UNIT 1 Groups

STRUCTURE PAGE NO.

1.1 Introduction 01

1.2 Objective 01

1.3 Normal and Subnormal Series 2-5

1.4 Composition Series 5-7

1.5 Jordan–Hölder theorem 7-11

1.6 Solvable Groups 11-13

1.7 Nilpotent Groups 13-15

1.8 Unit Summary/ Things to Remember 16

1.9 Assignments/ Activities 17

1.10 Check Your Progress 17

1.11 Points for Discussion/ Clarification 18

1.12 References 18

1.1 Introduction

The theory of groups was unified starting around 1880. Since then, the impact of group theory has been ever growing, giving rise to the birth of abstract algebra in the early 20th century, representation theory, and many more influential spin-off domains. The classification of finite simple groups is a vast body of work from the mid 20th century, classifying all the finite simple groups. These groups can be seen as the basic building blocks of all finite groups, in much the same way as the prime numbers are the basic building blocks of the natural numbers.

The classification of the finite simple groups is believed to classify all finite simple groups. These groups can be seen as the basic building blocks of all finite groups, in much the same way as the prime numbers are the basic building blocks of the natural numbers. The Jordan–Hölder theorem is a more precise way of stating this fact about finite groups. Jordan defines isomorphism of permutation groups and proves the theorem for permutation groups. In a huge collaborative effort, the classification of finite simple groups was accomplished in 1983 by Daniel Gorenstein.

1.2 Objective

After the completion of this unit one should able to :

 define normal, subnormal and composition series.  prove Jordan–Hölder theorem.  to prove fundamental theorem of Arithmetic as an application of Jordan–Hölder theorem.  relate solvable and nilpotent groups.  to show the solvability of finite p-group.  to analyze abstract and physical systems in which symmetry is present.

1.3 Normal and Subnormal series

1

An important part in group theory is played by subnormal, normal and central series. These series are related with the view of a series of a group G, which gives insight into the structure of G. The results hold for both abelian and non abelian groups.

Subgroup series: A subgroup series is a finite sequence of subgroups of a group G contained in each other such that:

{e}= G0 G1 G2  G3 ...... Gn G

or {e}=G0 G1 G2  G3  ......  Gn G ……………(1)

One also considers infinite chains of imbedded subgroups (increasing and decreasing), which may be indexed by a sequence of numbers or even by elements of an ordered set. Here n is called as the length of the series.

A subgroup series (1) is called subnormal series or sub invariant series.

If also each subgroup Gi , i = 0,1,2,……n, is normal in G, the subgroup series (1) is called a normal series or invariant series in G.

It is noted that for abelian groups the notions of subnormal and normal series coincide, since every subgroup is normal. So a normal series is a subnormal series, but the converse is not true in general. Sometimes the phrase "normal series" is used when the distinction between a normal series and a subnormal series is not important. Most of the authors use the term normal series, instead of subnormal series.

Examples:

1. {0} < 8 < 4 < : Normal series

2. {0} < 9 < 3 < : Normal series

3. : Subnormal series.

Since is not normal in . Here is the group of symmetries of the square denotes rotations and denotes mirror images.

2 Refinement of a normal series

A subnormal ( normal) series {1} = K0 = K1 = · · · = Km = G is a refinement of subnormal

( normal) series {1} = H0 = H1 = · · · = Hn = G if { Hi }  { Kj } . A refinement of a normal series is another normal series obtained by adding extra terms.

Examples:

1. Sn > An > 1 is a normal series for G = Sn for any n ≥ 1.

The associated subquotients are Sn /An ∼ /2 and An /1 = An .

For n = 4 this normal series has a refinement: S4 > A4 > K > 1.

2. The series { 0 } < 72 < 24 < 8 < 4 < is a refinement of a series

{ 0 } < 72 < 8 <

Schreier Refinement Theorem:

Any two subnormal series of a group have equivalent refinements.

Proof Consider the subnormal series of a group G.

G  G0  G1  G2  ...  Gs  (e) … (1)

G  H0  H1  H 2  ...  H1  (e) … (2)

Since for any i,j,k and l, Gi1 is a normal subgroup of Gi and H k1 is a normal subgroup of H k ' we get

Gi, j  Gi1 (Gi  H j ),(i  0,1,2,...,s 1; j  0,1,2,...,t) …(3)

H k,l  H k1 (H k G1 ),(k  0,1,2,...,t 1;l  0,1,2,...s) …(4) are subgroup of G.

3 Now H jl is normal subgroup of H j implies that Gi' j1 is normal subgroup ofGi' j' . Similarly H k 'l1 is normal subgroup of H k,l .

Since H t  (e)and H 0  G, we have Gi,t  Gi1 (Gi  Ht )  Gil (e)  Gi1 .

Also Gi,0  Gi1 (Gi  H0 )  Gi1 (Gi G)  Gi1Gi  Gi .

Thus Gi,t  Gi1  Gi1,0i  0,1,2,...s 1. …(5)

Similarly H k,s  H k1  H k1,0i  0,1,2,...,t 1 …(6)

Consider two series

G  G0  G0  G0,2  ...  G0,t ( G1  G1,0 )  G1,1  G1,2 ,0 …(7)  ...  G1,t ( G2  G2,0 )  ...  Gs 1,0  ...Gs1,t  Gs  (e)

G  H0  H0,0  H0,1  H0,2  ...  H0,s ( H1  H1,0 )  H1,1

 H1,2  ...  H1,s ( H 2  H 2,0 )  ...  Ht1,0  Ht1,0  Ht1,1  ...  Ht1,s  Ht  (e) ...... (8)

Both (7) and (8) have same number of terms.

Clearly G0 occurs in (7) and for each m 1,2,...s,as Gm  Gm1,t by (5), we see that each Gm occurs in (7) for all m. Thus (7) is a refinement of (1).

Similarly (8) is a refinement of (2).

Now by Zassenhaus theorem (Let A and B be any two subgroups of a group G, A* and B* be B *(B C) C *(C  B) normal subgroups of A and B respectively then  ) B *(B C*) C *(C  B*)

G G (G  H ) H (H G ) H We have r,s  r1 r s  s1 s r  s,r Gr,s1 Gr1(Gr  H s1 H s1(H s Gr1) H s,r1

For all r  0,1,2..., s 1 and for all s  0,1,2,...t 1.

4 Thus (7) and (8) are equivalent. This proves the theorem.

Lemma If G is a commutative group having a composition series then G is finite.

Proof We firstly show that a simple abelian group must be a cyclic group of prime order. This follows the fact that any subgroup of an abelian group is normal.

Now let G  G0  G1  G2  ...  Gs  (e) be a composition series of G.

Gs 1 Gs1 Since   Gs1 is simple abelian, o(Gs2 )  ps1 ' where ps1 is a prime number. Gs (e)

Gs2 Gs2  Further is simple abelian  o   ps2 , for some prime number ps2 . Gs1 Gs1 

Gs2  Thus o(Gs2 )  o o(Gs1 )  ps2 ps1. Gs1 

 Gi  Proceeding in this manner we get that G has p0 p1 p2 ...ps1 elements where pi  o  for Gi1  i  0,1,2,...,s 1

1.4 Composition series

If a group G has a normal subgroup N which is neither the trivial subgroup nor G itself, then the factor group G/N may be formed, and some aspects of the study of the structure of G may be broken down by studying the "smaller" groups G/N and N. If G has no such normal subgroup, then G is a simple group.

A composition series of a group G is a subnormal series

{1} = H0 = H1= H2 = · · · = Hn= G

5 with strict inclusions, such that each Hi is a maximal normal subgroup of Hi+1. Equivalently, a composition series is a subnormal series such that each factor group Hi+1 / Hi is simple. The factor groups are called composition factors.

A subnormal series is a composition series if and only if it is of maximal length. That is, there are no additional subgroups which can be "inserted" into a composition series. Thus, A subnormal series {1} = H0 = H1 = H2 = · · · = Hn = G is a composition series if and only if

(1) H0 < H1 < H2 < · · · < Hn = G. (2) For each i = 0, . . . , n - 1, we have that there is no normal subgroup of Hi+1 that lies strictly between Hi and Hi+1. The length n (the number of subgroups in the chain, not including the identity) of the series is called the composition length.

If a composition series exists for a group G, then any subnormal series of G can be refined to a composition series, informally, by inserting subgroups into the series up to maximality. Every finite group has a composition series, but not every infinite group has one. For example, the infinite cyclic group has no composition series.

Example: Let G = 6 = {[0], [1], [2], [3], [4], [5]} be the cyclic subgroup of order 6 and consider the subgroups H = [2] = {[2], [4], [0]} and K = [3] = {[3], [0]}. This gives rise to the two subnormal series {[0]} = H = G, {[0]} = K = G. In fact these are both composition series.

The first has composition factors H/{[0]} 3 and G/H 2, whereas the latter series has composition factors K/{[0]} 2 and G/K 3. Notice that the composition factors turn out to be the same (up to order).

Note: For infinite groups existence of a composition series is some kind of finiteness condition. If G has a composition series, then any subnormal series has a refinement which is a composition series.

Isomorphic normal series

Two normal series

e Gs1 Gs  ......  G2 G1 G and e  Ht1  Ht  ......  H2  H1  H

6 are called isomorphic or equivalent if there exists a 1-1 correspondence between the factors of the two series (thus s=t) such that the corresponding factors are isomorphic.

Theorem Every finite group has a composition series.

Proof Take the longest possible normal series. Then the sub quotients are all simple. If one sub quotient Hi /Hi+1 is not simple then it has a nontrivial normal subgroup which, by the correspondence theorem, gives a group K between Hi and Hi+1 extending the normal series:

Hi ⊳ K ⊳ Hi+1 .

A composition factor is a special case of a sub quotient of G which is defined to be a quotient of a subgroup of G (i.e., H/N where N ⊴ H ≤ G).

Thus a finite group has at least one composition series.

1.5 Jordan-Hölder Theorem

A group may have more than one composition series. However, the Jordan–Hölder theorem (named after Camille Jordan and Otto Hölder) states that any two composition series of a given group are equivalent. That is, they have the same composition length and the same composition factors, up to permutation and isomorphism. The Jordan–Hölder theorem is also true for transfinite ascending composition series, but not transfinite descending composition series .

Theorem

(1) Every finite group G having at least two elements has a composition series.

(2) If (e)  G0  G1  G2  ...  Gk  G

and (e)  H0  H1  H 2  ...  H1  G

G H are two composition series for a group G than k  l and i  0,1,2,...,k 1, i1   (i)1 for some Gi H (i) permutation  of the set  0,1,2...,k 1 (i.e. any two composition series for G are equivalent).

7 Proof: (1) Suppose we are given two composition series for a group G:

(e)  G0  G1  G2  ...  Gk  G ------1) and (e)  H0  H1  H 2  ...  H1  G ------2)

G H then k  l and i  0,1,2,...,k 1, i1   (i)1 for some permutation  of the set Gi H (i) 0,1,2...,k 1 (i.e. any two composition series for G are equivalent).

Now We shall apply induction on o(G).

Let n= o(G). In case n=2, (e) = G0  G1  G is the only composition series for G as G1 /G0  G and G being a cyclic group of prime order, is simple. So the theorem holds for n  2.

Let every group of order  n have a composition series and let G be a group of order n  2.

If G is simple, then G has no proper normal subgroup. Consequently (e) = G0  G1  G is the only composition series for G.

If G is not simple and let N be a proper normal subgroup of G. Since G is finite, there exists only finitely many proper normal subgroups of G containing N and let M be one such subgroup having largest number of elements. Then M is a maximal normal subgroup of G. Clearly G/M is a simple group and M  G . So that o(M)  n.Hence by the induction hypothesis M has a composition series,

(e)  M 0  M1  M 2  ...  M1  M.

Now consider the series

(e)  M 0  M1  M 2  ...  M t  M  G …(1)

M By definition of a composition series, each i1 is a simple group for all i  0,1,2,...,t 1. M i

8 G G Also  is simple. M t M

Consequently (l) is a composition series for G.

(2) Suppose

(e)  G0  G1  G2  ...  Gk1  Gk  G …(2)

and (e)  H0  H1  H 2  ...  Hl1  Hl  G …(3) are two composition series for the finite group G. Again we shall apply induction on o(G).Let o(G)  n. If n=2 then as seen before G has only one composition series, so 2nd part of theorem holds for groups of order 2.

Let us suppose that o(G)  n  2 and let the result be true for all groups of order <  n.We consider two cases (I)Gk1  Hl1 and (II)Gk1  Hl1 .

Case I Gk1  Hl1

Evidently (e)  G0  G1  G2  ...  Gk1 …(4)

and (e)  H0  H1  H 2  ...  Hl1  Gk1 …(5)

are two composition series for the group Gk1 whose order is less than n. So by the induction hypothesis series (4) and (5) are equivalent. This gives k 1  l 1i.e.,k  l and further series (2) and (3) are equivalents as

G G G H k    l . This case now follows. Gk1 Gk1 H l1 H l1

Case II Gk1  Hl1.

As Gk1 and H l1 are maximal normal subgroups of G,Q  Gk1  Hl1 is normal subgroup of Gk1 as well as of H l1 and as Gk1  Hl1'Q  Gk1 and Q  H l1. Further Gk1H l1 is a normal subgroup

9 of G. Now Gk1  Gk1Hl1.Since Gk1 and H l1 are maximal normal subgroups of

G,Gk1Hl1  G. (If can‟t be equal to Gk1 and H l1 simultaneously.)

Let (e)  Q0  Q1  Q2  ...  Qm  Q be a composition series for the group Q, then we claim that

(e)  Q0  Q1  Q2  ...  Qm ( Q)  Qk1  Gk  G …(6)

and (e)  Q0  Q1  Q2  ...  Qm ( Q)  Hl1 H1  G …(7)

G H both are composition series for G. For this purpose it is sufficient to prove that k 1 and l1 are Q Q

G G G H G G simple. Now k1  k1  k1 l1  ; but is simple. Q Gk1  H l1 H l1 H l1 H l1

G Hence k 1 is simple. Q

H G G G Similarly l1  is simple. Again (6) and (7) are equivalent, since k1  and Q Gk1 Q H l1 G H  l1 (as seen above) and each of (6) and (7) is of length M  2. Gk1 Q

Now by Case (I), series(2) and(6) are equivalent, so k  m  2. Again by Case(1), series(3) and (7) are equivalent, therefore l  m  2.Hence k  l.Also as series (6) and (7) are equivalent; the series (2) and (3) are equivalent. This completes the proof for the Case (II).

Note: This theorem can also be proved by Schreier refinement theorem.

Application of Jordan-Hölder Theorem

We use the Jordan-Hölder Theorem to prove the uniqueness part of the Fundamental Theorem of Arithmetic. The Fundamental Theorem of Arithmetic states that every positive integer not equal to a prime can be factored uniquely (up to order) into a product of primes.

First, we claim that such a factorization exists. Indeed, suppose n is composite (i.e., n>1and n is not a prime). Then an easy induction shows that n has a prime divisor of p and we can write n = pn1, where n1 is an integer satisfying n1

10 factor p1, and n1=p1n2 here n2integers n > n1 > n2 > n3…………… cannot continue indefinitely. We now have that n= p p1 p2 p3……… pj is a product of primes. This proves the existence claim.

On the basis of the Jordan-Hölder Theorem, we can easily show the other part of the Fundamental Theorem of Arithmetic, i.e., apart from order of the factors, the representation of n as product of primes is unique. To do this suppose that n= p1 p2 p3……… ps and n= q1 q2 q3……… qt where the pi and qj are primes. Then denoting, as usual, by Ck the cyclic group of order k, we have

and

as two composition series for Cn. But the Jordan-Hölder Theorem implies these must be equivalent; hence we must have s=t and by suitably arranging pi = qi, 1≤ i ≤ s. Thus we have established the unique factorization theorem for positive integers as an application of the Jordan-Hölder Theroem.

1.6 Solvable Groups

The group G is said to be solvable or soluble if there exists a finite chain of subgroups

G = N0 N1 · · · Nn such that (i) Ni is a normal subgroup in Ni-1 for i = 1,2, . . . ,n,

(ii) Ni-1 / Ni is abelian for i = 1,2, . . . ,n, and

(iii) Nn = {e}.

Examples: Any abelian group is solvable. A simple group is solvable if and only if it is cyclic of prime order. A4 is solvable. Indeed, the subgroup {e, (12)(34), (13)(24), (14)(23)}, isomorphic to 2 x 2 , is normal in A4, with quotient isomorphic to 3. Since 2 x 2 and 3 are abelian, A4 is solvable. Its composition factors are 2 and 3. As a corollary, one deduces that S4 is also solvable, since S4/A4 is isomorphic to 2 . An is not solvable for n > 4, because it is simple and not abelian.

Then Sn is not solvable for n>4 either.

11 Any abelian group is solvable even if it is infinite. Another interesting example is the symmetric group S4 which has the solvable series: S4 ⊳ A4 ⊳ K ⊳ 1 with quotients S4 /A4 ∼ /2, A4 /K ∼ /3 and K/1 = K ∼ /2 × /2.

Properties of solvable groups

 If G is solvable, and H is a subgroup of G, then H is solvable.

 If G is solvable, and H is a normal subgroup of G, then G/H is solvable.

 If G is solvable, and there is a homomorphism : G H, then H is solvable.

 If H and G/H are solvable, then so is G.

Theorem Let p be a prime number. Any finite p-group is solvable.

Proof Let G be a finite p-group(p is a prime number) then o(G)=pn, for some n  0. If n=l G is abelian and so it is solvable. So let n>l, we apply induction on n. suppose the result holds for groups of order pm with m

Supersolvable groups

As a strengthening of solvability, a group G is called supersolvable (or supersoluble) if it has an invariant normal series whose factors are all cyclic. Since a normal series has finite length by definition, uncountable groups are not supersolvable. The alternating group A4 is an example of a finite solvable group that is not supersolvable.

We now establish a criterion for the solvability of a group by means of commutator subgroups.

Commutator Subgroup

Let G be a group. An element g G is called a commutator if g = aba-1b-1 for elements a,b G. The smallest subgroup that contains all commutators of G is called the commutator subgroup. The subgroup (G' )' is called the second derived subgroup of G.

We define G(k) inductively as (G(k-1))', and call it the k th derived subgroup.

12 Proposition Let G be a group with commutator subgroup G', then

(a) The subgroup G' is normal in G, and the factor group G/G' is abelian.

(b) If N is any normal subgroup of G, then the factor group G/N is abelian if and only if G' N.

Theorem A group G is solvable if and only if G(n) = {e} for some positive integer n.

Proof Let G be solvable and let G  G2  G1  G2  ...  Gn  (e) be a solvable series. We prove

(k) (0) inductively that G  Gk k. clearly G  G  G0 .

(k) (k1) (k) Suppose for some k,G  Gk' this gives G  [G ]' Gk '.

1 (k1) As Gk / Gk1 is abelian, we have Gk  Gk1. Consequently G  Gk1. Hence by induction

(k) (n) (n) G  Gk k  0,1,2,...n In particular G  Gn . Hence G  (e) since Gn  (e). Conversely let

G(n)  (e) for some n, then G  G0  G(1)  G(2)  ...  G(n)  (e) is a subnormal series for G, such that G(i) /G(i1  G(i) /[G(i) ]' is abelian since for any group H,H/H’ is always abelian, hence G is solvable.

Corollary S n is not solvable for n  5.

Proof An is a simple non-commutative, so An ' (l). However An is simple, so its only normal subgroups are An and (l).

(2) (k) Consequently An ' An  An  (An ')' An ' An . In general An  An positive integers k, thus

(k) An  (e)k. Hence is not solvable.

As now is a subgroup of Sn' Sn is also not solvable for n  5, since subgroup of a solvable group is solvable.

1.7 Nilpotent Groups

A nilpotent group is a group that is "almost abelian". This idea is motivated by the fact that nilpotent groups are solvable, and for finite nilpotent groups, two elements having relatively prime orders must commute. It is also true that finite nilpotent groups are super solvable.

13 Nilpotent groups arise in Galois theory, as well as in the classification of groups. They also appear prominently in the classification of Lie groups. Analogous terms are used for Lie algebras including nilpotent, lower central series, and upper central series.

The following are equivalent definitions for a nilpotent group:

A nilpotent group is one that has a central series of finite length.

A nilpotent group is one whose lower central series terminates in the trivial subgroup after finitely many steps.

A nilpotent group is one whose upper central series terminates in the whole group after finitely many steps.

For a nilpotent group, the smallest n such that G has a central series of length n is called the nilpotency class of G and G is said to be nilpotent of class n. Equivalently, the nilpotency class of G equals the length of the lower central series or upper central series (the minimum n such that the nth term is the trivial subgroup, respectively whole group). If a group has nilpotency class at most m, then it is sometimes called a nil-m group.

The trivial group is the unique group of nilpotency class 0, and groups of nilpotency class 1 are exactly non-trivial abelian groups.

Properties of nilpotent groups

 Since each successive factor group i+1/ i is abelian, and the series is finite, every nilpotent group is a solvable group with a relatively simple structure.

 Every subgroup of a nilpotent group of class n is nilpotent of class at most n; in addition, if f is a homomorphism of a nilpotent group of class n, then the image of f is nilpotent of class at most n.

 If G is a nilpotent group and if H is a proper subgroup of G, then H is a proper normal

subgroup of NG(H) (the normalizer of H in G).

 Every maximal proper subgroup of a nilpotent group is normal.

 Nilpotent group is the direct product of its Sylow subgroups.

The last statement can be extended to infinite groups: If G is a nilpotent group, then every Sylow subgroup Gp of G is normal, and the direct sum of these Sylow subgroups is the subgroup of all elements of finite order in G

Note that:

14 Given a prime number p, a p-group (also p-primary group) is a group such that for each element g of the group there exists a nonnegative integer n such that g to the power pn is equal to the identity element. (In other words, g has order pn).

A subgroup of a finite group is termed a Sylow subgroup if it is a p-Sylow subgroup for some prime number p.

Nilpotent is Solvable

Given a central series for G, build a corresponding normal series, in reverse order, as follows.

A trivial homomorphism maps G onto G, with kernel e. Thus C0 = G, and N0 = e.

Moving to C1, a homomorphism maps G onto C1, and its kernel is the center of G. Let N1 be this kernel, the center of G.

A second homomorphism maps C1 onto C2. Combine this with the first homomorphism to get a map from G onto C2. Let N2 be the kernel of this map. Continue all the way to Ck = e, whence Nk = G.

Since each Ni is the kernel of a homomorphism, each Ni is normal in G. Conjugate Ni by the elements in Ni+1, which all belong to G, and the result is always Ni. Thus Ni is normal in Ni+1, and we have a normal series for G, running in reverse order.

The factor group Ni+1/Ni is the center of Ci, which is abelian. All factor groups are abelian, the series is solvable, and the group is solvable. Every nilpotent group is solvable.

