Msc Mathalgebra (1)

UNIT I GROUP THEORY 1 LESSON -1 ANOTHER COUNTING PRINCIPLE CONTENTS : 1.0 Aims and Objectives 1.1 Introduction 1.2 Equivalence Relation 1.3 Conjugacy Relation 1.4 Cauchys Theorem 1.5 Let us sum-up 1.6 Lesson-end Activities 1.7 References 1.0 AIMS AND OBJECTIVES

In this lesson we introduce a conjugacy relation in a group in order to derive the class equation of a finite group. We verify the equation in the case of 3S , the permutation group on three symbols. As an application of class equation, we prove an important theorem due to Cauchy.

After going through this lesson, you will be able to: (i) Define conjugacy relation in a group. (ii) Prove conjugacy relation is an equivalence relation. (iii) Obtain the class equation of a finite group. (iv) Prove Cauchys theorem. (v) Define the partition of a positive integer n. (vi) Obtain the conjugate classes of nS . 1.1 INTRODUCTION Counting principle is a powerful tool for deriving certain theorems in algebra. The process of counting the elements in two different ways and then comparing the two, yields the desired conclusions. We define a conjugacy relation in a group to derive the class equation of a finite group. As an application of this equation, we prove that cauchys

Theorem which asserts the existence of an element of order p in G whenever )(Gop , for a

given prime p. The number of distinct conjugate classes of nS is obtained as the number of partitions of n.

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Group Theory 2

1.2 EQUIVALENCE RELATION Definition The Cartesian product A x B of two sets A and B is the set{(a,b) : a A, and bB}. Defintion A subset R of A x A is called a relation on A. If (a,b) R, we write a b. Definition A relation on a set A is said to be an equivalence relation on A if for all a,b,c in A, (1) a a (reflexivity) (2) a b implies b a (symmetry) (3) a b and b c imply a c (transitivity) Definition If is an equivalence relation on A and a A, the equivalence class of a is the set C (a) = { x : x a } Theorem Any two equivalence classes on a set are disjoint or identical Theorem If is an equivalence relation on a set A, then A is the union of its distinct equivalence classes The proofs of the above theorems are left as exercise. 1.3 CONJUGACY RELATION Definiton Let G be a group and a, b G. We say a is conjugate of b, if there exists an element g G such that a = g-1 bg. We write a b. Lemma In a group G, the conjugacy relation is an equivalence relation. Proof : Reflexivity : Since a = e-1 ae, where e is identity element of G, a a. Symmetry : If a b, then $ x G such that b = x-1 ax.

Then (x-1)-1 b x -1 = (x -1) -1 (x -1 ax) x -1 i.e., x b x -1 = x (x -1 ax) x -1 .


3 Algebra = (x x -1) a (xx -1) . = a \ a = x bx -1 \ b a Transitivity : Let a b and b c. Then b = x -1 ax, for some x G and c = y by, for some y G. c = y (x ax) y = (y x ) a (xy) = (xy) a (xy) \ a c Therefore conjugacy is an equivalence relation. Definition If a belongs to the group G, the equivalence class of a under the conjugacy relation is C (a) = {x : x a }. It is called the conjucate class of a. If G is a finite group, then the number of elements in C (a) is donated by c a. Lemma If a belongs to aroup G, then N(a) = { x G : xa = ax } is a subgroup of G and is called normaliser of a. Proof : Let x,y G Then xa = ax ----- (1) ya = ay ----- (2) Therefore, (xy) a = x (ya), by associativity = x (ay), by (2) = (xa) y, by associativity = (ax) y, by (1) = a (xy), by associativity.

Therefore xy N (a). From (1), x (xa) x = x (ax) x i.e., (x x) (a x ) = (x a) (x x ) i.e., ax = x a

Therefore x N (a). Hence N (a) is a subgroup of G.


Group Theory 4 Theorem

If G is a finite group, then ca=

))(()(aNO

Go

Proof : First we show that if x and y are in the same right coset of N(a) in G, then they give rise to same conjugate of a.

Let x, y N (a) g, where g G. Then for some 1n N (a) x = 1n g ----- (1) and for some 2n N (a), y = 1n g ----- (2) From (1), g = 1n x ----- (3) From (2), g = 1n y ----- (4) From (3) and (4), 1n y = 1n x \ y = 2n 1n x = nx ----- (5) Where n = 2n 1n N (a). Then na = an ----- (6) \ y ay = (nx) a (nx), by (5) = x n (an) x = x n (na) x, by (6) = x e a x = x a x

\ x and y give the same conjugate of a. Next we show that if x and y are in different right cosets of N (a) in G, then they give rise to different conjugates of a, that is x ax y ay. Suppose

x a x = y ay then y(x a x) x = y (y ay) x (i.e). yx a = ayx \ yx N(a). Then yx = n for some n N(a).


5 Algebra

\y = nx ------- (7)

Let y N (a) g. Then for some 1n N (a), y = 'n g ------- (8)

from (7) and (8) nx = 'n g \ x = 1-n 'n g where 1-n 'n N(a) \ x N (a) g. Hence x and y are in the same right coset of N (a) in G. This is a contradiction. Therefore x and y are in different right costs of N (a) in G, they yield different conjugates of a. Hence there is a one-one correspondence between the conjugates of a and right cosets of N (a). The number of right cosets of N (a) is equal to

))(()(aNO

GO . Hence

ca = ))(()(aNO

GO

Corollary If G is a finite group, then O(G) =

))(()(aNO

GO

Where the sum runs over one element a in each conjugate class Proof : Since the conjugate classes are disjoint and their union is G, O (G) = aC Where the sum runs over one element a in each conjugate class. But by the above theorem ca = ))((

)(aNO

GO

Hence we get,

O (G) = ))((

)(aNO

GO

This equation is called class equation of G. Note : If G is abelian group and a G, then C (a) = { xa x : x G} = { x x a : x G}

= {a} \ ca = 1


Group Theory 6 Example We know that nS the set of n symbols form a group under composition of maps. This group is called symmetric group of n symbols. Now s3 = { e, (1,2), (1,3), (2,3), (1,2,3), (1,3,2)} where e is the identity element

The conjugate class of e is C (e) = { x ex : x G} = { x x : x G} = { e }

The conjugate class of

213123

= ( 1,2 ) is

C (1,2) = { e (1,2)e, (1,3) (1,2) (1,3), (2,3) (1,2), (2,3), (1,2,3) (1,2) (1,2,3), (1,3,2) (1,2) (1,3,2) }. = { (1,2), (1,3), (2,3)} Similarly, C (1,2,3) = {(1,2,3), (1,3,2)} We note that the conjugate classes C(e), C(1,2), C(1,2,3) are pairwise disjoint and their union is s3 Also ce = 1, c )2,1( = 3, c )3,2,1( = 2 O( s3 ) = ce + c )2,1( + c )3,2,1(

We can verify that N(1,2) = { e, (1,2) } N(1,2,3) = { e, (1,2,3), (1,3,2) } N(e) = s3

))(()( 3

eNOO S =

66 = 1 = eC

))2,1(()( 3

NOO S =

26 = 3 = C )2,1(

))3,2,1((

)( 3NOO S =

36 = 2 = C )3,2,1(

Also O( s3 ) =

))(()( 3

eNOO S +

))2,1(()( 3

NOO S +

))3,2,1(()( 3

NOO S

Which is the class equation forS 3 .


7 Algebra Note : If G is a group then Z, the set of all elements of G which commute with all the elements of the group G, is a subgroup and is called center of G. Lemma Let Z be the center of the group G. Then a Z. iff N(a) = G. Proof : Let a Z. Then a commutes with all the elements of G. Therefore G N(a) ------- (1) Also, N(a) G ------- (2) From (1) and (2) , N(a) = G Conversely, let N(a) = G Then, xa = ax for every xG Therefore, a Z. Theorem If O(G) = np where p is a prime number, then Z(G) {e } Proof : Let a G. Then N(a) is a subgroup of G. Then by Lagranges theorem, O(N(a)) divides O(G) = np . Therefore, O(N(a)) = anp Where an is a positive integer By previous lemma, a Z. iff O(N(a)) = O(G), that is, iff an = n. The number of elements a for which N(a) = G is O(Z). Therefore the class equation of G is

O(G) = =GaN )(

))((

)(aNO

GO + GaN )(

))((

)(aNO

GO

=O(Z(G)) + GaN )(

))((

)(aNO

GO

Therefore,

O(Z(G)) = O(G) - GaN )(

))((

)(aNO

GO

= np - GaN )(

an

n

pp ------- (1)

Since an < n, p divides ann

pp

\ p divides each term of the sum and hence p is a divisor of the sum. \ p is a divisor of R.H.S. of (1)


Group Theory 8 \ p is a divisor of L.H.S. of (1) That is p / o (Z(G)) As e Z(G), o(Z(G)) 0. Therefore, O(Z(G)) is a positive integer divisible by p. \Z(G) { e } Corollary If O(G) = 2p where p is a prime number, then G is abelian. Proof : First we note that G is abelian iff Z(G) = G. Since O(G) = 2p by previous theorem, Z(G) { e } and hence O(Z(G)) must be either p or 2p . If o(Z(G)) = 2p , then O(G) = O(Z(G)) = 2p and Z(G) = G \G is abelian Let O(Z(G)) = p Let a G, a Z(G) Then N (a) is a subgroup of G, Z(G) N (a) But a Z(G) \ Z(G) N(a) \ N(a) = G a (G), which is a contradiction Hence O(Z(G)) p and O(Z(G)) = 2p only \ G = Z(G) \ G is abelian 1.4 CAUCHYS THEOREM Theorem If p is prime number and p/O(G), then G has an element of order p. Proof : The proof is by induction on the order of G. The result is true for groups of order 1. We assume the theorem is true for all subgroups T such that O(T) < O(G). Let H be a subgroup of G, H G and p/o (H). Then by induction hypothesis, there exists an element h H such that ph = e. Then as H G, h G and ph = e. The theorem is proved in this case. Let us assume that p does not divide o(H) for any subgroup H of G. In particular, If a Z (G), then N(a) G and hence p does not divide O(N(a)). The class equation is

O (G) = O(Z(G)) + GaN )(

))((

)(aNO

GO

\ O (Z(G)) = O(G) - GaN )(

))((

)(aNO

GO ------- (1)

Since p/o (G) and p*o (N(a)), p/ ))((

)(aNo

Go

\ p divides each term of the sum and hence p divides the sum.


