Galois Theory, Fifth Edition [5th Ed] (Instructor Solution Manual, Solutions) [5 ed.] 103210158X, 9781032101583

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Introduction

1

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 = 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 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 ±): mk 1 r = ±pm 1 · · · pk where the m j are integers. So 2mk

r2 = p12m1 · · · pk

1 Classical Algebra

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√ 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 ± 13 3 18 ± 5 13 = 2 so that q 3

√ 18 + 5 13 +

q 3

√ √ √ 3 + 13 3 − 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 + qα + rα 2 )−1 =

(p + qωα + rω 2 α 2 )(p + qω 2 α + rωα 2 ) 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 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 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 cos 3θ =

b 2a

b which we can solve using cos−1 because | 2a | ≤ 1, getting

θ=

1 b cos−1 3 2a

There are three possible values of θ , the other two being obtained by adding 2π 3 or 4π . Finally, eliminate θ to get 3   1 −1 b t = 2a cos cos 3 2a q 3q where a = −p 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 Specifically, suppose we want √ √to solve (u + v) = a + b for rational u, v given rational a, b. Then assuming b, v are irrational, we are led to u3 + 3uv = a √ √ (3u + v) v = b √ √ It follows easily that (u − v)3 = a − b, whence p 3 u2 − v = a2 − b 2

1 Classical Algebra

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Therefore we seek a rational solution u of the cubic p 3 4u3 − 3( a2 − b)u − a = 0 √ and then v = u2 − 3 a2 − b. 2 p3 , so the cubic is In our case a = − 2q , b = q4 + 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)

(2)

B(n) = B(n − 1) + A(n − 3)

(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) P(n) tends to a limit as n → ∞. Dividing the recurrence (4) by P(n − 4) we get P(n) P(n − 1) P(n − 2) =2 − +1 P(n − 4) P(n − 4) P(n − 4)

6 which we rewrite as P(n) P(n − 1) P(n − 2) P(n − 3) P(n − 1) P(n − 2) P(n − 3) P(n − 4) P(n − 1) P(n − 2) P(n − 3) P(n − 2) P(n − 3) =2 − +1 P(n − 2) P(n − 3) P(n − 4) P(n − 3) P(n − 4) As n → ∞, all the fractions tend to the same limit x, so x4 = 2x3 − x2 + 1 Finally, x must be the largest positive real root of this equation by the general theory of linear recurrences. Now x4 − 2x3 + x2 − 1 = (x2 − x + 1)(x2 − x − 1) These two quadratics have roots

√ √ 1±i 3 1± 5 2 , 2

respectively. The first has complex

roots; the largest real root of the second is x = the golden number. 1.12*

√ 1+ 5 2

= 1·618034 . . ., often called

FIGURE 2: Left: Central square in Calabi’s triangle. Right: Tilted square in Calabi’s triangle. By Pythagoras, d2 = 1 −

x2 4 − x2 = 4 4

By similar triangles in Figure 2 (left), x 2

h d −h = h h −2 2

h2 = (d − h)(x − h) = dx − hx − dh + h2 0 = dx − hx − dh h(x + d) = xd

(5)

By similar triangles in Figure 2 (right), h d 2x = x = 1−h d 2 hx = 2d − 2dh h(x + 2d) = 2d

(6)

1 Classical Algebra

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Eliminating h from ((5), (6)) we get x+d xd x = = x + 2d 2d 2 2(x + d) = x(x + 2d) 2x + 2d = x2 + 2dx d(2 − 2x) = x2 − 2x = x(x − 2) d 2 (2 − 2x)2 = x2 (x − 2)2 4 − x2 (2 − 2x)2 = x2 (x − 2)2 4 (4 − x2 )(2 − 2x)2 = x2 (x − 2)2 (x − 2)2 (x + 2) + x2 (x − 2) = 0 2x3 − 2x2 − 3x + 2 = 0 as required. Numerically, x ∼ 1·551. 1.13 We seek p, q, r, s such that x4 + ax3 + bx2 + cx + d = (x2 + px + q)2 − (rx + s)2 which leads to 2p = a 2

p + 2q − r2 = b 2pq − 2rs = c q2 − s2 = d Clearly we must set p = a/2. Set q = (b + r2 − a2 /4)/2 to solve the second equation for q in terms of r, and s=

