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AN INTRODUCTION TO FINITE GROUP THEORYAND GALOIS THEORY

AYUSH KUMAR TEWARI

Abstract. Finite Group Theory has been of utmost importantin the field of mathematics from its very beginning.Recently manyadvances have taken in this field which intrigue people from differ-ent areas like theoretical physics,quantum chemistry,etc regardingits applications in various fields etc.In this project report I havemainly dealt with basics of Finite Group Theory mainly requiredfor the study of Galois Theory, and later on go to the applicationsof galois theory in polynomials, geometry etc.

1. Introduction To Group Theory

1.1. Group. [] A set G which is closed under a given multiplicationis called a group if the following conditions (GROUP AXIOMS) aresatisfied :

G1 The set G is non empty .

G2 If a, b, c G then (ab)c = a(bc).

G3 There exist in G a element e such that

(1) For any element in a in G, ea = a

(2) For any element in a in G there exists an element a′ in G such that a′ a =e.

A group is said to be abelian if ab = ba ∀a, b ∈ G.

1.2. Sub Groups. [] We say that H is a subgroup of G, if:-

(1) H v G.

Date: June 10, 2014.1

2 AYUSH KUMAR TEWARI

(2) ab−1 ∈ H,∀a, b ∈ G.

1.3. Cosets. [] The coset of a subgroup H in G is defined as:-

aH = { ah | h ∈ H}

this is called the left coset of H.

and

Ha = { ha | h ∈ H}

this is called the right coset of H.

1.4. Normal Group. [] We say that a subgroup N of G is normal, if:-

g−1ng ∈ N

∀ g ∈ G and n ∈ N.

or equivalently,

g−1Ng = N

∀g ∈ G.

or equivalently,

Every right coset is equal to a left coset, i.e

aH = Ha , ∀g ∈ G

or equivalently,

The product of two left cosets(true also for right cosets) is again a leftcoset, i.e.

(aH)(bH) = abH , ∀a, b ∈ G

THE SUMMER PROJECT REPORT 3

1.5. Solvable Group. [] A group is said to be solvable ,if there exists a sequence of subgroups .

G=G0 ⊃ G1 ⊃ ......... ⊃ Gn = {e}

� Gi+1 C Gi and Gi/Gi+1 is abelian(0 ≤ i ≤ n).

Such a sequence is called a solvable series .

Remark Every abelian group is solvable.

Theorem 1.1. Sn is not solvable for n > 4.

We need the following

Lemma 1.2. If a subgroup G of Sn (n > 4) contains every 3− cycleand if HCG � G/H is abelian , , thenH contains every 3− cycle.

Proof. Let q:G → G/H be the natural homomorphism .Ifσ, τ ∈ G,

q(σ−1τ−1στ) = q(σ)−1q(τ)−1q(σ)q(τ) = e, since G/H is abelian.

Therefore σ−1τ−1στ ∈ H,∀σ, τ ∈ G.Let (i, j, k) be an arbitrary 3 −cycle .

Since n > 4, we can choose σ = (i, k, l) , τ = (j, k,m) wherei, j, k, l and m are all distinct .

Then

σ−1τ−1στ = (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 the above lemma

that Gi contains every 3− cycle for every i where1 ≤ i ≤ m.


But , for i = m this is not clearly possible.

1.6. Jordan Holder Series. [] A group is said to have a JordanHolder series, if there exists an sequence of subgroups .

G=G0 ⊃ G1 ⊃ ......... ⊃ Gn = {e}

� Gi+1 C Gi and Gi/Gi+1 is simple.

Lemma 1.3. Every finite group ’G’ has a Jordan Holmer series.

Let r be the largest possible number so that there is a chain

e = G0 ( G1 ( G2 ( ...................................Gr = G

with Gi BGi+1.We can always take r = 1, and (G0, G1) = (1, G), so such an r exists

and, since G is finite, there is a largest such r.

We claim that Gi+1/Gi

is simple. If not, let H be a normal subgroup of Gi/Gi+1and let π be

the projection map π : Gi → Gi/Gi+1.Then π−1(H)

is normal in Gi+1, and Gi

is normal in H. So

e = G0 ( G1 ( G2 ( ..............Gi ( π−1(H) ( Gi+1 ( ...... ( Gr =G

is a longer chain, contradiction.

Theorem 1.4 (Jordan Holder Theorem). If G=G0BG1............................BGn

and


G=G0 BG1............................BGm

are the two Jordan Holder Series for the same finite group , G , then

=⇒ n = m,

and

∃ a permutation σ(1, 2, 3............., n) = (1, 2, 3...................,m)

� Gi/Gi+1∼= Gσ(i)/Gσ(i+1


2. Introduction To Ring and Field Theory

2.1. Ring. [] A ring is a structure consisting of a non vacuous set ’R’together with two binary operations +, .

in R and two distinguished elements 0,1 ∈ R, such that

(1) (R,+,0) is an abelian group .

