TIFR pamphlet on Galois Theory

8/11/2019 TIFR pamphlet on Galois Theory

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GALOIS THEORY

M. PAVAMAN MURTHY

K.G. RAMANATHAN C. S. SESHADRI

U. SHUKLA R. SRIDHARAN

Tata Institute of Fundamental Research, Bombay

School of Mathematics

Tata Institute of Fundamental Research, Bombay

1965


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All communications pertaining to this series should be addressed to:

School of Mathematics

Tata Institute of Fundamental Research

Bombay 5, India.

c

Edited and published by K. Chandrasekharan for the Tata Institute of Fundamen-

tal Research, Bombay, and printed by R. Subbu at the Commercial Printing Press

Limited, 3438 Bank Street, Fort, Bombay, India.


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EDITORIAL NOTE

This little book on Galois Theory is the third in the series of Mathemati-cal pamphlets started in 1963. It represents a revised version of the notesof lectures given by M. Pavaman Murthy, K.G. Ramanathan, C.S. Se-shadri, U. Shukla and R. Sridharan, over 4 weeks in the summer of 1964,to an audience consisting of students and teachers, some of them newto algebra. The course evoked enthusiasm as well as interest. Specialthanks are due to the authors, and to Professor M.S. Narasimhan who, asChairman of the Committee on the Summer School, was responsible for

organizing it. I owe personal thanks to Professor Raghavan Narasimhanwho has done the actual editing of the work.

K. Chandrasekharan


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Contents

1 Groups 1

1.1 Groups and Homomorphisms . . . . . . . . . . . . . . . . 11.2 Subgroups and quotient groups . . . . . . . . . . . . . . . 41.3 Solvable groups . . . . . . . . . . . . . . . . . . . . . . . . 101.4 Symmetric groups and solvability . . . . . . . . . . . . . . 12

2 Rings and Vector Spaces 15

2.1 Rings and homomorphisms . . . . . . . . . . . . . . . . . 152.2 Ideals and quotient rings. . . . . . . . . . . . . . . . . . . 182.3 Polynomial rings. . . . . . . . . . . . . . . . . . . . . . . . 202.4 Vector spaces . . . . . . . . . . . . . . . . . . . . . . . . . 24

3 Field Extensions 29

3.1 Algebraic extension . . . . . . . . . . . . . . . . . . . . . . 293.2 Splitting fields and normal extensions . . . . . . . . . . . 313.3 Separable extensions . . . . . . . . . . . . . . . . . . . . . 353.4 Finite fields . . . . . . . . . . . . . . . . . . . . . . . . . . 383.5 Simplicity of finite separable extension . . . . . . . . . . . 39

4 Fundamental Theorem 41

5 Applications of Galois Theory 47

5.1 Cyclic extensions . . . . . . . . . . . . . . . . . . . . . . . 475.2 Solvability by radicals . . . . . . . . . . . . . . . . . . . . 50

5.3 Solvability of algebraic equations . . . . . . . . . . . . . . 535.4 Construction with ruler and compass . . . . . . . . . . . . 57

i


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Chapter 1

Groups

1.1 Groups and Homomorphisms

Definition 1.1 A group is a pair (G, ), where G is a set and: G G G is a map ((x, y) being denoted by xy ) satisfying.

(a) (xy)z= x(yz), x,y ,z G(associativity)

(b) there exists an elemente G such that ex = xe = x, x G,and

(c) ifx Gthere exists an element x G such thatxx= xx =e.

Remark 1.2 The map is called the group operation. We denote thegroup simply by G when the group operation is clear from the context.

Remark 1.3 The element e is unique. For, ife1 G such that e1x=xe1 =x, x G, we have, in particular, e= ee1 =e1. The element eis called the identity element ofG.

Remark 1.4 For xG the element x is unique. For, ifx G suchthat x x= xx =e, then we have

x =xe= x(xx) = (xx)x =ex =x.

The element x is called the inverse ofx and is denoted by x1.

1


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2 Chapter 1. Groups

Remark 1.5 In view of associativity, we define

xyz = (xy)z = x(yz), where x, y ,z G.More generally, the productx1x2 . . . , xnis well-defined, wherex1, x2,

. . . , xn G (proof by induction). For x G we set

(i) xn =xx . . . x(n factors), for n >0;

(ii) x0 =e;

(iii) xn = (x1)n, for n 00x = 0

nx = (n)(x), for n


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1.1. Groups and Homomorphisms 3

group operation being given by the composition of maps. It is called the

symmetric group of degree n and is denoted by Sn. If Sn we write

=

1 2 . . . n(1) (2) . . . (n)

For n 3, Sn is not abelian. The order ofSn is n!

Definition 1.10 Let G and G be groups. A map f: GG is calleda homomorphism iff(xy) =f(x)f(y), x, y G.

Remark 1.11 For a group G, the identity map IG: G G is a homo-morphism.

Remark 1.12 Iff: G G and g: G G are homomorphisms, thenthe map g f: G G is a homomorphism.

Definition 1.13 A homomorphism f: G G is called an isomorphism,if there exists a homomorphism g: G G such that g f = IG andf g= IG . We then write G G. An isomorphism f: G Gis calledan automorphism.

Remark 1.14 A homomorphism is an isomorphism if and only if it isone-one and onto.

Remark 1.15 A homomorphism maps the identity into the identityand the inverse of an element into the inverse of its image.

Example 1.16 The natural map q: Z Z/(m) given by q(r) = r,where r Z, is an onto homomorphism. For m= 0, it is not anisomorphism.

Example 1.17 The map f: Z Z given by f(n) = 2n is a one-onehomomorphism, which is not onto.

Example 1.18 The map f: R

R+ (the set of non-zero positive real

numbers) given by f(x) = ax, where a is a fixed real number greaterthan 1 and x R, is an isomorphism.

Example 1.19 Let G be a group and let aG. The map fa: GGgiven by fa(x) = axa

1, where x G is an automorphism, called theinner automorphism given bya.


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4 Chapter 1. Groups

1.2 Subgroups and quotient groups

Definition 1.20 A subgroup Hof a groupG is a non-empty subset HofG such that ifx, y H, then x1y H.

Remark 1.21 The identity e ofG belongs to H. Also, ifx H, thenx1 H. In factHbecomes a group under the induced group operation.

Remark 1.22 The inclusion map i: H G given by i(x) = x, wherex H, is a homomorphism.

Example 1.23 G and{e} are subgroups ofG.

Example 1.24 Z is a subgroup Q, Q is a subgroup of R and R is asubgroup ofC.

Example 1.25 The set of even integers is a subgroup ofZ.

Let f: G G be a homomorphism of groups. The set f(G) is asubgroup ofG. Ife denotes the identity ofG, the set{x G | f(x) =e} is a subgroup ofG, called thekerneloffand denoted by Kerf. Moregenerally, the inverse image of a subgroup ofG is a subgroup ofG.

The intersection of a family of subgroups of a group is a subgroup.

IfSis a subset of the group G, the intersection of the family of all thesubgroups ofG which contain S is called the subgroup generated by S.If S consists of just one element a, the subgroup generated by{a} iscalled thecyclic subgroup generated bya. It is easily seen that the cyclicsubgroup generated bya consists of the powers ofa. A groupG is calledcyclic if it coincides with the cyclic subgroup generated by an elementa G.

Example 1.26 Z is an infinite cyclic group generated by 1.

Example 1.27 Z/(m) is a cyclic group of order|m|, ifm = 0.

Proposition 1.28 Any subgroup H of Z is cyclic.

Proof: For, ifH= {0}, there is nothing to prove. IfH= {0}, let mbe the least positive integer inH. Now for anyn Zwe haven= qm+rwhere q, r Z and 0r < m. IfnH then r =n qmH, whichimplies thatr = 0. Hence H=mZ.


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1.2. Subgroups and quotient groups 5

Notation: For subsetsA and B of a group G, we set AB =

{ab

|a A, b B}. IfA = {a} (resp. B = {b}) we write aB (resp. Ab) for{a}B (resp. A{b}). IfA, B, C G, it is obvious that (AB)C=A(BC)and we write AB C for either of them.

LetG be a group and let Hbe a subgroup. LetR = RHdenote theequivalence relation in G defined as follows: xRy, if x1y H, wherex, y G. The equivalence class xH to which x belongs is called a leftcoset ofG modulo H. If the quotient set G/R consists of n elements,the n is said to be the index ofH in G and is denoted by [G: H].

Proposition 1.29 (Lagrange) LetHbe a subgroup of a finite group G.Then the order of G is the product of the order ofH and the index of

H inG. In particular, the order ofH is a divisor of the order ofG.

Proof: We first note that the map t: H xH given by t(h) = xh,whereh H, is a one-one onto map. Therefore the number of elementsin any left coset is equal to the order ofH. Since any two distinct leftcosets are disjoint and the index ofH in G is the number of left cosets,the theorem follows.

An element of a group Gis said to be ofordern if the cyclic subgroupgenerated by a is of order n.

Corollary 1.30 The order of an element ofG is a divisor of the order

ofG.

Corollary 1.31 A group of prime orderp is cyclic.

For, if a is any element different from the identity, the order of adividesp and so is equal to p.

Proposition 1.32 The following statements are equivalent:

(i) n is the order ofa;

(ii) n is the least positive integer such thatan =e;

(iii) an =e and ifam =e,thenn|m.Proof: (i) (ii). Since the cyclic subgroup generated by a is finite,it follows that the elements (ai)iZare not all distinct. Ifa

p =aq wherep > q, then apq =e. Thus there exists a positive integerm such thatam = e. Taking m to be least positive integer for which am = e, we


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6 Chapter 1. Groups

observe that the elements

{ai

| 0

i < m

} are distinct and form a

subgroup, which is the cyclic subgroup generated by a. Hence m = n.(ii) (iii). Letam = e. We have m = qn+r, where q, r Z and

0 r < n. This gives e = am = (aq)nar = ar. Since n is the leastpositive integer such that an =e, it follows thatr = 0. Hence m = qn.

(iii) (i). The elements (ai)0i q, then apq = e. Therefore, byhypothesis, n divides p q, which is impossible since p q is less thann. Clearly{ai | 0 i < n} is the cyclic subgroup generated by a.

The maximum of the orders of the elements of a finite abelian groupis called the exponentof the group.

Proposition 1.33 Ifm is the exponent of a finite abelian group G, thenthe order of every element ofG is a divisor ofm.

We first prove the following.

Lemma 1.34 Leta andb be elements of a group G such thatab= ba.Ifa andb are of ordersm andn respectively such that(m, n) = 1, thenab is of ordermn.