−c + 2pq −c + a(b + r2 − a2 /4)/2 −2c + a(b + r2 − a2 /4) = = 2r 2r 4r

to solve the third equation for s in terms of r. Finally, substitute all of this into the fourth equation: ((b + r2 − a2 /4)/2)2 − (

−2c + a(b + r2 − a2 /4) 2 ) −d = 0 4r

and multiply by r2 to remove denominators. This yields    4  1 6 b 3a2 4 3a a2 b2 ac − r + − + + − d r2 0= r + 4 2 16 64 4q 4 4   6 4 2 3 2 a a b a a c abc c + − + − − + − 256 32 16b2 16 4 4 which is a cubic in r2 . 1.14 (a) F. (b) T. (c) T. (d) T. (e) F. (f) F. (g) F. (h) T. (i) F. (j) F.

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2 The Fundamental Theorem of Algebra 2.1 Use induction on ∂ p. If p has no rational zeros then q = p and we are done. Otherwise, p has a zero α1 ∈ Q. By the Remainder Theorem, (t − α1 )|p, so p(t) = (t − α1 )s(t) with ∂ s = ∂ p − 1 < ∂ p. Inductively, s(t) = (t − α2 ) · · · (t − αr )q(t) where a has no rational zeros and the α j ∈ Q. Clearly p(β ) = 0 for rational β if and only if β = α j for some j, since q has no rational zeros. For uniqueness, suppose that also p(t) = (t − β1 ) · · · (t − βs )Q(t) where the β j ∈ Q and Q has no rational zeros. Then (t − α1 ) · · · (t − αr )q(t) = (t − β1 ) · · · (t − βs )Q(t) Cancelling any common linear factors we can assume that the αi and β j are distinct. If r > 0 then 0 = p(α1 ) = (α1 − β1 ) · · · (α1 − βs )Q(α1 ) so Q(α1 ) = 0, a contradiction. Therefore r = 0. Similarly s = 0, so q = Q and the result follows. 2.2 As an example, we prove the commutative law for addition. By definition, (an ) + (bn ) = (tn ), where tn = an + bn (bn ) + (an ) = (un ), where un = bn + an Therefore un = tn for all n, so (an ) + (bn ) = (bn ) + (an ). The associative law for addition is similar. The commutative law for multiplication follows from: (an )(bn ) = (tn ), where tn = an b0 + · · · + a0 bn (bn )(an ) = (un ), where un = bn a0 + · · · + b0 an Therefore un = tn for all n, so (an )(bn ) = (bn )(an ). The remaining laws can be checked in the same manner. Next, observe that θ (k + l) = (k + l, 0, 0, . . .) = (k, 0, 0, . . .) + (l, 0, 0, . . .) = θ (k) + θ (l)

2 The Fundamental Theorem of Algebra

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θ (kl) = (kl, 0, 0, . . .) = (k, 0, 0, . . .)(l, 0, 0, . . .) = θ (k)θ (l) Finally, θ (k) = 0 if and only if (k, 0, 0, . . .) = (0, 0, 0, . . .), which is true if and only if k = 0. Therefore θ is an isomorphism between C and θ (C). Identify a ∈ C with θ (a), and let t = (0, 1, 0, . . .). Then t 2 = (0, 0, 1, 0, . . .), t 3 = (0, 0, 0, 1, . . .), and inductively t N = (0, . . . , 0, 1, 0, . . .) | {z } N