However, there are plenty of solvable groups that are not nilpotent. S3 has no center, and no central series, even though it has an abelian subgroup 3 with an abelian factor group 2.

Theorem If G is a group of order pn, with p prime, and H is a proper subgroup of G, then the normalizer NG(H) strictly contains H.

Proof The proof is by induction on the order of G. If |G|=1 there is nothing to prove. Suppose |G|>1. Let denote the center of G. We know by previous stuff that | |>1. Note also that is contained in the normalizer of H. So, if is not a subset of H, then we are done. Suppose is a subset of H. Then, since is normal in G, we consider the subgroup H/ of the quotient group

G/ . Since |G/ | is strictly smaller than |G|, and is still a power of p, we can apply the inductive hypothesis. We conclude that the normalizer of H/ in G/ strictly contains H/ .

15 This result also holds in arbitrary nilpotent groups. Using a lemma called the Frattini argument (Let G be a finite group, and let H be a normal subgroup of G. If P is any Sylow subgroup of H, then G=HN(P), and [G:H] is a divisor of |N(P)|.), one can show that a finite group is nilpotent if and only if all of its Sylow subgroups are normal. Hence G is isomorphic to the direct product of its Sylow subgroups. So, a general finite nilpotent group is isomorphic to a direct product of p-groups for some collection of primes p. 1.8 Unit Summary/ Things to Remember

1 A subnormal series of a group G is a sequence of subgroups, each a normal subgroup of the next one.

G G G  G ...... G G In a standard notation {e}= 0 1 2 3 n

There is no requirement made that Gi be a normal subgroup of G, only a normal subgroup

of Gi+1. The quotient groups Gi+1/ Gi are called the factor groups of the series.

If in addition each Gi is normal in G, then the series is called a normal series.

2 Composition series is a subnormal series such that each factor group Gi+1 / Gi is simple. The factor groups are called composition factors.

3 Two subnormal series are said to be equivalent or isomorphic if there is a bijection between the sets of their factor groups such that the corresponding factor groups are isomorphic.

4 Jordan–Hölder theorem states that any two composition series of a given group are equivalent. That is, they have the same composition length and the same composition factors, up to permutation and isomorphism.

5 A solvable group, or soluble group, is one with a subnormal series whose factor groups are all abelian.

6 The derived series of a group G is the normal series G > G1 > ... > Gn > ... where G1 = [G,G], G2=[G1,G1], ... , Gn=[Gn-1,Gn-1]. Note that Gn/Gn+1 is the largest abelian quotient of Gn. G is solvable if and only if its derived series eventually reaches the identity - that is, Gn=1 for some n.

7 A nilpotent series is a subnormal series such that successive quotients are nilpotent.

8 A nilpotent series exists if and only if the group is solvable.

9 Nilpotent groups do not have the "hereditary property" that solvable groups have. That is, nilpotency of a normal subgroup N and the quotient G/N do not imply nilpotence of G,

16 although the converse does hold. The issue is that one needs a normal, rather than subnormal series, and a normal subgroup of N need not be normal in G.

1.9 Assignment/Activities

1 Prove that a finite p-group where p is a prime number is cyclic if and only if it has only one composition series. 2 Let a group G be direct product of two subgroups H and K. Show that G has a composition series if and only if each of H and K has a composition series. 3 Show that if N is a normal subgroup of G and G has a composition series than N has a composition series. 4 Prove that if G is a group that has a normal subgroup N such that both N and G/N are solvable, then G must be solvable. 5 Prove that every group of order p 2q, p,q are primes is solvable.

6 A group G is said to be nilpotent if it has a normal series G  G0  G1  G2  ...  Gs  (e)

Gi  G  such that  Z  i  0,1,2,..., s 1 then prove that every nilpotent group is solvable, Gi1 Gi1  however converse is not true.

1.10 Check Your Progress

2 Let G be a group having a composition series and H, a normal subgroup of G. Prove that G has a composition series, one of whose terms is H.

3 Determine a composition series of An (n≠4).

4 A finite group G is solvable if and only if it has a composition series, each of whose factor group is cyclic of prime order.

5 A subgroup and a quotient group of a nilpotent group are nilpotent. 6 Show that every nilpotent group is solvable. 7 Prove that every group of order are primes is solvable.

8 Show that a finite direct product of solvable groups is solvable.

17

1.11 Points for discussion / Clarification

At the end of the unit you may like to discuss or seek clarification on some points. If so, mention the same.

1. Points for discussion

______

______

______

2. Points for clarification

______

______- ______

______

1.12 References

1. D.J.S Robinson, A Course in the Theory of Groups, 2nd Edition, New York: Springer- Verlag, 1995. 2. J. S. Lomont, Applications of Finite Groups, New York: Dover, 1993. 3. John S. Rose, A Course on Group Theory. New York: Dover, 1994. 4. P.B. Bhattacharya, S.K. Jain and S.R. Nagpaul, Basic Abstract Algebra,(2ndEdition),Cambridge University Press, Indian Edition, 1997. 5. John B. Fraleigh, A first Course in Abstract Algebra, 7th Edition, Pearson Education, 2004.

18 UNIT 2

Canonical Forms

STRUCTURE PAGE NO.

2.1 Introduction 19

2.2 Objective 19

2.3 Similarity of linear transformations 20-21

2.4 Invariant Subspaces 22-24

2.5 Reduction to triangular forms 24-25

2.6 Nilpotent transformation and Index of nilpotency 25

2.7 Invariants of a nilpotent transformation 26-27

2.8 The Primary decomposition theorem 27-28

2.9 Jordan block and Jordan forms 29-42

2.10 Cyclic modules 42-44

2.11 Simple modules 44

2.12 Semi- Simple modules 44

2.13 Schur’s Lemma 45

2.14 Free modules 46

2.15 Unit Summary/ Things to Remember 46-48

2.16 Assignments/ Activities 48

2.17 Check Your Progress 49

2.18 Points for Discussion/ Clarification 50

2.19 References 50

19 2.1 Introduction

Cayley in 1858 published Memoir on the theory of matrices which is remarkable for containing the first abstract definition of a matrix. He shows that the coefficient arrays studied earlier for quadratic forms and for linear transformations are special cases of his general concept. Cayley gave a matrix algebra defining addition, multiplication, scalar multiplication and inverses. He gave an explicit construction of the inverse of a matrix in terms of the determinant of the matrix.

In 1870 the Jordan canonical form appeared in Treatise on substitutions and algebraic equations by Jordan. It appears in the context of a canonical form for linear substitutions over the finite field of order a prime.

Frobenius, in 1878, wrote an important work on matrices On linear substitutions and bilinear forms although he seemed unaware of Cayley's work. Frobenius in this paper deals with coefficients of forms and does not use the term matrix. However he proved important results on canonical matrices as representatives of equivalence classes of matrices. He cites Kronecker and Weierstrass as having considered special cases of his results in 1874 and 1868 respectively. Frobenius also proved the general result that a matrix satisfies its characteristic equation. This 1878 paper by Frobenius also contains the definition of the rank of a matrix which he used in his work on canonical forms and the definition of orthogonal matrices.

The nullity of a square matrix was defined by Sylvester in 1884. He defined the nullity of A, n(A), to be the largest i such that every minor of A of order n-i+1 is zero. Sylvester was interested in invariants of matrices, that is properties which are not changed by certain transformations.

2.2 Objective

After the completion of this unit students should be able to:

 make use of matrices in various domains.  make similar linear transformation.  define nilpotent transformation.  prove Primary Decomposition Theorem.  classify cyclic, simple, semi-simple and free modules.  prove Schur‟s lemma.  represent matrix in Jordan Forms.

20 2.3 Similarity of linear transformations

In mathematics, a linear transformation (also called a linear map, linear function or linear operator) is a function between two vector spaces that preserves the operations of vector addition and scalar multiplication. The expression "linear operator" is commonly used for linear maps from a vector space to itself (endomorphism). In the language of abstract algebra, a linear map is a homomorphism of vector spaces.

We know that if V and W be vector spaces over the same field F. A linear transformation is a function T:V W such that:

 T(v+w)=T(v)+T(w) for all v,w V  T( v)= T(v) for all v V , and F

The set of all linear maps V W is denoted by HomF(V,W) or Hom (V,W).

If V=W, then Hom (V,V) is an algebra over F. For convenience of notation we shall write Hom (V,V) as A(v).

T  A(V) is called right (left) invertible if there exists S  A(V) such that TS=I (ST = I) where I is the identity transformation on V where S is called a right (left) inverse of T.

Thus,

is called invertible or regular if T is both right and left invertible or is invertible then there exists S  A(V) such that ST  TS  I.

is said to be equivalent to if there exists invertible transformations

P,Q A(V) such that T  PSQ.

is said to be similar to if there exists a invertible transformation

1 P  A(V) such that T  P SP.

A linear transformation T on a vecror space V is said to be cyclic if there exists v V such that

V  F[T]v; in such a situation V is said to be a cyclic space relative to T.

Now we need to know some more terms:

21 Minimal Polynomial: The polynomial of least degree which both divides the characteristic polynomial of a matrix and has the same roots.

Monomial: One degree polynomial is called monomial.

Coprime polynomials: Two polynomials p(t) and q(t) are coprime if gcd(p(t), q(t)) = 1 (or any nonzero number). Specifically this means that c(t) divides p(t) and q(t) ⇒ c(t) is a nonzero constant.

Direct Sum: The direct sum of a family of objects Ai, with i ∈ I, is denoted by , and each Ai is called a direct summand of A.

Cyclic Linear Transformation: A linear transformation T on a vector space V is said to be cyclic if there exists v V such that V  F[T]v; in such a situation V is said to be a cyclic space relative to T.

Theorem: A linear transformation T :V V is invertible if and only if its minimal polynomial has non-zero constant term.

k k1 k2 Proof Let m(x)  x  a1x  a2 x ...  ak be the minimal polynomial of T. Then

T k  a T k1 ...  a T  a I  0 yields 1 k1 k

T(T k1  a T k2 ...  a I)...(T k1  a T k2 ...  a I)T  a I 1 k1 1 k1 k …(1)

1 k1 k2 Let T be regular. If ak  0, then as T exists, (1) implies that T  a1T ...  ak1I  0.

This contradicts the fact that deg m(x)  k and T cannot be a root of any non-zero polynomial over

F of degree < k. Hence ak  0

Conversely if ak  0, then (1) yields that TT' T'T  I where T' a1 (T k1  a T k2 ...  a I). k 1 k1

Hence T is invertible.

22 Lemma A vector space V is cyclic relative to a linear transformation T on V if and only if dimF V is same as the degree of the minimal polynomial of T.

Proof If V is cyclic relative to T, then there exists v V such that v V[T]v.

Then F[T]v  deg mv (x) where mv (x) is the minimal polynomial of v relative to T. Conversely let the degree of m(x), the minimal polynomial of T, be n where n  dimV.Now V has a vector v whose minimal polynomial relative to T is m(x) and hence dim F[T]v = n .

Consequently V  F[T]v. This proves that V is cyclic relative to T.

2.4 Invariant Subspaces

Let T be a linear transformation on a vector space V. A subspace W of V is said to be T-invariant subspace of V if for every x W,T(x)W, [i.e, T(W)  W].

Remark: For any T  A(V); V, (0) and T(V ) are always T-invariant subspaces of V.

Given a T-invariant subspace W of V, we can define a linear transformation T':W W such that T'(x)  T(x) for ever xW. T ' is said to be induced by T.

Now we have the following:

Lemma: Let V  V1 V2 and T be a linear transformation from V to V such that V1 and V2 are both

T-invariant subspaces. If Ti (i  1,2) is the linear transformation on Vi (i  1,2) induced by T, then the minimal polynomial f (x) of T is the LCM of the minimal polynomials of T1 and T2 .

Proof: Let fi (x) be the minimal polynomial of Ti Since = 0 and Ti is the restriction of T to

Vi , we also have f (Ti )  0 . Hence fi (x) | f (x) . Therefore the LCM g(x) of all these fi (x) also divides .

Now, for any x1 V1, g(T)x1  g(T1 )  0, as g(T1 )  0

23 Similarly g(T)x2  0 for every x2 V2 .

Hence [g(T)]V1  (0)and [g(T)]V2  (0).This yields that g(T)V  (0) as V V1 V2. Thus, g(T)  0.Consequently f (x) | g(x).Hence f (x) is an associate of g(x) . As a result of this is LCM of f1 (x) and f 2 (x) .

Theorem Let T be a cyclic linear transformation on a vector space V and be its minimal

a a a polynomial. If 1 2 t is the decomposition of into product of f (x) [ f1(x)] [ f2 (x)] ....[ ft (x)] powers of distinct irreducible monic polynomials fi (x) over F, then V  V1 V2 ...Vt for some

T-invariant subspaces Vi of V such that the restriction Ti of T to Vi is a cyclic linear transformation

a i of Vi with [ fi (x)] as its minimal polynomial.

Proof Since T is cyclic transformation on V,V  F[T]v for some v V and is the minimal polynomial of v relative to T.

a at a a a v  [ f (t)] 2 ...[ f (T)] vv  [ f (T)] 1[ f (T)] 3...[ f (T)] t v,..., Define 1 2 t 2 1 3 t a a a vt  [ f1 (T)] 1[ f 2 (T)] 2 ...[ ft1 (T)] t1 v.

a Then the minimal polynomial of v is 1 In general the minimal polynomial of v is 1 [ f1 (T)] . i

a [ fi (T)] i .Thus the minimal polynomial of

w  v1  v2 ...  vi …. (1)

a a a is [ f1 (x)] 1[ f 2 (x)] 2 ...[ ft (x)] t .Hence dim V =deg = dim F(T)w yields that

V  F[T]w …. (2)

Let Vi  F[T]vi (1  i  t).Since TF[T]vi  F(T)vi , we get that Vi is T-invariant space. So if Ti is

a the restriction of T to Vi , we get Vi  F[Ti ]vi ,a cyclic space relative toTi . Since [ fi (x)] i is the

a minimal polynomial of vi relative to Ti .we also get [ fi (x)] i is the minimal polynomial of vi

a relative to Ti Hence Ti also has [ fi (x)] i as its minimal polynomial.

Since w  v1  v2 ...  vt yields V  F[T]w  F[T]v1 ...  F[T]vt , we get

24 V  V1 V2 ... Vt … (3)

We claim that this sum is direct.

Let g1 (T)v1  g2 (T)v2 ...  gt (T)vt  0 for some gi (T)vi Vi … (4)

Then

a a [ f 2 (t)] 2 ...[ ft (t)] t [g1 (T)v1 ...  gt (T)vt ]  0 … (5)

a Since [ fi (T)] i vi  0 for all i, (5) gives that

a a g1 (T)[ f 2 (T)] 2 ...[ fi (T)] t vi  0.

a a a Consequently [ f (x)] | g (x), as [ f (x)] is relatively prime to each of[ f (x)] i ,(i  2). 1 1 1 1 1 i

Hence g1 (T)v1  0.Similarly g2 (T)v2  0,..., gt  0.Thus in (4) every term on the left is zero.

Hence V  V1 V2 ...Vt . This proves the theorem.

2.5 Reduction to triangular forms

Theorem: Let V be a finite-dimensional vector space over F and T be a linear operator on V.

Then T is triangulable if and only if mT is a product of linear polynomials over F:

To prove this theorem we must know the following lemma:

Let , V finite-dimensional vector space.

Suppose mT is a product of linear factors and Let W be a proper T-invariant subspace of V. Then there exists a vector such that for some eigen values c of T. (that is, the T-conductor of is linear.)

Start by applying the lemma to W  O to obtain 1 W and T1  a111 for some a11. Then apply the lemma to W1  span 1.and obtain 2 W1 such that (T  a22I)2  a121 for some

25 a12a22. Let W2 = span 1,2 . Then W2 is T-invariant and dim W2  2.Continue in this way to obtain  ,...., and W = span  ,....,  such that 1 i i 1 i

1. i Wi1;

2. Ti  a1i1  a2i2 .... ai1,i i1  aiiai

3. Wi is T-invariant, and dim Wi  i.

Thus we can continue to reach i = n. With respect to B  1,...,n 

a a a ... a   11 12 13 1n   0 a22 a23 ... a2n   0 0 a ... .a   33 3n  T B   . . . .     . . . .   . . . .     0 0 0 ... ann 

If T is triangulable, then the characteristic polynomial has the form

d1 dk T  x  c1  ...x  ck 

Thus is a product of linear factors, since mT | T. 2.6 Nilpotent transformation and Index of Nilpotency

In linear algebra, a nilpotent matrix is a square matrix N such that for some positive integer k. The smallest such k is sometimes called the degree of N. More generally, a nilpotent transformation is a linear transformation T of a vector space such that Tk = 0 for some positive integer k. A linear transformation T on a vector space V is said to be nilpotent of index k  1 if T k  0 but

k1 T  0 .

Analogously, definition of a nilpotent matrix can be given; It is immediate that a linear transformation T is nilpotent of index k if and only if its corresponding matrix is nilpotent of index k. For example any n n matrix, whose elements on and below the diagonal are zeros is nilpotent of index  n.

26 2.7 Invariants of a nilpotent transformation

The following theorem is helpful to know about the invariants of a nilpotent transformation Theorem If T  A(V) is nilpotent of index k, then there exists an ordered basis B of V such that the matrix of T relative to B is of the form

M O O O   K1   O M K 0... O  2  ......     O O O M   KR 

where k  k1  k2  ...  kr and for any positive integer t, M t is t t matrix given by

 0 1 0... 0     0 0 1... 0  M  ......  t    0 0 0... 1     0 0 0... 0 

Proof Since T k  0, the minimal polynomial m(x) of T divides x k , hence m(x)  xl for some

k1 k l  k. AsT  0,l  k 1. Hence l  k. So that m(x)  x . Let q1 (x),q2 (x),...,qr (x) be the elementary divisors of T. Since each of them divides the minimal polynomial

k ki m(x)  x ,qi (x)  x for some ki  k. so we can re-arrange qi (x)'s in such a way that

k ki k1  k2  ...  kr. As the LCM of there qi (x)'s is x , we get k  k1 . Now the matrix qi (x)  x relative to x is ki  ki matrix.

 0 1 0 ... 0     0 0 1 ... 0  M  ......  ki    0 0 0 ... 1     0 0 0 ... 0 

27 Thus, V has an ordered basis B such that the matrix of T relative to B is

M O O   k1   O M O  k2  O O M   kr 

Remark Since the elementary divisors of T above are uniquely determined, the integers k1,k2 ,...,kr in the above theorem are uniquely determined by T. These integers are called the invariants of the nilpotent linear transformation T.

If T is nilpotent, the subspace M of V, of dimension m, which is invariant under T, is called cyclic with respect to T if:

 MTm = (0) , MTm-1 (0) ;

 There is an element z such that z , zT,….. zTm-1 form a basis of M.

2.8 The Primary decomposition theorem

The primary decomposition theorem states that decomposition is determined by the minimal polynomial of α :

Let α:V→V be a linear operator whose minimal polynomial factorises into monic, coprime polynomials- mα(t)=p1(t)p2(t)

Then, V= W1⊕W2

Where the Wi are α-invariant subspaces such that pi is the minimum polynomial of α|Wi

Repeated application of this result (i.e., fully factorising mα into pairwise coprime factors) gives a more general version: if mα(t)=p1(t)...pk(t) as described, then

V=W1⊕...⊕Wk

With α-invariant Wi with corresponding minimal polynomial pi.

28 It may now be apparent that diagonalisation is a special case in which each Wi has a minimal polynomial which consists of a single factor, (t-λi); i.e. if mα=(t-λ1)...(t-λk) for distinct λi then α is diagonalisable with the λi as the diagonal entries.

Proof of the Primary Decomposition Theorem

The theorem makes two assertions; that we can construct α-invariant subspaces Wi based on the pi; and that the direct sum of these Wi constructs V.

For the first, a result about invariant subspaces is needed:

Lemma: If α, β:V→V are linear maps such that αβ = βα, then ker β is α-invariant.

Proof: Take w∈Ker β - we need to show that α(w) is also in ker β. Now β(α(w))=α(β(w)) by assumption, so β(α(w))=α(0) since w∈ker β , = 0 since α is a linear map. But if β(α(w))=0 then α(w) ∈ ker β, hence ker β is α-invariant. Given this result, we now take the Wi as Ker pi(α). Then since pi(α)α = αpi(α), it follows that ker pi = Wi is α-invariant.

We now seek to show that (i) V = W1 + W2, and (ii) W1∩W2 = {0} (that is, V decomposes as a direct sum of the Wi's.)

Using Euclid's Algorithm for polynomials, since the pi are coprime there are polynomials qi such that p1(t)q1(t) + p2(t)q2(t) =1.

So for any v ∈V, consider w1=p2(α)q2(α)v and w2=p1(α)q1(α)v. Then v= w1 + w2 by the above identity. We can confirm that w1∈W1: p1(α)w1=mα(α)q2(α)v = 0. Similarly, w2∈W2.

So we have (i). For (ii), let v ∈ W1∩W2. Then v = q1(α)p1(α)v + q2(α)p2(α)v = 0. So W1∩W2 = {0}.

Finally, for the claimed minimum polynomials, let mi be the minimal polynomial of α|Wi . We have that pi(α|Wi)=0, so the degree of pi is at least that of mi. This holds for each i.

However, p1(t)p2(t)=mα(t)=lcm{m1(t), m2(t)} so we obtain deg p1 + deg p2 = deg mα ≤ deg m1 + deg m2.

It follows that deg pi = deg mi for each i, and given monic pi it must be that mi=pi. The proof is complete.

29 2.9 Jordan block and Jordan forms

k k1 Let f (x)  x  ak1x ...a1x  a0 be any monic polynomial over a field F. Then the k  k matrix

 0 0 ... 0 a   0   1 0 ... 0 a1   0 1 ... 0 a   2  ......   0 0 ... 0 a   k2     0 0 ... 1 ak1 

is called the companion matrix of f (x) and is denoted by C[ f (x)].

Example 1. The companion matrix of f (x)  x3  3x2  2x 1 is

0 0 0    1 0  2   0 1  3

Example 2. Consider the polynomials f (x)  x4  5x2 1.

We can write it as f (x)  x4  0x3  5x2  0x 1.

Hence its companion matrix is

0 0 0 1   1 0 0 0 

0 1 0 5      0 0 1 0 

m m1 Theorem Let f (x)  x  am1x ...  a1x  a0 be any monic polynomial over a field F. Then three exists a linear Trans a linear transformation T on a vector space W of dimension m  deg f (x), such that the minimal polynomial of T is f (x).

30 Proof Let v1,v2 ,...vm )be an ordered basis of W. Define a linear transformation T on W such that

T(v1 )  v2 ,T(v2 )  v3 ,...,T(vm1 )  vm ,T(vm )  a0v1  a1v2 ...  am1vm .

