ALGEBRA I : FIELD THEORY AND GALOIS THEORY

J. K. VERMA

Lecture 1 : Overview

Objectives

(1) A historical sketch of main discoveries about formulas for roots of

polynomials.

(2) Problems of classical Greek geometry.

(3) Discussion about the main themes of the course.

Key words and phrases: Quadratic, cubic and quartic equations, du-

plication of cube, trisection of an angle, construction of regular polygons,

Galois group.

1. Introduction and Overview

In this course we will study fields, Galois theory of field extensions and

applications to geometry and theory of equations. We outline the main

topics that we will study.

The formula

x =−b±

√b2 − 4ac

2afor the roots of the quadratic equation ax2 + bx + c = 0 was known to

Babylonians. During the reign of King Hammurabi (1750 B. C.), Babylonian

mathematicians found methods of solving linear and quadratic equations in

one and two variables. They described algorithms to solve specific examples.

From these examples it is clear that they knew the formula for the roots of

quadratic equations.

In 1494, an Italian mathematician Franciscan Luca Pacioli published the

book Summa de Arithmetica, Geometria, Proportioni et Proportionalita con-

taining all that was known in that period in arithmetic, algebra, geometry

2

and trigonometry. Paciolo ended his book with a remark that solutions of

cubic equations seemed impossible.

Generations of mathematicians at the University of Bologna in Italy tried

to find solutions of cubic equations. This was the largest and one of the most

famous universities at the turn of sixteenth century in Europe. Scipio Del

Ferro at this univerisity solved the cubic but never published his findings.

In 1535, Niccolo Tartaglia, a mathematician from Venice proved in a

public demonstration that he could solve cubic equations. But He kept

his formula a secret. But a doctor from Milan Gerolamo Cardano obtained

these formulas from Tartaglia under an oath that he will keep them a secret.

Cardano wrote his textbook Ars Magna in 1545 which described Tartaglia’s

method and extended it to all cubic equations. It is easy to see that the

equation x3+ax2+bx+c = 0 is transformed into the equation x3+px+q = 0

by replacing x by x−a/3. Let x1, x2, x3 denote the roots of x3+ px+ q = 0,

δ = −(4p3 + 27q2) and w = −1/2− 1/2√−3. Let

y1 = x1 + w2x2 + wx3

y2 = x1 + wx2 + w2x3

Then Cardano’s formulas are

y1 =3

√

−272

+3

2

√−3δ

y2 =3

√

−272− 3

2

√−3δ

We also have x1 + x2 + x3 = 0. These three linear equations determine the

roots x1, x2, x3.

In 1536 Lodovico Ferrari entered Cardano’s house as a servant. Due to

his extraordinary mathematical abilities he became a mathematician under

Cardano’s guidance. Ferrari showed that a quartic equation can be reduced

to a cubic equation and therefore it can be solved by means of four arithmetic

operations and extraction of square and cube roots. We will derive the

formulas of Cardano and Ferrari later.

Some of the greatest mathematicians, e.g., Euler and Lagrange attempted

to find similar formula for the roots of quintic equations. Lagrange gave a

3

general method to solve equations of degree atmost four. But this method

did not work for quintic equations.

Mathematicians became skeptical about existence of such formulas for

equations of degree five and higher. Paolo Ruffini, born 1765 was a student

of Lagrange. He published several papers(1802, 1813) about insolvability of

general quintic equation. His proof was not complete. The first complete

proof was given by Neils Henrik Abel (1802-1829) in 1824. Abel also proved

that if the Galois group of the polynomial is commutative then the poly-

nomial is solvable by radicals. Commutative groups are called Abelian to

honour Abel for his deep work in many branches of mathematics.

Gauss made two fundamental contributions to the theory of equations. He

provided complete solution by means of radicals of the cyclotomic equation

xn − 1 = 0.

The roots of this equation are the complex numbers represented by the ver-

tices of a regular polygon of n sides centered at the origin. Gauss’ analysis

of the roots of cyclotomic equation led him to find a criterion of the con-

structibility of regular polygons of n sides. We will discuss this later.

The second contribution was first rigorous proof of of the Fundamental

Theorem of Algebra: Every polynomial with complex coefficients is a product

of linear factors with complex coefficients.

The most decisive results in the theory of equations were found by Evariste

Galois (1811-1832). Modern algebra began with the work of Galois. He in-

troduced the Galois group of a polynomial which connected field theory with

group theory. In 1829, Galois presented two papers to the Paris Academy of

Sciences. These were sent to Cauchy who lost them. In 1830 he sent another

paper to the Academy whose secretary was Fourier who died before he could

examine this paper. The manuscript has never been found. In 1830, Galois

published a summary of his results. The first theorem in this account is:

An equation of prime degree is solvable by radicals if and only if if two of

its roots are known then the others are rational functions of them. This

implies that a general equation of degree five cannot be solved by means of

radicals. The most decisive result is the solvability criterion: A polynomial

is solvable by radicals if and only if its Galois group is a solvable.

4

We will also study the solutions of several problems in Greek Geometry

using rudiments of field theory. In Euclidean Geometry, we carry out sev-

eral geometric constructions with a ruler (unmarked) and compass such as

bisection of line segments and angles, constructions of certain angles, tri-

angles, quadrilaterals and circles. Ancient Greeks posed the following four

problems:

(1) The Delian Problem : Construct the side of a cube of volume 2.

(2) The angle trisection problem : Divide a given angle in three

equal parts.

(3) Squaring a circle : Construct a square having same area as that

of a given circle.

(4) Constructible regular polygons : Find n for which regular poly-

gon of n sides can be constructed by ruler and compass and describe

their constructions.

The above problems remained open for almost 2200 years. The final solu-

tion employed techniques from abstract algebra and analysis. We will show

that it is impossible to construct side of a cube whose volume is 2 by ruler

and compass. The word Delian is derived from Delos which was a city in in

ancient Greece. It is said that almost a quarter of population of Delos died

of plague in 428 B.C. A delegation was sent to the oracle of Apollo at Delos

to enquire how the plague could be arrested. The oracle replied that the

cubical altar to the Sun God Apollo should be doubled. Instead of doubling

the volume the faithfuls doubled the sides of the cube thereby increasing the

volume eightfold.The second and the third problems also circulated among

Greek geometers around the same time. It is not known who solved the

Delian problem first. The angle trisection problem was solved by Gauss as a

special case of his remarkable solution of the fourth problem. Gauss, barely

19, provided a construction of the 17-sided regular polygon. He also charac-

terized n for which regular n−gons are constructible by ruler and compass.

Recall that a prime of the form 22m

+ 1 is called a Fermat prime. Gauss

proved that a regular n−gon is constructible if and only if n = 2rp1p2 . . . pg

where n ≥ 0 and p1, p2, . . . , pg are distinct Fermat primes. Gauss’s Theorem

solves the angle trisection problem. If 20o was constructible, then we can

construct a regular 18-gon. Since 32|18, we have a contradiction.

5

The values of n for which regular n−gons were known to be constructible

upto the time of Gauss were n = 2m, 2m.3, 2m.5 and 2m.15. No one was able

to construct a heptagon or a regular 17−gon.

In March, 1796 Gauss made his first mathematical discovery : construc-

tion of a 17-sided regular polygon by ruler and compass. He began noting

down his mathematical discoveries in a diary which he maintained for the

next 19 years. Gauss published “Disquisitiones Arithmaticae” in 1801

which has become a classic in mathematical literature. The last result of

this is his solution to the fourth problem. Gauss was very proud of this

discovery. He desired that a regular polygon of 17 sides be engraved on his

tombstone. This wish was not fulfilled. It was fulfilled when a monument

to Gauss was built in his birth place Braunschwig. Explicit construction of

17−sided regular polygon was given by Erchinger in 1800. In 1892 Richelot

and Schwendenwein constructed a regular 257−gon. Around 1900 Hermes

constructed a regular 65537-gon. The manuscript fills a box and it is found

in Gottingen. The construction has now been computerized. See an article

by Bishop in American Math. Monthly (1978).

Lindemann proved in 1882 that π is not a root of any polynomial with

rational coefficients. This proved the impossibility of squaring a circle.

6

Lecture 2 : Algebraic Extensions I

Objectives

(1) Main examples of fields to be studied.

(2) The minimal polynomial of an algebraic element.

(3) Simple field extensions and their degree.

Key words and phrases: Number field, function field, algebraic element,

transcendental element, irreducible polynomial of an algebraic element, al-

gebraic extension.

2. Algebraic Extensions

The main examples of fields that we consider are :

(1) Number fields: A number field F is a subfield of C. Any such field

contains the field Q of rational numbers.

(2) Finite fields : If K is a finite field, we consider ψ : Z → K,ψ(1) = 1.

Since K is finite, ker ψ 6= 0, hence it is a prime ideal of Z, say generated by

a prime number p. Hence Z/pZ := Fp is isomorphic to a subfield of K. The

finite field Fp is called the prime field of K.

(3) Function fields: Let x be an indeteminate and C(x) be the field of

rational functions, i.e. it consists of p(x)/q(x) where p(x), q(x) are poly-

nomials and q(x) 6= 0. Let f(x, y) ∈ C[x, y] be an irreducible polynomial.

Suppose f(x, y) is not a polynomial in x alone and write

f(x, y) = yn + a1(x)yn−1 + · · ·+ an(x), ai(x) ∈ C[x].

By Gauss’ lemma f(x, y) ∈ C(x)[y] is an irreducible polynomial. Thus

(f(x, y)) is a maximal ideal of C(x)[y]/(f(x, y)) is a field. K is called the

function field of the curve defined by f(x, y) = 0 in C2.

Characteristic of a field : Let R be a commutative ring with identity e.

Define the ring homomorphism f : Z → R by f(n) = ne. Then ker f = (n)

for some integer n. If n = 0, then Z is isomorphic to a subring of R. In this

case we say that R has characteristic zero. If R is a domain then Z/(n)

7

is a domain as it is isomorphic to a subring of R. Hence n is a prime number,

say p. Therefore the finite field Fp is isomorphic to a subfield of R. In this

case, we say that R has characteristic p. Thus any field F contains either

an isomorphic copy of Q or Fp.

Definition 2.1. (i) Let K be a subfield of a field F . We say F is an

extension field of K. We also say that K is a base field. We also write

this as F/K.

(ii) An element a ∈ F is called algebraic over K if there exists a nonzero

polynomial f(x) ∈ K[x] such that f(a) = 0. If every element of F is algebraic

over K then we say that F is an algebraic extension of K.

(iii) An element a ∈ F which is not algebraic over K is called a transcen-

dental element over K.

Example 2.2. It is known that the base e of the natural logarithm and π

are transcendental over Q. Since (πi)2 = −π2, πi is a root of x2−π2 ∈ R[x].

Hence πi is algebraic over R. However πi is not algebraic over Q. Thus the

property of being algebraic depends upon the base field.

Example 2.3. LetK be a finite field whose characteristic is a prime number

p. Then K has a subfield F with p elements. Since K is finite, it is a finite

dimensional F -vector space. If dimF K = n then K has pn elements. If a ∈K then the set {1, a, a2, . . . , an} is linearly dependent. Let b0, b1, . . . , bn ∈ F,not all zero, so that b0+b1a+ · · ·+bnan = 0. Hence a is a root of the nonzero

polynomial b0 + b1x+ · · ·+ bnxn. Therefore b is algebraic over F and hence

K/F is an algebraic extension.

Proposition 2.4. Let F/K be a field extension and α ∈ F be algebraic over

K. Then there exists a unique monic irreducible polynomial f(x) ∈ K[x]

such that f(α) = 0.

Proof. Define ψ : K[x]→ F by ψ(g(x)) = g(α). Since ψ is a ring homomor-

phism and α is algebraic, ker ψ = I is a nonzero ideal of K[x]. Since K[x]

is a PID and K[x]/I is isomorphic to a subfield of F , I is generated by an

irreducible polynomial h(x) ∈ K[x].. If g(α) = 0 then g(x) = h(x)h1(x) for

some polynomial h1(x) ∈ K[x]. If g is irreducible, then g = αh(x) for some

α ∈ K× = K \ {0}. If g and h are taken to be monic, then g = h. �

8

Definition 2.5. The irreducible monic polynomial in F [x] whose root is

α ∈ K is denoted by irr(α, F ) and it is called the irreducible monic poly-

nomial of α over F. The degree of irr (α, F ) is called the degree of α

and it is written as degF α.

Example 2.6. (i)√i ∈ C satisfies f(x) = x4 + 1 = 0. Show that f(x) =

irr(√i,Q). Consider the field Q(i) = smallest field containing Q and i. Then

irr (√i,Q(i)) = x2 − i.

(ii) Let p be a prime number and ζp = e2πi/p. Then xp − 1 = 0 is satisfied

by ζp. Since xp − 1 = (x − 1)(xp−1 + xp−2 + · · · + x + 1) and Φp(x) :=

xp−1 + xp−2 + · · ·+ x+ 1 is irreducible over Q, irr(ζp,Q) = Φp(x).

Simple field extensions: Let K ⊂ F be a field extension. Let α, β ∈ F

be transcendental. Define ψ : K[x] → F such that ψ(g(x)) = g(α). Then

kerψ = {0}. Thus K[x] ≃ K[α] and hence K(α) ≃ K(β) by an isomorphism

σ such that σ(α) = β and σ|K = idK . The situation is quite different for

algebraic elements.

Proposition 2.7. Let F ⊂ K be a field extension and α ∈ K be algebraic

over F and f(x) = irr (α, F ). Let n = deg f . Then

(i) F [α] = F (α) ≃ F [x]/(f(x)). (ii) dimF F (α) = n and {1, α, . . . , αn−1} is

an F - basis of F (α).

Proof. Consider the substitution homomorphism

ψ : F [x]→ F [α] such that ψ(x) = α, ψ|F = idF

Then kerψ = (f(x)) where f(x) = irr(α, F ). Hence F [x]/(f(x)) ≃ F [α].

since (f(x)) is a maximal ideal, F [α] is a field, so F [α] = F (α).

(ii) Let g(α) ∈ F [α] and g(x) = f(x)q(x) + r(x) where q, r ∈ F [x], and

deg r(x) < degf(x) or r(x) = 0. Then g(α) = r(α). Thus F [α] is an

F−vector space generated by 1, α, . . . , αn−1 where n = deg f(x). Suppose

that∑n−1

i=0aiα

i = 0. If ai are not all zero then∑n−1

i=0aix

i is a nonzero poly-

nomial of degree less than deg f(x) satisfied by α. This contradicts mini-

mality of deg f(x). Thus {1, α, α2, . . . , αn−1} is an F - vector space basis of

F [α]. Hence dimF F [α] = deg irr (α, F ).

�

9

Proposition 2.8. Let K/F be a field extension and α ∈ K be algebraic

over F. Then F (α)/F is an algebraic extension.

Proof. If β ∈ F (α) and β 6= 0 then {1, β, β2, . . . , βn} is a linearly dependent

subset of F (α) since dimF F (α) = n. Hence there exist a0, a1, . . . , an ∈ F

not all zero so that a0+a1β+· · ·+anβn = 0. Hence β is algebraic. Therefore

F (α)/F is an algebraic extension. �

Proposition 2.9. Let α, β ∈ K ⊇ F be algebraic over F . Then there exists

an F -isomorphism ψ : F (α) → F (β) such that ψ(α) = β if and only if

irr (α, F ) = irr (β, F ).

Proof. Let f(x) = irr (α, F ) and g(x) = irr (β, F ). Then ψ(f(α)) = f(β) =

0. Thus g(x)|f(x). Since g, f are monic and irreducible, g(x) = f(x).

Conversely, suppose irr (α, F ) = irr (β, F ). Then F (α) ≃ F [x]/(f(x)) ≃F (β) and the isomorphisms are F -isomorphisms. Hence F (α) and F (β) are

F -isomorphic. �

Proposition 2.10. Let F ⊆ K,K ′ be two field extensions of F . Let ψ :

K → K ′ be an F - isomorphism. Let α ∈ K be a root of f(x) ∈ F [x]. Then

ψ(α) is a root of f(x).

Proof. ψ(f(α)) = f(ψ(α)) = 0 �

Example 2.11. (i) Let f(x) = x3 − 2 ∈ Q[x]. By Eisenstein’s criterion

f(x) is irreducible over Q. The roots of f(x) are α, αw, αw2 where α is the

real cube root of 2 and w is the complex cube root of 1. Thus the fields

Q(α), Q(αw), Q(αw2) are Q−isomorphic.

(ii) Since irr (i,R) = x2 + 1, R[x]/(x2 + 1) = R(i) = C.

(iii) The polynomial f(x) = x2 + x + 1 is irreducible over F2. Hence K =

F2[x]/(f(x)) is a field which is a two dimensional F2−vector space. Hence

K is a field with four elements.

(iv) The polynomial g(x, y) = y3 − x(x+ 1)(x− 1) is irreducible in C(x)[y]

by Eisenstein’s criterion. Hence C(x)[y]/(g(x, y)) is a simple field extension

of the function field C(x).

10

Lecture 3 : Algebraic Extensions II

Objectives

(1) Degree of a field extension and its multiplicative nature.

(2) A field extension of finite degree is algebraic.

(3) Transitivity of algebraic extensions.

(4) Compositum of two fields.

Key words and phrases: Simple field extension, degree of a field exten-

sion, compositum of fields.

3. Degree of a field Extension

Definition 3.1. Let F ⊆ K be a field extension. The dimension of the

F -vector space K, denoted by [K : F ] is called the degree of the field

extension K/F.

For an algebraic element α ∈ K, dimF F (α) = deg irr(α, F ). If [K : F ] <∞,

then F ⊆ K is called a finite extension.

Proposition 3.2. A finite extension K/F is an algebraic extension.

Proof. Let [K : F ] = n and β ∈ K. Then 1, β, . . . , βn are linearly dependent

over F. Hence there exist a0, a1, . . . , an, not all zero in F such that a0+a1β+

· · ·+ anβn = 0. Let f(x) = a0 + a1x+ · · ·+ anx

n. Then β is a root of f(x).

Hence β is algebraic over F .

�

Corollary 3.3. Every irreducible polynomial over R has degree ≤ 2.

Proof. Let f(x) ∈ R[x] be irreducible and α ∈ C a root of f(x). Then

R[α] ⊆ C. If α ∈ R, deg f(x) = 1. If α /∈ R, then [R[α] : R] ≥ 2. Thus

C = R[α]. Since [C : R] = 2, deg f(x) = 2.

�

11

Example 3.4. (1) Since irr (i,R) = x2 + 1, [C : R] = 2 as C ≃ R(i).

(2) Since irr (ζp,Q) = xp−1 + xp−2 + · · ·+ x+ 1, [Q(ζp) : Q] = p− 1.

(3) Algebraic extension of a field may not be finite. Consider the chain of

fields Q ⊆ Q(21/2) ⊆ · · · ⊆ Q(21/2n

) ⊆ · · · . Their union K contains the

algebraic numbers αn = 21/2n

for all n and αn is a root of the irreducible

polynomial fn(x) = x2n−2. Hence [K : Q] ≥ 2n for all n. Thus [K : Q] =∞.

(4) Quadratic Extensions: If [K : F ] = 2 then K is called a quadratic

extension of F. Let α ∈ K \F then {1, α} is a basis of K over F . Hence α2 =

aα+b for some a, b ∈ F. Therefore f(x) = irr(α, F ) = x2−ax−b. The rootsof f(x) are (a±

√a2 + 4ab)/2 if char F 6= 2. Therefore K = F (

√a2 + 4b).

Definition 3.5. A chain of fields F1 ⊂ F2 ⊂ · · · ⊂ Fn is called a tower of

fields if Fi is a subfield of Fi+1, for all i = 1, 2, . . . , n− 1.

Proposition 3.6. If K ⊆ F ⊆ L is a tower of fields then

[L : F ][F : K] = [L : K].

Proof. If either F/K or L/F are infinite dimensional, then L/K is also

infinite dimensional. Thus we may assume that F/K and L/F are finite.

Suppose that [F : K] = m and [L : F ] = n. Let x1, x2, . . . , xn be a basis of

L over F and y1, y2, . . . , ym be a basis of F over K.

We claim that the set

B = {xjyj | i = 1, 2, . . . n, and j = 1, 2, . . . ,m}

is a vector space basis of L over K. Let z ∈ L. Thus z = f1x1 + · · ·+ fnxn,

for some f1, . . . , fn ∈ F . We write fi =∑m

j=1kijyj . Therefore

z =

n∑

l=1

xlfl =

n∑

l=1

m∑

j=1

xlkljyj .

Thus B generates L as a K- vector space. Suppose∑m

j=1

∑ni=1

aijxiyj = 0.

Thenn∑

i=1

m∑

j=1

aijyj

xi = 0.

Since x1 . . . , xn are F -linearly independent. Therefore∑n

j=1aijyj = 0 for

each i. By linear independence of y1, . . . , yn to see that all the aij = 0. �

12

Corollary 3.7. Let F ⊆ K be a finite field extension. Then deg irr(α, F )

divides [K : F ], for all α ∈ K.

Proof. Since F ⊆ F (α) ⊆ K, we have

[K : F ] = [K : F (α)][F (α) : F ].

Thus deg irr(α, F ) divides [K : F ]. �

Proposition 3.8. Let K/F be a field extension. If a1, a2, . . . , an ∈ K are

algebraic over F then F (a1, a2, . . . , an) is a finite algebraic extension of F.

Proof. Since ai is algebraic over F, it is algebraic over F (a1, a2, . . . , ai−1).

Thus [F (a1, a2, . . . , ai) : F (a1, a2, . . . , ai−1)] is finite for all i. Therefore the

field F (a1, a2, . . . , an) is a finite extension of F. Hence it is algebraic. �

Corollary 3.9. Let E/F and K/E be algebraic extensions. Then K/F is

an algebraic extension.

Proof. Let a ∈ K and let a be a root of f(x) = a0+a1x+· · ·+an−1xn−1+xn ∈

E[x]. Consider the field L = F (a0, a1, . . . , an−1). Then a is algebraic over

L. Hence L(a) is a finite extension of L. Since a0, a1, . . . , an−1 are algebraic

over F, L is a finite extension of F. Hence L(a) is a finite extension of F.

Hence a is algebraic over F. �

Corollary 3.10. Let K/F be a field extension. Then the set of elements of

K which are algebraic over F is a subfield of K.

Proof. Let a, b ∈ K be algebraic over F. Then F (a, b) is a finite extension of

F. Hence all elements of F (a, b) are algebraic over F. In particular, a± b, aband a/b if b 6= 0, are all algebraic over F. �

Compositum of fields: Let L/k be a field extensions and E/k and F/k be

intermediate field extensions. Then the smallest field containing E and F,

to be denoted by EF, is called the compositum of F and F. Suppose E =

k(a1, a2, . . . , an) and F is an extension of k. Then EF = F (a1, a2, . . . , an).

Example 3.11. Let m and n be co prime positive integers. Consider the

subfields E = Q(ζm) and F = Q(ζn) of C. Then the compositum of E and

13

F is Q(ζmn). Indeed, as m and n are coprime, there exist p, q ∈ N such that

mp+ nq = 1. Therefore

ζmn = exp(2πi/mn) = exp(2pπi/n) exp(2qπi/m) = (ζn)p(ζm)q.

We can estimate the degree of the compositum of two finite field exten-

sions in terms of their degrees.

Proposition 3.12. Let L/k be a field extension and E/k, F/k be interme-

diate finite extensions fields. Then

[EF : k] ≤ [E : k][F : k].

If [E : k] and [F : k] are coprime then equality holds.

Proof. Let x1, x2, . . . , xm and y1, y2, . . . , yn be bases of the k-vector spaces

E and F respectively. Then it is easy to see that E = k(x1, x2, . . . , xm)

and F = k(y1, y2, . . . yn). Therefore EF = k(x1, x2, . . . , xm; y1, y2, . . . yn).