for all N ∈ N. Therefore a0 + a1t + · · · + aN t N = (a0 , a1 , . . . , aN , 0, . . .) = (an ) since an = 0 for n > N. 2.3 Use similar calculations but express them in the standard notation a0 + a1t + · · · + aN t N for polynomials. 2.4 Let f (t) = t + 1, g(t) = −t. Then ∂ f = ∂ g = 1, but ∂ ( f + g) = 0. 2.5* Follow the hint. Consider the z j as independent indeterminates over C. Then D is a polynomial in the z j of total degree 0 + 1 + 2 + · · · + (n − 1) = 12 n(n − 1). Moreover, D vanishes whenever z j = zk for all j 6= k, and these linear polynomials have no common factor, so D is divisible by ∏ j 1. Since [L : K] is prime, L = K(α).√Thus √ √ √ √ √ L/K√is simple. 6.13 The key point is that 6 10 = 60 = 4 15, so 15 ∈ Q( 6, 10). The rest is routine. 6.14* We prove the statement by induction on n. We know that [Q(α1 , . . . , α j+1 ) : [Q(α1 , . . . , α j )] = 1 or 2 for all j, and the Tower Law implies that it is enough to prove that the degree is always 2. This is equivalent to proving that α j+1 6∈ Q(α1 , . . . , α j ). This can be proved by induction on j using the same argument as in Exercise 6.11. (The final step requires a proof that α1 · · · αn 6∈ K, which follows since a1 · · · an is not a square in K.) When K = Q, a product a j1 · · · a jk is a square if and only if every prime p occurs an even number of times, in total, in the prime factorisations of a j1 , . . . , a jk . 6.15* (a) Since the number of intermediate fields is finite, there exists an intermediate field M that is maximal subject to the condition that [M : K] is finite. If M 6= L there exists α ∈ L \ M. Let M 0 = M(α). Since α is algebraic over K, hence over M, we have [M 0 : M] finite. By the Tower Law, [M 0 : K] is finite. This contradicts maximality of M. Therefore M = L and [L : K] = [M : K] is finite. (b) Suppose that L = K(α1 , α2 ). Since K is an infinite field, there are infinitely many distinct elements of L of the form α1 + β α2 for β ∈ K. If all intermediate fields Jβ = K(α1 + β α2 ) are distinct, there must exist infinitely many intermediate fields, contrary to hypothesis. Therefore the Jβ are not distinct, so there exist β 6= γ ∈ K with Jβ = Jγ . Then Jβ contains (α1 + β α2 ) − (α1 + γα2 ) = (β − γ)α2 , so α2 ∈ Jβ . But then α1 = (α2 + β α2 ) − β α2 ∈ Jβ , so L = Jβ . (c) Since [L : K] is finite by (a) we can write L = K(α1 , . . . , αn ) for some finite n. We prove by induction on r that whenever 1 ≤ r ≤ n, and for all α1 , . . . , αr ∈ L, the extension K(α1 , . . . , αr )/K is simple. This is true when r = 1. Assume it true for r − 1. By (b) K(α1 , . . . , αr ) = K(δ , α3 , . . . , αr ) for suitable δ ∈ L. By induction, K(δ , α3 , . . . , αr )/K is simple. But this is K(α1 , . . . , αr )/K and the induction step is complete. Setting r = n we deduce that L/K is simple. (d) Conversely, let L = K(α) be simple algebraic, so that α is algebraic over K. Let m be it minimal polynomial over K, which is monic and irreducible over K. Possibly M is reducible over L, but it always has only a finite number of monic irreducible divisors d1 , . . . , ds over L. Suppose that M is an intermediate field. Let mM be the minimal polynomial of α over M. Then mM |m and is monic, so mM = d j for some j ≤ s. We claim that M is determined by mM : if so, the proof is complete. Write mM (t) = t r + ar−1t r−1 + · · · + a + 0, where the ai ∈ M. We claim that M = K(a0 , . . . , ar−1 ). Define M0 = K(a0 , . . . , ar−1 ). Then M0 ⊆ M, so mM is irreducible over M0 . Since L = M0 (α) the degree [L : M0 (α)] = ∂ mM = r. But [L : M] = ∂ mM = r for similar reasons. Therefore by the Tower Law [M : M0 ] = r/r = 1 so M = M0 as claimed. 6.16 (a) F. (b) T. (c) F. (d) T. (e) T. (f) F. (g) F. (h) T. (i) F.