Proof: We have (ab)mn =amnbmn =e. Therefore, ifd is the order ofab, then d|mn. Again, since (ab)d =e we havead =bd and soand =e.Therefore m|nd. Since (m, n) = 1 it follows that m|d. Similarly, n|d.Since (m, n) = 1, mn|d. This proves that ab is of order mn.Proof of the Proposition: Leta be an element of maximum orderm and let b be any element of order n. Let, if possible, n|m. Thenthere exists a prime number p such that if r (resp. s) is the greatestpower of p dividing n(resp. m), we have r > s. Then ap

s

(resp. bn/pr

)has order m/ps(resp. pr). Since (m/ps, pr) = 1 we see, by the abovelemma, that the element ap

s

bn/pr

is of order (m/ps)pr which is greaterthan m, since r > s. This contradicts the assumption on a. Hence theproposition.

Definition 1.35 A subgroup Hof a group G is said to be normal in G

ifxH x1

=H, x G.A subgroup H ofG is normal if and only ifxHx1 H,x G.For any subgroupHofG and anyx G,xHx1 is a subgroup ofG,

which is called aconjugateofH. By the definition of a normal subgroup,it follows that a subgroup H is normal if and only if all its conjugatescoincide with H.


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Letf: G

G be a homomorphism of groups. The inverse image of

a normal subgroup ofG is a normal subgroup ofG. Moreover, iff isonto, the image of a normal subgroup ofG is a normal subgroup ofG.

G and{e} are normal subgroups ofG. A subgroup ofG other thanG is called a propernormal subgroup. If a group has no proper normalsubgroup other than{e}, it is called a simple group.

Example 1.36 In an abelian group every subgroup is normal.

Example 1.37 An abelian group G(= {e}) is simple if and only if it iscyclic of prime order.

Example 1.38 The kernel of a homomorphism of groups f: G G isa normal subgroup ofG.

Example 1.39 If H and K are subgroups of G and if HK = KH,then HK is a subgroup ofG. The conditionHK = KH is satisfied ifeitherH or K is a normal subgroup ofG. If both H and Kare normalsubgroups ofG, then H K is also a normal subgroup ofG.

LetGbe a group andHbe a normal subgroup ofG. Consider the setG/R of left cosets ofG modulo H . We define an operation on G/R bysetting xH

yH=xyH, where x, y

G. We assert that this operation

is well defined. For, if x xH, and y yH, that is x = xh1, y =yh2, where h1, h2 H, then xy = xh1yh2 = xy(y1h1y)h2 xyH,since y1h1y H. It is easily seen thatG/R is a subgroup under thisoperation, the identity being H(= eH) and the inverse of xH beingx1H. We call this group the quotient group ofG byH and denote itbyG/H. The natural map q: GG/H given by q(x) =xH is clearlyan onto homomorphism and its kernel is H.

Example 1.40 The group Z/(m) of residue classes modulo m is thequotient group Z/mZ.

Letf: G G be an onto homomorphism with kernel H. We definea homomorphism f: G/H G by setting f(xH) =f(x), where x G.Clearly fis well defined and is an isomorphism ofG/H onto G and wehave f q=f where q: G G/His the natural map. Thus, upto anisomorphism , every homomorphic image of a group is a quotient group.This is usually called the fundamental theorem of homomorphisms.


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8 Chapter 1. Groups

Remark 1.41 LetG be a group and letHandKbe normal subgroups

of G such that K H. The homomorphism f: G/K G/H givenby f(xK) = xH, where x G is onto and clearly has H/Kas kernel.Hence (G/K)(H/K) G/H. ( First isomorphism theorem.)

Remark 1.42 Let H and K be subgroups of a group G and let Kbe normal in G. Then the homomorphism f: H HK/K given byf(h) =hK, whereh Hhas H Kas kernel and H/H K HK/K.(Second isomorphism theorem.)

Remark 1.43 LetG be a cyclic group generated by an elementa. Themap f: Z

G given by f(n) = an, where n

Z is an onto homomor-

phism. Therefore Z/Ker f G. Since Kerf =mZ for some m0, itfollows that every cyclic group is isomorphic to Z/mZ. It can now beeasily shown that subgroups and quotient groups of a cyclic group arecyclic.

Remark 1.44 If G is any group with identity e we have G/{e} G, G/G {e}.

LetG be a group and leta, b G. We say that a is conjugate tob, ifthere exists an element x G such that a= xbx1. It is easy to verifythat the relation of conjugacy is an equivalence relation. An equivalence

class is called a conjugacy class or a class of conjugate elements.

Remark 1.45 The subset{e} ofG is a conjugacy class.

Remark 1.46 Ifa is conjugate tob, thenam is conjugate tobm, wherem Z.

Remark 1.47 A subgroup H ofG is normal if and only if it is a unionof conjugacy classes.

LetKbe a subset of a group G. We define a subset NK ofG called

the normaliser of K in G as follows: NK ={x G| xK = Kx}. IfKconsists of just one element a we denote the normalizer of{a}by Naand call it the normalizer ofa. It is easy to verify that thenormalizerof a subsetK in G is a subgroup ofG.

Remark 1.48 A subgroup H is normal in Gif and only ifNH=G.


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Proposition 1.49 Let G be a finite group and let a

G. Then the

number of elements conjugate to a is equal to the index of the normalizerofa inG.

Proof: LetCdenote the conjugacy class to whicha belongs and con-sider the map : G Cgiven by (x) = xax1 where x G. This isan onto map. Ifn Na then (xn) = (xn)a(xn)1 = x(nan1)x1 =xax1 = (x), x G. This shows that any two elements of thesame left coset of G modulo Na have the same image in C. Con-versely, if x, y G and (x) = (y), then xax1 = yay1 giving(y1x)a(y1x)1 = a. This means y1x Na and so x and y be-long to the same left coset ofG modulo Na. Hence induces a one-one

correspondence between the left cosets ofG moduloNa and the distinctconjugates ofa.

Corollary 1.50 The number of elements conjugates to a is a divisor ofthe order of the group G. In particular, ifG is of orderpn, wherep is aprime, then a conjugacy class consists ofpi elements, where0 i n.

An element a G is said to be central if xa = ax, x G. Thesubset of all central elements ofG is called the centre ofG.

Remark 1.51 An element a of a group G is central if and only if the

subset{a} is a conjugacy class.

Remark 1.52 The centre of a group is a normal subgroup.

Proposition 1.53 IfG is a group of orderpn wherep is a prime andn 1 then the centre ofG has more than one element.

Proof: LetCi(1 i m) be the distinct conjugacy classes ofG andki the number of elements ofCi. We have ki|pn (Proposition 1.29 and1.49) so that ifki= 1, p|ki. Let C1 be the conjugacy class containingthe identity. If the centre ofG reduces to

{e

}, we have C1 =

{e

}and

ki > 1 for every i = 1. Further,

pn = 1 +i2

ki.

Since p|ki for i 2, this is impossible.


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10 Chapter 1. Groups

1.3 Solvable groups

Definition 1.54 A group G is said to be solvable if there exists a se-quence of subgroups.

G= G0 G1 Gn= {e}

such thatGi+1 is a normal subgroup ofGi and Gi/Gi+1 is abelian (0 i < n). Such a sequence is called a solvable series ofG.

Remark 1.55 Every abelian group is solvable.

Proposition 1.56 Any subgroup and any quotient group of a solvable

group is solvable. Conversely, if there is a normal subgroupHof a groupG such thatH andG/Hare solvable, thenG is solvable.

Proof: LetG be a solvable group having a solvable series

G= G0 G1 Gn= {e}.

IfH is a subgroup ofG, then

H=H G0 H G1 H Gn= {e}

is a solvable series ofH, sinceH

Gi+1 is a normal subgroup ofH

Gi

and (H Gi)/(H Gi+1) is isomorphic to a subgroup ofGi/Gi+1 andso abelian (0 i < n). (Compose the inclusion map H Gi Gi withthe natural map Gi Gi/Gi+1 and apply the fundamental theorem ofhomomorphisms.) Again, ifH is a normal subgroup ofG and q: GG/His the natural homomorphism, then

G/H=q(G0) q(G1) q(Gn) = {e}

is a solvable series ofG/H, since q(Gi+1) is a normal subgroup ofq(Gi)and q(Gi)/q(Gi+1) which is isomorphic to a quotient of Gi/Gi+1 isabelian (0 i < n).

Conversely, letHbe a normal subgroup ofG such thatH and G/Hare solvable. Let q= G G/Hbe the natural homomorphism. Let

H=H0 H1 Hn= {e}

andG/H=G0 G1 Gm= {e}


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1.3. Solvable groups 11

be solvable series for H and G/H respectively. Then it is trivial to see

that

G= q1(G0) q1(G1) q1(Gm) (=H=H0) H1 Hn= {e}.

is a solvable series for G.

Proposition 1.57 Any group of order pn where p is a prime, is solv-able.

Proof: We prove the proposition by induction onn. Forn = 0, theproposition is trivial. Let n

1 and assume that the proposition is true

for r < n. LetG be a group of order pn. Then by Proposition 1.53, thecentreCofG has order ps where s 1. Then the order ofG/C ispnsand n s < n. By the induction hypothesisG/C is solvable. Now theproposition follows from Proposition 1.56.

Proposition 1.58 A finite groupG is solvable if and only if there existsa sequence of subgroups

G= G0 G1 Gn= {e}such that Gi+1 is a normal subgroup of Gi and Gi/Gi+1 is cyclic, of

prime order(0 i < n).Proof: Suppose G is solvable and let

G= G0 G1 Gn= {e}be a solvable series of G. We shall interpose between Gi and Gi+1 asequence of subgroups

Gi = Hi,0 Hi,1 Hi,m= Gi+1such that Hi,j+1 is a normal subgroup of Hi,j and Hi,j/Hi,j+1 is ofprime order (0

j < m). For this, it is sufficient to show that given a

finite groupA and a normal subgroup B ofA such that A/B is abelian,there exists a normal subgroup N ofA containing B such that A/N isof prime order. Indeed let N be a maximal proper normal subgroupof A containing B. Clearly A/N is simple. However, A/N being ahomomorphic image ofA/B is abelian and hence of prime order.

The converse is trivial.


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1.4 Symmetric groups and solvability

Definition 1.59 LetSnbe the symmetric group of degree n. Anr-cycleis a permutation such that there exist rdistinct integers x1, . . . , xr (1 xi n) for which (x1) = x2, . . . , (xr1) = xr, (xr) = x1 and(x) = x for x=xi, 1ir. We then write = (x1, x2, . . . , xr). A2-cycle is called a transposition.

Remark 1.60 An r-cycle is an element of order r .

Remark 1.61 If = (x1, x2, . . . , xr) is an r-cycle and is any permu-tation, then 1 = (y1, y2, . . . , yr), where (xi) =yi, for 1 i r.

Proposition 1.62 Sn is generated by transpositions.