Now

2 m1 {v1,T(v1 ),T (v1 ),...,T (v1 )}  {v1,v2 ,...,vm}is a linearly independent set, the degree of the minimal polynomial of v1 relative to T is at least m and hence m, as dim W = m; also degree of the minimal polynomial of any vector or of a linear transformation cannot exceed the dimension of underlying vector space .Now the defining relations of T give

m 2 2 m1 T (v1 )  a0v1  a1T(v1 )  a1T (v1 )  a2T (v1 ) ...am1T (v1 ) i.e.,

m m1 m2 T  am1T  am2T ...  a1T  a0 I)v1  0.

m m1 m2 Hence f (x).  x  am1x  am2 x ...  a0 is the minimal polynomial of v1 relative to T and thus we get that f (x).is the minimal polynomial of T.

Theorem If the minimal polynomial f (x) of a linear transformation T on V is of degree equal to the dimension of V, then V has on ordered basis such that matrix of T relative to that basis is C( f (x)).

k k1 Proof Let f (x)  x  ak1x ...  a0 be the minimal polynomial of T  A(V) where dim V = K.

Now there exists a vector v1 ( 0) V such that f (x) is minimal polynomial of v1 relative to T.

2 k1 Hence {v1,T(v1 ),T (v1 ),...,T (v1 )}is a basis of V.

2 k1 Put v2  T(v1 ),v3  T (v1 )  T(v2 ),...,vk  T (v1 )  T(vk1 ).

Now (v1,v2 ,...,vk ) is an ordered basis of V such that T(v1 )  v2 ,T(v2 )  v3 ,...,T(vk1 )  vk and

k k1 T(vk )  T (v1 )  a0v1  a1T(v1 ) ...  ak1T (v1 ) as f (T)v1  0.

This gives that T(vk )  a0v1  a1v2 ...  ak1vk . The matrix of T relative to the ordered basis

(v1,v2 ,...vk ) is

31  0 0 ... 0 a   0   1 0 ... 0 a1   0 1 ... 0 a   2   C( f (x)). ......   0 0 ... 0 a   k2     0 0 ... 1 ak1 

Corollary Let A be an n n matrix over a field F. If the minimal polynomial f (x) of A is of degree n, then there exists a non-singular matrix P over F such that P1 AP is C( f (x)).

Proof Let V be an n dimensional vector space over F and B  (v1,v2 ,...vn ) be an ordered basis of V. Let T be a L.T. on V whose matrix relative to B is A. Then the minimal polynomial of T is also , since deg is n, there exists an ordered basis B  (w1,w2 ,...wn ) of V such that matrix of T relative to B‟ is A' C( f (x)).if P is the matrix of w1,w2 ,...wn ) relative to ( v1,v2 ,..vn ) then

A' P1 AP . Hence P1 AP  C( f (x)).

n Let T be a cyclic linear transformation on V and f (x)  x  an1 ...  a1x  a0 be its minimal polynomial. Then the companion matrix of f (x) viz,

 0 0 ... 0 a   0   1 0 ... 0 a1   0 1 ... 0 a  A   2  ......   0 0 ... 0 a   n2     0 0 ... 1 an1  is called Jordan Matrix of T.

Theorem Let f (x)  (q(x))k , be a monic polynomial over F of degree n with q(x) being any monic polynomial of degree m(i.e.n  mk). If T is cyclic linear transformation of V such that is the minimal polynomial of T then, relative to some ordered basis of V, the matrix of T is of the form

32 C N     . .   . .     . .     C N   . .  A     . .   . .     . .   . .     . .     C N 

Where C  C(q(x)) and N is the mm matrix

 0 0 ... 1     0 0 ... 0 

......       0 0 ... 0 

Proof Now for some v V,V  F[T]v.

Define

k1 k1 m1 k1 v1  [q(T)] v,v2  T[q(T)] v,vm  T [q(T)] v,

k2 k2 m1 k2 vm1  [q(T)] v,vm2  T[q(T)] v,...,v2m  T [q(T)] v, v  [q(T)]k3 v,v  T[q(T)]k3 v,...v  T m1[q(T)]k3 v, 2m1 2m2 3m …(1) ...... v m1 v(k1)m1  v,v(K 1)m2  T(v),...vkm  T (v).

Thus each of the vi 's is of the from gi (T)v with gi (x)  0 and deg gi (x)  n.Let for some

km   ai  F,ai vi  0 such that some ai  0.i.e., ai gi (T)v  0. Noticing that gi (x)'s are all of i1  i  different degrees, we can see that g(x)  ai gi (x)  0,deg g(x)  n and g(T)v  0.This contradicts the fact that the minimal polynomial of v relative to T is f (x) and of degree n. Hence the n vectors v1,v2 ,...vkm ( vn )are linearly independent.

m m1 Let q(x)  x  a ..  a1x  a0 . …(2)

33 The definitions of vi 's given above yield the following:

T (v1 )  v2 T (v )  v 2 3 ......

T(vm1 )  vm

m k1 T(vm )  T [q(T)] v

m k1 k  [T  q(T)][q(T)] v, since [q(T)] v  0

m1 k1  (a0 I  a1T  ...  am1T )[q(T)] v

k1 k1 m1 k1  a0[q(T)] v  a1T[q(T)] v  ...  am1T [q(T)] v

This implies that

T(vm )  a0v1  a1v2 ...  am1vm

T(vm1 )  vm2

T(vm2 )  vm3 ......

T(v2m1 )  v2m

m k2 T(v2m )  T [q(T)] v

 [q(T)]k1 v [T m  q(T)] [q(T)]k2 v .

m1 k2  v1  (a0 I  a1T ...  am1T )[q(T) ]v

This gives that

T(v2m )  v1  a0vm1  a1vm2 ...  am1v2m

34 Continuing in this manner, we get

T(v2m1 )  v2m2

T(v2m2 )  v2m3 ......

T(v3m1 )  v3m

T(v3m )  vm1  a0v2m1  ...  am1v3m ......

T(v(k 1)  v(k 1)m2 ......

T(vkm1 )  vkm

T(vkm )  v(k 2)m1  a0v(k 1)m1  a1v(k 1)m2  ...  am1vkm

nkm Using the fact that if T(vi )  a ji v j , then in the ith column of matrix (a ji ),the jth entry is the j1 coefficient of v j , all above equations yield that the matrix of T relative to (v1,v2 ,...vkm ) is

C N O O ... O O    O C N O ... O O  O O C N ... O O     ......    O O O O ... C N  O O O O ... O O     

0 0 ... 0 a   0  1 0 ... 0 a1  where C is mm matrix 0 1 ... 0 a   C(q(x))andN  2  0 0 ... 1 am1     

35 0 0 ... 0 1   0 0 ... 0 0 is the mm matrix       0 0 ... 0 0

This proves the theorem.

Jordan Matrix

Let q(x) be a monic polynomial of degree, say m, over a field F, for any positive integer k, the km km matrix is given by

C N O ... O O    O C N ... O O 

O O O ... C N      O O O ... O C 

Where C is the companion matrix of and N is the mm matrix

0 0 ... 0 1   0 0 ... 0 0

0 0 ... 0 0     0 0 ... 0 0 called the Jordan matrix of f (x)  (q(x))k relative to .

Example Consider f (x)  (x2 1)3 , f (x) is of degree 6 over Q.

Here q(x)  x2 1. The companion matrix of is

0 1 C    1 0 

We have a linear transformation whose minimal polynomial is f(x)=(x2 + 1)3 then there exists an 6 ordered basis B  (u1,u2 ,u3 ,u4 ,u5 ,u6 ) of Q such that the matrix of T relative to this basis is the companion matrix of f (x)  (x2 1)3  x6  3x4  3x2 1

36 0 0 0 0 0 1   1 0 0 0 0 0  0 1 0 0 0  3 i.e., A    0 0 1 0 0 0    0 0 0 1 0  3   0 0 0 0 1 0 

6 At the same time, there exists an ordered basis B' (v1,v2 ,v3 ,v4 ,v5.v6 ) of Q such that the matrix of T relative to B is

C N O    0 1 A' O C N  Where N =     0 0 O O C 

0 1 0 1 0 0    1 0 0 0 0 0  0 0 0 1 0 1  ThusA'   0 0 1 0 0 0    0 0 0 0 0 1   0 0 0 0 1 0 

So if P is the matrix of B’ relative to B then A' P1 AP. Hence A' and A are similar matrices, A' is the Jordan matrix of f (x) relative to x 2 1.

Example Let a  F and f (x)  (x  a)n .

n n n1 n n2 2 n n Now f (x)  x  C1x a C2 x a ...  (1) a where for

n(n 1)...(n  r 1) 1  r  n,nC  r 1,2,3..r

37 So the companion matrix of f (x) is

0 0 0 ... 0 (1)n1 a n    1 0 0 ... 0 (1)n n c a n1  A  1 0 1 0 ... 0 (1)n1 n c a n2   2   n  0 0 0 ... 1 c1a 

At the same time the companion matrix of q(x)  x  a is (a). If N  (1), then n n matrix

a 1 0 ... 0 0 C N 0 ... 0      0 a 1 ... 0 0  0 C N ... 0    A' 0 0 0 ... a 1  0 0 ... C N  =       0 0 0 ... 0 a  0 0 ... O C     

Since A and A‟ are the matrices of same linear transformation on F (n) , A and A’ are similar, i.e.,

1 there exists a non-singular matrix P  M n (F) such that A' P AP

Remark given a  F, the n n matrix

a 1 0 0 ... 0 0   0 a 1 0 ... 0 0 A  f (x)  (x  a)n 0 0 0 0 ... a 1 is such that is     0 0 0 0 ... 0 a

The above matrix f (x)  (x  a)n is such that every eigen value of A is a root of its minimal polynomial and f (x) has a as its only root, we conclude that a is the only eigen value of A. This matrix A is called a Jordan Block.

Thus we can immediately, say that the matrices

5 1 0 0 1 1 0    2 1   0 5 1 0    0 1 1  0 2 , , 0 0 5 1        0 0 1   0 0 0 5

38 have respectively 2, - 1, 5 as their only eigen values. So if we can show that a given matrix is similar to a Jordan Block ,it is very easy to determine its eigen values .

Theorem Let T be a cyclic linear transformation on a vector space V and

a a a f (x) ] f1 (x)] 1[ f 2 (x)]1 2...[ ft (x)] t be the minimal polynomial of T, where fi (x)'s are distinct irreducible monic polynomials over F and ai  1. Then there exists an ordered basis B of V such that the matrix of T relative to B is

 J   1     J   2       J t 

a Where each J i (i 1,2,...,t) is the Jordan matrix of [ fi (x)] i relative to fi (x),

Proof We can write

V  V1 V2 ...Vt (1)

for some T-invariant subspaces Vi of V such that Vi is cyclic relative to Ti , the restriction of T to

a Vi and [ fi (x)] i is the minimal polynomial of Ti We can find an ordered basis B  (w ,w ,...w )of V such that the matrix of T relative to B is the Jordan matrix J of i 1 2 ni i i i i

a 1 1 1 2 2 2 t t t [ f (x)] i relative to f (x), Now B  (w ,w ,..., w ,w ,w ,..., w ,...w , w ,..., w ) is an ordered i i 2 2 n1 1 2 n2 1 1 nt

i basis of V. Let J i  (blm ).Then

ni i i i i T(wm )  Ti (wm )  blm wl i,m (2) l1

Observe that T(wi ) is expressible as a linear combination of wi ,wi ,..., wi and the coefficients bi m 1 2 ni lm are coming from J i . Keeping this in mind one can easily convince oneself that the matrix of T relative B is

39  J   I  J  2           J t 

Following immediate

Corollary Let A M n (F) be such that its minimal polynomial f (x) is of degree n. If

a1 at f (x)  [ ft (x)] ...[ ft (x)] where fi (x)'s are irreducible monic polynomials over F and ai  1. Then there exists a non-singular matrix P such that

 J   1   J 2  P 1 AP           J t 

Where each J i is the ai  ai Jordan Block

a 1 0 ... 0   i   0 ai 1 ... 0   0 0 a ......   i   0 0 0 ... 1     0 0 0 ... ai 

a a Let T be a cyclic linear transformation on a vector space VF and f (x)  [ f1 (x)] 1...[ ft (x)] t , where fi (x)'s are distinct irreducible monic polynomials over F and ai  1, be the minimal polynomial of T. Then the natrix

 J   1   J 2  P 1 AP           J t 

40 ai Where J i the Jordan matrix of [ fi (x)] relative to fi (x), is called Classical Canonical matrix or Jordan Canonical form of T; the above matrix is also called Classical Canonical matrix of the polynomial f (x).

Definition Let T be a linear transformation on a vector space V and V1,V2, ...,Vt

be T-invariant subspaces of V such that V  V1 V2 ,...,Vt. If Ti is the restriction of T to Vi then T is called a direct sum of T1,T2 ,...Tt and we write T  T T2 ...Tt.

If Ai is the matrix of Ti relative to some ordered basis Bi of Vi and B is the ordered basis of V, which is the union of all the Bi 's such that in the ordering, the members of B1 come first, those of

B2 come next, those of B3 come next and so on. It can be easily seen that the matrix A of relative to B is

 A   1   A2           At 

we say that A is direct sum of A1, A2 ,..., At and write A  A1  A2 ... At.

Example Let T be a linear transformation on Q3 whose minimal polynomial is f (x)  (x2 1)(x 1).x2 1 and x 1 are distinct monic irreducible factors of f (x).The respective

0 1 companion matrices of these polynomials are J1    and J 2  (1).So the Classical 1 0  0 1 0   J O     1  1 0 0 Canonical matrix of T is   =    O J 2    0 0 1

41 Example Let T be a linear transformation on C5, whose minimal polynomial is f (x)  (x 1)2 (x  2)(x  3)2 . Since deg f (x)  dimC 5 ,T is a cyclic linear trans-formation. The respective Jordan matrices of (x 1)2 ,(x  2) and (x  3)2 are

1 1 3 1     So the Classical Canonical matrix of T is J z   , J 2  (2), J3    0 1 0 3

1 1 0 0 0    J1 O O  0 1 0 0 0   O O O 0 0 2 0 0  2  =      O O J 3  0 0 0 3 1   0 0 0 0 3

Notice that 1,2,3 coming in the diagonal, are the eigen values of T.

2 Example Consider the linear transformation T on Q defined by T(e1 )  e2 ,T(e2 )  3e1  2e2 where 2 (e1,e2 ) is the standard basis of Q . Since e1,T(e1 ) e1,e2 is linearly independent, the minimal 2 polynomial of e1 relative to T is at least of degree 2; hence 2, as it cannot exceed 2(= dim Q ). So T is a cyclic linear transformation. Now the matrix of T relative to (e1,e2 ) is

0 3  A    …(1) 1 2

Since A2  2A  3I' 0, and T is cyclic, we get f (x)  x2  2x  3  (x 1)(x  3) is the minimal polynomials of T. Now x 1 and x  3 are the distinct monic irreducible factors of f (x); their respective Jordan matrices (here just the companion matrices) are -1 and 3.

1 0 Hence the Classical Canonical matrix of T is   0 3

Let us determine an ordered basis of Q2 relative to which the matrix of T is the above classical canonical matrix. Define

u1  (T  3I)e1  T(e1 )  3I(e1 )  3e1  e2 …. (2)

42 u2  (T  I)e1  T(e1 )  I(e1 )  e1  e2 …. (3)

2 u1,u2 are linearly independent. So B  (u1,u2 ) is an ordered basis of Q .

Now T(u1 )  3T(e2 )  3e2  (3e1  2e2 )  3e1  e2 i.e., T(u1 )  u1.

1 0 Similarly T(u2 )  3u2 .Hence the matrix of T relative to (u1,u2 ) is A'   . 0 3

 3 1 Form (2) and (3) it is clear that the matrix of u1,u2 ) relative to (e1,e2 ) is P    . 1 1

It can be easily verified that A' P1 AP.

2.10 Cyclic modules

Let R be a ring, and let M be an abelian group. Then M is called a left R-module if there exists a scalar multiplication : R × M -> M denoted by (r,m)=rm, for all r R and all m M, such that for all r,r1, r2 R and all m, m1, m2 M, (i) r(m1 + m2) = r m1 + r m2 (ii) ( r1 + r2 ) m = r1 m + r2 m (iii) r1 ( r2 m ) = ( r1 r2 ) m (iv) 1 m = m .

Example 1 (Vector spaces over F are F-modules) If V is a vector space over a field F, then it is an abelian group under addition of vectors. The familiar rules for scalar multiplication are precisely those needed to show that V is a module over the ring F.

Example 2 (Abelian groups are Z-modules) If A is an abelian group with its operation denoted additively, then for any element x Z and any positive integer n, we have defined nx to be the sum of x with itself n times. This is extended to negative integers by taking sums of -x. With this familiar multiplication, it is easy to check that A becomes a Z-module.

Another way to show that A is a Z-module is to define a ring homomorphism :Z->End(A) by letting (n)=n1, for all n Z. This is the familiar mapping that is used to determine the characteristic of the ring End(A). The action of Z on A determined by this mapping is the same one used in the previous paragraph.

43 If M is a left R-module, then there is an obvious definition of a submodule of M: any subset of M that is a left R-module under the operations induced from M. The subset {0} is called the trivial submodule, and is denoted by (0). The module M is a submodule of itself, an improper submodule. It can be shown that if M is a left R-module, then a subset N M is a submodule if and only if it is nonempty, closed under sums, and closed under multiplication by elements of R.

If N is a submodule of RM, then we can form the factor group M/N. There is a natural multiplication defined on the cosets of N: for any r R and any x M, let r(x+N) = rx+N. If x+N=y+N, then x-y N, and so rx-ry=r(x-y) N, and this shows that scalar multiplication is well- defined. It follows that M/N is a left R-module.

Any submodule of RR is called a left ideal of R. A submodule of RR is called a right ideal of R, and it is clear that a subset of R is an ideal if and only if it is both a left ideal and a right ideal of R.

For any element m of the module M, we can construct the submodule

Rm = { x M | x = rm for some r R }.

This is the smallest submodule of M that contains m, so it is called the cyclic submodule generated by m. More generally, if X is any subset of M, then the intersection of all submodules of M which contain X is the smallest submodule of M which contains X. We will use the notation for this submodule, and call it the submodule generated by X. We must have Rx for all x X, and then it is not difficult to show that

< X > = x XRx.

The left R-module M is said to be finitely generated if there exist m1, m2, . . . , mn M such that

M= Rmi.

In this case, we say that { m1, m2, . . . , mn } is a set of generators for M. The module M is called cyclic if there exists m M such that M=Rm.

Let M and N be left R-modules. A function f:M -> N is called an R-homomorphism if f(m1 + m2) = f(m1) + f(m2) and f(rm) = rf(m) for all r R and all m, m1, m2 M. The set of all R-homomorphisms from M into N is denoted by

HomR(M,N) or Hom(RM,RN).

For an R-homomorphism f HomR(M,N) we define its kernel as ker(f) = { m M | f(m) = 0 }.

44 We say that f is an isomorphism if it is both one-to-one and onto. Elements of HomR(M,M) are called endomorphisms, and isomorphisms in HomR(M,M) are called automorphisms. The set of endomorphisms of RM will be denoted by EndR(M).

A submodule N of the left R-module M is called a maximal submodule if N M and for any submodule K with N K M, either N=K or K=M. Consistent with this terminology, a left ideal A of R is called a maximal left ideal if A R and for any left ideal B with A B R, either A=B or B=R. Thus A is maximal precisely when it is a maximal element in the set of proper left ideals of R, ordered by inclusion. It is an immediate consequence of that every left ideal of the ring R is contained in a maximal left ideal, by applying the proposition to the set X = {1}. Furthermore, any left ideal maximal with respect to not including 1 is in fact a maximal left ideal.

2.11 Simple modules

Let R be a ring, and let M be a left R-module. For any element m M, the left ideal

Ann(m) = { r R | r m = 0 } is called the annihilator of m.

The ideal Ann (M) = { r R | r m = 0 for all m M }is called the annihilator of M.

The module M is called faithful if Ann(M)=(0).

A nonzero module RM is called simple (or irreducible) if its only submodules are (0) and M.

We first note that a submodule N M is maximal if and only if M/N is a simple module. A submodule N M is called a minimal submodule if N (0) and for any submodule K with N K (0), either N=K or K=(0). With this terminology, a submodule N is minimal if and only if it is simple when considered as a module in its own right.

The following conditions hold for a left R-module M: (a) The module M is simple if and only if Rm=M, for each nonzero m M. (b) If M is simple, then Ann(m) is a maximal left ideal, for each nonzero m M. (c) If M is simple, then it has the structure of a left vector space over a division ring.

2.12 Semi- Simple modules

Let M be a left R-module. The sum of all minimal submodules of M is called the socle of M, and is denoted by Soc(M). The module M is called semi-simple if it can be expressed as a sum of minimal submodules.

45 A semisimple module R M behaves like a vector space in that any submodule splits off, or equivalently, that any submodule N has a complement N' such that N+N'=M and N N'=0.

Any submodule of a semisimple module has a complement that is a direct sum of minimal submodules.

The following conditions are equivalent for a module R M. (1) M is semisimple; (2) Soc (M) = M. (3) M is completely reducible; (4) M is isomorphic to a direct sum of simple modules.

2.13 Schur’s Lemma

Schur's lemma is an elementary but extremely useful statement in representation theory, an elementary observation about irreducible modules, which is nonetheless noteworthy because of its profound applications. In the group case it says that that if M and N are two finite-dimensional irreducible representations of a group G and φ is linear map from M to N that commutes with the action of the group, then either φ is invertible, or φ = 0. An important special case occurs when M = N and φ is a self-map. The lemma is named after Issai Schur who used it to prove Schur orthogonality relations and develop the basics of the representation theory of finite group.

Lemma:

1. Suppose M and N are simple R-modules, and  : M  N is a homomorphism. Then  is either the zero homomorphism or an isomorphism.

2. Suppose M is a simple R module. Then EndR (M) is a division ring.

Proof .

1. Suppose  is non-zero. Then we have to show that  is an isomorphism, i.e. is both injective and surjective.

We know that ker( ) is a submodule of M. It can‟t be the whole of M, because is non-zero. So (since M is simple) Ker ( ) must be 0; in other words, is injective.

2. We know that EndR (M) is a ring; so it suffices to show that every non-zero element of EndR

(M has an inverse. But if is a non-zero element of EndR (M), then is a non-zero homomorphism from M to M, so by the first part of Schur‟s Lemma is an isomorphism; so

has an inverse, and thus this is an isomorphism form M to M, and hence an element of EndR (M).

46 One of the most important consequences of Schur's lemma is the following:

Corollary Let V be a finite-dimensional, irreducible G -module taken over an algebraically closed field. Then, every G -module homomorphism f:V→V is equal to a scalar multiplication.