9 Algebra \ p divides R.H.S of (1) \ p divides o(Z(G)) As p does not divide order of any proper subgroup of G, we should have Z(G) = G. Then G is abelian. Then by cauchys theorem for abelian groups, there is an element a G, (a e) such that na = e Definition A finite sequence ( 1n , 2n . kn ) is called a partition of a positive integer n if o 1n

2n . kn and 1n + 2n + . kn = n. The number of partitions of n is denoted by p(n). Example We have 4 = 4, 4 = 1+3, 4 = 1+1+2, 4 = 1+1+1+1, 4 =2+2 \ p(4) = 5. Similarly p(1) = 1, p(2) = 2, p(3) = 3, p(5) = 7, p(6) = 11. Theorem The number of distinct conjugate classes of nS is p(n). Proof : Let s sn . Then s can be expressed as disjoint cycles. Let s = 1s . 2s . ks Where 1s , 2s ks are disjoint cycles of length respectively 1n , 2n .. kn , 1n 2n 3n kn , 1n + 2n + ..+ kn = n. We give an algorithm to obtain 1-q s q if q s Suppose s sends i to j, q sends i to s and j to t. Then (s) 1-q s q = (i) s q = (j) q = t. Let s and t be two partitions giving rise to the same partition ( 1n , 2n . kn ) of n.

Let s = (a1 ,a2 , ..... an1 ) ( b1 , b2 bn2 ) . ( x1 , x2 x kn ) t = (a1 ,a 2 .a 1n ) ( b 1 , b 2 .. b kn ) ...

(g 1 ,g 2 ..g kn )

Define q =

k

k

nnn

nnn xxbbbaaa

ggbbbaaa ..............,.......,

.....................,.......,

12121

12121

21

21

Then t = 1-q s q so that t and s are conjugate. Therefore corresponding to a partition on n there is a unique conjugate class. Next we show that any two conjugates define the same partition.


Group Theory 10 Let s = 1c 2c . kc where 1c , 2c ,. kc are disjoint cycles of lengths 1n ,

2n , 3n ,. kn Let 1s be the conjugate ofs . Then 1s = q s 1-q

= q 1c 2c 3c . kc 1-q

= (q 1c1-q ) (q 2c

1-q ) . (q kc1-q )

= 11c 1

2c . 1

kc Where 11c = q 1c

1-q , 12c = q 2c1-q , . kc' = q kc

1-q . Let 1c = ( 1a , 2a , ..... 1na ) Then

q 1c1-q = (q ( 1a ),q ( 2a ), ., q ( 1na ).

Therefore if 1c is of length 1n , then q 1C 1-q is also of length 1n .

Also 1c , 2c , ., kc are disjoint, q 1c 1-q , q 2c

1-q ,., q kc1-q are

also disjoint. \ s and 1-s define the same partition. \ Corresponding to a conjugate class, there is a unique partition of n. \ The number of distinct conjugate classes of sn is p(n). Example Number of district conjugate classes of s3 is 3. Also p(3) = 3 Since 3 = 3, 3 = 2+1, 3 = 1+1+1. 1.5 LET US SUM -UP

In this lesson, we have discussed the following concepts: (i) Conjugacy relation is an equivalence relation. (ii) The normaliser of an element of a group G is a sub-group of G. (iii) The class equation of G. (iv) Every group of prime square order is abelian. (v) Cauchys Theorem/ (vi) The number of distinct conjugate classes of nS is p(n).


1.6 LESSON END ACTIVITIES (1) Prove that the conjugacy relation is an equivalence relation. (2) Prove that a normaliser N(a) is a sub-group of G. (3) Show that ca=

))(()(aNO

Go

(4) Derive the class equation for 3S . (5) If O(G) = np where p is a prime number, then show that Z(G) {e}. (6) If O(G) = 2p where p is a prime number, then show that G is abelian. (7) State and prove cauchys theorem. (8) Show that the number of distinct conjugate classes of nS is p(n). 1.7 REFERENCES

1) A First Course in Abstract Algebra by J.B. Fraleigh, Narota Publishing House, New Delhi, 1988.

2) Topics in Algebra by I.N. Herstein Second Edition.


UNIT-I GROUP THEORY 12 LESSON -2 SYLOWS THEOREM CONTENTS: 2.0 Aims and Objectives 2.1 Introduction 2.2 Sylows Theorem 2.3 Let us sum-up 2.4 Lesson-end activities 2.5 References 2.0 AIMS AND OBJECTIVES In this lesson, we prove three important theorems contributed by sylow. After going through this lesson, you will be able to: (i) Prove Sylows Theorem (ii) Define p-sylow subgroup of G. (iii) Prove any two p-sylow subgroups of G are conjugate. (iv) Determine the number of p-sylow subgroup of G. 2.1 INTRODUCTION

According to Lagranges Theorem, the order of a subgroup always divides the order of a finite group. However, the converse need not be true sylows Theorem asserts the existence of sub-groups of prescribed order in arbitrary finite groups. We introduce p-sylow sub-groups in a finite group and device a method for finding the number of p-sylow subgroups of a finite group. 2.2 SYLOWS THEOREM Theorem (First Part of Sylows Theorem) If G is a finite group, p is a prime number and ap |o (G), then G has a subgroup of order ap . Proof : Here p is a prime number and ap |o (G). Let o(G) = ap m. We know


13 Algebra

kn c = 1.2.3)...1()1)...(1(

-+--

kkknnn

Let n = ap m where p is a prime number, rp /m and 1+rp m. Let k = ap .

ap m apc = .1).....).....(1()1()).....(2)(1(

ippppmpimpmpmpmp

--+-----

aaa

aaaaaa

Now we show that the power of p dividing ap m-i in the numerator is same as the power of p dividing ap -i in the denominator. Let kp |( ap -i). Then ap - i = a kp where k a . Then. -i = a kp - ap \ ap m-i = ap m + a kp - ap = (m-1) ap + a kp = kp [ (m-1) kp -a + a ] \ kp | ( ap m-i) Conversely kp | ( ap m-i) kp | ( ap -i) Therefore all the powers of p in the numerator and denominator cancel out except the power of p which divides m. \ As rp |m and 1+rp m we get rp | ap m apc but

1+rp ap m apc .

Let M~ be the set of all subsets of G which have ap elements.

Then M~ has ap m apc elements Define for 1M , 2M 1~M 1M 2M if $ g G

such that 1M = 2M g. We show that is an equivalence relation. Reflexivity: Since 1M = 1M e, 1M 1M Symmetry: If 1M 2M , $ g G such that 1M = 2M g Then 2M = 1M

1-g Therefore, 2M 1M . Transitivity : Let 1M 2M and 2M 3M . Then there exists 1g G such that 1M = 2M 1g and $ 2g G such that 2M = 3M 2g . Now, 1M = 2M 1g = ( 3M 2g ) 1g = 3M 2g 1g where 2g 1g G and hence 1M 3M . Therefore is an equivalence relation on M~ . Then there is atleast one equive ence class in M~ such that 1+rp does not divide the number of elements in this class. For if 1+rp divides the number of elements in each class then 1+rp divides the number of elements in M~ which is impossible since 1+rp ap m apc

Let { 1M , 2M ,.. 1M } be such an equivalence class in M where 1+rp t.

This class can be taken as { 1M , 1M 2g , 1M 3g ,. 1M tg }. where 2g , 3g . tg belong to G


Group Theory 14 Let H = {g G : 1M g = 1M }, Let 1g , 2g H. Then 1M 1g = 1M and 1M 2g = 1M . Therefore 1M 1g 2g = ( 1M 1g ) 2g = 1M 2g = 1M . \ 1g 2g H. Therefore H is a finite subset of G satisfying closure property. Hence H is a subgroup of G. We now claim that there is a one-one correspondence between { 1M , 2M .. tM } and the set { Hg : g G }. Let 1M 1g and 1M 2g be two elements in { 1M , 2M .. tM }. Then 1M 1g = 2M 2g 1M 1g

12

-g = 1M 1g

12

-g H H 1g = H 2g . Therefore our claim is proved. Now,

)()(

HoGo = number of right cosets of H in G

= number of elements in { 1M , 2M .. tM } = t \ o(G) = t o(H) \ t o(H) = o(G) = ap m. Since, 1+rp t and rp +a ap m = t o(H), it must follow that ap |o(H). and so o(H) ap ------ (1) For all hH, 1M h = 1M , by definition of H.

Therefore 1M has atleast o(H) distinct elements. Therefore, O( 1M ) o(H) But, o( 1M ) =

ap \ ap o(H) ----- (2) From (1) and (2), o(H) = ap . Corollary If mp | o(G) and 1+mp o(G), then G has a subgroup of order mp . This is a special case of the above theorem and is usually known as Sylows theorem. Definition A subgroup of order mp where mp | o(G) but 1+mp o(G) is called p-Sylow subgroup of G. Lemma Let n(k) be defined by )(knp | ( kp ) ! but )1( +knp ( kp )!


15 Algebra Then n(k) = 1 + p + 2p + .. + 1-kp ------ (1) Proof : If k = 1, ( kp )! = p! = 1.2.3. . (p-1) p. Then p divides p! but 2p p! Therefore n(1) = 1. Now kp ! = 1.2 .. p(p+1) .. 2p (2p+1) .. 3p (3p+1) .. 1-kp p. Therefore the factors in the expansion of kp ! that can contribute to the powers of p dividing kp ! are p, 2p, 3p, 1-kp p. Therefore n(k) is a power of p which divides p.(2p) . (3p) ( 1-kp p) = 1-

kpp . ( 1-kp )! Therefore n(k) = 1-kp + n(k-1) Therefore, n(k) n(k-1) = 1-kp . n(k-1) n(k-2) = 2-kp . n(2) n(1) = p n(1) = 1 Adding these up with cross cancellation, n(k) = 1 + p + 2p + .. + 1-kp Lemma pkS has a p Sylow subgroup Proof : The proof is by induction on k. If k = 1, then the element (1,2,3,.p) is in pS and is of order p and so generates a subgroup of order p. Since n(1) = 1, the result is proved for k = 1. Suppose that the result is true for k 1. We show that the result is true for k. Divide the integers 1,2,. kp into p clumps each with 1-kp elements as follows. {1,2,., 1-kp }, { 1-kp +1, 1-kp +2,..,2 1-kp } .{(p-1) 1-kp + 1, . kp } Then the permutation s defined by s = (1, 1-kp +1, ., (p-1) 1-kp +1) (2, 1-kp +2, ., (p-1) 1-kp +2) ., ., ., (j, 1-kp +j, .,(p-1) 1-kp +j) ., ., ., ( 1-kp , 2 1-kp ., kp )


Group Theory 16 has the following properties : (1) ps = e,(2) If t is a permutation that leaves all the elements of i fixed for i > 1-kp and hence affects only 1,2,. 1-kp then j-s t js moves only j 1-kp +1, j 1-kp +2,. (j+1) 1-kp . Consider A = {t : t pS ,t (i) = i if i >

1-kp } Let 1t , 2t A. Then 1t (i) = " >

1-kp 2t (i) = " >

1-kp Then 1t 2t (i) = 1t (i) " >

1-kp = " > 1-kp \ 1t 2t A. Therefore A is a subgroup of kpS and elements in A can carry out any permutation

on 1,2 1-kp .

