Galois Theory 5th Edition
ST
SOLUTIONS
UV
MANUAL
IA
_A
Ian Stewart
PP
────────────────────────────────────────────────────
Comprehensive Solutions Manual for Instructors
RO
and Students
VE
© Ian Stewart. All rights reserved. Reproduction or distribution without permission is
prohibited.
D?
?
© DreamsHub
, Solutions Manual for Galois Theory (5th Edition)
Ian Stewart
ISBN: 9781032101583
ST
UNIT 1: FOUNDATIONS OF ALGEBRA AND POLYNOMIAL THEORY
1. Classical Algebra
2. The Fundamental Theorem of Algebra
3. Factorisation of Polynomials
UV
UNIT 2: FIELD EXTENSIONS AND CONSTRUCTIONS
4. Field Extensions
5. Simple Extensions
6. The Degree of an Extension
7. Ruler-and-Compass Constructions
IA
UNIT 3: CORE IDEAS OF GALOIS THEORY
8. The Idea Behind Galois Theory
9. Normality and Separability
_A
10. Counting Principles
11. Field Automorphisms
12. The Galois Correspondence
13. Worked Examples
PP
UNIT 4: SOLVABILITY AND CLASSICAL RESULTS
14. Solubility and Simplicity
15. Solution by Radicals
UNIT 5: ABSTRACT ALGEBRA AND GALOIS GROUPS
RO
16. Abstract Rings and Fields
17. Abstract Field Extensions and Galois Groups
18. The General Polynomial Equation
UNIT 6: FINITE FIELDS AND GEOMETRIC APPLICATIONS
19. Finite Fields
VE
20. Regular Polygons
21. Circle Division
UNIT 7: ADVANCED TOPICS AND COMPUTATIONAL METHODS
22. Calculating Galois Groups
D?
23. Algebraically Closed Fields
24. Transcendental Numbers
UNIT 8: HISTORICAL CONTEXT AND FURTHER STUDY
25. What Did Galois Do or Know?
?
26. Further Directions
© MedConnoisseur
,Solutions Manual for FGalois Theory, 5e
by Ian Stewart (All Chapters)
Introduction 1
Introduction
This Solutions Manual contains solutions to all of the exercises in the Fifth Edi-
ST
tion of Galois Theory.
Many of the exercises have several different solutions, or can be solved using
several different methods. If your solution is different from the one presented here, it
may still be correct — unless it is the kind of question that has only one answer.
The written style is informal, and the main aim is to illustrate the key ideas in-
UV
volved in answering the questions. Instructors may need to fill in additional details
where these are straightforward, or explain assumed background material. On the
whole, I have emphasised ‘bare hands’ methods whenever possible, so some of the
exercises may have more elegant solutions that use higher-powered methods.
IA
Ian Stewart
Coventry January 2022
_A
1 Classical Algebra
1.1 Let u = x + iy ≡ (x, y), v = a + ib ≡ (a, b), w = p + iq ≡ (p, q). Then
PP
uv = (x, y)(a, b)
= (xa − yb, xb + ya)
= (ax − by, bx + ay)
= (a, b)(x, y)
= vu
RO
(uv)w = [(x, y)(a, b)](p, q)
= (xa − yb, xb + ya)(p, q)
= (xap − ybp − xbq − yaq, xaq − ybq + xbp + yap)
= (x, y)(ap − bq, aq + bp)
VE
= (x, y)[(a, b)(p, q)]
= (uv)w
1.2 (1) Changing the signs of a, b does not affect (a/b)2 , so we may assume a, b > 0.
(2) Any non-empty set of positive integers has a minimal element. Since b > 0 is
D?
an integer, the set of possible elements b has a minimal element.
?
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Distribution of this document is illegal extra per year?
