CONTENTS
1. Introduction
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
2. Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
2.1
Roots of complex polynomials of small degree . . . . . . . . . . . . . . . .
4
2.2
Automorphisms of rings and fields . . . . . . . . . . . . . . . . . . . . . . .
4
3. Field Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
3.1
Extension Fields
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
3.2
Transcendental Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
3.3
Roots of Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
3.4
Construction with Straight Edge and Campass . . . . . . . . . . . . . . . . 35
3.5
More about roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
3.6
The Elements Of Galois Theorey . . . . . . . . . . . . . . . . . . . . . . . 51
3.7
Solvability by Radicals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
3.8
Galois Groups over the Rationals. . . . . . . . . . . . . . . . . . . . . . . . 80
4. Selected Topics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84
4.1
Finite Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84
4.2
Wedderburn’s Theorem on Finite Division Rings . . . . . . . . . . . . . . . 94
1. INTRODUCTION
What is Galois Theory?
Much of early algebra centred around the search for explicit formul for roots of
polynomial equations in one or more unknowns. The solution of linear and quadratic
equations in one unknown was well understood in antiquity, while formul for the roots of
general real cubics and quartics was solved by the 16th century. These solutions involved
complex numbers rather than just real numbers. By the early 19th century no general
solution of a general polynomial equation by radicals (i.e., by repeatedly taking n-th roots
for various n) was found despite considerable effort by many outstanding mathematicians.
What is mildly miraculous is not that the solutions exist, but they can always be
expressed algebraically in terms of the coefficients and the basic algebraic operations,
√ √ √ √
+, , ×, , , 3 , 4 , 5 , · · · By the turn of the 19th century, no equivalent formula for
the solutions to a quintic (degree five) polynomial equation had materialised, and it was
Abels who had the crucial realisation: no such formula exists!
Such a statement can be interpreted in a number of ways. Does it mean that there
are always algebraic expressions for the roots of quintic polynomials, but their form is
too complex for one single formula to describe all the possibilities? It would therefore
be necessary to have a number, maybe even infinitely many, formulas. The reality turns
out to be far worse: there are specific polynomials, such as x5 4x + 2, whose solutions
cannot be expressed algebraically in any way whatsoever. There is no formula for the
roots of just this single polynomial, never mind all the others. A few decades after Abels
bombshell, Evariste Galois started thinking about the deeper problem: why dont these
formulae exist? Thus Galois theory was originally motivated by the desire to understand,
in a much more precise way than they hitherto had been, the solutions to polynomial
equations.
Eventually, the work of Abel and Galois led to a satisfactory framework for fully
understanding this problem and the realization that the general polynomial equation of
degree at least 5 could not always be solved by radicals. At a more profound level,
the algebraic structure of Galois extensions is mirrored in the subgroups of their Galois
groups, which allows the application of group theoretic ideas to the study of fields. This
3
Galois Correspondence is a powerful idea which can be generalized to apply to such diverse
topics as ring theory, algebraic number theory, algebraic geometry, differential equations
and algebraic topology. Because of this, Galois theory in its many manifestations is a
central topic in modern mathematics.
In this lecture notes we will focus on the following topics.
1. The solution of polynomial equations over a field, including relationships between
roots, methods of solutions and location of roots.
2. The structure of finite and algebraic extensions of fields and their automorphisms.
3. Classical topics such as ”squaring the circle”, ”duplication of the cube”,
constructible numbers and constructible polygons.
4. Applications of Galois theoretic ideas in Number Theory, the study of differential
equations and Algebraic Geometry.
2. PRELIMINARIES
2.1 Roots of complex polynomials of small degree
In this section we work within the field of complex numbers. Consider the ring F [x] of
all polynomials in the indeterminate x over the field F where F = R or F = C.
For any monic linear (degree 1) polynomial p(x) = x ± a the method of finding the
root is very familier to all and it will be given by x = ±a. Again, if the monic quadratic
polynomial over F, say, p(x) = x√2 + ax + b the method of finding the roots are very
2
familier and is given by x = −a ± 2a − 4b .
We will study now the 16th century method of finding roots of cubic polynomial
p(x) = x3 + a2 x2 + a1 x + a0 ∈ C[x] due to Jerone Cardan who seems to have obtained
some preliminariy versions from Niccola Tartaglia by some what disrespectable manner.
The monic cubic polynomial p(x) can be transformed into one with no quadratic term
by a change of variables x 7−→ x − a32 to give the equation
a22
a1 a2 2a22
a2
3
x − a0 +
+
∈ C[x].
q(x) = p(x − ) = x − a1 −
3
3
3
27
Clearly finding the roots of p(x) is equivalent to finiding roots of q(x).
2.2 Automorphisms of rings and fields
3. FIELD THEORY
3.1 Extension Fields
In this section, we study the concept of extension of a given field. Also, we discuss the
concepts of finite extension, algebraic extension and the concept of algebraic numbers.
We recall some basic defjnitions and results before giving the definition of an extension
field. A commutative ring with unity in which every non-zero element has a multiplicative
inverse is called a field. For example, the set of rationals Q, the set of reals R are fields
under usual addition and multiplication numbers. Similarly, the set of complex numbers
C is a field under addition and multiplication of complex numbers and the set of integers
modulo p, Jp is a field under addition and multiplication of integers modulo p where p
is a prime number. Any field having only finite number of elements is called a finite field
and those having infinte elements are called infinite fields.
Lemma 3.1.1: Every field is an integral domain.
Definition 3.1.2: A field K is said to be an extension of the field F if F is a subfield of
K.
Example 3.1.3:
1. (R, +, ·) is an extension of (Q, +, ·)
2. (C, +, ·) is an extension of (R, +, ·)
Theorem 3.1.4:
If L is a finite extension of K and K is a finite extension of F then L is a finite
extension of F. Moreover [L : F ] = [L : K][K : F ].
Proof:
To prove the theorem we have to find a basis of L over F.
6
Given that L is a finite extension of K. Hence the dimension of L as a vector space
over K is finite.
Assume that [L : K] = m.
Hence there exists a finite basis consisting of m elements for L treated as a vector
space over K.
Let v1 , v2 , · · · , vm be a basis of L over K.
Also given that K is a finite extension of F.
Hence dimension of K as a vector space over F is finite.
Let [K : f ] = m.
Then there exists a finite basis consisting of n elements for K treated as a vector
space over F.
Let w1 , w2 . · · · , wn be a basis for K over F.
Claim: {vi wj | 1 ≤ i ≤ m, 1 ≤ j ≤ n} is a basis for L over F.
Let t ∈ L, since L
K, {v1 , v2 , · · · , vm } ∈ L
t = k1 v1 + · · · + km vm
⊇
K
and {v1 , v2 , · · · , vm } is a basis of L over
· · · · · · (1)
where k1 , · · · , km ∈ K.
Now K ⊇ F and {w1 , · · · , wn } is a basis of K over F.
∴ If k1 , · · · , km ∈ K
ki = fi1 w1 + · · · + fin wn ,
· · · · · · (2) fij ∈ F and i = 1, · · · , m
Substituing (2) in (1),
t = (f11 w1 + · · · + fnn wn )v1 + · · · + (fm1 w1 + · · · + fmn wn )vm
= f11 v1 w1 + · · · + f1n v1 wn + · · · + fij vi wj + · · · fmn vm wn ,
fij ∈ F.
∴ {vi wj | i = 1, · · · m, j = 1, · · · n} spans L over F.
Now suppose fi1 v1 w1 + · · · + fin v1 wn + · · · + fij vi wj + · · · + fmn vm wn = 0, fij ∈ F.
To show fij = 0 ∀i, j.
From (fi1 w1 + · · · + fin wn )vi = 0, i = 1, · · · m
7
Now wi ∈ K and K ⊃ F
∴ ki = fi1 w1 + · · · + fin wn ∈ K
Now k1 v1 + · · · + km vm = 0 with k1 , · · · , km ∈ K.
But v1 , · · · , vm are lenearly independent as thay form a basis for L over K.
Hence fi1 = · · · = fim = 0 ∀i = 1, · · · , m.
ie, fij = 0 for i = 1, · · · , m, j = 1, · · · , n
⇒ {vi wj | i = 1, · · · , m, j = 1, · · · , n} forms a basis of L over F.
Since it has mn elements [L : F ] = mn.
ie, [L : F ] = [L : K][K : F ].
Definition 3.1.5:
Let K be an extension of fields F.
An element a ∈ K is said to be algebraic over F, if there exists elements
α0 , α1 , · · · , αn ∈ F not all zero such that α0 + α1 a + · · · + αn an = 0. (or)
If K is an extension of field F, then an element a ∈ K is said to be algebraic over
F, if a is a root of a non- zero polynomial p(x) = α0 + α1 x + α2 x2 + · · · + αn xn where
atleast one αi 6= 0 over F.
Definition 3.1.6:
1. A monic polynomial in x sap p(x) over F of lowest positive degree satisfied by an
element a ∈ K where K ⊃ F is called a minimal polynomial for a over F.
2. A polynomial in x is said to be monic if the coefficient of the term in highest degree
of x is 1.
Example 3.1.7:
1.
√
2 ∈ R is algebraic over Q for it satisfies the polynomial x2 − 2 ∈ Q[x].
x2 − 2 is the minimal polynomial over Q.
2. The element i ∈ C is algebraic over Q for it satisfies the polynomial x2 + 1 ∈ Q[x].
Also i ∈ C is algebraic over R as it satisfies x2 + 1 ∈ R[x].
8
3. The real number e and π are not algerbaic over Q. However, π is algebraic over
R, for it satisfies the polynomial (x − π) ∈ R[x].
p
√
1 + 3 ∈ R is algebraic over Q. For,
4.
p
√
√
√
if α = 1 + 3, α2 = 1 + 3. ∴ α2 − 1 = 3
ie, (α2 − 1)2 = 3
∴ α4 − 2α2 + 1 = 3.
p
√
ie, α4 − 2α2 − 2 = 0. Thus α = 1 + 3 is a zero of x4 − 2x2 − 2 ∈ Q[x].
Definition 3.1.8:
An element α ∈ E which is not algebraic over F is called transcendental over
F, F ⊂ E.
Result:
Let K be an extension of F.
Let a ∈ K be algebraic over F. Then
1. The minimal polynomial p(x) of 0 a0 over F is irreducible over F and
2. If q(x) ∈ F [x] and q(a) = 0, then p(x) | q(x).
Notation:
Let F be a given field and K be an extension of F. let a ∈ K by F (a) we men the
smallest subfield of K containing the both the field F and the element a F (a) is called
the subfield of K obtained by adjoiningthe element a to F.
Theorem 3.1.9:
Let K be an extension of F and let a ∈ K then a is algebraic over F if and only
if [F (a) : F ] < ∞.
Proof:
Assume that [F (a) : F ] < ∞
Let [F (a) : F ] = n then dimF F (a) = n.
Since dimF F (a) = n, any n + 1 elements of F (a) will be linearly dependent.
Hence the n + 1 element 1, a, a2 , · · · , an of F (a) are linearly dependent over F.
9
∴ there exist α0 , α1 , · · · , αn ∈ F not all zero such that
α0 + α1 a + α2 a2 + · · · + αn an = 0.
⇒ a satisfies a non-zero polynomial F (x) = α0 + α1 x + α2 x2 + · · · + αn xn over F.
ie, a ∈ K is algebraic over F.
Conversely , assume that a is algebraic over F.
To prove that [F (a) : F ] < ∞.
Assume that p(x) is the minimal polynomial of a over F.
ie, p(x) ∈ F [x] such that p(a) = 0 where degree of p(x) is minimum.
Let degree of p(x) = k then
p(x) = p0 + p1 x + p2 x2 + · · · + pk−1 xk−1 + xk where p0 , p1 , · · · , pk−1 ∈ F.
Let L = β0 + β1 a + · · · + βk−1 ak−1 | β0 , β1 , · · · , βk−1 ∈ F .
claim: L = F (a).
Given that p(a) = 0.
∴ p0 + p1 a + p2 a2 + · · · + pk−1 ak−1 = ak = 0.
⇒ ak = −(p0 + p1 a + p2 a2 + · · · + pk−1 ak−1 )
· · · · · · (1).
From (1) we can write any term in ak (or) higher powers of a as a n element in L.
⇒ the product of any two elements in L is again an element in L.
Let β0 + β1 a + · · · + βk−1 ak−1 6= 0 ∈ L.
∴ f (x) = β0 + β1 x + β2 x2 + · · · + βk−1 xk−1 ∈ F [x].
degree f (x) ≤ k − 1.
p(x) being the minimal polynomial satisfying by a is irreducible over F. and
degree of f (x) ≤ k − 1 < k = degree of p(x).
∴ g.c.d (p(x), f (x)) = 1.
We can find g(x), h(x) ∈ F [x] such that g(x)p(x) + h(x)f (x) = 1.
Put x = a,
But p(a) = 0
g(a)p(a) + h(a)f (a) = 1.
∴ h(a)f (a) = 1.
10
Using (1) h(a) can be written as an element of the form
r0 + r1 a + r2 a2 + · · · + rk−1 ak−1 which is an element in L.
⇒ f (a) = β0 + β1 a + β2 a2 + · · · + βk−1 ak−1 6= 0 ∈ L has an inverse h(a) in L.
Thus L is a field. Obviously L contains both F and a and moreover any field
containing F and a must contain L.
But by construction F (a) is the smallest field containing both F and a
∴ L = F (a)
Claim: [L : F ] = k.
ie, L is a vector space over F has a basis of k elements.
Also 1, a, a2 , · · · , ak−1 spans L over F.
Suppose that β0 + β1 a + · · · + βk−1 ak−1 = 0 where β0 , β1 , · · · , βk−1 ∈ F.
⇒ a is a root of the polynoimial β0 + β1 x + β2 x2 + · · · + βk−1 xk−1 ∈ F [x]
which is either zero (or) has degree ≤ k − 1 < k, the degree of the polynomial p(x)
of a.
⇒ β0 = 0, β1 = · · · = βk−1 = 0.
⇒ the elements 1, a, a2 , · · · , ak−1 are linearly independent over F.
Thus the set 1, a, a2 , · · · , ak−1 form a basis of L over F.
⇒ That the [F (a) : F ] = k, a finite number.
Thus F (a) is a finite extension of F.
Definition 3.1.10:
The element a in K is said to be algebraic of degree n over F if it satisfies a nonzero polynomial of degree n over F but no non-zero polynomial of lower degree that
is an element a in K is said to be algebraic of degree n over F if the degree of the
minimal polynomial satisfied by a over F is n.
Theorem 3.1.11: If a in K is algebraic of degree n over F then the degree of F (a)
over F is n.
ie, ([F (a) : F ] = n).
11
Converse part of previous theorem.
Theorem 3.1.12:
If a, b ∈ K are algebraic over F then a ± b, ab,
a
(b
b
6= 0) are all algebraic over F.
Proof:
a ∈ K is algebraic over F.
∴ [F (a) : F ] is finite.
Let [F (a) : F ] = m = deg p(x) where p(x) is the minimal monic polynomial over F
satisfied by a.
Since b ∈ K is algebraic over F, [F (b) : F ] is finite.
Let [F (b) : F ] = n = deg q(x) where q(x) is the minimal monic polynomial over F
satisfied by b.
Since F ⊆ F (a),
q(b) = 0.
q(x) can also be considered as a polynomial over F (a) and
∴ b is algberaic over F (a) and is of degree ≤ n.
ie, [F (a, b) : F (a)] ≤ n
F ⊆ F (a) ⊆ F (a, b)
By theorem 3.1.4,
[F (a, b) : F ] = [F (a, b) : F (a)] · [F (a) : F ]
≤m·n
ie, F (a, b) is the smallest subfield containg both F and the elements a and b.
∴ a ± b, ab,
a
(b
b
6= 0) are all elements in F (a, b).
Since F (a, b) is a finite extension of F
a ± b, ab,
a
(b
b
6= 0) are all algebraic over F.
⇒ algebraic elements in K form a subfield of K.
Corollary 3.1.13:
If a, b ∈ K are algebraic over F of degree m and n respectvely then a±b, ab,
0) are all algebraic over F of degree atmost mn.
Definition 3.1.14:
a
(b
b
6=
12
An extension K of F is said to be an algebraic extension of F if every element of
K is algebraic over F.
Theorem 3.1.15:
Algebraic extension of an algberaic extension is algebraic.
ie, If L is an algebraic extension of K and K is an algebraic extension of F then
L is an algebraic extension of F.
Proof:
Let u ∈ L, since L is an algebraic extension of K, u is algebraic over K.
ie, there is a polynomial p(x) = k0 + k1 x + k2 x2 + · · · + kn−1 xn−1 + xn where
k0 , k1 , k2 , · · · , kn−1 ∈ K such that p(a) = 0.
Now K is an algebraic extension of F.
Hence each Ki , i = 0.1, · · · , n − 1 is algebraic over F.
∴ [F (k0 , k1 , · · · , kn−1 ) : F ] is finite
· · · · · · (1).
Clearly, k0 , k1 , · · · , kn−1 ∈ F (k0 , k1 , · · · , kn−1 ).
Let F (k0 , k1 , · · · , kn−1 ) ∈ M
since it satisfies the minimal polynomial p(x) over M,
[M (a) : M ] is finite
· · · · · · (2).
(1) ⇒ [M : f ] is finite
· · · · · · (3).
By theorem 3.1.4 [M (a) : F ] = [M (u) : M ][M : F ]
⇒ [M (u) : F ] is finite.
ie, u is algebraic over F. Since this is true for any element u ∈ L, L is an algebraic
extension of F.
3.2 Transcendental Numbers
A real number x is called algebraic if it is the root of a polynomial with integer coefficients.
√
Examples of algebraic numbers are 21 (it is the root of the polynomial p(x) = 2x − 1, 2
which is the root of the polynomial p(x) = x2 − 2, the golden ratio (two quantities are
13
in the golden ratio if their ratio is the same as the ratio of their sum to the larger of the
two quantities) given by the roots of the polynomial p(x) = x2 − x − 1, and the single
real root of the quintic polynomial x5 − x + 1 (which cannot be expressed with radicals).
Definition 3.2.1: A real number that is not algebraic over Q is called a transcendental
number.
It is easy to see that every transcendental number is irrational, but, as we see above,
not every irrational number is transcendental. In 1874 Georg Cantor proved that there
are only countably many algebraic numbers. Thus there must be uncountably many
transcendental numbers! The vast, vast majority of real numbers are transcendental.
Despite the fact that almost every real number is transcendental, it is very difficult
to prove that a given number is transcendental. Joseph Liouville discovered the first
transcendental number in 1844:
∞
X
10−n! = 0.1100010000000000000000010 · · ·
n=1
In 1873 Charles Hermite proved that e is transcendental and nine years later Ferdinand
von Lindemann proved that π is transcendental. Some other transcendental numbers
√
are: eπ , 2 2 , ln(2), sin(1), and ii (yes, this is a real number: e−π/2 .
Definition 3.2.2: A subset S of a field E is said to be algebraically independent over a
subfield F if the elements of S donot satisfy any non-trivial polynomial over the subfield
F.
The LindemannWeierstrass theorem states the following:
If α1 , α2 , · · · , αn are algebraic numbers that are linearly independent over the field of
rationals Q, then eα1 , eα2 , · · · , eαn are algebraically independent over Q.
In other words the extension field Q(eα1 , eα2 , · · · , eαn ) has transcendence degree n
over Q.
An equivalent formulation: If α1 , α2 , · · · , αn are distinct algebraic numbers, then the
exponentials eα1 , eα2 , · · · , eαn are linearly independent over the algebraic numbers.
The theorem is named for Ferdinand von Lindemann and Karl Weierstrass.
Lindemann proved in 1882 that eα is transcendental for every non-zero algebraic number
α, thereby establishing that π is transcendental. Weierstrass proved the above more
general statement in 1885.
14
Suppose α is a non zero algebraic number; then {α} is a linearly independent set
over the rationals, and therefore by the first formulation of the theorem {eα } is an
algebraically independent set; or in other words eα is transcendental. In particular,
e1 = e is transcendental.
This proof that e is transcendental is based on Adolf Hurwitzs 1893 simplification of
Hermites proof.
Theorem 3.2.3:
The number e is transcendental.
Proof:
Let f (x) be a real polynomial of degree r.
Let f i (x) denote the ith derivative of f (x) with respect to x.
Consider F (x) = f (x) + f 1 (x) + f 2 (x) + · · · + f r (x)
· · · · · · (1).
(1) × e−x gives
e−x F (x) = e−x [f (x) + f 1 (x) + · · · + f r (x)].
Differentiate with respect to x,
d −x
d −x
[e F (x)] =
e [f (x) + f 1 (x) + · · · + f r (x)]
dx
dx
e−x
d
[F (x)] − e−x F (x) = e−x [f 1 (x) + f 2 (x) + · · · + f r (x) + f r+1 (x)]
dx
− e−x [f (x) + f 1 (x) + f 2 (x) + · · · + f r (x)]
e−x F 1 (x) − e−x F (x) = e−x f r+1 (x) − e−x f (x)
= 0 − e−x f (x)
= −e−x f (x)
For any positive interval k, consider the closed interval [0, k].
e−x F (x) is continuousin [0, k] and is differentiable in (0, k). Apllying Mean value
theorem to e−x F (x) we have,
e−k F (k) − F (0)
= e−kθk F (kθk ), 0 < k < 1.
k
ie,
e−k F (k) − F (0)
= −e−kθk f (kθk )
k
15
ie, e−k F (k) − F (0) = −ke−kθk f (kθk )
· · · · · · (2)
where θk depends on k and 0 < θk < 1.
(2) × ek ,
⇒ F (k) − F (0)ek = −ke(1−θk )K f (kθk )
· · · · · · (3).
Writing (3) explicitely for k = 1, 2, · · · , n we get
F (1) − F (0)e = −e(1−θ1 ) f (θ1 ) = 1
F (2) − F (0)e2 = −2 e(1−θ2 )2 f (2θ2 ) = 2
..
.
F (n) − F (0)en = −n e(1−θn )n f (nθn ) = n
Suppose that e is an algberaic number.
∴ e satisfies some relation of the form
cn en + cn−1 en−1 + · · · + c1 e + c0 = 0
· · · · · · (5).
where c0 , c1 , c2 , · · · , cn are all integers , c0 > 0.
In the relation (4) multiply the first by c1 , second by c2 , · · · , nth by cn respectvely
to get
c1 F (1) − c1 eF (0) = c1 1
c2 F (2) − c2 e2 F (0) = c2 2
..
.
cn F (n) − cn en F (0) = cn n
Adding the relation on both sides, we get
c1 F (1) + c2 F (2) + · · · + cn F (n) − F (0)
[c1 e + c2 e2 + · · · + cn en ] = c1 1 + c2 2 + · · · + cn n
· · · · · · (6).