\A ~ 1-kps .

By induction, A has a subgroup 1P of order )1( -knp . Let

T = 1P )( 11 ss P- )( 21

2 ss P- . )( 111 -- pp Pss

= 121 .......... -PPPP Where 1+iP =

1-s 1P1s

Each iP is isomorphic to 1P and so has order )1( -knP . Also elements in distinct 'iP s

influence nonoverlapping sets of integers and hence commute \T is a subgroup kpS .

Since P Pj = (e) if 0 i j p-1, we see that O(T) = o( 1P )p = )1( -kpnp . Since ps

= e and i-s 1Pis = ip we have

i-s T s = T. Let P = { js t : t T, 0 j p - 1}. Since s T and 1-s T s = T,

we have T is a subgroup of kpS and o(P) = po(T) = p. )1( -kpnp p = 1)1( +- pkpnp .

Also n(k-1) = 1 + p + .. + 2-kp , hence n (k-1) + 1 = (1+ p+ 2p + .. + 2-kp ) p + 1 = p + 2p + 3p + .. + 1-kp + 1 = 1 + p + 2p + .. + 1-kp = n(k) Therefore o(P) = nkp and so P is a p Sylow subgroup of kpS . Definition Let G be a group and A and B be two sub-groups of G. If x, y G, we define x ~ y if y = a x b for some a A and b B.


17 Algebra Lemma The relation ~ defined above is an equivalence relation and the equivalence class of x G is the set

A x B = {a x b :a A ,b B} Proof : (i) Since x = x e x " x G, x ~ x " x G

(ii) If x ~ y and y = a x b. Then x = 1-a y 1-b \Therefore y ~ x. (iii) If x ~ y and y ~ z then

y = a x b for some a A, b B and z = 1a y 1b for some 1a 1A , 1b B. Then z = 1a (axb) 1b = ( 1a a) x (b 1b ) Where 1a a A and b 1b B. \ x ~ z. \~ is an equivalence relation on G. The equivalence class of x is {y : y =a x b for some a A, b B} = A x B. Lemma If A and B are finite subgroups of G, then

O(A x B) = )(

)().(1- xBxAo

BoAo

Proof Define T : A x B A x B 1-x by (a x b) T = a x b 1-x

First we show T is one-one. Let (a x b) T =( 1a x 1b ) T.

Then ax b 1-x = 1a x 1b 1-x .

\ a x b = 1a x 1b \ T is one-one. Next we show that T is onto. Let a x b 1-x A x B 1-x . Then a x b A x B. and (a x b) T = a x b 1-x . Therefore T is onto. \ o(A x B) = o (A x B 1-x ) ------- (1) But as A and x B 1-x are subgroups of G,

o(A x B 1-x ) = )()().(

1

1

-

-

xBxAoxBxoAo ------- (2)

The map : B x B 1-x defined by (b) = x b 1-x " b B is one-one and onto. \o(x B 1-x ) = o (B) ------ (3) From (1), (2), (3) we get

O(AxB) = )(

)()(1- xBxAo

BoAo


Group Theory 18 Definition Two subgroups A and B of a group G are said to be conjugate if $ x G such that A = x B 1-x . Theorem (Second Part of Sylows Theorem) If G is a finite group, p a prime and np | o(G) but 1+np o(G), then any two subgroups of G of order np are conjugates. Proof : Let A and B be subgroups of G each of order np . G is the union of disjoint double cosets of A and B. G = A x B. By Lemma

o(A x B) =)(

)().(1- xBxAo

BoAo

If A x B 1-x , for every x G, then o (A x B 1-x ) = mp where m < n

o(A x B) = mpBoAo )().( = m

n

pp2 = mnp -2 and 2n m n + 1.

Since 1+np | o(A x B) for every x and since o(G) = o(A x B) we get the contradiction 1+np | o(G). Therefore A= g B 1-g for some g G. The theorem is proved. Lemma

The number of p-Sylow subgroups in G equals ))((

)(PNo

Go where P is any p Sylow

subgroup of G. In particular, this number is a divisor of o(G). Proof: We have the normaliser of a subgroup H of G is defined by N(H) = {x : x H = H x} = {x : x H 1-x = H} is a subgroup of G. Since the number of distinct conjugates x H 1-x of H in G is the index of N(H) in G, the proof of the lemma follows as p-Sylow subgroups are conjugate. Theorem (Third part of Sylows Theorem) The number of p-Sylow subgroups in G, for a given prime is of the form 1 + kp. Proof : Let P be a p-Sylow subgroups of G. We decompose G into double cosets of P and P then G = P x P


19 Algebra By Lemma

o(P x P) = )o(P

[o(P)])(

)().(1

2

1 -- =

xPxxPxPoPoPo

Therefore if P x P 1-x P, 1+np | o(P x P). Where o(P) = np . \ if x N (P) then P x P = P(Px) = 2P x = Px and so o(P x P) = np . Now o(G) =

)(PNxo(P x P) =

)(PNx o(P x P) ------ (1)

where each sum runs over one element from each double coset. However, if x N(P), since P x P = P x ,

)(PNx o(P x P) =

)(PNx o(P x ) = o(N(P)) ------ (2)

If x N(P), 1+np | o (P x P) and hence

)(PNx o(P x P) = 1+np u

Therefore o(G) = o(N(P)) + 1+np u Therefore

))((

)(PNo

Go = 1 + ))((

1

PNoupn+ ------- (3)

Now as N(P) is a subgroup of G, o(N(P)) | o(G) and hence ))((

)(PNo

Go is an ingeter.

Also since 1+np o(G), 1+np cannot divide o(N(P)). But then ))((

1n

PNoup + must be

divisible by p. Therefore we can write))((

1

PNoupn+ as kp where k is an integer. Therefore (3)

becomes

))((

)(PNo

Go = 1 + kp

2.3 LET US SUM-UP In this lesson we have discussed the following concepts : (i) If G is a finite group, p is prime and )(Go

Pn , then G has a sub-group of order nP .

(ii) kps has a psylow sub-group. (iii) Conjugacy of p-sylow sub-groups. (iv) Number of p-sylow sub groups in a given finite groups.


Group Theory 20 2.4 LESSON-END ACTIVITIES 1. Prove sylows theorem (three different parts). 2. Determine the p-sylow subgroups of a given group of finite order. Business 2.5 REFERENCES

1) Topics in Algebra by I.N. Herstein Second Edition Chapter - 2.


UNIT I GROUP THEORY 21 LESSON -3 DIRECT PRODUCTS CONTENTS: 3.0 Aims and Objectives 3.1 Introduction 3.2 External and internal direct products. 3.3 Let us sum-up 3.4 Lesson-end activities. 3.5 References 3.0 AIMS AND OBJECTIVES In this lesson we introduce two different products of groups and then we prove that they are isomorphic. After going through this lesson, you will be able to : (i) Construct a new group based on two given groups (ii) Define external and internal products of a group G. (iii) Prove the isomorphism between the two products. 3.1 INTRODUCTION Given two groups A and B, we obtain a new group G = AxB as the external direct product A and B. We show that G can also be obtained from internal construction. We obtain G as internal direct product of two normal subgroups A and B of G. Finally we prove the isomorphism between the two products.


22 Algebra 3.2 EXTERNAL AND INTERNAL DIRECT PRODUCTS DIRECT PRODUCTS Theorem Let A and B be any two groups. In A x B = {(a,b) : a A, b B} we define a binary operation by ( 1a , 1b ) ( 2a , 2b ) = ( 1a 2a , 1b 2b )" ( 1a , 1b ) A x B and ( 2a , 2b ) A x B. Then A x B is a group under this binary operation. Proof : Let ( 1a , 1b ) ( 2a , 2b ), ( 33 ,ba ) A x B Then ( 1a , 1b )[( 2a , 2b ) ( 33 ,ba )] = ( 1a , 1b ) ( 2a 3a , 2b 3b ) = ( 1a ( 2a 3a ), 1b ( 2b 3b )) = (( 1a 2a ) 3a , ( 1b 2b ) 3b ) \ Associativity is satisfied. Let e and f be the identify elements of A and B respectively. Then (e,f) A x B. If (a,b) A x B, then (a,b) (e,f) = (ae, bf) = (a,b) by identity in A and B. Similarly (e,f) (a,b) = (a,b) Therefore (e,f) is the identity element of A x B. Let (a,b) A x B where a A and b B Let 1-a be the inverse of a in A and 1-b be the inverse of b in B. They ( 1-a , 1-b ) A x B and (a,b) ( 1-a , 1-b ) = (e,f) also ( 1-a , 1-b ) (a,b) = (e,f) Therefore ( 1-a , 1-b ) is the inverse of (a,b) \ A x B is a group. It is called the external direct product of the groups A and B. Definition Let 321 ,, GGG , ., nG be any n groups. Let G = 1G x 2G x . x nG = {( 1g , 2g ,.., ng ) : ig iG }. Define a product in G by ( 1g , 2g ,.., ng ) (

11g ,

12g ,

. 1ng ) = ( 1g1

1g , 2g 21g , ., ng

1ng ). G is group under this binary operation. It is

called the external direct product of 1G , 2G , . nG . This is the generalization of external product of 2 groups seen above. Definition If G is a group and 1N , 2N , ., nN are normal subgroups of a group G such that (i) G = 1N 2N nN = { 1h , 2h . nh : ih iN = 1, 2, n} (ii) Given g G then g is expressed as g = 1m , 2m .. nm , im iN in a unique

way, then we say G is internal direct product of 1N , 2N , . nN .


Group Theory 23 Theorem If G is the external direct product of the groups A and B, then G is internal direct product of A and B where A = { (a,f) : a A } B = { (e,b) : b B}

e, f being the identity elements of A and B respectively and A ~ A , B ~ B .

Proof: Define : A A by (a) = (a,f), for all a A. Then

( 1a ) = ( 2a ) ( 1a ,f) = ( 2a ,f) 1a = 2a Therefore is one-one. Let (a,f) A . Then a A and (a) = (a,f) Therefore is onto. Also if 1a , 2a A, ( 1a 2a ) = ( 1a 2a , f) = ( 1a ,f) ( 2a ,f) = ( 1a ) ( 2a ) Therefore is a homomorphism.

