorization of

pOxover _ then

(1 ) (1 ) (1 )p fx O x g x h x will we a non trivial factorization of f(x).

Since this is impossible, hence we conclude that

pOx is irreducible over _.

Let us understand the importance of irreducible polynomials.

1. The irreducible polynomial plays exactly same role as prime plays in case of

integer. For example the fundamental theorem of arithmetic (or unique

factorization theorem) irreducible polynomial are treated as prime.

2. One more reason which makes irreducible polynomial important, which is

given in terms of next theorem. munotes.in

## Page 155

155Chapter 8: Irreducibility in Polynomial Ring

Theorem 5: Let F be a field and let

<> px Fx Then

px is a maximal ideal

in F[x] if and only if p(x) is irreducible over F.

Proof: Let assume that

px is maximal ideal in F[x].

Let assume that p(x) is reducible then p(x) = f(x) . g(x)

px f x px c f x º º

but

px is maximal ideal.

Hence it either

px f x or

<>fxF x if

px f x then

px and

fx are associate of each that therefore

px f xgx implies

g(x) is unit.

And if

<>fxF x this implies f(x) is unit.

px= is irreducible.

Conversely, let p(x) is irreducible.

Let assume I is ideal of F[x] such that

<>px I Fx

As F[x] is PID,

Therefore there exists

<>gxF x such that

Ig x

.[ ] px gx px f x gx f x Fx º º IRUVRPH

Now as p(x) irreducible.

() () ()px f x gx= implies either f(x) is unit or g(x) is unit.

If f(x) is unit p(x) and g(x) are associates of each other.

px gx I=

If g(x) is unit then

Ig x F x

Hence

px is maximal. munotes.in

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156 ALGEBRA II

Corollary 1: L e t F b e a f i e l d a n d l e t

<> ,px ax bx Fx . If p(x) i s

irreducible over F and

,( ) .pxax bx t h e n pxaxo r p xbx

Proof: Let since p(x) i s i r r e d u c i b l e , <>()Fx

px=is a field and therefore

integral domain.

Let

axa n d bx b e t h e i m a g e s o f a(x) a n d b(x) u n d e r t h e n a t u r a l

homomorphism from F[x] to <>.()Fx

px Since

pxaxbx .

.0 ax bx = in <>

()Fx

px, but then

0 ax or

<>Fx0S i n c e i s i n t e g r a l d o m a i n

bx¬ ®.

Therefore it follows that

.pxaxo r pxbx

Now we conclude this chapter by proving <>x] is unique factorization domain.

Theorem 5: Every non zero non unit polynomial in <>x] c a n b e w r i t t e n a s

12 1.. ...sm bb b p x p x where '

ibs are irreducible polynomials of degree zero and

'

ips are irreducible polynomials of positive degree.

More over if

12 1 12 1.. ... .. ..sm k bb b p x p x cc c g x q x where ''

ijbsa n dc s

are irreducible polynomial of degree zero and

and 'ijpxs q x s are irreducible

polynomial of positive degree then s = k, m=t and after renumbering

. andij i jpxq x b c

Proof: Let

fx be a nonzero non unit polynomial from <>.x] If deg

, fx

then

fx is constant and we are through, by Fundamental Theorem of Arithmetic.

If deg

, fx let b d e n o t e t h e c o n t e n t o f

fx, and let 12...sbb b b e t h e

factorization of b a s a p r o d u c t o f p r i m e s . T h e n

12 1... ,sfxb b b f x w h e r e

1fx belongs to <>,x]is primitive and deg

1 deg .fxf x munotes.in

## Page 157

157Chapter 8: Irreducibility in Polynomial Ring

Thus to prove the existence portion of the theorem it suffices to show that a

primitive polynomial

fx o f p o s i t i v e d e g r e e c a n b e w r i t t e n a s a p r o d u c t o f

irreducible polynomials of positive degree. We proceed by induction on deg

.fx

If deg

1, fx then

fx is already irreducible and we are done.

Now suppose that every primitive polynomial of degrees less than deg

fx can

be written as product or irreducible of positive degree. If

fxis irreducibles,

there is nothing to prove. Otherwise Let

.fxg x h x where both

gx

and

hx are primitive and have degree less than that of

.fx Thus by induction

both

gx and

hx can be written as a product of irreducibles of positive degree.

Clearly then

fx is also such a product.

To prove the uniqueness portion of the theorem, suppose that

12 1 1 2 k 1... ... c c ... c ...sm tfxb bb x x q xq x pp

where 'ibs a n d 'jcs are irreducible polynomial of zero and

'ipxs a n d

'jqx s are irreducible polynomials of positive degree.

12 12 Let .. & ...sk bb bb cc c c

Since

' 'andijpxs q xs are primitive therefore by Gauss Lemmas

12 ...m pxp x p x and

1...t qx qx are primitive.

Hence both b and c must equal plus or minus the content of

fx and therefore

are equal in absolute value. It their follows from the fundamental Theorem of

Arithmetic that s = k and after renumbering ib = c and1 < i < k.io

Thus by cancelling the constant terms in the two factorizations for we get,

12 1 ... ....mt pxp x p x qx qx .