We have the following diagram of field extensions:

L

EF

{{{{

{{{{

CCCC

CCCC

E

CCCC

CCCC

F

{{{{

{{{{

k

Since EF = E(y1, y2, . . . , yn) we have [EF : F ] ≤ n. Since the degree is

multiplicative in a tower of finite extensions, we have

[EF : k] = [EF : E][E : k] ≤ mn.

Since m and n both divide [EF : k], and (m,n) = 1, we get mn | [EF : k].

Hence [EF : k] = mn. �

16

Lecture 4 : Ruler and Compass Constructions I

Objectives

(1) Describe standard ruler and compass constructions.

(2) The field of constructible numbers is closed under taking square roots

of positive reals.

(3) Characterization of constructible real numbers via square root towers

of fields.

(4) The degree of a constructible real number is a power of 2.

(5) Impossibility of squaring the circle, trisection of angles and duplica-

tion of cubes by ruler and compass.

Key words and phrases: Ruler and compass constructions, constructible

real numbers, square root tower, trisection of an angle, duplication of cube,

squaring a circle.

4. Four Problems of Classical Greek Geometry

The four problems of classical Greek Geometry: duplication of cube, tri-

section of angles, squaring of circles and construction of regular polygons

can now be solved using the rudiments of algebraic extensions of fields. A

complete solution of the last problem about characterization of constructible

regular polygons will use fundamental theorem of Galois theory. This will

be discussed later. As we have remarked before, these problems remained

open for more than 2000 years. We will see that the language of field exten-

sions provides the right framework for discussion of these problems. Once

translated into this language, the solutions are obtained quickly.

First we will precisely formulate constructibility by ruler and compass and

the concept of constructible points, lines and constructible real numbers.

A real number is called constructible if it is the length of a line segment

connecting two constructible points. We will then see that a constructible

real number is algebraic over Q and its degree over Q is a power of two. This

criterion leads to solutions of the first three problems and a partial solution

of the fourth problem.

17

Constructible points, lines, circles and real numbers:

Given a finite set {P1, . . . , Pn} of points in the Cartesian plane R2, define

the set Sm inductively. Put S0 = {P1, . . . , Pn}. Suppose Sm has been de-

fined. Put Sm+1 = Sm ∪ Tm where Tm is the set of points of intersection of

lines passing through points in Sm and circles with center at one point in Sm

with radii equal to distance between points of Sm. Let S = ∪∞m=0Sm. We say

that S = C(P1, . . . , Pn) is the set of points constructible from P1, P2, . . . , Pn

by ruler and compass. A real number a is called constructible if |a| is the

distance between two constructible points. A line passing through two con-

structible points is called a constructible line. A circle is called constructible

if its center is constructible and its radius is a constructible real number.

We can reformulate the problems according to the definition of con-

structible points. Let P1 = (0, 0) and P2 = (0, 1). Is ( 3√2, 0) ∈ C(P1, P2) ?

This is the Delian problem. For the squaring of the circle problem, if there

exists a square with side a such that a2 = π then a =√π. So the problem

is asking whether (√π, 0) ∈ C(P1, P2). For the angle trisection problem,

set P3 = (cos θ, sin θ). The problem asks whether C(P1, P2, P3) contains

(cos θ/3, sin θ/3). The problem of construction of regular n-gons asks for

which values of n, (cos 2πn , sin 2π

n ) ∈ C(P1, P2).

Trisection of an angle with a marked ruler and compass

Figure 1. Trisection of an angle with a marked ruler and compass

Let ∠AOB = θ. Draw a unit circle centered at O. Suppose one end of a

ruler is E and point P is marked on the ruler such that EP = 1. Slide the

18

ruler in such a way that E is on X-axis and P is on the circle and the edge

passes through B. Then △DCO gives α + α + π − β = π. Hence β = 2α.

The △BOC gives 4α+ π − (θ + α) = π. Hence α = θ/3

Duplication of a cube with a marked ruler and compass

Figure 2. Duplication of a cube with a marked ruler

Use a ruler with one end point marked as E and a point marked as P with

EP = 1. Let AB be a segment of unit length. Draw the angles ∠BAD = 90o

and ∠BAE = 120o. We show that PB = 3√2. Let x be the length of PB

and z be the length of AP. Then x2 = z2 + 1. Since △QEB || △APB, we

get

x+ 1

a+ 1=

x

1and

a√3

z=

a+ 1

1.

Hence a = 1/x and√3/xz = x+ 1/x. We also have

√3x = x2z + xz and

hence z =√3x/x(x+1) =

√3/x+ 1. Since x2 = 3/(x+ 1)2+1, x4+2x3+

x2 = 3 + x2 + 2x + 1. Therefore x4 + 2x3 − 2x − 4 = (x + 2)(x3 − 2) = 0.

Therefore x = 3√2

Standard Constructions

(i) Bisecting a line segment: Suppose A and B are constructible points,

we show that the mid point of the line segment AB is also constructible

19

Figure 3. Bisection of a line segment

Figure 4. Bisection of an angle

Draw circles with centers A,B with radius AB. Then the intersection points

of these circles C,D are constructible. The mid point of AB is the inter-

section of CD and AB. It is the mid point since it is the intersection of

diagonals of the rhombus ACBD.

(ii)Bisection of an angle: Let A,O,B be three constructible points. They

determine the angle AOB.

20

Draw a circle with center O and radius OB. It meets OA at D. Then D is

constructible. Now bisect the segment BD at E. So E is also constructible.

Then line OE bisects ∠AOB.

(iii) Drawing a right angle:

Figure 5. Drawing a right angle

Suppose O,A are constructible points. We wish to draw a perpendicular

at O which is also a constructible line. Draw a circle C(O,OA). It meets

the extended line OA at Q. Draw circles C(Q,QA) and C(A,QA). These

intersect at B and C. The triangle QAB is isosceles. Hence ∠BOA = 90o

(iv) Dropping a perpendicular:

Suppose L is a constructible line and P a point outside this line which is

constructible. Then we can draw a perpendicular onto L from P which is

also constructible. Draw the circle C(P, r) where r is a large constructible

number so that C(P, r) meets 2 points Q and R. Draw circles at centers Q

and R of radius PQ. Join PS and take the intersection of QR and PS.

(v) Drawing a parallel line.

Suppose L is a constructible line and P is a constructible point outside L.

Drop a perpendicular PO on L. Now draw 90o on OP at P to get a parallel

line.

Algebraic properties of Constructible Real Numbers

21

Figure 6. Construction of a perpendicular onto a line

Proposition 4.1. A point P = (a, b) is constructible if and only if a and b

are constructible real numbers.

Proof. Drop a perpendicular from P to X and Y axes to get constructible

points A and B. So a and b are constructible real numbers. If a and b are

constructible then we can draw circles C(0, a) and C(0, b) to get A and B.

Now draw perpendicular at A and B to get P . �

Proposition 4.2. Constructible real numbers form a subfield of R.

Proof. It is easy to show that a ± b are constructible if a and b are so. To

show ab and a/b for b 6= 0 are constructible, use the constructions in the

figures below.

�

Proposition 4.3. If a is a positive constructible real number then so is√a

Proof. Since A = (a, 0) is constructible so is B = (a + 1, 0). Hence the

mid point C = (a+ 1/2, 0) is constructible. Draw the circle C with center

(0, a+12 ) and radius (a + 1)/2. Draw a perpendicular at A which meets the

circle at D. Since △ODA and △DBA are similar, x/a = 1/x we have

x =√a

�

22

Figure 7. Construction of√a

Figure 8. Construction of ab

Corollary 4.4. Let F ⊆ C be a subfield of the field C of constructible real

numbers. Let k > 0 ∈ F . Then F (√k) ⊆ C.

Proof. We need to show each number of F (√k) is constructible. Since an

arbitrary element of F (√k) is of the form a + b

√k where a, b ∈ C, it is

constructible since√k is constructible. �

Theorem 4.5. Let Q ⊂ F1 ⊂ F2 ⊂ · · · ⊂ Fn be a sequence of fields such

that

23

Figure 9. Construction of a/b

Fj+1 = Fj(√

bj) and 0 < bj ∈ Fj for j = 0, 1, . . . , n− 1.

Then all elements of Fn are constructible.

Proof. Apply induction on n.

�

Definition 4.6. A tower of fields as in Theorem 4.5 is called a square root

tower over Q.

Definition 4.7. Let F be a field. Then F 2 is called the plane of F . Let

a, b, c ∈ F then the set {(x, y) | ax+ by + c = 0} is called a line in F 2 and

the set {(x, y) | x2 + y2 + ax+ by + c = 0} is called a circle in F 2.

The proof of the next lemma is left as an exercise.

Lemma 4.8. Let F be a subfield of R. (i) The point of intersection, if any,

of two lines in F 2 belongs to F 2.

(ii) The points of intersection of a line and a circle or two circles in F 2 lies

in F 2 or F (√k)2 where 0 < k ∈ F.

Theorem 4.9. A real number a is constructible if and only if there exists

a square root tower Q ⊂ F1 ⊂ · · · ⊂ FN such that a ∈ FN .

Proof. We have already proved that numbers in FN are constructible. If a

is constructible then P = (a, 0) is a constructible point. We wish to show

24

(a, 0) ∈ F 2N where FN is the last field in a square root over Q. Beginning

with O = (0, 0) and I = (1, 0), the point P is constructed in finite number

of steps

S0 = {O, I} ⊂ S1 ⊂ · · ·Sm ⊂ · · ·

Let P ∈ Sm. Apply induction on m. If m = 0 then we are done. Let m > 0.

By induction Sm−1 ⊂ F 2N , where FN is the last field in a square root tower

over Q. The points in Sm are intersections of lines and circles in F 2N . Hence

they are in FN (√k)2 for more 0 < k ∈ FN . Therefore P is in the plane of

FN . �

Theorem 4.10. Suppose a is a constructible real number, then

[Q(a) : Q] = 2m

for some m ∈ N.

Proof. Let Q ⊂ F1 ⊂ · · · ⊂ FN be a square root tower over Q and a ∈ FN .

Then

[Q(a) : Q][FN : Q(a)] = [FN : Q] = 2N

Hence [Q(a) : Q] = 2m for some m. �

Corollary 4.11. It is impossible to duplicate a cube with ruler and compass.

Proof. The number α = 3√2 is a root of the irreducible polynomial x3 − 2

over Q. Hence [Q(α) : Q] = 3 which is not a power of two. Therefore α is

not constructible. �

Corollary 4.12. It is impossible to trisect an arbitrary angle θ with ruler

and compass.

Proof. Suppose the angle θ is given. We may assume P = (cos θ, sin θ)

is given along with O = (0, 0) and I = (1, 0). We wish to show that

(cos θ/3, sin θ/3) is not constructible. If so, then cos θ/3 and sin θ/3

are constructible real numbers. Using the identity cos 3θ = 4 cos3 θ− 3 cos θ

we get cosπ/3 = 4 cos3 π/9 − 3 cosπ/9. Therefore u = cosπ/9 satisfies

8u3− 6u− 1 = 0. Hence w3− 3w− 1 = 0 where w = 2u. As [Q(w) : Q] = 3,

u is not constructible. �

25

Corollary 4.13. It is impossible to square the unit circle by ruler and com-

pass.

Proof. Suppose it is possible to construct a segment a by ruler and compass

such that a2 = π. Then a =√π is algebraic over Q, hence so is π. But

π is transcendental over Q. Therefore√π is not constructible by ruler and

compass. �

26

Lecture 5 : Ruler and Compass Constructions II

Objectives

(1) Wantzel’s characterization of constructible regular p-polygons.

(2) Richmond’s construction of a regular pentagon.

(3) Gauss’ criterion of constructible regular polygons.

Key words and phrases: Fermat’s primes, constructible regular polygons,

Gauss’ criterion.

5. Constructible regular polygons

In this section we discuss constructibility by ruler and compass of regular

polygons. Gauss’ proved that a regular polygon of n sides is constructible

by ruler and compass if and only if n = 2mp1p2 . . . pr where m ∈ N and

p1, p2, . . . , pr are distinct Fermat primes. The number Fm = 22m

+ 1 is

called a Fermat prime whenever it is a prime. The known Fermat primes

are:

F0 = 3, F1 = 5, F2 = 17, F3 = 257 and F4 = 65537.

Fermat showed that Fm is a prime for m ≤ 4. Eisenstein conjectured that

there are infinitely many Fermat primes. This conjecture is still open. Euler

showed that F5 is divisible by 641.

Proposition 5.1 (1837, Wantzel). Let a regular polygon of n sides be con-

structible and p be an odd prime dividing n. Then p is a Fermat prime.

Proof. If p|n and a regular n-gon is constructible then a regular p-gon is

also constructible. Thus the point (cos 2π/p, sin 2π/p) is a constructible

point. Hence there exists a field F ⊇ Q such that [F : Q] = 2m and

cos 2π/p, sin 2π/p ∈ F . Then ζp = cos 2π/p + i sin 2π/p ∈ F (i) and hence

[Q(ζp) : Q] = 2s = p−1 and therefore p = 1+2s. It follows that s is a power

of 2. Hence p is a Fermat prime. �

27

Figure 10. Richmond’s construction of a regular pentagon

Construction of a pentagon by ruler and compass

We show that a pentagon is constructible by ruler and compass. The vertex

(cos 720, sin 720) of a regular pentagon corresponds to the complex number

z = e2πi/5 which is the root of the irreducible polynomial

Φ5(x) = x4 + x3 + x2 + x+ 1.

Therefore

z2 + z + 1 +1

z+

1

z2= 0.

Completing the square we get

(

z +1

z

)2

+

(

z +1

z

)

− 1 = 0.

Put z + 1/z = y to get y2 + y − 1 = 0 and z2 − zy + 1 = 0. Hence

y = (−1±√5)/2. The second equation gives z = (y ±

√

y2 − 4)/2. Clearly

y is constructible as it belongs to Q(√5) and we have a square root tower

Q ⊂ Q(√5) ⊂ Q(

√5,

√

y2 − 4).

Thus all roots of Φ5(x) are constructible real numbers.

Richmond’s construction of a regular pentagon (1893)

28

Draw a unit circle with center O. Draw a perpendicular OR at O. Let Q

be the mid point of OR. Join Q and P and then bisect ∠PQO. Let the

bisector meet OP at S. Construct a perpendicular at S and let T be its

intersection point with the circle. We show ∠TOP = 72o. It is enough to

show that OS = cos 72o. Note that

∠OQS = 90−θ2 ⇒ tan

(

45o − θ2

)

= OS1

2

⇒ OS = 12 tan(45

o − θ2).

Using tan θ = 12 = 2 tan θ/2

1−tan2 θ/2we get tan θ

2 =√5− 2. Therefore

OS = 12

(

tan 45o−tan θ/21−tan 45o tan(−θ/2)

)

=√5−14 = cos 72o.

Proposition 5.2. A heptagon is not constructible by ruler and compass.

Proof. Let θ = 2π/7 and ζ7 = cos θ + i sin θ. Then ζ7 is a root of the irre-

ducible polynomial

Φ7(x) = x6 + x5 + x4 + x3 + x2 + x+ 1.

Therefore [Q(ζ7) : Q] = 6. Using ζ7+ζ7 = 2 cos θ we get [Q(ζ7) : Q(cos θ)] =

2 and hence [Q(cos θ) : Q] = 3. Thus cos θ is not constructible. Therefore a

heptagon is not constructible by ruler and compass. �

Proposition 5.3. Let p be a prime number and ζp = cos 2π/p2+i sin 2π/p2.

Then

[Q(ζp) : Q] = p(p− 1).

Proof. Since ζp satisfies the equation

xp2 − 1 = (xp − 1)(xp(p−1) + xp(p−2) + · · ·+ (xp)2 + xp + 1) = 0

and ζpp 6= 1, ζp is root of f(x) = (xp)p−1 + (xp)p−2 + · · · + (xp)2 + xp + 1.

We show that f(x) ∈ Q[x] is irreducible. Put x = u+1 and use Eisenstein’s

criterion:

29

f(u+ 1) =

p∑

k=1

(u+ 1)p(p−k)

=

p∑

k=1

(up + 1 + pg(u))p−k

=

p∑

k=1

(up + 1)p−k + phk(u)

where hk(u) ∈ Z[u] has degree p2 − pk − 1. Since

∑pk=1(u

2 + 1)p−k = (up+1)p−1up = up(p−1) + pH(u),

f(u+ 1) =

p∑

k=1

(u+ 1)p(p−k) = up(p−1) + pG(u).

Since f(1) = p, the constant term of f(u + 1) is divisible by p and not by

p2. By Eisenstein’s criterion f(u + 1) and hence f(x) is irreducible. Thus

[Q(ζp) : Q] = p2 − p. �

Theorem 5.4 (Gauss). If a regular polygon of n sides is constructible then

n = 2rp1p2 . . . ps where p1, . . . , ps are distinct Fermat primes.

Proof. If p2n then p-gon is constructible. Hence [Q(cos 2π/p2) : Q] = 2u for

some positive integer u. Thus p(p− 1) = 2u, which is a contradiction. �

This finishes the proof of one half of Gauss’s constructibility criterion for reg-

ular polygons. We shall prove the other half after we prove the fundamental

theorem of Galois Theory.

34

Lecture 6 : Symmetric Polynomials I

Objectives

(1) Examples of symmetric polynomials.

(2) The fundamental theorem of symmetric polynomials.

(3) Newton’s identities for power sum symmetric polynomials.

Key words and phrases: Symmetric polynomial, symmetrization of a

monomial, power sum symmetric polynomials, Newton’s identities.

6. Symmetric Polynomials

Our next goal is to prove the Fundamental Theorem of Algebra : Ev-

ery polynomial of positive degree with complex coefficients has a complex

root. You must have seen its topological and complex analytic proofs. We

will present a proof which uses symmetric polynomials and the construc-

tion of the splitting field of a polynomial. We will learn about symmetric

polynomials in this section and splitting fields of polynomials in the next

section.

Let R be a commutative ring with identity and S = R[u1, u2, . . . , un] be the

polynomial ring in n variables over R. Let φ ∈ Sn, the symmetric group

of all permutations of {1, 2, . . . , n}. A permutation φ ∈ Sn gives rise to an

automorphism gφ : S → S, defined as

gφ(f(u1, . . . , un)) = f(uφ(1), . . . , uφ(n)).

Definition 6.1. A polynomial f ∈ S is called a symmetric polynomial

if for all φ ∈ Sn

f(u1, . . . , un) = f(uφ(1), . . . , uφ(n)).

Example 6.2. (1) Consider the general polynomial

f(x) = (x− u1)(x− u2) . . . (x− un)

= xn − σ1xn−1 + σ2x

n−2 + · · ·+ (−1)nσn

35

where

σ1 = u1 + · · ·+ un, σ2 =∑

i<j

uiuj , . . . , σn = u1u2 · · ·un.

It is easy to verify that σ1, . . . , σn are symmetric. These are called the

elementary symmetric polynomials in u1, u2, . . . , un.

(2) The symmetrization of a monomial uα1

1 . . . uαn

n is defined as

S(uα1

1 · · ·uαn

n ) =∑

α∈Sn

uα1

σ(1)uα2

σ(2) . . . uαn

σ(n).

It is clear that S(uα1

1 · · ·uαn

n ) is a symmetric polynomial. The symmetriza-

tion of u21u2 is

S(u21u2) = u21u2 + u21u3 + u22u3 + u23u1 + u23u2 + u22u1.

(3) For each k the polynomials wk = uk1 + uk2 + · · · + ukn are symmetric

polynomials.

(4) Let hm denote the sum of all monomials of degree m in u1, u2, . . . , un.

It is called the complete homogeneous symmetric polynomial of degree m.

Fundamental Theorem for symmetric polynomials

Example 6.3. Consider the symmetric polynomial

f(u1, u2, u3) = u21u2 + u21u3 + u22u1 + u22u3 + u23u1 + u23u2.

Then f(u1, u2, 0) = u21u2 + u22u1 = u1u2(u1 + u2) = σ01σ

02, where

σ01 = σ1(u1, u2, 0) = u1 + u2 and σ0

2 = σ2(u1, u2, 0) = u1u2.

Consider f − σ1σ2 = g. Then g|u3=0 = 0. Thus u3 | g. Since g is symmetric

u1u2u3 = σ3 | g. This gives f − σ1σ2 = −3u1u2u3 = −3σ3 and therefore

f = σ1σ2 − 3σ3.

Theorem 6.4 (Newton). Let R be a commutative ring. Then every sym-

metric polynomial in R[u1, u2, . . . , un] is a polynomial in the elementary

symmetric polynomials in a unique way. In other words if f(u1, u2, . . . , un)

is symmetric then there exists a unique polynomial g ∈ R[x1, . . . , xn] such

that

g(σ1, σ2, . . . σn) = f(u1, u2, . . . , un).

36

Proof. Apply induction on n. The n = 1 case is clear. Let the theorem be

true for symmetric polynomials in n − 1 variables. To prove the theorem

in R[u1, u2, . . . , un], apply induction on deg f . If deg f = 0 then f is a

constant. It is clear in this case. Consider f(u1, u2, . . . , un−1, 0) = f0 ∈R[u1, u2, . . . , un−1]. Then f0 is symmetric. By induction hypothesis we have

f0 = g(σ01, σ

02, . . . , σ

0n−1). Then f − g(σ1, σ2, . . . , σn−1) = f1 is symmetric

and f1(u1, . . . , un−1, 0) = 0. Thus un | f1 and hence σn | f1, by symmetry.

So f1 = σnh(u1, . . . , un). Since σn is not a zerodivisor in R[u1, . . . , un], h is

symmetric. Since deg h < deg f , by induction hypothesis h is a polynomial

in σ1, . . . , σn, hence f is so. Therefore f is a polynomial in σ1, . . . , σn.

Uniqueness : Use induction on n. the n = 1 case is obvious. Let us first

prove that the map

φ : S = R[z1, z2, . . . , zn]→ R[σ1, σ2, . . . , σn] such that

φ(zi) = σi, i = 1, 2, . . . , n and φ|R = idR

is an isomorphism. If it is not an isomorphism, we pick a nonzero polynomial

f(z1, z2, . . . , zn) ∈ S of least degree such that

f(σ1, σ2, . . . , σn) = 0.

Write f as a polynomial in zn with coefficients in R[z1, z2, . . . , zn−1] :

f(z1, z2, . . . , zn) = f0(z1, z2, . . . , zn−1) + · · ·+ fd(z1, z2, . . . , zn−1)zdn.

Then f0 6= 0. If so, then f = zng where g ∈ S. Then σng(σ1, . . . , σn) = 0.

Hence g(σ1, . . . , σn) = 0. This contradicts the minimality of deg f. Therefore

we have

0 = f0(σ1, . . . , σn−1) + · · ·+ fd(σ1, . . . , σn−1).σdn.

In this relation put un = 0 to get

f0((σ1)0, (σ2)0, . . . , (σn−1)0) = 0.

This is a nontrivial relation among the elementary symmetric polynomials

in u1, u2, . . . , un−1. This is a contradiction. �

Newton’s identities for power sum symmetric polynomials

By the Fundamental Theorem for symmetric polynomials the symmetric

polynomials wk = uk1 + · · · + ukn, k = 1, 2, 3, . . . are polynomials in the

37

elementary symmetric polynomials. Isaac Newton found identities which

express wk in terms of σ1, σ2, . . . , σn.

Theorem 6.5 (Newton).

wk = σ1wk−1 − σ2wk−2 + · · ·+ (−1)kσk−1w1 + (−1)k+1σkk if k ≤ n,

= σ1wk−1 − σ2wk−2 + · · ·+ (−1)n+1σnwk−n if k ≥ n.