(2) (R,·, 1) is a monoid, i.e it follows the closure and associative property

with respect to this operation .

(3) The distributive laws

a(b+c)= ab+ac

(b+c)a= ba+ca

hold ∀a, b and c ∈ R.

Remark S ⊂ R is called a subring of R if it forms

a ring with the induced operations from R .

Remark A ring R in which ab = ba , ∀a, b ∈ R is called commutative ring .

2.2. Integral Domain. A ring is called a domain if

R∗ → the set of all non zero elements of R ,

forms a sub-monoid of (R,·, 1).

i.e, the ring has no zero divisors.

if a.b = 0 =⇒ a = 0 or b = 0.

Remark R is called a division ring , if all non zero elements of thering R , form a subgroup of (R,·, 1).


2.3. Quotient Rings. If R is a ring, with a congruence relation ,

’ ≡′ where a+ b = (a+ b)ab = a.b,then (R,+, 0) and (R, ·, 1) form the quotient ring.

2.4. Ideal. If R is a ring,an ideal I of R is a subgroup of the additivegroup such that for any a ∈ R and any b ∈ I , ab

and ba ∈ I.

The quotient ring of R with respect to the ideal I , has the followingoperations of addition and multipliaction defined as

(1) (a+I)+(b+I) = (a+b+I)

(2) (a+I).(b+I) = ab +I

If S ⊂ R,< S >= intersection of all ideals of R containing S.

〈S〉 = Ideal generated by S.

TYPES OF IDEALS

(1) Maximal Ideal - A proper ideal I is a a maximal ideal if ∃!J, �J ⊃ I and J is a proper ideal

(2) Principle Ideal - An ideal which is generated by a single element iscalled a principal ideal .

eg. I = { ar | r ∈ R}

so I = < a >

SPECIAL ELEMENTS IN A RINGThe elements in a ring which are the divisors of 1 are called units.

If a=bε, where ε is a unit then a and b are called associate elements .


The asociates of an element a and the units of R are called improperdivisors of a.

An element is called irreducible if it is not a unit and if every divisorof a is improper.

An element p is called a prime ,if p divides a product ab , then pdivides at least one of the factors a, b .

TYPES OF DOMAINS

(1) Unique Factorisation Domian -An integral domain R is calleda UNIQUE FACTORIZATION DOMAIN (or ,briefly UFD),ifit satisfies

the following conditions:UF1. Every non unit of R is a finite product of irreducible

factors.UF2. The foregoing factorization is unique to within order

and unit factors .

(2) Principle Ideal Domain -An integral domain R is called a PRIN-CIPLE IDEAL DOMAIN (or,briefly PID) iF every ideal of itis principal .

(3) Euclidean Domain-A Euclidean domain E is an integral domainin which with every element a there is associated a definite inte-ger φ(a), provided the function φ satisfies the following conditions :

E1. If b divides a , then φ(b) ≤ φ(a)E2. For each pair of elements a,b in E , b6= 0,∃ elements q and r such that a =

bq + r and φ(r) < φ(a).

2.5. Field. A ring F is called a FIELD if the following conditions(FIELD AXIOMS) are satisfied :

F1 F has at aleast two elements .

F2 F has an identity .

F3 Every element of F different from zero has an inverse


2.6. Polynomial rings. The polynomial ring over a field F, is definedas a ring with elements as polynomials such that each polynomial ele-ment consists of coefficients taken from the ring R , and the + and· operations are defined as

if f(x) and g(x) ∈ R[x] then

and if f(x) = a0 + a1x+ a2x2.........anx

n

and if g(x) = b0 + b1x+ b2x2.........bmx

m

then the operations are defined as

f(x)+g(x) = (a0 + b0) + (a1 + b1)x+ ............................(am + bm)xm +............anx

n

f(x) ·g(x) = (a0 +a1x+a2x2.........anx

n) · (b0 + b1x+ b2x2.........bmx

m)

Proposition 2.1. Let K be a field , then K[x] is a principle 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.

Then for any f ∈ I, f = qg + r, where deg(r) < deg(g)

=⇒ f − qg ∈ I

=⇒ r ∈ I

and deg(r) < deg(g) (which is a contradiction)

therefore ∃ only one g which generates I .

So I = 〈g〉

Therefore K[x] is a principal ideal domain.

Lemma 2.2. Let ρ be an irreducible polynomial in K[x] , then K[x]/(ρ) is a field.In particular if ρ | gh , with g h ∈ K[x] , then ρ|g or ρ|h.

Proof. Let ∈ K[x]/ρ , with g 6= 0 .Let g ∈ K[x] represent g ,then g /∈ (ρ)and if g ∈ (ρ)


=⇒ g ∈ Ker(f) , where f : K[x]/(ρ)→ K[x], which =⇒ g = 0.Since K[x] is a PID , let I = (t)p ∈ I =⇒ p = wt and w 6= unit and w ∈ K[x]

since if w = unit , (t) = (p) = (I) and hence g ∈ (p) (which is a contradiction)

p is irreducible so ’t’ is a unit

if t is a unit =⇒ I = K[x]

=⇒ 1 = up+ vg, for some u and v ∈ K[x]

=⇒ v · g = 1, where v ∈ K[x]/(ρ) is the class of v .

so every element in K[x]/(ρ) has an inverse .