But from (5) , c1 e + c2 e2 + · · · + cn en = −c0 using this in (6),
c0 F (0) + c1 F (1) + c2 F (2) + cdots + cn F (n) = c1 1 + c2 2 + · · · + cn n
· · · · · · (7).
This relation (7) is true for any F (x) constructed from any polynomial f (x).
Hence let us consider the polynomial f (x) as the Hermite polynomial,
f (x) =
1
xp−1 (1
(p−1)!
− x)(2 − x)p · · · (n − x)p where p is a prime number.
Choose p such that p > n and p > c0 when expanded, we get
16
f (x) =
(n!)p p−1
x
(p−1)!
+
a0 xp
(p−1)!
+
a1 xp+1
(p−1)!
+ ···
· · · · · · (8)
where a0 , a1 , a2 , · · · are all integer.
Lemma 3.2.4:
If g(x) is a polynomial with integer coefficients and if pi is a prime number, then for
g(x)
] is an polunomial with integer coefficients each of which is divisible by
i ≥ p, dxd i [ (p−1)!
p.
Proof of the lemma:
Let g(x) = a0 + a1 x + a2 x2 + · · · + ap xp + ap+1 xp+1 + · · · + an xn
g(x)
(p−1)!
=
a0
(p−1)!
d
[ g(x) ]
dx (p−1)!
=
d2
[ g(x) ]
dx2 (p−1)!
=
+
a1
(p−1)!
a1
(p−1)!
+
2a2
(p−1)!
x+
2a2
(p−1)!
a2
(p−1)!
x2 + · · · +
x + ··· +
+ ··· +
p(p−1)
(p−1)!
pap
(p−1)!
ap
(p−1)!
xp−1 +
ap xp−2 +
(p+1)p
(p−1)!
xp +
ap+1
(p−1)!
(p+1)ap+1
(p−1)!
xp+1 + · · · +
xp + · · · +
ap−1 xp−1 + · · · +
an
(p−1)!
nan
(p−1)!
n(n−1)
(p−1)!
xn
xn−1
an xn−2
..
.
dp
[ g(x) ]
(p−1)! (p−1)!
=
p(p−1)(p−2)···1
(p−1)!
ap +
(p+1)p(p−1)···2
(p−1)!
ap+1 x + · · ·
= pap + (p + 1)p ap+1 x + · · ·
⇒ For i ≥ p,
divisible by p.
di
[ g(x) ]
dxi (p−1)!
is a polynomial in x with integer co-efficients which are
Now applying the above lemma to equation (8) we see that for i ≥ p, f i (x) is a
polynomial in x with integer coefficients which are divisible.
Thus for any integer j, f i (j) for i ≥ p is an integer divisible by p.
· · · · · · (9).
Also from the definition of the polynomial f (x), we see that f (x) has a root of
multiplicity p at x = 1, 2, · · · , n.
∴ for j = 1, 2, · · · , n f (j) = 0, f 1 (j) = 0, f 2 (j) = 0, · · · , f (p−1) (j) = 0.
By our choice
F (x) = f (x) + f i (x) + f 2 (x) + · · · + f p−1 (x) + f p (x) + · · · + f r (x).
Put x = j
F (j) = f (j) + f 1 (j) + f 2 (j) + · · · + f p−1 (j) + f p (j) + · · · + f r (j).
F (j) = f p (j) + f (p+1) (j) + · · · + f r (j)
17
using (9), F (j) is an integer divisible by p for j = 1, 2, · · · , n.
Also f (x) has a root of multiplicity at x = 0,
∴ f (0) = 0, f 1 (0) = 0, · · · , f (p−2) (0) = 0
For i ≥ p, f i (0) is also an integer divisible by p.
But f (p−1) (0) = (n!)p
Since p > n and p is a prime number p - (n!)p .
∴ f (p−1) (0) is an integer not divisible by p.
But F (0) = f (0) + f 1 (0) + · · · + f (p−2) (0) + f (p−1) (0) + f p (0) + · · · + f r (0).
= f p−1) (0) + f p (0) + · · · + f r (0)
∴ F (0) is an integer which is not divisible by p (ie) p - F (0).
Since c0 > 0 and p > c0 , p - F (0) where as p - F (j) for j = 1, 2, · · · n, we conclude
that c0 F (0)+c1 F (1)+c2 F (2)+· · ·+cn F (n) is an integer not divisible by p. · · · · · · (10).
For any i = 1, 2, · · · , n
i = −ie(1−θi )i f (iθi )
Using the form of f (x) we get
p−1 (1−iθ )p (2−iθ )p ···(n−iθ )p
i
i
i
i = −ie(1−θi )i (i−iθi )
|i | ≤
(p−1)!
where 0 < θi < 1.
en np (n!)p
(p−1)!
Taking limit as p → ∞ we get
en np (n!)p
(p−1)!
→ 0 (ie) |i | → 0 as p → ∞.
Hence we can find a prime number p such that p > c0 and p > n a nd large enough
so that |c1 1 + c1 2 + · · · + cn n | < 1.
But c1 1 + c2 2 + · · · + cn n = c0 F (0) + c1 F (1) + · · · + cn F (n) is an integer.
∴ c1 1 + c2 2 + · · · + cn n = 0.
(ie) c0 F (1) + c1 F (1) + · · · + cn F (n) = 0.
Now p | 0 but p - [c0 F (0) + c1 F (1) + · · · + cn F (n)]
This is a contradiction.
(by(10))
18
∴ e cannot be algebraic.
(ie) e must be a transcendental number.
3.3 Roots of Polynomials
Lemma 3.3.1: Remainder Theorem:
Let p(x) ∈ F (x). Let K be an extension of F. For any b ∈ K,
p(x) = q(x)(x − b) + P (b) where q(x) ∈ K[x] and deg q(x) = deg p(x) − 1.
proof:
Since K is an extension of F
(ie) since F ⊆ K,
F [x] ⊆ K[x] for some indeterminate x.
∴ p(x) ∈ F [x] ⇒ p(x) ∈ K[x].
Since b ∈ K, x + b ∈ K[x].
By division algorithm, we can find polynomials q(x), r(x) ∈ K[x] such that
p(x) = q(x)·(x−b)+r(x) · · · · · · (1) where r(x) = 0 (or) deg r(x) < deg (x−b) = 1.
r(x) = 0 (or) deg r(x) < deg (x − b) ⇒ r(x) = 0, deg r(x) = 0.
∴ r(x) reduce to a constant polynomial in K[x], say r.
∴ (1) becomes p(x) = q(x)(x − b) + r
· · · · · · (2).
Put x = b, p(b) = 0 + r (ie) r = p(b).
∴ (2) gives p(x) = q(x) · (x − b) + p(b)
· · · · · · (3).
F rom(1), deg p(x) = deg [q(x)(x − b)] + deg r(x)
= deg q(x) + deg (x − b) + 0
deg p(x) = deg q(x) + 1
⇒ deg q(x) = deg p(x) − 1
Corollary 3.3.2:
If b ∈ K is a root of p(x) ∈ F [x] then (x − b) | p(x).
Since b ∈ K is a root of p(x) ∈ F [x],
19
p(b) = 0
∴ the above theorem shows that p(x) = q(x) · (x − b)
⇒ (x − b) | p(x).
(ie ) (x − b) divides p(x).
Definition 3.3.3:
Let F be a field. Let p(x) ∈ F [x], x = a is said to be a multiple root of multiplicity
m if (x − a)m | p(x) but (x − a)m+1 - p(x).
Lemma 3.3.4: Fundamental theorem on Algebra:
A polynomial of degree n over a field F can have atmost n roots in any extension
field.
Proof:
We prove the theorem by induction on n.
Let n = 1 then
f (x) = α1 x + α0 ,
α1 6= 0,
α0 .α1 ∈ F.
∴ α1−1 ∈ F.
If a is a root, f (a) = 0.
α1 a + α0 = 0 ⇒ a =
−α0
α1
∈ F.
Thus f (x) has exactly one root in K = F ⊇ F, thus proving the theorem.
Assume now that the theorem is true for any polynomial over F of degree < n.
(ie) If p(x) is a polynomial o ver F of degree ≥ n − 1, then by induction we assume
that p(x) has atmost (n − 1) roots in some extension of F.
Let f (x) = α0 + α1 x + · · · + αn xn ,
αi ∈ F (1 ≤ i ≤ n), αn 6= 0 be a polynomial over F of degree n.
Let F ⊆ K. If f (x) has no roots in K, then the result is true.
So assume that f (x) has atleast one root a ∈ K of multiplicity m.
∴ (x − a)m | f (x) but (x − a)m+1 - f (x).
Now deg (x − a)m = m ≤ deg f (x) = n (ie ) m ≤ n.
20
Since (x − a)m | f (x) in K[x], we can find a q(x) in K[x] such that
f (x) = (x − a)m · q(x).
Also (x − a)m+1 - f (x) ⇒ (x − a)m+1 - (x − a)m · q(x0
⇒ (x − a) - q(x).
⇒ x = a is not a root of q(x).
deg f (x) = deg (x − a)m + deg q(x)
(ie) n = m + deg q(x)
∴ deg q(x) = n − m < n.
Let b 6= a be a root of p(x) in K
∴ b − a 6= 0 (b − a)m 6= 0
f (b) = 0.
(ie) (b − a)m q(b) = 0
But (b − a)m 6= 0 ∈ K, q(b) ∈ (K) and K is a field.
Since K has no zero divisor, q(b) = 0.
Thus any root of f (x) other then a in K will be a root of q(x) in K.
Since degree of q(x) has atmost (n − m) roots in K. Other than a.
Thus f (x) has atmost (n − m) roots other then a ∈ K.
This implies that f (x) has atmost m + (n − m) = n roots in K.
Hence the proof.
Theorem 3.3.5:
If p(x) is a polynomial in F [x] of degree n ≥ 1 and is irreducible over F, then there
is an extension E of F such that [E : F ] = n in which p(x) has a root.
Proof:
Let V = (p(x)), be the ideal of F [x] generated by p(x).
Since p(x) is irreducible over F, V is a maximal ideal iin F [x].
∴ E=
F [x]
V
is a field.
21
Any element in E will be of the form V + f (x) where f (x) ∈ F [x] and
Let us denote this element by f (x).
To prove the theorem, define σ : F → E by σ(a) = a ∀a ∈ F.
Claim: σ is isomorphism.
For any a, b ∈ F, σ(a + b) = a + b = V + (a + b)
= (V + a) + (V + b) = a + b = σ(a) + σ(b) and
= σ(ab) = ab = V + ab = (V + a)(V + b) = a · b
= σ(a) · σ(b)
∴ σ is a homomorphism.
Let σ(a) = σ(b)
⇒ a=b
⇒ V +a=V +b
⇒ a−b∈V
Since V = (p(x)), a − b = f (x) · p(x) for some P (x) ∈ F [x].
If f (x) 6= 0 deg (f (x) · p(x)) = deg f (x) + deg p(x)
≥n>0
But deg (f (x) · p(x)) = deg (a − b) = 0
which is a contradiction. Hence we conclude that f (x) = 0.
∴ a − b = 0 ⇒ a = b (ie) σ is 1 -1.
Thus σ is an isomorphism from F into E.
Inparticular σ can be considered as this restriction of a mapping
ψ : F [x] → E =
(ie), σ =
ψ
F
F [x]
V
to F.
meaning ψ(a) = σ(a) ∀a ∈ F.
Then we can identify F and F = σ(F ) = Im σ in E.
Thus E can be regarded as an extension of F.
claim: p(x) has a root in E.
22
Let p(x) = a0 + a1 x + a2 x2 + · · · + an xn , an 6= 0
a0 , a1 , · · · , an ∈ F ⊆ E.
∴ p(x) can be considered as an element in E.
(ie) p(x) = a0 + a1 x + a2 x2 + · · · + an xn
then p(x) = a0 + a1 x + a2 x2 + · · · + an xn
= a0 + a1 x + a2 x 2 + · · · + an x n
= p(x)
the zero element of E ⇒ x is a root of p(x).
This implies that p(x) has a root x, in E.
Claim: [E : F ] = n.
To prove this, we have to prove that
n
o
1, x, x2 , · · · , xn−1 is a basis of E over
F.
Let f (x) ∈ E. Then f (x) = V + f (x) for some f (x) ∈ F [x].
Now p(x), f (x) ∈ F [x]. By division algorithm we can find q(x), r(x) ∈ F [x]
such that f (x) = q(x) · p(x) + r(x) where r(x) = 0 (or) deg r(x) < deg p(x) = n.
f (x) = q(x)p(x) + r(x)
= q(x) · p(x) + r(x) = r(x) for q(x), p(x) ∈ V.
(ie) f (x) = r(x) = b0 + b1 x + · · · + bn−1 xn−1
= b0 + b1 x + · · · + bn−1 xn−1
where b0 , b1 , b2 , · · · , bn−1 ∈
F [x]
V
⇒ {1, x, x2 , · · · , xn−1 } Spans FV[x] over F.
n
o
To show that 1, x, x2 , · · · , xn−1 is a linearly independent set, consider
c0 + c1 x + c2 x2 + · · · + cn−1 xn−1 ∈ V = zero of
F [x]
V
c0 , c1 , c2 , · · · , cn−1 ∈ F.
⇒ c0 + c1 x + c2 x2 + · · · + cn−1 xn−1 ∈ V = (p(x))
⇒ c0 + c1 x + c2 x2 + · · · + cn−1 xn−1 is a multiple of p(x).
⇒ c0 + c1 x + c2 x2 + · · · + cn−1 xn−1 is a multiple of p(x).
23
But deg (c0 + c1 x + · · · + cn−1 xn−1 ) = n − 1 < n = deg p(x).
⇒ c0 + c1 x + · · · + cn−1 xn−1 = 0
⇒ c0 = 0, c1 = 0, · · · , cn−1 = 0.
(ie) c0 = c1 = · · · = cn−1 = zero of FV[x]
n
o
2
n−1
∴ 1, x, x , · · · , x
is a linearly independent set over F.
Thus
n
o
1, x, x2 , · · · , xn−1 is a basis for E over F.
∴ [E =
F [x]
V
: F ] = n = deg p(x).
Corollary 3.3.6:
Let f (x) ∈ F [x] then there is a finite extension E of F in which f (x) has a root
and [E : F ] ≤ deg f (x).
Proof:
If f (x) ∈ F [x] is irreducible.
Then by the above theorem, we can find as extension E of F in which f (x) has a
root and [E : F ] = deg f (x).
Hence assume that f (x) is irreducible.
∴ f (x) can be written uniquely as a product of irreducible polynomials.
Let p(x) be an irreducible factor of f (x).
(ie) f (x) = p(x) · q(x)
deg p(x) ≤ degf (x)
Let a be a root of p(x) in some extension E of F. (ie) p(a) = 0.
⇒ f (a) = p(a) · q(a) = 0.
∴ any root of p(x) will also be a root of f (x).
Since p(x) is irreducible over F, by the above theorem, we can find an extension E
of F such that [E : F ] = deg p(x) in which p(x) has a root. Hence there is an finite
extension E of F in which f (x) has a root and
[E : F ] deg p(x) ≤ deg f (x).
24
Theorem 3.3.7:
Let f (x) ∈ F [x] be of degree n ≥ 1 then there is an extension E of F of degree
atmost n! in which f (x) has n roots [ and so, a full complement of roots]
Proof:
We shall prove the theorem by induction on deg f (x) = n.
Let n = 1 then f (x) = α0 + α1 x ∈ F [x]
If f (0) = 0, α0 + α1 a = 0
⇒ a = − αα01 ∈ F.
∴ the extension E can be treated to be the field F itself.
[E : F ] = 1 ≤ (deg f (x))!
Thus the theorem is true if deg f (x) = 1.
Assume now that the theorem is true when deg f (x) ≤ n − 1.
(ie ) assume that if g(x) has a polynomial of degree (n − 1) then we can find an
extension E of F in which g(x) has all its roots and [E : F ] ≤ (n − 1)!.
Let deg f (x) = n, f (x) ∈ F [x].
By the corollary, we can find an extension E0 of F in which f (x) has a root, say
α such that [E0 : F ] ≤ deg f (x) = n · · · · · · (1).
f (x) = (x − α) · g(x)
g(x) ∈ E0 [x] and deg g(x) = n − 1.
By induction hypothesis, there is an extension E of E0 in which g(x) has all its
(n − 1) roots such that [E : F0 ] ≤ (n − 1)! · · · · · · (2).
But all the roots of g(x) will also be a root of f (x).
Thus f (x) has all its n roots in E
But F ⊆ E0 ⊆ E
∴ [E : F ] = [E : E0 ] · [E0 : F ] ≤ (n − 1)!n = n!.
Definition 3.3.8: Splitting Field:
Let F be a field and f (x) be polynomial over F of degree n ≥ 1 then there is a
25
finite extension E of in which f (x) has n roots. An extension E of F in which f (x)
has n roots and [E : F ] is minimum is called a splitting field of f (x) over F.
Remark 3.3.9:
For every polynomial f (x) ∈ F [x] there exists a splitting field.
Definition 3.3.10:
Let F and F 0 be two isomorphic field under τ : F → F 0 . Let E and E 0 be extension
fields of F and F 0 respectively. An isomorphism σ : E → E 0 is called a continuation of
the isomorphism
τ : F → F 0 if σ(a) = τ (a) ∀a ∈ F.
Lemma 3.3.11:
Let F and F 0 be two fields. Let τ : F → F 0 defined by τ (a) = a0 ∀a ∈ F be onto
isomorphism. Then there is an onto isomorphism τ ∗ : F [x] → F 0 [t] of rings such that
τ ∗ (a) = a0 ∀a ∈ F.
Proof:
Given that τ : F → F 0 defined by τ (a) = a0 ∀ a ∈ F is an onto isopmorphism of
fields.
If f (x) = a0 + a1 x + a2 x2 + · · · + an xn ∈ F [x].
Define τ ∗ : F [x] → F 0 [t] given by
τ ∗ (f (x)) = a00 + a01 t + a02 t2 + · · · + a0n tn = f 0 (t).
Claim: τ ∗ is an onto isomorphism of rings.
τ ∗ is one - to - one :
Let f (x) = a0 + a1 x + a2 x2 + · · · + an xn and
g(x) = b0 + b1 x + b2 x2 + · · · + bm xm ∈ F [x]
be such that τ ∗ (f (x)) = τ ∗ (g(x)).
⇒ a00 + a01 t + a02 t2 + · · · + a0n tn = b00 + b01 t + b02 t2 + · · · + bn tm
⇒ n = m and a0i = b0j ∀ i, j = 0, 1, 2, · · · , n.
∵ a0i , b0j ∈ F 0 , τ : F → F 0 is onto, there exists ai , bj ∈ F such that
26
a0i = τ (ai ) and b0j = τ (bj )
∵ a0i = b0j ,
τ (ai ) = τ (bj )
∵ τ is one - one, ai = bj
∀ i, j.
∀ i, j = 0, 1, 2, · · · , n
Now n = m and ai = bj
f (x) = g(x)
(ie) τ ∗ is one - to -one.
τ ∗ is a homomrphism:
Let f (x) = a0 + a1 x + a2 x2 + · · · + an xn and
g(x) = b0 + b1 x + b2 x2 + · · · + bm xm ∈ F [x] and m ≥ n.
f (x) + g(x) = (a0 + b0 ) + (a1 + b1 )x + · · · + (an + bn )xn + bn+1 xn+1 + · · · + bm xm .
τ ∗ [f (x) + g(x)] = (a0 + b0 )0 + (a1 + b1 )0 t + · · · + (an + bn )0 tn + · · · + b0m tm
= (a00 + a01 t + a02 t2 + · · · + a0n tn ) + (b00 + b01 t + · · · + b0m tm )
Since (ai + bj )0 = τ (ai + bj ) = τ (ai + τ (bj )
= a0i + b0j
∀ i, j
[∵ τ is a homomorphism]
= τ ∗ (f (x)) + τ ∗ (g(x))
f (x) · g(x) = a0 b0 + (a0 b1 + a1 b0 )x + · · · +
(a0 bn + a1 bn−1 + · · · + an b0 )xn + · · · + an bm xm+n
τ ∗ (f (X) · g(x)) = (a0 b0 )0 + (a0 b1 + a1 b0 )0 + · · · +
(a0 bn + a1 bn−1 + · · · + an b0 )0 tn + · · · + (an bm )tm+n
= (a00 + a01 t + · · · + a0n tn ) · (b00 + b01 t + · · · + b0m tm )
= τ ∗ (f (x)) · τ ∗ (g(x))
⇒ τ ∗ is a homomorphism.
τ ∗ is onto :
Let f 0 (t) = a00 + a01 t + · · · + a0n tn ∈ F 0 [t].
⇒ a0i ∈ F 0
∀ i = 0, 1, 2, · · · , n.
27
Since τ is onto, we can find ai ∈ F (i = 0, 1, 2, · · · , n) such that a0i = τ (ai )
∴ f 0 (t) = τ (a0 ) + τ (a1 )t + τ (a2 )t2 + cdots + τ (an tn
= tau∗ (a0 ) + τ ∗ (a1 τ ∗ (x) + τ ∗ (a2 )τ ∗ (x2 ) + · · · + τ ∗ (an )τ ∗ (xn
F or a0 = τ (a) = τ ∗ (a)
= τ ∗ (a0 + a1 x + a2 x2 + · · · + an xn )
[∵ τ ∗ is a homomorphism]
= τ ∗ f (x).
where f (x) = a0 + a1 x + a2 x2 + · · · + an xn ∈ F [x].
Thus, for a given f 0 (t) ∈ F 0 [t], we can find a f (x) ∈ F [x] such that τ ∗ (f (x)) = f 0 (t).
∴ τ ∗ is onto.
Lemma 3.3.12:
Let τ : F → F 0 be an isomorphism of fields. Let for all a ∈ F, τ (a) be denoted
by a0 . Then there is an isomorphism τ ∗ of ring F [x] onto the ring F 0 (t) such that
τ ∗ (a) = a0 ∀a ∈ F. If for f (x) ∈ F [x] τ ∗ (f (x)) ∈ F 0 [t] is denoted by f 0 (t) then there
is an isomorphism
τ ∗∗ :
F [x]
(f (x))
→
F 0 [t]
(f 0 (t))
an onto isomorphism of rings such that τ ∗∗ (g(x)) = g 0 (t) where
g(x) and g 0 (t) are the cosets of f (x) and f 0 (t) respectively.