Proof Since the ground field is algebraically closed, the linear transformation f:V→V has an eigen value; call it  . By definition, f−  1 is not invertible, and hence equal to zero by Schur's lemma. In other words, f=  , a scalar.

2.14 Free modules

The module M is called a free module if there exists a subset X M such that each element m M can be expressed uniquely as a finite sum m= ai xi, with a1, . . . , an R and x1, . . . , xn X.

We note that if N is a submodule of M such that N and M/N are finitely generated, then M is finitely generated. In fact, if x1, x2, . . . , xn generate N and y1+N, y2+N, . . . , ym+N generate M/N, then x1, . . . , xn, y1, . . . , ym generate M.

The module RR is the prototype of a free module, with generating set {1}. If RM is a module, and X M, we say that the set X is linearly independent if ai xi=0 implies ai=0 for i=1,...,n, for any distinct x1, x2, . . . , xn X and any a1, a2, . . . , an R. Then a linearly independent generating set for M is called a basis for M, and so M is a free module if and only if it has a basis.

Examples:

1. For any positive integer n, R n is a free R-modules.

2. The matrix ring M mn (R) is a free R-module with basis Eij ,i 1,...m, j 1,...,n.

3. The polynomial ring R[X ] is a free R-module with basis 1, X, X 2 ,....

4. The zero module is free with the empty set as basis.

2.15 Unit Summary/ Things to Remember

1. If V and W are finite-dimensional, and one has chosen bases in those spaces, then every linear map from V to W can be represented as a matrix; this is useful because it allows concrete

47 calculations. Conversely, matrices yield examples of linear maps: if A is a real m-by-n matrix, then the rule f(x) = Ax describes a linear map Rn → Rm.

2. Similarity of Linear Transformation: A mapping that associates with each linear transformation T on a vector space the linear transformation R-1PR that results when the coordinates of the space are subjected to a nonsingular linear transformation R.

3. An invariant subspace of a linear mapping T : V → V from some vector space V to itself is a subspace W of V such that T(W) is contained in W.

4. In linear algebra, a nilpotent matrix is a square matrix N such that for some positive integer k. The smallest such k is sometimes called the degree of N.

5. A linear transformation T on a vector space V is said to be nilpotent of index k  1 if T k  0 but

k1 T  0 .

6. A Jordan block over a ring R (whose identities are the zero 0 and one 1) is a matrix which is composed of 0 elements everywhere except for the diagonal, which is filled with a fixed element , and the entries above and right to the main diagonal are composed of unities of the ring.

6. Primary Decomposition Theorem: “Let m(x) be the minimal polynomial of T: V→ V , dim

V< such that m(x)= m1 (x) m2 (x) where gcd (m1, m2 ) =1then there exists T-invariant subspaces

V1 , V2 such that V= V1 V2.”

7. A cyclic module is a module over a ring which is generated by one element. A left R-module M is called cyclic if M can be generated by a single element i.e. M = (x) = R x for some x in M. Similarly, a right R-module N is cyclic, if N = y R for some y in N.

8. The simple modules over a ring R are the (left or right) modules over R which have no non-zero proper submodules. Equivalently, a module M is simple if and only if every cyclic submodule generated by a non-zero element of M equals M. The simple modules are precisely the modules of length 1; this is a reformulation of the definition. They are also called irreducible modules.

9. A module over a (not necessarily commutative) ring with unity is said to be semisimple (or completely reducible) if it is the direct sum of simple (irreducible) submodules.

48 10. Shur’s Lemma: If M and N are two simple modules over a ring R, then any homomorphism f: M → N of R-modules is either invertible or zero. In particular, the endomorphism ring of a simple module is a division ring.

11. A free module is a module with a free basis: a linearly independent generating set.

2.16 Assignments/ Activities

1. Find all possible Jordan canonical forms of a matrix with characteristic polynomial p(x) over C in each of the following cases:

(a) p(x)  (x 1)2 (x 1).

(b) p(x)  (x  2)3 (x  5)2 .

(c) p(x)  (x 1)(x  2)2 (x2 1).

2. Show that every sub module of the quotient module M/N can be expressed as (L+N)/N for some sub module L of M.

3. In the matrix ring M n (R), let M be the sub module generated by E11, the matrix with 1 in row 1, column 1, and 0‟s elsewhere. Thus M  {AE11 : A M n (R)}.Show that M consists of all matrices whose entries are zero except perhaps in column 1.

4. Continuing Problem 3, show that the annihilator of E11consists of all matrices whose first column is zero, but the annihilator of M is {0}.

5. If I is an ideal of the ring R, show that R/I is a cyclic R-module.

6. Let M be an R-module, and let I be an ideal of R. We wish to make M into an R/I-module via ( r  I)m  rm,r  R,m M.When will this be legal?

7. Assuming legality in Problem 6, let M 1 be the resulting R/I-module, and note that as sets,

M1  M. Let N be a subset of M and consider the following two statements:

(a) N is an R-sub module of M;

49 (b) N is an R/I- sub module of M1.

Can one of these statements be true and the other false?

2.17 Check Your Progress

1. Show that two transformations in A(V) are similar if and only if they have same families of elementary divisors.

m1 m2 mt 2. Let T  A(V) have f (x)  [ p1 (x) [p2 (x)] ...[ pt (x)] as its minimal polynomial, where

pi (x)'s are distinct monic irreducible polynomial and mi  1.Prove that the matrix of T in the Jordan Normal Form can be put in the form  R O O ... O   1   O R2 O ... O  A   ......       O O O O RI 

3. if has all its characteristic roots in F, show that the matrix of F in Jordan Normal Form, is triangular in the sense that all its entries below the diagonal are zero. 4. Give a proof, using matrix computation that if A is triangular n n matrix with entries

1,2 ,...,n on the diagonal, then (A  1I')(A  2 I')... (An I')  0 Where I‟ is the n n identity matrix. 5. Find the Jordan Normal Form of the matrix representing T in each the following. (i) T  A(Q2 ) having elementary divisors x  2, x  3

(ii) T  A(Q2 ) having elementary divisors x2  2x 1

(iii) T  A(Q3 ) having minimal polynomial x3  x 1

(iv) T  A(R4 ) having elementary divisors x2 , x2  x 1

6 6 (v) T  A(C ) having minimal polynomial x 1

50 2.18 Points for discussion / clarification

At the end of the unit you may like to discuss or seek clarification on some points. If so, mention the same.

1. Points for discussion

______

______- ______

2. Points for clarification

______

______- ______

1.19 References

6. I . N. Herstein, Topics in Algebra, Second Edition, John Wiley and Sons, 2006. 7. Lay, David C. , Linear Algebra and Its Applications (3rd ed.), Addison Wesley, 2005. 8. Rose John, A Course on Group Theory. New York: Dover, 1994. 9. P.B. Bhattacharya, S.K. Jain and S.R. Nagpaul, Basic Abstract Algebra,(2ndEdition),Cambridge University Press, Indian Edition, 1997. 10. Fraleigh B John, A first Course in Abstract Algebra, 7th Edition, Pearson Education, 2004. 11. Steven J. Leon, Linear Algebra With Applications (7th ed.), Pearson Prentice Hall, 2006.

51 UNIT-3 Field Theory

STRUCTURE PAGE NO.

3.1 Introduction 01

3.2 Objective 01

3.3 Extension Field 2-4

3.4 Algebraic and Transcendental Extension 4-5

3.5 Inseparable and Separable Extension 5-6

3.6 Normal Extension 06

3.7 Perfect & Finite Field 6-8

3.8 Primitive Element 0 9

3.9 Algebraic Closed Field 10-15

3.10 Automorphism of Field 10-15

3.11 Galois theory 15-16

3.12 Fundamental theorem of Galois theory 17-19

3.13 Solvability and Insolvability of polynomials by radicals 9-26

3.14 Unit Summary/ Things to Remember 26-27

3.15 Assignments/ Activities 27-28

3.16 Check Your Progress 28

3.17 Points for Discussion/ Clarification 29

3.18 References 29

52 3.1 Introduction

Fields are important objects of study in algebra since they provide a useful generalization of many number systems, such as the rational numbers, real numbers, and complex numbers. In particular, the usual rules of associativity, commutativity and distributivity hold.

When abstract algebra was first being developed, the definition of a field usually did not include commutativity of multiplication, and what we today call a field would have been called either a commutative field or a rational domain. In contemporary usage, a field is always commutative. In 1893, Heinrich M. Weber gave the first clear definition of an abstract field.

In 1910 Ernst Steinitz published the very influential paper Algebraische Theorie der Körper (German: Algebraic Theory of Fields). In this paper he axiomatically studies the properties of fields and defines many important field theoretic concepts like prime field, perfect field and the transcendence degree of an field extension.

Galois, was honored to be the first mathematician in linking group theory and field theory. Galois theory is named after him. However it was Emil Artin who first developed the relationship between groups and fields in great detail during 1928-1942.

The concept of field was used implicitly by Niels Henrik Abel and Évariste Galois in their work on the solvability of equations that there is no general formula expressing in terms of radicals the roots of a polynomial with rational coefficients of degree 5 or higher.

3.2 Objective

After the completion of this unit students should be able to:

 get the knowledge of finite fields that are used in number theory, Galois theory and coding theory  construct many codes as subspaces of vector spaces over finite fields.  know various types of field and their extensions .  define primitive element.  define Automorphism of extension.  prove Fundamental theory of Galois theory  know about rational polynomials and their roots.

53 3.3 Extension Fields

A field extension of a field F is pair ( k, ), where k is a field and  is a monomorphism of F into k.

Suppose E is a field and F is a subfield of E. The injection map i : F  E defined by i(x)  x for all x  F is a monomorphism. Consequently (E,i) is a field extension of F. In such a situation when i is a trivial type of mapping, we do not mention i and simply say that E is a field extension of F. In general case of a field extension (K, ) , K has a subfield  (F)isomorphic to F. We can identify any two isomorphic algebraic systems. Here also we shall normally identify F with  (F)and thus treat F as a subfield of K, then we shall simply say that K is a field extension of F. Henceforth F will be a field and K will be a field extension of F.

Now, as for any a  F, x  K,ax  K, we get an external composition F  K  K given by (a, x)  ax. One can immediately see that the additive group  K,  becomes a vector space over F relative to the external composition in K with respect to F defined above. Thus K must have a basis and dimension over F.

The dimension of K as a vector space over F is called the degree of K over F. In general [K : F] will denote the degree of K over F.

Example 1: Q[ 2]  a  b 2 a,b Q is field extension of Q, the subset 1, 2 forms a basis of Q[ 2] over Q. Consequently[Q[ 2 ]:Q=2.

Example 2: Consider an indeterminate x over a field F. Let K be the field of quotients of

2 n F[x].Then, as we know that for any a0 ,a1,a2 ,...an  F , a01 a1x  a2 x ...  an x  0 implies

2 3 n ai  0i,if follows that 1, x, x , x ,..., x ,... is an infinite subset of K which is linearly independent over F. Consequently [k : F] is infinite.

Hence K is said to be a finite or infinite extension of F according as the degree of K over F is finite or infinite.

54 Thus, in Example 1, Q[ 2] is a finite extension of Q and in Example 2, K is an infinite extension of F. Following theorem about degrees of field extensions is of great significance.

Theorem If K is a finite extension of F and L is a finite field extension of K then L is a finite extension of F and [L : F]  [L : K][K : F].

Proof Let [K : F]  n and [L : K]  m.Then we can find a basis x1, x2 ,..., xn of K over F and a basis y1, y2 , y3 ,..., ym of L over K. Consider the elements xi y j ;i 1,2,...,n; j 1,2,...m. If we can show that these mn elements of L are linearly independent over F and they generate L as a vector space over F, it will give that [L : F]  mn  [L : K][K : F] and the theorem will follow. For their linear independence let  aij xi y j  0 for some aij  F. That gives i, j m  n  aij xi y j  0.Since y1, y2 ,..., ym are linearly independent over K and aij xi  K for every j, j1  i1 

n we get aij xi  0j  1,2,...,m.This gives aij  0i, j as x1, x2 ,..., xn are linearly independent over i1

F. Hence, the elements xi y j are linearly independent over F. Consider any x  L. Then

m x  a j y j for some, a j  K, j 1,2,...,m. j1

n Again for each j,a j  aij xi for some aij  F, as x1, x2 ,...xn is a basis of K over F. Consequently i1

m n x  aij xi y j with aij  Fi, j. j1 i1

Hence, the mn elements xi y j ,i 1,2...,n, j 1,2,...m form a basis of L over F.

Subfield generated by a Subset

Let S be a subset of a field K, then a subfield K' of K is said to be generated by S if

(i)S  K'.

(ii) for any subfield L of K, S  L implies K' L.

55 Notation Subfield generated by a subset S will be denoted by  S  .

Essentially 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 of K generated by F  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 and its members are a1,a2 ,....,an then we write

F(S)  F(a1,a2 ,...an ).

A field K is said to be finitely generated over F if there exists a finite number of elements

a1,a2 ,...an in K such that K  F(a1,a2 ,...,an ).

In particular if K is generated by a single element over F, then K is called a simple extension of

F .Thus Q[ 2] = Q ( 2) is a simple extension of Q. Consider any a  K and let F(a) be the

subfield of K generated by a over F. Then for any a0 ,a1,a2 ,...,an  F,

2 n a0  a1a  a2a ...  ana  F(a). This means that F[a]  F(a).Consequently the field of quotients T of F(a).is also contained in F(a).However F  T and a T therefore F(a)  T and we get T  F(a),i,e, F(a) is the field of quotients of the subring of K generated by F a . This discussion can be extended to any set S and we can say F  S. Let us remark

n that for any f (x)  a0  a1x ...  an x in F[x] and for any a  K , f (a) denotes

n a0  a1a ...  ana It can also be verified that the mapping f (x)  f (a) is a ring homomorphism of F(x) into K.

3.4 Algebraic and Transcendental Extension

The extension K of F is called an Algebraic Extension of F if every element in K is algebraic over F

Field extensions which are not algebraic, i.e. which contain transcendental elements, are called transcendental.

56 For example, the field extension R/Q, that is the field of real numbers as an extension of the field of rational numbers, is transcendental, while the field extensions C/R and Q(√2)/Q are algebraic, where C is the field of complex numbers.

All transcendental extensions are of infinite degree. This in turn implies that all finite extensions are algebraic. The converse is not true however: there are infinite extensions which are algebraic. For instance, the field of all algebraic numbers is an infinite algebraic extension of the rational numbers.

If „a‟ is algebraic over K, then K[a], the set of all polynomials in „a‟ with coefficients in K, is not only a ring but a field: an algebraic extension of K which has finite degree over K. In the special case where K = Q is the field of rational numbers, Q[a] is an example of an algebraic number field.

Theorem Every finite extension of a field is algebraic extension.

Proof Let K be a finite extension of a field F. Let K : F] = n. For any a  K, F(a) is a subfield of k. But [F(a) : F]| n, so it is finite. Hence a is algebraic over F. Consequently K is algebraic extension of F.

3.5 Separable and Separable Extensions

If an element a of a field extension K of F is algebraic over F, then a is said to be separable (inseparable) over F, if the minimal polynomial of a over F is separable (inseparable).

An algebraic extension K of a field F is a said to be a separable extension, if every element of K is separable over F, otherwise K is said to be an inseparable extension.

As observed before, every polynomial over a field of characteristic zero is separable; we see that every algebraic extension of a field of characteristic zero is a separable extension.

2 However, if we take F  Z2 (t) and K be the splitting field of x  t , then K is a finite extension of F as [k : F]  2. But K is an inseparable extension as x 2  t has a repeated root in K. Notice that F is an infinite field of finite characteristic. We show that any algebraic extension of a finite field is separable.

Theorem Any algebraic extension field F is a separable extension.

Proof Let f (x) be any irreducible polynomial over F. Suppose f (x) is inseparable over F, so

p p 2 p mp f (x)  f [x ](Lemma 13.44), f (x)  0  B1x  2 x ...  m x for some

57 p i  F,(0  i  m.) As a  a ,a  F is an automorphism of F we can find ai  F such that

p i  ai i.

p p p p 2 p p mp Consequently f (x)  a0  a1 x  a2 x ...  am x

2 m p  (a0  a1x  a2x ...  amx )

This implies f (x) is not irreducible. This is a contradiction. Hence f (x) is separable. Thus in particular if K is any algebraic extension of F, the minimal polynomial of each element of K over F is separable. Hence by definition K is a separable extension of F.

3.6 Normal Extensions

An algebraic field extension L/K is said to be normal if L is the splitting field of a family of polynomials in K[X].

The normality of L/K is equivalent to each of the following properties:

Let Ka be an algebraic closure of K containing L. Every embedding σ of L in Ka which restricts to the identity on K, satisfies σ(L) = L.

Every irreducible polynomial in K[X] which has a root in L factors into linear factors in L[X].

Other Properties :

Let L be an extension of a field K, then

If L is a normal extension of K and if E is an intermediate extension (i.e., L ⊃ E ⊃ K), then L is also a normal extension of E.

If E and F are normal extensions of K contained in L, then the compositum EF and E ∩ F are also normal extensions of K.

If K is a field and L is an algebraic extension of K, then there is some algebraic extension M of L such that M is a normal extension of K.

3.7 Perfect and Finite Field

A field F is called Perfect if all finite extensions of F are separable.

58 There exists a simple criterion for perfectness:

A field F is perfect if and only if F has characteristic 0, or F has a nonzero characteristic p, and every element of F has a p-th root in F.

Fields having only a finite number of elements are called Finite Fields. Such fields do exist, for the ring of integers modulo any prime p.

Theorem Let F be an infinite field and E be some field extension of F. Let a,b  E be algebraic over F such they are separable over F. Then there exists c  K such that F(c)  F(a,b) and c  a  ab for some a  F.

Proof Let f (x) and g(x) be the minimal polynomial over F of a and b respectively and let m,n be their respective degrees. Let K be the splitting field of f (x)g(x) over E. Then a,b  K. Clearly every root of f (x) is a root of f (x)g(x), so K contains a splitting field of f (x).Similar is the case for g(x).since a,b are separable over F, f (x) has m distinct roots a1  a,a2 ,a3 ,...,am in K and g(x) has n distinct roots b1  b,b2 ,....bn in K.

ai  a For 2  i  m,2  j  n.Define ij   K. b  b j

These ij 's are finite in number. As F has in finite number of elements, clearly we can find an a( 0)  F such that a  iji, j, 2.

Then a(b  b j )  ai  ai, j  2,i,e.,ai  ab j  a  abi, j  2.

Now put c  a  ab  F(a,b), so F(c)  (a,b).we show that F(c)  F(a,b).

Since c  F(c) and every coefficient of f (x) is in F so also in K, we get that the polynomial h(x)  f (c  ax) K[x]. Further deg h(x)  deg f (x)  m (why?). Now h(b)  f (c  ab)  f (a)  0.Suppose that for some j  2,h(b j )  0 then f (c  ab j )  0

So that c  ab j  ai for some i since only a1,a2 ,....an are that roots of f (x).If i  1then ai  a gives us that c  ab j  a,i.e.ab j  a  ab  a  bj  b. This is a contradiction. So i  2.In that case we shall get

59 ab  a  ab j  ai does not divide h(x). Now x  b is a factor of h(x) over K. As b is also a root of g(x) in K, x  b is a common factor of g(x).

We claim that x  b is their HCF. As f (x) has no multiple root, (x  b)2 g(x).Since g(x)  (x  b)(x  b)(x  bn ) and each of (x  b j )( j  2) does not divide h(x), it follows that x  b is the

HCF of g(x) and h(x). Now h(x) F(c)[x]as c  F(c) and a  F.Also g(x) F(c)[x]. Let g1 (x) be the minimal polynomial of b over F(c).Then g1 (x) | g(x), g1 (x) | h(x) over F(c) and hence over K also, as F(c)  K.

So g1 (x) | (x  b) over K as x  b is the HCF of g(x) and h(x) over K. However, g1 (x) is of positive degree and is monic so we must have This implies x  b  g1 (x) F(c)[x]. b  F(c).

Therefore a  c  abF(c)

Thus F(a,b)  F(c).Hence F(c)  F(a,b).

Theorem Any finite separable extension of an infinite field is a simple extension.

Proof Let F be an infinite field and K a finite field extension of F. There exist finite number of elements a1,a2 ,...,an in K such that K  F(a1,a2 ,...,an ). We prove the result by induction on n, If n  1, then K is already simple. Suppose n  1and theorem holds for all finite separable extension of F generated by less than n elements. Let K1  F(a1,a2 ,...an1 )  F(b).

Thus K  F(b,an ) that gives K  F(c) for some cF(b,an ). This proves the theorem.

Theorem Finite fields having the same number of elements are isomorphic.

n Proof Let K1 and K 2 be two fields each having q elements. By Theorem 13.52 q  p for some nN and prime number P, further characteristic of each of K1 and K 2 is p. Let

P1 and P2 be the prime subfields of K1 and K 2 respectively. As P1  Z /  p  and

q P2  Z /  p , P1  P2 . Now by Lemma 13.53, K1 is splitting field of x  x over P1 and is

q splitting field of y  y over P2 . Hence by Theorem 13.35, K1  K2

60 Notation A finite field of q elements is denoted by GF(q).

3.8 Primitive Element

A field extension L/K is called a simple extension if there exists an element θ in L with L = K(θ).

The element θ is called a primitive element, or generating element, for the extension; we also say that L is generated over K by θ.

A primitive element of a finite field is a generator of the field's multiplicative group. When said at greater length: In the realm of finite fields, a stricter definition of primitive element is used. The multiplicative group of a finite field is cyclic, and an element is called a primitive element if and only if it is a generator for the multiplicative group. The distinction is that the earlier definition requires that every element of the field be a quotient of polynomials in the primitive element, but within the realm of finite fields the requirement is that every nonzero element be a pure power.

The primitive element theorem provides a characterization of the finite field extensions which are simple and thus can be generated by the adjunction of a single primitive element.

A field extension L / K is finite and has a primitive element if and only if there are only finitely many intermediate fields F with K ⊆ F ⊆ L.

In this form, the theorem is somewhat unwieldy and rarely used. An important corollary states:

Every finite separable extension L / K has a primitive element.

In more concrete language, every separable extension L / K of finite degree n is generated by a single element x satisfying a polynomial equation of degree n, xn + c1xn−1 + ... + cn = 0, with coefficients in K. The primitive element x provides a basis [1, x, x2, ..., x n−1] for L as a vector space over K.

This corollary applies to algebraic number fields, which are finite extensions of the rational numbers Q, since Q has characteristic 0 and therefore every extension over Q is separable.

For non-separable extensions, one can at least state the following:

If the degree [L : K] is a prime number, then L / K has a primitive element.