Therefore , K[x]/(ρ) is a field .


POLYNOMIAL EXTENSIONS OF UNIQUE FACTORISATIONDOMAINS

If D is a UFD ,and if we consider an element of the ring D[x] , f(x),such that

f(x) = a0 + a1x+ a2x2 + ............................+ anx

n

then we define the content c(f) of f(x) as (a0, a1, ........., an)(6= 0)

If d = c(f), we can write ai = da′i, 0 ≤ i ≤ n and f(x) = df1(x)

where

f1(x) = a′0 + a

′1x+ a

′2x

2 + .........................+ a′nx

n

and

=⇒ c(f1) = 1and all polynomials with c(f) = 1 are termed as primitive

polynomials .

Theorem 2.3. If D is a unique factorisation domain , then so is D[x]

We will first prove three lemmas which will be followed by the proofof the theorem,

Lemma 2.4. Let D be a UFD ,F the quotient field of D , and f(x) 6= 0 ∈F [x].Thenf(x) = γf1(x) where γ ∈ F and f1 is a primitive polynomial in D[x] .Moreover ,this factorisation is unique up to unit multipliers in D.

Proof. Let f(x) = α0 + α1x+ α2x2 + ................+n x

n where the αi ∈F and αn 6= 0.

We can write αi = aib−1

i, ai, bi ∈ D.Then if b =∏bi , bf(x) ∈

D[x] so bf(x) = cf1(x) where

f1(x) ∈ D[x] and is primitive .Then f(x) = γf1(x)whereγ = cb−1 ∈F.Now let f(x) = δ f2(x) whereδ ∈ F and f2(x) ∈ D[x] and is primitive .Then δ = de−1, d, e ∈ D .

Hence we have cb−1f1(x) = de−1f2(x) and cef1(x) = bdf2(x) .The implies that f1(x) ≈ f2(x) and ce ≈ bd , since

f1 and f2 are primitive polynomials .Then we have bd = uce ,


u is a unit in D , and de−1 = ucb−1.Hence δ = uγ as required .

Corollary 2.5. Let f(x) and g(x) be primitive polynomials in D[x] andassume these are associates in F[x] . Then they are associates in D[x].

Proof. We are given that f(x) = αg(x), α 6= 0 in F.

Then the uniqueness part of the previous lemma shows that αis a unit in D .

The key lemma for the above theorem is the following

Lemma 2.6 (Gauss’ Lemma). The product of two primitive polynomi-als is primitive .

Proof. Suppose f(x) and g(x) are primitive but h(x) = f(x)g(x) is not .

Then there exists an irreducible element (and hence a prime) p ∈ D

such that p f(x) , p | g(x) or p | h(x).

We now observe that saying that p is a prime is equivalent to saying

that D ≡ mod D/(p) is a domain .

This is immediate from the definition . Hence D[x] is a domain.

We now apply the homomorphism of D[x] onto

D[x] sending a ∈ D onto its coset a = a+ (p) and x→ x.

This gives f(x)g(x) = h(x) = 0

but f(x) 6= 0, g(x) 6= 0.

This contradicts the fact thatD[x] is a domain and hence proves the lemma .


Lemma 2.7. If f(x) ∈ D[x] has positive degree and is irreducible in D[x] ,then f(x) is irreducible in F[x]

Proof. If f(x) ∈ D[x] has positive degree and is irreducible in D[x] then f(x) is primitive .×

Suppose that f(x) is reducible in F[x] : f(x) = φ1(x)φ2(x) where φi(x) ∈F [x] and deg(φi(x)) > 0.

We have φi(x) = αifi(x) where αi ∈ F and fi(x) is

primitive in D[x] . Then f(x) =α1α2f1(x)f2(x) and f1(x)f2(x) is primitive by Gauss′ lemma .×

It follows that f(x) and f1(x) and f1(x)f2(x) differ by a unit multiplier in D.

Since degfi(x) > 0 this contradicts the irreducibility of f(x) in D[x].

Now we will give a proof of the theorem

Proof. Let f(x) ∈ D[x] be non− zero and not a unit .

Then f(x) = df1(x) where d ∈ D and f1(x) is primitive .