Proof: We prove the proposition by induction on n. For n = 1, 2the assertion is trivial. Assume the proposition for n 1 and let Sn. If (n) = n, then by the induction hypothesis is a product oftranspositions. If(n) = k where k= n, the permutation = (k, n)is such that (n) = n and so is a product of transpositions. Therefore= (k, n) is also a product of transpositions.

Corollary 1.63 Sn is generated by the transpositions(1, n), (2, n), . . . ,(n

1, n).

For, (i, j) = (i, n)(j, n)(i, n).

Lemma 1.64 If a permutation can be expressed as a product ofmtrans-positions and also as a product ofn transpositions, thenm n is even.Proof: The mapf: Sn {1, 1} given by

f() =

1i


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1.4. Symmetric groups and solvability 13

It should be noted that, in view of the preceding lemma, the notion

of odd and even permutations is well defined.The set of even permutations is a normal subgroup of Sn and is

denoted by An. It is called the alternating group of degree n. We notethat the quotient group Sn/An is of order 2 ifn >1.

Remark 1.66 Sn is solvable for n 4. Forn = 1, 2 there is nothing toprove. For n = 3,

S3 A3 {e}is a solvable series ofS3. For n = 4,

S4

A4

V4

{e

}is a solvable series ofS4, where

V4= {e, (1, 2), (3, 4), (1, 3), (2, 4), (1, 4)(2, 3)}.It can be verified thatV4 is a normal subgroup ofA4 andA4/V4 is a

group of order 3 and so is cyclic. Also, V4 is abelian.

Theorem 1.67 Sn is not solvable forn >4.

We need the following

Lemma 1.68If a subgroupGofSn(n >4) contains every 3-cycle and ifHis a normal subgroup ofG such thatG/H is abelian, thenHcontains

every 3-cycle.

Proof: Let q: G G/H be the natural homomorphism. If , G, q(11) = q()1q()1q()q() = e, since G/H is abelian.Therefore 11 H, , G. Let (i,j,k) be an arbitrary 3-cycle. Since n > 4, we can choose = (i,k,l), = (j, k, m) wherei,j,k,l and m are all distinct. Then

11= (l ,k,i)(m,k,j)(i,k,l)(j, k, m) = (i,j,k) H,

Proof of Theorem: Let, if possible,Sn= G0 G1 Gm= {e}

be a solvable series. Since Sn contains every 3-cycle, it follows from theabove lemma thatGi contains every 3-cycle for everyi where 1 i m.But, for i = m this is clearly impossible.


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Chapter 2

Rings and Vector Spaces

2.1 Rings and homomorphisms

Definition 2.1 A ringA is a triple (A,,), whereA is a set and, mappings ofA A into A (we write (x, y) = x +y, (x, y) = xy , forx, y A) such that

(i) (A, ) is an abelian group;

(ii) x(yz) = (xy)z, for x,y , z A(associativity);

(iii) x(y + z) = xy + xz, (y +z)x = yx + zx, for every x, y, z A(distributivity);

(iv) there exists an element 1 A called the unit element such that1x= x1 =x, for every x A.

Remark 2.2 and are called the ring operations. is called theaddition and , the multiplicationin the ring A; we often write (A, +)for the abelian group (A, ).

Remark 2.3 The identity of (A, ) is called thezero element ofA and

is denoted by 0.

Remark 2.4 The unit element is unique.

Remark 2.5 The associative law for multiplication is valid for anynumber of elements.

15


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16 Chapter 2. Rings and Vector Spaces

Remark 2.6 For a, b

Awith ab = ba and any integern

0, we have

(a+b)n =

ni=0ni

aibni, where, for anyx Aand any positive integerm, we write xm =x . . . x(m times). We set x0 = 1.

Remark 2.7 For anya A, by (iii), the map x ax(resp. x xa) isa homomorphism of the group (A, +) into itself and hencea0 = 0 (resp.0a= 0).

Definition 2.8 A ringA is commutative ifxy = yx, for every x, y A.

Example 2.9 The set Z (resp. Q, R, C) of integers (resp. rationals,reals, complex numbers) is a commutative ring with the usual addition

and multiplication.

Example 2.10 The additive group of residue classes modulo an integerm is a ring, the multiplication being given by rs = rs, where r, sZ/(m).

Example 2.11 Let G be any abelian group. The set Hom(G, G) ofall homomorphisms of G into itself is a ring with respect to the ringoperations given by (f+g)(x) = f(x) +g(x), (f g)(x) = f(g(x)) forf, g Hom(G, G), x G.

Note that hom(G, G) is, in general, not commutative.LetA be a ring. AsubringB ofA is a sub-group of (A, +) such that

1 B and, for x, y B,xy B. We observe thatB is a ring under theinduced operations. The intersection of any family of subrings of A isagain a subring. Let Sbe a subset ofA. The intersection of the familyof all subrings ofAcontaining S is called the subring generated byS.

Unless otherwise stated, all the rings we shall consider hereafter willbe assumed commutative.

Let a, b A. We say that a divides b (or that a is a divisor ofb, notation a|b) if there exists c A such that b = ac. If a does notdivide b, we write a b. An element a

A is said to be a zero divisor

if there exists x A, x= 0 such that ax = 0. A is said to be anintegral domain ifA = {0}and it has no zero-divisors other than 0. Anelementa Ais said to be a unit inA if there exists a1 Asuch thataa1 = a1a = 1. Let A be an integral domain. A non-zero elementa A is called irreducible if it is not a unit and a = bc with b, c Aimplies that either b or c is a unit.


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2.1. Rings and homomorphisms 17

A commutative ringA is said to be a field ifA =A

{0

}is a group

under multiplication, so that every non-zero element is a unit. Clearly,a field contains at least two elements. A subring R of a field Kis calleda subfield ofK if the ring R is a field. Any intersection of subfields ofK is again a subfield. IfSis a subset ofK, then the intersection of allsubfields ofKcontaining S is called the subfield generated byS.

Example 2.12 Z is an integral domain.

Example 2.13 Z/(m) is a field if and only if m is a prime. For, ifm= rs, with|r|, |s| = 1, then rs = rs = 0, but r, s= 0. Hence Z/(m)is not an integral domain and a fortiori is not a field. Letm be prime

and r Z/(m) with r = 0. then (r, m) = 1, i.e. there exist s, t Zwithsr+ tm= 1. Obviously sr= 1. Hence Z/(m) is a field.

Example 2.14 Q, R, C are fields.

Let A, A be rings. A mapf: A A is called a homomorphism if(i)f(x+ y) =f(x) + f(y), (ii)f(xy) =f(x)f(y) for every x, y A, and(iii) f(1) = 1.

For any ring A, the identity map is a homomorphism. LetA, B,Cbe rings and let f: A B and g: B C be homomorphisms. Theng f: A Cis a homomorphism.

A homomorphism f: A A

is said to be an isomorphism if thereexists a homomorphismg: A Asuch thatg f=IA, f g= IA ; theringsA and A are then said to be isomorphicand we write A A. Anisomorphismf: A A is called an automorphism.

Remark 2.15 A homomorphism is an isomorphism if and only if it isone-one and onto.

Remark 2.16 Let f: A A be a ring homomorphism. IfB is a sub-ring ofA, then f(B) is a subring ofA.

Example 2.17 LetA be a ring and B a subring ofA. Then the inclu-sion i: B A is a one-one homomorphism.

Example 2.18 The natural map q: Z Z/(m) is a homomorphism.

Example 2.19 The map f: C Cgivenf(z) = z is an automorphismofC.


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Let A be an integral domain. Let A denote the set of non-zero

elements ofA. On the set A A we define the relation (a, b)(c, d)ifad= bc. Since A is an integral domain, it can be verified that this isan equivalence relation. We make the quotient set K = (A A)/a ring by defining the ring operations as follows a/b+ c/d = (ad+bc)/bd, (a/b)(c/d) = ac/bd, where a/b denotes the equivalence classcontaining (a, b). It is easily verified that these operations are well de-fined and theKis a ring. In fact, Kis a field, the inverse ofa/b, a = 0,beingb/a. K is called the fraction field ofA. The map i: A Kgivenby i(a) = a/1 is a 1-1 homomorphism. We shall identify A with thesubring i(A) ofK. Iff: A L is a one-one homomorphism ofA into afieldL, then fcan be extended in a unique manner to a one-one homo-

morphism f ofK into L, by defining f(a/b) =f(a)f(b)1 for b = 0. IfA is a subring ofL, the subfield generated by A is the quotient field ofA.

Example 2.20 Q is the quotient field ofZ.

2.2 Ideals and quotient rings.

Definition 2.21 Let A be a commutative ring. An ideal I of A is asubgroup of (A, +) such that for x I, a A, we have ax I.

For any ringA, the intersection of any of family if ideals ofAis againin ideal. Let Sbe a subset of A. The intersection of the family of allideals containing S is called the ideal generated by S. It is easily seenthat ifSis not empty, then this ideal consists precisely of all finite sumsof the form

ixi, i A, xi S. For a A, the ideal generated

by{a}, namely{xa| x A} is called the principal ideal generated bya and is denoted by{a}. An integral domain for which every ideal isprincipal is called a principal ideal domain.

Example 2.22{0} and A are ideals ofA.

Example 2.23 Z is a principal ideal domain, the ideals of Z beingprecisely subgroups ofZ.

Example 2.24 Let f: A A be a homomorphism, then ker f = {x A | f(x) = 0}is an ideal.


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2.2. Ideals and quotient rings. 19

Proposition 2.25 A commutative ringA is a field if and only if1

= 0

and it has no ideals other thanA and(0).

Proof: Let A be a field. Clearly 1= 0. Let Ibe a non-zero ideal ofA. Then there exists a I, a= 0. We have 1 = a1a I and henceA= I. Conversely, assume that 1 = 0 and that the only ideals ofA are(0) and A. Then for any a A, a= 0, (a) = A. Hence there existsb Awith ba= 1, i.e. A is a field.

Let Ibe an ideal ofA. The additive group A/Iis a ring called thequotient ringwith respect to the multiplication (x + I)(y + I) =xy + I,for x, y A. Since I is an ideal this multiplication is well-defined. Thenatural map q: A A/Iis an onto homomorphism with kernel I.

Example 2.26 The ring Z/(m) of residue classes modulo m is thequotient ring ofZby the principal ideal (m).

Let f: A A be an onto homomorphism. Let Ibe the kernel off.The homomorphismfinduces a homomorphism f: A/I A, by settingf(x + I) =f(x) for x A it is easy to verity that f is an isomorphismof rings and that f q= f where q is the natural map A A/I. Thisis called the fundamental theorem of homomorphisms for rings.