61 If the degree is not a prime number and the extension is not separable, one can give counterexamples. For example if K is Fp(T, U), the field of rational functions in two indeterminates T and U over the finite field with p elements, and L is obtained from K by adjoining a p-th root of T, and of U, then there is no primitive element for L over K. In fact one can see that for any α in L, the element αp lies in K. Therefore we have [L : K] = p2 but there is no element of L with degree p2 over K, as a primitive element must have.

3.9 Algebraic Closed Field

A field K is algebraically closed if every non-constant polynomial in K[X] has a root in K.

An extension field L of K is an algebraic closure of K if L is algebraically closed and every element of L is algebraic over K .Using the axiom of choice, one can show that any field has an algebraic closure. Moreover, any two algebraic closures of a field are isomorphic as fields, but not necessarily canonically isomorphic.

Theorem A finite field cannot be algebraically closed. Proof The proof proceeds by the method of contradiction. Assume that a field F is both finite and algebraically closed. Consider the polynomial p(x)=x2−x as a function from F to F There are two elements which any field (in particular, F must have -- the additive identity 0 and the multiplicative identity 1 ). The polynomial p maps both of these elements to 0 Since F is finite and the function p:F F is not one-to-one, the function cannot map onto F either, so there must exist an element a of F such that x2−x a for all x F In other words, the polynomial x2−x−a has no root in F so F could not be algebraically closed

Proposition Every algebraically closed field is perfect. Proof Let K be an algebraically closed field of prime characteristic p . Take a K . Then the polynomial −a admits a zero in K . It follows that a admits the p th root in K . Since a is arbitrary we have proved that the field K is perfect.

62 3.10 Automorphism of the Field

Let F be any field and K be any field extension of F. Then an automorphism  of K is said to be an F-automorphism if  (x)  x for every x  F i.e.,  leaves every element of F fixed.

Clearly, the identity automorphism of K is an F-automorphism.

1 Let 1, 2 by any two F-automorphisms of K. Then 1 2  Aut(K)as Aut(K) is a group. At

the same time for any x  K,1 (x)  x, 2 (x)  x. This implies

1 1 1 1 (x)  x 2 (x)  x  1 2 (x)  1[a2 (x)]  1 (x)  x. Hence 1 2 is also an F- automorphisms of K. Thus the set of all F-automorphisms of K is a subgroup of the group of all automorphisms of K.

Notation G(K, F) will denote the group of all F-automorphisms of K. G(K, F) is called the Galois group of K over F which we will discuss in the next section .

Corollary Any set of automorphisms of K is linearly independent over K.

Lemma The set of all automorphims of a field form a group under resultant composition.

Proof Let Aut (K) be the set of all automorphisms of K.

(i) Closure Let 1, 2  Aut(K).Then  1 and  2 are both 1-1 and onto mappings. So  1 2 is also onto mapping. Further for any x, y  K.

1 2 (x  y)  1[ 2 (x  y)]

  1[ 2 (x)  2 (y)]

  1[ 2 (x)]  1[ 2 (y)]

  1 2 (x)  1 2 (y)

1 2 (xy)  1[ 2 (xy)]

  1[ 2 (x) 2 (y)]

 { 1[ 2 (x)]}{ 1[ 2 (y)]}

 [ 1 2 (x)][ 1 2 (y)]

63 This shows that  1 2 is an automorphism of K.

Hence 1 2  Aut(K).

(ii) Associatively: If follows from the fact that the resultant composition of mappings, in general, is associative. (iii) Existence of Identity : The identity map I on K is identity of Aut(K).

(iv) Existence of Inverse: Consider any   Aut(K).Since  is 1-1 and onto,  1 (the in

verse of the mapping ) exists. For any x  K, 1 (x)  y if and only if  (y)  x.Now

1 1 consider x1, x2  K.Let  (x1 )  y, (x2 )  y2 . Then  (y1 )  x1,  (y2 )  x2 so that

(y1  y2 )  (y1 ) (y2 )  x1  x2 and (y1 y2 )  (y1 )(y2 )  x1x2 . Hence

1 1 1 1  (x1 ) (x2 ).Consequently  is an automorphism of Ki.e.,a  Aut(K). This proves that Aut(K)is a group.

Lemma Let K be any field extension of F and a  K be algebraic over F. Than for every F- automorphism  of G(K, F)(a)is a conjugate of a over F.

n n n1 n2 Proof Let f (x)  x  a  an  x  an 2 x ...  a0 be the minimal polynomial of a over F.

n n1 n2 Then a  an1  a  an2a ...  a0  0

n n1 n2 So 0  (0)(a  an1a  an2a ...  a0 )

n n1 n2  [(a)] (an1 )[(a) (an2 )[(a)] ... (a0 )

n n1  [(a)] ]  an1[(a)] ...  a0

Since (ai )  aii  0,1,2,...n 1.

This shows that  (a) is also a root of f (x).Hence  (a) is a conjugate of a over F.

64 Remark Let K  F(a1,a2 ,...an )be a finite algebraic extension with {a1,a2 ,...,an } as a basis of K. Then each x  K is expressible as x  a1a1  a2a2 ....anan for some ai  F.

Suppose  is any F-automorphism of K. Then

(x)  (a1 )(a1 ) (a2 )(a2 ) ... (an )(an )

 a1(a1 )  a2(a2 ) ...  an(an ).

So that  (x) is known if we know  (a1 ),(a2 ),...,(an ) i.e.,  is completely determined by the images of the basis elements of K. In fact, more generally, if K is finitely generated over F and a1,a2 ,...an is a set of generators of K over F then  is determined by  (a1 ),(a2 ),...(an ).

Theorem (Artin) Let G be a finite group of automorphisms of a field K; F0 , the fixed field under G. Then the degree of K over F0 is equal to the order of the group G.

Proof Let o(G)  n we show that

(i) if [K : F0 ] is m, then m  n and

(ii) if [K : F0 ] is m, then m  n.

(i) Suppose m  n.Let 1 1 2 ,... n be all the members of G. Let {x1, x2 ,...xm}be a basis of K over F0 .Consider the system of m linear homogenous equations.

1 (x j )u1  2 (x j )un ...  n (x j )un  0j 1,2,...,m … (1)

As the number of equations is less than the number of variables, the equation (1) has a non- trival solution, say (y1, y2 ,...yn )over K.

Then 1 (x j )y1  2 (x j )y2 ...  n (x j )yn  0j 1,2,...m ….(2)

Consider any x  K.Then x  a1x1  a2 x2 ...  am xm for some ai  F0 as {x1, x2 ,..., xm ) is a basis of K over F0 . In (2), multiplying jth equation by a j, adding the resulting equations and using the fact that  i (a j )  a ji, j we get

65 y11 (x)  y2 2 (x) ...  yn n (x)  0x  K.

So that y11  y2 2 ...  yn n ō with atleast one of y j  0. This is not possible. Hence m < n so m  n.

(ii) Suppose there exist n 1 linearly independent elements say x1, x2 ,..., xn1 of K over F0 . Consider the system of n liner homogenous equations in n 1 unknowns.

 j (x1 )u1  j (x2 )u2 ...  j (xn1 )  0 for j 1,2,...,n …(3)

As the number of variables is greater than the number of equations these homogenous equations have a non-trivial solution. Let (z1.z2 ,..., zn1 ) be a non-trivial solution of equations (3) with smallest number, say r, of non-zero components. We can renumber then and suppose that z j  0j  r 1.

Then (3) gives

 j (x1 )z1  j (x2 )z2 ...  j (xr )zr  0j 1,2,...n ….(4)

zi Dividing the equations by zr and setting z' j  i, we get zr

 j (x1 )z'1  j (x2 )x'2 ...  j (xr1 )z'r1  j (xr )  0j …(5)

Now for j 1, j  I, so 1 (xi )  xii. We get from (4)

x1z'1 x2 z2 '...  xr 1zr1 'xr  0 …(6)

If all of z1 ', z2 ',..., zr1 ' were in F0 . then (6) would give x1 ' x2 ,...xr are linearly dependent over F0 .

This would not be possible. Hence at least one of these, say, z1 ' is not in F0 . Notice further that r  1otherwise we would have got z1 ' 0.As z1 ' F0 there exists some  i G such that

 i (z1 ')  z1 '. Applying  i to (5) we get

 i [ j (x1 )z1 '] i [ j (x2 )z2 ']...  i [(xr1 )zr1 '] i j (xr )  0j 1,2,...,n.

66 However  iG  G so that every  k G is the form  i j , we get

 j (x1 ) i (z1 ')  j (x2 ) i (z2 ') ...  j (xr1 )  i (zr1 ') j (xr )  0 …(7) for all j 1,2,...,n.

Subtracting (7) from (5) we get

 j (x1 )[z1 ' i (z1 ')]  j (x2 )[z2 ' i (z2 ')]  ... …(8)  j (xr1 )[zr1 ' i (zr1 ')]  0j  1,2,...,n

Now we put tk  zk ' i (zk ')k 1,2,...,r 1.

and tk  0k  r,r 1,...,n 1

Then (8) gives

t  (x )t  (x )t ...  (x )t  (x )t ...  (x ) n1  0j 1,2,...,n; …(9) j 1 1 j 2 2 j r1 r1 j r r …… j n1

Further t1  0 as z1 '  i (z1 ')so that (t1,t2 ,...,tr1,0,0...,0)is a non-trivial solution of the system of equation (3). It has less than r non-zero components. But by our choice z1, z2 ,...zr ,0,0...,0)is a non- trivial solution (3), having smallest number of non-zero components and these are r in number.

Hence we get a contradiction. This prove that m  n.Hence m  n.i.e., [K : F0 ]  o(G). This completes the proof.

Remark 1 Let G be any group (not necessarily finite) of automorphism of a field K such that

[K : F0 ]  0m is finite, where F0 is the fixed field under G.

Remark 2 Let K be a finite extension of F and G be the group of all F automorphism of K. Then the F [K : F ]  [K : F]. fixed field 0 under G contains F, so 0 Hence o(G)  [K : F].

67 3.11 GALOIS THEORY

One of the most elegant theories in Abstract Algebra is the “Galois theory of fields”. This theory is an excellent composite of the theory of groups and the theory of algebraic field extensions. It has many applications to the Theory of Equations and Geometry. Its fundamental concepts and its applications are as under:

A finite extension K of a field F is said to be a Galois Extension of F if F is the fixed subfield of K under the group G(K, F) of all F-automorphisms of K.

Corollary Let K  F(a) be a simple finite separable extension of F. Then K is a splitting field of the minimal polynomial of a over F if and only if F is the fixed field under the group of all F- automorphisms of K.

Proof Let f (x) be the minimal polynomial of a over F and let deg f (x)  m.Then [K : F]  m. Let a1  a,a2 ,a3 ,...,ar be the distinct conjugates of a in K. Then K  F(ai )i 1,2,...,r. For each i, there exists an F-automorphism  i of K such that  i (a1 )  ai . Since a1 generates K over F, each

 i is uniquely a1 ( determined. Further for any F-automorphism  of K , as  (a1 ) is a conjugate of

(Lemma 14.5),  (a1 )  ai for some ai . From this it follows that    i .Hence the group

G(K, F) consists of 1, 2 ,..., r .Let F0 be the fixed field under G(K, F).Then by the above theorem [K : F0 ]  o[G(K, F)]  r. So F  F0 if and only if r  m. Hence F is the fixed field under G if and only if f (x) has all its m roots in K i.e., if and K is the splitting field of f (x) over F.

Corollary Let be a simple finite separable extension of F. Then K is a splitting field of the minimal polynomial of a over F if and only if F is the fixed field under the group of all F- automorphisms of K.

Proof Let be the minimal polynomial of a over F and let deg Then Let

be the distinct conjugates of a in K. Then For each i, there exists an F-automorphism of K such that Since generates K over F, each

is uniquely determined. Further for any F-automorphism of , as is a conjugate of

68 Lemma 14.5),  (a1 )  ai for some ai . From this it follows that    i .Hence the group

G(K, F) consists of 1, 2 ,..., r .Let F0 be the fixed field under G(K, F).Then by the above theorem [K : F0 ]  o[G(K, F)]  r. So F  F0 if and only if r  m. Hence F is the fixed field under G if and only if f (x) has all its m roots in K i.e., if and K is the splitting field of f (x) over F.

3.12 FUNDAMENTAL THEOREM OF GALOIS THEORY

Fundamental Theorem of Galois Theory Let K be a finite, normal, separable field extension of a field F and let F(K, F) be the Galois group of K over F. Then the correspondence E  G(K, E) where E is a subfield if K containing F is 1-1 between the families of the subfields of K containing F and the family of all subgroups of G(K, F), satisfying the following conditions.

Given any subfield E of K containing F and subgroup H of G(K, F)

(i) E  KG(K,E)

(ii) H  G(K, K H )

(iii) [K : E]  o[G(K, E)]and [E : F]= index of G(K, E)in G(K, F)

(iv) E is a normal extension of F if and only if G(K, E)is a normal subgroup of G(K, F).

(v) When E is a normal extension of F. then, G(E, F) is isomorphic

to G(K, F) | G(K, E).

Proof Since K is a finite extension of F and F  E  K, we get that K is a finite normal separable extension of E. So E is same as the fixed field KG(K ,E) .

Thus (i) follows by definition KH  {x  K |(x)  x  H}, each   H is a K H automorphism of K. So that H  G(K, K H ).However o(H)  [K : K H ] (Theorem 14.8). At the same time as K is a normal extension of K H ' Theorem 14.15 gives that K H is the fixed field under G(K, K H ). So

69 [K : K H ]  o[G(K, K H )].Thus o(H)  o(G(K, K H )] and consequently H  G(K, K H ).This proves (ii).

Now as K is a normal separable extension of E; [K : E  o[G(K, E)]. Thus o[G(K, F)]  [K : F]  [K : E][E : F]  o[G(K, E)][E : F] gives

o[G(K, F)] [E : F]  . This proves (iii). o[G(K, E)]

Let E be a normal extension of F. Consider any a  E, then the splitting field of the minimal polynomial of a over F is contained in E. That gives every conjugate of a over F in K is again in E. Since for any  G(K, F),(a) is a conjugate of a, we have  (a)  E. Thus for any

 G(K, E),[(a)]  (a)and hence ( 1 )(a)  a.This proves  1 G(K, E)for every  G(K, E)and  G(K, F).

Consequently G(K, E)is a normal subgroup of G(K, F). Coversely let be a normal subgroup of G(K, F) . Consider a  E. As K is a normal extension of F, K contains a splitting field say L of the minimal polynomial p(x) of a over F. Consider any root b of p(x) in L. Then b is a conjugate of a over F. So there exists an F-automorphism  of K such that  (a)  b.For any

 G(K, E).  1n G(K, E) so ( 1 )(a)  a.

Thus [(a)]  (a) G(K, E).However E is the fixed field under . This gives that b  (a) E.Hence L  E. This proves that E is a normal extension of F. Hence (iv) is proved.

Let E be a normal extension of F. E = F (a) for some a  E. For any  G(K, F),let  E denote the restriction of  to E. Since  (a)  E, we get (E)  E.As [(E) : F]  [E : F], we get

 (E)  E.Hence  E is an F-automorphism of E and so  E G(E, F).Define a mapping  : G(K, F)  G(E, F)

by ()   E G(K, F). Clearly for any  ,  G(K, F),( )E   EE .Hence  is a group of homomorphism. Consider any y G(E, F).Now y(a) is a conjugate of a over F. Thus there exists

70 an F-automorphism  of K such that (a)  y(a). Further as  and y are both identity on F and E is generated by a over F we get (x)  y(x)x  F(a)  E i.e. y   E  ().This proves that  G(K, F) is onto mapping. Hence G(E, F)  . Now   Ker if and only if  is identity on E i.e., Ker E if and only if (x)  xx  E i.e., if and only if  G(K, E). Hence Ker  G(K, E) and we G(K, F obtainG(K, F)  . G(K, E)

This proves (v).

Hence theorem is proved .

3.13 Solvability and Insolvability of Polynomials by radicals

Historically, the problem of solving polynomial equations by radicals has motivated a great deal of study in algebra. The quadratic formula gives a solution of the equation ax2 + bx + c = 0, where „a‟ is nonzero, expressed in terms of its coefficients, and using a square root. More generally, we say n that an equation anx + ... + a1x + a0 = 0 , an ≠ 0 is solvable by radicals if its solutions can be given in a form that involves sums, differences, products, or quotients of the coefficients an , ..., a1 , a0 , together with square roots, cube roots, etc., of such combinations of the coefficients.

Abel attacked the general problem of when a polynomial equation could be solved by radicals. His papers inspired Galois to formulate a theory of solvability of equations involving the structures we now know as groups and fields. Galois worked with fields to which all roots of the given equation had been adjoined. He then considered the set of all permutations of these roots that leave the coefficient field unchanged. The permutations form a group, called the Galois group of the equation. From the modern point of view, the permutations of the roots can be extended to automorphisms of the field, and form a group under composition of functions. Then an equation is solvable by radicals if and only if its Galois group is ``solvable''.

n n-1 Polynomials are functions of the type p(x)= anx + an-1x ... + a1x + a0 = 0 , an ≠ 0 The root(s) of a polynomial are the value(s) of x which satisfy p(x)=0.

Being able to solve for polynomial roots using radicals is not about finding a root, as this is known by the fundamental theorem of algebra that any polynomial of degree n has n complex roots, which need not be distinct. Solving a polynomial by radicals is the expression of all roots of a polynomial using only the four basic operations, addition, subtraction, multiplication and division, as well as the taking of radicals, on the arithmetical combinations of coefficients of any given polynomial.

71

Solving for polynomial roots by radicals, involves finding the general solution to the general form of a polynomial of some specific degree. The general solution to a polynomial of degree two, or a quadratic equation, for example, is by using the quadratic formula.

While the quadratic formula for solving the quadratic equations has been known for as long as two thousand years, cubic or quartic formulas were only known in the last five hundred years, and this problem has been worked on by many mathematicians, including Gauss.

This raises the question of how to solve polynomials of higher degree. The methods for solving the roots of polynomials of degrees three and four exists but are by far less well known and more complex as compared to the quadratic formula. All three known formulas will be mentioned in this paper.

However, there is no formula for any polynomial for degree five and above, and it will be proven that such formulas do not exist.

While it may be impossible to find a general formula for these polynomials, specific cases of polynomials which can be expressed by radicals will be investigated.

The purpose of this research is thus to find out if all polynomials can be solved by radicals and to prove the resultant findings about the solvability of polynomials.

RESULTS

Constant Polynomials

Constant polynomials of the form p(x) = c have no roots, unless c=0 , in which case there are infinitely many roots.

Linear Functions

These are functions of the form , with m nonzero, and have exactly one root, i.e.

Quadratic Functions

Quadratic equations can be solved using the Quadratic Formula:

For any quadratic equation of the form

72

Completing the square for will derive the quadratic formula.

Rearranging the terms of (2) we can get the quadratic formula.

Cubic Functions

Solving Cubic functions can be done using Cardano‟s method, which transforms the general cubic equation into a depressed cubic without x2 term.

This method is believed to have originated from Scipione del Ferro, and was later adapted by Niccolò Tartaglia, and published in Cardano‟s 1545 paper.

The method is as follows.

We begin with the general form of a polynomial of degree three.

Since it is easier to work with a polynomial of leading coefficient one, we can divide „a‟ out of the entire equation.

Substitute the following equation into (4)

The polynomial becomes

For a cubic polynomial of the form

Observe that

73

Equation (6) corresponds to equation (5) since we can let

Thus we can solve for y

where i∈{1,2,3} and is one of the 3rd roots of unity.

Thus the general solutions for the equation (4) are

Quartic Functions

Solving Quartic polynomials can be done using Ferrari‟s method, which transforms a quartic polynomial into a depressed quartic which has no x^3 term.

We begin with the general form of a quartic equation.

We can reduce all quartic polynomials to monic polynomials by dividing throughout by the leading coefficient, and replacing the coefficients of the other terms with a,b,c,d.

Substitute the following equation into (7)

to get a equation of the form

We can add 2zy^2+z^2 to the above equation, to obtain

Since we want the left side to be a square as well, we can let the discriminant of the quadratic on the RHS be 0.

74 Therefore,

Rearranging the terms we get a cubic in z,

We can thus find the root of this equation, and solve for y by substituting that value into (1) to get a quadratic in y2.

Solving the resultant quadratic in y2 gives the roots of the depressed quartic, from which we can derive x.

Theorem There exists a polynomial of degree 5 with rational coefficients that is not solvable by radicals.

Proof Let p(x) = x5 - 2x3 - 8x - 2. It is easy to check that the derivative of p(x) has two real roots, corresponding to one relative maximum and one relative minimum. Since the values of p(x) change sign between -2 and -1, between -1 and 0, and between 2 and 3, the polynomial has precisely three real roots.

There exists a splitting field F for p(x) that is contained in the field C of complex numbers. The polynomial p(x) is irreducible over the rational numbers by Eisenstein's criterion, and so adjoining a root of p(x) gives an extension of degree 5.

It follows from theorems in Galois theory and group theory that the Galois group of p(x) over the rationals must contain an element of order 5. Every element of the Galois group of p(x) gives a permutation of the roots, and so the Galois group turns out to be isomorphic to a subgroup of the symmetric group on 5 elements. This subgroup must contain an element of order 5, and it must also contain the transposition that corresponds to the element of the Galois group defined by complex conjugation. It can be shown that any subgroup contains both a transposition and a cycle of length 5 must be equal to all of the symmetric group. Therefore the Galois group of p(x) over the rationals must be isomorphic to the symmetric group on 5 elements.

The proof is completed by showing that this group is not a solvable group. With some hard work it can be shown that in the group the only candidate for a chain of normal subgroups as required in the definition of a solvable group is S5 > A5 > {e} . Since the alternating group on 5 elements is not abelian, it follows that the symmetric group on 5 elements is not a solvable group. Thus p(x) cannot be solved by radicals. This we had already discussed in block 1 section 1.6.

A general polynomial of degree greater than five cannot be solved by radicals, and this fact can be proved so using Galois Theory. However, many types of (non-general) polynomials of degrees

75 greater than five can be solved by radicals, and solutions by radicals have been found for polynomials of these types. For quadratic, cubic, and quartic equations, general solutions have been found, and they are respectively the quadratic formula, Cardano‟s method and Ferrari‟s method.

Quintic Functions

A general quintic polynomials are insolvable by radicals. This proof makes use of group theory and Galois Theory, and is unlike Abel‟s 1819 paper. We will use the result below:

Theorem A polynomial is solvable by radicals if and only if the Galois group of the splitting field of the polynomial f(x)∈F[x] is solvable.

Let be independent transcendental elements over Q. Consider

By Vieta‟s formulas, we know that are elementary symmetrical functions in yi .

Set and . Then the polynomial f(x) in F[x] has the E as its splitting field.

The proof of the insolvability of f(x) by radicals is as follows.