Proof. Let z, y be indeterminate. Then

(y − u1)(y − u2) · · · (y − un) = yn − σ1yn−1 + σ2y

n−2 + · · ·+ (−1)nσn

Put y = 1/z to get

(1− u1z)(1− u2z) · · · (1− unz) = 1− σ1z + σ2z2 + · · ·+ (−1)nσnzn := σ(z)

Consider the generating function of w1, w2, . . .

w(z) = w1z + w2z2 + w3z

3 + · · · =∞∑

k=1

wkzk

=

∞∑

k=1

n∑

i=1

uki zk =

n∑

i=1

∞∑

k=1

(uiz)k

=n

∑

i=1

uiz

1− uiz

Since σ(z) = (1− u1z) · · · (1− unz),

σ′(z) = −n

∑

i=1

uiσ(z)

1− uizand hence w(z) =

n∑

i=1

uiz

1− uiz=−zσ′(z)σ(z)

This implies that

w(z)σ(z) = −z(−σ1 + σ2(2z)− σ3(3z2) + · · ·+ (−1)nnσnzn−1)

= σ1z − 2σ2z2 + 3σ3z

3 + · · ·+ (−1)n+1nσnzn

if k ≤ n, equating the coefficient of zk we get

(−1)k+1kσk = wk − σ1wk−1 + wk − 2σ2 + · · ·+ (−1)kw1σk−1.

38

Hence

wk = σ1wk−1 − σ2wk−2 + · · ·+ (−1)k+1σkk.

If k > n, equate coefficient of zk to get

wk − wk−1σ2 − · · ·+ (−1)nσnwk−n = 0.

Therefore

wk = σ1wk−1 − σ2wk−2 + · · ·+ (−1)n+1σnwk−n.

�

39

Lecture 7 : Symmetric Polynomials II

Objectives

(1) Discriminant in terms of power-sum symmetric polynomials.

(2) Discriminant of a cubic.

(3) Existence of a splitting field of a polynomial.

(4) Fundamental theorem of algebra via symmetric polynomials.

Key words and phrases: Discriminant of a polynomial, splitting field,

fundamental theorem of algebra.

Discriminant of a polynomial: We discuss a method to calculate the

discriminant of a polynomial by employing Newton’s identities.

Definition 6.6. Let u1, u2, . . . , un, x be indeterminate and

f(x) = (x− u1)(x− u2) . . . (x− un).

The discriminant of f(x) is the symmetric function

disc (f(x)) = Πi<j(ui − uj)2

It is clear that f(x) has a repeated root if and only if disc (f) = 0.

Since disc (f) is a symmetric polynomial with integer coefficients, by the

fundamental theorem for symmetric polynomials, there exists a polynomial

g(X1, . . . , Xn) ∈ Z[X1, X2, . . . , Xn] such that disc (f) = g(σ1, σ2 . . . , σn).

The van der Monde matrix

M =

1 1 · · · 1

u1 u2 · · · un

u21 u22 · · · u2n...

......

...

un−11 un−12 · · · un−1n

has determinant detM = Πi>j(ui − uj). Hence

40

disc (f) = det(MM t) =

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

n w1 w2 · · · wn−1

w1 w2 w3 · · · wn

w2 w3 w4 · · · wn+1

......

......

...

wn−1 wn wn+1 · · · w2n−2

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

.

Example 6.7. Using Newton’s identities, we calculate the discriminant of

the polynomial p(x) = x3 + px+ q. We have σ1 = 0, σ2 = p, σ3 = −q and

MM t =

3 w1 w2

w1 w2 w3

w2 w3 w4

Newton’s identities in this case are

w1 = σ1 = 0

w2 = σ21 − 2σ2 = −2p

w3 = σ1w2 − σ2w1 + 3σ3 = −3q

w4 = σ1w3 − σ2w2 + σ3w1 = 2p2

Therefore

disc (f) = detMM t =

∣

∣

∣

∣

∣

∣

∣

3 0 −2p0 −2p −3q−2p −3q 2p2

∣

∣

∣

∣

∣

∣

∣

= −4p3 − 27q2.

7. The Splitting Field of a Polynomial

In this section we construct a field extension K/F which contains all the

roots of a given polynomial f(x) ∈ F [x]. For simplicity, we want K to be

the smallest field containing F with respect to this property.

Definition 7.1. Let F be a field and f(x) ∈ F [x] be a monic polynomial

of degree n. A field K ⊇ F is called a splitting field of f(x) over F

if there exist r1, r2, . . . , rn ∈ K so that f(x) = (x − r1) . . . (x − rn) and

K = F (r1, r2, . . . , rn).

Example 7.2. (i) Let f(x) = x2 + ax+ b ∈ F [x]. If f(x) is reducible then

F is a splitting field of f(x). If f(x) is irreducible then (f(x)) is a maximal

41

ideal of F [x]. Hence F (x)/(f(x)) ≃ F (r) is a field, where r = x+ (f(x)). If

s is another root of f(x), then s + r = −a, so s = −a − r ∈ F (r). Hence

F (r) is a splitting field of f(x) over F .

(ii) Consider the irreducible polynomial f(x) = x3 + x + 1 ∈ F2[x]. Let

r = x + (f(x)) ∈ F2[x]/(f(x)) = F2(r). Since [F2(r) : F2] = 3, F2(r) has 8

elements. A basis of the F2−vector space F2(r) is {1, r, r2}. Hence

F2(r) = {0, 1, r, r2, 1 + r, 1 + r2, r + r2, 1 + r + r2}

and we have the relation r3 = 1+r. Check that f(r2) = f(r4) = 0. Therefore

x3 + x+ 1 = (x+ r)(x+ r2)(x+ r4).

Thus Fr(r) is a splitting field of x3 + x+ 1 over F2.

We will later see that if f(x) ∈ Fq[x], where Fq is a finite field with q

elements, then Fq[x]/(f(x)) is a splitting field of f(x), if f(x) is an irreducible

polynomial over Fq.

Existence of Splitting field

Theorem 7.3. Let F be a field. Then any polynomial f(x) ∈ F [x] of positive

degree has a splitting field.

Proof. Apply induction on deg f. If deg f = 1 then F is the splitting field

of f over F. Suppose deg f > 1. If f(x) splits as a product of linear factors

in F [x] then F is the splitting field of f(x) over F. Suppose g(x) is an

irreducible factor of f(x) with deg g ≥ 2. Then r = x + (g(x)) ∈ K :=

F [x]/(g(x)) is a root of g(x) and hence of f(x). Since f(x) = (x − r)h(x)

for some h(x) ∈ K[x] and deg h(x) < deg f(x). By induction h(x) has a

splitting field L over K. Let r2, r3, . . . , rn ∈ L be the roots of h(x). Then

L = K(r2, r3, . . . , rn) = F (r1, r2, . . . , rn) is the required splitting field. �

We end this section by presenting a proof of the fundamental theorem of

algebra due to Gauss.

Theorem 7.4 (The Fundamental Theorem of Algebra). Every com-

plex polynomial of positive degree has a complex root.

42

Proof. We shall use the following facts:

(i) Every odd degree polynomial with real coefficients has a real root.

(ii) Every quadratic polynomial in C[x] has a complex root.

(iii) The fundamental theorem for symmetric polynomials.

(iv) Every polynomial f(x) has a splitting field.

(i) This is a consequence of the Intermediate Value Theorem.

(ii) It is enough to show that complex numbers have a complex square root.

Indeed, let z = a + bi ∈ C, where a, b ∈ R and (c + di)2 = a + bi. Then

c2 − d2 + 2cdi = a+ bi. Thus a = c2 − d2 and b = 2cd. Therefore

a2 + b2 = (c2 + d2)2

c2 + d2 =√

a2 + b2 ∈ R.

Therefore c2 = 12 [a +

√a2 + b2] ≥ 0 and d2 = 1

2 [√a2 + b2 − a] ≥ 0. Thus

c, d ∈ R.

The polynomial g(x) = f(x)f(x) ∈ R[x]. Here f denotes the polynomial

whose coefficients are conjugates of the coefficients of f(x). If g(x) has a

complex root z then either f(z) = 0 or f(z) = 0. If f(z) = 0, then f(z) = 0.

Thus by replacing f by g, we may assume that f(x) is a monic polynomial

with real coefficients.

Let d = deg f = 2nq, where q is odd. We apply induction on n. If n = 0,

then f is a real odd degree polynomial, hence it has a real root. Now let

n ≥ 1. Let K = C(α1, . . . , α2), be a splitting field of f(x), over C, where

α1, . . . , αd are the roots of f(x) in K. Consider the elements

yij = αi + αj + rαiαj ,

where r ∈ R is fixed and 1 ≤ i ≤ j ≤ d. There are(

d+12

)

such pairs (i, j).

Hence

deg h(x) =∏

1≤i≤j≤d

(x− yij) =

(

d+ 1

2

)

= 2n−1q(d+ 1).

The coefficients of h(x) are elementary symmetric polynomials in yij ’s. So

they are symmetric polynomials in α1, α2, . . . , αd. Hence they are polynomi-

als in the coefficients of f(x). Hence h(x) ∈ R[x]. By induction on n, h(x)

43

has a complex root say zr. Since all the roots of h(x) ∈ K and zr ∈ C ⊆ K,

zr = αi(r) + αj(r) + rαi(r)αj(r)

for some pair (i(r), j(r)) so that 1 ≤ i(r) ≤ j(r) ≤ d. Define

ϕ : R→ {(i, j) | 1 ≤ i(r) ≤ j(r) ≤ d} = P, ϕ(r) = (i(r), j(r)).

Since R is infinite and P is finite, there exists c 6= d ∈ R such that

(i(c), j(c)) = (i(d), j(d)) := (a, b). Therefore,

zc = αi(c) + αj(c) + rαi(c)αj(c) = αa + αb + cαaαb = zd = αa + αb + dαaαb.

Therefore (d− c)αaαb = zd − zc ∈ C. Hence αaαb ∈ C so that αa + αb ∈ C.

But αa and αb are roots of

x2 − (αa + αb)x+ αaαb ∈ C[x].

Hence αa, αb ∈ C. Therefore f(x) has a complex root. �

45

Lecture 8 : Algebraic Closure of a Field

Objectives

(1) Existence and isomorphisms of algebraic closures.

(2) Isomorphism of splitting fields of a polynomial.

Key words and phrases: algebraically closed field, algebraic closure, split-

ting field.

8. Algebraically Closed Fields

In the previous section we showed that all complex polynomials of positive

degree split in C[x] as products of linear polynomials in C[x]. While working

with polynomials with coefficients in a field F , it is desirable to have a field

extension K/F so that all polynomials in K[x] split as product of linear

polynomials in K[x].

Definition 8.1. A field F is called an algebraically closed field if every

polynomial f(x) ∈ F [x] of positive degree has a root in F.

It is easy to see that a field F is algebraically closed if and only if f(x) is

a product of linear factors in F [x]. The fundamental theorem of algebra

asserts that C is an algebraically closed field. Let us show that any field is

contained in an algebraically closed field.

Existence of algebraic closure

Theorem 8.2. Let k be a field. Then there exists an algebraically closed

field containing k.

Proof. (Artin) We construct a field K ⊇ k in which every polynomial of

positive degree in k[x] has a root. Let S be a set of indeterminates which

is in 1− 1 correspondence with set of all polynomials in k[x] of degree ≥ 1.

Let xf denote the indeterminate in S corresponding to f.

46

Let I = (f(xf ) | deg f ≥ 1) be the ideal generated by all the polynomials

f(xf ) ∈ k[S]. We claim that I is a proper ideal of k[S]. Suppose to the

contrary, I = k[S]. Then

1 = g1f1(xf1) + · · ·+ gnfn(xfn)(1)

for some g1, g2, . . . , gn ∈ k[S]. The polynomial g1, g2, . . . , gn involve only

finitely many variables. Put xfi = xi for i = 1, 2, . . . , n and let xn+1, . . . , xm

be the remaining variables in g1, g2, . . . , gn. Then

n∑

i=1

gi(x1, x2, . . . , xn, xn+1, . . . , xm)fi(xi) = 1.

Let E/k be an extension field in which the polynomials f1(x1), . . . , fn(xn)

have roots α1, . . . , αn respectively. Putting xn+1 = . . . = xm = 0 and xi = αi

for all i = 1, 2, . . . , n in the equation 1 we get a contradiction. Hence I is a

proper ideal of k[S]. Let m be a maximal ideal of k[S] containing I. Then

K1 = k[S]/m is a field. We claim that xf + m is a root of f(x). Indeed,

f(xf +m) = f(xf )+m = m. Thus each polynomial in k[x] has a root in K1.

Repeat the procedure on K1 to get K2 ⊃ K1 which has roots of all monic

polynomials in K1[x]. Let K = ∪∞

i=1Ki. Then K is a field. If f(x) ∈ K[x]

then f(x) ∈ Kn[x] for some n. Hence f(x) has a root in Kn+1 ⊆ K. Thus

K is algebraically closed. �

Corollary 8.3. Let F be a Field. Then there exists a field K ⊃ F such that

K is algebraically closed and K is algebraic over F.

Proof. Let L ⊃ F be an algebraically closed field. Then the field

K = {a ∈ L | a is algebraic over F}

is algebraically closed and it is algebraic over F. �

Definition 8.4. Let F be a field. An extension K/F is called an algebraic

closure of F if K is algebraically closed and K/F is an algebraic extension.

Isomorphism of algebraic closures

We now show that if E1 and E2 are algebraic closures of a field F then they

are F−isomorphic. As a consequence we also prove that any two splitting

47

fields of a polynomial f(x) ∈ F [x] are F−isomorphic. Extensions of embed-

dings of fields is one of the main observations in various arguments in Galois

theory. The next result prepares us for the theorem about isomorphism of

algebraic closures of a field.

Proposition 8.5. Let σ : k → L be an embedding of fields where L is

algebraically closed. Let α be algebraic over k and p(x) = irr(α, k). Let

p(x) =∑

aixi ∈ k[x] and pσ(x) =

∑σ(ai)x

i. Then τ → τ(α) is a bijection

between the sets

{τ : k(α)→ L | τ is an embedding and τ |k = σ} ←→ {β ∈ L | pσ(β) = 0}.

Proof. Let τ : k(α)→ L be an embedding extending σ. Then

τ(p(α)) = pσ(τ(α)) = 0.

Hence τ(α) is a root of pσ(x). Conversely let β ∈ L and pσ(β) = 0. Define

τ : k(α)→ L by τ(f(α)) = fσ(β). We show that τ is well defined.

Suppose f(α) = g(α). Then (f − g)(α) = 0, so p(x) | (f(x)− g(x)). Hence

pσ(x) | (f − g)σ(x). Thus pσ(β) = (fσ(β) − gσ(β) = 0. Hence fσ(β) =

τ(f(α)) = gσ(β) = τ(g(α)). Thus τ is well-defined. Suppose that fσ(β) =

τ(f(α)) = 0. Then pσ(x) | fσ(x). Since σ is an embedding, p(x) | f(x).

Thus f(α) = 0.

�

Proposition 8.6. Let σ : k → L be an embedding of fields where L is

algebraically closed. Let E be an algebraic extension of k. Then there exists

an embedding τ : E → L extending σ. If E is an algebraic closure of k and

L is an algebraic closure of σ(k) then τ is an isomorphism extending σ.

Proof. Consider the set

S = {(F, τ) | k ⊆ F ⊆ E are fields and τ : F → L such that τ |k = σ}.

Since (k, σ) ∈ S, it is nonempty. Let (F, τ) and (F ′, τ ′) ∈ S. Define

(F, τ) ≤ (F ′, τ ′) if and only if F ⊆ F ′ and τ ′|F = τ.

Then S is a partially ordered inductive set. Indeed, if {(Fα, τα)}α∈I is a

chain in S then F = ∪α∈IFα is a subfield of E. Define τ : F → L as

τ(x) = τα(x) if x ∈ Fα. Then τ is well-defined.

48

By Zorn’s Lemma there exists a maximal element (F, τ) ∈ S. We claim

that F = E. Suppose there exists α ∈ E \ F . Since α is algebraic over

F, τ : F → L can be extended to F (α)→ L. This contradicts maximality of

(F, τ). Thus E = F. Hence σ can be extended to an embedding of E into L.

Now suppose E is an algebraic closure of k and L is an algebraic closure of

σ(k). Since τ(E) is algebraically closed and L is algebraic over τ(E), L =

τ(E). Thus τ : E → L is an isomorphism. �

Theorem 8.7. If E1 and E2 are algebraic closures of a field k then they

are k−isomorphic.

Proof. The identity map k → E2 can be extended to τ : E1 → E2 by the

above proposition, τ is a k−isomorphism. �

Theorem 8.8. Let E and F be splitting fields of polynomial f(x) ∈ k[x]

where k is a field. Then they are k-isomorphic.

Proof. Let F a be an algebraic closure of F. Then it is also an algebraic

closure of k. Thus there exists an embedding τ : E → ka extending idk :

k → ka. Let f(x) = (x− α1) · · · (x− αn) be a factorization of f(x) in E[x].

Then

f τ (x) = (x− τ(α1)) · · · (x− τ(αn)) ∈ F a[x].

Thus F = k(τ(α1), . . . , τ(αn)) = τ(E) as ka contains a unique splitting field

of any polynomial in k[x]. �

54

Lecture 9 : Separable Extensions I

Objectives

(1) Criterion for multiple roots in terms of derivatives

(2) Irreducible polynomials are separable over fields of characteristic

zero.

(3) Characterization of perfect fields of positive characteristic,

Key words and phrases: Separable polynomial, separable element, sepa-

rable extensions, derivative of a polynomial, perfect fields.

9. Separable Extensions

Let F be a field. We have seen that the discriminant of a polynomial

f(x) ∈ F [x] vanishes if and only if f(x) has a repeated root. Calculation

of discriminant can be difficult. In this section we discuss an effective cri-

terion in terms of derivatives of polynomials whether certain root of f(x) is

repeated. We will also study fields F so that no irreducible polynomial in

F [x] has repeated roots.

Let E be a splitting field of a monic polynomial f(x) ∈ F [x] of degree n.

Write in E[x] the unique factorization of f(x).

f(x) = (x− r1)e1(x− r2)

e2 · · · (x− rg)en .

where r1, . . . , rg ∈ E and e1, e2, . . . , eg are positive integers.

Definition 9.1. The numbers e1, e2, . . . , en are called the multiplicities of

r1, r2, . . . , rn respectively. If ei = 1 for some i, then ri is called a simple

root. If ei > 1 then ri is called a multiple root. A polynomial f(x) with no

multiple roots is called a separable polynomial.

Proposition 9.2. The numbers of roots and their multiplicities are inde-

pendent of a splitting field chosen for f(x) over F.

55

Proof. Let E and K be splitting fields of f(x) over F. Then there is an

F−isomorphism σ : E → K. This isomorphism gives rise to an isomorphism

φσ : E[x]→ K[x], ϕσ

(

∑

i

aixi

)

=∑

i

σ (ai)xi.

Let f(x) =∏g

i=1(x− ri)ei be the unique factorization of f(x) ∈ E[x]. Then

φσ(f(x)) =∏g

i=1(x − σ(ri))ei . Since K[x] is UFD, σ(r1), . . . , σ(rg) are the

roots of φσ(f(x)) = f(x) with multiplicities e1, . . . , eg in K respectively. �

The derivative criterion for multiple roots

Let f(x) = a0 + a1x + · · · + anxn ∈ F [x]. We can define derivative of f(x)

without appealing to limits. This is preferable since F may not be equipped

with a distance function.

The derivative of f(x), is defined by f ′(x) :=∑m

i=0 iaixi−1. It is easy to

check that the usual formulas for (f(x)±g(x))′, (f(x)g(x))′ and (f(x)/g(x))′

where g(x) 6= 0 hold for derivatives of polynomials.

Theorem 9.3. Let f(x) ∈ F [x] be a monic polynomial.

(1) If f ′(x) = 0 then every root of f(x) is a multiple root.

(2) If f ′(x) 6= 0 then f(x) has simple roots if and only if gcd(f, f ′) = 1.

Proof. (1) Let f(x) = (x− r)g(x). Then

0 = f ′(x) = g(x) + (x− r)g′(x).

Thus g(x) = −(x− r)g′(x), so r is a root of g(x). Hence r is a multiple root.

(2) (⇐) Let gcd(f, f ′) = 1 and let r be a multiple root of f(x). Then

f(x) = (x− r)2g(x) in some splitting field E of f(x) over F. Thus

f ′(x) = (x− r)2g′(x) + 2(x− r)g(x).

Hence f ′(r) = 0. If d(x) = gcd(f(x), f ′(x)) ∈ F [x] then

d(x) = p(x)f(x) + q(x)f ′(x)

for some p(x), q(x) ∈ F [x]. Hence d(r) = 0. Therefore, deg d(x) ≥ 1, so

gcd(f, f ′) 6= 1, which is a contradiction. Therefore f(x) has only simple

roots.

56

(⇒). Let r1, r2, . . . , rn be the roots of f(x) and assume that they are simple.

Then

f(x) = (x− r1)(x− r2) · · · (x− rn) and f′(x) =

∑ni=1

f(x)(x−ri)

.

Therefore (x − ri) does not divide f ′(x) any i. Hence f and f ′ have no

common root. Therefore gcd(f, f ′) = 1.

�

Proposition 9.4. (1) Let f(x) ∈ F [x] be an irreducible polynomial. Then

f(x) is separable if and only if f ′ 6= 0.

(2) Irreducible monic polynomials over a field of characteristic zero are sep-

arable.

Proof. (1) (⇒) If f ′ = 0, then every root of f(x) is a multiple root.

(⇐) Suppose r is a multiple root of f(x). Then f ′(r) = 0. Since f(x)

is irreducible, f(x) | f ′(x). But this is a contradiction since deg f ′(x) <

deg f(x). Therefore f(x) is separable.

(2) If char F = 0, and f(x) is of positive degree, then f ′(x) 6= 0. �

Proposition 9.5. Let F be a field of positive characteristic p. Then xp−a ∈

F [x] is either irreducible in F [x] or a ∈ F p.

Proof. Suppose f(x) = xp − a = g(x)h(x) where 1 ≤ deg g = m < p

Let b be a root of f(x) in a splitting field E of f(x). Then a = bp, so

f(x) = (x− b)p. Hence b is also a root of g(x). Thus g(x) = (x− b)m Then

bm ∈ F. Since (p,m) = 1, there exists x, y ∈ Z such that px+my = 1. Hence

b = bpx+my = ax(bm)y ∈ F. Thus bp = a ∈ F p. �

Example 9.6. We construct an irreducible polynomial with a multiple root.

Let F = Fp(t) be the quotient field of the polynomial ring Fp[t]. Let f(x) =

xp − t ∈ F [x]. Then t /∈ F p. Suppose t is a pth power and

t =g(t)p

h(t)p=

(∑

i aiti)p

(∑

i biti)p

.

Then t(∑

bpi tip) =

∑

i api t

ip. Hence ai = bi = 0 for all i. Thus xp − t

is irreducible. Another way to see that xp − t is irreducible is to apply

57

Eisenstein’s Criterion with t as a prime element. Let E be a splitting field

of f(x) over F and u be a root of f(x). Then up = t so xp − t = (x − u)p,

Hence f(x) has only one root in E.

Proposition 9.7. Let f(x) ∈ F [x] where char F = p, be an irreducible

polynomial. If f(x) is not separable then there exists g(x) ∈ F [x] such that

f(x) = g(xp).

Proof. Since f(x) =∑

i aixi is irreducible and inseparable, we have f ′(x) =

∑

(iai)xi−1 = 0. Therefore i = pti for some ti ∈ N. Hence

f(x) =∑

aptixpti =

∑

api(xp)ti .

�

Perfect Fields

We have seen that irreducible polynomial over fields of characteristic 0 are

separable. But over a field of positive characteristic, irreducible polynomial

may not be separable. We now discuss a condition on a field F of positive

characteristic which will ensure that irreducible polynomials in F [x] are

separable.

Definition 9.8. Let F ⊆ K be a field extension. An algebraic element

α ∈ K is a called separable element over F if irr(α, F ) is separable.

We say K/F is a separable algebraic extension if each element of K

is separable. We say F is a perfect field if each algebraic extension is

separable.