Remark 2.27 LetKbe a field. Consider the homomorphismf: Z Kgiven byf(n) =n1(= 1 +

+ 1, ntimes). By the fundamental theorem

of homomorphisms, we have Z/ ker f f(Z). We know that ker f= (p)for some p 0. We call p the characteristic of the field K. Clearly

pa = 0 for every a K. Ifp= 0 then p is a prime. Obviously p= 1and ifp = rs, r >0, s > 0, then f(p) =f(r)f(s) = 0. Since K is a field,either f(r) = 0 orf(s) = 0 i.e. p|r or p|s. Since r 1, s 1, it followsthat either r = 1 or s = 1. Thus Z/(p) is a field and K contains asubfield isomorphic to Z/(p). Ifp = 0, then the one-one homomorphismf: Z Kcan be extended to an isomorphism of Q onto a subfield ofK. Hence K contains a subfield isomorphic to Q. Thus we have thefollowing

Proposition 2.28 Every field contains a subfield isomorphic either tothe field of rational numbers or to the field of residue classes of integersmodulo a primep.

The fields Q, Z/(p) where p is a prime are called prime fields.

Ifk is a field of characteristic p >0, then we have,


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(a + b)p =

pi=0

pi

aibpi =ap + bp for a, b K.

The mapping x xp is a one-one homomorphism ofK into K.Let K be a finite field (i.e. K has a finite number of elements);

then the characteristic p ofK is > 0. The mapping xxp is then anautomorphism ofK.

2.3 Polynomial rings.

Let A be a ring. Consider the setR of sequences f = (a0, a1, . . . , an),where, an Aand an= 0 for all but a finite number ofn. Let f , g Rwhere f = (a0, . . . , an, . . .) and g = (b0, . . . , bn, . . .). We make R into aring by defining the ring operations as follows:

f+ g= (a0+ b0, . . . , an+ bn, . . .), f g= (c0, . . . , cn, . . .), cn=i+j=n

aibj ,

The unit element ofR is (1, 0, 0,...). The mappingf: A Rdefinedbyf(a) = (a, 0, . . .) is clearly a one-one homomorphism and we identifyA with the subring f(A) ofR. If we write X = (0, 1, 0, . . .), then, fori > 0, Xi = (d0, d1, . . .) where di = 1 and dj = 0 for j

= i. It is

easy to see that every f R can be written uniquely as a finite sumaiX

i, ai A. The ring R is denoted by A[X], and is called thepolynomial ring in one variableoverA. The elements ofA[X] are calledpolynomials.

Let A, B be rings and : A B a ring homomorphism. Then admits a unique extension to a ring homomorphism : A[X] B[X],with (X) =X. We have only to set (

aiX

i) =

(ai)Xi.

Let B be a ring and A a subring ofB. For any B we define amap : A[X]B by setting ( aiXi) = aii. It is easily verifiedthat is a ring homomorphism. We write (A[X]) = A[] and forf(X) = aiXi

A[X], we write (f) =f()(= aii). We say that is a root off iff() = 0.

Let f =

aiXi A[X], f= 0. Then we define the degree of f

(notation : deg f) to be the largest integer n such that an= 0. Wecall an the leading coefficient off. Ifan = 1, f is said to be a monicpolynomial. We have deg(f+g) max(deg f, deg g) if f= 0, g= 0and f+ g= 0. A polynomial of degree 1 is called a linear polynomial.


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2.3. Polynomial rings. 21

Remark 2.29 If A is an integral domain and f, g

A[X], with f

=

0, g= 0,then f g= 0, i.e. A[X] is an integral domain and we havedeg(f g) = deg f+ deg g.

Remark 2.30 If A is an integral domain, the the units of A[X] areprecisely the units ofA.

Proposition 2.31 (Euclidean Algorithm.) Let A be a commutativering. Let f, g A[X], and let g be a monic polynomial. Then thereexist q, r A[X] such that f = qg + r, where r = 0 or deg r < deg g.Furtherq andr are unique.

Proof: Suppose deg f = n, deg g = m. If n < m choose q = 0and r = f. If n m, we prove the lemma by induction on n. As-sume the lemma to be true for polynomials f of degree < n. We havedeg(f anXnmg) < n where an is the leading coefficient off and soby induction hypothesis,

f anXnmg= q1g+ r1with r1 = 0 or deg r1 < m. Hencef =qg+r where q=anX

nm +q1and r = r1.

We next prove the uniqueness. Suppose, further, thatf =qg+r,where r = 0 or deg r < m. Then (q

q)g = r

r. If r

= r then

q= q and since g is a monic polynomial deg(r r) m. This is acontradiction. Hence r =r. Since g is monic, it follows that q =q.

Remark 2.32 The above proposition is valid, more generally, if theleading coefficient ofg is a unit. In particular, ifA is a field, it is enoughto assume that g= 0.

Corollary 2.33 Let f A[X]. ThenA is a root off if and onlyifX dividesf.Proof: By the above lemma we have f =q(X ) +a with aA.Hence f() =q()0 + a, i.e. f() = 0 if and only ifX

divides f.

Let f A[X], f= 0. We say that A is a simple root of f if(X )|fand (X )2 f. If (X )r|f with r >1, we say that isa multiple root off.

Proposition 2.34 LetA be an integral domain and letfbe an elementofA[X] of degreen. Thenfhas at mostn roots.


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Proof: If

Ais a root off, we havef= (X

)g whereg

A[X]

and deg g = n 1. If A, = is a root off, we have 0 =f() =( )g(). Since A is an integral domain, we have g() = 0. Now theproposition follows by induction on n.

Proposition 2.35 Let K be a field. Then K[X] is a principal idealdomain.

Proof: Let I be a non-zero ideal of K[X]. Let g I be a non-zero polynomial of smallest degree in I. For any f I, f= 0, byProposition 2.31, we have f = qg+ r, where r = 0 or deg r < deg g.Since r = f qg I, we have r = 0. Hence I= (g).

Proposition 2.36 LetKbe a field. Any non-constant polynomialfK[X] can be expressed as a product f = c

mi=1pi, where c K and

pi are monic irreducible polynomials. Further, this expression is uniqueexcept for the order of factors.

Proof: We prove the existence by induction on n = deg f. Ifn = 0,f K and there is nothing to prove. Further if f is irreducible,then f = c(c1f), where c K is so chosen that c1f is a monicirreducible polynomial. If f is not irreducible, then f = gh wheredeg h= 0, deg g= 0. Since deg g < n, deg h < n, by induction hy-pothesis g = ar

i=1pi, h = bq

j=1qj

, where a, b

K and pi, qj

aremonic irreducible polynomials. Hence f=ab

ri=1pi

sj=1 qj.

To prove the uniqueness we need the following

Lemma 2.37 Letpbe an irreducible polynomial inK[X]. ThenK[X]/(p)is a field. In particular, ifp|gh, whereg, h K[X], thenp|g orp|h.Proof: Let g K[X]/(p) with g= 0. Let g K[X] represent g.Then g (p). Consider the idealIgenerated by p and g. Since K[X]is a principal ideal domain, we have I = (t), for some tK[X]. Since

p I, p= wt, w K[X]. We assert thatw is not a unit. For otherwiseI = (p) and hence g

(p) and this contradicts our hypothesis. Sincep

is irreducible, t is a unit. HenceI = K[X] and 1 = up+vg , for someu, v K[X]. Thus vg = 1 where v K[X]/(p) is the class ofv. Thisproves the lemma.

Assume that f = cr

i=1pi = csj=1p

j with c, c

K and pi, pjmonic irreducible polynomials in K[X]. It is evident thatc = c. Hencewe have

ri=1pi=

sj=1p

j. We shall prove the uniqueness by induction


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2.3. Polynomial rings. 23

on r. Since pr divides the productsj=1p

j , by the above lemma pr

divides one of the factors of

sj=1pj . We may assume pr|ps. Sincepr, psare monic irreducible polynomials we have pr = p

s. Hence

r1i=1pi =s1

j=1pj. By induction hypothesis, the uniqueness follows.

Proposition 2.38 (Gauss.) Any non-constant irreducible polynomialinZ[X] is also an irreducible polynomial inQ[X].

Proof: Letfbe a non-constant irreducible polynomial in Z[X]. Thenthe g.c.d. of the coefficients of f is 1. If possible let f = gh whereg, h

Q[X] and deg g >0, deg h >0. Then we have df = g h, where

d Z, d >0 and g , h Z[X], deg g >0, deg h >0. Letd1 (resp.d2)be the g.c.d. of the coefficients ofg (resp. h). Since the g.c.d. of thecoefficients of f is 1, it follows that d1d2|d. Hence we may assumewithout loss of generality that the g.c.d. of the coefficients ofg (resp.h) is 1. Let p be a prime factor ofd. Let : Z[X] Z/(p)[X] be thering homomorphism given by(

aiX

i) =

aiXi, where ai is the class

of ai. We have 0 = (df) = (g)(h). Since Z/(p)[X] is an integral

domain, either (g) = 0 or(h) = 0, i.e. p divides all the coefficientsof either g or ha contradiction. Henced= 1, i.e. f =g h. Since fis irreducible in Z[X], either g or h is a unit. This is a contradiction.

Proposition 2.39 (Eisensteins irreducibility criterion). Letf =a0+a1X+ +anXn Z[X]. Suppose ai 0(mod p) for i < n, an0(mod p) and a0 0(mod p2);(here p is a prime). Then f is irre-ducible inQ[X].

Proof: We may assume that the g.c.d. of the coefficients of f is1. By virtue of the above proposition it is enough to prove that f isirreducible in Z[X]. Let if possible f = gh, where g, h Z[X] anddeg g > 0, deg h > 0. Let : Z[X] Z/(p)[X] be the ring homomor-phism given by ( biXi) = biXi, where bi is the residue class ofbi.Putting (u) = u for u Z[X] we have f = gh. Since f = anXn, itfollows by uniqueness of factorization in Z/(p)[X] that g = bXl, h =cXnl. Since an 0(mod p),we have l = deg g = deg g > 0 andnl = deg h = deg h > 0. Hence the constant terms of g and hare divisible by p which implies a0 0(mod p2). This contradicts theassumption on a0.


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2.4 Vector spaces

Definition 2.40 A vector space over a field Kis a triple (V, +, ) where(1) (V, +) is an abelian group,(2) : K V V (we write (, x) =x) is such that

(a) (x + y) =x + y,

(b) ( + )x= x + x,

(c) (x) = ()x,

(d) 1x= x,

where , K, x,y V.

Remark 2.41 The elements of K are called scalars and the elementsofVare called vectors, is called the scalar multiplication.

Remark 2.42 x= 0 if and only if = 0 or x= 0. For (a) (resp.(b))implies 0 = 0(resp. 0x= 0). On the other hand, ifx= 0 and = 0,we have 0 =1(x) = (1)x= 1x= x.

Example 2.43 Let K be a field and k a subfield of K. Then K is avector space over k if we set (, x) =x for

k, x

K.