\ A ~ A

Similarly B ~ B

Let ( 1a ,f), ( 2a ,f) A Then. ( 1a ,f) ( 2a ,f)

1- = ( 1a ,f) ( 2a1- ,f)

= ( 1a 2a1- , f) A

Therefore A is a subgroup of G. Also ( 2a ,f) ( 1a ,f) ( 2a ,f)1- = ( 2a 1a 2a

1- ,f) A Therefore A is a normal subgroup of G. Similarly B is a normal subgroup of G. Now we show that G = A B Let g = (a,b) G. Then a = (a,f) A , b = (e,b) B and (a,b) = (a,f) (e,b) = a .b The representation is unique since if g is also equal to x y where x = (x,f), y = (e,y) Then g = (x,f) (e,y) = (x,y). But g = (a,b) \ x = a, y = b. \ The representation is unique. Therefore G is internal direct product of A and B .


24 Algebra Lemma Suppose G is internal direct product of 1N , 2N , .. nN . Then for i j, iN jN = {e} and if a . iN , b jN then ab = ba. Proof: G is internal direct product of 1N , 2N , .. nN Then 1N , 2N , .. nN are normal subgroups of G. Let i j and x iN jN . Then x iN and x jN . As x iN , x = 1e 2e . 1-ie x 1+ie . je .. ne ------ (1) Where te = e and x is viewed as an element of iN As x jN , x = 1e 2e . ie . 1-je x 1+je . ne ------ (2) Where te = e and x is viewed as an element of jN But every element g G has a unique representation of the form g = 1m 2m . nm , im iN . \ The two decompositions (1) and (2) are equal. In the first decomposition i th entry is x and in the second decomposition ith entry is e. Therefore x = e. \ iN jN = {e} Let a iN , b jN and i j. Then ab 1-a jN as jN is normal subgroup

\ ab 1-a 1-b jN ------ (3) Similarly since 1-a iN ,b

1-a 1-b iN . \ ab 1-a 1-b iN ------ (4) From (3) and (4), ab 1-a 1-b iN jN .

\ ab 1-a 1-b = e \ ab = ba. Theorem Let G be a group and G is the internal direct product of 1N , 2N , .. nN . Let T = 1N x 2N x . x nN be the external direct product of 1N , 2N , .. nN Then G and T are isomorphic. Proof: G = { 1b . 2b . nb : ib iN } T = {( 1b , 2b . nb ) : ib iN } Define y : T G by y (( 1b , 2b . nb )) = 1b 2b . nb where each ib iN .

If yG, then y = 1b 2b . nb , ib iN . Let x = ( 1b , 2b . nb ). Then xT and y (x) = Y (( 1b , 2b . nb ))

= 1b . 2b . nb = y


Group Theory 25

\ Y is onto. Let Y (x) = Y (y) where x = ( 1a , 2a ,.., na ) and y = ( 1b , 2b . nb ). Then Y (( 1a , 2a ,.., na )) = Y (( 1b , 2b . nb )) \ 1a 2a . na = 1b 2b ,. nb . \ 1a = 1b , 2a = 2b , . na = nb . \ x = y \ Y is one-one. Also, Y (xy) = Y ( 1a 1b , 2a 2b , ., na nb ) = 1a 1b 2a 2b na nb = 1a 2a .. na 1b 2b .. nb (as ia jb = jb ia ). = Y (x) Y (y) \y is a homomorphism i, G N T. 3.3 LET US SUM-UP 1. Let A and B be any two groups. In AxB we define a linary operation by ( 1a , 1b ) ( 2a , 2b ) = ( 1a 2a , 1b 2b ). Then AxB is a group under this linary operation. 2. If G is a group and 1N , 2N are normal subgroups of G such that G = 1N 2N = { 1h 2h / 1h 1N , 2h 2N } and every element of G can be uniquely expressed as a product of elements of 1N and 2N then G is the internal direct product of 1N and 2N . 3. If G = AxB then A and B are normal subgroups of G 4. External and internal direct products are isomorphic. 3.4 LESSON END ACTIVITIES 1. Obtained the external direct product of two groups A and B. 2. If G is the internal direct product of 1N , 2N , . nN then prove that for i j, iN jN = {e} and if a jN then ab = ba. 3. If G is the internal direct product of 1N , 2N , . nN and T is the external direct product of 1N , 2N , . nN then G and T are isomorphic. 3.5 REFERENCES

1) Topics in Algebra by I.N. Herstein Second Edition.


Unit II RING THEORY

26 Algebra

LESSON -4 EUCLIDEAN RINGS

CONTENTS 4.0 Aims and Objectives 4.1 Introduction 4.2 Euclidean Rings 4.3 A particular Euclidean Ring 4.4 Let us sum-up 4.5 Lesson end activities.

4.6 References

4.0 AIMS AND OBJECTIVES

In this lesson we introduce a particular class of rings namely euclidean rings and we discuss some important properties of euclidean rings.

After going through this lesson, you will be able to:

(1) Define euclidean ring.

(2) Derive euclidean ring is a principal ideal ring

(3) Determine the g.c.d of any two elements of a euclidean ring.

(4) Prove Unique Factorization Theorem.

(5) Describe the structure of maximal ideals in a euclidean ring

(6) Prove Gaussian integers is an euclidean ring.

(7) Prove Fermats Theorem.

4.1 INTRODUCTION

The class of rings we propose to study now is motivated by several existing examples- the ring of integers, the Gaussian integers and polynomial rings. Euclidean rings constitute a special category of rings, in which there is defined a nonnegative d value (an integer) for each


Ring Theory 27

and every element of the ring. After introducing some important properties of this particular Euclidean ring, we shall prove the Unique Factorization Theorem, the structure of maximal ideals, and Fermats Theorem. 4.2 EUCLIDEAN RINGS

Definition: An integral domain R is said to be a Euclidean ring if for for a 0 in R there is defined a non-negative integer d (a) such that

1. for all a, b R , both non zero, d (a) d (ab)

2. for any a,b R, both non zero, there exist t, r R such that a = tb + r where either r = 0 or d (r) < d (b).

Examples : 1. Any field F, with d(a) = 1 for all a F - {0}. Then is an Enclidean domain

For a.b F {0} then d(a) = 1, d(ab) = 1 So that d (a) d (ab) Let a, b F with b 0. Them bexits and a = qb + r where q = ab , r = 0 \ F is an Euclidean domain.

2. The ring of Gaussion integer Z [i] is an Euclidean domain

Z [i] = {a + ib /a, b Z}. For any a + ib 0 in Z [i], define d(a + ib) = a + b Clearly d (a + ib) is a non negative integer.

(i)Let x = a + ib, y = m + in be any two non zero elements of Z [i. Then

=d (xy) = d [( a + ib) (m + in ) ] = d [(am bn) + I (bm + an)]

=(am bn) + (bm + an )

= (a + b) (m + n )

a + b (Q m + n 1).

= d (x)

\d (x) d (xy)


28 Algebra

(ii) Let x,y Z [i] with y 0. To show that there exists t, r z[i] such that x = ty +r where either r = 0 or d ( r) < d(y)

Let x = a + ib, y = m + in (m, n 0)

Then yx =

inmiba

++ =

n2m2)inm)(iba(

+

-+ = p + iq, where

P = n +

+m

bnam q = n +

-m

anbm

Choose integers , m , such that

| - p| 21 , | - q |

21 (1)

Put t = + i and r = x ty

, m Z [i], r Z [i], Also x = ty + r

To prove that either r = 0 or d(r) < d(y)

If r 0 then d(r) = d(x - ty)

ytyxd

-=

= d {[(p +iq)] ( + i m ] } (m + in)

= ( ) ([ )22 ml -+- qp ( )22 nm +

( )22 nm +

+

41

41 from (1)

< 22 nm +

= d(y)

d(r) < d(y)

Hence Z[i] is an Euclidean domain.


Ring Theory 29

Definition: An integral domain R with unit element is a principal ideal ring if every ideal A in R is of the from A = (a) for some a R. An integral domain R is called a Principal Ideal Domain, if every ideal of R is a principal ideal. THEROREM: Let R be a Euclidean ring and let A be an ideal of R. Then there exist an element a0 A such that A consists exactly of all xa0 as x ranges over R.(Every Euclidean ring is a Principal ideal Domain)

PROOF: First we prove that every ideal in R is generated by an element of R.

(ie) If A is an ideal in R, then there exists an element a0 in R such that A = < 0a >={r 0a /rR}.

If A is the zero ideal in R, then A= < 0 >

Hence we assume A {0}.

Let 0d = min {d(a)/a A,a 0} and 0a be an element of A with d( 0a ) = 0d (By well ordering principle)

To show that A = < 0a >

Now for any chosen element a A there exists q,r R

Such that a = q 0a + r where either r = 0 or d(r) < d ( 0a )

\ r = a | q 0a

Now, a A, q 0a A (QA is an ideal)

a - q 0a A

r A

If r 0 then d(r) < d (a0 )

Also r A, r o d (r) < d0

\ d (r) < d (a0 ) contradicts d (a0 ) being the least element ofd0

Thus r = 0.


30 Algebra

\ a =q a0 (a0 )

and hence A (a0 )

Again b (a0 ) b =a0 r for some r R,

Also, a0 A a0 rA (Q A is an ideal)

b A

\ (a0 ) A A = (a0 )

COROLLARY : A Euclidean ring possesses a unit element.

PROOF: Let R be Euclidean ring.

We know that every ideal of Euclidean domain is principal R itself is an ideal of R.

\ R is a principal ideal.

Hence there exists u0 R such that R = ( u0 )

Thus every elements in R is a multiple of u0

In particular u0 = u0 c for some c R.

We show that c is the unit element of R

If a R then a = x u0 for some x R.

Hence ac = (x u0 ) = x ( u0 c) = x u0 = a \ c is a unit element of R.

Definition: If a 0 and b are in a Commutative ring R then a is said to divide b if there exists c R such that b = ac.

We write a/b to mean a divides b.

REMARKS:

1. If a/b and b/c then a/c

2. If a/b and a/c then a / ( b c )

3. If a/b then a/bx for all x R

Definition: An integral Domain R is called a Principal ideal Domain ( P.I.D) if every ideal of R is a principal ideal.


Ring Theory 31 Examples:

1. Every field is a P.I.D. For, the only ideal of a field F are {0} and F itself.

2. Since every ideal of Z is of the from nZ for some nZ, each ideal is a principal ideal. \ z is a P.I.D.

Definition: If a, b R then d R is said to be a greatest common divisor of a and b if.

1. d /a and d /b

2. whenever c /a and c /b then c /d

Greatest common divisor of a and b is denoted by ( a,b )

Example: In Z,g.c.d of 18 and 24 is 6; another g.c.d is - 6.

LEMMA: Let R be Euclidean ring. Then any two elements a and b in R have a greatest common division d. Moreover, d = a + b for some , R.

PROOF: Let A = {ra + sb /r,s R }.

We claim that A is an ideal of R,For, suppose x,y A.

\ x = 1r a + 1s b, y = 2r a + 2s b and x y = ( 1r 2r ) a + ( 1s 2s )b A.