Inparticular τ ∗∗ (a) = a0 ∀ a ∈ F and τ ∗∗ (x + (f (x))) = t + (f 0 (t))
Proof:
Given that τ : F → F 0 is an onto isomorphism of fields.
Let τ (a) be denoted by a0 ∀ a ∈ F.
τ ∗ : F [x] → F 0 [t] is defined as follows.
If f (x) = a0 + a1 x + a2 x2 + · · · + an xn ∈ F [x],
τ ∗ (f (x)) = a00 + a01 t + a02 t2 + · · · + a0n tn = f 0 (t).
By previous lemma τ ∗ is an onto isomorphism of rings such that
τ ∗ (a) = a0 = τ (a) ∀ a ∈ F.
Let g(x) = g(x) + (f (x)) ∀ g(x) ∈ F [x].
Define τ ∗∗ :
F [x]
(f (x))
→
F 0 [t]
(f 0 (t))
given by τ ∗∗ (g(x)) = g 0 (t)
where g 0 (t) = g 0 (t) + (f 0 (t)).
28
τ ∗∗ is well defined:
Let g(x) = h(x) where g(x), h(x) ∈ F [x]
(ie) g(x) + (f (x)) = h(x) + (f (x))
⇒ g(x) − h(x) ∈ (f (x))
∴ g(x) − h(x) = α(x) · f (x).
But τ ∗ is an onto isomorphism of rings.
τ ∗ (g(x) − h(x)) = τ ∗ (g(x)) − τ ∗ (h(x))
= g 0 (t) − h0 (t)
But τ ∗ (g(x) − h(x)) = τ ∗ (α(x) · f (x))
= τ ∗ (α(x)) · τ ∗ (f (x))
= α0 (t) · f 0 (t)
∴ g 0 (t) − h0 (t) = α0 (t) · f 0 (t)
(ie) g 0 (t) − h0 (t) is a multiple of f 0 (t).
⇒ g 0 (t) − h0 (t) ∈ (f 0 (t))
⇒ g 0 (t) + (f 0 (t)) = h0 (t) + (f 0 (t))
(ie) g 0 (t) = h0 (t)
(ie) τ ∗∗ (g(x)) = τ ∗∗ (h(x))
⇒ τ ∗∗ is well defined.
τ ∗∗ is one - one :
Retracing the above steps, we find that τ ∗∗ is one - one .
τ ∗∗ is onto:
Let g 0 (t) ∈
F 0 [t]
(f 0 (t))
(ie) g 0 (t) = g 0 (t) + (f 0 (t)), g 0 (t) ∈ F 0 [t]
τ ∗ is onto implies there exists an element g(x) ∈ F [x]
such that τ ∗ (g(x)) = g 0 (t).
For this g(x) ∈ F [x], g(x) = g(x) + (f (x)) ∈
and τ ∗∗ (g(x)) = g 0 (t).
F [x]
(f (x))
29
⇒ τ ∗∗ is onto.
τ ∗∗ is homomorphism:
Let g(x), h(x) ∈
F [x]
(f (x))
g(x) + h(x) = g(x) + (f (x)) + h(x) + (f (x))
= (g(x) + h(x)) + (f (x))
= g(x) + h(x)
τ ∗∗ [g(x) + h(x)] = τ ∗∗ [g(x) + h(x)]
= g 0 (t) + h0 (t)
= g 0 (t) + h0 (t)
= τ ∗∗ (g(x)) + τ ∗∗ (h(x))
similarly g(x) · h(x) = [g(x) + (f (x))] · [h(x) + (f (x))]
= g(x) · h(x) + (f (x))
= g(x) · h(x)
∴ τ ∗∗ (g(x) · h(x)) = τ ∗∗ (g(x) · h(x)
= g 0 (t) · h0 (t)
= g 0 (t) · h0 (t)
= τ ∗∗ (g(x)) · τ ∗∗ (h(x))
⇒ τ ∗∗ is a homomorphism.
Thus τ ∗∗ :
F [x]
(f (x))
→
F 0 [t]
(f 0 (t))
is an onto isomorphism of rings.
For any a ∈ F, we can identify a with the coset a + (f (x)) in
In this way, F can be included in
F [x]
.
(f (x))
F [x]
.
(f (x))
Similarly for any a0 ∈ F 0 we can identify a0 with the coset a0 + (f 0 (t)) ini
In this way F 0 can be included in
F 0 [t]
.
(f 0 (t))
τ ∗∗ [a + (f (x))] = a0 + (f 0 (t)) = τ (a) + (f 0 (t))
τ ∗∗ (a) = a0 = τ (a) ∀ a ∈ F,
By choosing g(x) = x, we see that
τ ∗∗ (g(x) + (f (x))) = τ ∗∗ (x + (f (x)))
= t + (f 0 (t))
F 0 [t]
.
(f 0 (t))
30
Corollary 3.3.13:
If f (x) ∈ F [x] is irreducible, then τ ∗∗ becomes an isomorphism of fields.
Proof:
If f (x) is irreducible, since τ ∗ is an isomorphism ,
τ ∗ (f (x)) = f 0 (t) is also irreducible.
∴ (f (x)) and (f 0 (t)) are maximal ideals in F [x] and F 0 [t] respectvely.
Hence
F [x]
(f (x))
and
F 0 [t]
(f 0 (t))
are fields respectvely.
By the above theorem, τ ∗∗ is an onto isomorphism of fields.
Theorem 3.3.14:
Let F, F 0 be fields and τ : F → F 0 be an onto isomorphism (of F onto F 0 ). Let
p(x) be irreducible in F [x] and let v be a root of p(x) in some extension field E of
F. If w is a root of p0 (t) in some extension field E 0 of F 0 then there is an isomorphism
σ of F (v) onto F 0 (w) such that
1. σ(v) = w and
2. σ(a) = a0 = τ (a) ∀ a ∈ F.
Proof:
We know that
τ ∗∗ :
F [x]
(p(x))
→
F 0 [t]
(p0 (t))
is an onto isomorphism of fields.
Now define τ : F (v) →
F [x]
(p(x))
as follows:
If deg p(x) = n, then F (v) is a finite dimensional vector space over F with basis
1, v, v 2 , · · · , v n−1 then
ψ(a0 + a1 v + a2 v 2 + · · · + an−1 v n−1 ) = a0 + a1 x + · · · + an−1 xn−1
where f (x) ∈
F [x]
(p(x))
denotes the coset f (x) + (p(x)), f (x) ∈ F [x].
Thus ψ(f (v)) = f (x)
Let f (x) = a0 + a1 x + a2 x2 + · · · + an−1 xn−1 and
g(x) = b0 + b1 x + b2 x2 + · · · + bn−1 xn−1
31
so that f (v) and g(v) ∈ F (v).
∴ f (v) + g(v) = (a0 + b0 ) + (a1 + b1 )v + · · · + (an−1 + bn−1 )v n−1
∴ ψ[f (v) + g(v)] = f (x) + g(x) = {f (x) + g(x)} + (p(x))
= {f (x) + (p(x))} + {g(x) + (p(x))}
= f (x) + g(x)
= ψ(f (v)) · ψ(g(v))
To show ψ[f (v) − g(v)] = ψ(f (v)) · ψ(g(v)).
Now by division algorithm, forall f (x), g(x) ∈ F [x] there exist q(x), r(x) ∈ F [x] such
that
f (x) − g(x) = q(x) · p(x) + r(x)
· · · · · · (1).
where r(x) = 0 (or) deg r(x) < deg p(x).
Note that here we have used the given cindition that p(x) is irreducible over F.
∴ f (v) · g(v) = q(v) · p(v) + f (v) = r(v)
Since p(v) = 0 as v is given to be root of p(x).
∴ ψ(f (v) · g(v)) = ψ(r(v)) = r(x)
· · · · · · (2).
Also ψ(f (v)) = f (x) a nd ψ(g(x)) = g(x)
ψ(f (v)) · ψ(g(v)) = f (x) · g(x)
= q(x) · p(x) + r(x)
= q(x) · p(x) + r(x)
= r(x)
· · · · · · (3)
[∵ p(x) ∈ (p(x)) and (p(x)) isthe zero of
From (2) and (3) we get
ψ(f (v) · g(v)) = ψ(f (v)) · ψ(g(v))
⇒ ψ is a field homomorphism.
To show ψ is one - one :
Let ψ(f (v)) = ψ(g(v))
⇒ f (x) = g(x) ⇒ f (x) − g(x) is a multiple pf p(x).
Since deg (f (x) − g(x)) ≤ n − 1 and deg (p(x)) = n, p(x) is irreducible,
F [x]
]
(p(x))
32
we get f (x) − g(x) = 0 ⇒ ai = bi ∀ i = 0, 1, · · · , n − 1.
∴ f (v) = g(v) ⇒ ψ is one - one.
Clearly ψ is onto, For any element in
F [x]
(p(x))
is of the form f (x) + (p(x)).
If f (x) has a root in F say a, then f (a) ∈ F (v). Since f (a) = 0.
⇒ q(a) · p(a) + r(a) = 0 where r(a) = 0 (or) deg (r(a)) < deg p(a) ⇒ a is also a
root of p(x) and hence is an element in F (v).
If not, (ie) if a0 is not a root of f (x) ∈ F [x] then we can find an extension K of F
in which f (x) has a root. However F (v) ⊂ K and also f (v) ∈ F (v).
0
[t]
and show that ψ 0 is an onto
Similarly we define an mapping ψ 0 : F (w) → (pF0 (t))
isomorphism of fields. Since ψ 0 is an onto isomorphism of fields, (ψ 0 )−1 exists and is also
0 [t]
an isomorphism of the fields (pF0 (t))
onto F 0 (w).
Now define a morphism σ = ψτ ∗∗ (ψ 0 )−1 .
Then σ is an isomorphism of F (v) onto F 0 (w).
N ow σ(v) = [ψ τ ∗∗ (ψ 0 )−1 ](v)
= [τ ∗∗ (ψ 0 )−1 ] x
= (ψ 0 )−1 [τ ∗∗ (x)] = (ψ 0 )−1 (t) = w.
Thus σ(v) = w.
Let a ∈ F then σ(a) = [ψτ ∗∗ (ψ 0 )−1 )] (a)
= [τ ∗∗ (ψ 0 )−1 ]ψ(a) = [τ ∗∗ (ψ 0 )−1 ] a
= (ψ 0 )−1 [τ ∗∗ (a)] = (ψ 0 )−1 (a) = a0 .
Thus σ(a) = a0 = τ (a) ∀ a ∈ F.
Hence the theorem.
Corollary 3.3.15:
If p(x) ∈ F [x] is irreducible over F and if a, b are two roots of p(x) then F (a) is
isomorphic to F (b) by an isomorphism which takes a onto b and which leaves element
of F fixed.
Proof:
In the above theorem we can take the fields F 0 = F and p0 (t) = p(x) and
v = a, w = b.
33
∴ The result follows.
Theorem 3.3.16:
Let τ : F → F 0 be an onto isomorphism of fields. Let f (x), f 0 (t) be the polynomials
over F and F 0 respectively. If E and E 0 are splitting fields of f (x) and f 0 (t) over
F and F 0 respectively then there exists an onto isomorphism ϕ : E → E 0 such that
ϕ(a) = a0 = τ (a) ∀ a ∈ F.
Proof:
Given:
1. τ : F → F 0 is an onto isomorphism of fields.
2. f (x), f 0 (t) are polynomials over F and F 0 respectively.
3. E and E 0 are splitting fields of f (x) and f 0 (t) over F and F 0 respectively.
To prove there exists an onto isomorphism ϕ : E → E 0
such that ϕ(a) = a0 = τ (a) ∀ a ∈ F.
Since E is a splitting field of f (x) over F. E is a finite extension of F ( containing
the all the roots of f (x) )
Let [E : F ] = n (finite)
We shall prove the thoerem by induction on n.
Let n = 1
∴E=F
∴ f (x) ∈ F [x] splits into linear factor in F itself.
Since τ ∗ : F [x] → F 0 [t] is an onto isomorphism f 0 (t) = τ ∗ (f (x)) splits into linear
factors over F 0 and therefore E 0 = F 0 .
By taking ϕ = τ, the result follows.
Assume now that the theorem is true for all cases where [E : F ] < n.
Assume now that [E : F ] = n > 1.
If f (x) ∈ F [x] is irreducible over F by τ ∗
f 0 (t) ∈ F 0 [t] is also irreducible over F 0 .
Then by the above theorem, if a ∈ Esupset is a root of f (x) and if a ∈ E ⊃ F 0
is a root of f 0 (t), then there is an onto isomorphism σ : F (v) → F 0 (w) such that
σ(a) = a0 = τ (a) ∀ a ∈ F.
34
considering f (x) and f 0 (t) as polynomials over F (v) and F 0 (w) respectively. If
v1 (6= v) ∈ E and w1 (6= w) ∈ E 0 is another root of f (x) and f 0 (t) respectively .
By the theorem 3.3.14 there is an onto isomorphism σ1 : F (v, v1 ) → F 0 (w, w1 ) such
that σ1 (a) = σ(a) = a0 = τ (a) ∀ a ∈ F.
Continuing in this procedure for each root of f (x) and f 0 (t) respectively.
If E and E 0 are splitting fields of f (x) and f 0 (t) over F and F 0 respectively then
there is an onto isomorphism ϕ : E → E 0 such that ϕ(a) = a0 = τ (a) ∀ a ∈ F.
When f (x) ∈ F [x] is not irreducible, f (x) can be factorized into irreducible factors
over F.
Let p(x) be an irreducible factor of f (x) of degree r > 1.
Similarly f 0 (t) ∈ F 0 [t] is not irreducible over F 0 and therefore f 0 (t) can be factorized
into irreducible factors over F.
Let p0 (t) be an irreducible factor of f 0 (t).
Since E is a splitting field of f (x) over F. E contains all the roots of f (x) and
therefore contains all the roots of p(x). Hence there exists an element v ∈ E such
that p(x) = 0
∴ [F (v) : F ] = r (finite)
Similarly E 0 is a splitting field of f 0 (t) over F 0 .
⇒ E 0 contains all the roots of f 0 (t) and therefore contains all the roots of p0 (t).
Hence there exists an element w ∈ E such that p0 (w) = 0.
∴ [F 0 (w) : F ] = r (finite)
Thus by theorem 3.3.14 there exists an onto isomorphism σ : F (v) → F 0 (w) such
that σ(a) = a0 = τ (a) ∀ a ∈ F.
Now F ⊆ F (v) and f (x) ∈ F [x] ⊆ F (v)[x]
∴ E is also a splitting field of f (x), considered as a polynomial over F (v).
Similarly E 0 is also a splitting field of f 0 (t) considered as a polynomial over
F 0 (w) ⊇ F 0
Now F ⊆ F (v) ⊆ E and F 0 ⊆ F 0 (w) ⊆ E 0
By theorem 3.1.4, [E : F ] = [E : F (v)] · [F (v) : F ]
35
(ie) n = [E : F (v)] · r
∴ [E : F (v)] =
n
r
<n
∴ By induction hypothesis there exists an onto isomorphism ϕ : E → E 0
such that ϕ(a) = σ(a) = a0 = τ (a) ∀ a ∈ F.
This completes the proof of the theorem.
Corollary 3.3.17:
Any two splitting fields of the same polynomial over a field F are isomorphic and the
isomorphism leaves every element of F fixed. (ie) Splitting field of a polynomial over a
field is unique upto isomorphism.
Proof:
By taking F 0 = F, τ =|F and t = x in the above theorem.
3.4 Construction with Straight Edge and Campass
Definition 3.4.1: A real number α is said to be a Constructible number, if for a given
fundamental unit length, we can construct a line segment of lenth |α| by the use of
straight edge and compass alone.
Note : Using a straight edge and compass it is possible to eract a perpendicular to
a given line at a known point on the line and also, we can find a line passing through a
given point and parallel to a given line.
Result :1
If α and β are Constructible real numbers, then so are α 6= β, αβ and
α
β
(β 6= 0).
36
Proof: Since α and β are Constructible, there exist line segments of length |α| and
~ and OB
~ be the line segments of length respectvely .
|β|. Let OA
r
O
r
r
A
O
r
B
~ with the straight edge using the compass, start at
Now extend the line segments OA
~ of length |α| by off on the extension of length |β|
O if the segment OA
r
O
r
r
A
C
~ of length |α + β|. To construct α − β, from O we
This gives a line segment OC
~ of length |α|. This gives a line
lay off a length β by compass on the line segment OA
~ of length |α − β|.
segment CA
r
O
r
r
C
A
when both α and β are not positive, then dividing the above discussion into several
cases depending upon the signs of α and β. We see that α 6= β are Constructible.
To construct αβ.
~ be the line segment of length |α|. Consider a line l passing through O not
Let OA
~ Mark the points P and B on l such that OP
~ and OB
~ are of lengths
containing OA.
1 and |β| respectvely. Join P and A now construct a line l0 through B parallel to P~A
~ extended at the point Q. By similarity of the triangle OP A and OBQ,
intersecting OA
we have
1
|β|
=
~
α
|OA|
~ = |α||β| in OQ
~ is a line segment of length αβ.
⇒ |OA|
To construct
α
,
β
β 6= 0.
~ be the line segment of length |alpha|. Through O find a line l not containing
Let OA
~ Mark the points P and B on l of lengths 1 and |β| respectively. Join BA.
~ Through
OA.
37
~ intersecting the line segment OA
~ at Q. Since 4OP Q
P draw a line l0 parallel to BA,
similar to 4OBA, we have
~
~
|OA|
|OA|
=
~ |
~
|OP
|OB|
~ is a line segment of length
⇒ OQ
i.e
~
|α|
|OA|
=
1
|β|
α
.
β
Corollary 3.4.2: The set W of all Constructible real numbers forms a subfield of the field
of real numbers.
Note: Since 1 ∈ W, W must contain F, the field of rational numbers.
Definition 3.4.3:
Let F be any subfield of the field of real numbers. Consider all the points (x, y) in
the real Euclidean plane where x, y ∈ F. then PF = {(x, y)|x, y ∈ F } is called the plane
of F.
Result: 2 Let F be any subfield of the field of real numbers. Any straight line
joining two points in the plane of F has an equation of the form ax + by + c = 0 with
a, b, c ∈ F.
Proof:
Let (x1 , y1 ) and (x2 , y2 ) be two points in the plane of F. Then (x1 , y1 ) and (x2 , y2 )
are also the points in the real Euclidean plane. From geomentry, we know that, the
equation of the straight line joining two points (x1 , y1 ) and (x2 , y2 ) is given by
x − x1
y − y1
=
x1 − x2
y1 − y2
⇒ (y1 − y2 )(x − x1 ) = (x1 − x2 )(y − y1 ).
i.e, (y1 − y2 )x + (x2 − x1 )y + (y2 − y1 )x1 + (x1 − x2 )y1 = 0.
Taking a = y1 − y2 , b = x2 − x1 and c = (y2 − y1 )x1 + (x1 − x2 )y1 the above equation
takes the form ax + by + c = 0. Since x1 , x2 , y1 , y2 ∈ F, F is a field a, b, c ∈ F.
Result: 3
Let F ⊆ R be a subfield. A circle in the plane of F has an equation of the form
x + y 2 + ax + by + c = 0, a, b, c ∈ F.
2
Proof:
38
Let (x1 , y1 ) ∈ PF be the centre of the circle. Let r ∈ F be its radius. Then the
equation of the circle is given by (x − x1 )2 + (y − y1 )2 = r2 .
i.e, x2 + y 2 − 2xx1 − 2yy1 + x21 + y12 − r2 = 0
i.e, x2 + y 2 + ax + by + c = 0
Where a = −2x1 , b = −2y1 , c = x21 + y12 − r2 ∈ F.
Note:
By a circle in F we mean any circle having as center a point in the plane of F and
having as radius as element of F.
Result: 4
Two lines in the plane of F, F being any subfield of R, which intersect in the real
plane will intersect at a point in the plane of F.
Proof:
By result 2, any line in the plane of F has an equation of the form ax + by + c =
0, a, b, c ∈ F. Let ax + by + c = 0 and a1 x + b1 y + c1 = 0 be two lines in F, where
a, b, c, a1 , b1 , c1 ∈ F.
Let (x1 , y1 ) be the intersection of the two lines, then we have ax1 + by1 + c = 0 and
a1 x1 + b1 y1 + c1 = 0.
Soilving for x1 , y1 we have
x1
y1
1
=
=
bc1 − b1 c
ca1 − ac1
ab1 − a1 b
∴ x1 =
bc1 −b1 c
ab1 −a1 b
and y1 =
ca1 −ac1
ab1 −a1 b
∈ F.
i.e, (x1 , y1 ) ∈ plane of F.
Result: 5
Let F ⊆ R, be a subfield of R. If a plane and a circle in F intersect in the real
√
plane, then they intersect either in a point in the plane of F or in the plane of F ( ν)
for some ν ≥ 0, ν ∈ F.
Proof:
A line and a circle in F are given by equations of the form
ax + by + c = 0 · · · · · · (1) and
x 2 + y 2 + a1 x + b 1 y + c 1 = 0
· · · · · · (2) where a1 , b1 , c1 , a, b, c ∈ F.
39
Consider (1) and (2) do not intersect in the plane of F. The points of intersection of
(1) and (2) are given by solving (1) and (2) simultaneously,
From (1), y = − ax+c
b
Using in (2), we get x2 + ( ax+c
)2 + ax1 + b1 (− ax+c
) + c1 = 0 · · · · · · (3).
b
b
This gives a quadratic equation in x. This equation has two roots.
Since (1) and (2) do not intersect in F, equation (3) has a root
√
ν1 ∈
/ F, then we extend the field F to F ( ν1 ).
√
ν1 ≥ 0 where
√
√
Similarly, a value for y corresponding to x = ν1 will be an element of F ( ν1 ).
√
Thus (1) and (2) intersect in the plane of F ( ν) for some ν > 0, ν ∈ F.
Result: 6
The intersection of two circles in F can be realized as the intersection of a line in F
and a circle in F.
proof:
Let x2 + y 2 + a1 x + b1 y + c1 = 0 · · · · · · (1) and
x2 + y 2 + a2 x + b2 y + c2 = 0 · · · · · · (2) where a1 , b1 , c1 , a2 , b2 , c2 ∈ F be two circles
in F. which intersect in F. Then their equation of their common chord is given by
(a1 − a2 )x + (b1 − b2 )y + c1 − c2 = 0 · · · · · · (3) which is a line in F.
Thus the intersection of (1) and (2) can be considered as the intersection of (1) and
(3), thus proving the result. The point of intersection of (1) and (3) and hence of (1) and
√
(2) may either in F or in F ( ν) for some ν > 0, ν ∈ F.
Result: 7
Let F ⊆ R be a subfiled of R. Let ν ∈ F be positive . Then
an intersection of lines and circles in F.