Example 2.44 For any field K, the set Kn of all ordered n-tuples(1, . . . , n), i K, is a vector space over K if we set

(1, . . . , n) + (1, . . . , n) = (1+ 1, . . . , n+ n)

and(1, . . . , n) = (1, . . . , n).

Example 2.45 For any field K, the ring K[X] becomes a vector spaceover K if we set (, f) =f, K, f K[X].

A subset W of V is called a subspace of V if W is a subgroup of(V, +) and xW for K, xW. Then W is a vector space overKunder the induced operations. We say that a subspace W is apropersubspace ofV ifW=V.

Example 2.46 (0) and Vare subspaces ofV.


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2.4. Vector spaces 25

Example 2.47 IfKis a field, any ideal in K[X] is a subspace ofK[X].

Example 2.48 The intersection of a family of subspaces ofV is a sub-space of V. IfS is a subset ofV, the intersection W of the family ofall subspaces containing S is called the subspace generated by S andS is called a set of generators of the subspace W. It is easy to seethat if S is not empty, W consists precisely of elements of the formn

i=1 ixi, i K, xi S, n >0; ifS= , then W = {0}.Let V, W be vector spaces over a field K. A map f: V W is

called aK-linear map iffis a homomorphism of (V, +) into (W, +) andf(x) =f(x) for K, x V. A K-linear map f: V W is called

an isomorphism if there exists a K-linear map g: W V such thatg f =IV, f g= IW. It is easy to see that aK-linear mapf: V Wis an isomorphism if and only iffis one-one and onto.

Example 2.49 The map pi: Kn K defined by pi(x1, . . . , xn) =xi is

a K-linear map. The maps pi are called projections.

Example 2.50 The map zz is an R-linear isomorphism of the vec-tor space Conto C.

LetVbe a vector space and Wa subspace ofV. The additive groupV /Wbecomes a vector space if we set x = x for x V/W, K.The vector space V /W is called the quotient space of V by W. Thenatural map q: V V /W is a K-linear map.

Remark 2.51 It is easy to see that iff: V W is a linear map, thenker fis a subspace ofVand that iffis onto, it induces a linear isomor-phism f ofV / ker f onto W.

Let xi, 1 i n be elements ofV. We say that they are linearlyindependent if

1in ixi = 0, i K implies i = 0 for every i, 1

i n.A subset S of V is said to be linearly independent if every finite

subset of elements of S is linearly independent. A set consisting of a

single non-zero element ofVis linearly independent. Note that a subsetof a linearly independent set is linearly independent. S is said to belinearly dependentif it is not linearly independent.

Definition 2.52 A subset SofV is called a base (or a K-base) ofV ifSis linearly independent and generates V.


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Clearly S is a base ofV if and only if every element v

V can be

uniquely written as finite sum v =

isi, i K, si S.

Example 2.53 The set{1, x , x2, . . .} is a base for the vector spaceK[X].

Example 2.54 For any field K, the elements ei = (i1, . . . , in), 1i n, form a base for the vector space Kn, whereij = 0 for i =j , andij = 1 for i = j .

Proposition 2.55 LetVbe a vector space. Letxi, 1 i m, generate

V. If S is any linearly independent set of V, then S contains at mostm elements.

Proof: We shall prove the proposition by induction onm. Ifm= 0,then V = 0 and S =. Let m > 0 and let y1, . . . , yn be finitely manyelements ofS. LetV be the subspace ofV generated by x2, . . . , xm. Ifyi V, for 1 i n, then by the induction hypothesis, n m 1.Otherwise yi V for some i, say for i = 1. Then y1=

ixi, 1= 0,

so that x1 =1y1+

2im ixi, i K. yi iy1 V . Clearly theelements yi iy1, 2 i n, are linearly independent and hence bythe induction hypothesis, n 1 m 1. Thus Sconsists of at most melements.

Corollary 2.56 If{x1, . . . , xm} and{y1, . . . , yn} are bases of V, thenm= n.

We say that a vector space is ofdimensionn(notation : dim V =n)if there exists a base ofV consisting ofn elements. If dim V =n, thenby Corollary 2.56 every base ofVconsists ofn elements.

Corollary 2.57 Let W be a subspace of V with dim V = n. Then Whas a base consisting of at mostn elements, i.e. dim W dim V. IfWis a proper subspace ofV, thendim W


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2.4. Vector spaces 27

Proposition 2.58 Let V be a vector space over an infinite field K.

ThenV is not the union of a finite number of proper subspaces.

Proof: We shall prove that givenn proper subspaces (Vi), 1 i n,there exists an x V, x ni=1 Vi. We shall prove this by induction onn. Ifn = 1, choose x V1. Assume there exists ane Vi, 1 i n1.Ife Vn, there is nothing to prove. Supposee Vn. Choose f Vn.Then e+ f Vn for every K. We claim that there exists a0 K such that e+0f Vi 1 i n. For otherwise, since K isinfinite, there exist , K, = such that e+f, e+f Vi,for some i < n. Then ( )f Vi, i.e. f Vi. Hence e Vi. This isa contradiction.


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Chapter 3

Field Extensions

3.1 Algebraic extension

LetKbe a field and k a subfield ofK. Then Kwill be called an exten-sion ofKand written K/k. Two extensions K/k, K/k are said to bek-isomorphicif there exists an isomorphismofKontoK such that|kis the identity. is then called a k-isomorphism. Let1, . . . , n K.We denote by k(1, . . . , n) (resp.k[1, . . . , n]) the subfield (resp. sub-ring) of K generated by k and 1, . . . , n. Clearly k[1, . . . , n] is asubring of k(1, . . . , n). Clearly, any k(1, . . . , n) can be writ-ten as = f(1, . . . , n)

g(1, . . . , n) where f(1, . . . , n) (resp. g(1, . . . , n)) =

i1...inai1...in

i11 nin (resp. =

j1...jn

bj1...jns11 . . .

snn ) with ai1...in,

bj1...jn k and g(1, . . . , n)= 0. ForK, the field k() is called asimple extension ofk . The map : k[X] k[] defined by (g) =g()for any g k[X] is clearly an onto homomorphism of rings.Case(i) Ker = (0), i.e. is not a root of any non-zero polynomialover k. We say in this case that is transcendental over k. Then theone-one homomorphism : k[X] k() can be extended to a one-onehomomorphism of k(X) the quotient field of k[X], onto a subfield ofk(). However, this subfield contains and k, and it must therefore

coincide with k(). Hence k() is isomorphic to k(X).Case (ii) Ker = (0), i.e. is root of a non-zero polynomial. Wesay then that is algebraic over k. Since every ideal in k[X] is aprincipal ideal (Chapter 2 Proposition 2.25), we have Ker = (f) forsome f k[X]. Clearly, f is not a constant. The polynomial f isirreducible. In fact, suppose f =ghwhere deg g and deg hare both less

29


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3.2. Splitting fields and normal extensions 31

suppose that1im1jn

sijij = 0 where sij

K. We must then have

1im siji = 0 for every j. This implies that sij = 0 for every i, jwith 1 i m, 1 j n. Thus the ij are linearly independent.

Corollary 3.4 LetK/k be any extension and 1, . . . , n K be alge-braic overk. Thenk(1, . . . , n)/k is a finite extension.

Proof: In fact, the corollary has already been proved for n = 1(Proposition 3.1). Let us assume by induction thatk(1, . . . , n1)/kisa finite extension. Thenk(1, . . . , n) =k(1, . . . , n1)(n) is a finiteextension of k(1, . . . , n1). The corollary now follows from Proposi-tion 3.3.

It follows from Corollary 3.4, that if , K are algebraic, then + , 1( = 0) and are algebraic.

Corollary 3.5 If K/k and L/Kare algebraic extensions, thenL/k isalgebraic.

Proof: Let Land let 0in aii = 0 with a0, . . . , an K, ai=0 for at least one i. Clearly is algebraic over k(a0, . . . , an). Sincek(a0, . . . , an)/k is finite by Corollary 3.4, it follows from Proposition3.3that k(, a0, . . . , an)/kis finite. Hence is algebraic over k, by Propo-sition 3.2.

Example 3.6 Any field is an extension of its prime field.

Example 3.7 C/R is an extension of degree 2.

Example 3.8 For any field k, k(X)/k is not algebraic; here k(X) isthe fraction field ofk[X]. In fact X is transcendental over k .

3.2 Splitting fields and normal extensions

Definition 3.9 Letk be a field and let f k[X]. An extension K/k iscalled a splitting field off if

(i) f(X) =c

1in(X i); i K, c k;(ii) K=k(1, . . . , n).

Proposition 3.10 Any nonconstant polynomial f k[X] has a split-ting field.


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32 Chapter 3. Field Extensions

Proof: Letf

k[X] be an irreducible polynomial. Then k[X]/(f) is

a field (see Chapter 2). The map a a of k into k[X]/(f), where ais the coset of a in k[X]/(f), is clearly a one-one homomorphism andwe identify k with its image. Thus k[X]/(f) is an extension ofk. Letq: k[X]k[X]/(f) denote the natural map and let q(X) =. Clearlyf() = 0. Let deg f=n. Since 1, , . . . , n1 form ak-base fork[X]/(f)(Proposition 3.1), we note that (k[X]/(f): k) = deg f.

We prove the proposition by induction on n = deg f. Ifn = 1, f is alinear polynomial andk is obviously a splitting field off. Let us assumen 2 and that for any polynomial g of degree n 1 over any field,there exists a finite extension in which g can be written as a productof linear factors. Let f be any polynomial of degree n. Let f1 be an

irreducible factor off. Let K1 = k() be an extension ofk such thatf1() = 0. Thenf() = 0 and therefore f= (X)g, whereg K1[X]with deg g = n 1 (corollary to Proposition 2.31) By induction, thereexists a finite extension ofK1 in which g can be written as a productof linear factors. Thus, there exists a finite extension K/k such thatf(X) = c

1in(X i); i K, c k. Then k(1, . . . , n) is a

splitting field off. This proves the proposition.Let k, k be fields and let : k k be an isomorphism. If we define

for any f =a0+ +anXn k[X], (f) =

0in (ai)Xi, we have

an isomorphism : k[X] k[X] such that |k = . We shall oftenwrite for . Iff

k[X] is irreducible, then (f) is irreducible in k [X]

and we have an induced isomorphism k[X]/(f) k[X]/((f)). Sincefor any root off, we have an isomorphism k[X]/(f) k(), in whichthe coset ofXis mapped on to , this implies that if (resp. ) is aroot off(resp. (f)), we have an isomorphism k()k() (in which is mapped onto ) and which coincides with on k.