Suppose on the contrary that is solvable for some polynomial of degree five p(x)∈F[x]

Consider the composition series of subgroups from to Gr = (1) :

⊵ ⊵ ⊵ ⊵ .

This corresponds to the extension fields of

F= ⊂ ⊂ ⊂ ⊂Fr .

Each extension is cyclic and Galois .

76 The composition series of is as follows: ⊳

Thus Gal(E/F) is not solvable. Hence f(x) is not solvable by radicals .

Special Solvable Cases

By the proof above, we know that it is impossible to solve all quintics by radicals, and thus no general solution can be found. However, there are many cases of quintics which are solvable by radicals.

Consider the general quintic polynomial

If a=0, then the quintic becomes a quartic polynomial, and is thus solvable by radicals using the aforementioned Ferrari‟s method. we know that a polynomial is solvable if and only if its Galois group is solvable.

Consider the cyclotomic polynomial x5-1=0 .

This equation is solvable in radicals as its splitting field is generated by the 5th roots of unity, so the resultant Galois group is solvable.

The roots of this equation are simply the 5th roots of unity,

where k∈{0,1,-1,-2,2} .

These roots of unity can be expressed by radicals.

Similarly, all equations of the form , where m is a constant, are solvable by radicals, since the roots are simply

Eisenstein‟s Irreducibility Criterion

77 For any polynomial of the form

∈Z

If for some prime number p,

are divisible by p, a0 is not divisible by p^2,

Then the polynomial is irreducible over , or it cannot be decomposed non-trivially into two polynomials of lesser degree with rational coefficients. In particular, it does not have root in.

If a quintic polynomial is reducible, then its roots can be expressed by radicals.

Also, any polynomial which can be decomposed into rational factors which are defined over Q can be solved, since we can express the roots of each of these factors in radicals.

Thus in general, polynomials of degree 5 or greater than 5 cannot be solved using radicals. Polynomials of degree 0,1,2,3,4 can be solved generally by radicals.

While polynomials of degree five or larger cannot be solved by radicals generally, there are many more specific types of polynomials f(x) that can. For instance, if where

each fi (x) is defined over Q, then f(x) can be solved by radical.

3.14 Unit Summary/ Things to Remember

1. A field extension of a field F is pair ( k, ), where k is a field and  is a monomorphism of F into k.

2. The extension K of F is called an Algebraic Extension of F if every element in K is algebraic over F

3. Field extensions which are not algebraic, i.e. which contain transcendental elements, are called transcendental.

78 4. Every finite extension of a field is algebraic extension.

5. An algebraic extension K of a field F is a said to be a separable extension, if every element of K is separable over F, otherwise K is said to be an inseparable extension.

6. Every polynomial over a field of characteristic zero is separable.

7. Any algebraic extension field F is a separable extension.

8. An algebraic field extension L/K is said to be normal if L is the splitting field of a family of polynomials in K[X].

9. A field F is called Perfect if all finite extensions of F are separable.

10. Fields having only a finite number of elements are called Finite Fields.

11. Any finite separable extension of an infinite field is a simple extension.

12. Finite fields having the same number of elements are isomorphic.

13. A field extension L/K is called a simple extension if there exists an element θ in L with L = K(θ) where the element θ is called a primitive element, or generating element.

14. A field K is algebraically closed if every non-constant polynomial in K[X] has a root in K.

15. Every algebraically closed field is perfect. Also A finite field cannot be algebraically

closed.

16. An automorphism  of K is said to be an F-automorphism if  (x)  x for every

x  F where F be any field and K be any field extension of F.

17. A finite extension K of a field F is said to be a Galois Extension of F if F is the fixed subfield of K under the group G(K, F) of all F-automorphisms of K.

18. An equation is solvable by radicals if and only if its Galois group is ``solvable''. 19. In general, polynomials of degree 5 or greater than 5 cannot be solved using radicals. However, there are many cases of 5 degree polynomials which are solvable by radicals.

79 3.15 Assignments/ Activities

1. Prove that if F is a field extension of K and K = FG for a finite group G of automorphisms of F, then there are only finitely many subfields between F and K. 2. 7. Let F be the splitting field over K of a separable polynomial. Prove that if Gal (F/K) is cyclic, then for each divisor d of [F:K] there is exactly one field E with K E F and [E:K] = d. 3. 8. Let F be a finite, normal extension of Q for which | Gal (F/Q) | = 8 and each element of Gal (F/Q) has order 2. Find the number of subfields of F that have degree 4 over Q. 4. 9. Let F be a finite, normal, separable extension of the field K. Suppose that the Galois group Gal (F/K) is isomorphic to D7. Find the number of distinct subfields between F and K. How many of these are normal extensions of K? 5. 10. Show that F = Q( , i) is normal over Q; find its Galois group over Q, and find all

intermediate fields between Q and F. 6. Let K be a field of characteristic zero, and let E be a radical extension of K. Then there exists an extension F of E that is a normal radical extension of K. 7. Let p be a prime number, let K be a field that contains all pth roots of unity, and let F be an extension of K. If [F:K] = |Gal(F/K)| = p, then F = K(u) for some u in F such that up is in K. 8. Let K be a field of characteristic zero that contains all nth roots of unity, let a be an element of K, and let F be the splitting field of xn-a over K. Then Gal(F/K) is a cyclic group whose order is a divisor of n. 9. Let F be the splitting field of xn - 1 over a field K of characteristic zero. Then Gal(F/K) is an abelian group. 10. Let f(x) be a polynomial over a field K of characteristic zero. The equation f(x) = 0 is solvable by radicals if and only if the Galois group of f(x) over K is solvable.

3.16 Check Your Progress

1. A polynomial of degree n over a field can have atmost n roots in any extension field. 2. Let K be a finite, separable field extension of a field F. Then K is a normal extension of F if and only if the fixed field under the Galois group G(K, F) is F itself in case K is a normal extension of F, [K : F]  0[G(K, F)].

3. Prove that any subgroup of S5 that contains both a transposition and a cycle of length 5 must be equal to S5 itself.

4. If F is the splitting field over K of a separable polynomial and G = Gal(F/K), then FG = K.

80 5. Let K be a finite field and let F be an extension of K with [F:K] = m. Then Gal(F/K) is a cyclic group of order m. 6. State and Prove Fundamental theorem of Galois Theory.

7. Let f(x) K[x] be a polynomial with no repeated roots and let F be a splitting field for f(x) over K. If : K -> L is a field isomorphism that maps f(x) to g(x) L[x] and E is a splitting field for g(x) over L, then there exist exactly [F:K] isomorphisms : F -> E such that (a) = (a) for all a in K.

3.17 Points for discussion / Clarification

At the end of the unit you may like to discuss or seek clarification on some points. If so, mention the same.

1. Points for discussion

______

______

______

2. Points for clarification

______

______- ______

3.18 References

12. I . N. Herstein, Topics in Algebra, Second Edition, John Wiley and Sons, 2006.

S.P.B. Bhattacharya, S.K. Jain and S.R. Nagpaul, Basic Abstract Algebra,(2ndEdition),Cambridge University Press, Indian Edition, 1997.ss

81 13. John S. Rose, A Course on Group Theory. New York: Dover, 1994. 14. P.B. Bhattacharya, S.K. Jain and S.R. Nagpaul, Basic Abstract Algebra,(2ndEdition),Cambridge University Press, Indian Edition, 1997.

UNIT-4 Rings and Modules

STRUCTURE

1.1 Introduction

82 1.2 Objective 1.3 Northerian and Artinian modules and ring 1.4 Hilbert Basis Theorem 1.5 Wed artin theorem 1.6 Uniform modules 1.7 Primary modules 1.8 Northern Lasker theorem 1.9 Smith Normal Form 1.10 Unit Summary/ Things to Remember 1.11 Assignments/ Activities 1.12 Check Your Progress 1.13 Points for Discussion/ Clarification 1.14 References

UNIT-5

STRUCTURE

1.1 Introduction 1.2 Objective

83 1.3 Structure theorem for finitely generated modules over a principal ideal domain and its applications to 1.4 Rational Canonical Form 1.5 Normal Jordan form to any field

In a vector space, the set of scalars forms a field and acts on the vectors by scalar multiplication, subject to certain formal laws such as the distributive law. In a module, the scalars need only be a ring, so the module concept represents a significant generalization. In commutative algebra, it is important that both ideals and quotient rings are modules, so that many arguments about ideals or quotient rings can be combined into a single argument about modules. In non-commutative algebra the distinction between left ideals, ideals, and modules becomes more pronounced, though some important ring theoretic conditions can be expressed either about left ideals or left modules.

Noetherian module is a module that satisfies the ascending chain condition on its submodules, where the submodules are partially ordered by inclusion.

Historically, Hilbert was the first mathematician to work with the properties of finitely generated submodules. He proved an important theorem known as Hilbert's basis theorem which says that any ideal in the multivariate polynomial ring of an arbitrary field is finitely generated. However, the property is named after Emmy Noether who was the first one to discover the true importance of the property.

Polynomial rings over fields have many special properties; properties that follow from the fact that polynomial rings are not, in some sense, "too large". Emmy Noether first discovered that the key property of polynomial rings is the ascending chain condition on ideals. Noetherian rings are named after her.

In mathematics, Hilbert's basis theorem states that every ideal in the ring of multivariate polynomials over a field is finitely generated. This can be translated into algebraic geometry as follows: every algebraic set over a field can be described as the set of common roots of finitely many polynomial equations. The theorem is named for the German mathematician David Hilbert who first proved it in 1888.

Hilbert produced an innovative proof by contradiction using mathematical induction; his method does not give an algorithm to produce the finitely many basis polynomials for a given ideal: it only shows that they must exist. One can determine basis polynomials using the method of Gröbner bases. [edit] Proof

The following more general statement will be proved: if R is a left (respectively right) Noetherian ring then the polynomial ring R[X] is also a left (respectively right) Noetherian ring.

84 Let I be an ideal in R[X] and assume for a contradiction that I is not finitely generated. Inductively construct a sequence f1, f2, ... of elements of I such that fi+1 has minimal degree in I \ Ji, where Ji is the ideal generated by f1, ..., fi. Let ai be the leading coefficient of fi and let J be the ideal of R generated by a1, a2, ... Since R is Noetherian there exists an integer N such that J is generated by a1, n ..., aN, so in particular aN+1 = u1a1 + ... + uNaN for some u1, ..., uN in R. Now consider g = u1f1x 1 + ... n + uNfNx N where ni = deg fN+1 − deg fi. Because deg g = deg fN+1 and the leading coefficients of g and fN+1 agree, the difference fN+1 − g has degree strictly less than the degree of fN+1, contradicting the choice of fN+1. Therefore I is finitely generated, and the proof is complete.

A constructive proof also exists: Given an ideal I of R[X], let L be the set of leading coefficients of the elements of I. Then L is an ideal in R so is finitely generated by a1,... ,an in L, and pick f1,... ,fn in I such that the leading coefficient of fi is ai. Let di be the degree of fi and let N be the maximum of the di. Now for each k = 0, ..., N − 1 let Lk be the set of leading coefficients of elements of I with k k degree at most k. Then again, Lk is an ideal in R, so is finitely generated by a 1,... ,a mk say. As k k before, let f i in I have leading coefficient a i. Let H be the ideal in R[X] generated by the fi and the k f i. Then surely H is contained in I and assume there is an element f in I not belonging to H, of least degree d, and leading coefficient a. If d is larger than or equal to N then a is in L so, a = r1a1 + ... + d−d d−d rnan and g = r1X 1f1 + ... + rnX nfn is of the same degree as f and has the same leading coefficient. Since g is in H, f − g is not, which contradicts the minimality of f. If on the other hand d is strictly d d smaller than N, then a is in Ld, so a = r1a 1 + ... + rmda md. A similar construction as above gives the same contradiction. Thus, I = H, which is finitely generated, and this finishes the proof. [edit] Mizar System http://en.wikipedia.org/wiki/Structure_theorem_for_finitely _generated_modules_over_a_principal_ideal_domain

In linear algebra, Jordan normal form (often called Jordan canonical form)[1] shows that a given square matrix M over a field K containing the eigenvalues of M can be transformed into a certain normal form by changing the basis. This normal form is almost diagonal in the sense that its only non-zero entries lie on the diagonal and the superdiagonal. This is made more precise in the Jordan– Chevalley decomposition. One can compare this result with the spectral theorem for normal matrices, which is a special case of the Jordan normal form.

It is named after Camille Jordan.

In abstract algebra, the Artin–Wedderburn theorem is a classification theorem for semisimple rings. The theorem states that a semisimple ring R is isomorphic to a product of ni-by-ni matrix rings over division rings Di, for some integers ni, both of which are uniquely determined up to permutation of the index i. In particular, any simple left or right Artinian ring is isomorphic to an n- by-n matrix ring over a division ring D, where both n and D are uniquely determined.

85 As a direct corollary, the Artin–Wedderburn theorem implies that every simple ring that is finite- dimensional over a division ring (a simple algebra) is a matrix ring. This is Joseph Wedderburn's original result. Emil Artin later generalized it to the case of Artinian rings.

The Artin–Wedderburn theorem reduces classifying simple rings over a division ring to classifying division rings that contain a given division ring. This in turn can be simplified: The center of D must be a field K. Therefore R is a K-algebra, and itself has K as its center. A finite-dimensional simple algebra R is thus a central simple algebra over K. Thus the Artin–Wedderburn theorem reduces the problem of classifying finite-dimensional central simple algebras to the problem of classifying division rings with given center.

86 UNIT-4 Noetherian and artinian modules and rings

STRUCTURE PAGE No.

4.1 Introduction 30

4.2 Objective 30

4.3 Northerian and Artinian modules and rings 31-38

4.4 Hilbert Basis Theorem 38-40

4.5 Wedderburn Artin theorem 40-44

4.6 Uniform and Primary modules 44-45

4.7 Northern Lasker theorem 46-47

4.8 Smith Normal Form over a PID and rank 47-51

4.9 Unit Summary/ Things to Remember 52

4.10 Assignments/ Activities 53

4.11 Check Your Progress 54-55

4.12 Points for Discussion/ Clarification 55-56

4.13 References 56

87

4.1 Introduction

Commutative algebra is the branch of abstract algebra that studies commutative rings, their modules, and modules over such rings. Module Theory, began with Richard Dedekind's work on modules, itself based on the earlier work of Ernst Kummer and Leopold Kronecker. Later, David Hilbert introduced the term ring to generalize the earlier term number ring. Hilbert introduced a more abstract approach to replace the more concrete and computationally oriented methods grounded in such things as complex analysis and classical invariant theory. In turn, Hilbert strongly influenced Emmy Noether, to whom we owe much of the abstract and axiomatic approach to the subject. Another important milestone was the work of Hilbert's student Emanuel Lasker, who introduced primary modules and proved the first version of the Lasker–Noether theorem.

In particular, the integers are the principal ideal domains, so one can always calculate the Smith normal form of an integer matrix. The Smith normal form is very useful for working with finitely generated modules over a principal ideal domain, and in particular for deducing the structure of a quotient of a free module.

4.2 Objective

After the completion of this unit one should able to :

 understand the central notions of commutative algebra  describe uniform and primary modules.  to prove well known theorems : Hilbert Basis Theorem, Northern Lasker theorem and Wedderburn Artin theorem.  to understand representation theory, and the theory of Banach algebras.

88  to state and prove that every matrix is equivalent to diagonal matrix; Smith normal form.

4.3 Northerian and Artinian modules and rings

A ring in which every strictly ascending chain of right (left) modules is finite, is called a Right (Left) noetherian ring.

Equivalently one can define a right (left) noetherian ring to be a ring in which for every infinite ascending chain M1  M 2  M3  ...of right (left) modules there exists a positive integer n such that M m  M nm  n.

Remark A right (left) noetherian ring is also known as a ring with ACC (i.e., Ascending Chain Condition) on right (left) modules.

A ring in which ACC holds for right as well as left modules is called a noetherian ring.

Example 1 Every finite ring is clearly both right and left noetherian.

Example 2 Consider a division ring D. Since the only right modules of D are (0) and D itself. D is right noetherian. For similar reasons D is also left noetherian.

If in a ring R every non-empty set of right (left) modules, partially ordered by inclusion relation, has a maximum element; we say that maximum condition holds for right (left) modules of R.

Remark As the results for left noetherian rings are in complete analogy with those for right Noetherian rings, we shall confine ourselves to right Noetherian rings only.

A right module I of a ring R is said to be finitely generated if it is generated by a finite subset of R.

Theorem For any ring R the following statements are equivalent:-

a. R is right noetherian.

89 b. Maximum condition holds for right modules of R. c. Every right module of R is finitely generated.

Lemma A homomorphic image of a right noetherian ring is right noetherian.

Proof Let S be a homomorphic image of a right noetherian ring R. Then S  R | M for some module M of R. So it is sufficient to prove that R/M is right noetherian.

Let J1  J 2  J 3  ...be an ascending chain of right modules of R/M. Now each J i is of the form

Ki / M where K i a right module of R containing M. Also Ji  Ji1  Ki  Ki1.

So the above ascending chain gived rise to ascending chain.

K1  K2  K3  ...of right modules of R. But as R is right noetherian there exists a positive integer n such that Km  Knm  n. This implies that J m  J nm  n.Hence R/M is right noetherian.

Theorem Ler M be an module of a ring R then R is right noetherian if and only if both M and R/M are right noetherian.

Proof If R is right noetherian then M is right noetherian and R/M is also right noetherian.

Conversely let both M and R/M be right noetherian and let J1  J 2  J 3 ,...be an ascending chain of right modules of R. Then J1  M  J2  M  J3  M ,...is an ascending chain of right modules of R contained on M. Since M is a right noetherian there exists a positive integer n such that Jm  M  Jn  Mm  n.

Again (J1  M) / M  (J2  M)M  (J3  M) / M  ...is an ascending chain of right modules of R/M. As R/M is right noetherian there exists a posiyive integer t such that

(Jm  M) / M  (Jt  M) / M m  t.If r  max(n,t) then Jm  M  Jr  M

and (Jm  M) / M  (Jr  M) / M m  r. But then Jm  M  Jr  M m  r. We claim that

J m  J rm  r. Now for all m  r, Jm  Jm  (Jm  M)

90  Jm (Jr  M), as Jm  M  Jr  M m  r,

 Jr  Jm  M as J r  J mm  r and by Modular law,

 Jr  Jr  M since Jm  M  Jr  M,

 J r as Jr  M  Jr .

Hence J m  J rm  r. As a consequence R is right noetherian.

A ring in which every strictly descending chain of right (left) modules is finite is called a right (left) artinian ring.

Alternatively one can define a right (left) artinian ring to be a ring in which for every infinite descending chain of right (left) modules M1  M 2  M3  ...there exists a positive integer n such that M m  M nm  n.

Remark A right (left) artinian ring is also known as a ring with DCC (Descending Chain Condition) on right (left) modules.

A ring in which DCC holds for right as well as left modules is called an artinian ring.

Example 1 Every division ring D is right artinian as its only right modules are (0) and D itself. Because of similar reason D is also left artinian.

Example 2 Every finite ring is both right and left Artinian.

Example 3 Consider the set Z(P ) of all rational numbers of the form m/ P n such that

0  m/ Pn 1,where P is a fixed prime number, m is an arbitrary positive integer and n runs through all non-negative integers. Then Z( p  ) is an abelian group under addition modulo 1. We make Z( p  ) , a ring by definig ab  0a,b . Note that each subgroup of is an module of . Now let M be any proper module of and let k be the least non-negative integer such that for some positive integer a,a / pk1 M.Notice that k is positive and p does not

91 divide a otherwise we shall have b / pk1 M some positive integer b, which is against the choice of k.

Now 0,1/ pk1,2/ pk1,...,(pk1 1) / pk1 M.We contend that M consists of precisely these elements. If our contention is not correct, there exists an element c / pi  M where c is a positive integer, i  k and p does not divide c. Since (c, p) 1, we can find integer‟s r and s such that cr  ps 1.

Now (rpik )c / Pi  rc/ Pk and s / Pk1  sp/ p k , each reduced modulo 1 lies in M. it follows that rc/ pk  sp/ pk  (rc  sp) / pk 1/ pk M, a contradiction to the choice of k.

k1 k1 k1 k1 Hence module M {0,1/ p ,2/ p ,...,(p 1) / p }.We denote this module by M k 1. Thus we

 conclude that the only proper modules of Z( p ) are of the form M j1 for each positive integer J.

Since each proper module of has finite number of elements, DCC holds. Hence is an Artinian ring.

This ring is not noetherian, as it contains infinte properly ascending chain of modules.

M1  M 2  M3  ......  M n  ......

If in a ring R, every non-empty collection of all right (left) modules of R, partially ordered by inclusion relation, has a minimal element we say that minimum condition holds for right (left) modules of R.

Note that by a minimal element of a partially ordered non-empty set F under the relation , we mean an element A F such that there exists no B  F satisfying B  A.In other words A is a minimal element of F if and only B  A  B  A.As in the case of maximal element, a set may possess more than one minimal element.

As the results for left artinian rings are completely analogous to those of right Aritinian rings we shall henceforth discuss only right artinian rings.

Theorem For a ring R, the following statements are equivalent.

92 a. R is right artinian. b. Minimum condition holds for right modules of R.

Proof (1)  (2) Let F be a non-empty set of right modules of R. Let I i be and element of F. If

I1 is not a minimal element, we can find another right module I 2 in F or R such that Ii  I 2 . If F has no minimal element this process can be repeated indefinitely giving rise to infinite strictly descending chain I1  I 2  I 2 ...of right modules of R in F. This is a contradiction to the fact that R is right Artinian. Hence F has a minimal element.

(2) (1). Let I1  I 2  I3  ...be a descending chain of right modules of R. Consider

F  {Ii /i 1,2,3,...}. Then F is non-empty as Ii  F.By (2), F has a minimal element, say I n , for some positive integer n. Now m  n, I m  I n . If I m  I n then as I n is a minimal element of F,

I m  I nm  n.Consequently R is right Artinian.

Now we give examples of ring which are right noetherian.(artinian) but not left Noetherian.(Artinian).

a b  Example Let R    a  Z;b,c Q  0 c 

Then R is a ring under matrix addition and multiplication. We claim that R is right noetherian but not left noetherian.

0 m / 2k   For any non-negative integer k, consider the set A    m  Z . k    0 0  

0 1/ 2k1  Then A is a left module of R. Further A  A as m/ 2k  2m/ 2k1and    A k k k1   k. 0 0 

Thus we get a non-terminating strictly ascending chain A0  A1  A2  A3  ...of left modules of R. This shows that R is not left noetherian.

To prove that R is right noetherian.it is sufficient to establish that each non-zero right module of R is finitely generated.

93 Let A[ (0)] be a right module of R and let ae11  e12  ye22  A where eij denotes the matrix with

„1‟ in (i, j)th position and zero elsewhere, and a  Z,, y Q.