√
ν can be realized as
Proof:
~ be the line segment of length |OR|
~ = ν. Let us find a point P
Let ν > 0, let OR
~ extended so that |OP
~ | = 1. Find the midpoint of P~R. Draw a semicircle P QR
on OR
with P~R as diameter. Erect a perpendicular to P~R at O intersection the semicircle at
Q. Then the triangles OP Q and OQR are similar
Now ∠OP Q = ∠OQR ;
⇒
~
|OQ|
~
|OR|
=
~ |
|OP
~
|OQ|
∠OQP = ∠ORP .
~ 2 = |OR||
~ OP
~ | = 1.ν = ν.
⇒ |OR|
40
~ is of length √ν. Thus we can consider for ν > 0 ∈ F, √ν can be realized
Thus |OQ|
~ and a semicircle P OR.
as an intersection of line OQ
Theorem 3.4.4: The real number α is Constructible if and only if we can find a finite
number of real numbers λ1 , λ2 , λ3 , · · · , λn such that
1. λ21 ∈ F0 , the field of rationals.
2. λ2i ∈ F0 (λ1 , λ2 , · · · , λi−1 ) for i = 1, 2, · · · , n.
such that α ∈ F0 (λ1 , λ2 , · · · , λn ).
Proof:
Let α ∈ R be Constructible.
Consider the field F0 of all rationals.
By our discussions above, we see that lines ane circles in the plane of F0 lead us to
√
points either in F0 or in a quadratic extension of F0 , say F0 ( λ1 ) for some λ1 ∈ F0 .
√
Now consider the field F0 ( λ1 ) ⊇ F0 . Using the same discussions, we see that the
√
√ √
lines and circles in the plane of F0 ( λ1 ) intersect in points in the plane of F0 ( λ1 , λ2 )
√
where λ2 (> 0) ∈ F0 ( λ1 ) extending this we see that a poin t is Constructible from
F0 if we can find real numbers λ1 , λ2 , · · · , λn such that λ21 ∈ F0 , λ22 ∈ F0 (λ1 ), λ23 ∈
F0 (λ1 , λ2 ), · · · , λ2n ∈ F0 (λ1 , λ2 , · · · , λn−1 ) such thatthe point is in F0 (λ1 , λ2 , · · · , λn ).
√
Conversely, if α ∈ F0 , is such that α > 0 is real, then by result 7, we can realize α as
an intersection of lines and circles in F0 . Thus a point α is Constructible from F0 if and
only if we can find a finite number of real numbers λ1 , λ2 , · · · , λn such that [F0 (λ1 ) : f0 ] =
1 or 2.
[F0 (λ1 , λ2 , · · · λi ) : F0 (λ1 , λ2 , · · · , λi−1 )] = 1 or 2 for i = 1, 2, · · · , n. and α ∈
F0 (λ1 , λ2 , · · · , λn ).
Corollary 3.4.5: If α is Constructible , then α lies in some extension of the rationals of
degree a power of 2.
Proof:
Let α ∈ R be Constructible.Then by theorem 4.1.1 we can find a finite number of
real numbers λ1 , λ2 , · · · , λn such that
41
[F0 (λ1 , λ2 , · · · , λi ) : F0 (λ1 , λ2 , · · · , λi−1 )] = 1 or 2 i = 1, 2, · · · , n and α ∈
F0 (λ1 , λ2 , · · · , λn ).
Thus F0 ⊂ F0 (λ! ) ⊂ F0 (λ1 , λ2 ) ⊂ · · · ⊂ F0 (λ1 , λ2 , · · · , λn ).
∴ By theorem 5.1.1 we have
[F0 (λ1 , λ2 , · · · , λn ) : F0 ] = [F0 (λ1 , λ2 , · · · , λn : F0 (λ1 , λ2 , · · · , λn−1 )]
· · · × [F0 (λ1 , λ2 , · · · , λi ) : F0 (λ1 , λ2 , · · · , λi−1 )]
· · · × [F0 (λ1 : F0 ]
Since each term in the product on the right hand side is either 1 or 2 we get
[F0 (λ1 , λ2 , · · · , λn : F0 ] = 2r for some r > 0, r ≤ n.
This proves the corollaary.
Corollary 3.4.6: If the real number α satisfies an irreducible polynomial over the field of
rational numbers of degree k and if k is not a power of 2, then α is not Constructible.
Proof: Suppose α is Constructible,then by Corollary 4.1.2, of the theorem 4.1.1 there
is a subfield k of real field such that α ∈ k and [k : F0 ] = 2r for some r > 0. But
F0 ⊂ k. Thus we have a subfield F0 (α) of k where k is a finite extension of F0 . Hence
by corollary to theorem 5.1.1 [F0 (α) : F0 ] | [k : F0 ]
i.e, [f0 (α) : F0 ] | 2r ⇒ [F0 (α) : F0 ] is also a power of 2 · · · · · · (1).
By hypothesis α satisfies an irrreducible polynomial of degree k over F0 . Hence by
theorem 5.1.3 [F0 (α) : F0 ] = k.
Thus, by (1) k must be a power of 2 which is a contradiction to the hypothesis that
k is not a power of 2. Hence our assumption is wrong and we conclude that α is not
Constructible.
Theorem 3.4.7: It is impossible, by straight edge and copass alone, to trisect 60o
Proof:
We note that an angle θ is Constructible if and only if a line segment of length |cosθ|
is Constructible.
Now cos60o =
1
2
which is Constructible number. Hence 60o is a constructible angle.
We know that cos3θ = 4cos3 θ − 3cosθ · · · · · · (1).
42
Let θ = 20o and α = cosθ = cos20o .
Then (1) ⇒
1
2
= 4α3 − 3α
⇒ 8α3 − 6α = 1
In otherwords, α is a root of the polynomial 8x3 − 6x − 1 over the rational filed.
Claim: The polynomial f (x) = 8x3 − 6x − 1 is irreducible over F0 .
f (x) = 8x3 − 6x − 1
∴ f (x − 1) = 8(x − 1)3 = 6(x − 1) − 1
= 8(x3 − 3x2 + 3x − 1) − 6x + 6 − 1
= 8x3 − 24x2 + 18x − 3
Now f (x) will be irreducible over F0 if and only if f (x − 1) is irreducible over F0 .
Now f (x − 1) = 8x3 − 24x2 + 18x − 3 ∈ Z[x]. Also 3 is a prime such that 3 divides
each of the coefficients of f (x − 1) except the coefficient of x3 which is 8.Also 32 does
not divide the constant term which is -3.
Hence by Eisenstein criterion, f (x−1) is irreducible over F0 which in turn establishes
that f (x) is irreducible over F0 .
Since degf (x) = 3 6= 2r , for some r, by corollary 4.1.2 to theorem 4.1.1 α = cos20o
is not Constructible.
i.e, 20o is not a Constructible angle.
⇒ 60o cannot be trisected.
Theorem 3.4.8: By straight edge and compass it is impossible to duplicate the cube.
i.e, given a side of a cube, it is not always possible to construct with a straight edge
and a compass alone the side of a cube that has double the volume of the original cube.
Proof:
Let the original cube be a unit cube.
Hence the volume of the cube is 1. If we construct a cube of volume 2, i.e double the
√
volume of the original cube, then each side of the new cube would be 3 2. i.e, the side
α of the new cube is such that α3 = 2 ⇒ α is a root of the polynomial x3 − 2 which is
irreducible over F0
For, f (x) = x3 − 2
43
Here a0 = −2, a1 = 0, a2 = 0, a3 = 1.
Then by Eisenstein criterion p divides a0 , p divides a1 , p divides a2 , p does not
divide a3 , p2 does not divide a0 .
∴ f (x) is irreducible over F0 .
Hence by corollary 4.1.4 to theorem 4.1.1, α is not Constructible.
Hence the theorem.
Result: 8
Squaring the circle is impossible. That is, given a circle it is not always possible to
construct with straight edge and a compass a square having area equal to the area of the
given circle.
Proof: Consider the unit circle so that its area is π. Now we have to construct a
√
√
square whose side is π. Since π is transcendental and therefore π is transcendental
over F0 . we get the result.
Remark 3.4.9: The regular n− gon is Constructible for n ≥ 3 if and only if the angle 2π
n
is Constructible. That is, the regular n− gon is constructible for n ≥ 3 if and only if a
line segment of length cos 2π
is Constructible.
n
Result: 9
The regular p− gon can be constructed only for primes p of the form 2s + 1.
Proof:
Consider the construction of the regular p− gon. By the remark above, then we have
to construct the complex number
z = cos 2π
+ isin 2π
But z p = 1
p
p
⇒ z p − 1 = 0.
i.e, (z − 1)(z p−1 + z p−2 + · · · + 1) = 0.
(∵ z 6= 1, z − 1 6= 0)
z p−1 +z p−2 +· · ·+1 = 0 But xp−1 = (x−1)(xp−1 +xp−2 +· · ·+1) and xp−1 +xp−2 +· · ·+1
is irreducible in Q[x].
Thus, z satisfies an irreducible polynomial over Q of degree p − 1.
Hence [Q(z) : Q] = p − 1.
44
Thus z is constructible, by corollary 4.1.2 to theorem 4.1.1 if p − 1 = 2s for some
s > 0.
That is, z is constructible if p = 2s + 1 is a prime. In otherwords, a regular p− gon
is Constructible only for the primes p = 2s + 1, s > 0 is an integer.
Note:
We know that 2s + 1 will be a prime if s = 2t for some non-negative integer t. Hence
a regular p− gon is Constructible for some p if p is of the form 22t + 1.
Theorem 3.4.10: It is impossible to construct a regular septagon by straight edge and
compass.
Proof:
Since 7 6= 2s + 1, by result 9, 7− gon is not Constructible.
3.5 More about roots
Definition 3.5.1: Let f (x) = a0 + a1 x + · · · + an xn ∈ F [x]. The derivative of f (x) written
as f 0 (x) is the polynomial f 0 (x) = a1 + 2a2 + · · · + nan xn−1 ∈ F [x].
Note:
If f (x) = a0 , a constant polynomial, f 0 (x) = 0.
Definition 3.5.2: A field F is said to be of characteristic 0 if ma 6= 0 ∀a 6= 0 in F and
m > 0 an integer. If ma = 0 for some m > 0 and some a 6= 0 in F, then F is said
to be of finite characteristic. Then it is defined to be the smallest positive integer p such
that p − a = 0 ∀a ∈ F.
Note:
If F is of finite characteristic then its characteristic is a prime.
Result: 1
Let f (x) ∈ F [x] and let f 0 (x) = 0. Then
1. f (x) = a constant if the characteristic of F = 0
2. f (x) = g(xp ) if the characteristic of F = p, a prime p 6= 0.
45
Proof:
Let the characteristic of F = 0.
Assume f (x) is not a constant.
Then degf (x) > 0.
Let deg(f (x)) = n > 0.
Then f (x) = a0 + a1 x + a2 x2 + · · · + an xn ,
a0 , a1 , · · · , an ∈ F and an 6= 0.
∴ f 0 (x) = a1 + 2a2 x + · · · + nan xn−1 6= 0 for n 6= 0, an 6= 0 and therefore nan 6= 0
for characteristic of F = 0.
This is a contradiction to tha hypothesis.
Hence f (x) must be a constant.
Let the characteristic of F = p where p 6= 0.
Suppose f (x) contains a term bxnp+r , where 0 < r < p and b 6= 0.
Then f 0 (x) will contain a term
b(np + r)xnp+r−1 = brxnp+r−1 , b 6= 0, r 6= 0.
⇒ f 0 (x) 6= 0 which is a contraction to f 0 (x) = 0.
Hence there can be no such term as bxnp+r .
This forces us to conclude that every term must be a term in a power of x which is
a multiple of p, for p is a prime.
Thus f (x) = g(xp ).
Lemma 3.5.3: For any f (x), g(x) ∈ F [x] and any α ∈ F
1. (f (x) + g(x))0 = f 0 (x) + g 0 (x)
2. (αf (x))0 = αf 0 (x)
3. (f (x) · g(x))0 = f 0 (x)g(x) + f (x)g 0 (x)
Proof:
Let f (x) = a0 + a1 x + · · · + an xn and g(x) = b0 + b1 x + · · · + bm xm ,
m > n.
46
f (x) + g(x) = (a0 + b0 )(a1 + b1 )x + · · · + (an + bn )xn + bn+1 xn+1 + · · · + bm xm .
∴ (f (x)+g(x))0 = (a1 +b1 )+2(a2 +b2 )x+· · ·+n(an +bn )xn−1 +(n+1)bn+1 xn +· · ·+mbm xm−1 .
(f (x) + g(x))0 = (a1 + 2a2 + · · · + nan xn−1 ) + (b1 + 2b2 x + · · · + nbn xn−1 + mbm xm−1 )
= f 0 (x) + g 0 (x)
αf (x) = αa0 + αa1 x + · · · + αan xn
∴ (αf (x))0 = αa1 + 2αa2 x + · · · + nαan xn−1
= α(a1 + 2a2 x + · · · + nan xn−1 )
= αf 0 (x)
Let f (x) = xi , g(x) = xj , i, j > 0.
f (x) · g(x) = xi xj = xi+j
∴ (f (x) · g(x))0 = (i + j)xi+j−1
But f 0 (x)g(x) = ixi−1 xj = ixi+j−1 and
f (x)g 0 (x) = xi jxj−1 = jxi+j−1
∴ (f (x) · g(x))0 = f 0 (x)g(x) + f (x)g 0 (x).
Lemma 3.5.4: Let f (x) ∈ f [x]. Then f (x) has a multiple root if and only if f (x) and
f 0 (x) has a common factor of positive degree.
Proof:
Let f (x) have a multiple root α of order m ≥ 2 in the splitting field E of f (x)
over F. Then (x − α)m ∈ E[x] where α is a root of f (x) lying in E.
∴ f (x) = (x − α)m g(x), g(x) ∈ E[x]
f 0 (x) = m(x − α)m−1 g(x) + (x − α)m g 0 (x)
f 0 (x) = (x − α)m−1 [mg(x) + (x − α)g 0 (x)] where m − 1 ≥ 1
⇒ f (x) and f 0 (x) have a common factor of (x − α)m−1 of positive degree in E[x].
It follows that f (x) and f 0 (x) have a common factor of positive degree in F [x]. For
otherwise, f (x) and f 0 (x) are relatively prime in F [x].
Hence there exists a(x), b(x) ∈ F [x]
such that a(x)f (x) + b(x)f 0 (x) = 1 · · · · · · (1).
47
Also a(x), b(x) ∈ E[x] and from (1) it follows that f (x) and f 0 (x) are relatively
prime in E[x].
This contradicts the fact that (x − α)m−1 is a common factor of f (x) and f 0 (x).
Conversely, let f (x) and f 0 (x) have a non-trival common factor.
To show that f (x) has a multiple root in E.
Assume f (x) has no multiple roots. Let a1 , a2 , · · · , an be the n distinct roots of
f (x) in E.
∴ f (x) = k(x − a1 ) · · · (x − an ), k ∈ E
⇒ f 0 (x) = k
n
P
(x − a1 ) · · · (x − ai−1 )(x − ai )(x − ai+1 ) · · · (x − an )
i=1
∴ f 0 (ai ) = k(ai − ai ) · · · (ai − ai−1 )(ai − ai+1 ) · · · (ai − an )
6= 0
(∵ a1 , a2 , · · · , ai , · · · , an are distinct)
⇒ no ai is a root of f 0 (x).
∴ f (x), f 0 (x) have no common root and hence no non-trivial common factors.
This contradicts the hypothesis.
∴ f (x) has a multiple root.
Corollary 3.5.5: If f (x) ∈ F [x] is irreducible, then
1. If the charateristic of F = 0, f (x) has no multiple roots and
2. If the characteristic of F is p 6= 0, f (x) has a multiple root only if it is of the
form f (x) = g(xp )
Proof:
Since f (x) ∈ F [x] is irreducible, its only factors are 1 and f (x).
Assume that f (x) has a multiple root.
Then by lemma 3.5.4 f (x) and f 0 (x) has a non-trivial common factor in F [x].
Hence f (x) divides f 0 (x).
⇒ f 0 (x) = g(x)f (x)
But degf 0 (x) < degf (x),
g(x) = 0.
48
Hence f 0 (x) = 0.
Thus, the irreducible polynomial f (x) ∈ F [x] having a multiple root forces f 0 (x) = 0.
Now we apply a our result proved before lemma 3.5.3 when the characteristic of
F = 0, f 0 (x) = 0 for f (x) ∈ F [x].
⇒ f (x) = a constant which is a contradiction as it has no roots.
When the characteristic of F = p 6= 0, f (x) = g(xp ).
This proves the corollary.
Corollary 3.5.6: If F is a field of characteristic F = p 6= 0 then the polynomial
n
xp − x ∈ F [x] for n ≥ 1 has distinct roots.
Proof:
n
Let f (x) = xp − x
f 0 (x) = pn xp
n −1
−1
= −1
(∵ characteristic of F = p 6= 0)
Thus f (x) and f 0 (x) have no common factors.
⇒ f (x) has no multiple roots.
n
That is the roots of xp − x are distinct.
Remark 3.5.7:
Here after we assume the characteristic of the field F as 0.
Definition 3.5.8: The extension K of F is a simple extension of F if K = F (α) for
some α in K.
Theorem 3.5.9: If F is of characteristic 0 and if a, b are algebraic over F, then there
exists an element c ∈ F (a, b) such that F (a, b) = F (c).
Proof:
Let f (x), g(x) be the F − minimal polynomials of a and b respectvely.
Let degf (x) = m ; deg g(x) = n.
49
Let K be an extension of F in which both f (x) and g(x) split into linear factors (
i.e K is the splitting field of both f (x) and g(x) )
Since the characteristic of F = 0 and f (x) and g(x) are irreducible over F, the
roots of both f (x) and g(x) are distinct.
Let the roots of F (x) be a = a1 , a2 , · · · , am and the roots of
g(x) be b = b1 , b2 , · · · , bn .
Let j 6= 1. Then bj 6= b1 = b. Hence the equation
ai + λbj = a1 + λb1
λbi − λbj = ai − a1
λ(b1 − bj ) = ai − a1
has the solution for λ, say λ =
1 ≤ i ≤ m and 2 ≤ j ≤ n.
ai −a1
b1 −bj
which is uniquely determined in K ∀i, j where
Since the characteristic of F = 0 has an infinite number of elements. So it is possible
to find ν ∈ F such that ν is not any of the ν 0 s found.
∴ ai + νbj 6= a1 + νb1
∀i, j
1 ≤ i ≤ m and 2 ≤ j ≤ n.
Let c = a1 + νb1 = a + νb
Claim : F(a,b) = F(c)
Now c = a + νb and a + νb ∈ F (a, b)
⇒ c ∈ F (a, b) Hence F (c) ⊆ F (a, b)
To show F (a, b) ⊆ F (c), we will show a, b ∈ F (c)
Now b satisfies a polynomial g(x) over F which is also a polynomial over F (c).
Let h(x) = f (c − νx) be a polynomail over F (c).
∴ h(b) = f (c − νb) = f (a) = 0 for a is a root of f (x).
∴ b satisfies h(x) also. Thus we see that b satisfies both the polynomials g(x) and
h(x).
Thus it follows that h(x) and g(x) have a common factor (x − b) over K.
Since g(x) has no multible roots (i.e, we have abserved that the roots of g(x) are
dixtinct )
50
(x − b)2 divides g(x)
∴ ∀j ≥ 2, h(bj ) = f (c − νbj ) 6= 0.
This is true for, c − νbj 6= ai ∀i, j
1 ≤ i ≤ m, 2 ≤ j ≤ n and a0i s are distinct.
∴ x − bj is not a factor of h(x)
so the greatest common factor of h(x), g(x) over K is x − b. This shows that g(x)
and h(x) have a non-trivial greatest common divisor over K which must be a divisor of
x − b. Since the degree of x − b is 1 the greatest common divisor of g(x) and h(x) over
K is exactly x − b.
But then g(x), h(x) (which are polynomial over F (c) ) must have a non-trivial
common factor over F (c). It follows that x−b is a non-trivial common factor of h(x), g(x)
over F (c) ⇒ b ∈ F (c).
Since, a + νb = c and ν, c, b ∈ F (c)
a ∈ F (c) ⇒ a, b ∈ F (c) ⇒ F (a, b) ⊆ F (c).
Hence the claim that F (a, b) = F (c).
Corollary 3.5.10: Any finite extension of a field of characteristic 0 is a simple extension.
Proof:
Let K be a finite extension of a field F.
Let [K : F ] = n.
If n = 1, K = F, we are done
Let K 6= F and let {b1 , b2 , · · · , bn } be a basis of K as a vector space over F.
Now K is a finite extension of F.
⇒ K is an algebraic extension of F.
∴ b1 , b2 , · · · , bn are algebraic over F
Since b1 is algebraic over F and since K 6= F we can consider the subfield F (b1 ) of
K which is a finite extension of F.
Now b2 is algebraic over F ⇒ b2 is also algebraic over F (b1 ) ⊇ F. Hence we can
consider the subfield (F (b1 ))b2 = F (b1 , b2 ) of K which is also a finite extension over
F (b1 ) and hence over F.
51
Hence by theorem 3.5.9, we can find a c such that F (c) = F (b1 , b2 ).
Proceeding in this way, since b3 , b4 , · · · , bn ∈ K are algebraic over F.
We can found the subfiled F (b1 , b2 , · · · , bn ) of K.
Thus we have F (b1 , b2 , · · · , bn ) = K.
Hence we can find from the above theorem that there is an element c ∈
F (b1 , b2 , · · · , bn ) = K such that F (c) = F (b1 , b2 , · · · , bn ) = K.
i.e, K = F (c), c ∈ K.
⇒ K is a simple extension.
Definition 3.5.11: An element a in an extension K of F is called separable over F if it
satisfies a polynomial over F having no multiple roots.
Definition 3.5.12: An extension K of F is called separable over F if all its elements are
separable over F.
3.6 The Elements Of Galois Theorey
Automorphism:
A mapping σ of a field K onto itself such that σ(a + b) = σ(a) + σ(b) and
σ(ab) = σ(a) · σ(b) is called on automorphism of K.
Two automorphisms σ and τ of are said to be distinct if σ(a) 6= τ (a) for some
element a ∈ K. By Aut (K) we mean the set of all automorphisms of K.
Theorem 3.6.1:
If K is a field and if σ1 , σ2 · · · , σn are distinct automorphisms of K then it is
impossible to find elements a1 , a2 , · · · , an not all zero in K such that
a1 σ1 (u) + a2 σ2 (u) + · · · + an σn (u) = 0 ∀ u ∈ K.