If degf1(x) >0 then f1(x) is not a unit and if this is not irreducible

we have f(x) = f11(x)f12(x) where degf1i(x) > 0so degf1i(x) < degf1(x)

=⇒ f1i(x) is primitive .Hence by induction we can see that

f(x) = q1(x)q2(x)......................qt(x) where each qi is irreducible in D[x] ,and if d is a not a unit since D is a

UFD so d has a unique factorisation as

d = p1p2..............................ps

So by using the factorisation of d and f1(x) ,we can get factorisations for each element of D[x]

But for proving uniqueness of such factorisations

lets assume


Case - 1 - f(x) is primitive

Then irreducible factors of f(x) have positive degree

thus f(x) = q1(x)q2(x)...................qh(x) = q′1q

′2..................q

′

k

where each qi(x) and q′j(x) are irreducible of positive degree

∴ they are irreducible in F [x] by previous lemma

and since F is a UFD =⇒ h = k and by suitably ordering ,

we can assume that qi(x) and q′j(x) are associates in

F[x] then by the Corollary 2.5 =⇒ qi(x) ≈ q′i(x)inD[x]

Case - 2 - f(x) is not primitive

Since the irreducible factors are primitive , there product is also prim-itive ( Gauss’ lemma )

Therefore any factorisation of f(x) into irreducible elements in D[x] con-tains elements from D , and their product is the content of f(x).

By modifying this product through a unit multiplier we may assumethat this remains same for two factorisations

and since D is a UFD we can pair off the irredubile factors of f(x) be-longing to D into a associate pairs .

The product of the remaining pairs is a primitive polynomial for whichwe have already stated a proof in the Case - 1.

Hence , the proof is complete .

The result is also true in the case of more than one indeterminatesi.eif D is a UFD , then so is D[x1, x2, x3..............., xn].

Assumptions* Divisor chain condition D contains no infinite chain of sequencesa1, a2, .............. such that ai+1 is a proper factor of ai


** Primeness condition Every irreducible element of D is prime .

The Ring of Gaussian IntegersZ[√−1] = {m+ ni | m,n ∈ Z}

Z[i] is a subring of C and hence an integral domain , and we define afunction for every element

of this ring , called the norm, which is defined as

if r ∈ Z[i] � r = m+ ni, where m, n ∈ N

then , norm , N(r) = m2 + n2

so evidently if we take the norm as the euclidean function , then Z[i] isan euclidean domain .

Proposition 2.8. Z[√−3] is not a UFD .

Proof. Firstly Z[√−3] = {m+ n

√−3i | m,n ∈ N}

and with the definition of the norm function it is evident that it ismultiplicative ,

and if we take the element 4 ∈ Z[√−3]

then 4 = 2·2 = (1 +√−3)(1−

√−3)

clearly 2 and (1+√−3) and (1−

√−3) are irreducibles as they are prime and

so if we show that these elements in the two factorisations are not as-sociates

with respect to any unit of the ring then our proposition is proved.

so for an element to be a unit in the ring Z[√−3], it has to satisfy

m2 + 3n2 = 1

therefore possible solutions for this equation are m = ±1 and n = 0


so the only units in this ring are { 1 , -1 }

so the associates of any element r in the ring are r and -r .

so clearly the above two elements are not associates .

Hence, Z[√−3] is not a UFD.


3. Introduction to Galois Theory

3.1. Field Extensions. Let k and K be two fields , such that k is asubfield of K ,

then K is called a field extension of k.

If x1, x2, .......................xn are the fixed elements of K,

then K contains the ring k[x1, x2, ............, xn] (the least subring of Kwhich contains k and x1, x2, ............, xn)

and the quotient field is the least subfield of K

which contains k and x1, x2, ............, xn, and is denoted by k(x1, x2, ............, xn)

and it would be referred as the field generated over k by x1, x2, ............, xn.

An extension K is called finitely generated if K = k(x1, x2, ............, xn)

For α ∈ K , the field k(α) is called a simple extension of k.

The map φ : k[x]→ k[α] defined by φ(g) = g(α) for any

g ∈ k[x] is clearly an onto homomorphism of rings .

Case - 1 Ker φ = (0)

=⇒ α is a root of some non zero polynomial over k , and α is said to be algebraic.

Case - 2 Ker φ 6= 0

=⇒ α is not a root for any polynomial over k and α is said to be transcedental.

Theorem 3.1. If x is algebraic over k, then the field k(x) coincideswith the ring k[x] .Moreover, if minimal polynomial of of x over k is of degree n , then anyelement of k(x) has a unique expression of the form c0x

n−1 + c1xn−2 +

.............+ cn−1, ci ∈ k.

Proof. let f(X) be the minimal polynomial of x over k and let h(x)/g(x)

be an element of k(x) .


g(x) 6= o, f(X) does not divide g(X).

=⇒ f(x)andg(X) are relatively prime (f(X) is irreducible )

hence , 1 = g.c.d(f(x),g(x))

=⇒ A(X)f(X) +B(X)g(X) = 1, whereA(X)andB(X) ∈ k[X]

substituting x for X , we get B(x)g(x) = 1

=⇒ that g(x) is a unit in k[x] and this applies that h(x)g(x)∈ k[x].

this is the first part of the proof .