Similarly, for any u R, ux ( 1r a + 1s b) = (u 1r ) a + (u 1s ) b A

Thus A is an ideal of R.

Now, by theorem there exists an element d A such that A =. As every element of A is of the form ra + sb, for some r,s R, there exists element ,R such that d = a +b.

Now by corollary to theorem R has a unit element 1.

Thus a = 1. a + ob A: b =0.a + 1.b A

As every element of A is a multiple of d, it follows that d/a and d/b.

To show that d is a G.C.D. of a and b, suppose c /a and c /b

Then c / a and c/ b. This means c/ a + b (ie) c/d

Thus (a,b) =d = a + b

Definition: Let R be a Commutative ring unity. An element a R is a unit in R, if there exits an element bR such that ab = 1.


32 Algebra

Note: A unit in a ring is an element whose inverse is also in the ring. Also note that unit is different from unity.

LEMMA: Let R be integral domain with unit element and suppose that for a, b, R both a/b and b/a are true. Then a =ub,where u is a unit in R.

PROOF: Let R be an intergral domain and a,b R

Given that a/b and b/a.

a/b b = xa for some x R

b/a a = yb for some y R

Thus b = xa

= x (yb) = (xy) b

(ie) b = (xy) b b [1 - xy] = 0

Since R is an integral domain 1 xy = 0 xy =1

Thus y is a unit in R.

Definition: Let R be a Commutative ring with unit element. Two elements a and b in R are said to be associates if b = ua for some unit u in R.

LEMMA : Let R be a Euclidean ring and a,bR If b 0 is not a unit in R then d(a)

Ring Theory 33 Since R is an integral Domain we obtain.

1 bx = 0 bx = 1.

be is the unit in R which is a Contradiction.

\ Our assumption d (a) = d (ab) is wrong.

\ d (a) < d (ab).

Definitions: In the Euclidean ring R a non-unit pi is said tobe a prime element of R, if whenever pi = ab, where a, b are in R then on e of a or b is a unit in R.

LEMMA : Let R be a Euclidean ring. Then every element in R is either a unit in R or can be written as the product of a finite number of prime elements of R.

PROOF

Step 1: state and prove the previous lemma.

Step 2: An element a 0 is a unit d (a) = d(1) If part,

Suppose a is a unit. Then there exists b R such that ab = 1.

\ d (a) d (ab) = d (1) . (1)

By definition of an Euclidean domain

d (1) d (1.a) = d (a) a 0 in R

\d (a) d (1) .. (2)

From (1) & (2), d (1) = d(a)

Converse

Suppose d (a) = d (1)

Consider 1 abd a. Since R is Euclidean there exists q,r R such that 1 = qa + r, where either r = 0 or d (r) < d(a).

If r 0 then d (r)

34 Algebra

Step 3: Let a be non zero element of R.Proof is by induction on d (a).

If d (a) = d(1), then a is a unit (by step 2). Hence the lemma is true.

Let d (a) d (1)

Assume the lemma to be true for all elements x in R such that d (x) < d (a)

Suppose a is a prime element, there is nothing to prove. Suppose a is a not a prime element then there exists b R, c R non units such that a = bc.

By lemma, Since b and c are non- units d(b) < d (bc) = d (a) and d (c) < d (bc) = d (a).

Thus b,c R are such that d(b) < d (a) and d (c) < d (a).

\ By induction hypothesis b and c can be written as a product of a finite number of prime elements of R.

b = 1p 2p np , c = 1'p 2'p m'p where 'p s and

,'p s are prime elements of R.

\ a = bc = ( 1p 2p .. np ) ( 1'p m'p )

A product of a finite number of prime elements

This completes the proof.

Definition: In the Educlidean ring R, a and b in R said to be relatively prime if their greatest common divisor is a unit in R.

(ie) (a, b) = 1.

LEMMA : Let R be a Euclidean ring Suppose that for a,b,c, R a/bc, but (a,b) = 1. Then a/c.

PROOF: By Lemma the greatest common divisor of a and b is d which is of the form l a + m b i.e, d = l a + m b Here G.C.D is 1 then l a + m b = 1

Multiplying this relation by c, l ac + m c = c Since a/bc, we have bc =xa for some xR.

\c = l ac + m xa

= ( l c)a +( m x)a (by associativty)

= ( l c + m x) a

(i.e.) c is a multiple of a. \ a/c.


Ring Theory 35

LEMMA : If pi is a prime element in the Euclidean ring R and pi/ab where a, b, R then pi divides atleast one of a or b.

PROOF: Result: Suppose p does not divide a. Sine p is prime and p soes not divide a. then p and a are relatively prime.

P is a prime element in R and p does not divide a.

\ (p,a) = 1

Proof of Lemma: Assume that pi does not divide a then (pi, a) = 1 by lemma 2.7.5, pi must divide b.

Similarly pi does not divides b then (pi, b) = 1.

Q pi is a prime element, pi must divides a.

COROLLARY: If pi is a prime element in the Euclidean ring R and pi/ 1a 2a na then pi divides atleast one 1a 2a na .

PROOF: Let us prove the result using induction on n.

When n = 2 then p / 1a 2a

p / 1a or p / 2a (proved in the above lemma)

Assume the result for (n-1) products, If p does not divide 1a

then (p , 1a ) = 1

p / 1a 3a na

p divides any one ia (i 1) (by induction)

When (p , 3a ) = 1

then p / 2a 3a 4a na

p divides any one ia (i 2)

Thus p divides any one of ia .


36 Algebra THEOREM: UNIQUE FACTORIZATION THEOREM

Let R be a Euclidean ring and a 0 a non unit in R

Suppose that a = 1p 2p np

= 1'p 2'p . m'p

Where ip and 'j'p are prime elements of R. Then n =m and each ip 1 i n is an associate of some 'j'p ; 1 j m and conversely each k'p is an associate of some ip

PROOF: Given that a = 1p 2p np = 1'p 2'p . m'p

Where 1p , 2p , 'np , 1'p , 2'p , m'p prime elements of R

First let us prove n = m. To prove this we have to prove n m and m n.

Since 1p is a prime element. 1p is a divisor of 1p 2p . np .

(i.e.) 1p / 1p 2p np . But 1p 2p . np = 1'p 2'p .. m'p

\ 1p - 1'p 2'p . m'p

\ 1p divides atlest one 1'p 2'p m'p

By lemma 1p must divide 1'p .Since 1p 1'p are both prime elements of R and 1p / 1'p must be associates and 1'p = 1u 1p where 1u is a unit in R.

\ 1p 2p . np = 1'p 2'p . 'mp

= 1u 1p 2'p m'p (Q R is an integral domain canceling 1p on both sides)

2p np = 2'p 3'p m'p

Repeat the argument on this relation with 2p we get.

3p np = 1u 2u 3'p m'p

If m > n then after n steps L.H.S become 1 while the R.H.S. reduces to a product of a certain number of pi s.

But '1p'

2p '

mp are prime element of R.

So they are not units in R.

\ Product of some units and certain number of ,'p s cannot be equal to one.


Ring Theory 37 \ m cannot be greater than n. and \ m n ----- (1)

Similarly, interchanging the roles of ,'

jp s and ,

jp s.We get n m ----- (2)

Combining (1) and (2) n = m. Also in the above process we have shown that every jp is an associate of some

,

jp are associates of some jp . LEMMA : The ideal A = ( 0a ) is a maximal ideal of the Euclidean ring R if 0a is a prime element of R. PROOF : Let R be in an Euclidean ring. Step 1 : Suppose A = ( 0a ) is a maximal ideal, our aim is to prove 0a is prime in R.

Suppose 0a is not a prime element then A = ( 0a ) is not a maximal ideal. Let b, c R such that 0a = bc, neither b is a unit nor c is a unit. Let B = (b). Since 0a is a multiple of b from 0a = bc then 0a B and A B. We claim that A B and R. If B = R then 1B (Q R in the ring with unit element 1R) But every B is a multiple of b. \ 1 = bx for some x R. B is a unit in R, which is a contradiction. \B R. Now, assume that A = B, bB, bA \b = x 0a for some x R. But oa =bc and hence oa = x oa c xc = 1

c is unit in R. Which is a contradiction.

\ A B.

\ For any other ideal B R such that A B and A B, B R A cannot be maximal ideal of R.

\ oa is a prime element of R

Conversely, suppose oa is a prime element of R and U is an ideal of R such that A U R.

To prove A is maximal ideal it is enough to prove that


38 Algebra A = U or U = R.

Let oa A but A U

oa A U oa U.

But U = ( ou ) (by theorem)

Every element of U is a multiple of ou

\ oa also a multiple of ou

(i.e.) oa = x ou for some x R ..(1)

Q oa is a prime element it follows that either x or ou is a unit in R.

Case 1: ou is a unit in R, 1

0-u R.

For ou U and 1

0-u R

u 10-u U

1 U

Let r R,1 U then r.1 = 1. rU.

i.e., r R r U R U

Also U R \ R = U.

Case 2: x is a unit in R. Then its inverse 1-x R

The relation (1) becomes ou = 1-x 0a .

0a A, 1-x R. 1-x 0a A (Q A isan ideal)

(i.e) 0u A But 0u U U A.

Also A U A = U

\ A is the maximal ideal of R. Hence the proof.


Ring Theory 39

4.3 A PARTICULAR EUCLIDEAN RING

THEOREM : J[i] is a Euclidean ring. J[i] = {a + ib/a, bZ} called the set of Gaussian integers in an Euclidean domain.

PROOF :

Step 1: J[i] is an integral domain under usual addition and multilplication of complex numbers.

Let 1a + i 1b , 2a + i 2b J[i]

Then ( 1a + i 1b + 2a + i 2b ) = ( 1a + 2a ) + i ( 1b + 2b ) J[i]

( 1a + i 1b ) ( 2a + i 2b ) = 1a 2a - 1b 2b + i( 1a 2b + 2a 1b ) J[i]

\ J[i] is closed under addition and multiplication, since J[i] is a subset of complex number + and . are associate and commutative and distributive over +.

0 = 0 + 0i is the zero element 1 = 1 + 0i is the unit element (a + ib) + (-a ib) = 0 (i.e) (-a ib) is the additive inverse of (a + ib). \ J[i] is a commutative ring with unity. Let a + ib, c + id J[i] and let (a + ib) (c + id) = 0, a + ib 0

as a complex number a + ib 0 has its inverse 22 baiba

+-

\ 22))((

baibaiba

+++ (c + id) = 0

\ c + id = 0 \ J[i] is an integral domain. Step 2:

J [i] is an Euclidean domain

Define d as d (a + ib) = a + b

Q a + ib 0 atleast one of a or b is non-zero.

(a + b ) > 0.

In fact (a + b) 1 Q a, b are integers.