, 2
(3) We know that a2 = 2b2 . Then
(2b − a)2 − 2(a − b)2 = 4b2 − 4ab + a2 − 2(a2 − 2ab + b2 )
= 2b2 − a2 = 0
ST
(4) If 2b ≤ a then 4b2 ≤ a2 = 2b2 , a contradiction. If a ≤ b then 2a2 ≤ 2b2 = a2 ,
a contradiction.
(5) If a − b ≥ b then a ≥ 2b so a2 ≥ 4b2 = 2a2 , a contradiction. Now (3) contra-
dicts the minimality of b.
Note on the Greek approach.
UV
The ancient Greeks did not use algebra. They expressed them same underlying
idea in terms of a geometric figure, Figure 1.
IA
_A
√
FIGURE 1: Greek proof that 2 is irrational.
PP
Start with square ABCD and let CE = AB. Complete square AEFG. The rest of
the figure leads to a point H on AF. Clearly AC/AB = AF/AE. In modern notation,
let AB = b0 , AC = a0 . Since AB = HF = AB and BH = AC, we have AE = a0 + b0 = b,
0
say, and AF = a0 + 2b0 = a, say. Therefore a0 + b0 = b, b0 = a − b, and ab = ab0 .
√ 0 0
RO
√ a , b are also integers,
If 2 is rational, we can make a, b integers, in which case
and the same process of constructing rationals equal to 2 with ever-decreasing
numerators and denominators could be carried out. The Greeks didn’t argue the proof
quite that way: they observed that the ‘anthyphaeresis’ of AF and AE goes on forever.
This process was their version of what we now call the continued fraction expansion
(or the Euclidean algorithm, which is equivalent). It stops after finitely many steps if
and only if the initial ratio lies in Q. See Fowler (1987) pages 33–35.
VE
1.3 A nonzero rational can be written uniquely, up to order, as a produce of prime
powers (with a sign ±):
mk
r = ±pm 1
1 · · · pk
where the m j are integers. So
2mk
r2 = p12m1 · · · pk
D?
?
Downloaded by: tutorsection | Want to earn $1.236
Distribution of this document is illegal extra per year?
ST
SOLUTIONS
UV
MANUAL
IA
_A
Ian Stewart
PP
────────────────────────────────────────────────────
Comprehensive Solutions Manual for Instructors
RO
and Students
VE
© Ian Stewart. All rights reserved. Reproduction or distribution without permission is
prohibited.
D?
?
© DreamsHub
, Solutions Manual for Galois Theory (5th Edition)
Ian Stewart
ISBN: 9781032101583
ST
UNIT 1: FOUNDATIONS OF ALGEBRA AND POLYNOMIAL THEORY
1. Classical Algebra
2. The Fundamental Theorem of Algebra
3. Factorisation of Polynomials
UV
UNIT 2: FIELD EXTENSIONS AND CONSTRUCTIONS
4. Field Extensions
5. Simple Extensions
6. The Degree of an Extension
7. Ruler-and-Compass Constructions
IA
UNIT 3: CORE IDEAS OF GALOIS THEORY
8. The Idea Behind Galois Theory
9. Normality and Separability
_A
10. Counting Principles
11. Field Automorphisms
12. The Galois Correspondence
13. Worked Examples
PP
UNIT 4: SOLVABILITY AND CLASSICAL RESULTS
14. Solubility and Simplicity
15. Solution by Radicals
UNIT 5: ABSTRACT ALGEBRA AND GALOIS GROUPS
RO
16. Abstract Rings and Fields
17. Abstract Field Extensions and Galois Groups
18. The General Polynomial Equation
UNIT 6: FINITE FIELDS AND GEOMETRIC APPLICATIONS
19. Finite Fields
VE
20. Regular Polygons
21. Circle Division
UNIT 7: ADVANCED TOPICS AND COMPUTATIONAL METHODS
22. Calculating Galois Groups
D?
23. Algebraically Closed Fields
24. Transcendental Numbers
UNIT 8: HISTORICAL CONTEXT AND FURTHER STUDY
25. What Did Galois Do or Know?
?