Proof:
Given σ1 , σ2 , · · · , σn are distinct automorphisms of K.
Assume that it is possible to find elements a1 , a2 , · · · , an ∈ K not all zero such that
a1 σ1 (u) + a2 σ2 (u) + · · · + an σn (u) = 0 ∀ u ∈ K.
52
Deleting the zeros among a0i s we can find a relation with a few non-zero terms.
Renumbering we can find a minimal relation.
a1 σ1 (u) + a2 σ2 (u) + · · · + am σm (u) = 0 ∀ u ∈ K
· · · · · · (1).
where a1 , a2 , · · · , am are all distinct from 0 m < n.
If m = 1,
a1 σ1 (u) = 0 ∀u ∈ K.
Since σ1 is an automorphism of K, σ1 (u) = 0 ⇔ u = 0.
∴ a1 σ1 (u) = 0 ∀ u ∈ K.
⇒ a1 = 0 contradicting our assumption.
We assume that m > 1.
Since σ1 , σ2 , · · · , σm are distinct, automorphisms, we can find a c ∈ K such that
σ1 (c) 6= σi (c) · · · · · · (2) forall j 6= 1, 1 < i ≤ m.
c ∈ K, u ∈ K ⇒ cu ∈ K ∀ u ∈ K.
Since (1) is true for all element in K, (1) holds for cu also.
(ie) a1 σ1 (cu) + a2 σ2 (cu) + · · · + am σm (cu) = 0 ∀ u ∈ K
· · · · · · (3).
Since σi (1 ≤ i ≤ m) are automorphisms, σi (cu) = σi (c) · σi (u) (1 ≤ i ≤ m).
∴ (3) becomes
a1 σ1 (c) · σ1 (u) + a2 σ2 (c) · σ2 (u) + · · · + am σm (c) · σm (u) = 0
· · · · · · (4).
Multipling (1) by σ1 (c) we get
a1 σ1 (c)σ1 (u) + a2 σ1 (c)σ2 (u) + · · · + am σ1 (c)σm (u) = 0
· · · · · · (5).
(4) - (5) gives
a2 (σ2 (c)−σ1 (c))σ2 (u)+a3 (σ3 (c)−σ1 (c))σ3 (u)+· · ·+am (σm (c)−σ1 (c))σm (u) = 0 ∀u ∈ K.
Taking bi = ai (σi (c) − σ1 (c)) the above relation becomes,
b2 σ2 (u) + b3 σ3 (u) + · · · + bm σm (u) = 0 ∀ u ∈ K
· · · · · · (6).
But bi 6= 0 as σi (c) 6= σ1 (c) ∀ i 6= 1 by (2) and ai 6= 0 ∀ i, 2 ≤ i ≤ m.
Thus (6) is a relation with non-zeros bi 0 s having m − 1 terms contradicting the
minimaltity of the relation (1).
Hence we conclude that it is impossible to find elements a1 , a2 , · · · , an not all zero in
K such that
53
a1 σ1 (u) + a2 σ2 (u) + · · · + an σn (u) = 0 ∀ u ∈ K.
Hence the proof.
Definition 3.6.2:
Let K be a field. Let G be a subgroup of the group Aut(K) of all automorphisms
of K. The fixed field of G, denoted by KG is defined to be the set
KG = {a ∈ K | σ(a) = a ∀ a ∈ G}
Lemma 3.6.3:
The fixed field KG of G is a subfield of K.
Proof:
Let a, b ∈ KG such that σ(a) = a and σ(b) = b ∀σ ∈ G.
σ(a ± b) = σ(a) ± σ(b) [∵ σ is an automorphisms ]
=a±b
∴ a ± b ∈ KG
σ(ab) = σ(a) · σ(b) = ab
∴ ab ∈ KG
Let b 6= 0 ∈ KG
σ(b−1 ) = (σ(b))−1 = b−1
⇒ b−1 ∈ KG
Thus any non-zero b ∈ KG has an inverse in KG .
∴ KG is a subfield of K.
Definition 3.6.4:
Let K be a field and F be a subfield of K, then the group of automorphisms of K
relation to F is defined as the set
G(K, F ) = {σ ∈ Aut(K) | σ(a) = a ∀ a ∈ K}
Lemma 3.6.5:
54
G(K, F ) is a subgroup of the group of all automorphisms of K.
Proof:
σ(a) = a, τ (a) = a ∀ a ∈ F
∴ (στ −1 )(a) = τ −1 (σ(a))
= τ −1 (a)
= a
[∵ τ (a) = a ⇒ τ −1 (a) = a]
∴ στ −1 ∈ G(K, F ).
Example 3.6.6:
Let R be the field of all real numbers C, the field of all complex numbers.
To find G(C, R) and FG (C, R).
Let σ ∈ G(C, R) then σ is an automorphisms of C
such that σ(x) = x ∀ x ∈ R.
Now i · i = −1
∴ σ(i · i) = σ(−1)
⇒ σ(i) · σ(i) = −1
(ie) (σ(i))2 = −1 ⇒ σ(i) = i (or) −i
Let a + ib ∈ C, a, b ∈ R
σ(a + ib) = σ(a) + σ(i) · σ(b) = a + σ(i)b
Case(i) :
σ(i) = i then σ(a + ib) = a + ib ∀a, b ∈ R
⇒ σ is the identity automorphism on C.
Case(ii) :
σ(i) = −i, σ(a + ib) = a − ib ∀a, b ∈ R.
Clearly, σ is an automorphisms on C.
We shall denote this automorphisms σ by τ.
∴ G(C, R) = {1, τ } ⊆ Aut(C)
To find the fixed field of G(C, R).
55
Let FG(C,R) be the fixed field of G(C, R) then
FG(C,R) = {x ∈ C | σ(x) = x ∀σ ∈ G(C, R)}
Clearly R ⊆ FG(C,R) ⊆ C
Now, G(C, R) = {1, τ }
x ∈ FG(C,R) ⇔ σ(x) = x ∀ σ ∈ G(C, R).
Let x = a + ib ∈ FG(C,R) ⇔ 1(a + ib) = a + ib
and τ (a) = τ (a + ib) = a − ib = a + ib ⇔ b = 0
(ie) x = a ∈ R ⇒ FG(C,R) ⊆ R
But R ⊆ FG(C,R)
∴ FG(C,R) = R.
Theorem 3.6.7:
If K is a finite extension of F, (ie) [K : F ] is finite, then G(K, F ) is a finite group
and its order O(G(K, F )) satisfies O(G(K, F )) ≤ [K : F ].
Proof:
Let [K : F ] = n (finite)
To prove : O(G(K, F )) ≤ n
Assume that O(G(K, F )) > n
Let σ1 , σ2 , · · · , σn+1 be the (n + 1) distinct automorphisms of K relative to F.
Now [K : F ] = n and hence there is a basis u1 , u2 , · · · , un of the vector space K
over F.
Consider n linear homogeneous equations ini (n + 1) variables x1 , x2 , · · · , xn+1 in
K namely,
x1 σ1 (u1 ) + x2 σ2 (u1 ) + · · · + xn+1 σn+1 (u1 ) = 0
x1 σ1 (u2 ) + x2 σ2 (u2 ) + · · · + xn+1 σn+1 (u2 ) = 0
..
.
x1 σ1 (un ) + x2 σ2 (un ) + · · · + xn+1 σn+1 (un ) = 0
Since the number of variables is greater than the number of equations there is a non-trivial
solution.
56
(ie) a1 , a2 , · · · , an+1 ∈ K not all zero such that the set of equations (I) holds.
Let u ∈ K,
∴ u = α1 u1 + · · · + αn un for some α1 , α2 , · · · , αn ∈ F.
x1 σ1 (u) + x2 σ2 (u) + · · · + xn+1 σn+1 (u) =
n+1
X
xi σi (u)
i=1
=
n+1
X
xi σi (α1 u1 + · · · + αn un )
i=1
=
n+1
X
xi [α1 σi (u1 ) + · · · + αn σi (un )]
i=1
= α1 [x1 σ1 (u1 ) + · · · + xn+1 σn+1 (un )]
+α2 [x2 σ1 (u1 ) + · · · + xn+1 σn+1 (un )] + · · ·
+αn [x1 σ1 (u1 ) + · · · + xn+1 σn=1 (un )]
= 0
[f rom (1)]
(ie) x1 σ1 (u) + x2 σ2 (u) + cdots + xn+1 σn+1 (u) = 0 ∀u ∈ K,
x1 , x2 , · · · , xn+1 ∈ K.
Thus we have found x1 , x2 , · · · , xn+1 ∈ K not all zero such that forall u ∈ K
x1 σ1 (u) + · · · + xn+1 σn+1 (u) = 0 which is impossible. [ By theorem 3.6.1 ]
Hence our assumption O(G(K, F )) > n is wrong and the theorem is proved.
(ie) O(G(K, F )) ≤ n = [K : F ] showing that G(K, F ) is a finite group.
Notation:
The field F (x1 , x2 , · · · , xn ), the field of rational functions in n variables x1 , x2 , · · · , xn
over a field F.
Let F be a field, then it is a commutative ring with unit element. By the polynomial
ring in the inderminate x over F we mean.
F [x] = {a0 + a1 x + · · · + am xm | a0 , a1 , · · · , am ∈ F }
Then F [x] is also a commutative ring with unit element.
The ring of polynomials in the n− variables x1 , x2 , · · · , xn over F,
F [x1 , x2 , · · · , xn ] is defined as follows:
Let F1 = F [x1 ], F2 = F1 [x2 ] · · · Fn = Fn−1 [xn ]
(the polynomial ring in x2 over F1 )
Fn is called the ring of polynomials in x1 , x2 , · · · , xn over F. Its elements are of the
57
form
P
ai1 ,i2 ,··· ,im xi11 · · · xinn
where the equality and addition are defined coefficient wise and where multiplication
is defined by use of the distribution laws and the law of exponents
(xi11 · · · xinn )(xj11 xj22 · · · xjnn ) = (xi11 +j1 · · · xinn +jn )
The field of rational functions in x1 , x2 , · · · , xn say F (x1 , x2 , · · · , xn ) over F is
defined as the ring of all quotients of such polynomials.
Let Sn be the symmetric group order n. Let σ ∈ Sn . For any integer i, 1 ≤ i ≤ n,
Let σ(i) be the image of i under σ.
Now σ ∈ Sn induces a mapping,
σ 0 : F (x1 , x2 , · · · , xn ) → F (x1 , x2 , · · · , xn )
such that r(x1 , x2 , · · · , xn ) → r(xσ(1) , · · · , xσ(n) )
Claim:
σ 0 is an automorphism of F (x1 , x2 , · · · , xn ) and σ 0 (α) = α ∀ α ∈ F.
(ie) σ 0 ∈ G(F (x1 , x2 , · · · , xn ), F ) suxh automorphisms σ 0 are also called by σ itself.
Hence we can conclude that
Sn ⊆ G(F (x1 , x2 , · · · , xn ), F )
For example , consider Q(x1 , x2 , x3 ) S3 ⊆ G(Q(x1 , x2 , x3 ), Q)
we can take for example σ = (1, x) then
1 2
x1
σ( 2
+ 2x2 − 3x3
( 12 x22 + 2x1 − 3x3 )
)
=
x1 + 2x2 − x23
(x2 + 2x1 − x23 )
Notation:
Now Sn ⊆ G(F (x1 , x2 , · · · , xn ), F )
Let S be the fixed field of Sn .
(ie) S = {r(x1 , x2 , · · · , xn ) ∈ F (x1 , x2 , · · · , xn ) | σ(r(x1 , x2 , · · · , xn )) = r(x1 , · · · , xn ∀ σ ∈ Sn }
By definition F ⊆ S ⊆ F (x1 , x2 , · · · , xn ) the elements of S are called symmetric rational
functions.
Equations of symmetric rational functions.
Let n = 3 and F = Q
58
1.
x21 +x22 +x23 −2x1 x2 x3
x1 +x2 +x3
2.
x21 +x22 +x23 +x1 x2 +x2 x3 +x3 x1
5
∈S
3. x1 + x2 + x3
4. x1 x2 + x2 x3 + x3 x1
5. x1 x2 x3
Definition 3.6.8: Elementary Symmetric rational functions in x1 , x2 , · · · , xn :
The following elements of S, the fixed field of Sn are called the elementary symmetric
rational functions in x1 , x2 , · · · , xn .
a1 =
n
P
xi , a 2 =
i=1
n
P
xi xj , a 3 =
i<j=1
n
P
xi xj xk , · · · , an = x1 x2 · · · xn
i<j<k=1
It is easy to see that
Sn ⊆ G[F (x1 , x2 , · · · , xn ), F (a1 , a2 , · · · , an )]
⊆ G(F (x1 , x2 , · · · , xn ), F )
F ⊆ F (a1 , · · · , an ) ⊆ S ⊆ F (x1 , x2 , · · · , xn )
Theorem 3.6.9:
Let F be a field and let F (x1 , x2 , · · · , xn ) be the field of rational functions in
x1 , x2 , · · · , xn over F. suppose that S is the field of symmetric rational functions, then
1. [F (x1 , x2 , · · · , xn ) : S] = n!
2. G(F (x1 , x2 , · · · , xn ), S) = Sn , the symmetric group of degree n.
3. If a1 , a2 , · · · , an are the elementary symmetric functions in x1 , x2 , · · · , xn then
S = F (a1 , a2 , · · · , an ).
4. F (x1 , x2 , · · · , xn ) is the splitting field over F (a1 , a2 , · · · , an ) = S of the polynomial
tn − a1 tn−1 + a2 tn−2 + · · · + (−1)n an .
Proof:
Given that S is the field of symmetric rational functions. Hence by definition
S = {r(x1 , x2 , · · · , xn ) ∈ F (x1 , x2 , · · · , xn ) | σ(r(x1 , x2 , · · · , xn )) =
r(xσ(1) , xσ(2) , · · · , xσ(n) ) = r(x1 , x2 , · · · , xn ) ∀ σ ∈ Sn
59
(ie) Every σ ∈ Sn leaves element of S fixed.
Also Sn leaves the elements of F fixed (ie) for any σ ∈ Sn σ(α) = α ∀ α ∈ F and
P
P
P
P
σ(a1 ) = σ( xi ) = (σ(xi ) = xσ(i) = xj = a1 ,
σ9a2 ) = a2 , · · · , σ(an ) = an .
⇒ σ(r(a1 , a2 , · · · , an )) = r(a1 , a2 , · · · , an ) [ai ∈ F (x1 , x2 , · · · , xn )]
∴ r(a1 , a2 , · · · , an ) ∈ S
Since any element in F (a1 , a2 , · · · an ) is of the form r(a1 , a2 , · · · , an ) and
r(a1 , a2 , · · · , an ) ∈ S
⇒ F (a1 , a2 , · · · , , an ) ⊆ S ⊆ F (x1 , x2 , · · · , xn )
By theorem, Finite extension of a finite extension is finite we get
[F (x1 , x2 , · · · , xn ) : F (a1 , a2 , · · · , an )] = [F (x1 , x2 , · · · , xn : S]·[S : F (a1 , a2 , · · · , an )] · · · · · · (1).
⇒ [F (x1 , x2 , · · · , xn : S] divides [F (x1 , x2 , · · · , xn : F (a1 , a2 , · · · , an )]
⇒ [F (x1 , x2 , · · · , xn : S] ≤ [F (x1 , x2 , · · · , xn ) : F (a1 , a2 , · · · , an )]
· · · · · · (2).
Since Sn is a group of automorphisms of F (x1 , x2 , · · · , xn ) leaving elements of S
fixed, Sn ⊆ G(F (x1 , x2 , · · · , xn ), S)
∴ O(Sn ) ≤ O(G(F (x1 , x2 , · · · , xn ), S))
(ie)
n! ≤ O(G(F (x1 , x2 , · · · , xn ), S))
· · · · · · (3).
By theorem 3.6.7 (ie) O(G(K, F )) ≤ [K : F ]
∴ O(G(F (x1 , x2 , · · · , xn , S)) ≤ [F (x1 , x2 , · · · , xn ) : S]
Therefore by using (3), n! ≤ [F (x1 , x2 , · · · xn ) : S]
· · · · · · (4).
· · · · · · (5).
x1 , x2 , · · · , xn ∈ F (x1 , x2 , · · · , xn ), For any x
(x − x1 ), (x − x2 ), (x − x3 ), · · · , (x − xn ) are factors in F (x1 , x2 , · · · , xn )
∴ f (x) = (x − x1 )(x − x2 ) · · · (x − xn )
X
X
= xn − (
xi )xn−1 + (
xi xj )xn−2 + · · · + (−1)n x1 · · · xn
= xn − a1 xn−1 + a2 xn−2 + · · · + (−1)n an
Thus we observe that the polynomial f (x) = xn − a1 xn−1 + a2 xn−2 + · · · + (−1)n an
over F (a1 , a2 , · · · , an ) splits into linear factors in F (x1 , x2 , · · · xn ).
60
But no proper subfield of F (x1 , x2 , · · · , xn ) can split f (x), for, such a subfield will
contain both F and each of the root x1 , x2 , · · · , xn contradicting the construction of
F (x1 , x2 , · · · , xn ) [ because, by definition F (x1 , x2 , · · · , xn ) is the smallest subfield that
contains both F and x1 , x2 , · · · , xn ]
⇒ F (x1 , x2 , · · · , xn ) is the splitting filed of f (x) over F (a1 , a2 , · · · , an ).
Thus proving (4).
∴ [F (x1 , x2 , · · · , xn ) : F (a1 , a2 , · · · , an )] ≤ (deg f (x))
· · · · · · (6)
= n!
(2) and (6) ⇒ [F (x1 , x2 , · · · , xn ) : S] ≤ n!
· · · · · · (7).
Therefore (5) and (7) gives [F (x1 , x2 , · · · , xn ) : S] = n!
· · · · · · (8).
Thus proving (1).
Using (8) in (4) ⇒ O(G(F (x1 , x2 , · · · , xn ), S)) ≤ n!
· · · · · · (9).
(3) and (9) ⇒ O(G(F (x1 , x2 . · · · , xn ), S)) = n! = O(Sn ).
Since Sn ⊆ G(F (x1 , x2 , · · · , xn ), S),
G(F (x1 , x2 , · · · , xn ), S) = Sn .
Thus proving (2).
By (6) [F (x1 , x2 , · · · , xn ) : F (a1 , a2 , · · · , an )] ≤ n!
Using (1),
[F (x1 , x2 , · · · , xn ) : S][S : F (a1 , a2 , · · · , an )] ≤ n!
But (8) ⇒ n![S : F (a1 , a2 , · · · an )] ≤ n!.
⇒ [S : F (a1 , a2 , · · · , an )] = 1
⇒ S = F (a1 , a2 , · · · , an )
Thus proving (3).
Normal Extension :
Definition 3.6.10:
Let F be a filed of characteristic zero. A finite extension of F, K is said to be a
normal extension, if the fixed field of G(K, F ) is F.
(ie) FG(K,F ) = F
61
Example 3.6.11:
Q(i) is a normal extension of Q.
Proof:
Take F as the field of rationals Q and K as the field
K = Q(i) = {a + bi | a, b ∈ Q}
Then Q ⊆ Q(i) and
[Q(i) : Q] = 2 = degree of the minimal polynomial x2 + 1 of i
To find G(Q(i), Q) :
Let σ ∈ G(Q(i), Q) then σ is an automorphisms of Q(i) such that σ(x) = x ∀ x ∈
Q.
Now i · i = −1
∴ σ(i · i)σ(−1)
⇒ σ(i) · σ(i) = −1.
(ie) (σ(i))2 = −1 ⇒ σ(i) = i (or) −i
Let a + ib ∈ Q(i),
a, b ∈ Q.
σ(a + ib) = σ(a) = σ(i)σ(b) = a + σ(i)b
Case (i) :
σ(i) = i then σ(a + ib) = a + ib ∀a, b ∈ Q
⇒ σ is the identity automorphism on Q(i).
Case(ii):
σ(i) = −i, σ(a + ib) = a − ib ∀ a, b ∈ Q
Clearly, σ is an automorphism on Q(i) we shall denote this automorphism σ by τ
∴ G(Q(i), Q) = {1, τ } ⊆ aut(Q(i))
To find the fixed filed of G(Q(i), Q)
Let FG(Q(i),Q) be the fixed field of G(Q(i), Q) then
FG(Q(i),Q) = {x ∈ Q(i) | σ(x) = x ∀ σ ∈ G(Q(i), Q)}
Clearly , Q ⊆ FG(Q(i),Q) ⊆ Q(i)
62
Now G(Q(i), Q) = {1, τ }
x ∈ FG(Q(i),Q) ⇔ σ(x) = x ∀ σ ∈ G(Q(i), Q)
And τ (x) = τ (a + ib) = a − ib = a + ib
⇔ b=0
[ For x ∈ FG(Q(i),Q) τ (x) = x ]
(ie) x = a ∈ Q ⇒ FG(Q(i),Q) ⊆ Q
But Q ⊆ FG(Q(i),Q) .
∴ FG(Q(i),Q) = Q.
Example 3.6.12:
√
Q( 3 2) is not a normal extension of Q.
Proof:
F = Q,
√
K = Q( 3 2)
Since x3 − 2 is the Q− minimal polynomial of
n
o
√
1
2
Q( 3 2) has a basis 1, a 3 , 2 3 over Q.
√
3
2,
o
n
√
1
2
3
3
3
(ie) Q( 2) = a + b2 + c2 | a, b, c ∈ Q
√
and [Q( 3 2) : Q] = 3.
√
√
G(Q( 3 2), Q) = σ ∈ Aut(Q( 3 2)) | σ(a) = a ∀ a ∈ G
= {1}
√
∴ Q ⊆ Q( 3 2)
For any a ∈ Q,
√
σ(a0 ∈ Q( 3 2)
1
2
∴ σ(a) = a + b2 3 + c2 3 = a
⇒ σ(a) = a ⇔ b = c = 0
n
o
√
√
3
3
3
FG(Q( √
=
x
∈
Q(
2)
|
σ(x)
=
x
∀σ
∈
G(Q(
2),
Q)
2),Q)
n
o
√
3
= a ∈ Q( 2)
√
3
= Q( 2)
√
⇒ Q( 3 2) is not a normal extension of Q.
63
Theorem 3.6.13:
Let K be an normal extension of F a nd Let H be a subgroup of G(K, F ) and FH
be a fixed field of H. Then
1. [K : FH ] = O(H)
2. G(K, FH ) = H
Proof:
By definition G(K, F ) = {σ ∈ Aut(K) | σ(x) = x ∀ x ∈ F }
Since K is a normal extension of F.