24

7 Ruler-and-Compass Constructions 7.1 Clearly P0 = {0, 1} is closed under complex conjugation. Assume Pn is closed under complex conjugation. Let z ∈ Pn+1 . Then z is constructed from a set of points S ⊆ Pn . Perform the same construction on S¯ = {¯z : z ∈ S} to obtain z¯. Then z¯ ∈ Pn+1 . 7.2 (a) See Figure 6 (left). Given line AB, draw circle centre A radius AB and circle centre B radius BA. Let them meet in C, D. Then CD is the required perpendicular bisector.

FIGURE 6: Left: Finding the perpendicular bisector. Right: Trisecting a line. (b) See Figure 6 (right). Given line AB, draw any line BC. Draw an arc of a circle, centre B, meeting BC at X. Draw an arc of a circle with radius BX, centre X, meeting BC at Y. Draw an arc of a circle with radius BX, centre Y, meeting BC at Z. Join ZA. Draw XP parallel to ZA meeting AB at PQ. Draw YQ parallel to ZA meeting AB at Q. Then P, Q are the required trisection points. (For a construction of the parallel to a given line through a given point, see Exercise 20.1 below.) (c) Use the obvious generalisation of (b), marking out n equal steps along line BC. (d) See Figure 7 (left). Let A lie on a circle centre O. Produce OA to form a line OAB and let the circle centre A radius AO cut OB at C. Use (a) to construct the perpendicular bisector PQ of OC. Then PQ is the tangent to the circle at the point A. (e) See Figure 7 (right). Let the circles have centres P, Q and suppose that they cut the line segment PQ at A, B respectively. Draw a circle centre A, radius BQ, cutting PQ at C. Draw a circle centre P radius PC. Construct the midpoint M of PQ using (a). Draw the semicircle centre M through P and Q, meeting the circle centre P radius PC at X. Draw the line QX. Produce QY to cut the larger circle at Y and draw YZ parallel to XQ. Then YZ is one of the required common tangents. To find the other common tangent, replace the semicircle shown by the other half of the same circle and continue as before. 7.3 With a little coordinate geometry we can obtain ‘estimates’ that are best possible and therefore exact in ‘typical’ cases where the coordinates are not special.

7 Ruler-and-Compass Constructions

25

FIGURE 7: Left: Finding a tangent to a circle. Right: Finding the common tangent to two circles. (a) The degree is 1. (A line at right angles to one of slope a has slope −1/a so no square roots are needed). (b) The degree is 1. (c) The degree is 1. (d) The degree is 1. (Reason as in (a).) (e) The degree is ≤ 2. The key step is to construct point X. Suppose that P = (0, 0) and Q = (a, 0). Let the radii of the initial two circles be PY = r, QZ = s. Then PX = r − s. We know that ∠PXQ is a right angle. Therefore if X = (x, y) we have x2 + y2 = (r − s)2 y y = −1 x x−a p leading to x = r2 /a. This does not involve a square root, but y = r 1 − r2 /a2 does, unless r and a take special values. So the degree is ≤ 2. 7.4 Let ∠EDY = φ and draw OE as in Figure 8 (left). Triangles OXE and EOD are isosceles. Therefore ∠EOD = φ and ∠OEX = ∠OXE = 2φ . Therefore ∠XOE = π − 4φ . Since θ + (π − 4φ ) + φ = π, we have θ = 3φ so φ = θ /3.

FIGURE 8: Left: Trisecting an angle with a marked ruler. Right: Construction of a regular 15-gon.