(i) d (x) 0 V x J [i]


40 Algebra

(ii) Let x = a + ib, y= c + id J [i]

xy = (a + ib) (c + id) = (ac - bd) + i (ad + bc)

d (xy) = (ac bd) + (ad + bc)

= (ac + bd + ad + bc) = (a + b) (c + d)

= d (x) d (y) \ d (x) d (xy) for y 0 (Qd (y) 1.)

(iii) To prove the division algorithm in J [i]

Given x, y J [i], y 0 $ q, r J [i] such that x = qy + r where either r = 0 or d (r) < d (y).

Case 1: Let = n (a + ve integer) and x = c+ id

Apply divition algorithm in the ring of integers to c and n there exists u and v such

that c = nu + v where u and v are integer satisfying |v| 21 n.

Similarly d = 1nu + 1v where 1u 1v are integers satisfying | 1v | 21 n.

\ x = c + id = n (u + 1iu ) + (v +i 1v )

\ x =qn + r where q = u + 1iu and r = v + i 1v

Here d (r) = d (v + i 1v ) < 4n2 +

4n2 =

21

n2 < d (n) \ If x = c + id and y = n $ q, r J [i] such that x =qn + r with d (r) < d (n) ---------- (1)

Case 2: Let x =c + id , y 0 are in J [i]

Let y = a + ib, then y-y = (a + ib) (a - ib) = a +b = n, a + ve integer.

Applying case 1 to -y x and n we get

-y x = qn + r with d (r)

41 Algebra

-y x - qy

-y = r with d (r)

Ring Theory 42 = (a + b) ( + g) \ (a + b) / p and hence a + b = 1,p, p.

Since a + ib is not unit in J [i],d (a + ib ) 1

If 22 ba + = 2p then 2f + 2y = 1 which is not possible since f + ig is not a unit in J[i].

\ a + b = p (i.e.) p = a sum of squares of two integers.

LEMMA : If p is a prime number of the form 4n + 1 then we can solve the congruence x - 1 (mod p).

PROOF: Since p = 4n + 1, 21p- is an even intger.

Let x = 1.2.3 21p- In this product there are n even number of terms.

By Wilsons therem, (p -1)! + 1 0 (mod p)

(i.e.)1.221p-

+

21p

+

23p . (p 1) - 1 (modp)(1)

Now 21p+ = p-

21p- -

-

21p (modp).

23p+ = p-

23p- -

-

23p (mod p).

.. ..

P -1 - 1 (mod p) \ (1) becomes

x.

--

2)1p(

--

2)3p( (-2) (-1) - 1 (mod p)

(i.e.) x. 1.223p- .

21p- . 2

1

)1(-

-p

- 1(mod p)

(i.e) x - 1 (mod p)

-=-

-

1)1( 21p

Q

Thus x - 1 (mod p)


43 Algebra Hence the proof. THEOREM (FERMAT). If p is a prime number of the from 4n + 1 then p = a + b for some integers a, b. PROOF: By Lemma there exists an integer x such that x = - 1 (mod p). We can choose x so that 0 x p -1.

If x > 2p put y = p x

Then y = ( p x) = p - 2px + x x (mod p) (i.e.) y = x (mod p) (Qx 1-x (mod p)

Since y = p x and x > 2p ,|y|

2p .

\ we can assume that |x| 2p and x - 1(mod p)

\ x + 1 is a multiple of p. x + 1 = mp, m is an integer.

Now mp = x +1 < 4p2 + 1 < p

\ m < p and p c \ By lemma there exists integers a, b such that p = a + b

4.4 LET US SUM-UP (1) Definition and examples of Euclidean ring (2) Proving every Euclidean ring is a principal ideal ring. (3) Determination of g.c.d of any two elements. (4) Unique Factorization Theorem for Euclidean rings. (5) Proving the Gaussian integers is an Euclidean ring. (6) Fermats theorem in number theory


Ring Theory 44 4.5 LESSON END ACTIVITIES

(1) Prove that a Euclidean ring possesses a unit element. (2) Prove that an ideal is a maximal ideal if and only if it is generated by a prime element. (3) Prove that if an ideal U of a ring R contains a unit of R, then U = R. (4) Find all the unit in J[i]. 1.7 REFERENCES

1) Topics in Algebra by I.N. Herstein Second Edition Chapter 3.


45 Algebra

LESSON -5 POLYNOMIAL RINGS

CONTENTS : 5.0 Aims and Objective 5.1 Introduction 5.2 Poly nomial rings 5.3 Poly nomial over the Rational Field 5.4 Let us Sum-up 5.5 Lesson end activities

5.6 References

5.0 AIMS AND OBJECTIVES

In this lesson we introduce another class of rings namely polynomial rings. We prove some important properties of polynomial rings and we conclude the lesson by providing an important theorem to determine the irreducibility of polynomials. After going through this lesson, you will be able to: (1) Define the structure of polynomial rings (2) Prove a polynomial ring is an Euclidean ring (3) Prove the Eisenstein Criterion for irreducibility of polynomials. 5.1 INTRODUCTION Let F be a field. The ring of Polynomials in the indeterminate, x, we mean the set of all symbols oa + 1a x +. + mm xa , where m can be any nonnegative integer and coefficients

oa , 1a .. ma are all in F. This ring is denoted by F[x]. We introduce the basic algebraic operations to claim that F[x] is a ring. After proving the Division Algorithm in F[x], we will show that F[x] is a Euclidean ring. Finally by proving Gauss Lemma, we shall establish a powerful tool for verifying irreducibility (Factorization) of a polynomial. 5.2 POLYNOMIAL RINGS Definition: If p (x) = oa + 1a x +. + mm xa and


Ring Theory 46 q(x) = ob + 1b x +. +

nn xb are in F[x], then p (x) = q(x) if and only if for every integer i 0,

ia = ib .

Definition : If p (x) = oa + 1a x +. + mm xa and q (x) = ob + 1b x +. +

nn xb are both in F[x] then

p (x) + q (x) = 1c + 1c x + + 1ctx where for each i, ic = ia + ib .

Definition : If p (x) = oa + 1a x +. + mm xa and q (x) = ob + 1b x +. +

nn xb then p (x) q (x) = 0c + 1c x + + kc

kx where

tc = ta ob + 1-ta 1b + + oa tb

Example:

Let p(x) = 1 + x - x, q (x) = 2 + x + x

Here oa = 1, 1a = 1, 2a = -1, 3a = 4a = .. = 0

ob = 2, 1b = 0, 2b = 1, 3b = 1, 4b = 5b = = 0

Thus oc = oa ob = 1.2 = 2 1c = 1a ob + oa 1b = 1.2 +1.0 = 2

2c = 2a ob + 1a 1b + oa 2b = (-1) 2 + 1.0 + 1.1 = - 1 3c = 3a ob + 2a 1b + 1a 2b + oa 3b = 0 (2) + (-1)(0) + 1.1 + 1.1 = 2 4c = 4a ob + 3a 1b + 2a 2b + 1a 3b + oa 4b = 0 (2) + 0 (0) + (-1) (1) +(1)(1) + (1)(0)= 0 5c = 5a ob + 4a 1b + 3a 2b + 2a 3b + 1a 4b + oa 5b = 1

6c = 6a ob + 5a 1b + 4a 2b + 3a 3b + 2a 4b + 1a 5b + oa 6b = 0

7c = 8c = . = 0

\ Accroding to our definition,

(1+ x-x)(2+ x+x x) = oc + 1c x +.=2+2x- x + 2x - 5x

Definition: If (x) = oa + 1a x + + nn xa 0 and na 0 then the degree of (x) written as deg (x), is n.

LEMMA : If (x), g (x) are two non-zero elements of (x) then deg ( (x) g(x)) =deg (x) + deg g (x)


47 Algebra

PROOF: Suppose (x) = =

m

0i

ii xa = oa + 1a x

mm xa and

g(x) = =

n

i 0

ij xb = ob + 1b x++ nn xb with ma 0 and nb 0.

This mean deg (x) = m and deg g (x) = n

We have (x) g(x) = oc + 1c ++ k

k xc where

ic = oa ib + 1a 1-ib +. + ia ob . Now mnc + = oa mnb + + 1a 1-+mnb + . + 1-na 1+mb + na mb + 1+na 1-nb + + 0ba mn+

= na mb since ai = 0 v i> n and jb = 0 j>m.

If i> n+ m, then ic = sum of term of the from jijba -

\ i> n + m i = j + (i-j) > n + m.

j > n or i- j > m

ja = 0 or jib - = 0

ja jib - = 0

ic = 0

\ (x) g(x) = 0c + 1c x + .. + mnc +mnx +

deg ((x) g(x)) = m + n = deg (x) +deg g (x).

COROLLARY: If (x), g(x) are non zero elements in (x) then deg (x) deg (x)g (x)

PROOF: From the above lemma,

deg [ (x) g (x)] = deg (x) + deg g(x) If deg g(x) 0 then deg (x) deg [ (x) g (x)]

Corollary : F[x] is an integral domain.

Proof : We have to show that F[x] is a commutative ring.


Ring Theory 48

If (x)= =

n

k 0

kkca g(x) =

=

m

k 0

kk xb are in F [x]

define (x) + g(x) = +

=

mn

k 0

( ka + kb ) kx

(x) g(x) = +

=

mn

k 0

kk xc where kc =

=0iia ikb -

Since all but a finite number of s'ka and s'kb are zeros it follows that only a finite number of

ka + kb can be non-zero and only a finite number of 'kc s can be non zero.