Any field of characteristic zero is perfect. By the previous example Fp(t) is

not perfect. This is basically due to t not being a pth power in Fp(t).

Theorem 9.9. Let F be a field of positive characteristic p. Then F is perfect

if and only if

F = F p = {ap | a ∈ F}.

Proof. Suppose a ∈ F \F p. Then xp−a ∈ F [x] is irreducible and inseparable.

Hence F is not perfect.

58

(⇐) Let F = F p and f(x) ∈ F [x] be an irreducible polynomial. If f(x) is

inseparable, then f(x) = g(xp) =∑

ai(xp)i =

∑

(bi)p(xp)i = (

∑

bixi)p for

some bi ∈ F. This contradicts irreducibility of f(x). Hence f(x) is separable.

�

Corollary 9.10. Every finite field is perfect.

Proof. Let |F | = pn. By Lagrange theorem applied to the multiplicative

group F× we get αpn−1 = 1 for all α ∈ F×. Hence αpn = α for all α ∈ F.

Therefore α = (αpn−1

)p. �

59

Lecture 10 : Separable Extensions II

Objectives

(1) Roots of an irreducible polynomial have equal multiplicity.

(2) Separable finite algebraic extensions and separable degree.

(3) Transitivity of separable extensions

Key words and phrases: Separable degree, extensions of embeddings.

Proposition 9.11. Let F be a field and f(x) ∈ F [x] be a monic irreducible

polynomial. Then all roots of f(x) have equal multiplicity. If char F = 0

then all roots of f(x) are simple and if char F = p > 0 then all roots of

f(x) have multiplicity pn for some n.

Proof. Let α, β be roots of f(x) in F . Consider the F -isomorphism σ :

F (α) → F (β) given by σ(α) = β. Then σ can be extended to an automor-

phism of F . Let f(x) = (x−α)mh(x) where h(x) ∈ F [x] and α is not a root

of h(x). Then f(x) = τ(f(x)) = (x− β)mτ(h(x)). Hence the multiplicity of

β is at least m. By symmetry we conclude that both roots have the same

multiplicity.

We know that irreducible polynomials are separable if char F = 0. Let

char F = p > 0. Suppose f(x) has roots of positive multiplicity. Then there

exists a separable polynomial g(x) so that f(x) = g(xpn

). Let r1, r2, . . . , rg

be distinct roots of f(x) in F . Then

f(x) = (x− r1)pn(x− r2)

pn . . . (x− rg)pn

�

Separability and extensions of embeddings

Let E = k(a) be an algebraic extension of a field k. Let p(x) = irr(a, k).

We have seen that if σ : k → L is an embedding of fields where L is alge-

braically closed then the number of embeddings τ : E → L extending σ is

equal to the number of distinct roots of pσ(x) in L. Hence if p(x) is separa-

ble, then the number of extensions of σ to embeddings of E into L is [E : k].

Conversely, if σ has [E : k] extensions, then for any such extension τ, τ(α) is

60

a root of pσ(x). Hence p(x) is separable. We now discuss this phenomenon

for finite algebraic extensions.

Let σ : F → L be an embedding of fields where L is an algebraic closure

of σ(F ). Let τ : F → L′ be an embedding of fields where L′ is an algebraic

closure of τ(F ). Let E be an algebraic extension of F. Let Sσ (resp. Sτ )

denote the set of extensions of σ (resp. τ) to embeddings of E into L (resp.

L′). Consider the following diagram of fields and embeddings:

L′ Lλ

oo

τ∗(E) Eτ∗

ooσ∗

// σ∗(E)

τ(F ) Fτ

ooσ

// σ(F )

Let λ be an extension of the embedding τ ◦ σ−1 : σ(F ) → τ(F ) to an

isomorphism λ : L→ L′.

Theorem 9.12. The map

ψ : Sσ → Sτ , ψ(σ∗) = λ ◦ σ∗

is a bijection.

Proof. If σ∗ ∈ Sσ then for any x ∈ F we have

λ ◦ σ∗(x) = λ ◦ σ(x) = τ ◦ σ−1(σ(x)) = τ(x).

Hence λ ◦ σ∗ is an extension of τ to an embedding of E into L′. Hence λ

induces a mapping ψ : Sσ → Sτ defined by ψ(σ∗) = λ ◦ σ∗. Since λ is an

isomorphism, ψ is a bijection.

�

Definition 9.13. If E/F is an algebraic extension then the cardinality of

Sσ is called the separable degree of E/F and it is denoted by [E : F ]s.

Proposition 9.14. Let k ⊆ F ⊆ E be a tower of finite algebraic extensions.

Then [E : k]s ≤ [E : k] and

[E : k]s = [E : F ]s[F : k]s.

61

Proof. First we show that the separable degree is multiplicative in a tower

of field extensions. Let σ : k → L be an embedding into an algebraically

closed field L. Let (σi)i∈I be distinct extensions of σ to embeddings of F

into L. Each σi has [E : F ]s extensions to embeddings of E into L. Let these

be (τij). Hence (τij) has cardinality [F : k]s[E : k]s. If γ : E → L is an

embedding extending σ, then γ|F is an extension of σ to an embedding of F

into L. Hence γ|F = τij . This proves the multiplicativity of separable degree

in a tower of field extensions.

Since E/k is finite, there exist elements a1, a2, . . . , an such that

k ⊂ k(a1) ⊂ k(a1, a2) ⊂ · · · ⊂ k(a1, a2, . . . , an).

Each step in the above tower is a simple algebraic extension. Hence the

separable degree of each step is atmost its degree. Since the separable degree

and degree of a field extension are multiplicative, we have [E : k]s ≤ [E : k].

�

Corollary 9.15. Let k ⊂ F ⊂ K be a tower of finite extensions. Then

[E : k]s = [E : k] if and only if the corresponding equality holds in each step

of the tower.

Theorem 9.16. Let E/k be a finite extension. Then E/k is separable if

and only if [E : k]s = [E : k].

Proof. Let E/k be finite separable extension. Then E = k(a1, a2, . . . , an)

for some a1, a2, . . . , an ∈ E. Then each ai is separable over k. Hence ai is

separable over k(a1, a2, . . . , ai−1) for i = 1, 2, . . . , n.

[k(a1, a2, . . . , ai) : k(a1, a2, . . . , ai−1)]s = k(a1, a2, . . . , ai) : k(a1, a2, . . . , ai−1)].

for i = 1, 2, . . . , n, whence [E : k]s = [E : k].

Conversely let [E : k]s = [E : k]. Then using the fact that the separable

degree and degree are multiplicative and the separable degree is atmost the

degree, we conclude that for any a ∈ E, [E : k(a)]s[k(a) : k]s = [E : k]s =

[E : k]. Hence [k(a) : k]s = [k(a) : k]. Hence irr(a, k) is separable. Thus E/k

is a separable extension.

�

63

Lecture 11 : Finite Fields I

Objectives

(1) Existence and uniqueness of finite fields.

(2) Algebraic closure of a finite field.

(3) Finite subgroup of the multiplicative group of a field is cyclic.

(4) Gauss’ formula for the number of monic irreducible polynomials of

a given degree over a finite field.

Key words and phrases: Finite field, Gauss’ formula for irreducible poly-

nomials,

10. Finite Fields

A finite field F of prime characteristic p contains a prime field Fp. Since F

is a finite dimensional vector space over Fp, |F | = pn, where n = [F : Fp].

We usually write pn = q.

Proposition 10.1. Let K and L be finite fields of cardinality q = pn, where

p is a prime number. Then K and L are isomorphic.

Proof. Since |K×| = q − 1, by Lagrange’s theorem xq−1 = 1 for all x ∈ K×.

Thus every element of K is a root of the polynomial xq − x = f(x). Hence

K is a splitting field of f(x) over Fp. Since any two splitting fields of f(x)

over Fp are isomorphic, K ≃ L. �

Notation: We shall denote a finite field with pn elements by Fpn .

Corollary 10.2 (Wilson). Let p be a prime number. Then

(p− 1)! ≡ −1( mod p).

64

Proof. The assertion is clear for p = 2. Let p be odd. Since Fp is the set of

roots of xpn

− x, taking n = 1 we get

xp−1 − 1 = (x− 1)(x− 2) . . . (x− (p− 1)).

Putting x = 0 we obtain (p− 1)! ≡ −1( mod p). �

Proposition 10.3. For any prime p and any n ∈ N, there exists a finite

field of cardinality pn. An algebraic closure Fap of Fp has a unique subfield

with pn elements for every n ∈ N and

Fap =

⋃

n∈N

Fpn .

Proof. Let q = pn. Then Fap contains a unique splitting field of xq−x = f(x)

over Fp. Let

K = {α ∈ Fαp | f(α) = 0}.

Then K is a field. The polynomial xq − x is separable since its derivative is

−1. Hence K has q elements. Therefore K is the required finite field with q

elements. Let a be algebraic over Fp and [Fp(a) : Fp] = n. Then Fp(a) has

pn elements. Hence a ∈ Fpn . Thus Fap =

⋃

n∈N Fpn . �

Theorem 10.4. Let U be a finite subgroup of the multiplicative group F×

of a field F. Then U is cyclic.

Proof. Let |U | = n. By Lagrange’s Theorem xn = 1 for all x ∈ U . Since

U is an abelian group, by the structure theorem for finite abelian groups,

there exist d1, d2, . . . , dr ∈ N such that n = d1d2 . . . dr, d1 | d2 | · · · | dr, and

U ≃ Z/d1Z× Z/d2Z× · · · × Z/drZ.

Thus each x ∈ U satisfies xdr − 1 = 0. But xdr − 1 has atmost dr roots.

Thus n = dr and so U is cyclic. �

Counting irreducible polynomials over finite fields

Let Nq(n) denote the number of irreducible polynomials of degree n over a

finite field Fq.We derive a formula, due to Gauss, for Nq(n). Let α be a cyclic

generator of F×qn . Then Fqn = Fq(α) and deg irr(α, Fq) = [Fqn : Fq] = n.

65

Hence Nq(n) ≥ 1. Recall the Mobius inversion formula. Let f, g : N→ N be

functions so that

f(n) =∑

d|n

g(d).

Then

g(n) =∑

d|n

f(n/d)µ(d),

where µ is the Mobius function µ : N→ N defined as

µ(n) =

1 for n = 1

(−1)r if n = p1 . . . pr, where p1, . . . , pr are distinct primes,

0 if p2|n for some prime p.

Theorem 10.5 (Gauss). The number of irreducible monic polynomials of

degree n over Fq is given by

Nq(n) =1

n

∑

d|n

µ(d)qn/d.

Proof. Let f(x) be a monic irreducible polynomial in Fq[x]. We show that

f(x) | xqn

− x if and only if deg f | n.

Suppose f(x) | xqn

− x. Then Fqn contains all the roots of f(x). Let α be a

root of f(x). Then Fq(α) ⊆ Fqn . Thus [Fq(α) : Fq] = d = deg f | n.

Conversely, let d = deg f | n. Let f(α) = 0, where α ∈ Faq . Let β be another

root of f(x) in Faq . Then there exists an embedding σ : Fq(α) → Fq(β)

such that σ(α) = β. Since Fq(α) is a splitting field of xqd

− x over Fq, so

is σ(Fq(α)) = Fq(β). But Faq has only one splitting field of xq

d

− x, hence

β ∈ Fq(α). Thus f(x) | xqn − x.

Notice that xqn

− x is a separable polynomial. Hence

xqn

− x =∏

d|n

f(d)1 (x) · · · f

(d)Nq(d)

(x),

where f(d)1 (x), . . . , f

(d)Nq(d)

(x) are all the degrees d irreducible polynomials over

Fq. Equate degrees to get qn =∑

d|n dNq(d). By Mobius inversion

nNq(n) =∑

d|n

µ(n/d)qd.

�

71

Lecture 12 : The Primitive Element Theorem

Objectives

(1) Factorization of polynomials over finite fields.

(2) The Primitive element theorem.

(3) Finite separable extensions have a primitive element.

Key words and phrases: Primitive element, finite separable extensions,

factorization.

Example 10.6. We know that the polynomial xpn − x is the product of

all the degree d monic irreducible polynomials in Fp[x] where d | n. Thisis useful for constructing irreducible polynomials over Fp. Let us factorize

x16−x over F2. The irreducible quadratic polynomials are factors of x4−x =

x(x+1)(x2+x+1). Hence there is only one quadratic irreducible polynomial

over F2. The cubic irreducible are factors of

x8 − x = x(x7 + 1) = x(x+ 1)(x6 + x5 + x4 + x3 + x2 + x+ 1).

By Gauss’ formula N2(3) = 2. Therefore the irreducible cubics over F2 are

x3+x2+1 and x3+x+1. By Gauss’ formula, we count irreducible quartics

over F2 :

4N2(4) =∑

d|4

µ(4/d)2d = µ(4)2 + µ(2)22 + µ(1)24 = −4 + 16 = 12.

Hence N2(4) = 3. These quartics are factors of x16 − x. The irreducible

factors of this polynomial have degrees 1, 2 and 4. Therefore the irreducible

quartics are factors of

x16 − x

x(x+ 1)(x2 + x+ 1)= (x4 + x+ 1)(x4 + x3 + 1)(x4 + x3 + x2 + x+ 1).

We end this section by an interesting application of finite fields.

Proposition 10.7. The polynomial x4 + 1 is irreducible in Z[x] but it is

reducible over Fp for every p.

72

Proof. Let f(x) = x4+1. Then f(x+1) is irreducible over Z by Eisenstein’s

criterion. For p = 2, we have x4 + 1 = (x + 1)4. Now let p be odd. Then

8 | p2 − 1. Hence

x4 + 1 | x8 − 1 | xp2−1 − 1 | xp2 − x.

The splitting field of xp2 − x over Fp is the finite field F = Fp2 . Hence

[F : Fp] = 2. Therefore the roots of x4+1 in F have degree 1 or 2. Therefore

x4 + 1 cannot have a cubic or quartic irreducible factor over Fp. Hence it is

reducible over Fp for each prime p.

�

11. The Primitive Element Theorem

Since F×qn is a cyclic group, Fqn = Fq(α) where α is a generator of F×qn .We say

that α is primitive element of the field extension Fq ⊂ Fqn . In this section

we discuss existence of primitive elements in finite algebraic field extensions.

We will show that in a finite separable extension, primitive elements always

exist.

Definition 11.1. Let E/F be a field extension. An element α ∈ E is called

a primitive element of E over F if E = F (α).

Example 11.2. (1) Let f(x) = x3 − 2, α = 3√2 and ω = e2πi/3. Then

Q(α, ω) is a splitting field of f(x). Moreover [Q(α, ω) : Q] = 6. Since

Q(α) ⊆ R, α+ ω /∈ Q(α).

Q(α, ω)

LLLLLLLLLL

LLLLLLLLLLQ

wwww

wwww

ww

Q(α) Q(α+ ω)

rrrrrrrrrrrr

HHHH

HHHH

HH

Qid

Q

We know that the number of ways id : Q → Q can be extended to

an embedding σ : Q(α + ω) → Q is deg irr(α + ω,Q) = [Q(α + ω) : Q].

Since degree irr(ω,Q(α)) = 2, id : Q(α) → Q can be extended in two

73

ways: ω → ω2 or ω → ω. Restriction of this embedding to Q(α + ω) maps

α + ω to α + ω2 or α + ω. In a similar way we can embed Q(α + ω) onto

Q(αω + ω),Q(αω + ω2), Q(αω2 + ω2) and Q(αω2 + ω). Thus [Q(α + ω) :

Q] = 6. So Q(α, ω) = Q(α+ ω). Therefore α+ ω is a primitive element.

(2) An algebraic extension need not have a primitive element. Let field k be

a field with char(k) = p and let u, v be indeterminates. Let E = k(u, v) and

F = k(up, vp). Then f(u, v)p ∈ F for any f(u, v) ∈ E. But [E : F ] = p2. If

y ∈ E is a primitive element of E/F then deg irr(y, F ) = p2. But yp ∈ F.

This is a contradiction.

Theorem 11.3 (The Primitive Element Theorem). Let E/k be a finite

extension.

(1) There is a primitive element for E/k if and only if the number of

intermediate subfields F such that k ⊂ F ⊂ E is finite.

(2) If E/k is a finite and separable extension then it has a primitive element.

Proof. (1) If k is a finite field then E is finite and hence E× is a cyclic group.

Thus E/k has a primitive element.

Let k be infinite and let E/k have finitely many intermediate fields. Suppose

α, β ∈ E. As c varies in k, k(α+ cβ) varies over finitely many intermediate

subfields of E/k. Hence, there are c1 6= c2 ∈ k such that k(α + c1β) =

k(α+ c2β) := L. Thus (c1 − c2)β ∈ L. Therefore β ∈ L. Hence α ∈ L. Thus

k(α, β) = k(α+c1β). Proceed inductively to show that E = k(α1, . . . , αn) =

k(α1 + c2α2 + · · ·+ αncn) for some c2, . . . , cn ∈ k.

Conversely, let E = k(α) for some α ∈ E and f(x) = irr(α, k). Let k ⊂ F ⊂E be a tower of fields. Set hF = irr(α, F ). Then hF | f(x) as F varies over

all the intermediate subfields of E/k.

Since hF is irreducible over F , it is also irreducible over F0, a subfield of F

generated by the coefficients of hF (x) over k. Since deg hF (x) = [E : F ] =

[E : F0], it follows that F = F0. Since there are finitely many divisors of

f(x), there can be only finitely many intermediate fields of E/k.

(2) Now let E/k be a finite separable extension. Then E = k(α1, α2, . . . , αn).

To show that E/k has a primitive element it is enough to find a primitive

74

element when n = 2 and then apply induction on n. So let E = k(α, β). We

look for a primitive element of the form α+ cβ where c ∈ k.

Let [E : k] = n. If α+cβ generates E/k, then α+cβ must have n conjugates

(images of α + cβ under the action of n embeddings of E into ka). Hence

there exist n k-embeddings σ1, σ2, . . . , σn : E → k. which map α + cβ to

n distinct roots of p(x) = irr(α + cβ), k) in k. Thus α + cβ is a primitive

element if and only if there exist n embeddings σ1, . . . σn : E → k such that

σi(α+ cβ) 6= σj(α+ cβ), for all i 6= j, if and only if∏

i<j

(σi(α)− σj(α)) + c(σi(β)− σj(β)) 6= 0

if and only if c is not a root of the polynomial

f(x) =∏

i<j

(σi(α)− σj(α)) + x(σi(β)− σj(β)).

Since k is infinite and f(x) has finitely many roots, such a c exists. �

79

Lecture 13 : Normal Extensions

Objectives

(1) Normal extensions and their examples.

(2) Characterization of normal extensions in terms of embeddings and

splitting fields.

Key words and phrases: Normal extensions, Galois extensions, Galois

group of a Galois extension.

12. Normal extensions

Suppose F is a field and E is a splitting field of f(x) ∈ F [x]. Let r1, r2, . . . , rnbe distinct roots of f(x) in F . Then E = F (r1, r2, . . . , rn). Suppose that

a ∈ E and g(x) = irr (a, F ). Let b ∈ F be another root of g(x). Then the

map σ : F (a) → F (b) given by σ(a) = b and σ(c) = c for all c ∈ F is an

F -embedding.

Let τ : E → F be an extension of σ. Then τ(ri) = rj for each i and some

j. Hence τ(E) ⊆ E. Since E is a finite dimensional F -vector space and τ is

injective, it is also surjective. Hence τ(E) = E. Therefore b ∈ E. This showsthat E contains splitting fields of irr (a, F ) for all a ∈ E. This property is

the defining condition for normal algebraic extensions.

Definition 12.1. An algebraic extension E/F is called a normal exten-

sion if whenever f(x) ∈ F [x] is irreducible and has a root in E then f(x)

splits into linear factors in E[x].

Example 12.2. (1) The algebraic closure F a of a field F is a normal ex-

tension of F.

(2) Let ζ be a primitive nth root of unity in C. Then Q(ζ) is a normal ex-

tension of Q. An element α ∈ Q(ζ) is of the form g(ζ) for some g(x) ∈ Q[x].

If β is a root of irr (α,Q) then there is a Q−isomorphism φ : Q(α)→ Q(β)

such that φ(α) = β. The isomorphism φ can be extended to an embedding

Φ : Q(ζ) → Q. But Φ(ζ) = ζm for some m. Hence β = φ(f(ζ)) = f(ζm) ∈

80

Q(ζ). Hence Q(ζ)/Q is a normal extension. Alternatively, note that Q(ζ) is

a splitting field of xn − 1 over Q.

(3) Every quadratic extension E/F is normal. Let a ∈ E\F. Then irr (a, F ) =

f(x) = x2 + bx + c for some b, c ∈ F. Let f(x) = (x − a)(x − s) for some

s ∈ E. Hence E is a splitting field of f(x) over F. Hence E/F is a normal

extension.,

(4) The extensions Q( 4√2)/Q(

√2) and Q(

√2)/Q are normal but the ex-

tension Q( 4√2)/Q is not a normal extension since the complex roots of

irr ( 4√2,Q) are not in Q( 4

√2).

(5) If E/F is a normal extension and K is an intermediate subfield of E/F

then E/K is a normal extension.

Lemma 12.3. Let E/F be an algebraic extension. Let σ : E → E be an

F−embedding, then σ is an automorphism of E.

Proof. We need to prove that σ(E) = E. Let a ∈ E and p(x) = irr (a, F ).

Let K be the subfield of E generated by the roots of p(x) in E. Then K is

a finite dimensional F -vector space. Since σ is an F−embedding, it maps

roots of p(x) to its roots. Hence σ(K) ⊆ K. Since σ is an injective F−linear map of the F−vector space K, dimF K = dimF σ(K). Hence σ is

surjective. �

Theorem 12.4. Let E/F be an algebraic extension such that E ⊂ F a. Then

the following conditions are equivalent:

(1) Every F -embedding σ : E → F a is an automorphism of E.

(2) E is a splitting field of a family of polynomials in F [x].

(3) E/F is a normal extension.

81

Proof.

F a

Eσ

// σ(E) = E

F (a)τ

// F (b)

Fid

// F

(1)⇒ (2): Let a ∈ E and pa(x) = irr (a, F ). If b ∈ F a is a root of pa(x) then

there is an F -isomorphism τ : F (a)→ F (b). The embedding τ : F (a)→ F a

can be extended to an embedding σ : E → F a. But σ(E) = E. Hence b ∈ E.Thus all roots of pa(x) are in E. Hence E is a splitting field of the family of

polynomials (pa(x)) a∈E .

(2) ⇒ (3): Let E be a splitting field of (pi(x))i∈I of polynomials in F [x].

Let a ∈ E and f(x) = irr (a, F ). Let b be any other root of f(x) in F a.

Then there is an F -isomorphism τ : F (a) → F (b) so that τ(a) = b. The

map τ can be extended to an F -embedding σ : E → F a. But σ maps roots

of (pi(x))i∈I to their roots. Hence σ(E) ⊂ E. Hence b ∈ E. Thus f(x) splitsinto linear factors in E[x].

(3) ⇒ (1) : Let σ : E → F a be an F -embedding. Let a ∈ E. Then

p(x) = irr (a, F ) splits into linear factors in E[x]. Since σ(a) is a root of

p(x), σ(a) ∈ E. Hence σ(E) ⊆ E. By Lemma 12.3, σ(E) = E. �

Proposition 12.5. Let E1, E2 be subfields of a field E. Let E1, E2 be normal

extensions of F. Then E1E2/F and E1 ∩ E2/F are normal.

Proof. Let E1 and E2 be normal extensions of F. Let σ : E1E2 → F a be

an F -embedding. Then σ(E1E2) = σ(E1)σ(E2) = E1E2. Similarly observe

that σ(E1 ∩ E2) = σ(E1) ∩ σ(E2) = E1 ∩ E2. Hence E1 ∩ E2 is a normal

extension of F.

