Instructor Solutions Manual for Galois Theory, Fifth Edition 5th edition
Introduction
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Introduction This Solutions Manual contains solutions to all of the exercises in the Fifth Edition 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 involved 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) contradicts 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 = b′, AC ′= a′. Since AB = HF = AB and BH = AC, we have AE = a′ + b′ = b, √ AF = a + 2b′ = a, say. Therefore a′ + b′ = b, b′ = a — b, and ba = ba′ . say, and If 2 is rational, we can make a b integers, in which case a′ b′ , √ , are also integers, 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 ±): m m r = ±p1 1 · · · pk k where the mj are integers. So r2 = p12m1 · · · pk2mk
1 Classical Algebra √ Now q = r if and only if q = r2, and all exponents 2mj are even. √ √ 1.4 * Clearly 18 ± 325 = 18 ± 5 13. A little experiment shows that √ !3 3 ± 13 √ = 18 ± 5 13 2
3
(The factor 12 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 + rα2, where p, q, r∈ Q. Clearly K is closed under addition and subtraction. Since α3 = 2 we also have α4 = 2α, and it follows easily that K is closed under multiplication. Tedious but elementary calculations, or computer algebra, show that (p + qα + rα2)(p + qωα + rω2α2)(p + qω2α + rωα2) = p3 + 2(q3 — 3pqr) + 4r3 (1) so that (p + qα + rα 2)—1 =
(p + qωα + rω2α2)(p + qω2α + rωα2) 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 + rα2 = 0 or p + qωα + rω2α2 = 0 or p + qω2α + rωα2 = 0. The required result follows since 1, α, α2 are linearly independent over Q. 1.6 The map is one-to-one since it is linear in (p, q, r) and p + qω2α + rωα2 = 0 implies p = q = r = 0. Compute (p + qα + rα2)(a + bα + cα2) = (pa + 2qc + 2rb) + (pb + qa + 2rc)α + (pc + qb + ra)α2 and compare with (p + qωα + rω2α2)(a + bωα + cω2α2)
= (pa + 2qc + 2rb) + (pb + qa + 2rc)ωα + (pc + qb + ra)ω2α2
4 The coefficients are there same in both formulas, so products are preserved as required. 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 /∈ 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 = —a2b, and the cubic becomes t3 — 3a2t = a2b The inequality becomes a > |b|/2. Substitute t = 2a cos θ , and observe that t3 — 3a2t = 8a3 cos3 θ — 6a3 cos θ = 2a3 cos 3θ The cubic thus reduces to
cos 3θ =
b 2a
which we can solve using cos—1 because |2ab | ≤ 1, getting θ=
1 3
cos—1
b 2a
There are three possible values of θ , the other two being obtained by adding 23π or 4π . Finally, eliminate θ to get 3 t = 2a cos
1
b cos—1 3 2a
q where a = —3p , b = 3pq . 1.9 By inspecti√ on 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 b for rational u, v given Specifically, suppose we w √a n t√to solve (u + v) = a + rational a, b. Then assuming b, v are irrational, we are led to u3 + 3uv = a √ √ (3u2 + v) v = b √ √ It follows easily that (u — v)3 = a — b, whence √ 3 u2 — v = a2 — b
1 Classical Algebra
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Therefore we seek a rational solution u of the cubic √ 3 4u3 — 3 ( a2 — b)u — a = 0 √ and then v = u2 — 3 a2 — b. 2 3 In our case a = — q , b = q + p , so the cubic is 2
4
27
4u3 + pu = —
q 2
which is equivalent to x3 + px + q = 0 with x = 2u.
1.11 * Let A(n) be the number of permissible sequences of length n ending in 0, and let B(n) be the number of permissible sequences of length n ending in 1. We claim that
A(n) = A(n — 1) + B(n — 1) B(n) = B(n — 1) + A(n — 3)
(2) (3)
Equation (2) holds because every permissible sequence of length n ending in 0 is uniquely of the form S· 0 where S is a permissible sequence of length— n 1. Equation (3) holds because every permissible sequence of length n ending in 1 is either of the form S ·1 where S is a permissible sequence of length n—1 ending with 1, or T 1· where S is a permissible sequence of length n—1 ending with 011 (which is not permissible). But sequences of the latter form are precisely those of the form U · 11 where U is a permissible sequence of length n — 3 ending with 0. Clearly P(n) = A(n) + B(n) From ((2), (3)) 0 = [A(n)— A(n — 1)— B(n — 1)] + [B(n — 1) + A(n — 3)— B(n)] But
B(n — 1) = A(n)— A(n — 1)
and eliminating B(n — 1) leads to A(n) = 2A(n — 1)— A(n — 2) + A(n — 4) Similarly
B(n) = 2B(n — 1)— B(n — 2) + B(n — 4)
Adding and using (2) we get P(n) = 2P(n — 1)— P(n — 2) + P(n — 4)
(4)
The theory of linear recurrences, see for example Slomson (1991) chapter 6, now tells us that P(n+1) tends to a limit as n → ∞. Dividing the recurrence (4) by P(n — 4) P(n) we get P(n) P(n — 1) P(n — 2) =2 — +1 P(n — 4) P(n — 4) P(n — 4)