Now let y = g(x) ∈ k(x), where g(X) ∈ k[X],

since we know that k[X] is a Euclidean Domain , therefore by thedivision algorithm

we can find y = r(x) = c0xn−1+c1x

n−2+.............+cn−1, where n is the degree of f and ci ∈k .

if ∃ r1(x) ∈ k[X], such that deg (r1(x)) ≤ n− 1, � y = r1(x),

=⇒ x is a root of the polynomial r(x)− r1(x),

and this polynomial is either of degree zero or degree < n,

therefore its the zero polynomial .

This , completes the proof .

Remark The result is true for more than one indeterminates also ,

and a slightly different variant of this proof is used for stating

the Hilbert Nullstellensatz to study affine varities in algebraic geometry.

Proposition 3.2. Let α ∈ K be algebraic over k and let n

denote the degree of its minimal polynomial .

Then the k-vector space k(α) has dimension n over k .

Proof. Let 1,α, α2, ...............αn−1 form a basis for k(α).


α cannot satisfy a polynomial of degree < n, therefore the elements

1,α, α2, ...........αn−1 are linearly independent over k .Moreover it is clear by induction that

any polynomial in k[x] can be expressed in terms of 1,α, α2, ...........αn−1.

If K is a finite extension of k , then (K : k) = n (n is called the degreeof K over k) .

Proposition 3.3. Any finite extension K/k is an algebraic extension.

Proof. For any α ∈ K, α 6= 0 there exists a non zero integer

such that 1,α, α2, ...........αn−1 are linearly independent over k

and hence there exists a1, a2, .................., an ∈ k, with ai 6= 0

for at least one i such thatn∑i=1

aiαi = 0 i.e.

α is a root of the non− zero polynomialn∑i=1

aiXi.

3.2. Splitting Field. Let k be a field and let f ∈ k[x].

An extension of K/k is called a splitting field of f if

(i) f(X) = cn∏i=1

(X − αi);αi ∈ K, c ∈ k

(ii) K = k(α1, α2, .................., αn).

Proposition 3.4. Any non-constant polynomial f ∈ k[X] has a splitting field.

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

a field . The map a → a of k into k[X]/(f) , where a

is the coset of a in k[X]/(f), is clearly a one-one homomorphism . Thusk[X]/(f) is an extension of k.

Let q:k[X]→ k[X]/(f), and q(X) = α and f(α) = 0 and let deg(f) =n


1,α, α2, ...........αn−1 form a basis for k[X]/(f) , we note that (k[X]/(f) :k) = degf.

We prove the proposition by induction on n= deg f. If n = 1,

f is a linear polynomial and k is the splitting field of f . Let us assumen ≥ 2 and

that for any polynomial g of degree n-1 over any field , ∃ a finite extension

in which g can be written as a product of linear factors . Let f be anypolynomial of degree n ,

Let f1 be an irreducible factor of f .Let K1 = k(α) be an extension of k such that f1(α) =0.

Then f(α) = 0 and therefore f = (X−α)g where g ∈ K1[X] with degree g =n− 1 , and by induction hypothesis

a finite extension K1 in which g can be expressed as product of linear factors.

Thus ∃ a finite extension K/k such that f(X) = cn∏i=1

(X−αi) ;αi ∈

K and c ∈ k .

=⇒ k(α1, α2, ...................., αn) is the splitting field of f

Hence, proved .

Result Any two splitting fields of a polynomial are isomorphic i.e. thesplitting field fo a polynomial is unique ” up to k-isomorphism ”.

Proposition 3.5. Let K be the splitting field of polynomial f over klet φ be an irreducible polynomial over k .If φ has a root in K

then φ is a product of linear factors in K .Conversely , if K/k is a finite extension

which is such that any irreducible polynomial

over k having a root in K is a product of linear factors in K , then Kis the splitting field of some polynomial over k .

Proof. Let K = k(α1, α2, ...................., αn) be the splitting field of f where

α1, α2, ....αn are the roots of f .Let β ∈ K be are root of φ


and let L be the splitting field of φ over K.

Let β′be any root ofφ in L .We have a k− isomorphism σ : k(β) ≈

k(β′)such that σ(β) = β

′.

Since the splitting field off (resp. σ(f) = f)over k(β)(resp.k(β

′) is K(β) = k(β, α1, α2, ......

.......αn) , σ can be extended to a k−isomorphism of K onto K(β′) .Since K is a splittingfield

=⇒ this k isomorphism is an automorphism of K , i.e. K =K(β

′) orβ

′ ∈ K.

Suppose that K/k is a finite extension such that any irreducible poly-nomial which has a root in K is a product of

linear factors in K˙ Let K = k(α1, α2, ...................., αn)

and let f1, f2..................fn denote

the the minimal polynomials of α1, α2, ................αnrespectively.Clearly ,K is the

splitting field ofn∏i=1

fi.

3.3. Normal extensions. A normal extension K/k is

an algebraic extension such that any irreducible polynomial

over k which has a root in K is a product of linear factors in K.