FG(K,F ) = {x ∈ K | σ(x) = x ∀ σ ∈ G(K, F )} = F
Since H ⊆ G(K, F ) and F ⊆ K for all σ ∈ H, σ(x) = x ∀ x ∈ F.
FH = {x ∈ K | σ(x) = x, ∀σ ∈ H}
∴ F ⊆ FH
Again for all σ ∈ H, σ(x) = x ∀x ∈ FH
∴ H ⊆ G(K, FH )
O(H) ≤ O(G(k, FH ))
By theorem, 3.6.7 O(G(K, FH )) ≤ [K : FH ]
· · · · · · (1).
Combining the above inequalites we get
O(H) ≤ [K : FH ]
· · · · · · (2).
Claim: [K : FH ] ≤ O(H)
Since FH is a subfield of K and F ⊆ FH ,
we have F ⊆ FH ⊆ K
Since K is a normal extension of F, K is a finite extension of F.
∴ [K : F ] is finite.
Since [K : F ] = [K : FH ][FH : F ]
⇒ [K : FH ] is finite.
By hypothesis Characteristic F = 0 and F ⊆ FH
⇒ Characteristic FH = 0
64
Since FH is a field of Characteristic 0 and K is a finite extension of FH , K is a
simple extension of FH .
[ By theorem 3.5.9 ]
∴ K = FH (a) for some a ∈ K.
∴ a satisfies some irreducible polynomial over FH of degree [K : FH ].
Let O(H) = h.
H = {σ1 , σ2 , · · · , σn } where σ1 is the identity.
Consider the elementary symmetric expressions in σ1 (a) = a, σ2 (a), · · · , σn (a) namely
α1 =
h
X
σi (a)
i=1
α2 =
h
X
σi (a)σj (a)
i<j≤1
..
.
αn = σ1 (a)σ2 (a) · · · σn (a)
for all σ ∈ H, σ(α1 ) = α1 , σ(α2 ) = α2 , · · · σ(αn ) = αn
∴ α1 , α 2 , · · · αn ∈ F H .
Now consider the polynomial
f (x) = (x − σ1 (a))(x − σ2 (a)) · · · (x − σn (a))
= xh − α1 xh−1 + α2 xh−2 + · · · + (−1)n αn over FH
Since σ1 is the identity element of H, σ1 (a) = a
∴ f (a) = 0.
⇒ deg a over FH ≤ deg f (x) = h
∴ [FH (a) : FH ] ≤ h
(ie) [K : FH ] ≤ h = O(H)
(ie) [K : FH ] ≤ O(H)
(2) and (3)
· · · · · · (3)
⇒ [K : FH ] = O(H).
Also (1) , (2) and (3) ⇒ O(H) ≤ O(G(K, FH )) ≤ [K : FH ] ≤ O(H)
∴ O(H) = O(G(K, FH ))
65
Since H ⊆ G(K, FH )
H = G(K, FH )
Theorem 3.6.14:
K is a normal extension of F if and only if K is the splitting field of some polynomial
over F.
Proof:
Let K be a normal extension of F.
∴ FG(K,F ) = F
∴ [K : F ] is finite since Characteristic F is zero.
Any finite extension of a field of Characteristic zero is a simple extension.
By corollary to theorem 3.5.9, K is a simple extension of F.
(ie) K = F (a) for some a ∈ K.
By theorem 3.6.13,
O(G(K, F )) = n, finite.
Let G(K, F ) = (σ1 , σ2 , · · · , σn ), σ1 is the identity.
Consider the polynomial
p(x) = (x − σ1 (a))(x − σ2 (a)) · · · (x − σn (a))
= xn − α1 xn−1 + α2 xn−2 + · · · + (−1)n αn
where α1 =
n
P
σi (a), α2 =
i=1
n
P
σi (a)σj (a), · · · , αn = σ1 (a)σ2 (a) · · · σn (a)
i<j=1
Clearly α1 , · · · , αn are invariant under any σ ∈ G(K, F ).
Hence α1 , · · · , αn are in the fixed field of G(K, F ) which is F itself.
Since K is a normal extension of F.
ie
αi ∈ F
1≤i≤n
∴ p(x) ∈ F [x].
Now σ1 (a) = a is a root of p(x) and hence any s− field of p(x) must contain a and
hence must contain F (a).
But K = F (a). As all the roots of p(x) are in K, the s− field of p(x) must be
66
contained in K. Thus K is the s− field of p(x) over F.
Conversely, Let K be the splitting field of f (x) in F [x].
∴ K is a finite extension of F.
ie [K : F ] = n (finite)
We shall first prove the following lemma
Lemma 3.6.15:
If p(x) is an irreducible factor of f (x) in F [x] and if β1 , · · · , βr are the roots of
p(x), then for all i, 1 ≤ i ≤ r there exists σi ∈ G(K, F ) such that σi (β1 ) = βi .
(ie)σ1 (β1 ) = β1 , σ2 (β1 ) = β2 , · · · , σr (β1 ) = βr .
Proof:
Since every root of p(x) is a root of f (x), it must lie in K.
Let β1 and βi be any two roots.
By theorem 3.3.14, there is an isomorphism of F1 = F (β1 ) onto F10 = F (βi )
such that τ (β1 ) = βi and τ (α) = α ∀ α ∈ F.
Now K is the splitting field of f (x) considered as a polynomial over F1 and is also
the splitting field of f (x) considered as a polynomial over F10 .
Hence by theorem 3.3.16, there is an isomorphism (automorphism) σi of K onto
such that σi (x) = τ (β1 ) = βi
⇒ σi (x) = τ (x) = x ∀ x ∈ F.
(ie) σi ∈ G(K, F ) and σi (β1 ) = βi .
Hence the lemma.
Now we proceed the proof of the converse of the theorem by induction on n = [K : F ].
If n = 1 then K = F and G(K, F ) = {1F } and the fixed filed of G(K, F ) is F
itself.
∴ K is a normal extension of F.
Assume the result to be true for all extensions of degree < n over F.
Let now [K : F ] = n > 1.
67
It follows that f (x) does not split into linear factors over F. Hence f (x) has an
irreducible factor p(x) ∈ F [x] of degree r > 1.
The roots of p(x), namely β1 , · · · , βr are all distinct and are in K.
ie β1 , · · · , βr are all roots of f (x).
Consider F (β1 ). Then [F (β1 ) : F ] = r.
Also F ⊆ F (β1 ) ⊆ K
Now K is the splitting field of f (x).
Considered as a polynomial over F (β1 ).
∴ [K : F (β1 )] =
[K:F ]
[F (β1 ):F ]
=
n
r
< n.
Hence by induction hypothesis, K is a normal extension of F (β1 ).
Let θ ∈ FG(K,F ) , the fixed field of G(K, F ).
Since F ⊆ F (β1 ).
G(K, F (β1 )) ⊆ G(K, F )
FG(K,F (β1 )) ⊇ FG(K,F )
∴ θ ∈ FG(K,F (β1 )) = F (β1 ) for K is a normal extension of F (β1 ).
(ie) θ ∈ F (β1 )
∴ θ = λ0 + λ1 β1 + · · · + λr−1 β1r−1
· · · · · · (1),
λ0 , λ1 , · · · , λr−1 ∈ F.
Since β1 , · · · , βr are the roots of p(x), and p(x) is irreducible over F.
By lemma 3.6.15, there exist σ1 , · · · , σr ∈ G(K, F )
such that σi (β1 ) = βi
∀i = 1, · · · , r.
Since θ ∈ FG(K,F ) and σi ∈ G(K, F ) σi (θ) = θ, (i = 1, · · · , r)
N ow θ = σi (θ) = σi (λ0 + λ1 β1 + · · · + λr−1 βr−1 )
= σi (λ0 ) + λi σi (β1 ) + · · · + λr−1 σi (β1r−1 )
= λ0 + λ1 βi + · · · + λr−1 βir−1
⇒ λr−1 βir−1 + · · · + λ1 βi + (λ0 − θ) = 0,
i = 1, 2, · · · , r.
∴ βi (i = 1, · · · , r) are distinct roots of the polynomial
68
g(x) = λr−1 xr−1 + · · · + λ1 x + (λ0 − θ) over K.
⇒ Thus we see that the polynomial g(x) ∈ K[x] of degree (r − 1) has r distinct
roots in K.
This shows that λr−1 = λr−2 = · · · = λr = λ0 − θ = 0.
(ie) θ = λ0 ∈ F
Thus we have shown that if θ ∈ FG(K,F ) then θ ∈ F.
∴ FG(K,F ) ⊆ F.
But FG(K,F ) ⊆ F
ie, FG(K,F ) = F ⇒ K is a normal extension of F.
Hence the theorem.
Definition 3.6.16:
Galois extension means normal extension. Let K be the splitting field of a polynomial
f (x) over F. By the Galois group of f (x), we mean G(K, F ).
Theorem 3.6.17: Fundamental Theorem of Galois Theorey
Let K be the splitting field of f (x) over F, there exists a one-to-one correspondence
between the subgroups H of G(K, F ) which associates with every subgroup H of
G(K, F ), the fixed field FH such that
1. L is a normal extension if and only if G(K, L) is a normal subgroup of G(K, F ).
L⊇F
∴ G(K, L) ⊆ G(K, F )
2. L = FG(K,L)
3. H = G(K, FH )
4. [K : L] = (G(K, L))
5. [L : F ] =
O(G(K,F ))
)(G(K,L))
6. When L is a normal extension of F, then G(L, F ) ∼
=
G(K,F )
G(K,L)
Proof:
To prove the stated one-one correspondence it is enough to prove.
69
1. For all extension L of F, contained in K,
L = FG(K,L) .
2. For all subgrouop H of G(K, F ), G(K, FH ) = H
1. Since K is the splitting field of f (x) over F, K is also the splitting field of f (x)
considerede as a polynomial over L which contains F.
F ⊆ L ⊆ K.
∴ K is a normal extension of L.
By theorem 3.6.14 . Hence by definition L = FG(K,L) .
2. Since K is a normal extension of F, then by theorem 3.6.13 for a given subgroup
H of G(K, F ) and FH be a fixed field of H then H = G(K, FH ).
Moreover any subgroup of G(K, f ) arises iin the form G(K, L) where the
association L with G(K, L) maps the set of all subgroups of G(K, F ).
This correspondence in one-to-one for if G(K, L1 ) = G(K, L2 ) then by
L1 = FG(K,L1 ) = FG(K,L2 ) = L2 .
3. H is a subgroup of G(K, F )
∴ O(H) = [K : FH ] by theorem 3.6.13 (2).
But O(G(K, FH )) = [K : FH ]
∴ O(H) = O(G(K, FH ))
[ ∵ K is a normal extension of F, Hence H = G(K, FH )]
But H ⊆ G(K, FH )
Hence H = G(K, FH ).
4. If L is an extension of F contained in K by the one-to-one correspondence there
exists a subgroup G(K, L) of G(K, F )
Since K is a normal extension of L,
∴ O(G(K, L)) = [K : L]
L = FH .
[∵ K is a normal extension of L ]
5. Now [K : F ] = O(G(K, F ))
But [K : F ] = [K : L][L : F ]
⇒ [L : F ] =
[K:F ]
[K:L]
(ie) [L : F ] =
O(G(K,F ))
(O(G(K,L))
To prove that L is a normal extension if and onle if G(K, L) is a normal subgroup
of G(K, F ) of F.
70
First we observe that L is a normal extension of F if and only if for every
σ ∈ G(K, F ), σ(L) ⊆ L.
K being the splitting field of f (x) over F.
⇒ K is a normal extension of F.
⇒ K is finite extension of F.
Since F ⊆ L ⊆ K, L is also a finite extension.
Hence by theorem 3.5.9 L = F (a).
thus if σ(L) ⊆ L then σ(a) ∈ L ∀ σ ∈ G(K, F ).
Q
⇒ L is a splitting field of p(x) =
(x − σ(a))
σ∈G(K,F )
which has coefficients iin F.
As a splitting field, by thoerem 3.6.14 L is a normal extension of F.
Conversely,
if L is a normal extension of F, L is a finite extension of F where Characteristic
F = 0.
∴ L = F (a) where the minimal polynomial of a p(x) over F has all its roots in
L.
However , for any σ ∈ G(k, F ), σ(a) is also a root of p(x).
⇒ σ(a) ∈ L.
Since L = F (a) is generated by a over F,
σ(L) ⊆ L ∀ σ ∈ G(K, F ).
Thus L is a normal extension of F if and only if for any σ ∈ G(K, F ),
τ ∈ G(K, L) and t ∈ L, σ(t) ∈ L and therefore τ (σ(t)) = σ(t).
(ie) if and only if (σ −1 τ σ)(t) = t.
⇒ L is normal over F if and only if σ −1 G(K, L)σ ⊂ G(K.L) ∀ σ ∈ G(K, F ).
⇒ G(K, L) is a normal subgroup of G(K, F ).
6. If L is normal over F, given σ ∈ G(K, F ).
Since σ(L) ⊆ L, σ induces an automorphism σL of L
defined by σL (t) = σ(t) ∀t ∈ L.
Note that σL is the restriction of σ ∈ G(K, F ) to L.
Since σL leaves every element of F fixed σL ∈ G(L, F ).
Thus define a mapping φ : G(K, F ) → G(L, F )
by φ(σ) = σL ∀ σ ∈ G(K, F ).
71
Claim: φ is well defined.
For this we have to show that σL ∈ G(L, F ) for σ ∈ G(K, F ) (ie) σ(L) ⊆ L.
Since L is a normal extension of L, L = F (a) for some a ∈ L and L is the
splitting field of the minimal polynomial of p(x) of a over F.
If σ ∈ G(K, F ), p(σ(a)) = σ(p(a)) = σ(0) = 0.
∴ σ(a) is also a root of p(x)
⇒ σ(a) ⊆ L.
⇒ σ(F (a)) ⊆ L. (ie) σ(L) ⊆ L.
Hence the claim.
If σ, τ ∈ G(K, F ),
(σ ◦ τ )L = σL ◦ τL .
∴ φ is a homomorphism of groups.
Now σ ∈ Ker φ ⇔ φ(σ) = identity of G(L, F ).
(ie) σL = identity automorphism of L.
⇔ σ ∈ G(K, L) → Ker φ = G(K, L).
By theorem 2.7.1 ” Let φ be a homomorphism of G onto G with kernal K. Then
G ∼
G. ”
K =
φ(G(K, F )) ∼
=
G(K,F )
.
G(K,L)
Since G(K, L) is a normal subgroup of G(K, F ).
)
O( G(K,F
) = [L : F ] by part (3).
G(K,L)
= O(G(L, F )) by theorem 3.6.13
Thus φ(G(K, F )) in G(L, F ) is all of G(L, F ) and hence G(L, F ) ∼
=
G(K,F )
G(K,L)
Hence the theorem is completely proved.
3.7 Solvability by Radicals
Consider the general polynomial of degree 2 over a field F, p(x) = x2 +a1 x+a2 , a1 , a2 ∈
F. We can consider p(x) as a particular polynomial over F (a1 , a2 ) of rational functions
in the two variables a1 and a2 over F . Consider now the extension of F (a1 , a2 ) by
adjoining ω where ω 2 = a21 − 4a2 ∈ F (a1 , a2 ). We can see that p(x) has all its roots in
this extension field. There is a formula expressing the roots of p(x) in terms of a1 , a2
and their square roots of rational functions of a1 , a2 :
r
a21
−a1
x=
+
− a2
2
4
72
This formula was known to Indians (1500 BC) in vedic period, to the Balylonians (400
BC algorithmic to euclid (300 BC geometric) and to Brahmagwpk ( 6 th century AD)
allowing negative quantities using letters for unknown).
Cubic Equations:
We next consider a cubic equation x3 + a1 x2 + a2 x + a3 = 0. It has three complex
solutions . This was first solved by dal Ferro in1515 (atleast in some special cases). He
kept his methods secretly for 11 years before passing his knowledge to his student Fior. By
1535 Tartaglia had a general solution and defeated Fior iin a public competition. Cardano
convinced Tartaglia to divulge his solution and, breaking an oath of secrecy, published it
in his volume ARS Magna in 1545.
While somewhat messy, the solutions of the equation can be given explicitely by
Cardands formulas combinations of square roots and cube roots of rational functions in
a1 , a2 , a3 :
Let p = a2 −
r
Let P =
3
a21
3
− 2q
and q =
+
q
r
Q=
3
− 2q
−
p3
27
q
p3
27
+
+
2a31
27
q2
4
−
a1 a2
3
+ a3 .
and
q2
4
with cube roots chose properly.
Then the roots of the equations are P + Q − a31 , ωP + ω 2 Q − a31 where w 6= 1 is
a cube root of 1. The above formulas only serve to illustrace for us that by adjoining a
certain square root and then a cube root to F (a1 , a2 , a3 ) we search a field in which p(x)
has all its roots.
Quartic Equations:
For a fourth degree polynomials, by using rational operations and square roots, we
can reduce the problem to that of solving a certain cubic equation. Hence, in this case
two, a formula can be given expressing the roots in terms of combinations of radicals of
rational functions of the coefficients. These formulas were given by Cardano’s student
Ferrari. There is an even more complicated formulas, attributed to Descartes.
Quintic Equations:
what is Midly Miraculus is not that the solutions exists for quintic polynomial
equations ( polynomial equations of 5 th degree ) but they can be expressed algebraically
3 ,
in terms of the coefficients and the basic algebraic ooprations : +, −, ×, ÷, √, √
··· .
73
By the turn of 19 th century, no equivalent formula for the solutions to a quintic
polynomial equations had Materialised. It was Abel who had the crucial realisation no
such formula exists. Such a statement can be interpreted in a number of ways. Does it
mean that there are always algebraic expressions for the roots of quintic polynomials, but
their form is too complex for one single formula to describe all possibilities ? It would
therefore be necessary to have a number of formulas, may be even infinitely many. The
reality turns out to be far worse there are specific polynomials such as x5 − 4x + 2 whose
solution can not be expressed algebraically in any way whatsoever. there is no formula
for the roots of these polynomials.
A few decades after Abel, Evariste Galois started thinking a deeper problem why
dont these formula exists? Thus Galois theory was originally motivated by the desire to
understand, in a much more precise way than they hitherto had been, the solution to
general polynomial of degree greater than or equal to 5.
Definition 3.7.1:
Let F be a given field and let p(x) ∈ F [x]. Let K be the splitting field of p(x) over
F. Then p(x) is said to be solvable by radicals over F, if we can find a finite sequance
of fields F ⊆ F1 = F (ω1 ) ( F2 = F1 (ω2 ) ( · · · ( Fk = Fk−1 (ωk ) · · · · · · (1). such that all
the roots of p(x) lie in Fk and K ⊆ Fk . A tower of subfields in (1) will be called a root
tower over F for K.
Remark 3.7.2:
If such an Fk can be found, without loss of generality,we can assume it to be a normal
extension of F.
By the general polynomial p(x) of degree n over F, we mean the particular
polynomial
p(x) = xn + a1 xn−1 + a2 xn−2 + · · · + an
over F (a1 , a2 , a3 , · · · , an ), the field of rational functions in the n variables a1 , a2 , · · · , an
over F.
Now any general polynomial of degree n,
p(x) = xn + a1 xn−1 + a2 xn−2 + · · · + an
over F is solvable by radicals over F, if it is solvable by radicals over F (a1 , a2 , · · · , an ).
Definition 3.7.3: Let f (x) ∈ F [x]. Let K be a splitting field of f (x) over F. The Galois
group of f (x), or of the equation f (x) = 0 is defined to be the Galois group of the
splitting field of K over F, G(K, F ) and is denoted by Gf .
74
Definition 3.7.4: A group G is said to be solvable if we can find a chain of subgroups
G = N0 ⊃ N1 ⊃ N2 ⊃ · · · ⊇ Nk (e), where each Ni is a normal subgroup of Ni−1 such
that every factor group Ni−1 |Ni is abelian.
Example 3.7.5:
1. Every abelian group is solvable.
G = N0 ⊇ N1 = (e).
2. The symmetric group S3 of degree 3.
For take N1 = {(1), (123), (132)} is a normal subgroup of S3 , S3 = N0 , N2 = (e)
N1
1
are both abelian group as O( NS31 ) = 2 and O( N
) = 3
such that NS31 and N
N2
2
respectvely and S3 = N0 ⊇ N1 ⊇ N2 = (e).
Definition 3.7.6: Let G be a grouop and let a, b ∈ G. The commutator subgroup of G,
denoted by G0 , is a subgroup generated by all the commutators aba−1 b−1 in G where
a, b ∈ G. That is , the commutator subgrouop of G, G0 is that subgroup of G whose
elements are finite products of elements of the form aba−1 b−1 , a, b ∈ G.
Result 1: The commutator subgroup G0 of G is a normal subgroup of G.
Proof:
Let a ∈ G0 and x ∈ G. Then
xax−1 = (xax−1 a−1 )a.
Now xax−1 a−1 ∈ G0 and a ∈ G0 .
Since G0 is a subgroup if G, xax−1 = (xax−1 a−1 )a ∈ G0 .
⇒ G0 is a normal subgroup of G.
Result 2: Let G be a group consider the set S = {aba−1 b−1 |a, b ∈ G} . Let
G0 = {s1 s2 · · · sm |si ∈ S, marbitrary} . Then G0 is a subgroup of G.
Proof:
Let a, b ∈ G0 .
∴ a = s1 s2 · · · sm and b = s01 s02 · · · s0n where si , s0j ∈ S. (1 ≤ i ≤ m, 1 ≤ j ≤ n).
N ow ab−1 = (s1 s2 · · · sm )(s01 s02 · · · s0n )−1
0−1
0−1
= s1 s2 · · · sm s0−1
n sn−1 · · · s1
75
∵ s0j ∈ S, s−1
j ∈ S, 1 ≤ j ≤ n.
Hence ab−1 is again a product of a finite number of elements of S.
⇒ ab−1 ∈ G0 ⇒ G0 is a subgroup of G.
Result 3: The group
G
G0
is abelian.
Proof:
Let X, Y ∈
∴ X = aG0 , Y = bG0 where a, b ∈ G. Then
G
.
G0
X · Y = (aG0 )(bG0 ) = (ab)G0 = ba(a−1 b−1 ab)G0
= ba(a−1 b−1 (a−1 )−1 (b−1 )−1 )G0
= baG0
= bG0 · aG0 = Y · X
⇒
G
G0
is abelian.
Result 4: If N is a normal subgroup of G such that
G
N
is abelian then N ⊇ G0 .
Proof:
Let a, b ∈ G. Then
(aN )(bN ) = (bN )(aN )
(∵
G
is abelian)
N
⇒ (ab)N = (ba)N
∴ (a−1 b−1 ab)N = N ⇒ a−1 b−1 ab ∈ N.