In particular, ifk =k and = identity, then for any two roots ,

offwe have a k-isomorphismk() k() which maps onto .Definition 3.11 Let K/k be an algebraic extension. Two elements, Kare said to be conjugates overkif there exists ak-isomorphismofk() onto k() which maps onto .

Proposition 3.12 LetK/k be an algebraic extension and let, K.Then and are conjugates overk if and only if they have the sameminimal polynomial overk.

Proof: Ifand have the same minimal polynomial onerk, we haveproved above that and are conjugates. Conversely, suppose that


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3.2. Splitting fields and normal extensions 33

, are conjugates overk. Letfandg denote the minimal polynomials

of and respectively. Let : k()k() be a k-isomorphism suchthat () =. We have

0 =(f()) =f(()) =f().

Thus g|f. Since fis monic and irreducible we have f=g.

Proposition 3.13 Let k, k be fields and let : k k be an isomor-phism. Letfbe any polynomial with coefficients ink, and letK, K besplitting fields off and(f) respectively. Then there exists an isomor-phism: K K such that|k= .Proof: We proceed by induction on the degree off. If deg f= 0 thereis nothing to prove. Let deg f 1 and letf1be an irreducible factor off.Let be a root off1 and let

be any root of(f1). By what we haveseen above, can be extended to an isomorphism 1: k() k()such that 1() =

. Let f = (X )g, with g k()[X]. Then(f) = (X )1(g). The field K (resp. K) is a splitting field ofthe polynomialg (resp. 1(g) overk() (resp. k

()). By induction, 1admits an extension to an isomorphism : K K. We clearly have|k= .

In particular, taking k = k and = identity, we obtain the

Corollary 3.14 Any two splitting fields of a polynomial are isomorphic,i.e. the splitting field of a polynomial is unique up to k-isomorphism.)

In view of the above corollary, we can talk of the splitting field ofa polynomial f over k .

Let k be a field and let K, K be extensions ofk. A one-one homo-morphism: K K such that|k= identity is called ak-isomorphismofK into K.

Proposition 3.15 LetKbe the splitting field of a polynomialfover afieldk and letL/K be any extension. Then, for any k-isomorphismofK into L, we have(K) =K.

Proof: For, let1, . . . , n be the roots off in K. Since (i) is alsoa root of f and since f can have at most n roots in L, we must have(i) = j for some j. Since is one-one, it permutes the elements1, . . . , n. Hence (K) =K.


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3.3. Separable extensions 35

splitting field ofi (1

i

n). Hence there exists a k-isomorphism of

Ni (and hence Ki) onto L.

Example 3.19 Let be a root of X3 2 Q[X]. Then Q()/Q isnota normal extension.

Example 3.20 The fieldC of complex numbers is such that every non-constant polynomial with coefficients in C has a root in C (Funda-mental Theorem of Algebra), i.e. C contains a splitting field of anypolynomial in C[X]. Such fields are said to be algebraically closed.

3.3 Separable extensions

Definition 3.21 Letk be a field. An irreducible polynomial f k[X]is called separable if all its roots (in the splitting field) are simple. Oth-erwise, f is called inseparable. A non-constant polynomial f k [X] iscalled separable if all its irreducible factors are separable.

Let K/k be an algebraic extension. An element K is calledseparable overk if the minimal polynomial of over k is separable. Anelement K which is not separable will be called inseparable. Analgebraic extension K/k will be called separable if all its elements are

separable over k, otherwise it is called inseparable.If Kis separable overk, is separable over anyLwithK L

k. In fact the minimal polynomial g of over L divides the minimalpolynomial f of over k. Since f is separable, it follows that g isseparable.

Before treating separable extensions further, we introduce some re-sults on roots of polynomials.

Let f k[X] and let f = 0in aiXi. We define the derivative off, denoted by f, byf =

1in iaiX

i1. The following properties arethen easily verified; we suppose that f , g k[X] and a k.

(i) Iff k, then f

= 0,

(ii) (f+ g) =f + g,

(iii) (f g) =f g + fg,

(iv) (af) =af.


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Let

k be a root of a polynomial f. Let f= (X

)g. We then

havef = (X )g + g.

This gives g () =f().

Proposition 3.22 Let f k[X] be a non-constant polynomial with as a root. Then is a multiple root if and only iff() = 0.

Proof: Let f = (X )g. Clearly, is a multiple root off if andonly ifg () = 0. Since f() =g(), the proposition follows.

Corollary 3.23 Letf be an irreducible polynomial. Thenfhas a mul-

tiple root if and only iff = 0.

Proof: We may suppose thatf is monic. Let be a root off. Bythe above corollary is a multiple root offif and only if it is a root off. Since f is the minimal polynomial of, this is the case if and onlyiff|f. Iff = 0, we have deg f 0, is inseparable if and only if there exists g k[X]such thatf(X) =g(Xp).

Supposefis inseparable. Letf=

0in aiXi. In view of Corollary3.23, we must have

1in iaiX

i1 = 0, which implies that iai = 0 for1in. Ifk is of characteristic 0, this implies that ai = 0 for i1.Ifk is of characteristic p = 0, and ifai= 0, we have p|i.

Remark 3.25 Let k be a field of characteristic 0. Then any algebraicextension ofk is separable.

Remark 3.26 Letk be a field of characteristicp = 0 having an elementsuch that the polynomialf=Xphas no root in k. Then we assertthatXp

is an irreducible polynomial overk which is inseparable over

k. Let1, 2 be two roots of this polynomial (in a splitting field). Thenp1 =

p2 = and hence 1 =2. Thus all the roots of this polynomial

are equal, say, to . Let g be the minimal polynomial of. Ifh is anymonic irreducible factor off, we haveh() = 0 and hence g = h. Thereis thus an integeri such thatf=gi. This equation implies thatp = ni,where n = degree ofg . Since g is not linear n = 1. Hence i = 1.


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3.3. Separable extensions 37

In particular, letk(x) be the field of rational functions in one variable

x over a field of characteristic p= 0. Then Xp x is an irreducibleinseparable polynomial overk(x). For, ifXpxhas a root ink(x) thereexist g, hk[X] with x= (g/h)p,i.e. xhp =gp. But this implies that

p deg h+ 1 =p deg g, which is impossible. Thus there exist inseparablealgebraic extensions.

Lemma 3.27 Let K/k and L/K be finite extensions. Let N/k be afinite normal extension of k such that L is a subfield of N. Let m bethe number of (distinct) k-isomorphisms of K into N and let n be thenumber of (distinct) K-isomorphisms ofL into N. Then the number ofdistinctk-isomorphisms ofL into N ismn.

Proof: Let (i)1im be the distinct k-isomorphisms of K into N,and let (j)1jn be the distinct K-isomorphisms ofL into N, For 1 i m let i be an extension of i to an automorphism of N; (thisexists by Proposition 3.13). We assert that i j are distinct. Supposei j(x) = i s(x) for every xL. Therefore, for everyxK, wegeti(x) =r(x), which implies thati = r. Therefore j(x) =s(x) forevery x Lwhich implies thatj =s. Let be any k-isomorphisms ofLinto N, Clearly |K=i for some i.Then (i)1 |K= identity. Thus(i)

1 = j for some j . Hence = i j.Proposition 3.28 LetK/k be an extension of degreenandN/k a finite

normal extension such thatKis a subfield ofN. Then there are at mostn distinctk-isomorphisms ofK into N.

Proof: We prove the proposition by induction on (K: k). If (K: k) =1, there is nothing to prove. Assume that (K: k) > 1. Choose Kwith k. Then [K: k()]< (K: k). Hence, by induction, the numberof distinctk()-isomorphisms ofK intoNis at most [K: k()]. On theother hand, for any K, the number of (distinct) conjugates of isat most equal to the degree of the minimal polynomial of. Since anyk-isomorphism ofk() into N takes into a conjugate of and, givena conjugate N, there exists a unique k-isomorphism ofk() into Nwhich mapson , it follows that the number of distinctk-isomorphismofk() into N is at most [k(): k]. The proposition now follows fromthe lemma above.

Proposition 3.29 A finite extensionK/k of degreenis separable if andonly if for any finite normal extensionN/k such thatK is a subfield ofN, there aren distinctk-isomorphisms ofK into N.


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Proof: LetK/k be separable. LetKbe a subfield of any finite normal

extension N ofk. We prove the assertion by induction on n. Ifn = 1,there is nothing to prove. Let n > 1. Choose K, k. Then[K: k()] < n and since K/k()is separable, the number of distinctk()-isomorphisms of K into N is precisely [K: k()]. On the otherhand, since is separable, all the roots of its minimal polynomial aresimple and hence the number of distinct k-isomorphisms ofk() intoNis [k(): k]. The lemma above now proves the assertion.

Conversely, suppose K/k has n-distinct isomorphisms into a finitenormal extension N/k containing K as a subfield. Let K. ByProposition 3.28, the number of distinct k()-isomorphisms of K intoN is at most [K : k()]. Also, the number ofk-isomorphisms ofk()

into N is at most [k() : k]. Since n = [K : k()][k() : k], it followsthat [k() :k] = number of distinct k-isomorphisms ofk() into N, i.e.all the conjugates of are distinct. Hence is separable.

Corollary 3.30 If K/k is a finite separable extension and L/K is afinite separable extension, thenL/k is separable.

In view of the lemma, it follows that the number of distinct k-isomorphisms of L into any normal extension N/k containing L as asubfield is equal to (L: K)(K :k) = (L: k). Hence L/k is separable.

Corollary 3.31 If 1 . . . , n Kare separable elements overk, thenk(1, . . . , n)/k is separable.

3.4 Finite fields

Let F be a finite field of characteristic p= 0. We know then thatF/(Z/(p)) is a finite extension. Let (F : Z/(p)) = n. Let 1, . . . , nF be a Z/(p)-base for F. Then every element of F can be uniquelyexpressed as

1in aii; ai Z/(p). Since Z/(p) has p elements,

it follows that F has pn elements. Now F = F

{0

} is a group of

order pn 1 and hence any non-zero element of F is a root of thepolynomial Xp

n1 1. Thus any element ofFsatisfies the polynomialXp

n X Z/(p)[X]. Since F has pn elements, it follows that F is thesplitting field of the polynomial Xp

n Xover Z/(p). Since all the rootsof this polynomial are distinct, it follows that F is a separable extensionofZ/(p). In view of the uniqueness of splitting fields, it follows thatany


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3.5. Simplicity of finite separable extension 39

two finite fields with the same number of elements are isomorphic. It is

also clear that any algebraic extension of a finite field is separable.

Proposition 3.32 For any finite fieldF, F =F{0} is a cyclic group.Proof: Let be an element ofF of maximum order, say, n. Thenn = 1 for every F (Proposition 1.33, Chapter 1). Since thepolynomial Xn 1 has at most n roots, it follows that the order ofF n. However 1, , . . . , n1 F. Hence F is generated by .