3.4. Separable Extensions. Let k be a field.

An irreducible polynomial f ∈ K[x] is called separable if all

its roots(in the splitting field) are simple i.e the multiplicity of each isroot is one .

A non-constant polynomial f ∈ K[x] is seperable

if all its irreducible factors are separable.

Let K/k be an algebraic extension.An element α ∈ K

is called separable over k if the minimal polynomial of α over k is seperable.


An algebraic extension K/k is called separable

if all its elements are separable over k.

Let f ∈ K[x] and f =∑

0≤i≤n aixi,

we define the derivative as f′and

f′=

∑0≤i≤n iaix

i−1.The following are

some of its properties ,

(i) If f ∈ K then f′= 0

(ii) (f + g)′= f

′+ g

′

(iii) (fg)′= fg

′+ f

′g

(iv) (af)′= af

′

Proposition 3.6. Let f ∈ k[x] be a non constant polynomial with α as a root .Then α is a multiple root if and only if f

′(α) = 0.

Proof. Let f = (x-α)g. Clearly , α is a multiple root iff g(α) = 0.

f′(α) = g(α), the propositionfollows.

Corollary 3.7. Let f be any irreducible polynomial. Then f has a

multiple root iff f′= 0.

Corollary 3.8. Any irreducible polynomial f over a field of character-istic 0 is separable .

An irreducible polynomial f over a field k of characteristic p > 0 , isinseparable iff

if there exists g ∈ k[x] suchthatf(x) = g(xp).

Remark Let k be a field of characteristic p 6= 0having an element α such thatthe polynomial f = xp − α has no roots in k .

Then we assert that xp − α is an irreducible polynomial over k

which is inseparable over k .Let β1, β2 be two roots of this polynomial (in a splitting field).


Then βp1 = βp2 = α and hence β1 = β2.

Thus all roots of this polynomial are the same and say equal to β.

Let g be the minimal polynomial of β.If h is any monic irreducible factor off,

we have h(β) = 0and hence g = h .There is thus an integer i such that f =gi.This equation

implies that p = ni where n = degree of g . Since g is not linear , n6= 1 . Hence i = 1.

In particular , let k(x) be the field of rational functions in one

variable x over a field of characteristic of p 6= 0.Then

xp−x is an irreducible inseparable polynomial over k(x).For if xp−x has a root

in k(x) ∃ g, h ∈ k[x] with x = (g/h)p, i.e.xhp = gp.

But this implies that p·degh+ 1 = p · g, which is impossible .

Thus there exist inseparable extensions .


4. FUNDAMENTALS OF GALOIS THEORY

If A is the set of all automorphisms between a group G ,

then we know that it also forms a group under the operation of com-position, such that

φ : A X A → A

φ(σ1, σ2) = σ1σ2 = σ1 • σ2.

We say that G is a group of automorphisms if G ⊂ A.

Let k be a subfield of K . Then we call G(K/k) is the subset of Aformed the k-automorphisms of K .

4.1. Galois Extension. An extension is called a galois extension ifits finite,normal and separable.

For a galois extension K , G(K/k) is called the galois group.

The set of elements x ∈ K, such that σ(x) = x∀ σ ∈ G,

is a subfield of K called the fixed field of G.

Proposition 4.1. Let K/k be a Galois extension . Then G(K/k)isa finite group of order(K : k) and k coincides with the fixed field ofG(K/k).

Proof. We know that any finite extension over a field is of order(K:k)by previous propositions ,

to show that it coincides with the fixed field of G(K/k), where we mayassume that K 6= k.Now if αis an element of K not belonging to k , there exists an element β ∈ K,

α 6= β, such that α and β are conjugates over k

since K/k is normal and separable .

=⇒ k(α) ≈ k(β)and since this isomorphism can be extended to an

automorphism of K , ∃ an element

σ in G(K/k) such that σ(α) = β


=⇒ that fixed field of G(K/k) is k.

Theorem 4.2. Let H be a finite group of automorphisms of a field K.Then if k is a field of H , K/k is a Galois extension and H =G(K/k).

Proof. Let σ1, σ2......σn be the elements of H .Let α ∈ K , and β1, β2, ........, βm be the

distinct elements

among σ1(α), σ2(α), ...............σn(α).

Now if σ ∈ H, σ(β1), ........σ(βm) are again distinct,

σ is an automorphism .Further σσ1, .....σσn

is a permutation of σ1, ....., σn

σ(β1), ........σ(βm) is a permutation of β1, β2, .......βm.

Now if we take f ∈ K[x] as f =∏m

i=1(x− βj) .We have

σ(f) =∏m

i=1(x− σ(βi)) =∏m

i=1(x− βi) = f ∀σ ∈ H.

=⇒ if f is a polynomial over k , all roots of f lie in K and are distinct

=⇒ f is separable over k .Further f is irreducible over k and g is the minimal polynomial

of α over k , we have g(σi(α)) = σi(g(α)) = 0.