�

82

Lecture 14 : Galois group of a Galois Extension I

Objectives

(1) Galois extension and the Galois group of a Galois extension.

(2) Galois group of a finite extension of finite fields and quadratic ex-

tensions.

(3) Galois groups of biquadratic extension.

(4) Galois group of a separable cubic polynomial.

(5) Fundamental Theorem of Galois theory (FTGT).

Keywords and phrases: Biquadratic and cubic extensions, fundamental

theorem of Galois Theory.

13. The Galois Group of a Field Extension

Definition 13.1. A field extension E/F is called a Galois extension if it

is normal and separable. The Galois group of a Galois extension E/F

denoted by G(E/F ) or Gal(E/F ) is the group of all F -automorphisms of E

under composition of maps.

Proposition 13.2. The Galois group of the Galois extension Fqn/Fq is a

cyclic group of order n generated by the Frobenious automorphism φ : Fqn →Fqn , defined as φ(a) = aq.

Proof. Note that φ is an Fq-automorphism since any a ∈ Fq is a root of

xq − x. Let G = 〈φ〉. Then φn(x) = xqn

= x. Therefore |G| ≤ n. Suppose

|G| = d. Then φd = id, so φ(x) = xqd

= x. But xqd −x has atmost qd roots.

Thus d = n.

We now show that G(Fqn/Fq) = 〈φ〉. Since Fqn/Fq is a separable extension,

[Fqn : Fq]s = n. Hence the number of Fq-automorphisms of Fqn is n whence

〈φ〉 = G(Fqn/Fq). �

Example 13.3. Quadratic extensions: Let K/F be a separable qua-

dratic extension. Then for any α ∈ K \ F we have irr (α, F ) = f(x) =

x2 + bx + c. Let β be another root of f(x). Then α + β = −b and αβ = c

and f(x) = (x − α)(x − β) ∈ K[x]. Hence K/F is a normal extension.

83

Let σ : K = F (α) → K be a K- automorphism different from idF . Then

σ(α) = β. Thus G(K/F ) = {idF , σ} is a group of order 2.

Example 13.4. Biquadratic extensions: A field extension K/F is called

biquadratic if [K : F ] = 4 and K is generated by roots of two irreducible

quadratic separable polynomials. Let K = F (α, β) and irr(α, F ) = x2 − a

and irr(β, F ) = x2 − b.

F (α, β)

IIIIIIIII

uuuuuuuuu

F (α)

JJJJJJJJJJF (β)

tttttttttt

F

Since [F (α, β) : F ] = 4, x2 − a is irreducible over F (β) and x2 − b is

irreducible over F (α). Any F - automorphism of K maps α to α or −α and

β to β or −β. Let σ(α) = −α, σ(β) = β and τ(α) = α, τ(β) = −β. Thenστ = τσ and σ2 = τ2 = id. Therefore

G(K/F ) = {id, σ, τ, στ = τσ}

is the Klein 4-group.

Example 13.5. The Galois group of a separable cubic : Let F be

a field of char 6= 2, 3. Consider an irreducible cubic polynomial f(x) =

x3 + px + q ∈ F [x]. Thus f(x) has no root in F . Let us observe that f(x)

is separable over F . Since f ′(x) = 3x2 + p we have

f =x

3(3x2 + p) +

2p

3x+ q

and hence

gcd(f, f ′) =

(

2p

3x+ q, 3x2 + p

)

.

Since f has no root in F, 2px/3 + q does not divide f(x). Hence (f, f ′) = 1

and so f(x) is separable. Thus a splitting field E of f must have degree 3 or

6. Let E = F (α1, α2, α3) where α1, α2, α3 are the roots of f(x) in E. Then

any F - automorphism σ permutes the roots α1, α2, α3.

84

Define ψ : G(E/F ) → S3 by ψ(σ) = pσ where pσ is the corresponding

permutation. It is easy to check that ψ is an injective group homomorphism.

Hence G(E/F ) ≃ S3 or A3. Let us see how disc (f(x)) determines the Galois

group. We identify G(E/F ) with a subgroup of S3. Let

δ = (α1 − α2)(α2 − α3)(α1 − α3).

Then δ2 = disc (f(x)) = −(4p3 + 27q2) ∈ F . Hence [F (δ) : F ] ≤ 2. If

disc (f(x)) is not perfect square in F then 2 | [E : F ]. hence G(E/F ) = S3.

If disc (f) is a square in F then δ ∈ F and hence G(K/F ) cannot have any

odd permutations since these do not fix δ. Thus G(E/F ) = A3. For example,

if f(x) = x3 + x + 1, then disc (f) = −31. Therefore G(E/F ) = S3. If

f(x) = x3 − 3x+ 1, then G(E/F ) = A3 as disc (f) = 34.

14. The Fundamental Theorem of Galois Theory

Let F be a field. We know that a Splitting field E of a polynomial

f(x) ∈ F [x] is a normal extension of F. If f(x) is separable then E/F is

separable. Thus a splitting field of a separable polynomial f(x) ∈ F [x] is aGalois extension of F. Conversely if E/F is a finite Galois extension then

by the Primitive Element Theorem there is an a ∈ E such that E = F (a).

Since E/F is normal, E is a splitting field of irr (a, F ). Thus a finite

extension E/F is Galois if and only if E is a splitting field of a separable

polynomial f(x) over F. We say in this case that G(E/F ) is the Galois

group of f(x). Since any two splitting fields of f(x) are F -isomorphic, we

write G(E/F ) = Gf .

Definition 14.1. Let G be a group of automorphism of a field E. Then

EG = {a ∈ E | σ(a) = a for all σ ∈ G}

is called the fixed field of G acting on E.

Theorem 14.2 (Fundamental Theorem of Galois Theory (FTGT)).

Let E/F be a finite Galois extension. Consider the sets:

I = {K | K is an intermediate field of E/F} and G = {H | H < G(E/F )}.

(i) The maps:

K 7→ G(E/K) and H 7→ EH

85

give a one-to-one correspondence, called the Galois correspondence be-

tween I and G.

(ii) K/F is Galois if and only if G(E/K)⊳G(E/F ) and in this case

G(K/F ) ≃ G(E/F )G(E/K) .

(iii) [E : K] = |G(E/K)|.

The FTGT will be proved in several steps. We shall prove parts of it for

infinite Galois extensions.

Theorem 14.3. Let E/F be a Galois extension with G = G(E/F ). Then

(1) F = EG.

(2) Let K be an intermediate subfield of E/F. Then E/K is Galois and the

map K 7→ G(E/K) is an injective map from I to G.

Proof. (1) Let a ∈ EG. Let σ : F (a) → F be an F -embedding. Let τ :

E → F be an extension of σ. Since E/F is Galois, τ is an automorphism

of E. Hence τ(a) = a. Therefore [F (a) : F ]s = 1. But E/F is separable, so

F (a)/F is also separable. Thus [F (a) : F ]s = [F (a) : F ] = 1. So a ∈ F.(2) Let K be an intermediate subfield of E/F. Then E/K is separable as

E/F is so. Let σ : E → K = F be a K-embedding. Then it is also an

F -embedding. As E/F is normal, σ is an automorphism of E. Thus E/K

is a Galois extension. Let H = G(E/K). Then by (1), we have K = EH .

Let K and K ′ be intermediate subfields of E/F. If H = G(E/K) and H ′ =

G(E/K ′) then K = EH and K ′ = EH′ . Hence the map K 7→ G(E/K) is an

injective map.

�

86

Lecture 15 : Galois group of a Galois Extension II

Objectives

(1) Artin’s Theorem about fixed field of a finite group of automorphisms.

(2) Behavior of Galois group under isomorphisms.

(3) Normal subgroups of the Galois groups and their fixed fields.

Keywords and phrases: Fixed field, Galois correspondence, normal sub-

groups of Galois group.

The next theorem is perhaps the most important ingredient of the Funda-

mental Theorem of Galois Theory (FTGT). We will need the following

Lemma 14.4. Let E/F be a separable algebraic extension. suppose that for

all α ∈ E, deg irr (α, F ) ≤ n. Then [E : F ] ≤ n.

Proof. Let β ∈ E be such that deg irr(β, F ) is maximal among deg irr(α, F )

for α ∈ E. We claim that E = F (β). Suppose E 6= F (β) and choose

α ∈ E \F (β). Then F (α, β) is a finite separable extension. By the Primitive

Element Theorem, there exists η ∈ F (α, β) such that F (α, β) = F (η). But

then deg η > deg β. �

The above lemma is not true without separability assumption. For example,

degF α ≤ p for all α ∈ k(u, v), where F = k(up, vp), where k is a field of

char p > 0. But [k(u, v) : k(up, vp)] = p2.

Theorem 14.5 (Emil Artin). Let E be a field and G a finite group of

automorphisms of E. Then

(1) E/EG is a finite Galois extension.

(2) G(E/EG) = G. (3) [E : EG] = |G|.

Proof. (1) Let α ∈ E and G = {σ1, σ2, . . . , σn} and S = {σ1(α), . . . , σn(α)}.Suppose |S| = r. Without loss of generality let S = {σ1(α), . . . , σr(α)}. If

87

τ ∈ G then τσ1(α), . . . , τσr(α) are distinct. Hence S = τ(S). So τ restricted

to S is a permutation of S. Consider the polynomial

f(x) = (x− σ1(α))(x− σ1(α))(x− σ2(α)) · · · (x− σr(α)).

The coefficient of f(x) are elementary symmetric functions of σ1(α) . . . , σr(α).

Since τ(S) = S these elementary symmetric functions are in EG. Thus

f(x) ∈ EG[x] is a separable polynomial and f(α) = 0. Hence E/EG is a sep-

arable and normal extension. Moreover for all α ∈ E, deg irr (α,EG) ≤ |G|.Hence [E : EG] ≤ |G|. Thus E/EG is a finite Galois extension.

(2) and (3) : Since E/EG is a finite separable extension, [E : EG] is the

number of EG-embeddings of E → Ea. These embeddings are automor-

phisms of E as E/EG is a normal extension. Using (1) and the fact that

G ⊆ G(E/EG), we get

|G| ≤ |G(E/EG)| = [E : EG] ≤ |G|.

Thus |G| = |G(E/EG)| = [E : EG] and so G = G(E/EG). �

Theorem 14.6. Let E/F be a Galois extension with Galois group G. Let

K1 and K2 be intermediate subfields of E/F and H1 = G(E/K1), H2 =

G(E/K2). Let (H1, H2) denote the smallest subgroup containing H1 and H2.

Then

K1K2 = EH1∩H2 , K1 ∩K2 = E(H1,H2), and K1 ⊆ K2 ⇐⇒ H1 ⊇ H2.

Proof. Since E/Ki is Galois for i = 1, 2, we have Ki = EHi ⊂ EH1∩H2 for

i = 1, 2. Therefore K1K2 ⊂ EH1∩H2 . Conversely, if σ ∈ G fixes K1K2 then it

fixes K1 and K2, consequently σ ∈ H1 ∩H2. Hence G(E/K1K2) ⊆ H1 ∩H2.

Hence K1K2 ⊇ EH1∩H2 . The remaining statements are obvious. �

Behavior of Galois groups under isomorphisms

Proposition 14.7. Let E/F be a Galois extension. Let λ : E → λ(E) be

an isomorphism of fields. Then

(1) λ(E)/λ(F ) is a Galois extension.

(2) G(λ(E)/λ(F )) = λG(E/F )λ−1 ≃ G(E/F ).

88

Proof. (1) Since E/F is Galois, E is a splitting field of a family of separable

polynomials {fi(x) ∈ F [x] | i ∈ Λ}. Then λ(E) is a splitting field of of the

family of polynomials: {fλi (x) ∈ λ(F )[x] | i ∈ Λ}. Hence λ(E) is a Galois

extension of λ(F ).

(2) Define ψ : G(E/F )→ G(λE/λF ) by the rule σ 7→ λσλ−1.

λ(E)λ−1

// Eσ

// Eλ

// λ(E)

λ(F )λ−1

// F // Fλ

// λ(F )

The inverse of ψ is given by the rule τ 7→ λ−1τλ. Hence ψ is an isomorphism.

�

Theorem 14.8. Let E/F be a Galois extension. Let K be an intermediate

subfield of E/F . Then

(1) K/F is Galois if and only if G(E/K)⊳G(E/F ).

(2) If K/F is Galois, then G(K/F ) ≃ G(E/F )/G(E/K)

Proof. (1) and (2) : Let K/F be Galois. Define

ψ : G(E/F )→ G(K/F ) by ψ(σ) = σ|K .

Since K is a normal extension of F, σ|K ∈ G(K/F ). Since

Kernel ψ = {σ ∈ G(E/F ) | σ|K = idK} = G(E/K),

G(E/K) is a normal subgroup of G(E/F ).

Conversely, let G(E/K)⊳G(E/F ). Let λ : E → E be any F -automorphism.

We show that λK = K. Now

λG(E/K)λ−1 = G(E/λK) = G(E/K),

provided λ ∈ G(E/F ). Thus λK = K. Let σ : K → F a be an F -embedding.

Then σ can be extended to an embedding τ : E → F a. Since E/F is Galois,

τ(E) = E. Thus σ(K) = K. Hence K/F is Galois.

�

96

Lecture 16 : Applications and Illustrations of the FTGT

Objectives

(1) Fundamental theorem of algebra via FTGT.

(2) Gauss’ criterion for constructible regular polygons.

(3) Symmetric rational functions.

(4) Galois group of some binomials.

Keywords and phrases: Fundamental theorem of algebra, constructible

regular polygons, symmetric rational functions.

15. Applications and Illustrations of the FTGT

The Galois correspondence between the set of subfields of a finite Galois

extension E/F and the set of subgroups of the Galois group G(E/F ) con-

verts problems about roots of a separable polynomial to problems about the

Galois group of its splitting field. We shall see that difficult problems about

polynomials are converted into much simpler problems about finite groups.

The Galois correspondence is perhaps the first example of a well-established

technique in mathematics: find a suitable formulation for a problem in one

branch of mathematics in another branch where the problem becomes much

easier to solve.

We will see that the Galois correspondence is powerful enough to provide

new ways to prove old results and solve new problems as well. This will

be demonstrated here by giving a new proof of the fundamental theorem of

algebra. We will also finish the proof of Gauss’ criterion for constructibility

of regular polygons. We shall derive an expression for cos 2π/17 in terms

of square roots which proves that a seventeen sided regular polygon is con-

structible by ruler and compass.

We will provide concrete examples of Galois correspondence for some poly-

nomials. In later sections we will derive formulas for the roots of cubic and

quartic polynomials as a consequence of the Galois correspondence. Let us

begin by proving:

97

The Fundamental Theorem of Algebra

Theorem 15.1. The field of complex numbers is algebraically closed.

Proof. Let f(x) =∑

aixi ∈ C[x]. Write f(x) =

∑

aixi where − denotes the

complex conjugation. Then g(x) = f(x)f(x) ∈ R[x]. Hence it is enough to

prove g(x) has a complex root.

The splitting field E of g(x) over C is a splitting field of (x2 + 1)g(x) over

R. Hence E/R is a Galois extension. Since 2 | [E : R], the Galois group

G = G(E/R) has a 2- Sylow subgroup say S. If S < G then E ⊃ ES ⊃ R.

We know [E : ES ] = |S|. Thus [ES : R] is odd. But R admits no proper odd

degree algebraic extensions. Hence S = G. Thus G is a 2-group. If |G| = 2,

then E = C and we are done. If |G| = 4, then [E : C] = 2. But C admits

no quadratic extension. Thus |G| ≥ 8. Let H < G(E/C) of index 2. Then

[EH : C] = 2, which is a contradiction. Hence E = C. �

Gauss’ Criterion for Constructible Regular Polygons

Lemma 15.2. Let m,n be coprime natural numbers. If regular polygons of

m sides and n sides are constructible then so is a regular mn-gon.

Proof. There exist integers x, y so that xm+ yn = 1. Hence

2π

mn=

2πx

n+

2πy

m.

Since 2πx/n and 2πy/m are constructible, so is 2π/mn. �

Proposition 15.3. Let ζ be a complex primitive pth root of unity where p

is a prime number. Then G(Q(ζ)/Q) is a cyclic group of order p− 1.

Proof. If σ ∈ G, then σ restricted to the cyclic group U = (ζ) is an auto-

morphism. Hence σ(ζ) = ζiσ for some i = 1, 2, . . . , p − 1. Define a group

homomorphism ψ : G → U(Z/pZ) = {1, 2, . . . , p − 1} by ψ(σ) = iσ. It is

easy to see that ψ is an isomorphism. �

Theorem 15.4 (Gauss). A regular polygon of n sides is constructible if

and only if n = 2rp1p2 . . . ps where r ∈ N and p1, p2, . . . , ps are distinct

Fermat primes.

98

Proof. We have already proved the necessity. For sufficiency, note that by

the above lemma and the fact that angles can be bisected by ruler and

compass, it is enough to prove that if p is a Fermat prime then cos(2π/p)

is a constructible real number. Let ζ be a primitive pth root of unity. Then

[Q(ζ) : Q] = p − 1 = 2t for some t, the Galois group G = G(Q(ζ)/Q) is

cyclic of order 2t. Hence every intermediate subfield of Q(ζ)/Q is a Galois

extension of Q. In particular K = Q(cos 2π/p) is a Galois extension of Q

of degree 2t−1. Since G(K/Q) is a 2-group of order 2t−1, there a chain of

subgroups Gi having order 2i for i = 0, 1, . . . , t− 1. Hence

Q ⊂ KGt−2 ⊂ KGt−3 ⊂ · · · ⊂ KG0 = K

is a tower of real quadratic extensions terminating with K. Hence cos 2π/p

is a constructible real number. �

Example 15.5. Let K be a splitting field of x4 − 2 over Q. We find the

Galois group G = G(K/Q) and show how to find subfields of K/Q.

The polynomial f(x) = x4−2 is irreducible over Q by Eisenstein’s criterion.

Let a = 4√2 be the real 4th root of 2. Then the roots of f(x) in C are

a,−a, ia,−ia. The splitting field of f(x) over Q is Q(a, i) and [K : Q] = 8.

Hence G = G(K/Q) is a group of order 8. An automorphism in G maps a

to one of the four roots of f(x) and it maps i to either i or −i. Let τ be the

conjugation map and σ be defined by σ(a) = ia. Check that

o(σ) = 4, o(τ) = 2 and στστ = id.

The lattice of the subgroups of G is:

G

mmmmmmmmmmmmmm

QQQQQQQQQQQQQQQ

{1, σ2, τ, σ2τ}

qqqqqqqqqq

PPPPPPPPPPPP

{1, σ, σ2, σ3} {1, σ2, στ, σ3τ}

mmmmmmmmmmmmm

OOOOOOOOOOO

{1, τ}

WWWWWWWWWWWWWWWWWWWWWWWWWWWWWW {1, σ2τ}

PPPPPPPPPPPPPP

{1, σ2} {1, στ}

mmmmmmmmmmmmmmm

{1, σ3τ}

ggggggggggggggggggggggggggggggg

{1}

99

By Galois correspondence, there are 10 intermediate subfields ofK/Q. These

are all fixed fields of the subgroups displayed above. Set H = {1, σ, σ2, σ3}.Since [K : KH ] = o(H) = 4 we see that [KH : Q] = 2. Since i is fixed

by each element of H, we conclude that KH = Q(i). Set L = {1, τ}. Since[K : KL] = o(L) = 2, we see that [KL : Q] = 4. Since τ(a) = a, KL = Q(a).

Set M = {1, στ}. Since [K : KM ] = o(M) = 2, [KM : Q] = 4. The orbit

of a under the action of M is {a, ia}. Adding the elements of this orbit we

get b = a + ia. Hence a + ia ∈ KM . To find g(x) = irr (b,Q), we find all

the conjugates of b by applying the automorphisms in G. This way we see

that the orbit of b under the action of G is {b,−b, a − ia,−a + ia}. Hence

degQ(b) = 4. Hence KM = Q(b). The other fixed fields can be found by

similarly.

Example 15.6. We discuss the Galois group of xp − 2, where p is an odd

prime, We will show that it is isomorphic to the group

G =

{[

a b

0 1

]

: a, b ∈ Fp and a 6= 0

}

.

Put ω = e2πi/p and α = p√2. The roots of xp − 2 are α, αω, αω2, . . . , αωp−1.

Thus K = Spl(xp− 2,Q) = Q(α, ω) and [K : Q] = p(p− 1). If σ ∈ G(K/Q),

then σ(α) = αωi(σ) and σ(ω) = ωj(σ), where 1 ≤ j(σ) ≤ (p − 1) and

i(σ) = 0, 1, . . . , (p− 1). Define

ψ : G(K/Q)→ G by ψ(σ) =

[

j(σ) i(σ)

0 1

]

.

Define σ, τ ∈ G(K/Q) by

τ(α) = αωa, τ(ω) = ωb, σ(α) = αωc, and σ(ω) = ωd.

Therefore

ψ(σ) =

[

d c

0 1

]

, ψ(τ) =

[

b a

0 1

]

, ψ(σ)ψ(τ) =

[

bd c+ ad

0 1

]

.

Since

τσ(α) = τ(αωa) = αωc+ad

τσ(ω) = τ(ωd) = ωbd

100

we have

ψ(τσ) =

[

bd c+ ad

0 1

]

= ψ(τ)ψ(σ).

Therefore ψ is a group homomorphism. As

Ker ψ = {σ : d = 1 and c = 0} = {id},

we conclude that ψ is an isomorphism.

Example 15.7. Let x1, x2, . . . , xn be indeterminates over a field F . The

symmetric group Sn acts on E = F (x1, x2, . . . , xn), the fraction field of the

ring of polynomials F [x1, . . . , xn]. If σ ∈ Sn then φσ : E → E defined

by φσ(xi) = xσ(i) is an automorphism of E. If σ1, σ2 ∈ Sn then φσ1σ2=

φσ1φσ2

. Thus G = {φσ : σ ∈ Sn} is a group of automorphism of E and it is

isomorphic to Sn. Let x be a variable over E and consider the polynomial

ring E[x]. Then

g(x) = (x− x1)(x− x2) · · · (x− xn) ∈ E[x]

= xn − σ1xn−1 + σ2xn−2 − · · ·+ (−1)nσn

Where σi’s are the elementary symmetric functions of x1 . . . , xn. The auto-

morphism φσ : E → E can be extended to E[x] by fixing x which we again

denote by φσ. Therefore

φσ(g(x)) = (x− xσ(1))(x− xσ(2)) · · · (x− xσ(n)) = g(x)

Thus φσ(σi) = σi for all i = 1, 2, . . . , n. Hence F (σ1, σ2, · · · , σn) ⊂ EG.

Notice that E = F (σ1, . . . , σn, x1, . . . xn). So E is a splitting field of g(x)

over F (σ1, . . . , σn) and g(x) is separable. If π ∈ G(E/F (σ1, . . . , σn)) then

π permutes the roots of g(x), hence π = φσ for some σ. Thus G =

G(E/F (σ1 . . . , σn)). Therefore symmetric rational functions are rational

functions of symmetric functions.

101

Lecture 17 : Cyclotomic Extensions I

Objectives

(1) Roots of unity in a field.

(2) Galois group of xn − a over a field having nth roots of unity.

(3) Irreduciblilty of the cyclotomic polynomial Φn(x) over Q.

(4) A recursive formula for Φn(x).

Keywords and phrases : Roots of unity, Galois group of xn− a, cyclo-tomic polynomials.

16. Cyclotomic Extensions

Roots of unity in any field

Let F be a field. A root z ∈ F of xn − 1 is called an nth root of unity in

F. Roots of unity play important role in algebra and number theory. Their

analysis led Gauss to his first mathematical discovery: construction of a

regular polygon of 17 sides.

Suppose that char F = p and n = pem where (m, p) = 1. Then xn − 1 =

(xm − 1)pe

. By the derivative criterion, xm − 1 is separable. Hence the

splitting field of xn − 1 is equal to that of xm − 1. Therefore we consider

fields of chararacteristic zero or of characteristic p where (p, n) = 1.