Remark 3.33 LetFbe a finite field with 1 =a0, . . . , anas its elements.Then the polynomial f(X) = a0+

0in(X ai) has no root in F.

Thus a finite field is not algebraically closed.

3.5 Simplicity of finite separable extension

Theorem 3.34 Let K/k be a finite separable extension. Then thereexists an Ksuch thatK= k() (i.e. any finite separable extensionis simple).

Proof: Case (i) k is a finite field. Then K being a finite extensionof a finite field is finite. Hence K is a cyclic group by Proposition 3.32.Let be a generator. We then have K=k().Case (ii) k is an infinite field. Let (K : k) = n. Let N/k be a finitenormal extension containingKas a subfield. Since K/k is separable, itfollows by Proposition 3.29 that there exist n distinct k-isomorphisms1, . . . , n of K into N. For each i= j, let Vij ={x K| i(x) =j(x)}. ThenVij is clearly a subspace of the k-vector space K. Sinceby assumption,i=j fori =j , it follows thatVij is aproper subspaceofK. By Proposition 2.58, Chapter 2

i=jVij is a proper subset ofK.

Hence there exists an K such that i() =j() for i =j . Thus has n distinct conjugates, and we have (k() :k) =n. Thus K=k().


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42 Chapter 4. Fundamental Theorem

This is an immediate consequence of the results proved in the last

chapter. It follows from Proposition 3.15 and 3.29 of Chapter 3 thatG(K/k) is finite and order ofG = (K : k). To prove that k coincideswith the fixed field of G(K/k), we may assume that K= k. Nowif is an element of K not belonging to k, there exists an element K, = , such that and are conjugate over k, since K/k isnormal and separable (Sections 2 and 3, Chapter 3). Now k() andk() are k-isomorphic and since this isomorphism can be extended toa k-automorphism ofK (Proposition 3.13, Chapter 3), there exists anelement G(K/k) such that () = . This shows that the fixedfield ofG(K/k) is k .

The following theorem is in some sense a converse of Proposition 4.1.

Theorem 4.2 LetHbe a finite group of automorphisms of a fieldK.Then ifk is the field ofH, K/kis a Galois extension andH=G(K/k).

Let 1, . . . , n be the distinct elements ofH. Let be an elementofK, and 1, . . . , m be the distinct elements among 1(), . . . , n().Now ifis an element ofH, (1), . . . , (m) are again distinct, since isan automorphism. Further 1, . . . , n is a permutation of1, . . . , n,from which it follows that (1), . . . , (m) is a permutation of1, . . . , m.Consider the polynomial f K[X] defined by f = mi=1(X i). Wehave (f) =

mi=1(X

(i)) =

mi=1(X

i) = f, for every

H.

It follows that fis a polynomial over k . All the roots off lie in K andare distinct; therefore f is a separable polynomial over k. Further, fis irreducible over k. In fact, ifg is the minimal polynomial of overk, we have g(i()) = i(g()) = 0. Thus deg g deg f. Since g|f,we have f = g. Since f() = 0, is algebraic and separable over kand (k() : k) n = order ofH. It follows that K/k is an algebraic,separable extension.

Let N/k be a finite extension such that N is a subfield ofK. Thensince N/k is separable, N = k(), for some K (Theorem 3.34,Chapter 3.) Thus we have (N : k) n. We choose nowN such thatN/k is finite and (N : k) is maximum among all subfields of K con-

taining k and finite over k. We have N = k(). Let now be anyelement ofK. LetMbe the subfield ofKgenerated byN and . ThenM/k is finite (Corollary 3.4, Proposition 3.3, Chapter 3) and thereforeby the choice of N, we have (M: k) (N: k). But M contains N sothat (M: k) = (N: k)(M: N). (Proposition 3.3, Chapter 3). It followsthat (M : N) = 1 and thus M = N. Therefore every element of K


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43

belongs to N, i.e. K = N. We have therefore proved that K/k is a

finite separable extension. NowK=k() and since 1, . . . , n are dis-tinct k-automorphisms of K, 1(), . . . , n() are distinct. Thereforethe polynomial f =

ni=1(X i()) over k of degree n, is the min-

imal polynomial of over k and K is obviously the splitting field off. We have H G(K/k) and order ofG(K/k) = n (Proposition 3.28,Chapter 3). Therefore H = G(K/k). This concludes the proof of thetheorem.

Let K/k be a Galois extension. Let S(K/k) denote the set ofsubfields of K containing k and S(G) denote the set of subgroups ofG= G(K/k). We define mappings:

: S(K/k) S(G): S(G) S(K/k)

as follows: ifK1 is an element ofS(Kk), i.e. a subfield ofKcontainingk, K/K1 is also a Galois extension (the finitness and separability ofK/K1 are immediate; to show thatK/K1 is normal, we can use the factthatKis the splitting field of a polynomial f overk and a fortiori K isthe splitting field off overK1, so that the normality ofK/K1 follows).Now we set (K1) =G(K/K1). IfHis a subgroup ofG, we denote by(H) the fixed field ofH.

Theorem 4.3 (Fundamental theorem of Galois Theory.) The maps : S(G) S(G) and : S(K/k) S(K/k) are identity mappings.Proof: That : S(K/k) S(K/k) is the identity map, is equiv-alent to the statement that if K1/k is an extension such that K1 is asub-field ofK, thenK1 is the fixed field ofG(K/K1). This results fromProposition 4.1. That : S(G) S(G) is the identity map, is equiv-alent to the assertion that ifHis a subgroup ofG and K1 the fixed fieldofH, we have H=G(K/K1). This results from Theorem 4.2.

Corollary 4.4 LetK1/k be an extension such thatK1 is a subfield of

K. ThenK1/kis a Galois extension if and only ifG(K/K1)is a normalsubgroup ofG(K/k)and then there is a natural isomorphism ofG(K1/k)onto the quotient group G(K/k)/G(K/K1).

If is an element ofG(K/k), then (K1) is again a subfield ofK con-tainingkand G(K/(K1)) is the subgroup ofG(K/K1)

1 ofG(K/k),as is verified easily. NowK1/k is normal if and only if(K1) =K1 for


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45

Remark 4.6 In generalfneed not be onto; for example ifK1= K2, we

have K1K2 = K2. ThenG(K1K2/K2) reduces to one element and wecan chooseK1/Ksuch thatG(K1/k) consists of more than one element.

Theorem 4.7 Letk be a finite field consisting of qelements and K afinite extension of k. Then K/k is a Galois extension and the Galoisgroup G(K/k) is cyclic. If: K K is the mapping defined by(a) =aq, a K, is ak-automorphism and is a generator ofG(K/k).

Let p be the characteristic of k and Z/(p) the prime field of char-acteristic p. We have K k Z/(p). We have seen that K/Z/(p) isa Galois extension (Section 4, Chapter 3). Therefore K/k is a Galois

extension. Consider the mapping: K K defined by (a) = aq

. Ifa k, (a) =a since the multiplicative group k is of order (q 1) andaq1 = 1, ak, which gives aq =a. Ifa is the zero element ofk, wehave obviouslyaq =a. It is easily verified that is an automorphism ofK. Let m = (K : k). The groupG(K/k) is of order m and if we showthat the element is of order m, it follows that G(K/k) is cyclic andthat is a generator ofG(K/k). We know that the multiplicative groupK is cyclic (Proposition 3.32, Chapter 3); let be a generator ofK.Therefore the order of is (qm 1). We have () =q. Suppose that as an element ofG(K/k) is of order s i.e. q

s

= and s is the leastpositive integer with this property. It follows that s = m and therefore

the theorem is proved.


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46 Chapter 4. Fundamental Theorem


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Chapter 5

Applications of Galois

Theory

Notation LetKbe a field of characteristicpandma positive integer.We write [m, p] = 1 if either of the following conditions is satisfied:

(1) p= 0 and m arbitrary,

(2) p > 0 and m and p are coprime.

5.1 Cyclic extensions

LetKbe a field characteristicp and m an integer 1 such that [m, p] =1.

Consider the polynomial f = Xm 1 in K[X]. If is any rootof f, then f() = mm1 = 0 so that all the roots of f are distinct(Proposition 3.22, Chapter 3). Let1, . . . , m be the roots off. Theseare called the m-th roots of unity. They form an abelian group undermultiplication. Lett be the exponent of this abelian group. Thenti = 1for 1 i m (Proposition 1.33, Chapter 1). Since Xt 1 has only troots in its splitting field over K, we see that t = m. This means thatthe i (1im) form a cyclic group order m. Any generator of thisgroup is called aprimitivem-th root of unity. If is primitivem-th rootof unity, we have

f=Xm 1 =

0im1

(X i),

47


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48 Chapter 5. Applications of Galois Theory

and the field L = K() is clearly the splitting field of f over K. The

extension L/K is separable since all the roots off are distinct. ThusL/Kis a Galois extension.

Let G be the Galois group of L/K and G. If is a primitivemth root of unity, so is () and therefore () = where (, m) = 1and the integer is determined uniquely modulo m. LetRmdenote themultiplicative group of residue classes mod m which are prime to m.It is easily verified that the map , where is the residue class ofmod m, defines a homomorphismofGintoRm. If an element Gis such that () =, we have (i) =i, 0 i m 1, and therefore is the identity element e ofG. Thus ker = (e) i.e. G is isomorphicto a subgroup ofRm. We have therefore proved the following.

Proposition 5.1 LetL be the splitting field ofXm 1 overK. ThenL= K()where is a primitivem-th root of unity andL/K is a Galoisextension whose Galois group is isomorphic to s subgroup ofRm.

An extension F/E is called cyclic if it is a Galois extension withcyclic Galois group.

Remark 5.2 Let us assume that the integer m in Proposition 5.1 is aprime. Then Rm is cyclic (Proposition 3.32, Chapter 3). Hence G isalso cyclic, i.e. L/Kis cyclic.

Proposition 5.3 Let the field K contain all the m-th roots of unity.LetL be the splitting field of the polynomialf =Xm , K, overK. If L is a root of f, we have L = K() and L/K is a cyclicextension. Ifm is a prime, eitherL= K or(L: K) =m.

Proof: We have

f=

0im1

(X i)

whereis a primitivemth root of unity. It follows thatL = K(). Since[m, p] = 1, f is separable over Kand thus L/Kis a Galois extension.

LetG be the Galois group ofL/K. We have, for any G, () =i for some integer i, which is determined uniquely modulo m. It iseasily verified that the map i(modulo m) defines a homomorphism of G into Z/(m) and that ker = (e). Thus G is isomorphic toa subgroup of the cyclic group Z/(m) and hence G is cyclic. Ifm is aprime,Z/(m) has no subgroup other than (0) and Z/(m) so thatG= (e)


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5.1. Cyclic extensions 49

or G

Z/(m). From this it follows that either L= K or (L: K) =m

(Proposition 4.1, Chapter 4).