26 7.5 Yes, because 2π/15 is constructible by combining the constructions of 2π/3 and 2π/5. Indeed,   2π 1 2π 2π = − 15 2 3 5 leading to the construction in Figure 8 (right). Let ABC be an equilateral triangle and APQRS be a regular pentagon, both inscribed in a circle centre O. Bisect ∠POB at M. Then ∠POM is equal to 2π/15. The rest of the construction is easy: ‘step out’ 15 copies of ∠POM round the circle using a compass. 7.6 A regular 9-gon is constructible if and only if the angle 2π/9 is constructible. If this construction is possible we can trisect 2π/3, contrary to Theorem 7.14. 7.7 The identity we need is cos 5θ = 16 cos5 θ − 20 cos3 θ + 5 cos θ

(12)

Let θ = 2π/5 and x = 2 cos θ . Then x5 − 5x3 + 5x − 2 = 0 which factorises as (x − 2)(x2 +√x − 1)2 = 0. Therefore x2 + x − 1 = 0, so by the quadratic formula √ 5−1 5 from a right 2 cos 2π 5 = 2 . Using Pythagoras’s Theorem we can construct triangle whose sides are 1 and 2. Using standard constructions for adding or halving numbers we can construct cos 2π 5 ; now the regular pentagon can easily be completed. θ 7.8 Let t = cos 3 . Then t is a zero of 4t 3 − 3t − cos θ . If the angle θ can be trisected then [Q(cos θ ,t) : Q(cos θ )] is a power of 2. If 4t 3 − 3t − cos θ is irreducible over Q(cos θ ) the degree is 3. Otherwise it is 1 or 2, in which case the angle θ can be trisected. So the angle can be trisected if and only if the polynomial is reducible. 7.9 We repeat the figure for reference, Figure 9.

FIGURE 9: Ramanujan’s approximate squaring of the circle. We calculate the coordinates of the successive points constructed. Assume that O is the origin and OA = OB = 1. Write A = (Ax, Ay) and so on. Then:

7 Ruler-and-Compass Constructions

Ax = −1 Ay = 0 Bx = 1 By = 0 Ax 1 Mx = =− 2 2 Ay My = =0 2 Bx 2 Tx = 2 = 3 3 By Ty = 2 =0 3 2 Px = T x = 3 √ p 5 2 Py = 1 − Px = 3 √ q 5 2 2 PT = (Px − T x) + (Py − Ty) = 3 1 − PT2 13 Qx = = 2 18√ p 155 Qy = 1 − Qx2 = 18 Ax + Qx 5 Sx = =− 2 36 √ Ay + Qy 155 Sy = 2 36 5Qx Ax 47 Rx = + = 6 6 108 √ 5Qy Ay 5 155 Ry = + = 6 6 108 √ q 31 2 2 AS = (Ax − Sx) + (Ay − Sy) = 6 AS2 41 Dx = −1 = − 2 72√ p 3503 Dy = − 1 − Dx2 = 72 √ q 31 2 2 RS = (Rx − Sx) + (Ry − Sy) = 9

27

28 Cx = Ax = −1 √ 31 Cy = −RS = 9 √ q 113 2 2 BD = (Bx − Dx) + (By − Dy) = 6 q 3 BM = (Bx − Mx)2 + (By − My)2 = √ 2 BM 113 Ex = Bx + (Dx − Bx) = 1 − BD √ 8 31 BM (Dy − By) = − Ey = By + BD 8 √ q 355 BC = (Bx −Cx)2 + (By −Cy)2 = 9 BM 18 Xx = Bx + (Cx − Bx) = 1 − √ BD 113 r BM 31 Xy = By + (Cy − By) = − BD 113 r q 355 BX = (Bx − Xx)2 + (By − Xy)2 = 113 355 Since √ 113 is a close approximation to π, it follows that BX is a close approximation to π. Ramanujan’s own solution uses Euclidean geometry and is much more elegant: Let d be the diameter of the circle. Then: 5 2 RS2 = d 36 31 2 PS2 = d 36 PL = MN PK = PM, so 31 31 2 PK2 = d 2 PL2 = d 144 324