Let k be a field and (n, char k) = 1. Then xn − 1 is separable by the

derivative criterion. Let Z = {z1, z2, . . . , zn} be the set of its roots of in ka.

Then Z is a multiplicative subgroup of (ka)×. Hence it is cyclic. Any of

the ϕ(n) generators of Z is called a primitive nth root of unity. Let z

be any such generator. Then k(z) is a splitting field of xn − 1 over k. Let

Φn(x) = irr (z,Q). A primitive nth root of unity over Q is denoted by ζn.

Definition 16.1. A splitting field of xn − 1 over F is called a cyclotomic

field of order n over F .

102

Proposition 16.2. Let (char F, n) = 1 and f(x) = xn−1 ∈ F [x]. Then Gf

is isomorphic to a subgroup of U(n). In particular Gf is an abelian group

and o(Gf ) | ϕ(n).

Proof. As f(x) is separable, it has n distinct roots. Let {z1, z2, . . . , zn} = Z

be the set of roots of f(x) in F a and E = F (z1, z2, . . . , zn). Since Z ⊆ E× is

a subgroup, it is cyclic. The map ψ : G(E/F )→ Aut(Z) such that σ 7→ σ|Zis an injective group homomorphism. Since Aut(Z) ≃ {m | (m,n) = 1} :=U(n) is an abelian group, G(E/F ) is also an abelian group whose order

divides ϕ(n). �

Example 16.3. Let F = F2. Then x3− 1 = (x− 1)(x2+x+1). Any root z

of x2 + x+ 1 is a primitive cube root of unity over F. Hence [F (z) : F ] = 2.

To find the degree of a primitive seventh root of unity over F, consider the

factorization of x7 − 1 into irreducible polynomials over F :

x7 − 1 = (x− 1)(x3 + x2 + 1)(x3 + x+ 1).

Therefore there are 6 primitive 7th roots of unity over F with two minimal

polynomials. In contrast to this, we shall see that all the primitive nth

roots of unity over Q have the same irreducible polynomial called the nth

cyclotomic polynomial Φn(x).

Proposition 16.4. Let xn − a = f(x) ∈ F [x] and suppose F has n distinct

roots of xn − 1. Then Gf is a cyclic group and o(Gf ) divides n.

Proof. Let Z = {z1, z2, . . . , zn} ⊂ F be the set of roots of xn− 1. Let r be a

root of f(x) in a splitting field E of f(x). Then rz1, rz2, . . . , rzn are roots of

f(x). Thus E = F (r). Let σ, τ ∈ G(E/F ). Then σ(r) = zσr and τ(r) = zτr

for some zσ, zτ ∈ Z. Hence στ(r) = σ(zτr) = zτzσr. Define

ψ : G(E/F )→ Z such that ψ(σ) = zσ.

Then ψ is a group homomorphism. The map φ is clearly injective. Since Z

is a subgroup of F×, it is a cyclic group of order n. Hence |Gf | divides nand Gf is cyclic.

�

103

Theorem 16.5. (1) Φn(x) ∈ Z[x], (2) [Q(ζn) : Q] = ϕ(n) = degΦn(x) and

(3) G(Q(ζn)/Q) ≃ Un.

Proof. Let Φn(x) = irr (ζn,Q). Then xn − 1 = Φn(x)h(x), where h(x) is

monic in Q[x]. By Gauss’ Lemma Φn(x), h(x) ∈ Z[x]. We show that each

primitive nth root of unity is a root of Φn(x). Let p be a prime number and

(p, n) = 1. Suppose Φn(ζpn) 6= 0. Hence h(ζpn) = 0. Hence ζn is a root of

h(xp). Thus

h(xp) = Φn(x)g(x) for some monic g(x) ∈ Z[x].

Reduce mod p to get

(h(x))p = Φn(x)g(x) ,

where “–” denotes reduction of coefficients mod p. Hence Φn(x) and h(x)

have a common root mod p. But xn − 1 has distinct roots over Fp. Hence

ζpn is a root of Φn(x). Hence all primitive nth roots of unity are roots

of Φn(x). Since G = G(Q(ζn)/Q) is isomorphic to a subgroup of U(n),

[Q(ζn) : Q] = |G| ≤ ϕ(n). But deg Φn(x) ≥ ϕ(n). Hence |G| = ϕ(n). Hence

G ≃ U(n). �

Computation of Cyclotomic Polynomials

Let ζn be a primitive nth root of unity. Then the other roots of Φn(x) are

ζin such that (i, n) = 1. Thus

Φn(x) =∏

(i,n)=1

(x− ζin).

Since the roots of xn − 1 form a cyclic group of order n, the order of any

root divides n. Since Φd(x) =∏

o(z)=d(x− z), it follows that

xn − 1 =∏

d|n

Φd(x).

Therefore

Φn(x) =xn − 1

∏

d|n,d<nΦd(x).

104

This is a recursive formula for computation of Φn(x). First few cyclotomic

polynomials are:

Φ1(x) = x− 1

Φ2(x) =x2 − 1

Φ1(x)= x+ 1

Φ3(x) =x3 − 1

Φ1(x)= x2 + x+ 1

Φ4(x) =x4 − 1

Φ1(x)Φ2(x)= x2 + 1

Φ5(x) = x4 + x3 + x2 + x+ 1

Φ6(x) =x6 − 1

Φ1(x)Φ2(x)Φ3(x)= x2 − x+ 1

105

Lecture 18 : Cyclotomic Extensions II

Objectives

(1) Discriminant of Φp(x).

(2) Subfields of Q(ζp).

(3) Kronecker-Weber Theorem for quadratic extensions of Q.

(4) Algorithm for construction of primitive elements of subfields ofQ(ζp).

(5) Subfields of Q(ζ7), Q(ζ13) and Q(ζ17).

Keywords and phrases : Discriminant of Φp(x), Kronecker-Weber The-

orem, subfields of Q(ζp).

17. Subfields of Q(ζp)

A celebrated theorem of Kronecker and Weber states that a Galois extension

E of Q with abelian Galois group is contained in a cyclotomic extension (an

extension of Q obtained by adjoining roots of unity.) We will prove this

theorem for quadratic extensions of Q. For this purpose, we show that the

square root of the discriminant of Φp(x) is a primitive element of the unique

intermediate subfield of K of Q(ζp) so that [K : Q] = 2.

Lemma 17.1. Let p be an odd prime. Then disc (Φp(x)) = (−1)(p

2)pp−2.

Proof. Let ζp be a primitive pth root of unity. Since xp − 1 = Φp(x)(x− 1),

and pxp−1 = Φp(x) + (x− 1)Φp(x), we have for each i = 1, 2, . . . , p− 1,

p(ζip)p−1 = (ζip − 1)Φp(ζ

ip).

Therefore

p−1∏

i=1

Φp(ζip) =

p−1∏

i=1

p(ζip)p−1/(ζip − 1)

=pp−1

∏p−1i=1 (ζ

ip − 1)

=pp−1

(−1)p−1Φp(1)= pp−2.

106

Using the formula for discriminant in terms of derivatives, we get

disc (Φp(x)) = (−1)(p

2)pp−2

�

Proposition 17.2. The field Q(ζp) contains a unique quadratic extension

of Q, namely

Q(√

disc (Φp(x)))

= Q

(√

(−1)(p

2)p

)

which is real if p ≡ 1(mod 4) and complex if p ≡ 3(mod 4).

Proof. The Galois groupG ofQ(ζp) overQ is cyclic of order p−1.Hence there

is a unique subgroup of G having index 2. Thus there is a unique subfield

of Q(ζp) which is a quadratic extension of Q. As√

disc (Φp(x)) ∈ Q(ζp)\Qit generates the unique quadratic subfield of Q(ζp). �

Corollary 17.3. Every quadratic extension of Q is contained in a cyclo-

tomic extension.

Proof. If p ≡ 3(mod 4), then Q(√−p) ⊆ Q(ζp) and if p ≡ 1(mod 4) then

Q(√p) ⊆ Q(ζp). A quadratic extension of Q is of the form Q(

√d) where

d is a squarefree integer. Suppose d = ±p1p2 . . . pr where p1, p2, . . . , pr are

distinct primes. Then Q(√d) ⊆ Q(ζp1 , ζp2 , . . . , ζpr , i).

�

Proposition 17.4. Let L ⊂ Q(ζp) be a subfield with [Q(ζp) : L] = 2. Then

L = Q(ζp + ζ−1p ).

Proof. As ζp is a root of x2− (ζp+ ζ−1p )x+1 = 0,

[

Q(ζp) : Q(ζp + ζ−1p )]

≤ 2.

Since L = Q(ζp + ζ−1p ) ⊆ R, we conclude that [Q(ζp) : L] = 2. �

Proposition 17.5. Let p be a prime number. Let ζ be a primitive pth

root of unity. Let H be a subgroup of G = G(Q(ζ)/Q) = U(p). Put βH =∑

σ∈H σ(ζ). Then

EH = Q(βH).

Proof. Let τ ∈ H. Since H is finite, H = {τσ | σ ∈ H}. Hence τ(βH) = βH

for all τ ∈ H. Hence Q(βH) ⊆ Q(ζ)H . Let τ /∈ H.We show that τ(βH) 6= βH .

107

The set

B = {1, ζ, ζ2, . . . , ζp−2}is a basis of the Q-vector space Q(ζ). If τ(βH) = βH then ζ = τσ(ζ) for

some σ ∈ H. Hence τσ = 1 and so τ−1 = σ. Therefore τ ∈ H which is

a contradiction. If Q(βH) 6= Q(ζ)H , then by FTGT, there is a subgroup

M > H such that Q(βH) = Q(ζ)M ( Q(ζ)H . Hence βH is fixed by an

element τ ∈M \H. This is a contradiction.

�

Example 17.6. Let p = 7 and ζ7 = w. Then[

Q(w + w−1) : Q]

= 3 Let us

find the irreducible polynomial of w + w−1 = w + w6. To do this find the

orbit of w+w6 under the action of the Galois group G = G(Q(w)/Q). G is

is generated by the automorphism σ(w) = w2. Hence The orbit of w + w6

under the action of G is {β1 = w+w6, β2 = w2+w5, β3 = w4+w3}. Hence

irr (w + w6,Q) =3∏

i=1

(x− βi) = x3 + x2 − 2x− 1.

Example 17.7. Put ζ13 = ζ. We list all subfields of E = Q(ζ) by using

the procedure in the proposition above. Since Galois group G of the Galois

extension E/Q is cyclic of oder 12 it has proper subgroups of orders 2, 3, 4,

and 6. The automorphism σ(ζ) = ζ2 generates G. The action of powers of

σ on ζ is described in the table:

i 1 2 3 4 5 6 7 8 9 10 11

σi(ζ) = ζ2 ζ4 ζ8 ζ3 ζ6 ζ12 ζ11 ζ9 ζ5 ζ10 ζ7

The unique quadratic extension of Q in E is Q(√13). The unique subfield of

degree 6 is Q(ζ + ζ12). The subgroup H of oder 4 is generated by σ3. Hence

H = {σ3, σ6, σ9, id}. Hence a primitive element of the degree 3 extension of

Q in E is

βH = ζ + σ3(ζ) + σ6(ζ) + σ9(ζ) = ζ + ζ8 + ζ12 + ζ5.

The subgroup K of G of order 3 is generated by σ4. Hence a primitive

element of Q(ζ)K is

βK = ζ + σ4(ζ) + σ8(ζ) = ζ + ζ3 + ζ9.

Hence the poset of intermediate subfields of Q(ζ) is

108

E = Q(ζ)

ooooooooooo

OOOOOOOOOOO

EK = Q(βK) Q(ζ + ζ−1)

Q(√13)

OOOOOOOOOOOOO

EH = Q(βH)

ooooooooooooo

Q

Example 17.8. Let E be the splitting field of x17 − 1 over Q generated by

a primitive seventeenth root z of 1. So irr(z,Q) = x16 + x15 + · · · + x + 1

and E = Q(z). Therefore [E : Q] = 16. Thus |G(E/Q)| = |U(17)| = 16.

The multiplicative group of units mod 17 can be generated by 3+(17). Thus

η : z → z3 is a generator of G(E/Q) = {η, η2, . . . , η16 = 1}. The subgroups

of G and their orders are:

G = G1 = 〈η〉 ⊃ G2 = 〈η2〉 ⊃ G3 = 〈η4〉 ⊃ G4 = 〈η8〉 ⊃ {id}

|G1| = 16, |G2| = 8, |G3| = 4, and |G4| = 2.

The chain of intermediate subfields is:

EG = Q ⊂ EG2 ⊂ EG3 ⊂ EG4 ⊂ E.

We determine the generators for these fixed fields. Note that

η(z) = z3, η2(z) = z32

, . . . , ηi(z) = z3i

.

Let

x1 =

8∑

i=1

(η2)i(z), y1 =

4∑

i=1

(η4)i(z) and z1 =

2∑

i=1

(η8)i(z).

The fixed fields are

EG2 = Q(x1) ⊂ EG3 = Q(x1, y1) ⊂ EG4 = Q(x1, y1, z1).

110

Lecture 19 : Abelian and Cyclic Extensions

Objectives

(1) Infinitude of primes p ≡ 1 ( mod n ).

(2) Inverse Galois problem for finite abelian groups.

(3) Structure of some cyclic extensions.

Keywords and phrases : Primes of the form p ≡ 1 ( mod n ), abelian

extension, cyclic extension, inverse Galois problem.

18. The Inverse Galois Problem for Finite Abelian Groups

A Galois extension E/F is called abelian (resp. cyclic) if G(E/F ) is

abelian (resp. cyclic). In this section we will show that any finite abelian

group is the Galois group of a Galois extension of Q. In other words, any

finite abelian group is the Galois group of a polynomial with rational coef-

ficients. A proof of this theorem requires the theorem from number theory

that there are infinitely many primes p ≡ 1( mod n). We shall prove this

using cyclotomic polynomials. This is a special case of Dirichlet’s theorem

about infinitude of primes in the arithmetic progression a+nb where a, b are

coprime natural numbers and n = 1, 2, 3, . . . . We will also construct cyclic

extensions of fields having enough roots of unity.

Lemma 18.1. Let p be a prime number and n be relatively prime to p. Let

Φn(x) have a root in Fp. Then p ≡ 1 (mod n).

Proof. Let k ∈ Z, k ∈ Fp and Φn(k) = 0. Then p | Φn(k). Hence p | kn − 1.

Thus kn ≡ 1(mod p). We claim that o(k) = n in the group (Fp)×. Suppose

o(k) = m < n. Then km = 1. Hence

xn − 1 =∏

d|n

Φd(x) = Φn(x)∏

d<n

Φd(x)

= Φn(x)∏

d|m

Φd(x)h(x)

= Φn(x)(xm − 1)h(x)

111

Hence kn − 1 = Φn(k)(km − 1)h(k). This means xn − 1 has a multiple root

in Fp. This is a contradiction. Hence o(k) = n. Hence n | p − 1. Thus

p ≡ 1 mod n. �

Theorem 18.2. There are infinitely many primes p ≡ 1 (mod n).

Proof. Suppose to the contrary, p1, p2, . . . , pg are all such primes. Let m =

np1p2 . . . pg. Since Φm(x) ∈ Z[x], is monic, limx→∞Φm(mx) = ∞. Hence

there exists k such that Φm(mk) ≥ 2. Let p be a prime factor of Φm(mk).

Then p | (mk)m − 1. Hence p does not divide mk. Hence (p, n) = 1 and

p 6= p1, . . . , pn. Moreover Φm(mk) = 0. Hence p ≡ 1(mod n). This is a

contradiction. �

Theorem 18.3. Let G be a finite abelian group. Then there is a Galois

extension K/Q such that G(K/Q) = G.

Proof. We may assume that |G| ≥ 2. Then G ≃ Cn1× · · · × Cnk

. where

|G| = n = n1n2 . . . nk and n1|n2| · · · |nk. There exist infinitely many primes

pi ≡ 1 (mod ni) for i = 1, 2, . . . , k. We can find subgroups H1 < U(p1), H2 <

U(p2), . . . , Hk < U(pk) such that

U(p1)

H1≃ Cn1

,U(p2)

H2≃ Cn2

, . . . ,U(pk)

Hk≃ Cnk

.

U(p1)× U(p2)× · · · × U(pk)

H1 ×H2 × · · · ×Hk≃ Cn1

× · · · × Cnk.

Let H < U(n) and H ≃ H1×H2×· · ·×Hk. ThenU(n)H ≃ G. By the FTGT

G(Q(ζn)H/Q) =

U(n)

H≃ G.

�

19. Cyclic Galois Extensions

In this section we discuss cyclic extensions of degree n if F has a primitive

nth root of unity or when F has characteristic p > 0 and E/F has degree p.

There is no simple description of cyclic extensions of Q or fields devoid of

roots of unity. We need a theorem of Dedekind about linear independence

over K of automorphims of a field K.

112

Definition 19.1. Let G be a group and K a field. By a character of

G in K we mean a homomorphism χ : G → K×. We say that characters

χ1, χ2, . . . , χn : G → K× are linearly independent if for a1, . . . , an ∈ K

a1χ1 + a2χ2 + · · ·+ anχn = 0 if and only if ai = 0 for i = 1, 2, . . . , n.

Theorem 19.2 (Dedekind). Let χ1, χ2, . . . , χn be distinct characters of a

group G in a field K. Then χ1, χ2, . . . , χn are linearly independent.

Proof. Apply induction on n. If n = 1, then χ1 : G→ K× is clearly linearly

independent. Let n ≥ 2. Let n be the smallest positive integer such that

there exist a1, . . . , an ∈ K, not all zero with

a1χ1 + · · ·+ anχn = 0.(2)

Then ai 6= 0, for all i. Since χ1 6= χ2, there exists z ∈ G such that χ1(z) 6=

χ2(z). Hence for all x ∈ G,

a1χ1(xz) + a2χ2(xz) + · · ·+ anχn(xz) = 0(3)

a1χ1(z)χ1 + a2χ2(z)χ2 + · · ·+ anχn(z)χn = 0.(4)

Multiply (2) by χ1(z) and subtract (4) to get the relation :

(χ1(z)−χ2(z))a2χ2 + (χ1(z)−χ3(z))a3χ3 + · · ·+ (χ1(z)−χn(z))anχn = 0.

The above relation has smaller length, which is a contradiction.

�

Lemma 19.3. Let F be a field containing a primitive nth root of unity ζ.

Suppose that E/F is a Galois extension of degree n and G = G(E/F ) = (σ).

Then ζ is an eigenvalue of σ.

Proof. The field E is an n-dimensional F -vector space. Since σ has order n,

σ satisfies xn − 1 = 0. If σ is a root of a polynomial f(x) ∈ F [x] of degree

m < n then σ, σ2, . . . , σm are linearly dependent over F. This contradicts

Dedekind’s Theorem. Hence the minimal and the characteristic polynomials

of σ are equal to xn − 1. Hence ζ is an eigenvalue of σ.

�

We now describe the structure of cyclic extensions of degree n over a field

having a primitive nth root of unity.

113

Theorem 19.4. Let E/F be a cyclic extension of degree n with G =

G(E/F ) = (σ) and let ζ ∈ F be a primitive nth root of unity. Then there

exists a b ∈ F so that E = F (a) where an = b.

Proof. Since ζ is an eigenvalue of σ, there exists an eigenvector a ∈ E× so

that σ(a) = ζa. Hence σi(a) = ζia for all i = 1, 2, . . . , n. Hence a has at

least n conjugates in E. As E/F is a Galois extension of degree n, and E

contains a splitting field of f(x) = irr (a, F ), it follows that E = F (a) and

an ∈ F since σ(an) = ζnan = an. �

Intermediate subfields of a cyclic Galois extension

Let E/F be a cyclic Galois extension of degree n where F has a primitive nth

root of unity. We have proved that E = F (a) where an ∈ F. The number of

subgroups of the Galois group G = G(E/F ) is d(n), the number of divisors

of n. Each of these subgroups is cyclic. Hence there are d(n) intermediate

subfields of E/F. We show that they are F (ad) where d is a divisor of n.

Proposition 19.5. Let E/F be a cyclic Galois extension of degree n where

F has a primitive nth root of unity. Let E = F (a) where an ∈ F. Then The

intermediate subfields of E/F are F (ad) where d is a divisor of n.

Proof. The Galois group G has unique subgroup of order d for every divisor

d of n. Hence E/F has a unique subfield of degree d for each divisor d of n.

Consider the subfield K = F (ad). Then a is a root of xd − ad ∈ K[x]. Thus

[E : F (ad)] ≤ d. Since an ∈ F, we have (ad)n/d ∈ F. Hence [F (ad) : F ] ≤ n/d.

It follows that [E : F (ad)] = d. Hence the intermediate subfields of E/F are

F (ad) where d varies over the divisors of n.

�

114

Lecture 20 : Cyclic Extensions and Solvable Groups

Objectives

(1) Cyclic extensions of degree p over fields with characteristics p.

(2) Solvable groups.

(3) Simplicity of Sn and An.

Keywords and phrases : Cyclic extensions, solvable groups, commuta-

tor, simple groups.

Structure of cyclic Galois extensions over fields of characteristic p

Let F be a field of positive characteristic p. We discuss the structure of

Galois extensions of F of degree p. Consider the map δ : F → F defined by

δ(a) = ap−a. Then δ is a homomorphism of the additive group F. Moreover

Fp = Ker δ. Note that δ−1(a) = {a+ i | i = 0, 1, . . . , p− 1}.

Theorem 19.6 (Artin-Schreier). (1) Let E/F be a cyclic Galois exten-

sion of degree p where char F = p, a prime number. Then E = F (a) where

a is a root of xp − x− b for some b ∈ F.

(2) Suppose that a /∈ F p − F. Then f(x) = xp − x− a is irreducible over F

and a splitting field of f(x) over F is cyclic of degree p.

Proof. (1) Let G = G(E/F ) = (σ) and let T : E → E be the linear map of

the F -vector space E defined by T (a) = σ(a)− a. Then

Ker T = {a ∈ E | σ(a) = a} = F.

Since T p = (σ − id)p = σp − id = 0, we have Im (T p−1) ⊂ Ker T = F. If

T p−1 = 0 then there is a nontrivial F -linear relation among σp−1, σp−2, . . . , σ, id.

This contradicts Dedekind’s theorem. Hence Im T p−1 = Ker T = F. Let

b ∈ E so that T p−1(b) = 1. Set α = T p−2(b). Then T (α) = σ(α) − α = 1.

Hence σ(α) = α+1. Thus σi(α) = α+ i for all i = 1, 2, . . . , p− 1. Therefore

E = F (α).

Since σ(αp − α) = (α + 1)p − (α + 1) = αp − α, The element a = αp − α ∈

EG = F. Hence α is a root of xp − x− a.

115

(2) Conversely, suppose that a /∈ F p−F. Then we show that f(x) = xp−x−a

is irreducible over F. If α is a root of f(x) then α+ i is a root of f(x) for all

i = 1, 2, . . . , p− 1. Hence E = F (α) is a splitting field of f(x). If we assume

that f(x) is irreducible over F then [E : F ] = p and the Galois group is

generated by the automorphims σ(α) = α+ 1.

Suppose that f(x) = g1(x)g2(x) . . . gn(x) where each gi is irreducible over

F. If β is a root of gi then E = F (β) as shown above. Hence each gi(x) has

same degree r and so deg f(x) = p = rn. Thus r = p and n = 1. Hence f(x)

is irreducible over F.

�

20. Solvable groups

Definition 20.1. Let G be a group. A sequence of subgroups

G = G0 ⊇ G1 ⊇ G2 ⊇ · · · ⊇ Gs+1 = 1(5)

is called a normal series for G if Gi is a normal subgroup of Gi−1 for

i = 1, 2, . . . , s. The normal series (5) is called abelian (resp. cyclic ) if

the quotients Gi/Gi+1 are abelian (resp. cyclic) for i = 0, 1, . . . s. A group

having an abelian series is called a solvable group.