∴ N ⊃ G0 .
Let G be a group and G0 be its commutator subgroup. Considering G0 itself
as a group, we can find its commutator subgroup (G0 )0 = G(2) generated by all the
elements (a0 )−1 (b0 )−1 a0 b0 where a0 , b0 ∈ G0 . Clearly G(2) is a subgroup of G0 and hence
a subgroup of G. Infact, G(2) is a normal subgroup of G0 and also a normal subgroup
of G. Continuing this process, we can define the higher commutator subgroups G(k) as
(k−1)
G(k) = (G(k−1) )0 . Each G(k) is a normal subgroup of G(k−1) as well as of G and GG(k)
is an abelian group.
Lemma 3.7.7: G is solvable if and only if G(k) = (e) for some integer k.
Proof:
76
Let G(k) = (e) for some integer k.
Let N0 = G, N1 = G1 , N2 = G(2) , · · · , Nk = G(k) = (e).
Then G = N0 ⊇ N1 ⊇ N2 ⊇ · · · ⊇ Nk = (e)
where each Ni is normal in Ni−1 as well as in G. Also
Ni−1
Ni
=
G(i−1)
G(i)
=
G(i−1)
(G(i−1) )0
is abelian.
Hence G is solvable.
Conversely, if G is solvable,
G = N0 ⊇ N1 ⊇ N2 ⊇ · · · ⊇ Nk = (e)
each Ni is nprmal in Ni−1 and
Ni−1
Ni
is abelian.
0
Then Ni−1
⊆ Ni . Thus
N1 ⊃ N00 = G1 , N2 ⊃ N10 ⊇ (G0 )0 = G2 , N3 ⊇ N20 ⊇ (G(2) )0 = G(3) , · · · , (e) = Nk ⊇ G(k) .
∴ G(k) = (e).
Corollary 3.7.8: If G is a solvable group and if G is a homomorphic image of G, then
G is solvable.
Proof:
Since G is the homomorphic image of G, (G)(k) is the homomorphic image of G(k)
But G(k) = (e) for some k as G is a solvable group.
(k)
∴G
= (e) for the same k.
⇒ G is also solvable.
Lemma 3.7.9: Let G = Sn , the symmetric group of degree n, where n ≥ 5. then
G(k) , k = 1, 2, · · · contains every 3-cycle of Sn .
Proof:
First note that if N is a normal subgroup of G, then N 0 is also a normal subgroup
of G.
Claim: If N contains every 3-cycles in G, N 0 must contain every 3-cycle.
Let a = (1 2 3) and b = (1 4 5) ∈ N.
77
∴ a−1 b−1 ab = (3 2 1)(5 4 1)(1 2 3)(1 4 5) = (1 4 2) ∈ N 0 .
Since N 0 is a normal subgroup of G = Sn , for any permutation π ∈ Sn ,
π −1 (1 4 2)π ∈ N 0 .
Choose aπ ∈ Sn such that |pi(1) = i1 , π(4) = i2 , π(2) = i3 where i1 , i2 , i3 are any three
distinct integers in 1, 2, · · · , n. Then
π −1 (1 4 2) π = (i1 i2 i3 ) ∈ N 0 .
⇒ N 0 contains all 3-cycles, thus proving the claim.
Since G is normal in G and contains all 3- cycles, by the claim above G0 contains
all 3-cycles.
Since G0 is normal in G and contains all 3-cycles, G(2) contains all 3-cycles.
Continuing in this way, we get G(k) contains all 3-cycles for arbitrary K.
Theorem 3.7.10: Sn is not solvable for n ≥ 5.
Proof:
Let G = Sn ,
n ≥ 5.
By lemma 3.7.9, the k th commutator subgroup G(k) contains all 3-cycles in Sn for
any k.
∴ G(k) 6= (e) for any K.
Hence by lemma 3.7.7, G can not be solvable for n ≥ 5.
This theorem will be used tp prove that a general polynomial of degree n, n ≥ 5 is
not solvable by radicals over F.
Lemma 3.7.11: Let the field F have all the nth roots of unity. Let a 6= 0 ∈ F and let
f (x) = xn − a ∈ F [x]. Let k be the splitting field of f (x) over F. Then
1. k = F (u), where u is any root of xn − a.
2. the Galois group of xn − a over F is abelian.
proof:
Since F contains all the nth roots of unity,
78
ξ=e
2πi
n
∈ F.
Note that ξ n = 1 but ξ n 6= 1 for 0 < m < n.
1
If u ∈ k is any root of xn − a, u = a n · 1
∴ u, ξu, ξ 2 u, · · · , ξ n−1 u are all roots of xn − a.
claim: u, ξu, ξ 2 u, · · · , ξ n−1 u are all distinct.
If ξ i u = ξ i u
for 0 ≤ i < j < n
then ξ i u − ξ j u = 0 ⇒ (ξ i − ξ j )u = 0.
∵ u 6= 0, ξ i − ξ j = 0 ⇒ ξ i = ξ j ⇒ ξ j−1 = 1
with 0 < j − i < n contradicting that ξ is the nth roots of unity.
Hence the claim.
Since ξ ∈ F, u, ξu, ξ 2 u, · · · , ξ n−1 u ∈ F (u).
⇒ F (u) splits the polynomial xn − a into linear factors over F.
Since any proper subfield of F (u) which contains F also should contain u, no proper
subfield of F (u) can split xn − a.
Thus F (u) is the splitting field of f (x) = xn − a.
By the uniqueness of the splitting field of a polynomial, k = F (u).
Let σ, τ be any two elements in the Galois group of xn − a.
Then σ, τ are automorphisms of k = F (u) which leaves the element of F fixed.
∴ u is a root of xn − a ⇒ σ(u) and τ (u) are also roots of xn − a.
Hence σ(u) = ξ i u and τ (u) = ξ j u for some i and j.
∴ (στ )(u) = σ(τ (u)) = σ(ξ j u) = ξ j σ(u) = ξ j ξ i u.
i.e (στ )(u) = ξ i+j u.
Also (τ σ)(u) = τ (σ(u)) = τ (ξ i u) = ξ i τ (u) = ξ i ξ j u.
i.e (τ σ)(u) = ξ i+j u
∀ u.
Thus στ and τ σ agree on u and on F . Hence στ and τ σ agree on all of F (u) = k.
⇒ the Galois group of x − a over F is abelian.
79
Theorem 3.7.12: Let F be a field containing all the nth roots of unity for every integer
n. If p(x) ∈ F [x] is solvable by radicals over F, then the Galois group over F of p(x)
is solvable.
Proof:
Let K be the splitting field of p(x) over f and let G(K, F ) be the Galois group of
p(x) over F.
Since p(x) is solvable by radicals, by definition,
F ⊃ F1 = F (ω1 ) ⊆ F2 = F1 (ω2 ) ⊂ · · · ⊂ Fk = Fk−1 (ωk )
where ω1r1 ∈ F, ω2r2 ∈ F1 , · · · , ωkrk ∈ Fk−1 and K ⊆ Fk .
Since K is the splitting field of p(x) and Fk contains all the roots of p(x) such that
K ⊆ Fk , we can assume Fk is normal extension of F. Since Fk can be considered as the
splitting field of p(x) over any Fi , Fk is a normal extension of each Fi .
Lemma 3.7.11 says that when F has all the nth roots of unity, then adjoining one
root of p(x) to F gives the whole splitting filed of p(x) and hance the field obtained is
a normal extension of F.
Thus by lemma 3.7.11, each Fi is normal over Fi−1 .
By theorem 5.6.6, G(Fk , F1 ) is a normal subgroup of G(Fk , Fi−1 ).
Consider,
G(Fk , F ) ⊇ G(Fk , F1 ) ⊇ G(Fk , F2 ) ⊇ · · · ⊇ G(Fk , Fk−1 ) ⊇ (e)
· · · · · · (1).
Since Fi is a normal extension of Fi−1 , by the fundamental theorem of Galois theorey,
we get
G(Fk , Fi−1 )
G(Fi , Fi−1 ) ≡
G(Fk , Fi )
But by lemma 3.7.11,
chain (1) is abelian.
G(Fi , Fi−1 ) is an abelian group. Thus each quotient group of
⇒ the group G(Fk , F ) is solvable.
Now K is the splitting field of p(x) over F and K ⊆ Fk .
∴ K is a normal extension of F.
⇒ G(Fk , K) is a normal subgroup of G(Fk , F ), by theorem 5.6.6 and
G(K, F ) ≡
G(Fk , F )
G(Fk , K)
80
Thus G(K, F ) is a homomorphic image of G(Fk , F ).
But G(Fk , F ) is a solvable grouop.
⇒ G(K, F ) is a solvable group.
Thus the Galois group G(K, F ) of p(x) over F is solvable.
Note:
1. The converse of the above theorem is also true. That is, if the Galois group of p(x)
over F is solvable, then p(x) is solvable by radicals over F.
2. The above theorem and its converse hold even if F does not contain roots of unity.
Theorem 3.7.13: Abel’s theorem
The general polynomial of degree n ≥ 5 is not solvable by radicals.
Proof:
By definition, a general polynomial of, degree n over F is of the form
p(x) = xn + a1 xn−1 + · · · + an .
p(x) is said to be solvable by radicals over F, if it is solvable by radicals over
F (a1 , a2 , · · · , an ),
where F (a1 , a2 , · · · , an ) is the of rational functions in the variables a1 , a2 , · · · , an .
Then the galois groups P (x) over F (a1 , a2 , · · · , an ), by theorem 5.6.3, is Sn , the
symmetric group of degree n.
But by theorem 3.8.1, Sn for n ≥ 5 is not solvable.
Hence, by theorem 3.7.12,
when n ≥ 5.
p(x) is not solvable by radicals over F (a1 , a2 , · · · , an )
3.8 Galois Groups over the Rationals.
Theorem 3.8.1:
Let q(x) be an irreducible polynomil of degree p, p a prime, over the field Q of
rational numbers. Suppose that q(x) has exactly two non real roots in the field of complex
81
numbers. Then the Galois group of q(x) over Q is Sp , the symmetric group of degree
p. Thus the splitting field of q(x) over Q has degree p! over Q.
Proof:
Let K be the splitting field of the polynomial q(x) over Q.
Let α be a root of q(x) in K. Then, since q(x) is irreducible over Q, by theorem
5.1.3, as α is algebraic of degree p over Q, we have [Q(α) : Q] = p.
Now K is a finite extension of Q as K is a splitting field of p(x) over Q and
K ⊇ Q(α) ⊇ Q.
Hence, [K : Q] = [K : Q(α) · [Q(α) : Q]
= [K : Q(α)] · p.
⇒ p divides [K : Q].
If G is the Galois group of K over Q, by theorem 5.6.4, O(G) = [K : Q]
∴ p divides O(G).
Hence by Cauchy’s theorem, G has an element σ, σ ∈ Sp and O(σ) = p.
We know that the polynomials with rational coefficient have all their has exactly two
non-real roots.
Let α1 , α2 be the two nonreal roots.
∴ α1 = α2 and α2 = α1 for if the complex number α is a root of q(x), then α, its
complex conjugate is also a root of q(x).
Let α3 , α4 , · · · , αp be the other roots. Since αi , i = 3, 4, · · · , p are real, αi = αi for
i ≥ 3.
Thus the complex conjugate mapping that takes K into itaelf, is an automorphism
τ of K over Q. That is, τ interchaniges α1 and α2 and leaves all the other roots αi
for i ≥ 3 of q(x) fixed.
That is, τ (α1 ) = α2 , τ (α2 ) = α1 and τ (αi ) = αi for i ≥ 3.
Now the elements of G take all roots of q(x) into roots q(x). Thus, the elements of
G induces permutations of α1 , α2 , · · · , αp .
In this way we imbed G in Sp .
i.e G ⊇ Sp .
82
Now the automorphisms τ on K over Q interchanging α1 and α2 and leaving the
other roots of q(x) fixed is the transposition (1 2 ).
Further, we have obtained an element σ ∈ G such that O(σ) = p.
∴ G ⊇ Sp , σ ∈ Sp and O(σ) = p.
But the only elements of order p in Sp are p− cycles. Thus σ must be a p− cycle.
∴ G as a subgroup of Sp , contains a transposition and a p− cycle.
Now any permutation in p can be written as a product of disjoint cycles. Since any
cycle (1 2 · · · r) can be written as the product of transpositions (1 2), (1 3), · · · , (1 r).
That is, (1 2 · · · r) = (1 2)(2 3) · · · (1 r).
∴ each cycle and hence each σ ∈ Sp can be written as a product of transpositions.
Thus Sp is generated by its subset of transposition.
If 1 ≤ j < k < p, we have (j, k + 1) = (k, k + 1)(j, k)(k, k + 1).
⇒ (j, k + 1) can be obtained from (j, k) with the help of the transposition (k, k + 1)
of successive points. This shows that Sp is generated by the transpositions (k, k + 1) of
successive points 1 ≤ k ≤ p.
Again, for 0 ≤ r ≤ p − 2 we have
(1 2 · · · p)r (1 2) (1 · · · p)−r = (r + 1, r + 2)
⇒ Sp is generated by the transposition (1 2) and the p− cycle (1 2 · · · p − 1 p).
Thus σ and τ generate Sp .
But σ τ ∈ G. ∴ the subgroup generated by σ and tau must be in G. That is
sp ⊆ G.
Hence G = Sp .
This proves that the Galois grouop G of the irreducible polynomial q(x) of degree p
is Sp .
∵ K is the splitting field of q(x) over Q,
[K : Q] = O(Sp ) = p!
This completes the proof of the theorem.
Problem:
83
Show that the polynomial q(x) = x3 − 3x − 3 over Q irreducible and have exactly
two non real roots. Find the Galois group of p(x) over Q.
Solution:
q(x) = x3 − 3x − 3.
Here deg q(x) = p = 3.
a0 = −3, a1 = −3, a2 = 0, a3 = 1.
p divides a0 , p divides a1 , p divides a2 , but p does not divides a3 and also p2
does not divides a0 .
p | a0 , p | a1 , p | a2 , but p - a3 and also p2 - a0 .
hence by eisensteins criteria, q(x) is an irreducible polynomial over Q.
Let us now plot the graph y = q(x) = x3 − 3x − 3.
x −2 −1 0
1
2 2.1 2.15 2.25 2.5
y −5 −1 −3 −5 −1 0 0.4 1.6 5.1
y = x3 − 3x − 3.
dy
dx
d2 y
dx2
= 3x2 − 3.
= 6x.
dy
dx
= 0 ⇒ 3x2 − 3 = 0 ⇒ x2 − 1 = 0 ⇒ x = ±1.
2
2
d y
d y
( dx
( dx
2 )x=1 = 6 > 0
2 )x=−1 = −6 < 0.
The graph of y has a maximum at x = −1 and a minimum at x = 1.
Also y = q(x) = x3 − 3x − 3 crosses the x− axis exactly once.
Hence q(x) has exactly one real root.
⇒ the other two roots must be non-real roots.
Thus q(x) = x3 − 3x − 3 is an irreducible polynomial of degree p = 3 a prime over Q
and has exactly two non-real roots in the field of complex numbers. Hence by Theorem
3.8.1, the Galois group of q(x) over Q is S3 and the splitting field of q(x) over Q has
degree 3! = 6.
Moreover, S3 is a 8owable group.
Hence by theorem 3.7.12, the polynomial q(x) is solvable by radicals.
4. SELECTED TOPICS
4.1 Finite Fields
Lemma 4.1.1: Let F be a finite field having q elements. if K is a finite containing F
then K has q n elements for some positive integer n where n = [K : F ].
Proof:
K, being a finite containing F, is a finite extension of F.
Let n = [K : F ].
Then K can be considered as a vector space over F of dimension n. Then K has a
finite basis say v1 , v2 , · · · , vn over the field F.
Hence v ∈ K has a unique representation
v = α1 v1 + α2 v2 + · · · + αn vn where αi ∈ F
Since F has q elements for each i,
(1 ≤ i ≤ n).
i = 1, 2, · · · , n αi has q choices.
Thus the number of elements in K is equal to q n .
Corollary 4.1.2: Let F be a finite field. Then F has pm elements where p is a prime
such that the characteristic of F = p.
Proof:
Assume that the finite field F has q elements.
Then by corollaryto Lagranges theorem (If G is a finite group and a ∈ G then
aO(G) = e )
q · 1 = 1 + 1 + 1 + · · · + 1 = 0 additive identity.
⇒ F is a field of finite characteristic .
hence we can find a prime p such that the characteristic of F = p.
85
Since the characteristic of F = p, F contains a subfield F0 which is isomorphic to
Jp , the field of integers modulo the prime p.
Thus F is a finite field containing the field F0 having p elements. Hence by the
above lemma F has pm elements where m = [F : F0 ].
Corollary 4.1.3:
m
If the finite field F has pm elements then every a ∈ F satisfies ap = a.
Proof:
If a = 0, the zero element of F, then the result is true trivially.
Hence assume that a 6= 0 ∈ F. Now the set F ∗ of all non-zero elements of F form a
group under multiplication and O(F ∗ ) = pm − 1.
Since a ∈ F ∗ , by corollaryto Lagranges theorem (If G is a finite group and a ∈ G
then aO(G) = e )
We have aO(F
ap
∗)
m −1
=1
=1
m
⇒ ap = a.
Lemma 4.1.4:
m
If the finite field F has pm elements then the polynomial xp − x ∈ F [x] splits into
m
Linear factors in F [x] as xp − x = π (x − λ).
λ∈F
Proof:
m
Let f (x) = xp − x.
Then f (x) is a polynomial of degree pm over F.
By fundamental theorem of algebra ( A polynomial of degree n over a field ccan have
atmost n roots in any extension field) f (x) can have atmost pm roots.
m
By the previous corollary, we see that every a ∈ F, satisfies the relation ap = a,
i.e, every a ∈ F is a root of the polynomial f (x).
m
Thus we see that the polynomial f (x) = xp − x has all the pm roots in F only.
Again if a is a root of the polynomial f (x) then we see that (x − a) divides f (x).
86
m
Thus we have, f (x) = xp − x
= π (x − a)
a∈F
Corollary 4.1.5: If the field F has pm elements then F is the splitting filed of the
m
polynomial xp − x.
Proof:
m
By the above lemma the polynomial xp − x splits into linear factors in F. if there
m
is a subfield of F in which the polynomial xp − x splits into linear factors.
Then such subfield must have pm elements. Since F itself is a field having pm
elements.
m
We have to prove that F is the smallest field in which the polynomial xp − x splits
m
into linear factors. Thus F becomes the splitting field of xp − x.
Lemma 4.1.6: any two finite fields having the same number of elements are isomorphic.
Proof:
Let F and F 0 be two finite fields having q elements. Then by corollary 4.1.2 to
lemma 4.1.1 [ Let F be a finite field, then F has pm elements where the prime number
p is the characteristic of F ] both F and F 0 has q = pm elements for some prime p.
Then by corollary to lemma 4.1.4 both F and F 0 are the splitting fields of the
m
polynomial xp − x. By the uniqueness theorem of splitting filed F and F 0 are
isomorphic.
Lemma 4.1.7:
For every prime p and every positive integer m there exists a field having pm
elements.
Proof:
Let Jp denote the field of integers modulo p.
m
Now xp − x is a polynomial over Jp .
By the fundamental theorem of algebra, this polynomial will have atmost pm roots.
m
Let K be the splitting field of the polynomial xp − x over the field Jp .
87
m
Let F = a ∈ K | ap − a = 0
m
ie, F is the set of all elements of K which are the roots of the polynomial xp − x.
m
If we denote the polynomial xp − x as f (x),
m
if f (x) = xp − x then
f 0 (x) = pm xp
m −1
−1
f 0 (x) = −1
Since f (x) and f 0 (x) have no non-trivial common factors.
m
f (x) = xp − x have no mutiple roots.
ie, The roots of f (x) are all distinct.
Thus the set F has pm elements.
Claim: F is a subfield of K
Let a, b ∈ F.
m
ap − a = 0
m
bp − b = 0
m
or
ap = a
or
bp = b
m
Expanding (a ± b)p
pm
(a + b)
= a
pm
m
pm
1
+
m
= ap + b p
as a binomial series, we have
!
a
pm −1
b+
pm
2
m
!
ap
m −2
m
b2 + · · · + bp
(∵ characteristic of (Jp ) = p)
= a+b
(a − b)p
m
m
= ap −
m
= ap ± b p
pm
1
!
ap
m −1
m
b+
pm
2
!
m −2
ap
m
m
(∵ characteristic of Jp = p, (−1)p = ±1)
= a−b
⇒ a ± b ∈ F whenever a, b ∈ F
m
m
m
b2 + · · · + (−1)p bp
m
(ab)p = ap bp = ab
⇒ ab ∈ F.
F is a subfield of K.
m
by definition of F. is a splitting filed of the polynomial xp − x.
88
Since any two splitting fields of the same polynomial are isomorphic, F and K are
m
splitting fields pf xp − x and since F is a subfield of K we have
F =K
Thus for any prime p we have determinie a field having pm elements for some positive
integer m.
Theorem 4.1.8:
For every prime number p and for every positive integer m there exists a unique field
having pm elements.
Proof:
Let Jp denote the field of integers modulo p.
m
Now xp − x is a polynomial over Jp .
By the fundamental theorem of algebra, this polynomial will have atmost pm roots.
m
Let K be the splitting field of the polynomial xp − x over the field Jp .
m
Let F = a ∈ K | ap − a = 0
m
ie, F is the set of all elements of K which are the roots of the polynomial xp − x.
m
If we denote the polynomial xp − x as f (x),
m
if f (x) = xp − x then
f 0 (x) = pm xp
m −1
−1
f 0 (x) = −1
Since f (x) and f 0 (x) have no non-trivial common factors.
m
f (x) = xp − x have no mutiple roots.
ie, The roots of f (x) are all distinct.
Thus the set F has pm elements.
Claim: F is a subfield of K
Let a, b ∈ F.
m
ap − a = 0
m
bp − b = 0
m
or
ap = a
or
bp = b
m
89
Expanding (a ± b)p
pm
(a + b)
= a
pm
m
pm
1
+
m
= ap + b p
as a binomial series, we have
!
a
pm −1
b+
pm
2
m
!
m −2
ap
b2 + · · · + bp
m
(∵ characteristic of (Jp ) = p)
= a+b
pm
(a − b)
= a
pm
−
m
= ap ± b p
pm
1
!
a
pm −1
m
b+
pm
2
!
m −2
ap
m
m
b2 + · · · + (−1)p bp
m
(∵ characteristic of Jp = p, (−1)p = ±1)
= a−b
⇒ a ± b ∈ F whenever a, b ∈ F
m
m
m
(ab)p = ap bp = ab
⇒ ab ∈ F.