By viewing

ipx and

jqx a s e l e m e n t o f _[x] & noting that

112pxpx p x ….

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## Page 158

158 ALGEBRA II

Therefore

11 ....t px qx qx , and by corollary of previous theorem

1 j px q x f o r s o m e j . b y r e n u m b e r i n g w e g e t ,

11px qx

11qx fx px º but as both

1qx and

1px are irreducible therefore

fx is unit in []x_say1f

11 1 , qx fpx= b u t b o t h

1qx and

1px are primitive which implies

11 f r, so 11() ()qx px r.

Also after canceling we get

23 2 mtpxp x p x q x q x "" . After repeating above argument if m

< t then we get 1 on left side and non constant polynomial in right which leads to

contradiction. Hence m = t and

iipx qx .

8.5 Summary

1) Let D be an integral domain. A polynomial p(x) in D[x] is said to be irreducible

polynomial i f w h e n e v e r p(x) = f(x) g(x) t h e n e i t h e r f(x) or g(x) i s u n i t ( i . e .

constant polynomial) in D[x].

2) A non zero – non unit polynomial of D[x] which is not irreducible is said to

reducible .

3) The content of a non zero polynomial 1

1 .. ,nn

nn oax a x a

where

ia] is the greatest common divisor of 11,, . . , .nn oaa a a

4) Gauss Lemma: The product of two primitive polynomials is Primitive.

5) Eisenstein’s Criterion: Let

1

1nn

nnfxa x ax

<>..oax ] if

there is a prime p such that 2

1 †, , . . . , b u t †nn o oPaP a P a P a then f(x) is

irreducible over _.

6) Let F be a field and let

<> px Fx Then

px is a maximal ideal in F[x]

if and only if p(x) is irreducible over F.

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## Page 159

159Chapter 8: Irreducibility in Polynomial Ring

8.6 Unit and Exercises

1) Show that 2x+1 a n d 2x+x+4 are irreducible polynomial in =>[@.

Show that <>

=[

[ and <>

=[

[[ are fields having 121 elements.

Solution :- Let

21 fx x

0f= 1, f 1 = 2, f 2 = 5, f 3 = 10

f4= 1 7 = 6 (mod 11),

f5 = 2 6=4 (mod 11),

f6= 3 7 = 4 (mod 11) ,

f7 = 5 0 = 6 (mod 11),

I (mod 11),

f9= 5 (mod 11),

f10 = 2 (mod 11).

Hence f(x) is irreducible over <>11x]

fx =is maximal ideal in <>11x]

<>11

()xfx=§] is field.

Now for any

<>11 hx x], by division algorithm

2hx = gx x+ 1 + rx

where

rx oor deg

rx = 1

11 ,, rx a x ba b= ]

h2x = g x x + 1 + ax + b =

22 211 1 hx x gx x a x b x 2ax + b + x +1 <><>\^2 11

11 1: ,()xax b x a b xfx = §]] munotes.in

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160 ALGEBRA II

= There are 11 choice for a and 11 choice for b. Hence total 121 polynomials are

there of the form ax + b .

Hence <>11

2121.

1x

x

]

Similarly we can prove that <>11

2121.

4x

xx

]

2) Show that

<>2

5 3x +4x+3 x ]ෛ f a c t o r s a s

3x+2 x+4 a n d

4x+1 2x+3 Is <>5x] a UFD?

Justify your answer. Is the above factorization is unique upto multiplication by a

unit?

Solution:

234 3 fx x x

03 1 1 0 0ff

22 232 4 3 21 28 3 1 48 3 43 ( m o d 5 )xx x x x xx x x =

Similarly

2241 23 8 1 43 3 43 ( m o d 5 )xx x x x x

Hence

234 3 3 2 4 4 1 2 3xx x x x x

=Since

3x+2 x+4 and

4x+1 2x+3 are two factors of 234 3xx .

4 1(mod 5)

1 4 (mod 5)w

'

41 4 ( m o d 5 )

4( m o d 5 )xx

x=

4x+1 =is associate of

< >5x+4 in x ] Similarly 2x+3 = - (3x+2) (mod 5).

H e n c e 2 x + 3 i s a s s o c i a t e o f ( 3 x + 2 ) i n <>5x]

= Hence the above factorization is unique upto multiplication by a unit.

5'] is field <>5x=] us UFD. munotes.in

## Page 161

161Chapter 8: Irreducibility in Polynomial Ring

3. Determine which of the following polynomials are irreducible in the indicated

rings.

( i ) <>2

2 1xxi nx ]

( i i ) <>4

52 xx i n]

( i i i ) <>4210 1xxi n x ]

4. Show that if

fx is maximal ideal in <>x\, then f(x) is irreducible.

5. Let F be a field. Show that the <>

/ Fx f x is field if and only if f(x) is

irreducible over F.

6. Show that the only maximal ideals of <>x\ are of the form

(x-a), 2ao r x b x c \ where b,c \ with 24 ba cR

7. Show that maximal ideals of <>x^ are x-D where .B^

8. Show that if D is UFD then D[x] is also (Proof is on similar line of proving

<>x] is UFD)

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