Example 20.2. (1) Any abelian group is solvable.

(2) The group S3 is solvable since S3 ⊃ A3 ⊃ 1 is an abelian series.

(4) The group S4 is solvable since

S4 ⊃ A4 ⊃ V4 ⊃ 1

is an abelian series where V4 = {(1), (12)(34), (13)(24), (14)(23)}.

Proposition 20.3. Any group G of order pn where p is a prime number is

solvable.

Proof. Apply induction on n. If n = 1 then G is cyclic and hence solvable.

Let n ≥ 2. Let C be the center of G. We know that o(C) > 1. Hence

o(G/C) < o(G). By induction, we have an abelian series

G/C ⊃ G1/C ⊃ G2/C ⊃ · · · ⊃ Gs/C = 1

Since (Gi/C)/(Gi+1/C) ≃ Gi/Gi+1 for all i, we have an abelian series:

G ⊃ G1 ⊃ G2 ⊃ · · · ⊃ Gs ⊃ C ⊃ 1.

116

Thus G is solvable.

�

Definition 20.4. Let G be a group. The commutator [g, h] of g, h ∈ G

is defined as [g, h] = g−1h−1gh. The derived subgroup of G denoted by

G′ is the subgroup generated by all the commutators in G. The kth derived

subgroup of G is defined inductively as G(k) = (G(k−1))′.

Proposition 20.5. Let f : G→ H be a homomorphism of groups.

(1) f(G′) ⊆ H ′. If f is onto then f(G′) = H ′.

(2) If K ⊳G then K ′ ⊳G. In particular G′ ⊳G.

(3) If K ⊳G then G/K is abelian if and only if G′ ⊆ K.

Proof. (1). Let g, h ∈ G. Then f([g, h]) = f(g)−1f(h)−1f(g)f(h) = [f(g), f(h)].

Hence f(G′) ⊆ H ′. It is clear that equality holds true if f is onto.

(2) Let a ∈ G. The inner automorphism ia : G→ G restricts to an automor-

phism of K as K ⊳G. Hence ia(K′) = K ′. Therefore K ′ ⊳G. Since G⊳G,

we have G′ ⊳G.

(3) Let K ⊳ G. Then G/K is abelian ⇔ for all g, h ∈ G, ghK = hgK ⇔

h−1g−1hg ∈ K for all g, h ∈ G ⇔ G′ ⊆ K. �

Proposition 20.6. A group G is solvable if and only if G(s) = 1 for some

s ∈ N.

Proof. Let G be solvable. Then there is an abelian series for G

1⊳G1 ⊳G2 ⊳ · · ·⊳Gs = G.

We show by induction on s that G(s) = 1. If s = 1, then G is abelian. Hence

[g, h] = 1 for all g, h ∈ G. Hence G′ = 1. Now let s > 1. Then

1⊳G1 ⊳G2 ⊳ · · ·⊳Gs−1

is an abelian series for Gs−1. Hence G(s−1)s−1 = 1. Since G/Gs−1 is abelian,

G′ ⊆ Gs−1. Hence

G(s) = (G′)(s−1) ⊆ G(s−1)s−1 = 1.

Conversely suppose that G(s) = 1 for some s. Then

G⊲G(1)⊲G(2)

⊲ · · ·⊲G(s) = 1

is an abelian series for G. Thus G is solvable. �

117

Proposition 20.7. Let G be a group and H be a subgroup.

(1) If G is solvable then so is H.

(2) If f : G → H is a surjective group homomorphism and G is solvable

then H is so.

(3) If K ⊳G and G/K are solvable then G is solvable.

Proof. (1) If G is solvable then G(s) = 1 for some s. Since H(s) ⊂ G(s) = 1,

we have H(s) = 1. Thus H is solvable.

(2) Let G(s) = 1. Since f is surjective, f(G(s)) = H(s) = 1. Hence H is

solvable.

(3) Let K ⊳G and K and G/K be solvable. Then there exist s and t such

that K(s) = 1 and (G/K)(t) = 1. Hence G(t) ⊂ K. Thus G(t+s) ⊂ K(s) = 1.

Hence G is solvable.

�

Lemma 20.8. The group An is generated by 3-cycles. If n ≥ 5 then all

3-cycles are conjugates in An.

Proof. Let σ be an even permutation. Let (ij)(rs) occur in a decomposi-

tion of σ as a product of transpositions. If (ij) and (rs) are disjoint then

(ij)(rs) = (ijr)(rsj). If j = r then (ir)(rs) = (irs). Hence every even per-

mutation is a product of 3-cycles. Now suppose that n ≥ 5. Let σ be any

permutation and (j1j2 . . . jp) be a p-cycle. Then

σ(j1j2 . . . jp)σ−1 = (σ(j1)σ(j2) . . . σ(jp)).

Let (ijk) and (rst) be any two 3-cycles. Define γ by γ(i) = r, γ(j) =

s, γ(k) = t and let γ(u) = u for any u 6= i, j, k. Then

γ(ijk)γ−1 = (γ(i)γ(j)γ(k)) = (rst).

If γ is odd then put σ = (ij)γ. Then σ is even and

σ(ijk)σ−1 = (ij)γ(ijk)γ−1(ij) = (rst).

�

Theorem 20.9. The groups Sn and An are not solvable for n ≥ 5.

118

Proof. Since Sn/An is abelian, S′n ⊂ An. Note that since n ≥ 5, any 3-cycle

is a commutator in view of :

[(jkv), (ikr)] = (vkj)(rki)(jkv)(ikr) = (vkj)(jiv) = (ikj).

Therefore S′n = A′n = An. Thus S(s)n = A

(s)n = An for all s. Hence An and

Sn are not solvable for n ≥ 5. �

Theorem 20.10 (Galois). The alternating group An is simple for n ≥ 5.

Proof. (S. Lang) Suppose An is not simple for n ≥ 5. Let N be a proper

normal subgroup of An for some n ≥ 5. Let σ 6= 1 be a permutation in N

that has maximum number of fixed points. We say that j is a fixed point of

σ if σ(j) = j. Consider a decomposition of σ as a product of disjoint cycles

of length at least two: σ = τ1τ2 . . . τg. Suppose the length of each τj is two.

Since σ is an even permutation, g ≥ 2. Suppose that σ = (ij)(rs)τ3 . . . τg.

Let k be different from i, j, r, s and set τ = (rsk). Consider the commutator

γ = [σ, τ ] = σ−1τ−1στ. Then 1 6= γ ∈ N. Moreover γ(i) = i and γ(j) = j.

This is a contradiction since σ has maximum number of fixed points among

the permutations in N \ {1}.

Now suppose that for some a, τa = (ijk...) has length at least 3. If σ = (ijk)

then N has a 3-cycle and hence N = An. If σ is not a 3-cycle then σ must

move at least two other elements r, s besides i, j, k. Put τ = (rsk) and

consider γ = [σ, τ ]. Then 1 6= γ ∈ N. Moreover γ(j) = j and γ fixes all the

elements that σ fixes. This is a contradiction. �

123

Lecture 21 : Galois Groups of Composite Extensions

Objectives

(1) Galois group of composite extensions

(2) Galois closure of a separable field extension.

Keywords and Phrases : Composite extensions, normal closure.

21. Galois groups of composite extensions

LetK be a field andK be an algebraic closure ofK. Let E,F be intermediate

subfields of K/K. Recall that the compositum of E and F denoted by

EF is the smallest subfield of K containing E and F. In this section we

will discuss Galois groups of composite extensions and normal closure of an

algebraic field extension.

Proposition 21.1. If E/K is a Galois extension and F/K is a field exten-

sion, then EF/F is Galois. If F/K is Galois then EF/K and E ∩F/K are

Galois.

Proof.

K

EF

FFFF

FFFF

F

xxxx

xxxx

x

E

FFFF

FFFF

F F

xxxx

xxxx

x

E ∩ F

K

Consider the diagram above. As E/K is a separable and normal extension,

it is a splitting field of a family {fi(x)} of separable polynomials over K.

Then EF/F is the splitting field of the same family of polynomials. Hence

124

EF/F is a Galois extension. If F/K is Galois then it is a splitting field of

a family of polynomials {gj(x)} over K. Hence EF/K is a splitting field of

the polynomials {fi(x)} ∪ {gj(x)}. Hence EF/K is Galois.

Now we show that if E/K and F/K are Galois then E ∩ F/K are Galois.

Let σ : E ∩F → K be a K-embedding. Let τ : EF → K be an extension of

σ. Then τ(E) = E and τ(F ) = F since E/K and F/K are Galois. Therefore

τ(E ∩ F ) ⊆ E ∩ F. Since E ∩ F/K is algebraic, τ(E ∩ F ) = E ∩ F. Hence

E ∩ F/K is a Galois extension.

�

Proposition 21.2. Let E/K be a Galois extension and F/K be a field

extension so that E,F ⊂ K. Then the map ψ : G(EF/F ) → G(E/K)

defined by ψ(σ) = σ|E is injective and it induces an isomorphism:

G(EF/F ) ≃ G(E/E ∩ F ).

Proof. Since σ is an F -automorphism of EF, it is also a K-automorphism.

Hence σ|E ∈ G(E/K). If σ|E = idE then σ = idEF . Hence ψ is an injective

group homomorphism.

The image of ψ is a subgroupH ofG(E/K). By Artin’s TheoremG(E/EH) =

H. Hence E ∩F ⊂ EH . Let a ∈ E \ (E ∩F ). Then a ∈ EF \F. Hence there

is a σ ∈ G(EF/F ) so that σ(a) 6= a. Hence a /∈ EH . Therefore EH = E ∩F

and we conclude that G(E/E ∩ F ) = H ≃ G(EF/F ). �

Corollary 21.3. Let E/K be a finite Galois extension and F as above.

Then

[EF : F ] = [E : E ∩ F ].

In particular, [EF : K] = [E : K][F : K] if and only if E ∩ F = K.

Proof. Since G(EF/F ) ≃ G(E/E ∩ F ), we obtain

|G(EF/F )| = [EF : F ] = |G(E/E ∩ F )| = [E : E ∩ F ].

Therefore we have:

[EF : K] = [E : E ∩ F ][F : K] =[E : K][F : K]

[E ∩ F : K].

The conclusion follows from the equation above. �

125

Theorem 21.4. Let E/K and F/K be finite Galois extensions so that

E,F ⊂ K. Then the homomorphism

ψ : G(EF/K) −→ G(E/K)×G(F/K), ψ(σ) = (σ|E , σ|F )

is injective. If E ∩ F = K then ψ is an isomorphism.

Proof. It is clear that ψ is a group homomorphism. The kernel of ψ consists

of σ ∈ G(EF/K) so that σ(a) = a for all a ∈ E and for all a ∈ F. Hence

such σ = idEF . Thus ψ is injective.

Suppose that E ∩ F = K. Then by Corollary 21.3,

|G(EF/K)| = [EF : E ∩ F ] = [F : K][E : K] = |G(E/K)||G(F/K)|.

This shows that ψ is an isomorphism. �

The Normal Closure of an Algebraic Extension

Let K/F be an algebraic extension and K ⊂ F . The normal closure of

K/F in K is the splitting field N over F of the polynomials { irr (a, F ) | a ∈

K}. It is clear that N is a normal extension of F containing K. Moreover

any normal extension N ′ ⊂ F of F containing K must contain the splitting

fields of { irr (a, F ) | a ∈ K}. Hence N = N ′. If K = F (a1, a2, . . . , an) then

N is the splitting field of the polynomials irr (ai, F ) for all i = 1, 2, . . . , n.

If K/F is separable then N/F is a separable extension as it is obtained by

adjoining roots of separable polynomials over F. Hence the normal closure

of K/F when K/F is separable, is a Galois extension.

Let K/F be a separable extension that is not normal. Let N be a normal

closure of K/F. Put H = G(N/K). Then K = NH . Let H ′ < H be a normal

subgroup of G = G(N/F ). Then NH′ > NH = K and NH′/F is a normal

extension of K. Thus NH′ = N by minimality of N. Hence H ′ = (id).

126

Lecture 22 : Solvability by Radicals

Objectives

(1) Radical extensions.

(2) Solvability by radicals and solvable Galois groups.

(3) A quintic polynomial which is not solvable by radicals.

Keywords and Phrases : Radical extensions, solvable Galois groups,

insolvable quintic.

22. Solvability by radicals

Let F be a field and f(x) ∈ F [x]. If there is a formula for the roots of f(x)

which involves the field operations and extraction of roots, then we say f(x)

is solvable by radicals over F. This can be made precise in field theory by

introducing the notion of a radical extension.

Definition 22.1. A field extension K/F is called a simple radical exten-

sion of F if K = F (a) where an ∈ F for some positive integer n. We say

that K/F is a radical extension if there is a sequence of field extensions

F = F0 ⊂ F1 ⊂ F2 ⊂ · · · ⊂ Fn = K

such that each Fi is a simple radical extension of Fi−1 for 1 = 1, 2, . . . , n. A

polynomial f(x) ∈ F [x] is called solvable by radicals over F if a splitting

field of f(x) over F is contained in a radical extension of F .

Proposition 22.2. Let E/F be a separable radical extension. Let L ⊃ E

be the smallest Galois extension of F so that L ⊂ F a. Then L is a radical

extension of F .

Proof. Since E/F is separable and [E : F ] = n, there are n F -embedding of

E into F a :

σ1, σ2, . . . , σn : E −→ F a.

Then L = σ1(E)σ2(E) · · ·σn(E) is the smallest Galois extension of F con-

taining E. Indeed, let E = F (a). Then the roots of fa(x) = irr (a, F )

in F a are σi(a) for i = 1, 2, . . . , n. Hence L = F (σ1(a), σ2(a), . . . , σn(a)) is

127

the splitting field of fa(x) over F. Since σi(E) ≃ E, σi(E)/F is a radical

extension for each i = 1, 2, . . . , n. Hence L/F is a radical extension. �

Theorem 22.3. Suppose char F = 0. If f(x) ∈ F [x] is solvable by radicals

then Gf is a solvable group.

Proof. Let F = F0 ⊂ F1 ⊂ . . . ⊂ Fr = E be a sequence of simple radical

extension with Fi = Fi−1(ai) such that ani

i ∈ Fi−1, i = 1, 2, . . . , r and E

contains a splitting field K of f(x) over F. We may assume E/F is Galois

by the above proposition. Let n = n1n2 . . . nr and M be the splitting field

of xn − 1 over E.

E = Fr E(w) =Mr =M (e) = Hr

Fr−1

���

Fr−1(w) =Mr−1

���

G(Mr/Mr−1) = Hr−1

���

K

�����������������

FFFFFFFFF E = F1 F1(w) =M1 G(M1/M1) = H1

F = F0 F0(w) =M0 G(Mr/M0) = H0

Let w be a primitive nth root of unity. Then F (w) has primitive nthi root of

unity for i = 1, 2, . . . , r. Since E/F is Galois, E is a splitting field of some

polynomial g(x) over F . Then M is a splitting field of (xn − 1)g(x) over F .

Thus M is Galois over F . By the FTGT, G(K/F ) ≃ G(M/F )/G(M/K).

Hence it is enough to prove that G(M/F ) is solvable.

Each Mi/Mi−1 is a Galois extension. Hence Hi ⊲ Hi−1 for i = 1, 2, . . . r.

Moreover

Hi−1/Hi ≃ G(Mi/Mi−1).

Since Mi = Mi−1(ai) where ani

i ∈ Mi−1 and Mi−1 has a primitive nthi root

of unity, the group Hi−1/Hi is cyclic. Thus G(Mr/F ) is a solvable group.

Hence Gf is a solvable group. �

We will now construct a quintic f(x) ∈ Z[x] which is not solvable by radicals.

128

Proposition 22.4. A subgroup of S5 containing a 5-cycle and a transposi-

tion is S5.

Proof. By renumbering we may assume G contains σ = (12) and τ =

(12345). Then G has τ(12)τ−1 = (23), τ(23)τ−1 = (34), τ(34)τ−1 = (45).

It is easy to show that 〈(12), (23), (34), (45)〉 = S5. �

Any irreducible quintic f(x) ∈ Q[x] which has exactly 3 real roots is the

polynomial we are looking for. Gf has an element of order 5 and the con-

jugation automorphism gives an element of order 2 in Gf . The polynomial

x(x2 − 4)(x2 + 4) = x5 − 16x = g(x) has exactly 3 real roots 0, 2,−2. Since

g(−1) = 15, g(1) = −15, f(x) = g(x) + 2 = x5 − 16x+ 2 have exactly 3 real

roots and it is irreducible over Q, Thus f(x) = 0 is not solvable by radicals

over Q.

Theorem 22.5 (Galois). Suppose F is a field of characteristic zero and

f(x) ∈ F [x]. If Gf solvable then f(x) is solvable by radicals over F.

L = K(w) (e) = Hk = G(L/E)

Ek−1 = LHk−1

���

Hk−1

���

E1 = LH1 H1

E = LH0 = F (w) H0 = G(L/E)

Proof. Let K be a splitting field of f(x) over F and [K : F ] = n. Let L be

a splitting field of xn− 1 over K and w be a primitive nth root of unity over

K. Then L = K(w). Put E = F (w). Then L is a splitting field of f(x) over

E. Since H = G(L/E) embeds into G(K/F ) H is also a solvable group. It

is enough to show f(x) is solvable by radicals over E. Consider an abelian

series for H.

H = H0 ⊲ H1 ⊲ · · · ⊲ Hk = (1)

By refining this we may assume Hi/Hi+1 is cyclic of order pi+1 for i =

129

0, 1, . . . , k − 1 where p1, p2, . . . , pk are primes numbers. Let Ei = LHi for

1, 2 . . . k. Then [Ei : Ei−1] = |Hi−1/Hi| = pi. Since Ei−1 has a primitive pthi

root of unity for i = 1, 2, . . . , k, Ei/Ei−1 is a simple radical extension. Hence

L/F is a radical extension. Thus f(x) is solvable by radicals over F. �

Example 22.6. In this example we show that a splitting field of E over

a field F of a polynomial f(x) ∈ F [x] solvable by radicals need not be a

radical extension of F. Consider the polynomial f(x) = x3 − 3x+ 1 ∈ Q[x].

Let E be a splitting field of f(x) over Q. We argue that E is not a radical

extension of Q. Reducing f(x) modulo 2, we see that the reduced polynomial

has no root in F2. Hence f(x) is irreducible over Q. The discriminant of f(x)

is 81. Hence Gf = A3 and therefore f(x) is solvable by radicals by Galois’

theorem. Suppose that E/Q is a radical extension. Since [E : Q] = 3, there

is no proper intermediate subfield of E/Q. So E = Q(a) where an ∈ Q, for

some n. Let g(x) = irr (a,Q). Then E is a splitting field of g(x). Moreover

g(x) | xn−an. Hence any root r of g(x) satisfies rn = an. Since f(x) is a real

root, we may assume that E = Q(r). Hence r/a is a real nth root of unity.

Hence r = ±a. Hence g(x) has only two roots. This is a contradiction as

g(x) is a separable cubic polynomial.

131

Lecture 23 : Solutions of Cubic and Quartic Equations

Objectives

(1) Cardano’s method for roots of cubic equations.

(2) Lagrange’s method for roots of quartic equations.

(3) Ferrari’s method for roots of quartic equations.

Keywords and Phrases : Cubic equations, quartic equations.

23. Solutions of cubic and quartic equations

In this section we present algorithms for finding roots of cubic and quartic

polynomials over any field F of characteristic different from 2 and 3. This

is to make sure that irreducible cubics and quartics are separable.

Cubic polynomials

Cardano published Tartaglia’s method to find roots of cubic polynomials

in 1545. This is known as Cardano’s method. We may assume that the

given cubic is of the form f(x) = x3 + px + q since a general cubic can be

transformed into this form without changing its splitting field. One begins

by introducing two unknowns u and v. Put x = u+ v into f(x) = 0 to get

u3 + v3 + 3u2v + 3uv2 + p(u+ v) + q = u3 + v3 + q + (3uv + p)(u+ v) = 0.

We set u3 + v3 + q = 0 and 3uv + p = 0. Hence v = −p/3u. Put this into

the first equation to get

u6 + qu3 − p3/27 = 0.

This is a quadratic equation in u3. Put D = −(4p3+27q2). By the quadratic

formula we get

u3 =−q ±

√

q2 + (4p3/27)

2= −q

2±√

−D/108.

Set A = −q/2 +√

−D/108 and B = −q/2−√

−D/108. By symmetry of u

and v, we set u3 = A and v3 = B. Let ω be a primitive cube root of unity.

Then

u =3√A, ω

3√A, ω2 3

√A, and v =

3√B, ω

3√B, ω2 3

√B.

132

We must choose cube roots of A and B in such a way that 3√A 3√B = −p/3.

Having chosen these we see that the three roots of f(x) are

3√A+

3√B, ω

3√A+ ω2 3

√B, ω2 3

√A+ ω

3√B.

Example 23.1. Consider the cubic f(x) = x3−3x+1. Reducing modulo 2,

we see that f(x) is irreducible over Q. The discriminant of f(x) is D = −81.

Hence

A = −q/2 +√

−D/108 = exp(2πi/3), and B = exp(−2πi/3).

Substitute these values of A and B into the formula for the roots, we see

that the three roots of f(x) are 2 cos(2π/9), 2 cos(8π/9) and 2 cos(14π/9).

Let f(x) = x3 + px+ q ∈ R[x]. If disc (f) < 0, then cube roots of A and

B can be chosen to be real. In this case

r1 =3√A+

3√B ∈ R,

r2 = −3√A+ 3

√B

2+ i

√3

(

3√A− 3

√B

2

)

,

r3 = r2.

If D = disc (f(x)) > 0 then A = −q/2 + i√

D/108 and B = A. Suppose

that 3√A = a+ ib then due to uv = −p/3 we have 3

√B = a− ib. Hence the

roots of f(x) are r1 = 2a, r2 = −a− b√3 and r3 = −a+ b

√3.

Notice that in this case, all the roots are real. However, they are expressed

in terms of complex numbers. It can be proved that the roots cannot be ex-

presseed in terms of real radicals. Historically, this is called the irreducible

case. This fact forced mathematicians to accept complex numbers as a valid

mathematical constructs.

Quartic polynomials

We present Lagrange’s method for the roots of a quartic polynomials. We

continue with the assumption that F has characteristics different from 2, 3.

Consider a general quartic polynomial f(x) = x4+ax3+bx2+cx+d.We put

y = x− a/4 to get the polynomial g(y) = y4 + py2 + qy+ r. Let r1, r2, r3, r4

be roots of g(y). Consider the quantities

θ1 = (r1 + r2)(r3 + r4), θ2 = (r1 + r3)(r2 + r4), θ3 = (r1 + r4)(r2 + r3).

133

The cubic polynomial whose roots are θ1, θ2 and θ3 is called the resolvent

cubic of the quartic polynomial. It turns out to be the polynomial

h(x) = x3 − 2px2 + (p2 − 4r)x+ q2.

Using the relation r1 + r2 + r3 + r4 = 0 we get

r1 + r2 =√

−θ1 r3 + r4 = −√

−θ1r1 + r3 =

√

−θ2 r2 + r4 = −√

−θ2r1 + r4 =

√

−θ3 r2 + r3 = −√

−θ3

One can show that√−θ1

√−θ2

√−θ3 = −q. Hence two of the square roots

determine the third. Adding the three equations on the left and using the

fact that r1 + r2 + r3 + r4 = 0, we get

2r1 =√

−θ1 +√

−θ2 +√

−θ3.

2r2 =√

−θ1 −√

−θ2 −√

−θ3.

2r3 = −√

−θ1 +√

−θ2 −√

−θ3.

2r4 = −√

−θ1 −√

−θ2 +√

−θ3.