∴ deg g ≥ degf . g|f =⇒ g = f.

f(α) = 0 and α is algebraic and separable over k , then (k(α) : k) ≤n = order of H.

=⇒ K/k is an algebraic , separable extension .

Let N/k be a finite extension such that N is a subfield of K.

then N=k(β), (since N is separable), forsome β ∈ K.

=⇒ (N : k) ≤ n.Lets choose N such that N is finite and

(N:k) is maximum among all subfields of K containing k

and finite over k .We have N = k(α).Let now θ be any element of K.


Let M be the subfield of K generated by N and θ.

Then M/k is finite and therefore by the choice

of N , we have (M:k) ≤ (N : k).But M contains N so that (M : k) =(N : k)(M : N).

=⇒ (M : N) = 1 and thus M = N.

=⇒ K = N.We have therefore proved that K/k is a finite separable extension.

Now , f =∏n

i−1(x− σi(α))over k of degree n

, is the minimal polynomial of α over k and K is obviously the splitting field of f.

We have H ⊂ G(K/k) and |G(K/k)| = n.

∴ H = G(K/k).

Let K/k be a galois extension. Let S(K/k) denote the set of subfields

of K containing k and S(G) denote the set of subgroups of G = G(K/k). We define

Φ : S(K/k)→ S(G)

Ψ : S(G)→ S(K/k)

Theorem 4.3 (Fundamental Theorem of Galois Theory). The mapsΦ ◦ Ψ : S(G) → S(G) and Ψ◦Φ : S(K/k)→ S(K/k) are identity mappings .

Proof. Ψ ◦ Φ : S(K/k)→ S(K/k) is identity mapping,

is equivalent to saying that if K1/k is an extension such that K1 is a sub−field of K1 ,

then K1 is the fixed field of G(K/K1).That Φ◦Ψ : S(G)→ S(G) is the identity map

is the equivalent to the assertion that if H is a subgroup of G andK1 the fixed field of H ,

we have H = G(K/K1).


5. AN IMPORTANT RESULT IN ALGEBRAICGEOMETRY - ACTION OF ELEMENTS OF THE

GROUP SO3(R) ON POINTS OF A UNIT SPHERE IN 3SPACE AND ENUMERATING ITS FINITE SUBGROUPS

SO3(R) = {u | uT · u = 1 and ||u|| = 1}

S2(R) = {∑

(x, y, z) | x2 + y2 + z2 = 1}

so lets consider an element A belong to some finite subgroup of SO3(R)

� Au = λufor some λ >0 and u ∈6= 0, then

〈Au,Au〉 = 〈(λu, λu〉

= (λ)2 < u, u >

= (λ)2

and

〈Au,Au〉 = 〈u, (AtA)u〉

= 〈u, u〉(AT = A−1)

= 1

=⇒ (λ)2 = 1

=⇒ λ = 1 or λ = −1.

Leonhard Euler established in the eighteenth century that every proper

rotation in 3-space which is not the identity I is rotation around anaxis,

i.e. it leaves fixed not only the origin O but every point on a certainstraight line through

O, the axis l. It is sufficient to consider the two-dimensional sphere∑of unit radius around instead of the three

-dimensional space; for every rotation carries∑

into itself and thus

is a one-to- one mapping of∑

into itself .


Every proper rotation 6= I has two fixed points on∑

which are antipodes of each other namely the points

where the axis l pierces the sphere .

Given a finite group T of proper rotations of order N, we consider the

fixed points of the N 1 operations of T which are different from I.

We call them poles.If ’p’ is a pole than

gp = p for some g ∈ the finite subgroup of SO3(R).

and we define points equivalent to a pole as the equivalence class C ,

C = { q | q = gp, g ∈ H ⊂ SO3(R)}

and we define the set of operators which fix a pole as Cp

Cp = {g | gp = p}

Lemma 5.1. The image of a pole under the action of an element ofthe group SO3(R) is again a pole.

Proof. Lets consider a pole ’p’

then ∃ some g � gp = p.

we have to show that gp is also a pole ,

=⇒ there must exist some h , � h(gp) = gp

(g−1hg)p = p

∴ such an h exists , and hence the image of every pole is a pole itself˙

Lemma 5.2. All elements in Cp are conjugate to an element inCq.

Proof. Let S ∈ Cp, such that S : p→ p

and let L be an arbitrary element of the group , such that

L: p → q Then if we define the element T = L−1SL, then


t: q → q, i.e.T ∈ Cq.

It is evident that the set of elements in Cpforms a

cyclic subgroup of the finite subgroup of SO3(R)

and order of the group Cp can be defined as µp,

called the multiplicity of the pole ,

and it can also be defined alternately as the number

of operations of the group fixing a pole p.

Corollary 5.3. Every pole has the same multiplity .

Proof. Since by the lemma every image of a pole is a pole in itself

so for any two arbitrary poles if q ∈ [p],

=⇒ [q] = [p].