Proposition 5.4 Letm be a prime andKcontain all them-th roots ofunity. LetL/Kbe a cyclic extension such that(L: K) =m. Then thereexists an element Ksuch thatL is the splitting field ofXm overK.

We require the following

Lemma 5.5 Let be a primitivem-th root of unity, m being a prime.Then, ifa is an integer

0im1

ia =

0 if m |am if m | a.

Proof of the Lemma If m|a, ia = 1 for every integer i. Hence0im1

ia = m. Ifm|a, = a is again a primitive mth root ofunity since m is prime. Therefore

0im1

ia =

0im1 i = sum

of the roots of the polynomial Xm 1. Hence 0im1 ia = 0.Proof of Proposition Since L/K is separable, L = K() for some L (Theorem in Section 3.5, Chapter 3). Let f be the minimalpolynomial of over K. Thenf splits into linear factors over L since

L/K is normal; let f = (X 1) (X m), = 1. Let be agenerator of the Galois group ofL/K. We can assume, without loss ofgenerality that (i) = i+1 for 1 i m 1 and that (m) = 1.Letk L, 1 k m, be defined as follows1

k =

0im1

kii+1.

By the above lemma, we have

1km

k =

0im1i+1

1km

ki

=m1.

Furtherm=

1im i and therefore belongs to K. Since m1 is notin K, it follows that there exists a k, 1 k m 1 such that k K.

1The expression for k is called the Lagrange resolvent.


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Let = k. Now

() =

0im1

ki(i+1)

=

0im2

kii+2+ k(m1)1

= k 0im1

kii+1= k.

so that (m) = (())m =m. Since generates G, (m) =m forevery G. Thusm = K. Since (K(): K) divides m which isa prime, (K(): K) = 1 or m. Since

K, (K(): K) = m, so that

L= K(). It follows thatL is the splitting field ofXm overK, andour proposition is proved.

Corollary 5.6 Let K be of characteristic= 2 and L/K an extensionsuch that (L: K) = 2. Then there exists an element L such that2 K andL= K().

The proof is immediate.

Remark 5.7 Propositions 5.1 and 5.3 do not remain valid if we dropthe condition that Kall the mth roots of unity.

5.2 Solvability by radicals

Let K be a field of characteristic p. An extension L/K is said to bea simple radical extension if there exists an element in L such that = m K, [m, p] = 1 and L = K(). We sometimes write = 1/mand call a simple radical over K. An extension L/K is said to bea radical extension if there exist subfields Ki(1in) containingKsuch thatK1= K, Kn= L, Ki+1 Ki and Ki+1/Ki is a simple radicalextension. Any element ofL is called a radical overK.

IfM/L and L/Kare radical extensions then M/K is a radical ex-

tension. We note that a simple radical extension is finite and separableand therefore any radical extension is finite and separable. Let L/Kbea radical extension,N/L any extension andFa subfield ofNcontainingK. Then it is easily seen thatLF/Fis a radical extension, where LF isthe field generated by L and F in N. From this it follows that ifL/Kis an extension and Li (i = 1, 2) are subfields of L containing K such


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5.2. Solvability by radicals 51

thatLi/Kis radical (i= 1, 2) thenL1L2/Kis a radical extension, since

L1L2/L1 and L1/K are radical extensions. Now ifLi(1il) are afinite number of subfields ofL containing Ksuch that Li/K is radicalfor 1 i l, it is clear by induction on l that (L1L2 Ll)/K is againa radical extension.

Proposition 5.8 LetL/Kbe a radical extension. Then there exists anextensionM/L such thatM/Kis a Galois radical extension.

Proof: Clearly it is sufficient to prove that there is a Galois radicalextension M /Kand a K-isomorphism ofLonto a subfield ofM.

The proof is by induction on (L: K). If (L: K) = 1, there is nothing

to prove. Suppose that (L: K) = n > 1. Then there exists a radicalextension L1/K such that L = L1(),

m = a L1, [m, p] = 1 andL =L1. Since (L1: K)< n, by the induction hypothesis, there exists aGalois radical extensionM1/K such that L1 is a subfield ofM1. LetGbe the Galois group ofM1/K. Set

f=G

(Xm (a)).

Clearly f K[X]. Let M be the splitting field of f over M1. It isobvious that M/M1 is a radical extension an it follows that M/K is aradical extension. FurtherM/Kis a Galois extension, for ifM

1 is the

splitting field of a polynomial over K, then M is the splitting fieldover Kof the polynomial f. The inclusion mapping ofL1 intoM1 canbe extended to an isomorphism ofL = L1() into a subfield ofM (cf.Section 2 of Chapter 3). This completes the proof of the proposition.

Proposition 5.9 LetL/Kbe a Galois radical extension. Then the Ga-lois group ofL/K is solvable.

Proof: By the definition of a radical extension, there exist subfieldsKi(1 i n) of L, such that Kn = L, K1 = K and Ki+1 =Ki(i),

mii = ai

Ki, [mi, p] = 1 ( 1

i

n

1). Let m =

1in1 mi. Clearly [m, p] = 1. Let L be the splitting field over K ofa polynomial f K[X]. IfM is splitting field over L of the polynomial= (Xm 1)f, it is also the splitting field of over K. Since is sep-arable overK, it follows that M /K is a Galois extension. LetFbe thesubfield ofMgenerated byKand the roots ofXm1. LetFi(1 i n)be the subfield ofMgenerated by F and Ki. We setF0 =K. Clearly


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F1 =F, Fn =M and Fi+1 =Fi(i) (1

i

n

1). Since F contains

all the mth roots of unity, it follows that Fi+1/Fi(1 i n 1) is acyclic extension (Proposition 5.3).

We assert that the Galois group G ofM/K is solvable. Let Gi bethe subgroups of G having Fi, 1 i n, as fixed fields. We haveG = G0 G1 G2 Gn ={e}. Since Fi+1/Fi(0i n 1)are normal, it follows that Gi+1 is a normal subgroup of Gi and thatGi/Gi+1can be identified with the Galois group ofFi+1/Fi, 0 i n1(Corollary to Theorem 4.3, Chapter 4). By Proposition 5.1 G0/G1 isabelian and we have seen above that Gi/Gi+1(1 i n 1) is cyclic.Thus we have a solvable series forG (Section 3, Chapter 1) so that G issolvable.

Since the Galois group of L/K can be identified with a quotientgroup of G (Corollaryto Theorem 4.3, Chapter 4), it follows that theGalois group ofL/Kis solvable (Proposition 1.56).

Proposition 5.10 LetL/Kbe a Galois extension of degreen such thatG= G(L/K) is solvable. Suppose that [n, p] = 1. Then there exists anextensionM/L such thatM/Kis a radical extension.

Proof: We prove the proposition by induction on the order ofG. IfG ={e}, there is nothing to prove. Otherwise, there is a normal sub-group G1ofG such thatG/G1 is cyclic of prime order (and the order of

G1is then strictly less than that ofG; Proposition 1.58, Chapter 1). LetL1 be the fixed field ofG1. Then L/L1 and L1/Kare Galois extensionwith Galois groupG1 andG/G1 respectively (Theorem 4.3, Chapter 4).Letm = order ofG/G1. Then [m, p] = 1. LetNbe the splitting field ofXm 1 over L and F the subfield ofNgenerated by Kand the rootsofXm 1. Let L1F (resp. LF) be the subfield ofNgenerated by L1and F (resp. L and F).

NowL1F/Fis a Galois extension and G(L1F/F) is isomorphic to asubgroup of G(L1/K) (Proposition 4.5, Chapter 4). Since m is primeG(L1f /F) ={e} or it is a cyclic group of order m. Since F containsall the mth roots of unity, it follows that L1F/F is a simple radical

extension (Proposition 5.4.)The extension LF/L1F is Galois and G(LF/L1F) is isomorphic to

a subgroup of G(L/L1) (Proposition 4.5, Chapter 4), since LF is thesubfield ofNgenerated by L and L1F. It follows that G(LF/L1F) issolvable (Proposition 1.56, Chapter 1) and that ifr is its order [r, p] = 1.Then, by the induction hypothesis, there is an extension M/LF such


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5.3. Solvability of algebraic equations 53

that M/L1F is a radical extension. Since F/K is a radical extension,

it follows that M /K is a radical extension. This completes the proof ofthe proposition.

Let f K[X]. Thenf is said to be solvable by radicals over K ifthe splitting field off over K is a subfield of a radical extension overK. It is easily seen thatfis solvable by radicals over K if and only ifevery irreducible factor off is solvable by radicals over K.

Theorem 5.11 Let f K[X] and L be the splitting field of f overK. Suppose that [n, p] = 1 where n = (L: K). Then L/K is a Galoisextension andfis solvable by radicals if and only ifG(L/K)is solvable.

Proof: Let be any element ofL. Letg be the minimal polynomialof over K and m its degree. Since m = (K(): K), we have m|n and[m, p] = 1. Thereforeg is separable over K. Thus is separable overK and it follows that L/K is a Galois extension. Suppose now thatG(L/K) is solvable. By Proposition 5.10, it follows that there existsan extension M/L such that M/K is a radical extension. Converselylet M/L be an extension such that M/K is a radical extension. UsingProposition 5.8, we may assume thatM/Kis a Galois radical extension.By Proposition 5.9, we see that G(M/K) is solvable. SinceG(L/K) isisomorphic to a quotient group ofG(M/K), it follows that G(L/K) issolvable.

Remark 5.12 If in the above theorem, fis such that [m!, p] = 1 wherem= deg f, it follows that [n, p] = 1.

5.3 Solvability of algebraic equations

LetHbe a subgroup of the symmetric group Snon the set {x1, . . . , xn}.We say that H is transitive if given xi, xj there exists H such that(xi) =xj.

Letf K[X] and suppose that it is separable over K. LetL be thesplitting field off overK. Then L/K is a Galois extension. The group

G= G(L/K) is called the group of the polynomial f over K.Suppose now that f is irreducible. Let1, . . . , n be its roots. For

any rootioffand G, (i) is again a root off, so that(i) =jfor some j . Thus induces a permutation of the set{1, . . . , n}.

The mapofGinto the permutation groupSnas defined above,is clearly an isomorphism ofGonto a subgroup ofSn. We identifyGwith


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its image under this mapping and thus regard Gas a permutation group.

The subgroupGofSnis transitive, since given any two roots i, j thereis a K-automorphism of L such that (i) = j (Proposition 3.13,Chapter 3).

Theorem 5.13 Let the characteristicp of the fieldKbe different from2 and3 andf K[X] be of degree 4. Thenf is solvable by radicalsoverK.

Proof: We can assume that f is irreducible. If n = deg

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