This shows that the roots of the resolvent cubic determine the roots of the

quartic.

Proposition 23.2. The discriminant of the quartic g(y) = y4+py2+qy+r

and its resolvent cubic h(x) = x3 − 2px2 + (p2 − 4r)x+ q2 are equal.

Proof. The differences of the roots of the resolvent cubic are:

θ1−θ2 = (r2−r3)(r4−r1), θ1−θ3 = (r2−r4)(r3−r1), θ2−θ3 = (r3−r4)(r2−r1).

Hence the quartic and the resolvent cubic have same discriminant. �

Remark 23.3. In the literature, we find that the term resolvent cubic is

also used for the cubic whose roots are

t1 = r1r2 + r3r4, t2 = r1r3 + r2r4, and t3 = r1r4 + r2r3.

It can be shown that this cubic is r(x) = x3− px2− 4rx+4pr− q2 and h(x)

and r(x) have equal discriminant and the same splitting field over F.

134

Ferrari’s method for solving quartic equations

Consider the general quartic equation

x4 + bx3 + cx2 + dx+ e = 0.

Rewrite this as x4 + bx3 = −cx2 − dx− e. Now complete the square to get(

x2 +bx

2

)2

=

(

b2

4− c

)

x2 − dx− e.

Let y be another variable and consider the equation:(

x2 +bx

2+y

2

)2

=

(

b2

4− c

)

x2 − dx− e+ y

(

x2 +bx

2

)

+y2

4

= x2(

b2

4− c+ y

)

+ x

(

by

2− d

)

+y2

4− e(6)

The right hand side of the last equation is a square of a linear polynomial

in x if and only if its discriminant is zero. i.e.

(12by − d)2 − 4(14y2 − e)(14b

2 − c+ y) = 0.

Therefore

y3 − cy2 + (bd− 4e)y − b2e+ 4ce− d2 = 0.

Let y be any root of this cubic and substitute it in the equation (6) to get

x2 +1

2bx+

1

2y = ±mx+ n(7)

Notice that the roots of the equation (7) are the roots of the given quartic.

Proposition 23.4. Let x1, x2, x3, x4 be the roots of

f = x4 + bx3 + cx2 + dx+ e = 0.

Then y1 = x1x2 + x3x4, y2 = x1x3 + x2x4, y3 = x1x4 + x2x3 are roots of

resolvent cubic g(y) = y3 − cy2 + (bd− 4e)y − b2e+ 4ce− d2.

135

Lecture 24 : Galois Groups of Quartic Polynomials

Objectives

(1) Galois group as a group of permutations.

(2) Irreducibility and transitivity.

(3) Galois groups of quartics.

Keywords and phrases : Transitive subgroups of S4 Galois groups of

quartics, irreducibilty and transitivity.

24. Galois Groups of Quartic Polynomials

Galois group as a group of Permutations

Let f(x) ∈ F [x] be a monic polynomial with distinct roots r1, r2, . . . , rn. Let

E = F (r1, r2, . . . , rn) and G = G(E/F ). Any σ ∈ G permutes the roots of

f(x). Define ψ : G = G(E/F ) → Sn by ψ(σ) = σ|R. Then ψ is an injective

group homomorphism. The subgroup ψ(G) is called the Galois group of

f(x), and it is denoted by Gf . By the FTGT, there is an intermediate

subfield of E/F corresponding to Gf ∩An.

Theorem 24.1. Let F be a field of characteristic 6= 2 and f(x) ∈ F [x], a

monic polynomial of positive degree with distinct roots r1, r2, . . . , rn ∈ F a.

Put E = F (r1, r2, . . . , rn). Put δ = Π1≤i<j≤n(ri − rj). Then

EGf∩An = F (δ).

Proof. Any transposition acting on δ maps δ to −δ. Hence all permutations

in Gf ∩An fix δ. Thus F (δ) ⊆ EGf∩An . Let |Gf/Gf ∩An| = d. Then d ≤ 2.

If d = 1 then Gf ∩ An = Gf and so Gf ⊆ An. Thus δ ∈ F . Let d = 2.

Then Gf ∩ An 6= Gf . So Gf has an odd permutation. Hence δ /∈ F . Thus

EGf∩An = F (δ).

�

Definition 24.2. A subgroup H ⊂ Sn is called a transitive subgroup if for

any i 6= j ∈ {1, 2, . . . , n}, there exists σ ∈ H such that σ(i) = j.

136

Theorem 24.3. Let f(x) ∈ F [x] be a polynomial of degree n with n distinct

roots r1, r2, . . . , rn in F a. Then f(x) is irreducible if and only if Gf is a

transitive subgroup of Sn.

Proof. (⇐) Suppose Gf is a transitive subgroup of Sn. If f(x) is reducible

in F [x] then f(x) = g(x)h(x) for some g, h ∈ F [x] of positive degree. Let

g(r) = h(s) = 0 where r, s ∈ F a. Let σ ∈ Gf be a permutation which maps

r to s. We may assume that g(x) is irreducible. But then s has to be a root

of g(x). Since f(x) has no repeated roots, h(x) is a constant.

(⇒) Suppose f(x) is irreducible. Let r, s be roots of f(x). Then there exists

an F -isomorphism σ : F (r) → F (s) such that σ(r) = s. It can be extended

to an automorphism of F (r1, . . . , rn). Hence Gf is transitive.

�

Transitive Subgroups of S4

Let H be a transitive subgroup of Sn. The orbit of action of H on [n] is

[n]. Thus n = | orbit (1)| = |H|/| stab (1)|. Hence n | |H|. The orders of

possible Galois groups of irreducible separable quartics are 4, 8, 12 and 24.

These groups are listed below.

(1) C4 = {(1234), (13)(24), (1432), (1)}.

A cyclic group of order 4 has two 4-cycles. There are six 4-cycles in

S4. Thus there are three transitive cyclic subgroups of order 4.

(2) Klein 4 -group V = {(1), (12)(34), (14)(32), (13)(24)} is a normal

subgroup of S4.

(3) There are 3- Sylow subgroups of order 8. They are all isomorphic to

D4. These are H1 = 〈V, (13)〉, H2 = 〈V, (12)〉, H3 = 〈V, (14)〉.(4) A4 is the only subgroup of order 12 and it is normal in S4.

(5) S4 is the only subgroup of order 24.

Calculation of Galois group of quartic polynomials

Let F be a field of char 6= 2, 3. Let f(x) = x4+ b1x3+ b2x

2+ b3x+ b4 ∈ F [x]

be separable. By the change y = x + b14 we may assume that there is no

x3 term. This change does not alter the Galois group and the discriminant.

So let f(x) = x4 + bx2 + cx + d ∈ F [x] be an irreducible polynomial with

137

roots r1, r2, r3, r4 in a splitting field E of f(x) over F. We write Gf ⊂ S4.

So Gf ≃ G(E/F ). Set

t = {t1 = r1r2 + r3r4, t2 = r1r3 + r2r4, t3 = r1r4 + r2r4}.

Proposition 24.4. EGf∩V = F (t) and G(F (t)/F ) =Gf

Gf∩V .

Proof. Clearly, F (t1, t2, t3) ⊆ EGf∩V . The element t1 is fixed by H1 =

〈(12), V )〉, a dihedral group of order 8 in S4. Moreover

S4 = H1 ∪ (13)H1 ∪ (14)H1.

ThusH1 is the stabilizer of t1. Similarly,H2 = Stab (t2) = 〈(13), V )〉, H3 =

Stab (t3) = 〈(14), V )〉. Since V = H1 ∩ H2 ∩ H3, if σ ∈ Gf fixes t1, t2, t3

then σ ∈ V. Hence G(E/F (t)) ⊆ Gf ∩ V which gives F (t) ⊇ EGf∩V . We

know that F (t1, t2, t3) is the splitting field of the resolvent cubic over F,

hence it is Galois. Thus G(F (t)/F ) ≃ Gf

Gf∩V . �

Proposition 24.5. The resolvent cubic of a separable irreducible quartic

has a root in F if and only if Gf ⊆ D4.

Proof. Let t1 ∈ F . Then G(E/F (t1)) = Gf = Gf ∩ H1 ⇒ Gf ⊆ H1.

Conversely if Gf ⊂ Hi for some i say i = 1, then each σ ∈ Gf fixes t1 and

hence t1 ∈ EGf = F . �

Theorem 24.6. Let f(x) be an irreducible separable quartic over a field F

of char F 6= 2 and E = F (r1, r2, r3, r4) be a splitting field where r1, . . . , r4

are the roots of f(x). Let r(x) denote resolvent cubic of f(x).

(1) If r(x) is irreducible in F [x] and disc (r(x)) /∈ F 2 then Gf ≃ S4.

(2) If r(x) is irreducible in F [x] and disc (r(x)) ∈ F 2 then Gf ≃ A4.

(3) If r(x) splits completely in F [x] then Gf ≃ V.

(4) Let r(x) have one root in F. Then

(a) If f(x) is irreducible over F (t) then Gf ≃ D4.

(b) If f(x) is reducible over F (t) then Gf ≃ C4.

Proof. Since f(x) is irreducible over F, Gf is a transitive subgroup of S4.

Hence |Gf | = 4, 8, 12, or 24, |Gf∩V | = 1, 2 or 4, and |Gf/Gf∩V | = |Gr(x)| =1, 2, 3, 6. Thus |Gf ∩ V | > 1. We also have |V ∩Gf | × | Gf

V ∩Gf| = |Gf |. Thus

{2, 4} × {1, 2, 3, 6} = {4, 8, 12, 24}.

138

(1) If r(x) is irreducible over F and disc (r(x)) ∈ F 2 then Gr(x) ≃ A3.

Hence |Gf/Gf ∩ V | = 3. Hence |Gf | = 12 and therefore Gf ≃ A4.

(2) If r(x) is irreducible over F and disc (r(x)) is not a square in F, then

Gr(x) ≃ S3. Hence |Gf/Gf∩V | = 6. Thus |Gf | = 12 or 24. If |Gf | = 12 then

Gf ≃ A4 and |Gf/Gf ∩ V | = 3 which is a contradiction. Hence Gf ≃ S4.

(3) If r(x) has all its roots in F, then EGf∩V = F = EGf . Thus Gf ⊆ V .

Since 4 | |Gf |, Gf = V .

(4) Now let r(x) have exactly one root in F . Then [F (t) : F ] = 2 =

|Gf/Gf ∩ V |. Thus |Gf | = 4 or 8.

(a) Suppose f(x) is irreducible over F (t). Then

[E : F (t)] = |Gf ∩ V | ≥ 4 ⇒ |Gf | = 8 ⇒ Gf ≃ D4.

(b) Suppose f(x) is reducible over F (t). If Gf ≃ D4 then

[E : F ] = 8 ⇒ [E : F (t)] = 4.

Hence G(E/F (t)) = V which is transitive. Hence f(x) is irreducible over

F (t). This is a contradiction. So |Gf | = 4. If Gf = V then Gr(x) =

Gf/Gf ∩ V = {1}. But |Gr(x)| = 2. Thus Gf ≃ C4. �

Example 24.7. (1) (Gf = V ) Let f(x) = x4+1 ∈ Q[x]. Then the resolvent

cubic is r(x) = x(x− 2)(x+ 2). Since f(x) is irreducible over Q, Gf = V .

(2) (Gf = C4) Consider f(x) = x4 + 5x2 + 5 which is irreducible over Q by

Eisenstein criterion. Then

r(x) = x3 − 5x2 − 20x+ 100 = (x− 5)(x− 2√5)(x+

√5).

Thus t1 = 5, t2 = 2√5, t3 = −2

√5. Hence F (t) = Q(

√5) and

x4 + 5x2 + 5 =(

x2 + 5+√5

2

)(

x2 − 5−√5

2

)

.

Therefore f(x) is reducible over F (t). Thus Gf ≃ C4.

(3) (Gf = S4) Consider f(x) = x4 − x+1. Then f(x) is irreducible modulo

2, and hence it is irreducible over Q. The resolvent cubic r(x) = x3− 4x− 1

is irreducible over Q and disc (r(x)) = 229 /∈ Q2. Hence Gf = S4.

139

(4) (Gf = D4) The polynomial f(x) = x4 − 3 is irreducible over Q and

r(x) = x(x+ i2√3)(x− i2

√3). Therefore F (t) = Q(i

√3). Hence

f(x) = (x2 −√3)(x2 +

√3) = (x− i

4√3)(x+ i

4√3)(x+

4√3)(x− 4

√3).

Thus f(x) has no root in Q(i√3). The splitting field of f(x) over Q is

Q(i, 4√3) which is a degree 8 extension of Q. Hence Gf = D4.

(5) (Gf = A4) Let f(x) = x4 − 8x+ 12. Then r(x) = x3 − 48x− 64. Using

Eisenstein’s criterion, f(x) is irreducible over Q. Since disc (r(x)) = 21234

is a perfect square in Q, Gf = A4.

Example 24.8. Let p be a prime number and f(x) = x4 + px + p. Then

r(x) = x3−4px−p2. Possible roots of r(x) in Q are ±1, ±p, ±p2. Check that

±1, ±p2 are not roots for any p. But r(p) = p2(p−5) and r(−p) = p2(3−p).Hence r(x) has a rational root if and only if p = 3, 5. For p 6= 3, 5, the

resolvent cubic is irreducible over Q. Check that disc (f(x)) = p3(256−27p)

is never a perfect square in Q. Let G be the Galois group of f(x). Then

G = S4 if p 6= 3, 5. If p = 3 then r(x) = (x + 3)(x2 − 3x − 3). Hence the

splitting field L of r(x) over Q is Q(√21). Check that x4+3x+3 is irreducible

over Q(√21). Hence G = D4. The p = 5 case has been considered in the

previous example.

145

Lecture 25 : Norm, Trace and Hilbert’s Theorem 90

Objectives

(1) The norm and the trace function.

(2) Multiplicative form of Hilbert’s Theorem 90.

(3) Cyclic extensions of degree n.

(4) Additive version of Hilbert’s 90.

(5) Cyclic extensions of prime degree: Artin-Schreier Theorem.

Keywords and phrases: Norm, trace, Hilbert’s theorem 90, cyclic exten-

sions, Artin-Schreier Theorem.

25. Norm, Trace and Hilbert’s Theorem 90

Definition 25.1. Let E/F be a finite separable extension of degree n. Let

σ1, . . . , σn be the F -embeddings : E → F a. For any a ∈ E, define the norm

and trace of a by,

NE/K(a) = σ1(a)σ2(a) · · ·σn(a)

TrE/K(a) = σ1(a) + · · ·+ σn(a).

Example 25.2. Let m be a square free integer. Consider the quadratic

extension E = Q(√m) of Q. The Galois group G = G(E/Q) consists of

identity map and the automorphism σ(a +√m) = a − b

√m. Therefore

Tr(a+ b√m) = 2a and N(a+ b

√m) = a2 −mb2.

Proposition 25.3. (1) NE/K : E× → F× is a group homomorphism.

(2) Let E ⊃ K ⊃ F be a tower of finite separable extensions. Then

NE/F = NK/F ◦NE/K , T rE/F = TrK/F ◦ TrE/K

(3) If E = F (a) and irr (a, F ) = xn + an−1xn−1 + · · ·+ a0 then

NE/F (a) = (−1)na0, and TrE/F (a) = −an−1.(4) Tr : E → F is a surjective F -linear map.

Proof. (1) NE/F (ab) = NE/F (a)NE/F (b) for all a, b ∈ E is clear.

Let L = σ1(F ) . . . σn(F ). Then L/F is a Galois extension. Let a ∈ E×.

Then NE/F (a) is fixed under all σ ∈ G(L/F ), thus it is in F×.

146

(2) Let {τj} be the family of F -embeddings : K → F a and {σi} be the

family of all K-embeddings of : E → F a. Each τj can be extended to an

automorphism of F a. Let this extension be denoted by τj . Then {τjσi} is

the family of all F -embeddings of E → F a. For any x ∈ E,

NK/F ◦NE/K(x) = NK/F

(

n∏

i=1

σi(x)

)

=m∏

j=1

n∏

i=1

τjσi(x) = NE/F (x).

For any x ∈ E we have

TrK/F ◦ TrE/K(x) = TrK/F

(

n∑

i=1

σi(x)

)

=

m∑

j=1

n∑

i=1

τjσi(x) = TrE/F (x).

(3) Suppose E = F (a) and f(x) = xn + an−1xn−1 + · · · + an = irr (a, F )

and f(x) = (x − a1)(x − a2) · · · (x − an) where a1, . . . , an are all the roots

in F a of f(x). Each ai = σ(a1) for some F -embedding σ : E → F a. Thus

NE/F (a) = (−1)nan and TrE/F (a) = −an−1.(4) TrE/k(a) = σ1(a) + · · · + σn(a). By Dedekind’s theorem on characters,

σ1 + · · · + σn is not a zero map. Since TrE/F is a linear map of F -vector

spaces, it is surjective. �

Proposition 25.4. Let E/F be a finite separable extension of degree n and

a ∈ E. Let ma : E → E be the F -linear map defined as ma(x) = ax for all

x ∈ E. Then

NE/F (a) = det(ma) and TrE/F (a) = Tr(ma).

Proof. Let K = F (a) and f(x) = irr (a, F ) = xd+ad−1xd−1+ · · ·+a1x+a0.

Then 1, a, a2, . . . , ad−1 is an F -basis for K. Let v1, v2, . . . , ve be a K-basis of

E. Then {viaj | i = 1, 2, . . . , e; j = 0, 1, . . . , d − 1} is an F -basis of E. We

order this basis as :

B = {v1, av1, a2v1, . . . , ad−1v1; . . . ; , ve, ave, a2ve, . . . , ad−1ve}.

Consider the matrix

A =

0 0 0 . . . 0 −a01 0 0 . . . 0 −a10 1 0 . . . 0 −a2...

......

......

0 0 0 . . . 1 −ad−1

.

147

Then the characteristic polynomial of A is f(x). The matrix of ma with

respect to B is the n× n matrix:

A 0 0 . . . 0

0 A 0 . . . 0...

......

...

0 0 0 . . . A

.

Therefore detma = (detA)e and Trma = eTrA. Therefore

NE/F (a) = NK/F ◦NE/K(a) = NK/F (ae) = (detA)e = detma,

Tr E/F (a) = Tr K/F ◦ Tr E/K(a) = Tr K/F (ea) = e Tr A = Tr ma.

�

Proposition 25.5. Let E/F be a finite separable extension. Then

(1) the map ϕ : E × E → F given by ϕ(x, y) = Tr(xy) is bilinear.

(2)The map Tx : E → F given by Tx(y) = Tr(xy) is an F -linear map.

(3) The map ψ : E → Hom(E,F ) given by ψ(x) = Trx is an isomorphism.

Proof. It is easy to see (1) and (2). For (3), if ψ(x) = Trx = 0 then Trx(y) =

Tr(xy) = 0 for all y ∈ E. Hence for any e ∈ E, Trx(x−1e) = Tr(e) = 0.

Thus Tr is the zero functional. This is a contradiction. Hence ψ is an

injective linear map. Since dimE = dimHom(E,F ), we conclude that ψ is

an isomorphism.

�

Theorem 25.6 (Hilbert’s Theorem 90 (multiplicative form)). Let

E/F be a cyclic extension. Let G(E/F ) = (σ). Then for β ∈ E,

NE/F (β) = 1 if and only if β =α

σ(α)for some α ∈ E×.

Proof. Let [E : F ] = n. If β = ασ(α) , then

NE/F (β) = βσ(β) · · ·σn−1(β) = ασ(α)σ(α)σ2(α)

· · · σn−1(α)α = 1.

Conversely, suppose NE/k(β) = 1. Consider

id+ βσ + βσ(β)σ2 + βσ(β)σ2(β)σ3 + · · ·+ βσ(β) · · ·σn−2(β)σn−1

is a nonzero map from E → F due to Dedekind’s independence theorem.

Let θ ∈ K be such that

148

α = θ + βσ(θ) + βσ(β)σ2(θ) + · · ·+ βσ(β) · · ·σn−2(β)σn−1(θ) 6= 0.

Then

βσ(α) = βσ(θ) + βσ(β)σ2(θ) + · · ·+ βσ(β)σ2(β) · · ·σn−1(β)θ = α.

Therefore β = ασ(α) .

�

Theorem 25.7. Let k be a field, n a positive integer coprime with char k

and assume k has a primitive nth root w of 1. Let E/k be cyclic extension

of degree n. Then E is splitting field of xn − a ∈ k[x].

Proof. Let G(E/k) = (σ). Then NE/k(w−1) = w−n = 1. By Hilbert’s the-

orem 90, there exists α ∈ E such that σ(α) = wα. Thus σi(α) = wiα for

i = 1, ..., n. Hence α has n distinct conjugates in E. Since [E : k] = n,

E = k(α). Since σ(αn) = (wα)n = αn := a ∈ EG = k. Thus E is a splitting

field of xn − a.�

We now discuss the additive form of Hilbert 90 and its application to cyclic

extension of degree p, where p is prime and is equal to the characteristic of

the base field.

Theorem 25.8 (Additive form of Hilbert’s Theorem 90). Let E/k be

a cyclic extension of degree n with Galois group G. Let G = 〈σ〉. Then for

β ∈ E

TrE/k(β) = 0 if and only if β = α− σ(α) for some α ∈ E.

Proof. Let β = α− σ(α). Then Tr(β) = Tr(α)− Tr(σ(α)) = 0.

Let Tr(β) = 0. Since Tr : E → k is a nonzero map, there exits θ ∈ E such

that Tr(θ) 6= 0. For the element

α = 1Tr(θ) [βθ + (β + σ(β))σ(θ) + · · ·+ (β + σ(β) + · · ·+ σn−2(β))σn−2(θ)],

σ(α) = 1Tr(θ) [σ(β)σ(θ) + (σ(β) + σ2(β))σ2(θ) + · · · + (σ(β) + σ2(β) + · · · +

σn−1(β))σn−1(θ)]

As Tr(β) = 0, α− σ(α) = 1Tr(θ) [βθ + βσ(θ) + · · ·+ βσn−1(θ)] = β. �

149

Theorem 25.9 (Artin-Schreier). Let k be a field of char p > 0. Let E/k

be a cyclic extension of degree p. Then E is a splitting field of xp − x − a

for some a ∈ E and E = k(α) where αp − α = a for some α ∈ E.

Proof. Let E/k be cyclic of degree p. Then Tr(−1) = 0. Hence there exists

α ∈ E such that α − σ(α) = −1 where 〈σ〉 = G(E/k). Thus σ(α) = α + 1.

Hence σi(α) = a + i for i = 1, 2, . . . , p. Since char k = p, the elements

α, α + 1, ..., α + p − 1 are distinct. Hence [k(α) : k] = p and E = k(α). As

σ(αp−α) = (σ(α))p−σ(α) = (α+1)p−(α+1) = αp−α, αp−α ∈ E(σ) = k.

Let a = αp − α ∈ k. Then α satisfies f(x) = xp − x − a = 0. The roots of

f(x) are α, α+ 1, . . . , α+ p− 1. Thus E is a splitting field of f(x).

�

Example 25.10. (Pythagorean Triples) Let us find all Pythagorean

triples (x, y, z) such that x2 + y2 = z2 where x, y, z ∈ N. Hence x2/z2 +

y2/z2 = N(x/z + iy/z) = 1. Let us apply Hilbert’s theorem 90 to the cyclic

extension Q(i)/Q. The Galois group of this extension is cyclic of order 2

generated by the conjugation automorphism. Hence N(a+ ib) = a2+ b2. So

there exists α = c+ id ∈ Q(i) such that

x/z + iy/z = (c+ id)/(c− id) = (c2 − d2 + 2icd)/(c2 + d2).

Thus x/z = (c2 − d2)/(c2 + d2) and y/z = 2cd/(c2 + d2). Putting c = s/u

and d = t/u where s, t, u ∈ N, we get

x = s2 − t2, y = 2st, z = s2 + t2.