F is a subfield of K.
m
By definition of F. is a splitting filed of the polynomial xp − x.
Since any two splitting fields of the same polynomial are isomorphic, F and K are
m
splitting fields pf xp − x and since F is a subfield of K we have
F =K
Thus for any prime p we have determine a field having pm elements for some positive
integer m.
Now let F and F 0 be two finite fields having pm elements for some prime p.
Then by corollary to lemma 4.1.4 both F and F 0 are the splitting fields of the
m
polynomial xp − x. By uniqueness theorem of splitting field F and F 0 are isomorphic.
Thus for any prime p we have a unique field having pm elements.
Lemma 4.1.9: Let G be a finite abelian group such that the relation xn = e is satisfied
by atmost n elements of G for every integer n. Then G is a cyclic group.
Proof:
Case (i) :
Let O(G) be a power of some prime number q.
90
ie, let O(G) = q m , q is a prime.
Choose a ∈ G such that O(a) is as large as possible.
since O(a) | O(G) = q m
⇒ O(a) = q s , s ≤ m
s
ie, aq = e for some integer s
Consider the elements e, a, a2 , · · · , aq
s −1
in G.
We claim that all the q s elements are distinct for if ai = aj for some i 6= j, 0 <
i, j < q s ai−j = e.
O(a) | i − j
⇒ qs | i − j
a contraction, i − j < q s as i, j < q s .
s
s
Since aq = e, a satisfies a relation xq = e.
Thus we see that the elements e, a, a2 , · · · , aq
s
the equation xq = e.
s −1
are all the q s distinct solutions of
Let b be some element in G.
Since O(b) | O(G) = q m ,
O(b) = q r for some integer r.
s
r
Now bq = (bq )q
s−r
=e
s
⇒ b satisfies the relation xq = e.
s
s
since e, a, a2 , · · · , aq −1 are all the only possible solutions of xq = e, b = ai for some
i which implies G is cyclic.
Case (ii):
Let O(G) be a composite number.
Let O(G) = q1 q2 · · · qk .
By direct product theorem, the finite group G can be written as the direct product
of Sylow’s subgroups.
Assume that G = Sq1 Sq2 · · · Sqk where qi is a prime divisor of O(G) and Sqi is a
Sylow subgroup of G from i varying 1 to k (i = 1 to k).
91
Hence if g ∈ G, g can be uniquely written as g = s1 s2 · · · sk where si ∈ Sqi (1 ≤
i ≤ k).
If xn = e in any Sqi then we have xn = e in G. Thus the finite group Sqi (1 ≤ i ≤ k)
is such that xn = e is satiisfied by atmost n elements of Sqi .
Hence by case(i) each Sqi is a cyclic group.
Let ai be the generator of the cyclic group Sqi (1 ≤ i ≤ k).
Claim: c = a1 a2 cdots ak generates G.
Let O(c) = m.
Since O(c) | O(G), m | O(G)
Since O(c) = m,
· · · · · · (1).
cm = e
ie, (a1 a2 · · · ak )m = e.
m
m
Since G is abelian, am
1 a2 · · · ak = e
· · · · · · (2).
By the uniqueness of the representation in G, as a product of elements from Sqi
(2) ⇒ am
i = e for each i = 1 to k
⇒ O(ai ) | m for each i, 1 ≤ i ≤ k.
Since Sqi is a cyclic group generated by ai , O(Sqi ) = O(ai ).
Hence we have O(Sqi ) | m for each i 1 ≤ i ≤ k
∴ O(G) = O(Sq1 ) O(Sq2 ) · · · O(Sqk )
= O(a1 ) O(a2 ) · · · O(ak ) | m
⇒ O(G) | m
· · · · · · (3).
(1) and (3) ⇒ O(G) = m = O(c)
∴ G is cyclic group generated by the element c = a1 a2 · · · ak .
Lemma 4.1.10:
Let K be a field. Let G be a finite subgroup of the multiplicative group of non-zero
elements of K. Then G is a cyclic group.
Proof:
Consider the ring of polynomials over K in the variable x, say K[x]. Any polynomial
92
of degree n over K will have atmost n roots in K.
∴ we can consider the particular polynomial of degree n over K, say the polynomial
x − 1.
n
This polynomial xn − 1 has atmost n roots, which are of course non-zero in the field
K.
Since G is given to be a finite subgroup of the group of non-zero elements of K we
can consider that these n non-zero roots of xn − 1 will be in G.
Thus G is a finite group such that atmost n elements of G satisfy the relation
x = 1.
n
Hence by the lemma 4.1.9, G is a cyclic group.
Theorem 4.1.11:
The multiplicative group of non-zero elements of a finite field is a cyclic group.
Proof:
Let F be a finite field. Let G be a finite subgroup of the multiplicative group of
non-zero elements of F.
Consider the ring of polynomials over F will have atmost n roots in F.
∴ we can consider the particular polynomial of degree n over F, say the polynomial
xn − 1.
This polynomial xn − 1 has atmost n roots, which are ofcourse non-zero in the field
F.
Since G is given to be a finite subgroup of the group of non-zero elements of F.
We can consider that these n non-zero roots of xn − 1 will be in G. Thus G is a
finite group such that atmost n elements of G satisfy the relation xn = 1.
Hence by the lemma 4.1.9, G is a cyclic group.
Lemma 4.1.12:
If F is a finite field and α 6= 0, β 6= 0 are two elements of F then we can find
elements a and b in F such that 1 + αa2 + βb2 = 0.
Proof:
93
Since F is a finite field its haracteristic must be a prime number.
Case(i):
Characteristic of F = 2
Then the field F will have 2n elements.
n
By corollary 4.1.2 to lemma 4.1.1 every element a ∈ F satisfies therelation a2 = a
which implies that every element in F is a square.
Thus for some a ∈ F we have α−1 = a2 .
By choosing b = 0 we get
1 + αa2 + βb2 = 1 + αα−1 + 0
= 1+1=2
[since characteristic of F = 2]
= 0
Case(ii):
Characteristic of F = p, odd p.
The field F has pn elements.
Consider the set Wα = {1 + αx2 | x ∈ F }
If 1 + αx2 = 1 + αy 2 then α(x2 − y 2 ) = 0.
Since α 6= 0 we have x2 − y 2 = 0 (or)
x2 = y 2
i.e, x = ±y
Thus for x 6= 0 we get one element in Wα from each pair x or −x and when
x = 0, 1 ∈ Wα .
This implies that the number of elements in Wα will be equals to
1+pn −1
2
=
pn +1
.
2
Now consider the set Wβ = {−βx2 | x ∈ F }
Similarly Wβ has
pn +1
2
elements.
Since both Wα and Wβ contain more than half the elements of F. They must have
some elements in common.
ie, Wα ∩ Wβ 6= φ.
Let c ∈ Wα ∩ Wβ
94
c ∈ Wα and c ∈ Wbeta .
c ∈ Wα ⇒ c = 1 + αa2 for some a ∈ F.
c ∈ Wβ ⇒ c = −βb2 for some b ∈ F.
1 + αa2 = −βb2
⇒ 1 + αa2 + βb2 = 0.
4.2 Wedderburn’s Theorem on Finite Division Rings
Lemma 4.2.1:
Let R be a ring and let a ∈ R. Let Ta be the mapping of R into itself defined by
xTam = xam −maxam−1 +
m(m − 1) 2 m−2 m(m − 1)(m − 2) 3 m−3
a xa
−
a xa
+ · · · +(−1)m am x.
2
3!
Proof:
We shall prove this by induction
xTa = xa − ax
xTa2 = (xTa )Ta = (xa − ax)Ta
= (xa − ax)a − a(xa − ax)
= xa2 − axa − axa + a2 x
= xa2 − 2axa + a2 x
xTa3 = (xTa2 )Ta
= (xa2 − 2axa + a2 x)Ta
= (xa2 − 2axa + a2 x)a − a(xa2 − 2axa + a2 x)
= xa3 − 2axa2 + a2 xa − axa2 + 2a2 xa − a3 x
= xa3 − 3axa2 + 3a2 xa − a3 x
Assume that the result is true for m
xTam = xam − maxam−1 +
m(m − 1) 2 m−2
a xa
+ · · · + (−1)m am x
2
95
We have to prove that the result is true for m + 1
xTam+1 = (xTam )Ta
m(m − 1) 2 m−2
a xa
+ · · · + (−1)m am x)Ta
2
m(m − 1) 2 m−2
= (xam − maxam−1 +
a xa
+ · · · + (−1)m am x)a
2
m(m − 1) 2 m−2
− a(xam − maxam−1 +
a xa
+ · · · + (−1)m am x)
2
m(m − 1) 2 m−1
= xam+1 − maxam +
a xa
+ · · · + (−1)m am xa
2
m(m − 1) 3 m−2
− axam + ma2 xam−1 −
a xa
+ · · · + (−1)m+1 am+1 x
2
m(m + 1) 2 m−1
= xam+1 − (m + 1)axam +
a xa
+ · · · + (−1)m+1 am+1 x
2
= (xam − maxam−1 +
⇒ the result is true for all values.
Corollary 4.2.2:
If R is a ring in which px = 0
m
m
pm
xa + (−1)p ap x.
∀x ∈ R where p is a prime, then xTap
Proof:
Case(i) : Let p = 2
xTa2 = (xTa )Ta = (xa − ax)Ta
= (xa − ax)a − a(xa − ax)
= xa2 − axa − axa + a2 x
= xa2 − 2axa + a2 x
= xa2 + a2 x
(∵ 2(axa) = 0)
xTa4 = (xTa2 )Ta2
= (xa2 + a2 x)Ta2
= (xa2 + a2 x)a2 + a2 (xa2 + a2 x)
= xa4 + a2 xa2 + a2 xa2 + a4 x
= xa4 + 2a2 xa2 + a4 x
= xa4 + a4 x
continuing in this way we have
m
m
m
xTa2 = xa2 + a2 x.
(∵ 2a2 xa2 = 0)
m
=
96
case(ii):
When p is an odd prime.
By lemma 4.2.1,
p(p−1) 2
a xap−2
2!
xTap = xap − paxap−1 +
Since p |
p(p−1)(p−2)···(p−i+1)
i!
−
p(p−1)(p−2) 3
a xap−3
3!
+ · · · + (−1)p ap x
for i < p and x = 0 ∀x ∈ R.
xTap = xap − ap x
xTap
2
2
= xap − p2 axap
2 −1
+
p2 (p2 − 1) 2 p2 −2
a xa
2!
p2 (p2 − 1)(p2 − 2) 3 p2 −3
2
−
a xa
+ · · · − ap x
3!
Since p | p(p−1)(p−2)···(p−i+1)
for i < p and x = 0 ∀x ∈ R all the intermediate terms
i!
vanish and we have
2
2
2
xTap = xap − ap x.
Lemma 4.2.3:
Let D be a division fing of characteristic p > 2 with centre Z.
Let P = {0, 1, 2, · · · , (p − 1)} be the subfield of Z isomorphic to the filed Jp . Suppose
n
that a ∈ D, a ∈
/ Z satisfies ap = a for some n ≥ 1. Then there exists an element
x ∈ D such that
1. xax−1 6= a
2. xax−1 ∈ P (a), the field obtained by adjoining the element a to the filed P.
In particular, xax−1 = ai 6= a for some integer i.
Proof:
For a ∈ D and a ∈
/ Z define Ta : D → D by yTa = ya − ay
n
Also a satisfies ap = a.
n
ie, a satisfies the polynomial xp − x ∈ P [x].
⇒ a is algebraic over P.
⇒ P (a) is a finite extension of P.
∀y ∈ D.
97
∴ P (a) will have pn elements.
1
All these pn elements of P (a) satisfy u1 = u.
yTap
n
n
n
n
= yap + (−1)p ap y
n
= ya + (−1)p ay
n
when p is an odd prime we get yTap = ya − ay = yTa .
n
Thus Tap = Ta .
Let λ ∈ P (a).
(λx)Ta = (λx)(a) − a(λx)
= λ(xa) − λ(ax)
[∵ λ commutes with a]
= λ(xa − ax)
ie, (λx)Ta = λ(xTa ) · · · · · · (1). if we consider λI is a mapping from D into itself defined
by x(λI) = λx then (1) implies (xλI)Ta = (xTa )(λI).
ie, x[(λI)Ta ] = x[Ta (λI)] ∀λ ∈ P (a)
⇒ The mapping λI : D → D commutes with Ta ∀λ ∈ P (a).
n
By lemma 4.2.1, up − u =
Π (u − λ)
λ∈P (a)
If for every λ 6= 0 in P (a), the mapping Ta − λI annihilates every y 6= 0 in D,
0 = y · 0 = y[Ta (Ta − λ1 I)(Ta − λ2 I) · · · (Ta − λk I)]
= yTa
⇒ Ta = 0
⇒ yTa = 0 (or) ya − ay = 0
⇒ a ∈ Z contradicting the hypothesis.
∴ there exists aλ 6= 0 in P (a) and an element x 6= 0 in D
x(Ta − λI) = 0
⇒ xTa − x(λI) = 0
⇒ xa − ax − λx = 0
⇒ xa = ax + λx
⇒ xax−1 = a + λ 6= a
(∵ λ 6= 0)
Since P (a) is generated by a and since λ ∈ P (a) we have xax−1 ∈ P (a).
98
Corollary 4.2.4: In lemma 4.2.3 xax−1 = ai 6= a for some integer i.
proof:
Assume that the element a is of order s.
∴ as = 1.
Thus a satisfies the polynomial us − 1.
Now the roots of us − 1, which are distinct are given by 1, a, a2 , · · · , as−1 .
However, (xax−1 )s = xas x−1 = 1.
Since xax−1 ∈ P (a) and satisfy the polynomial us − 1 we have xax−1 = ai for some
integer i.
Theorem 4.2.5: (Wedderburn) a finite division ring is a commutative field.
Proof:
Let D be a finite division ring.
Let Z be its center. If Z has q elements then D has q n elements.
Our aim is to prove that D = Z (or) n = 1.
We shall prove the theorem by induction on the number of elements in D.
Assume that the theorem is true for all division ring whose number of elements is less
than that of D.
Claim: 1
If a, b ∈ D such that bt a = abt (t > 1) with ab 6= ba then bt ∈ Z.
Proof:
Define the set N (bt ) as follows N (bt ) = {x ∈ D | xbt = bt x}
We shall first prove that N (bt ) is a subdivision ring.
Since 1 ∈ D is such that 1 · bt = bt · 1 = bt .
1 ∈ N (bt )
∴ N (bt ) 6= φ.
Let x, y ∈ N (bt )
99
By definition, xbt = bt x; ybt = bt y
(x − y)bt = xbt − ybt
= bt x − bt y
= bt (x − y)
∴ (x − y) ∈ N (bt )
(xy)bt = x(ybt )
= x(bt y)
= (xbt )y
= (bt x)y
= bt xy
⇒ xy ∈ N (bt )
xbt = bt x ⇒ x−1 (xbt )x−1 = x−1 (bt x)x−1
⇒ bt x−1 = x−1 bt
⇒ x−1 ∈ N (bt )
∴ N (bt ) is a subdivision ring.
Since N (bt ) is a finite division ring with fewerelements than D, by our induction
hypothesis N (bt ) is a commutative ring whenever N (bt ) 6= D.
However both a and b ∈ N (bt ) and ab 6= ba implies that N (bt ) is not commutative
contradicting our conclusion.
Hence N (bt ) must be equals to D.
∴ bt ∈ Z.
Every non- zero element in D has finite order.
∴ some positive of the element falls in the centre Z [ for w ∈ D let the order of w
relative to Z be the smallest positive integer m(w) such that wm(w) ∈ Z. ]
Let a ∈ D, a ∈
/ Z be such that the order of the element a relative to Z, denoted by
r, be the minimal possible order.
Claim: 2
r is a prime number.
Proof:
Let r = r1 r2 ,
1 < r1 , r2 < r
100
By definition, a is chosen such that a ∈ D,
positive power)
∴ ar1 ∈
/Z
a ∈
/ Z,
ar ∈ Z ( r is the smallest
(∵ r1 < r).
(ar1 )r2 = ar1 r2 = ar ∈ Z
∴ order of ar1 relative to Z is r2 < r = order of a relative to Z
∴ r is not a composite number (or) r is a prime.
By lemma 4.2.3, there is an element x ∈ D such that xax−1 = ai 6= a
N ow x2 ax−2 = x(xax−1 )x−1
= xai x−1
= (xax−1 )i
= ai
Similarly xr−1 ax−(r−1) = ai
2
r−1
However by claim 2, r is a prime number .
since r is prime ir−1 = 1 + u0 r
∴ ai
r−1
= a1+u0 r
= a · au0 r = λa
∴ xr−1 ax−(r−1) = ai
r−1
where λ = au0 r ∈ Z
= λa
⇒ xr−1 a = λaxr−1
Since x ∈
/ Z and since r is the minimal order of a relative to Z,
Since xa 6= ax, xr−1 a 6= axr−1 ,
xr−1 ∈
/ Z.
λ 6= 1.
Let b = xr−1 then bab−1 = xr−1 ax−(r−1) = λa
∴ λr ar = (λa)r = (bab−1 )r = bar b−1 = ar
(∵ ar ∈ Z)
Since ar ∈ Z, λr = 1.
Claim:3
If y ∈ D, then y = λi for some i whenever y r = 1.
Proof:
Since y r = 1, y satisfies the polynomial ur − 1. However in the field Z(y), the
polynomial ur − 1 has atmost r roots, say 1, λ, λ2 , · · · , λr−1 for λ is of the prime order
r.
101
Since y is a root of ur −1 and all the r distinct roots of ur −1, 1, λ, λ2 , · · · , λr−1 , y =
λi for some i.
Since λr = 1,
br = λr br
= (λb)r
= (aba−1 )r
br = abr a−1
⇒ br a = abr .
⇒ a commutes with br but not with b
⇒ br ∈ Z.
Since the multiplicative group of non-zero elements of the centre Z is cyclic of ν is
a generator of this cyclic group and since ar , br ∈ Z, we have
ar = ν j and br = ν k for some j and k.
If r | j then j = rs
∴ ar = ν j = ν rs
⇒ ( νas )r = 1
⇒
a
νs
= λi
(∵ 1 = y r = λi = y)
⇒ a = λi ν s ∈ Z
⇒ a∈Z
which is a contradiction
[∵ a ∈
/ Z]
∴ r - j.
Similarly we can prove that r - k.
Let a1 = ak and b1 = bj
102
Since ba = λab, we have
a1 b 1 = ak b j
= ak−1 (ab)bj−1
= ak−1 (λ−1 ba)bj−1
= λ−1 (ak−1 b)abj−1
= λ−1 (ak−2 (ab))(ab)bj−2
= λ−1 ak−2 λ−1 (ba)λ−1 (ba)bj−2
= λ−3 ak−2 (ba)(ba)bj−2
a1 b1 = λ−3 ak−3 (ab)(ab)(ab)bj−3
Continuing in this way we get
a1 b1 = λ−jk b1 a1 .
ie, a1 b1 = µb1 a1 where µ = λ−jk .
Since O(λ) = r, r - j, r - k
λjk 6= 1
ie, µ 6= 1
However µr = 1.
Thus we have to determine two elements a1 and b1 such that
1. ar1 , br1 ∈ Z
2. a1 b1 = µb1 a1 , µ 6= 1 ∈ Z
3. µr = 1
103
r
We shall compute (a−1
1 b1 )
2
−1
(a−1
= (a−1
1 b1 )
1 b1 )(a1 b1 )
−1
= a−1
1 (b1 a1 )b1
−1
= a−1
1 (µa1 b1 )b1
2
= µa−2
1 b1
3
2 −1
(a−1
= (a−1
1 b1 )
1 b1 ) (a1 b1 )
2 −1
= µa−2
1 b 1 a1 b 1
−1
= µa−2
1 b1 (b1 a1 )b1
3
−1
(a−1
= µa−2
1 b1 )
1 b1 (µa1 b1 )b1
−1 2
= µ2 a−2
1 (b1 a1 )b1
3
= µ3 a−3
1 b1
3
= µ1+2 a−3
1 b1
Continuing, we get
r
r
(a−1
= µ1+2+···+(r−1) a−r
1 b1 )
1 b1
= µ
r(r−1)
2
r
a−r
1 b1
= µ
r(r−1)
2
r
(a−1
1 b1 )
(∵ ar1 , br1 ∈ Z)
r
If r is an odd prime, whenever (a−1
1 b1 ) = 1 we have µ
r(r−1)
2
= 1.
µr = 1
−1
r
r
(a−1
1 b1 ) = 1 ⇒ a1 b1 is a solution of y = 1
i
∴ a−1
1 b1 = λ for some i
⇒ b1 = λi a1
1
Now a1 b1 = µb1 a1 ⇒ b1 a−1
1 = µa1 b1
Since b1 = λi a1 we get λi = µλi
⇒ µ = 1 a contradiction to µ 6= 1
This contradiction forces us to conclude that D = Z.
ie, D is a commutative division ring or simply a field.
If the prime r = 2, we have the two elements a1 , b1 ∈ D such that
1. a21 , b21 ∈ Z
104
2. a1 b1 = µb1 a1
3. µ2 = 1 with µ 6= 1
The contradiction (3) ⇒ µ = −1
⇒ a1 b1 = −b1 a1
⇒ a1 b 1 + b 1 a1 = 0
⇒ characteristic of D 6= 2.
∴ By lemma 4.1.12, for 1, −b21 6= 0 ∈ D.
We can find elements τ, η such that 1 + s2 − b1 η 2 = 0.
(a1 + τ b1 + ηa1 b1 )2 = (a1 + τ b1 + ηa1 b1 )(a1 + τ b1 + ηa1 b1 )
= a21 + τ a1 b1 + ηa21 b1 + τ b1 a1 + τ 2 b21 + τ ηb1 a1 b1 + ηa1 b1 a1 + ητ a1 b21 + η 2 a1 b1 a1 b
= a21 + τ 2 b21 = η 2 a21 b21
= b21 (1 + τ 2 − η 2 b21 )
= b21 (0)
= 0
∴ a1 + τ b1 + ηa1 b1 = 0.
0 6= 2a21 = a1 (a1 + τ b1 + ηa1 b1 ) + (a1 + τ b1 + ηa1 b1 )a1
= 0.
This contradiction forces us to conclude that D = Z.
∴ D is a commutative division ring.
BIBLIOGRAPHY
[Br] R. Brauer, On algebras which are connected with the semisimple continuous
groups, Ann. of Math. 38(1937), 854-872.