Note also that eij ekl  eij if j = k and eij ekl  0 if j  k. Two cases arise.

Case 1 a  0.Let  be the least positive integer such that e11  be12  ce22  A for some b,c Q.Then we claim that either A is generated by matrices e11,e12,e22 or by e11 and e12. Now

 1  (e11  be12  ce22) A  e12  A  e12 e22  e12  A  be12  e12(be22)  A.  

Also e11  be12  ce22  A  (e11  be12  ce22)e11  A  e11  A.

Thus we get ce22  A. In case c  O for all  and b,A is generated by e11,e12. In other case  1  e22  (ce22) e22   A, and A is generated by se12,e12, and e22.  2 

Case 2 a  0.If there exists be12  ce22  A such that all elements is A are of the type (be12  ce22) for some  Q, then A is generated by single matrix be12  ce22. Otherwise there exist b1,c1 Q such that b1e12  c1e22  A but b1c  bc1. Then  e12be22  A  be12  A  ce22  A.

Either c = 0 or e22  A.Thus is either generated by e12 and e22 or by e12 alone. Hence A is finitely generated. Consequently R is right Noetherian.

a b  Example Let R    a Q;b,c  R  0 c 

Clearly R is a ring under matrix addition and multiplication. We claim that R is right artinian but not left artinian R, as a vector space over Q, is of infinite dimension, there exists real numbers a1,a2 ,a3....,an ,....which are linearly independent over Q. For each positive integer k, put

94 0 a Ak    a  Subspace of R generated by ak ,ak1ak2 .... It can be easily checked that Ak is 0 0 a left module of R. Further Ak  Ak1, as ak e12  Ak1. Since the set {a1,a2 ,a3 ,...}is infinite, we get an infinite strictly descending chain of left modules A1  A2  A3  ...of R. This establishes the fact that R is not left Artinian.

Let A1  A2 be two modules of R and ae11  e12  ye22  A2 where a Q,, y  R and eij 's as defined in previous example. (eij 's are called matrix units). If a  0 then as proved in previous example either ae11  A2 ,e12  A2 ,e22  A2 or ae11  A2 ,e12  A2 and y  0.In the former situation

(ae11)(1/ a)e11  A2i.e.,e11  A2

and so A2  R,which is absurd as A1  A2. In the next situation A2 is generated by e11 and e12

Consider now A  A  A where A is a non-zero right module of R. 1 2 3 , 3 Now let

0  ae11  be12  A3. If a  0 then (ae11  be12)(1/ a)e11  A3  e11  A3  e12  e11

e12  A3  A2  A3 , a contradiction. Hence a  0.Hence b  0

 1  and e12  (be12) e22   A3. Consequently A3 is generated e12. This at once prove that A3 is  b  minimal right module of R. Hence the strictly descending chain of right modules

A1  A2  A3  (0) is finite.

Finally if a  0 then e12  ye22  A2. If all other elements of A2 are of the type (e12  e22) for some   R then A2 is minimal right module and A1  A2 (0) is the strictly descending chain of right modules. Otherwise as seen in previous example is generated by either e12 and e22 or by e12 only.

In case is generated by e12 only; it is minimal right module of R and we are done. In the case

is generated by and e22, a similar argument will show that if A1  A2  A3 for some right

95 module A3 of R, then either A3 is generated by e12 or by de12  fe22 for some d, f  R. In each case is a minimal right module of R.

Consequently R does not admit any infinite strictly descending chain of right modules. This establishes the fact R is right Artinian.

Example Z is noetherian ring but not artinian. In fact for any positive integer n, the strictly descending chain  n  2n  4n ...of modules of Z is infinite.

We now give a proof of a famous theorem of Hilbert.

4.4 Hilbert’s Basis Theorem

Hilbert's basis theorem states that every ideal in the ring of multivariate polynomials over a field is finitely generated. This can be translated into algebraic geometry as follows: every algebraic set over a field can be described as the set of common roots of finitely many polynomial equations. The theorem is named for the German mathematician David Hilbert who first proved it in 1888.

Statement: If R is a right noetherian ring with unity then R[x], the ring of polynomials over R, is right noetherian.

Proof It is sufficient to prove that every non-zero right module of is finitely generated.

Let I be a non-zero ring of For each integer k  0, define I k  {a  R | a  0 and there exists a

k1 k polynomial a0  a1x ...  ak1x  ax }{0}.

I k is a right module of R and I k  I k1 for all integers k  0.

Since R is right Noetherian there exists a positive integer n such that I m  I n  n.Also each I i , being a right module of right Noetherian ring R, is finitely generated. Let I  a ,a ,a ,...,a > i i1 i2 i3 imi

96 for all i  0,1,2,...n where aij is the leading coefficient of a polynomial fij  I, of degree i. We claim that I is generated by m0  m1  m2 ...  mn polynomials. f , f ,..., f , f , f ,..., f ,..., f , f ,..., f . 01 02 0m0 11 12 1m1 n1 n2 nmn

Let J  f , f ,..., f ,... f  . As each f  I, J  I. 01 02 n1 nmn r ij

2 s Let f  0 R[x]be such that f  I, than f  c0  c1x  c2 x ...cs x ,cs  0. We shall apply inductions on s. For s  0, f  c  I and by definition of f 's,a  f ,a  f ,...,a  f 0 0 ij 01 01 02 02 0m0 om0 are elements of I 0 which generate I 0 so i0  J, we get c0  J.This gives F  J.

Supoose now that all non-zero polynomials in I of degree

Let s  n.The leading coefficient cs of f belong to I s .

But implies that c  I i.e., c  a b  a b  a b ...  a b s n s n1 1 n2 2 n3 3 nmn mn for some b ,b ,b ,...b  R …(1) 1 2 3 mn

The polynomial g  f  ( f b  f b ...  f b )x sn is zero or of degree less than s as the n1 1 n2 2 nmn mn coefficient of x s in g is equal to c  (a b  a b ... a b ) by (1). s n1 1 n2 2 nmn mn

If g  0 then g  J, if g  0 then by induction hypothesis g  J and so in each case f  g  ( f b  f b ... f b )x sn  J. n1 n n2 2 nmn mn

If s  n then c  I  c  a d  a d ...  a d for some d ,d ,...,d  R. s s s s1 1 s2 2 sms ms 1 2 ms

The polynomial h  f  ( f d  f d ...  f d ) is either zero or of degree < s as the s1 1 s2 2 sms ms coefficient of x s in h is equal to c  (a d  a d ...  a d )  0.Hence once again h  J, so s s1 1 s2 2 sms ms f  J.

97 Thus in each case every non-zero polynomial f which is in I is also in J. This gives that I  J,which in turn implies that I  J consequently I is finitely generated. Hence R[x], is right Noetherian.

Example Let F be the ring of all real-valued functions on R. For any real number r > 0, we define Mr {f F | f (x)  0  r  x  r}. then M r is a module of F.

The strictly descending chain of modules M1  M2  M3...and the strictly ascending chain of modules M1  M1/ 2  M1/ 3...never terminate. Hence F is neither noetherian nor artinian.

Now we show by examples that a subring of Noetherian (artinian) ring need not be Noetherian (Artinian)

Example Consider Q [x], as Q is noetherian, Q[x] in Noetherian by Hilbert‟s Basis Theorem.

Let R  { f Q[x]| Constant term of f  Z}.

R is a subting of Q [x]. The strictly ascending chain  x  x/ 2  x/ 4 ... of modules of R never terminate. Hence R is not Noetherian.

The following example shows that Hilbert‟s Basis Theorem fails to hold for Artinian rings.

Example Let F be a field. Then F is Artinian. Consider F[x]the strictly descending chain

 x  x2  x3 ...of modules if F[x]is infinite. Hence F[x] is not Artinian.

4.5 Wedderburn Artin theorem

The Wedderburn Artin theorem implies that every simple ring that is finite-dimensional over a division ring (a simple algebra) is a matrix ring. This is Joseph Wedderburn's original result. Emil Artin later generalized it to the case of Artinian rings.

Statement : Let R be a left (or right) artinian ring with unity and no nonzero nilpotent modules. Then R is isomorphic to a finite direct sum of matrix rings over division rings. Proof We first establish that each nonzero legt module in R is of the form Re for some idempotent e. So let A be any nonzero left module. By virtue of the dcc on left modules in R,A contains a

98 minimal left module M. By Lemma 3.1 either M 2  (0) or M = Re for some idempotent e. If

M 2  (0), then (MR)2  (0); so, by hypothesis, MR  (0), which gives M  (0), a contradiction. Hence, M  Re.This yields that each nonzero left module contains a nonzero idempotent. Consider now a family F of left modules, namely,

F  {R(1 e)  A | e is a nonzero idempotent in A}.

Clearly, F is nonempty. Because R is left artinian, F has a minimal member, say R(1 e)  A.We claim R(1 e)  A  (0).Otherwise, there exists a nonzero idempotent e1 R(1 e)  A. Clearly, e1e  0. Set e' e  e1  ee1.It is easy to verify that e'e' e' and e1e' 0.It is also obvious that

R(1 e')  A  R(1 e)  A.But e1e' 0 gives e1  R(1 e')  A and e1  R(1 e)  A. Hence, R(1 e)  A  R(1 e)  A, a contradiction to the minimality of R(1 e)  A in F This establishes our claim that R(1 e)  A  (0). Next, let a  A.Then a(1 e) R(1 e)  A  (0). Thus, a  ae. Then A  Re  Ae  A proves that A = Re, as asserted.

Let S be the sum of all minimal left modules in R. Then S = Re for some idempotent e. If R(1 e) then A  Re R(1 e)  0, a contradiction. Hence, R(1 e)  0, which proves that

R  S  i Ai , where (Ai )i  A, is the family of all minimal left modules in R. Then there exists a subfamily (Ai ),i  A', of the family of the minimal left modules such that R  iA' Ai. Let 1  e ...  e ,0  e  A ,i  A'. Then R  Re ... Re . After re-indexing if i1 in i j i j j i1 in necessary, we may write R  Re1 ... Re n , a direct sum of minimal left modules.

In the family of minimal left modules Re1,...,Re n , choose a largest subfamily consisting of all minimal left modules that are not isomprphic to each other as left R-modules. After renumbering if necessary, let this subfamily be Re1,...Re k.

Suppose the number of left modules in the family (Rei )1 i  n, that are isomorphic to is Rei is ni . Then

n1 summands n2 summands nk summands

99   

R  Re1 ...   Re 2 ... ... Re k ..., where each set of brackets contains pairwise isomorphic minimal left modules, and no minimal left module in any pair of bracketd is isomorphic to a minimal left module in another pair. By observing that HomR (Rei ,Re j )  0,i  j,1 i, j  k, and recalling Schur‟s lemma in unit 2 that HomR

(Rei ,Rei )  Di ,a division ring we get:

n1  n1 block

D1 . . . D1     .   .     .  D . . . D   1 1   n2  n2 block   D . . . D   2 2   . .   . .     . .    D . . . D Hom (R, R)   2 2  R  .     .   .     nk  nk block   D . . . D   k k   . .     . .   . .     Dk . . . Dk       

(D1)n   1 (D   2)n2   .     .  (D )  (D ) ... (D )   1 n1 2 n2 k nk .  .    (D )  K nk     

oP But since HomR (R, R)  R as rings and the opposite ring of a division ring is a division ring, R is a finite direct dum of matrix rings over division rings.

100 Because the matrix rings over division rings are both right and left noetherian and artinian, and a finite direct sum of noetherian and artinian rings is again noetherian and artinian we get the result that if R be a left (or right) artinian ring with unity and no nonzero nilpotent ideals. Then R is also right and left artinian (noetherian) as not every ring that is artinian on one side is artinian on the other side.

The following example gives a ring that is left artinian but not right artinian.

Q Q 0 Q Let R   .If A    , then A is an module of R, and as a left R-module it is simple. 0 0 0 0 Q 0 Thus, A is left artinian. Also, the quotient ring R / A    , which is a field. Therefore R/A as an 0 0 R/A-module (or as an R-module) is artinian. Because A is an artinian left R-module, we obtain that R as a left R-module is artinian; that is, R is a left artinian ring, but R is not right artinian. For there exists a strictly descending chain

0 2 0 22  0 23           ... of right modules of R. 0 0 0 0 0 0      

As an application of the Wedderburn-Artin theorem, we prove the following useful result for a certain class of group algebras.

Maschke's theorem – The theorem is named after Heinrich Maschke. It states that:

If F is the field of complex and G is a finite group, then F(G)  F ... F for some positive n1 nk integers n1,...,nk.

Proof We first prove thet F(G) has no nonzero nilpotent modules. Let G  {g1  e, g2 ,...gn ), and

1 x  ai gi  F(G).Set x*  ai gi , where a i denotes the complex conjugate of ai . Then

n n 2 xx*  | ai | i gi i1 i2

101 n for some   F , Hence, xx*  0 implies | a |2  0 so each a  0;that is, x = 0. Thus, i i1 i i xx*  0 implies x  0.Let A be a nilpotent module in F(G).Let .a  A Then aa* A. so aa * is nilpotent, say (aa*)r  0.(We may assume r is even.) Set b  (aa*)r / 2 . Then b  b*. Thus, bb*  0, which gives (aa*)r / 2  b  0.Proceeding like this, we get aa*  0.Hence, a  0, which proves that A = (0). Hence, F(G) has no nonzero nilpotent modules.

Further, F(G) is a finite-dimensional algebra with unity over the field F. Therefore, F(G) is an artinian ring. Then by the Wedderburn-Artin theorem,

F(G)  (1) ... (k) Dn1 Dnk, where D(i) ,1 i  k, are division rings. Now each (i) is a finite-dimensional algebra over K. Let Dni

i (i) 2 [D : K]  n, and a  D . Then 1,a,a ,...,an are linearly dependent over K. Thus, there exist

n a0 ,a1,...,an (not all zero) in K such that a0  a1a ...  ana  0.But since K is algebraically closed,

n (i) a0  a1x ...  an x  K[x] has all its roots in K. Hence, a  K, which shows that D  K  F and completes the proof.

4.6 Uniform and Primary Modules

A nonzero module M is called uniform module if any two nonzero sub modules of M have nonzero intersection. If U and V are uniform modules, we say U is sub isomorphic to V and write U  V provided U and V contain nonzero isomorphic sub modules.

A module M is called primary module if each nonzero sub module of M has uniform sub module and any two uniform sub modules of M are sub isomorphic.

We note that Z as a Z-module is uniform and primary. Indeed, any uniform module must be primary. Another example of a uniform module is a commutative integral domain regarded as a module over itself.

102

Theorem Let M be a noetherian module or any module over a noetherian ring. Then each nonzero sub module contains a uniform module.

Proof Let 0  x  M.It is enough to show xR contains a uniform sub module. If M is noetherian the sub module xR is also noetherian. But if R is noetherian than xR, the homomorphism image of R, is noetherian.

For convenience, call ,as usual, a nonzero sub module N of M large if N  K  0 for all nonzero sub modules K of M.

Consider now the family F of all sub modules of xR which are not large. Clearly 0 F. Since xR is noetherian. F has a maximal member K, say. Because K is not large, hence K U  0 for some nonzero sub module U of xR. We claim U is uniform. Otherwise, there exit sub modules A,B of U such that A B  (0).But then (K  A)  B  (0).For if x (K  A)  B,then x  k  a  b for some k  K,a  A,b B.This gives k  b  a  K U  (0).Hence k  0 and b  a  0, which further yields b  a  A B  (0). Thus b = 0 = a, proving x = 0; that is, (K  A)  B  (0).However, this yields a contradiction to the musicality of K. This shows U is uniform, completing the proof.

If R is a commutative noetherian ring and P is a prime module of R, then P is said to be associated with the module M if R/P embeds in M, or equivalently P = r(x) for some x  M, where r(x)  {a  R | xa  0}denotes the annihilator of x.

A module M is called P-primary for some prime module P if P is the only prime module associated with M.

Remark. If R is a commutative noetherian ring and P is a prime module of R, then an R-module is P-primary if and only if each nonzero sub module of M is sub isomorphic to R/P.

103

4.7 Noether-Lasker Theorem

The Lasker–Noether theorem states that every Noetherian ring is a Lasker ring, which means that every ideal can be written as an intersection of finitely many primary ideals (which are related to, but not quite the same as, powers of prime ideals). The theorem was first proven by the world chess champion Emanuel Lasker (1905) for the special case of polynomial rings and convergent power series rings, and was proven in its full generality in a brilliant paper by Emmy Noether (1921).

The Lasker-Noether theorem is an extension of the fundamental theorem of arithmetic, and more generally the fundamental theorem of finitely generated abelian groups to all Noetherian rings.

Statement

Let M be a finitely generated module over a commutative noetherian ring R. Then there exists a finite family N1,..., N1 of sub modules of M such that

l l (a) i1 Ni  0and i1 Ni  0 for all 1  i0  l. ii0

(b) Each quotient M / Ni is a Pi" primary module for some prime module Pi .

(c) All Pi are distinct, 1 i  l.

N P P (d) The primary component i is unique if and only if i does not contain i for any j  i.

We can understand Noether lasker theorem with the help of following theorem:

Let M be a nonzero finitely generated module over a commutative noetherian. ring R. Then there are only a finite number of primes associated with M.

Proof Consider the family F consisting of the direct sums of cyclic uniform sub modules of M. F is not empty and partial order F by   iI xi R   jj y j R if and only if I  J and xi R  yi R for

104 i  I.By Zorn‟s Lemma, F has a maximal member K   jj x j R, say. Since M is noetherian, K

t is finitely generated and let K   x R. there exist x a  x R such that r(x a )  P , the  j1 j j j j j j j

t prime ideal associated with x R Set x'  x a and K'  x' R. Let Q  r(x) be an j j j j  j1 j associated prime ideal of M. We shall show Q  Pj for some j, 1 j  l.

Since K is a maximal member of F, K as well as K’ has the property that each intersects nontrivially with any nonzero sub module L of M. Now let 0  y  xR  K'.

l Write y  x' b  xb.We claim r(x' b )  r(x' )whenever x' b  0.  j1 j j j j j j j Clearly r(x' )  r(x' b ). Let x' b c  0.This implies b c r(x' )  P and so c  P j j j j j j j j j since b j  Pj .Hence, c  r(x' j ).

l Furthermore, we note Q  r(x)  r(y)   j1r(x' j bj )   Pj , omitting those arising from x' b  0, j j

Q  Pj for some j.

Now the proof of Noether – Lasker theorem is a consequence of the above results.

Let {Ui},1 i  l be uniform sub modules obtained as in the proof of above theorem.

Choose N i to be a maximal member in the family {K  M | K contains no sub module sub isomorphic to U i }with the choice if N1,...., Nl' (a), (b), and (c) follow directly.

4.8 Smith Normal Form over a PID and rank

The Smith normal form is a normal form that can be defined for any matrix (not necessarily square) with entries in a principal ideal domain (PID). The Smith normal form of a matrix is diagonal, and can be obtained from the original matrix by multiplying on the left and right by invertible square matrices.

105 We now set up some basic machinery to be used in connection with the Smith normal form and its applications. Assume that M is a free Z-module of rank n, with basis x1,..., xn , and that K is a sub module of M with finitely many generators u1,....,um .(We say that K is finitely generated.) We change to a new basis y1,...., yn via Y  PX , where X [respectively Y] is a column vector with components xi [respectively yi ].Since X and Y are bases, the n n matrix P must be invertible, and we need to be very clear on what this means. If the determinant of P is nonzero, we can construct P 1 , for example by the “adjoint divide by determinant” formula given in Cramer‟s rule.

But the underlying ring is Z, not Q, so we require that the coefficients of P 1 be integers. Similarly, we are going to change generators of K via V  QU, where Q is an invertible mm matrix and U is a column vector with components ui .

The generators of K are linear combinations of basis elements, so we have an equation of the form U  AX , where A is an m n matrix called the relations matrix. Thus

V  QU  QAX  QAP1Y.

So the new relations matrix is B  QAP1

Thus B is obtained from A by-and post multiplying by invertible matrices, and we say that A and B are equivalent. We will see that two matrices are equivalent if they have the same Smith normal form. The point we wish to emphasize now is that if we know the matrix P, we can compute the new basis Y, and if we know the matrix Q, we can compute the new system of generators V. In our applications, P and Q will be constructed by elementary row and column operations.

Now we are going to describe a procedure that is very similar to reduction of a matrix to echelon form. The result is that every matrix over a principal ideal domain is equivalent to a matrix in Smith normal form. Explicitly, the Smith matrix has nonzero entries only on the main diagonal. The main diagonal entries are, from the top, a1,....,ar (possibly followed by zeros), where the ai are nonzero and divides ai 1 for all i.

Let‟s start with the following matrix:

106  0 0 22 0     2 2  6  4    2 2 6 8 

Now we assume a free Z-module M with basis x1, x2 , x3 , x4 , and a sub module K generated by u1,u2 ,u3 , where u1  22x3 ,u2  2x1  2x2  6x3  4x4 ,u3  2x1  2x2  6x3  8x4 . The first step is to bring the smallest positive integer to the 1-1 position. Thus interchange rows 1 and 3 to obtain

 2 2 6 8     2 2  6  4    0 0 22 0 

Since all entries in column 1, and similarly in row 1, are divisible by 2, we can pivot about the 1-1 position; in other words, use the 1-1 entry to produce zeros. Thus add row 1 to row 2 to get 2 2 6 8   0 4 0 4   0 0 22 0

Add - 1 times column 1 to column 2, then add -3 times column 1 to column 3, and add -4 times column 1 to column 4. The result is

2 0 0 0   0 4 0 4   0 0 22 0

Now we have “peeled off” the first row and column, and we bring the smallest positive integer to the 2-2 position. It‟s already there, so no action is required. Furthermore, the 2-2 element is a multiple of the 1-1 element, so again no action is required. Pivoting about the 2-2 position, we add -1 times column 2 to column 4, and we have

2 0 0 0   0 4 0 0   0 0 22 0

107 Now we have peeled off the two rows and columns, and we bring the smallest positive integer to the 3-3 position.. But 22 is not a multiple of 4, so we have more work to do. Add row 3 to row 2 to get

2 0 0 0   0 4 22 0   0 0 22 0

Again we pivot about the 2-2 position; 4 does not divide 22, but if we add -5 times column 2 to column 3, we have

2 0 0 0   0 4 2 0   0 0 22 0

Interchange columns 2 and 3 get

2 0 0 0   0 2 4 0   0 22 0 0

Add -11 times row 2 to row 3 to obtain

2 0 0 0   0 2 4 0   0 0  44 0

Finally, add -2 times column 2 to column 3, and then (as a convenience to get rid of the minus sign) multiply row (or column) 3 by – 1; the result is

2 0 0 0   0 2 0 0   0 0 44 0 which is the Smith normal form of the original matrix. Although we had to backtrack to produce a new pivot element in the 2-2 position, the new element is smaller than the old one (since it is a

108 remainder after division by the original number). Thus we cannot go into an infinite loop, and the algorithm will indeed terminate in a finite number of steps. Now we have the following interpretation.

