Introduction 1
Introduction
This Solutions Manual contains solutions to all of the exercises in the Fifth Edi-
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-
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.
Ian Stewart
Coventry January 2022
1 Classical Algebra
1.1 Let u = x + iy ≡ (x, y), v = a + ib ≡ (a, b), w = p + iq ≡ (p, q). Then
uv = (x, y)(a, b)
= (xa − yb, xb + ya)
= (ax − by, bx + ay)
= (a, b)(x, y)
= vu
(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)
= (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
an integer, the set of possible elements b has a minimal element.
,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
(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.
The ancient Greeks did not use algebra. They expressed them same underlying
idea in terms of a geometric figure, Figure 1.
√
FIGURE 1: Greek proof that 2 is irrational.
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
√ 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.
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
, 1 Classical Algebra 3
√
Now q = r if and only if q = r2 , and all exponents 2m j are even.
√ √
1.4* Clearly 18 ± 325 = 18 ± 5 13. A little experiment shows that
√ !3
3 ± 13 √
= 18 ± 5 13
2
(The factor 21 is the only real surprise here: it occurs because 13 is of the form 4n + 1,
but it would take us too far afield to explain why.) At any rate,
√
√
q
3 3 ± 13
18 ± 5 13 =
2
so that
√ √
√ √
q q
3 3 3 + 13 3 − 13
18 + 5 13 + 18 − 5 13 = +
2 2
3 3
= + =3
2 2
1.5 Let K be the set of all p + qα + rα 2 , where p, q, r ∈ Q. Clearly K is closed under
addition and subtraction. Since α 3 = 2 we also have α 4 = 2α, and it follows easily
that K is closed under multiplication.
Tedious but elementary calculations, or computer algebra, show that
(p + qα + rα 2 )(p + qωα + rω 2 α 2 )(p + qω 2 α + rωα 2 ) = p3 + 2(q3 − 3pqr) + 4r3
(1)
so that
(p + qωα + rω 2 α 2 )(p + qω 2 α + rωα 2 )
(p + qα + rα 2 )−1 =
p3 + 2(q3 − 3pqr) + 4r3
implying closure under inverses, hence division.
However, it is necessary to check that p3 + 2(q3 − 3pqr) + 4r3 = 0 in rational
numbers implies p = q = r = 0. By (1) p3 + 2(q3 − 3pqr) + 4r3 = 0 implies that
p + qα + rα 2 = 0 or p + qωα + rω 2 α 2 = 0 or p + qω 2 α + rωα 2 = 0. The required
result follows since 1, α, α 2 are linearly independent over Q.
1.6 The map is one-to-one since it is linear in (p, q, r) and p + qω 2 α + rωα 2 = 0
implies p = q = r = 0. Compute
(p + qα + rα 2 )(a + bα + cα 2 )
= (pa + 2qc + 2rb) + (pb + qa + 2rc)α + (pc + qb + ra)α 2
and compare with
(p + qωα + rω 2 α 2 )(a + bωα + cω 2 α 2 )
= (pa + 2qc + 2rb) + (pb + qa + 2rc)ωα + (pc + qb + ra)ω 2 α 2
, 4
The coefficients are there same in both formulas, so products are preserved as re-
quired. Thus the map is a monomorphism.
All maps are onto their image. But the image here is not Q(α) because Q(α) ⊆ R,
but ω 6∈ R. So the map is not an automorphism.
1.7 Observe that √
(2 ± i)3 = 2 ± 11i = 2 ± −121
and (2 + i) + (2 − i) = 4.
1.8 The inequality 27pq2 + 4p3 < 0 implies that p < 0, so we can find a, b such that
p = −3a2 , q = −a2 b, and the cubic becomes
t 3 − 3a2t = a2 b
The inequality becomes a > |b|/2. Substitute t = 2a cos θ , and observe that
t 3 − 3a2t = 8a3 cos3 θ − 6a3 cos θ = 2a3 cos 3θ
The cubic thus reduces to
b
cos 3θ =
2a
which we can solve using cos−1 because | 2a
b
| ≤ 1, getting
1 b
θ= cos−1
3 2a
There are three possible values of θ , the other two being obtained by adding 2π 3 or
4π
3 . Finally, eliminate θ to get
1 −1 b
t = 2a cos cos
3 2a
q
where a = −p 3q
3 ,b = p .
1.9 By inspection √ one root is√t = 4. Factoring out t − 4 leads to a quadratic whose
roots are −2 + 3 and −2 − 3.
1.10 If you carry out the algebra, it turns out that trying to solve for α and β leads
back to the original cubic equation. Unless the solutions are obvious (in which case
the method is pointless) no progress is made. √ √
3
√ √to solve (u + v) = a + b for rational u, v given
Specifically, suppose we want
rational a, b. Then assuming b, v are irrational, we are led to
u3 + 3uv = a
2 √ √
(3u + v) v = b
√ √
It follows easily that (u − v)3 = a − b, whence
p3
u2 − v = a2 − b
Introduction
This Solutions Manual contains solutions to all of the exercises in the Fifth Edi-
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-
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.