Therefore , each pole has the same multiplicity .

Proposition 5.4. If N represents the number of operation in the finitesubgroup of SO3(R), and nc represents the cardinality of the equivalence class of a pole then

N = nc · µc, whereµc = µp = µ

Proof. Lets define a Group action of the finite subgroup of SO3(R)say

’G’ on the set of poles say ’X’

G x X → X

( g ,p ) → gp

Hence by the Orbit-Stabiliser Theorem we can state that

| G | / | Gp |= |Gp|, where Gp is the orbit and the Gp isthe

stabiliser of the action, but we can see that the orbit

of the action is essentially the equivalence class of the pole p

and the stabiliser is essentially the set Cp

=⇒ Gp ≈ C and Gp ≈ Cp


| G | / | Cp |= |C|

N/µp = nc

=⇒ N = ncµc(µp = µc)

After these preparations let us now consider all pairs (S, p)

consisting of an operation S 6= I of the group G and a point p left fixed by S or

what is the same of any pole p and any operation S 6= I of the group leaving p

fixed˙ There are two way to enumerate such pairs

(1) In the first case we fix the poles and count the number of operators.

In this case ,

µp − 1 are the number of operations except identity which

fix a pole , therefore number of such pairs =∑p

µp − 1

and X = ∪{Ci}, where each Ci is the equivalence class of each pole =∑c

nc(µc − 1)

(2) By fixing the elements of the group and counting the number ofpoles.

In this case as each operator 6= I,

has two fixed points associated with it , so number of such pairs =2(N-1)

so equating these two numbers gives

2(N-1) =∑c

nc(µc − 1)

On taking the equation N = ncµc into account,

division by N yields the relation


2 - 2/N =∑c

{1− 1/µc}

Case 1 -The most trivial case is the one in which the group G consistsof the identity only;

then N = 1, and there are no poles. N is at least 2 and hence the leftside of our

equation is at least 1, but less than 2. The first fact makes itimpossible for the sum to the right to consist of one term only.

Hence there are at least two classes C. But certainly not more than 3.For as each µc is at least 2,

the sum to the right would at least be 2 if it consisted of 4 or moreterms. Consequently we have either two or

three classes of equivalent poles.

Case 2 - In this case our equation gives

2/N = 1/µ1 + 1/µ2

or 2 = N/µ1 +N/µ2

But two positive integersn1 = N/µ1, n2 = N/µ2 can have the sum 2 only ifeach equals1.

: µ1 = µ2 = N n1 = n2 = 1

Hence each of the two classes of equivalent poles consists of one poleof

multiplicity N. What we find here is the cyclic group of rotationsaround a (vertical) axis of order N.

Case 3 - In this case we have

1/µ1 + 1/µ2 + 1/µ3 = 1 + 2/N

Arrange the multiplicities µ in ascending order µ1 < µ2 < µ3.

Not all three numbersµ1 µ2 andµ3 can be greater than 2 for then the left


side would give a result that is≤ 1/3 + 1/3 + 1/3 = 1, contrary to the value of the right side

Hence µ1 = 2

1/µ2 + 1/µ3 = 1/2 + 2/N

Not both numbers µ2, µ3 can be >4, for then the left sum would be ≤ 1/2 .Therefore µ2 = 2 or 3

First alternative µ1 = µ2 = 2,

N = 2µ3.

Second alternative µ1 = 2, µ2 = 3;

1/µ3 = 1/6 + 2/N

Set µ3 = n.

Case 3 -We have two classes of poles of multiplicity 2 each consistingof n poles,

and one class consisting of two poles of multiplicity n. It is easily seenthat these

conditions are fulfilled by the dihedral groupDn and by this group only

For the second alternative we have, in view ofµ3 > µ2 = 3, the following three

possibilities: µ3 = 3, N = 12;≈ A4.

µ3 = 4, N = 24;≈ S4.

µ3 = 5, N = 60,≈ S5

So these are the only finite subgroups ofSO3(R), and each group represents a 3− d structure

through its actions on the vertex of those figures

(1)A4 represents the tetrahedron and its dual , but since tetrahedron

is a self dual figure , so A4 represents only tetrahedron .

(2) S4 represents the octahedron and its dual the cube .


(3) S5 represents the icosahedron and its dual , the dodecahedron.


6. BIBLIOGRAPHY

• BASIC ALGEBRA by Nathan Jacobson

• COMMUTATIVE ALGEBRA by Pierre D Samuels andOscar Zariski

• ALGEBRA by Serge Lang

• SYMMETRY by Hermann Weyl

• LECTURES ON ICOSAHEDRON AND SOLUTION OFTHE FIFTH DEGREE by Felix Klein

• ALGEBRA - FIELDS AND GALOIS THEORY by FalkoLorenz

• GALOIS THEORY by M.Pavaman Murthy, K.G Ramanathan,C.S. Seshadri,U.Shukla, R.Sridharan

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