We have a new basis y1, y2 , y3 , y4 for M, and new generators v1,v2 ,v3 for K, where v  2y ,v  2y and v  44y . In fact since the v 's are nonzero multiples of the corresponding 1 1 2 2 3 3 j y j 's, they are linearly independent, and consequently form a basis of K.

The above discussion indicates that the Euclidean algorithm guarantees that the Smith normal form can be computed in finitely many steps. Therefore the Smith procedure can be carried out in any Euclidean domain. In fact we can generalize to a principal ideal domain. Suppose that at a particular stage of the computation, the element „a‟ occupies the 1-1 position of the Smith matrix S, and the element b is in row 1, column 2. To use a as a pivot to eliminate b, let d be the greatest common divisor of a and b, and let r and s be elements if R such that ar  bs  d . We post multiply the Smith matrix by a matrix T of the following form r b / d 0 0 0    s  a / d 0 0 0 0 0 1 0 0   0 0 0 1 0   0 0 0 0 1

The 2 2 matrix in the upper left hand corner has determinant -1, and is therefore invertible over R. The element in the 1-1 position of ST is ar  bs  d, and the element in the 1-2 position is ab / d  ba / d  0, as desired. We have replaced the pivot element a by a divisor d, and this will decrease the number if prime factors, guaranteeing the finite termination of the algorithm. Similarly, if b were in the 2-1 position, we would premultiply S by the transpose of T; thus in the upper left hand corner we would have

 r s    b / d  a / d 

109

4.9 Unit Summary/ Things to Remember

1 A ring in which every strictly ascending chain of right (left) modules is finite, is called a Right (Left) noetherian ring.

2 A ring in which every strictly descending chain of right (left) modules is finite is called a right (left) artinian ring. 3 Hilbert’s Basis Theorem: If R is a right noetherian ring with unity then R[x], the ring of polynomials over R, is right noetherian. 4 Wedderburn Artin theorem: Let R be a left (or right) artinian ring with unity and no nonzero nilpotent modules. Then R is isomorphic to a finite direct sum of matrix rings over division rings. 5 A nonzero module M is called uniform module if any two nonzero sub modules of M have nonzero intersection. 6 A module M is called primary module if each nonzero sub module of M has uniform sub module and any two uniform sub modules of M are sub isomorphic. 7 Let M be a nonzero finitely generated module over a commutative noetherian. ring R. Then there are only a finite number of primes associated with M.

8 Lasker-Noether Decomposition Theorem: Let R be a commutative, Noetherian ring, and

let I be an ideal of R. There exist primary ideals { Qi } with I = Qi, such that no Qi

contains the intersection of the other primary ideals, and the ideals Qi have distinct associated primes. Furthermore, in any such representation of I as an intersection of primary ideals, there must be n ideals, and the set of their associated prime ideals must be the same. 9 Every matrix over a principal ideal domain is equivalent to a matrix in Smith normal form.

110

4.10 Assignments/ Activities

1. Prove that every commutative Artinian ring possesses a finite number of proper prime modules.

2. Let R be the direct sum of rings R , R2 ,...Rn .Show that R is right noetherian.(artinian) if and

only if each Ri is right noetherian.(artinian). 3. If in a noetherian. ring R every module generated by two elements is principal show that R is a principal Module Ring. Let A be the matrix

  2 3 0     3 3 0   12 12 6

over the integers. Find the Smith normal form of A.

4 Show that every principal left ideal ring is noetherian.

5 The nonzero components ai of the Smith normal form S of A are called the invariant factors of A. Show that the invariant factors of A are unique (up to associates). 6 Prove that the intersection of all prime ideals in a noetherian ring is nilpotent.

7 Show that a square matrix P over the integers has an inverse with integer entries if and only if P is unimodular, that is, the determinant of P is 1.

8 Let V be the direct sum of the R-modules V1,....,Vn , and let W be the direct sum of R-

modules W1,....,Wm . Indicate how a module homomorphism V to W can be represented by a matrix. (The entries of the matrix need not be elements of R.)

111 9 Show that if V n is the direct sum of n copies of the R-module V, then we have a ring

n isomorphism EndR (V )  M n (EndR (V )).

4.11 Check Your Progress

1. Prove that if an integral domain with unity, has finite number of modules, and then it is a field.

2. If R is right noetherian.(artinian) prove that Rn , the ring of n n matrices over R is right noetherian (artinian). 3. If in a noetherian. ring R every module generated by two elements is principal show that R is a principal Module Ring. 4. Show that two m n matrices are equivalent if and only if they have the same invariant factors, i.e. if and only if they have the same Smith normal form.

5. State and Prove Hilbert‟s Basis Theorem. 6. Recall that when a matrix over a field is reduced to row-echelon form (only row operations are involved), a pivot column is followed by non-pivot columns whose entries are zero in all rows below the pivot element. When a similar computation is carried out over the integers, or more generally over a Euclidean domain, the resulting matrix is said to be in Hermite normal form. Let

 6 4 13 5    A   9 6 0 7    12 8 1 12

Carry out the following sequence of steps:

112 1. Add -1 times row 1 to row 2

2. Interchange rows 1 and 2

3. Add -2 times row 1 to row 2, and then add -4 times row 1 to row 3

4. Add -1 times row 2 to row 3

5. Interchange rows 2 and 3

6. Add -3 times row 3 to row 3

7. Interchange rows 2 and 3

8. Add -4 times row 2 to row 3

9. Add 5 times tow 2 to row 1

10. Add 2 times row 3 to row 1, and then add row 3 to row 2

7. Continuing Problem 6, consider the simultaneous equations

6x  4y 13z  5,9x  6y  7,12x 8y  z 12 (mod m)

For which values of m  2 will the equations be consistent?

i. Show that if R is regarded as an R-module, then EndR (R) is isomorphic to the opposite

ring Ro.

ii. Let R be a ring, and let f  EndR (R). Show that for some r  R we have f (x)  xr for all x  R.

o iii. Let M be a free R-module of rank n. Show that EndR (M )  M n (R ), a ring isomorphism.

4.12 Points for discussion / Clarification

At the end of the unit you may like to discuss or seek clarification on some points. If so, mention the same.

113

1. Points for discussion

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2. Points for clarification

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4.13 References

15. D.J.S Robinson, A Course in the Theory of Groups, 2nd Edition, New York: Springer-Verlag,

1995.

16. J. S. Lomont, Applications of Finite Groups, New York: Dover, 1993.

17. John S. Rose, A Course on Group Theory. New York: Dover, 1994.

114 18. P.B. Bhattacharya, S.K. Jain and S.R. Nagpaul, Basic Abstract Algebra,

(2ndEdition),Cambridge University Press, Indian Edition, 1997.

19. John B. Fraleigh, A first Course in Abstract Algebra, 7th Edition, Pearson Education, 2004.

115 UNIT-5

Fundamental Structure theorem for finitely generated modules

STRUCTURE PAGE NO.

5.1 Introduction 57

5.2 Objective 57

5.3 Structure theorem for finitely generated modules over PID 57-60

5.4 Application to finitely generated abelian groups 60-61

5.5 Rational Canonical Form 61-64

5.6 Generalized Jordan form over any field 64-65

5.7 Unit Summary/ Things to Remember 65-66

5.8 Assignments/ Activities 67

5.9 Check Your Progress 67-68

5.10 Points for Discussion/ Clarification 68-69

5.11 References 69

116

5.1 Introduction

The structure theorem for finitely generated modules over a principal ideal domain is a generalization of the fundamental theorem of finitely generated abelian groups and roughly states that finitely generated modules can be uniquely decomposed in much the same way that integers have a prime factorization. The result provides a simple framework to understand various canonical form results for square matrices over fields.

5.2 Objective

After the completion of this unit one should able to:

 generalize the fundamental theorem of finitely generated abelian groups.  decompose finitely generated modules into a direct sum of cyclic modules.  describe the applications of Structure theorem for finitely generated modules over a principal ideal domain  generalize Jordan Form over any field.  find all possible Jordan Canonical Forms of a matrix.

5.3 Structure theorem for finitely generated module over a PID (particular ideal domain)

117 To prove the theorem one must know the following definitions:

An element x of an R-module M is called a torsion element if there exists a nonzero element r  R such that rx  0.

A nonzero element x of an R-module M is called a torsion-free element if rx  0,r  R, implies r  0 .

If R be a principal ideal domain, and M be an R-module. Then T or M  {x  M | x is torsion} is a sub module of M .

Theorem Let R be a principal ideal domain, and let M be any finitely generated R-module. Then

s M  R  R | Ra1 ... R | Ra r ,

a direct sum of cyclic modules, where the ai are nonzero no units and ai / ai1,i 1,...,r 1.

Proof Because M is a finitely generated R-module, M  Rn / K Further, K  Rm ,m  n . Let  be

m m this isomorphism form R to K. Thus, K  (R ) . Let (e1,...,em ) be a basis of . Let us write

a11 a   21

 .  n (e1 )     R  .   .    an1 

......

118 a1m  a   2m 

 .  n (em )     R .  .   .    anm 

m m Then (R )  AR , where A  (aij ) is an n m matrix.

Choose invertible matrices P and Q of order n n and mm, respectively, such that

PAQ  diag (a1,a2 ,....,ak ,0,....,0), where a1 | a2 | ... | ak . Then

R n R n R n PR n R n M      K (R m ) AR m PAQR m PAQR m

a1   a 0   2       .  R  .  R     n  ak   .   R / .    0  .      0 .   .     .  R      .     .   0 

119 R  R .... R R R  =  .... Ra  Ra .... Ra Ra Ra 1 2 k 1 2 1

nk copies R  =   R .... R Ra k

R R = ....  R s Ra Ra u k 1

[by deleting the zero terms if any corresponding to those ai 's that are units]

R R  ....  R s (by renumbering if necessary). Ra Ra 1 r 1

Because for any ideal I [including (o)], R/I is a cyclic R-module.

Hence Proved.

Theorem Let M be a finitely generated module over a principal ideal domain R. Then M  F  T or M, where (i) F  R s for some nonnegative integer s, and (ii) T or M  R / R ... R / R , a1 ar where ai are nonzero nonunit elements in R such that a1 | a2 | ... | ar .

Proof By the structure theorem for finitely generated modules over a PID,

s M  R  R / Ra1 ... R / Ra r , where ai are nonzero nonunit elements in R such that a1 | a2 | ... | ar . It then follows that

M  F T,

s where F  R , and T  R / Ra1 ... R / Ra r .Clearly, arT  0. Thus, T  T or M.

120 Next, let x  T or M. Then x  x1  x2 , x1  F, x2 T . So x1  x  x2 T or M, and, hence, rx1  0

s s for some 0  r  R.If (y1..., ys )  R is the image of x1 under the isomorphism F  R , then r(y1,..., ys )  0.Hence, ryi  0,1 i  s.Therefore, each yi  0, because r  0. this yields x1  0 and proves that x T,and, thus, T= T or M. This completes the proof.

5.4 Application to finitely generated abelian groups

Let A be a finitely generated abelian group. Then

s A  Z  Z / a1Z ... Z / ar Z (1)

where s is a nonnegative integer and ai are nonzero non units in Z, such that

a1 | a2 | ... | ar . (2)

Further, the decomposition (1) of A subject to the condition (2) is unique. In particular, if A is

n generated by (x1..., xn ) subject to aij x j  0,1  i  m, j1 then (n  r) copies



A  Z ... Z  Z / a1Z ... Z / ar Z, where a1,....,ar are the invariant factors of the m n matrix

A  (aij ).

5.5 Rational Canonical Form

The fundamental theorem of finitely generated modules over a PID has interesting applications in obtaining important canonical form for square matrices over a field. Given a n n matrix A over a

121 field F, our effort will be to choose an invertible matrix P such that P1 AP is in a desired canonical form. This problem is equivalent to finding a suitable basis of a vector space F n such that the matrix of the linear mapping x  Ax with respect to the new basis is in the required canonical form.

1 a Not every matrix is similar to a diagonal matrix. For example, let A   ,a  0, be a matrix 0 1 over R. Suppose there exists an in invertible matrix P such that

1 d1 0  P AP   .  0 d 2 

Clearly, Det (P1 AP  xI )  det(A  xI ).

Thus,

d  x o 1 x a 1  . 0 d 2  x 0 1 x

This implies d1d2  1and d1  d2  2, so d1 1  d2 .

Hence, P 1 AP 1; that is, A = I, a contradiction.

However, simple canonical forms are highly important in linear algebra because they provide an adequate foundation for the study of deeper properties of matrices. As an application of the fundamental theorem for finitely generated modules over a PID, we show that every matrix A over a field is similar to a matrix

122 B1   B   2   .     .   .     Bs 

Where Bi matrices of the form

0 0 ... 0 *   1 0 ... 0 * 0 1 ... 0 *    . . . ..  . . . .    . . . .   0 0 ... 1 *

Let V be a vector space over a field F, and let T :V V be a linear mapping, We can make V an

m F[x]-module by defining the action of any polynomial f (x)  a0  a1x ...  am x on any vector

m i i  V as f (x)  a0  a1 (T ) ...  am (T  ), where T  stands for T ( ).Clearly, this action of F[x]on V is the extension of the action of F such that x  T.First we note the following simple fact.

If V is a finite- dimensional vector space over F, then V is a finitely generated F[x]-module.

For if {1,... m}is a basis of V over F, then {1,... m}is a set of generators of V as -module.

Theorem Let T  HomF (V,V). Then there exists a basis of V with respect to which the matrix of T is

123 B1 0   B   2   .  A    ,  .   0 .     Br 

where Bi is the companion matrix of a certain unique polynomial fi (x),i 1,...r, such that f1 (x) | f 2 (x) | ... | f r (x).

This form of A is called the rational canonical form of the matrix of T. The uniqueness of the decomposition of V subject to f1 (x) | f 2 (x) | ... | f r (x) shows that T has a unique representation by a matrix in a rational canonical form.

The polynomials f1 (x),..., f r (x) are the invariant factors of A  xI , but they are also called the in invariant factors of A.

Example

The rational canonical form of the 6 6matrix A whose invariant factors are

(x  3),(x  3)(x 1),(x  3)(x 1)2 is

3     0  3   1 4     0 0 3   1 0  7    0 1 5 

5.6 Generalized Jordan form over any field

124 This is the another application of the fundamental theorem of finitely generated modules over a PID. Every matrix is similar to an “almost diagonal” matrix of the form

i 0 ... 0  J  1  ... 0   1   i   .   0 1       .  ,  . . . .   .  . . .         J 5   . . .     0 0 ... i 

But first we obtain a canonical form over any field, of which the Jordan canonical form is a special case.

Let V be a vector space of dimension n over F. Let T  HomF (V, V). Because HomF (V, V) is an n 2 -dimensional vector space over F, the list 1, T,T 2 ,...,T n2 is linearly dependent, and, hence, there

n2 a ,a ,...a 2  F a  a T ...  a 2 T  0 exist 0 1 n (not all zero) such that 0 1 n .

Therefore, T satisfies a nonzero polynomial over F.

Example 1 For a 33 matrix with invariant factors (x  2)and (x 2  4), the elementary divisors are (x  2),(x  2), and (x  2), so the Jordan canonical form is-

2       2        2

Example 2

125 For a 6 6 matrix with invariant factors (x  2)2 ,(x  3)2 (x  2)2 the elementary divisors are

(x  2)2 ,(x  2)2 , and (x  3)2 , so the Jordan canonical form is

 2 0     1  2    2 0     1  2    3 0     1  3

5.7 Unit Summary/ Things to Remember

1. An element x of an R-module M is called a torsion element if there exists a nonzero element r  R such that rx  0.

2. A nonzero element x of an R-module M is called a torsion-free element if rx  0,r  R, implies r  0 .

3. Let M be a finitely generated module over a principal ideal domain R. Then M  F  T or M,

where (i) F  R s for some nonnegative integer s, and (ii) T or M  R / R ... R / R , a1 ar

where ai are nonzero non unit elements in R such that a1 | a2 | ... | ar .

4. Rational canonical form of a square matrix A is a canonical form for matrices that reflects the structure of the minimal polynomial of A and provides a means of detecting whether another matrix B is similar to A without extending the base field F.

5. Rational canonical form is generally calculated by means of a similarity transformation.

6. The Jordan form is determined by the set of elementary divisors.

7. If a polynomial q(x) is a power of a monic irreducible polynomial p(x), then the Jordan matrix of relative to p(x) will be simply called the Jordan Matrix of .

126 8. For any A M n (C), Jordan Normal Form of the matrix A is of the type

 J O O ... O   1   O J 2 O ... O   O l J ... O   3   ...     O O O ... J m 

where each ji is of the form

a 1 0 ... 0 0   1   0 ai 1 ... 0 0   0 0 a ......   i   ......   0 0 0 ... a 1   i     0 0 0 ... 0 ai 

Further each ai is an eigen value of A.

5.8 Assignments/ Activities

1. Let A be an n n matrix. Then there exists an invertible matrix P such that P 1 AP is a

direct sum of generalized Jordan blocks J i , is unique except for ordering.

2. Find the abelian group generated by{x1, x2 , x3 ) subject to

5x1 9x2 5x3  0, 2x1  4x2  2x3  0, x1  x2  3x3  0.

3. The abelian group generated by x1 and x2 subject to 2x1  0,3x2  0 is isomorphic to Z /(6).

127 4. Consider the polynomial (x 1),(x 1)2 ,(x  i)2 ,(x  i)2 over C. Find their respective Jordan matrices (Jordan Blocks).

5. Find rational canonical forms of the following matrices over Q:

1 1  2 4 1 5 7 2 4 0 0 1 2 2 (a)0 4 3 (b)1 4 1 (c) .     0 0 1 3 0 0 1 3 8 3   0 0 0 0

6. Reduce the following matrix A to rational canonical form:

 3 2 0    A   1 0 1   1  3  2

5.9 Check Your Progress

1. State and Prove Structure theorem for finitely generated modules over PID .

2. Compute the invariants and write down the structures of the Abe liar group with generators x1,...., xn subject to the following relations:

(a) n  2; x1  x2  0.

(b) n  3;3x1  2x2  0, x1  x3  0,x1  3x2  2x3  0.

(c) n  3;2x2  x3  0,3x1  8x2  3x3  0 and 2x1  4x2  x3  0.

3. Find invariant factors, elementary divisors, and the Jordan canonical form of the matrices

 1  5  2 4  0 4 2   2    0 5 4 4 (a) 3 8 3  . (b) .  4  8  2 0 0 5 3     0 0 0 4

128 4. Find all possible Jordan canonical forms of a matrix with characteristic polynomial p(x) over C in each of the following cases:

(a) p(x)  (x 1)2 (x 1).

(b) p(x)  (x  2)3 (x  5)2 .

(c) p(x)  (x 1)(x  2)2 (x2 1).

5.10 Points for Discussion/ Clarification

At the end of the unit you may like to discuss or seek clarification on some points. If so, mention the same.

1. Points for discussion

______

______- ______

2. Points for clarification

______

______- ______

______

5.11 References

20. I . N. Herstein, Topics in Algebra, Second Edition, John Wiley and Sons, 2006. 21. Lay, David C. , Linear Algebra and Its Applications (3rd ed.), Addison Wesley, 2005.

129 22. Rose John, A Course on Group Theory. New York: Dover, 1994. 23. P.B. Bhattacharya, S.K. Jain and S.R. Nagpaul, Basic Abstract Algebra,(2ndEdition),Cambridge University Press, Indian Edition, 1997. 24. Fraleigh B John, A first Course in Abstract Algebra, 7th Edition, Pearson Education, 2004. 25. Steven J. Leon, Linear Algebra With Applications (7th ed.), Pearson Prentice Hall, 2006.

130

JORDAN NORMAL FORM

 Let A  (aij ) be an n n matrix over a field F and V be an n-dimensional vector space over F with

B  (b1,b2 ,...,bn ) as its ordered basis. Then the linear transformation T on V defined by

n T(b j )  aijbi ,1  j  n admits A as its matrix relative to B. Let q1 (x), i1

 q2 (x),...,qm (x)be the elementary divisors of T. Then V  W1 W2 ...Wm ,T  T1

 T2 ...Tm for some cyclic T-invariant subspaces Wi (1  i  m) such that the restriction Ti of

T to Wi has qi (x) as its minimal polynomial (Theorem 16.54). Since each Wi is cylic Ti -space, by

Theorem 16.36 has basis Bi such that matrix of relative to Bi is the Jordan Matrix J i of

qi (x) relative to the irreducible polynomial of which qi (x) is a power.

' ' '  T  T1 T2 ...Tm Yields that V has a basis B' (b1,b2 ,...,bn ) such that the matrix of

T relative to B' is J  J1  J 2 ... J m. Then A is similar to J (Corollary 12.40). We call J, a Jordan Normal Form of matrix A. Since the elementary divisors of T are uniquely determined (Theorem 16.54), the Jordan Normal Form of A is uniquely determined except for the orders in

' which J i s are written.

 Theorem 16.57

 Proof Let qi (x)...qm (x) be the elementary divisors of A. Since over C only irreducible polynomial

are of degree one, and each qi (x) is a power of an irreducible polynomial, we have

ki qi (x)  (x  ai ) for some ai C and some positive integer ki. Then the Jordan matrix of

qi (x) is the ki  ki matrix.

131 a 1 0 ... 0 0   i   0 ai 1 ... 0 0   ......     J i   0 0 0 ... ai 1   0 0 0 ... 0 a   i     

 Hence the Jordan normal form of A is

 J 0 ... O   i   0 J 2 ... O   J  J ... J  1 2 m  ......       O O ... J m 

 The next part follows form Corollary 16.56

1 1   i 1  i 1   Example 23 J1  (1), J 2   , J 3    and J 4     0 1 0 i  0  i

 Then 7 7 matrix.

 1 0 0 0 0 0 0     0 1 1 0 0 0 0   J1 O O O     0 0 1 0 0 0 0    O J 2 O O  =    O O J O   0 0 0 i 1 0 0   3      0 0 0 0 i 0 0  O O O J 4     0 0 0 0 0 i 1     0 0 0 0 0 0  i 

 is a Jordan Normal Form and its elementary divisors are the above given polynomial . Further, -1, i, - i are its distinct eigen values.

Definition.

2.2

Problem

132 Rational canonical form

Problem

Some authors further decompose each Bi as the direct sum of matrices, each of which is a companion matrix of a polynomial that is a power of an irreducible polynomial, and call the resulting form rational canonical form. The powers of the irreducible polynomials corresponding to each block in the form thus obtained are then called the elementary divisors of A

5 Generalized Jordan form over any field

Problems

1.

133