Ian Stewart
Coventry January 2022
1 Classical Algebra
1.1 Let u = x + iy ≡ (x, y), v = a + ib ≡ (a, b), w = p + iq ≡ (p, q). Then
uv = (x, y)(a, b)
= (xa − yb, xb + ya)
= (ax − by, bx + ay)
= (a, b)(x, y)
= vu
(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)
= (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
an integer, the set of possible elements b has a minimal element.
,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
(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.
The ancient Greeks did not use algebra. They expressed them same underlying
idea in terms of a geometric figure, Figure 1.
√
FIGURE 1: Greek proof that 2 is irrational.
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
√ 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.
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
, 1 Classical Algebra 3
√
Now q = r if and only if q = r2 , and all exponents 2m j are even.
√ √
1.4* Clearly 18 ± 325 = 18 ± 5 13. A little experiment shows that
√ !3
3 ± 13 √
= 18 ± 5 13
2
(The factor 21 is the only real surprise here: it occurs because 13 is of the form 4n + 1,
but it would take us too far afield to explain why.) At any rate,
√
√
q
3 3 ± 13
18 ± 5 13 =
2
so that
√ √
√ √
q q
3 3 3 + 13 3 − 13
18 + 5 13 + 18 − 5 13 = +
2 2
3 3
= + =3
2 2
1.5 Let K be the set of all p + qα + rα 2 , where p, q, r ∈ Q. Clearly K is closed under
addition and subtraction. Since α 3 = 2 we also have α 4 = 2α, and it follows easily
that K is closed under multiplication.
Tedious but elementary calculations, or computer algebra, show that
(p + qα + rα 2 )(p + qωα + rω 2 α 2 )(p + qω 2 α + rωα 2 ) = p3 + 2(q3 − 3pqr) + 4r3
(1)
so that
(p + qωα + rω 2 α 2 )(p + qω 2 α + rωα 2 )
(p + qα + rα 2 )−1 =
p3 + 2(q3 − 3pqr) + 4r3
implying closure under inverses, hence division.
However, it is necessary to check that p3 + 2(q3 − 3pqr) + 4r3 = 0 in rational
numbers implies p = q = r = 0. By (1) p3 + 2(q3 − 3pqr) + 4r3 = 0 implies that
p + qα + rα 2 = 0 or p + qωα + rω 2 α 2 = 0 or p + qω 2 α + rωα 2 = 0. The required
result follows since 1, α, α 2 are linearly independent over Q.
1.6 The map is one-to-one since it is linear in (p, q, r) and p + qω 2 α + rωα 2 = 0
implies p = q = r = 0. Compute
(p + qα + rα 2 )(a + bα + cα 2 )
= (pa + 2qc + 2rb) + (pb + qa + 2rc)α + (pc + qb + ra)α 2
and compare with
(p + qωα + rω 2 α 2 )(a + bωα + cω 2 α 2 )
= (pa + 2qc + 2rb) + (pb + qa + 2rc)ωα + (pc + qb + ra)ω 2 α 2
, 4
The coefficients are there same in both formulas, so products are preserved as re-
quired. Thus the map is a monomorphism.
All maps are onto their image. But the image here is not Q(α) because Q(α) ⊆ R,
but ω 6∈ R. So the map is not an automorphism.
1.7 Observe that √
(2 ± i)3 = 2 ± 11i = 2 ± −121
and (2 + i) + (2 − i) = 4.
1.8 The inequality 27pq2 + 4p3 < 0 implies that p < 0, so we can find a, b such that
p = −3a2 , q = −a2 b, and the cubic becomes
t 3 − 3a2t = a2 b
The inequality becomes a > |b|/2. Substitute t = 2a cos θ , and observe that
t 3 − 3a2t = 8a3 cos3 θ − 6a3 cos θ = 2a3 cos 3θ
The cubic thus reduces to
b
cos 3θ =
2a
which we can solve using cos−1 because | 2a
b
| ≤ 1, getting
1 b
θ= cos−1
3 2a
There are three possible values of θ , the other two being obtained by adding 2π 3 or
4π
3 . Finally, eliminate θ to get
1 −1 b
t = 2a cos cos
3 2a
q
where a = −p 3q
3 ,b = p .
1.9 By inspection √ one root is√t = 4. Factoring out t − 4 leads to a quadratic whose
roots are −2 + 3 and −2 − 3.
1.10 If you carry out the algebra, it turns out that trying to solve for α and β leads
back to the original cubic equation. Unless the solutions are obvious (in which case
the method is pointless) no progress is made. √ √
3
√ √to solve (u + v) = a + b for rational u, v given
Specifically, suppose we want
rational a, b. Then assuming b, v are irrational, we are led to
u3 + 3uv = a
2 √ √
(3u + v) v = b
√ √
It follows easily that (u − v)3 = a − b, whence
p3
u2 − v = a2 − b