Hence (x) + g(x) F [x] and (x) g(x) F[x] With addition and multiplication defined on F[x]. We show that F [x] is a commutative ring with unity We can verify that [f(x) g(x)] h(x) = f(x) [g(x) h(x)] and [f(x) + g(x)] = f(x) h(x) + g(x) h(x) for any f(x), g(x), h(x), F[x]. F[x] inherits commutativity from F and 1F treated as a constand polynomial is the unity in F (x). Thus F(x) is a commutative ring with unity. Now suppose (x) g (x) = 0, (x),g(x) F [x] If both (x)and g(x) are constant polynomials then from the fact that F is an integral domain it follows that either (x) = 0 or g(x)= 0. If one of the polynomials say (x) is non constant, then (x) g(x) = 0 only if g(x) = 0. Thus F(x) has no zero divisors and hence F [x] is an integral Domain. LEMMA : (The Division Algorithm) Given two polynomials (x) and g(x) o in F[x] then there exist two polynomials t(x) and r (x) in F(x) such that (x) = t(x) g(x) + r(x) where r (x) = 0 or deg r(x) < deg g(x). PROOF: By induction on degree of (x)

1. If deg (x) < deg(x),take t(x) = 0,r(x) = (x) we get the result. 2. Assume deg (x) deg g (x)

Let (x) = 0a + 1a x +. +

mm xa , ma 0

g(x) = 0b + 1b x +.+n

n xb , nb 0 and m n suppose m = 1,if n = 0,g(x) = ob , ob 0 a unit


49 Algebra

\ (x) = 0a + 1a x = ob ( oa

10

-b + 1a 10b - x) and r (x) = 0

(post multiply by 10-b & premultiply by ob )

If n = 1, g(x) = 0b + 1b x , 1b 0 \ (x) = 0a + 1a x = 1a 10b - ( 0b + 1b x )+( oa - 1a ob 10b - )

(x) = g(x) t(x) + r(x) and r = oa - 1a ob 10b - with degr r (x) = 0

Ring Theory 50

We show that d is an Euclidean function on F [x]

We know that deg (x) is a non-negative integer

(i) Let (x) and g(x) be any two non-zero elements in F[x] Then d[(x) g(x)] = deg [(x) g(x)]

= deg (x) + deg g (x) deg (x) (Q deg g(x) > 0)

= d[ (x) ] \ d[(x)] d[ (x) g(x) ]

(i) By division algorithm, if (x),g (x), F[x] with g (x) 0 then there exist g(x), r(x), F[x] such that

(x) = q (x) g(x) + r(x) where either r(x) = 0 or deg r(x)< deg g(x) (i.e) either r(x) = 0 or d [r(x)] < d [g(x)]

Thus d satisfies all the properties of an Euclidean ring. \ F[x] is an Euclidean domain. LEMMA: F(x) is a principal ideal ring or principal ideal domain. PROOF: F(x) is an Euclidean ring, every Euclidean ring is a principal ideal ring. LEMMA 2.9.4: Given two polynomials (x),g(x) in F(x) they have a greatest common divisor d(x) which can be realized as d(x) = (x) (x) + (x) g(x) PROOF: Let I = { s(x) (x) +t(x) g(x)/ s(x), t(x) F[x] } Then I f since (x) I Let d(x) be a polynomial of least degree in I. Then d(x) = l (x) (x) + (x) g(x) for some l (x) (x) F[x] We show that d(x) is a g.c.d of (x) and g (x)

Now by division algorithm $ q(x),r(x), F[x] such that (x) = q(x) d(x) + r(x), Where either r(x) = 0 or deg r(x)

51 Algebra

If r(x) 0 then deg r(x) < deg d(x) But r(x) I and d(x) is a polynomial of least degree in I. Hence we get a contradiction. \ r(x) = 0 \ d(x) / (x) Similarly d(x)/ g(x)

Again, c(x)/ (x), c(x) /g (x) c(x)/[ l (x) (x) + m (x) g(x)] c(x)/ d(x) \ d(x) is a g.c.d of (x) and g (x) Hence the proof. Definition: A polynomial p(x) is said to be irreducible over F if whenever p(x) = a(x) b(x) with a(x),b(x) F[x], then one of a(x) or b(x) has degree 0 (ie a constant). Remark: Note that in the above definition we say irreducible over R and not just irreducible .A polynomial which is irreducible over a domain R may be reducible in another domain R For example, x + 1 is irreducible over R, but is reducible over C (x + 1 = (x+i) (x- i)) LEMMA : Any polynomial in F(x) can be written in a unque manner as a product of irreducible polynomials in F(x). PROOF: Let deg (x) = n 1 Proof is by induction on n.

If n = 1, then (x) is irreducible over F and hence there is nothing to prove. Assume that the theorem is true for any non-constant polynomial of degree less than n.. Let (x) be a polynomial of degree n If (x) itself is irreducible over F, then there is nothing to prove Otherwise let (x) = g(x)h(x), where 1 deg g(x)

Ring Theory 52 Where ip (x) and )x(q j are irreducible polynomials over F. Now 1p (x) divides 1p (x) . )x(pr \ 1p (x) divides 1q (x) sq (x) Hence 1p (x) divides some )x(q j

1p (x) divides 1q (x)(say) \ 1q (x) = 1u 1p (x) where 1u is a unit

\ 1p (x) 2p (x). )x(pr = 1u 1p (x) 2q (x) )x(qs

Canceling, 1p (x) on both sides,

2p (x) )x(pr = 1u 2q (x). )x(qs

Repeat the process with 2p (x). Proceeding after r steps we arrive at

1 = 1u 2u . 1+rr qu (x) . )x(qs Q )x(q j are irreducible, we must have r = s so that we get 1 = 1u ru ru Hence the proof. LEMMA : The ideal I = (p(x)) in F[x] is a maximal ideal if p(x) is irreducible over F. PROOF : If part: suppose I = (p(x)) is a maximal ideal. Let p(x) be reducible over F. Then p(x) = (x) g(x), where 1 deg (x) < deg p(x) and 1 deg g(x) < deg p(x). Consider the ideal J = ((x))

We prove that I # J

# F [x]

Now, a(x) I a(x) = p(x) h(x)for some h(x) F(x) a (x) = (x) g(x) h(x)

a (x) = (x)[ g(x) h(x)] a (x) J \ I J

Clearly, J F [x] Suppose I = J, then (x) J (x) I


53 Algebra

\ (x) =p(x) r(x) F[x]

(ie) (x) = (x)[g(x) r(x)] or 1 = g(x) r (x) \ g(x) is of degree 0 which is a contradiction. \ I J Suppose J = F[x] Then 1F[x] 1J \ 1 = f(x) s(x) for some s(x) F[x] \ f(x) is of degree 0 which is a contradiction \ J F [x] Thus, I

# J

# F[x]. This contradicts I being a maximal ideal in F[x]

\ p(x) must be irreducible over F. Only if: Suppose p(x) is irreducible over F. To prove I is a maximal ideal in F[x] Let J be an ideal in F[x] such that I

# J

# F[x]

We prove J = F[x] F[x] is a P.I.D and J is an ideal in F[x]. \J is a principal ideal. J = (f(x)) (say) for some for f(x) F [x] with f(x)I (If f(x)I, then J = 1) Now I

# J p(x)J

p(x) = f(x) g(x) for some g(x) F[x] But p(x) is irreducible over F. \ either f(x) or g(x) is of degree 0. If g(x) is of degree 0, then it is a constant polynomial. (ie) it is unit. \ f(x) = p(x) [g(x)] 1- I which is a contradiction. \ f(x) must be of degree 0 and hence it must be a unit. Since f(x) J, it follows that J = F[x]. Thus I is a maximal ideal in F[x]. 5.3 Polynomials over the Rational field. Definition: The Polynomial (x) = oa + 1a x ++ na xn , where oa , 1a na are integers is said to be primitive, y the greatest common divisor of oa , 1a na is 1. Lemma: If f(x) and g(x) are primitive polynomials then f(x) g(x) is a primitire polynomial.

Proof: Let f(x) = oa + 1a x +. + na xn : g(x) = ob + 1b x +. + mb xm . Suppose the lemma is false, then all the coefficients of f(x) g(x) is divisible by some integer larger than 1 and


Ring Theory 54 hence by some prime number p. Since f(x) is primitive, p does not divide some coefficient

ia . Let ja be the first coefficient of f(x) which is not divisible by p. Similarly let kb be the first co efficient of g(x) which p doesnt divide. In f(x) g(x), the coefficient of kjx - is kjc - which is given by kjc - = ia kb + ( 1+ja 1-kb + 2+ja 2-kb +.+ kja + ob ) + ( 1=ja 1-kb + 2-ja 2-kb +.+ oa kjb + )______ (1) By the choice of kb , we have p| 1-kb , p| 2-kb ,.. p | ob p| ( 1+ja 1-kb + 2+ja 2-kb +..+ kja + ob ) Similarly by the choice of ja , we have p| 1-ja , p| 2-ja ,. p| oa p| ( 1-ja 1+kb + 2-ja 2+kb +. + oa kjb + ) By assumption p| kjc + (as p divides all the coefficients of f(x) g(x))

Hence from (1), p| ja kb which is not true since p ja and p kb this proves the lemma. Definition: The content of the polynomial f(x) = oa + 1a x +.+ na xn where the as are integers, is the greatest common divisor of the integers oa 1a . na . Note: Given any polynomial p(x) with integer coefficients, if can be written as p(x)= dq(x) where d is content of p(x) and q(x) is a primitive polynomial. Theorem (Gauss Lemma) If the primitive polynomial f(x) can be factored as the product of two polynomial having rational coefficients, it can be factored as the product of two polynomial having integer coefficients. Proof: Suppose that f(x) = u(x)v(x) where u(x) and v(x) have rational coefficients. By clearing of denominatiotors and taking out common factors in u(x) v(x), we can write f(x) =

ba l (x) m (x) where a and b are integers, l (x) and m (x) have integer coefficient and are

primitive .Thus b f(x) = a l (x) m (x) since f(x) is primitive, the content of the left-hand side is b. since l (x) and m (x) are primitive the content of the right side is a. Thergere a =b and hence f(x) = l (x) m (x) where l (x) and m (x) have integer coefficients. Definition: A polynomial is said to be integer monic if all its coefficients are integer and its highest coefficient is 1. Note: An integer monic polynomial is of the form mx + 1a 1-nx +.+ na where the 1a s are integer. An integer monic polynomial is always primitive Theorem (The Eisenstein criterion)


55 Algebra Let f(x) = oa + x1a + 2a

2x +. + na nx be a polynomial with integer coefficients. Suppose

that p na , p| 1a , p | 2a ......p| oa ,2p 0a for some prime number p then f(x) is irreducible over

the rationals. Proof: We may assume that f(x) is primitive (otherwise, f(x) = d. g(x)) where g(x) is primitive and we can prove the result for g(x)) If f(x) can be factored as a product of two rational polynomials then by Gauss Lemma, it can be factored as the product of two polynomials having integer coefficient thus if we assume that f(x) is reducible then f(x) = ( ob + 1b x + .+ rb

rx ) ( 0c + 1c x +.+ scsx )

Where the bs and cs are integer and where r > o and s > o on comparing the coefficients we get oa = ob 0c , 1a = ob 1c + 1b 0c . ka = kb 0c + kb 1c +..+ ob kc etc., Since p| oa , p must divide one of ob or 0c . Since

2p oa , p cannot divide ob and 0c suppose p| ob but p 0c .since f(x) is primitive, all its coefficient are not divisible by p and hence all the co-efficient ob , 1b , 2b ,. rb are not divisible by p. Let kb be the first coefficient which is not divisible by p then p| 1-kb , p| 2-kb ,.. p| ob now ka = kb 0c + kb 1c +..+ ob kc . By the hypothesis of the theorem p| ka and by the choice of kb , p| 1-kb , 2-kb ,. p | ob so that p|( kb + 1c + 2-kb 2c + ob kc ) p| kb 0c . But p kb and p 0c , this contradiction proves that f(x) is irreducible. 5.4 LET US SUM-UP 1. The structure of F[x] 2. Division Algorithm for F[x] 3. F[x] is a Euclidean ring 4. Gauss Lemma 5. The Eisenstein Criterion 5. 5 LESSON END ACTIVITIES 1. If p is a prime number prove that the polynomial nx -p is irreducible over the rationals 2. Prove that the polynomial 1 + x + 2x + .. + 1-px , where p is a prime number, irreducible over the field of rational numbers. 5.6 REFERENCES

1) Topics in Algebra by I.V. Herstein (Second Edition)..