26. Further Directions
© MedConnoisseur
,Solutions Manual for FGalois Theory, 5e
by Ian Stewart (All Chapters)
Introduction 1
Introduction
This Solutions Manual contains solutions to all of the exercises in the Fifth Edi-
ST
tion of Galois Theory.
Many of the exercises have several different solutions, or can be solved using
several different methods. If your solution is different from the one presented here, it
may still be correct — unless it is the kind of question that has only one answer.
The written style is informal, and the main aim is to illustrate the key ideas in-
UV
volved in answering the questions. Instructors may need to fill in additional details
where these are straightforward, or explain assumed background material. On the
whole, I have emphasised ‘bare hands’ methods whenever possible, so some of the
exercises may have more elegant solutions that use higher-powered methods.
IA
Ian Stewart
Coventry January 2022
_A
1 Classical Algebra
1.1 Let u = x + iy ≡ (x, y), v = a + ib ≡ (a, b), w = p + iq ≡ (p, q). Then
PP
uv = (x, y)(a, b)
= (xa − yb, xb + ya)
= (ax − by, bx + ay)
= (a, b)(x, y)
= vu
RO
(uv)w = [(x, y)(a, b)](p, q)
= (xa − yb, xb + ya)(p, q)
= (xap − ybp − xbq − yaq, xaq − ybq + xbp + yap)
= (x, y)(ap − bq, aq + bp)
VE
= (x, y)[(a, b)(p, q)]
= (uv)w
1.2 (1) Changing the signs of a, b does not affect (a/b)2 , so we may assume a, b > 0.
(2) Any non-empty set of positive integers has a minimal element. Since b > 0 is
D?
an integer, the set of possible elements b has a minimal element.
?
Downloaded by: tutorsection | Want to earn $1.236
Distribution of this document is illegal extra per year?
, 2
(3) We know that a2 = 2b2 . Then
(2b − a)2 − 2(a − b)2 = 4b2 − 4ab + a2 − 2(a2 − 2ab + b2 )
= 2b2 − a2 = 0
ST
(4) If 2b ≤ a then 4b2 ≤ a2 = 2b2 , a contradiction. If a ≤ b then 2a2 ≤ 2b2 = a2 ,
a contradiction.
(5) If a − b ≥ b then a ≥ 2b so a2 ≥ 4b2 = 2a2 , a contradiction. Now (3) contra-
dicts the minimality of b.
Note on the Greek approach.
UV
The ancient Greeks did not use algebra. They expressed them same underlying
idea in terms of a geometric figure, Figure 1.
IA
_A
√
FIGURE 1: Greek proof that 2 is irrational.
PP
Start with square ABCD and let CE = AB. Complete square AEFG. The rest of
the figure leads to a point H on AF. Clearly AC/AB = AF/AE. In modern notation,
let AB = b0 , AC = a0 . Since AB = HF = AB and BH = AC, we have AE = a0 + b0 = b,
0
say, and AF = a0 + 2b0 = a, say. Therefore a0 + b0 = b, b0 = a − b, and ab = ab0 .
√ 0 0
RO
√ a , b are also integers,
If 2 is rational, we can make a, b integers, in which case
and the same process of constructing rationals equal to 2 with ever-decreasing
numerators and denominators could be carried out. The Greeks didn’t argue the proof
quite that way: they observed that the ‘anthyphaeresis’ of AF and AE goes on forever.
This process was their version of what we now call the continued fraction expansion
(or the Euclidean algorithm, which is equivalent). It stops after finitely many steps if
and only if the initial ratio lies in Q. See Fowler (1987) pages 33–35.
VE
1.3 A nonzero rational can be written uniquely, up to order, as a produce of prime
powers (with a sign ±):
mk
r = ±pm 1
1 · · · pk
where the m j are integers. So
2mk
r2 = p12m1 · · · pk
D?
?
Downloaded by: tutorsection | Want to earn $1.236
Distribution of this document is illegal extra per year?