UNIT III FIELD THEORY 56 Algebra LESSON 6 EXTENSION FIELDS CONTENTS : 6.0 Aims and Objectives 6.1 Introduction 6.2 Extension of Fields 6.3 Let us Sum-up 6.4 Lesson end activities 6.5 References 6.0 AIMS AND OBJECTIVES In this lesson we study about a field containing a given field, which is called the extension field. This field extension play and important role in the theory of equations and theory of numbers. After going through this lesson, you will be able to : (1) Define extension field, degree of extension, finite extension (2) Prove finite extension of a finite extension is a finite extension (3) Define F(a) the smallest field containing a and F. (4) Prove the algebraic elements in K form a subfield of K. 6.1 INTRODUCTION Any field K which contains the given field F is called an extension field of F. The extension field K can be regarded as a vector space over F. If the dimension of K over F is finite then we define K as a finite extension over F. We also introduce algebraic extension over F. 6.2 EXTENSION FIELDS Definition Let F be a field. A field K is said to be an extension of F if F K. Equivalently, a field K is an extension of field F, if F is subfield of K. Note : Throughout this lesson, F will denote a given field and K, a field extension of F.


Fields 57 Example We know that R, the set of reals, Q the set of rationals and C the set of complex numbers are fields under addition + and multiplication. And Q R C: So R is an extension of Q and C is an extension o R. Note: If K is an extension of F, then under the field operations in K, K is vector space over F. Definition The degree of an extension K over F is the dimension of K as a vector space over F and is denoted by [K : F] Definition If [K:F] is finite, then K is said to be a finite extension of F. Example Q ( 2 ) = {a+b 2 : a, b Q} where Q is the set of rationals, is an extension of Q. The set {1, 2 } is a basis of Q( 2) over Q. Therefore [Q ( 2): Q]=2. Hence Q( 2) is a finite extension of Q. Theorem: K is a finite extension of F and L is a finite extension of K, then L is a finite extension of F and [L : F] =[[L : K] [K : F] Proof: Let [L : K] = m and [K : F] =n. We have to prove that [L : F] =mn.

Let { 1v 2v ., mv } be a basis of L over K and { 1w , 2w , , nw } be a basis of K over F.

The theorem will be proved if we show that the mn elements iv jw , i = 1,2 m, j=1, 2,., n form a basis of L over F. For this we have to show that these mn elements are linearly independent over F and every element of L is a linear combination of these mn elements. Let t L. As { 1v 2v ., mv } is a basis of L over K,

t = 1k 1v + 2k 2v +. . . .+ mk mv .. (1) where 1k , 2k ,. . . . mk belong to K.

As{ 1w , 2w , , nw } is a basis of K over F , every element in K is a linear combination of the elements 1w , 2w ,,.. nw Let 1k ,= 11f 1w + 12f 2w +. . . . .+ nf1 nw 2k = 21f 1w + 22f 2w +. . . . .+ nf2 nw . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. (2) ik = 11f 1w + 2if 2w + . . . . .+ inf nw


58 Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

mk = mf , 1w + 2mf 2w + . . . . . mnf nw where ijf are in F.

Substituting for 1k , 2k , . . . . ., ik , . . . . . mk from (2) in (1) we get t = ( 11f 1w + 12f 2w + . . . . + nf1 w) 1v + ( 21f 1w + 22f 2w + .. . . + nf2 nw ) 2v

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . + ( 11f 1w + 2if 2w + . . . .+ inf nw ) iv . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . + ( 1mf 1w + 2mf 2w + . . .. .+ mnf nw ) mv = 11f 1v 1w + 12f 1v 2w + . . . .+ nf1 iv nw + 21f 2v 1w + 22f 2v 2w + . . . .+ nf2 2v nw + . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . + . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . + 1if iv 1w + 2if iv vi 2w + . . . . .+ inf iv nw + . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . + 1mf mv 1w + 2mf mv 2w + . . . .+ mnf mv nw

Therefore tL is a linear combination of iv , jw , i = 1, 2, . . . .m, j = 1,2, . . . . n over F. Next, we prove that the elements iv jw , i = 1, 2. . . . .m, j=1, 2, . . . . ,m are linearly

independent. Suppose 11f 1v 1w + 12f 1v 2w + . . . .+ nf1 1v nw

+ 21f 2v 1w + 22f 2v 2w + . . . .+ nf2 2v nw + . . . + 11f iv 1w + 12f iv 2w + . . . . .+ inf iv nw + . . .

+ 1mf mv 1w + 2mf mv 2w + . . . .+ mnf mv nw = 0 . (3) where ijf are in F. Regrouping (3) we get

( 11f 1w + 12f 2w + . . . . .+ nf1 nw ) 1v = ( 21f 1w + 22f 2w + . . . . .+ nf2 nw ) 2v + . . .

+ ( 1if 1w + 2if 2w + . . . . . .+ inf nw ) iv + . . . + ( 1mf 1w + 2mf 2w + . . . . . + mnf nw ) mv = 0 . (4)


Fields 59

Let 1if 1w + 2if 2w + . . . . .+ inf nw = ik . (5) for i = 1, 2, . . . . m. Then ik , i =1, 2, . . . ., m are in k and from (4) and (5) we get

1k 1v + 2k 2v + . . .+ ik iv + . . .+ mk iv = 0. But 1v 2v , . . . . mv are linearly independent over K Therefore ik = 0 for i = 1,2 . . . .m. That is,

1if 1w + 2if 2w + . . . . .+ inf nw = 0 .. (5) for i = 1 , 2 . . . .m But { 1w , 2w ..... nw } is a basis of K over F and hence are linearly independent Hence (5) gives

11f = 0, 12f = 0, . . . . . inf = 0 for i = 1, 2, . . . . , m

That is ijf = 0 for i = 1, 2 , . . . . .m, j = 1, 2 , . . . . n.

Hence iv jw , i = 1, 2, . . . . .m, j = 1, 2, . . . . ., n are linearly independent and hence from a basis of L over F.

Therefore [L : F] = mn That is [L : F] = [L : K] [K : F]

Corollary If L is a finite extension of F and K is a subfield of L which contains F then [K : F] divides [L : F]. Proof: Here L, K, F are fields and L K F. Let [L : F] be finite. Let 1u 2u , . . . . . . . . ru in be L be linearly independent over K. Then we claim that

1u 2u , . . . . . . . . ru , are linearly independent over F also. For, if possible let them be linearly dependent over F. Then scalarsf 1 ,f 2 , . . . . . , f r , not all zero exist in F such that

1f 1u + 2f 2u + . . . .+ rf ru = 0. But as F K, 1f , 2f , . . . . . rf are in K also . Therefore 1u 2u , . . . . . . . . ru are

linearly dependent over K also . This is a contradiction. Therefore 1u 2u , . ru in L are linearly independent over F if they are linearly independent over K.

Now, [L : F] is finite. The basis vectors of L (K) are linearly independent in L(F) and hence are finite. Therefore [L: K] is also finite. Hence [K: F] is finite as K is a subspace L over F.

By the previous theorem [L: F] = [L: K] [K: F]

Therefore F]:[KF]:L[ = [L: K] is a finite positive integer

Hence [K: F] divides [L: F]


60 Algebra Definition Let K be an extension field of F. Then an element aK is said to be algebraic over F if there exist element n10 ,....,, aaa in F, not all zero, such that n1n1n0 ...a a+a+aa - = 0. That is, aK is algebraic over F if there exists a nonzero polynomial p(x) F[x] such that p(a) = 0. Definition

Let K be an extension of F and let a K. Let M be the collection of all subfields of K each of which contains both F and a. M is nonempty since K itself is an element of M. The intersection of any number of subfields of K is again a subfield of K. Thus, the intersection of all those subfield of K which are members of M is a subfield of K. It is donated by F (a). F (a) contains both F and a. Every subfield of K in M contains F (a). Hence F (a) is the smallest subfield of K containing both F and a. F (a) is called the subfield of K obtained by adjoining a to F. Definition Let S be a subset of a field K. Then a subfield K1 of K is said to be generated by S if (i) S K1 (ii) for any subfield L of K, S L implies K1 L.

Notation: The subfield generated by S will be denoted by < S>. The subfield generated by S is the intersection of all subfields of K which contain S. Now, let K be a field extension of F and S be any subset of K, then the Subfield K generated by F U S is said to be the subfield of K generated by S over F and this subfield is denoted by F(S). However, if S is a finite set { naaa ....., 21 } we write F(S) = F( naaa ....., 21 ). Definition A field K is said to be finitely generated if there exists a finite number of elements naaa ....., 21 in K such that K = F ( naaa ....., 21 ). Definition

If K is generated by a single element over F, then K is called a simple extension of F.

Note: We had already seen that if K is an extension of the field F and aK, then F(a) is the smallest subfield of K containing both F and a . Thus F(a) is a simple extension of F. Definition A nonzero polynomial f(x) in F[x] is said to be a monic polynomial over F if the coefficient of the highest power of x in f(x) is equal to 1, the unity of F. Example f(x) = 2x - 2x + 3 is a monic polynomial in Q[x].


Fields 61

Theorem If an element aK is algebraic over F, then there exists a unique monic polynomial p(x) of positive degree over F such that (1) p(a) = 0 (2) if for any f(x) F[x], f(a) = o then p(x) divides f(x) . Proof: Since a is algebraic over F, a is a root of some nonzero polynomial t(x) =

nn xaxaxaa .....

2110 +++ with a , F and an 0. Without loss of generality, let us assume

that t(x) is the nonzero polynomial over F of smallest degree such that t(a) = 0. Now degree of t(x) is n and na 0 Hence 1n -a F. We write 1n -a t(x) = p(x) = .x...xxx 01n11n2n2n1n1n1n1nn aa+aa++aa+aa+ --------

As t(a) = 0, p(a) = 0. Therefore p(x) is the required monic nonzero polynomial of smallest degree n such that p(a) = 0. We have to prove that p(x) divides f(x) if f(a) = 0. By division algorithm we can find polynomials q(x), r(x), in F[x] such that f(x) = p(x) q(x) + r(x) where r(x) = 0 or deg r(x) < deg p(x) then f(a) = p(a) q(a) + r(a). As f(a) = 0, p(a) = 0 we get r(a) = 0.

\ r (x) = 0 for if r(x) 0, we get a contradiction to the fact the p(

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