Hence 113 2 d 144 355 2 RL2 = PR2 + PL2 = d 113

RK2 = PR2 − PK2 =

But RK RC 3 = = RL RD 2 3 RC = d 4

r

113 355

7 Ruler-and-Compass Constructions

29

so d RD = 2

r

355 113

7.10 Let ∠AOB = θ , ∠AOE = φ . Then EF OE OB sin θ = OB(2 + cos θ ) sin θ = 2 + cos θ

sin φ =

Therefore φ = θ /3 if and only if sin

θ sin θ = 3 2 + cos θ

(13)

Let x = θ3 . Then cos θ = cos3 x − 3 cos x sin2 x cos θ = 3 cos2 x sin x − sin3 x For simplicity, write s = sin x, c = cos x. Therefore (13) is true if and only if s=

3c2 s − s3 2 + c3 − 3cs2

which simplifies to (c − 1)(4c2 − 3) = 0 q so that c = 1, ± 34 , which is in general false. A numerical plot of   sin θ θ arcsin − 2 + cos θ 3 for the range 0 ≤ θ ≤ π2 is shown in Figure 10 (left). Numerically the error appears to be at most 0 · 006. Routine calculations (find the local maximum of the graph) verify this rigorously. 7.11 Let O, A, B be points in P0 . We have to show that it is possible to draw a circle centre O radius AB using a restricted version of the ‘compass’ operation, in which the only permitted circles are those whose centre lies in P0 and which pass through some other point of P0 . If we can do this, the the restricted operation implies that we can carry out the one defined in Section 7.2. Clearly the operation defined in Section 7.2 implies that we can carry out the restricted operation. See Figure 10 (right). Construct the midpoint M of OB. (To do so, draw the

30

FIGURE 10: Left: Numerical plot of error. Right: Restricted use of compass suffices. circle centre O passing through B and the circle centre B passing through O. The line joining the two points where these circles meet cuts OB at M.) Draw the circle centre M passing though A. Then line AM, suitably produced, meets this circle at a point X 6= A. Then OX = AB (we have completed the parallelogram OABX) and the circle centre O passing through X has radius AB as required. 7.12 Given an angle θ , bisect it twice to get θ /4 and add the two angles. Bisect twice more to get θ /64 and add that. Continue and take the limit as the number of steps tends to infinity. The result is θ /3. √ 7.13 The construction is possible if and only if n 2 is constructible from Q. Since t n − 2 is irreducible over Q (Eisenstein) the degree is n. Therefore n must be a power of 2. p √ n/2 √ Conversely, if n = 2k then n 2 = 2, so by induction on k the construction is possible using ruler and compass. 7.14

FIGURE 11: Duplicating the cube using a marked ruler. In Figure 11 join BD. We have XY = AB = 1. Let BY = x. The triangles XYA and XBD are similar since YAE is parallel to BD by hexagon geometry. The cosine rule applied to triangle XBA yields BX2 = AX2 + AB2 − 2.AX.AB cos ∠XAB Now ∠XAB =

2π 3

so cos ∠XAB = − 21 . Therefore (x + 1)2 = y2 + 1 + y

7 Ruler-and-Compass Constructions Since AD = 2, we have

x 1

31

= 2y . Therefore y = 2x , so 4 2 + x2 x x4 + 2x3 = 4 + 2x x2 + 2x =

x3 (x + 2) = 2(x + 2) x3 = = 2 √ since x 6= −2. The only real solution is x = 3 2 as required. 7.15 Figure 11 shows three positions in which the marked ruler can be placed. The θ 4π associated angles are θ3 , θ3 + 2π 3 , and 3 + 3 . The proofs are similar to Exercise 7.3.

FIGURE 12: Three positions for trisection using a marked ruler. 7.16 Let ζ = e2πi/11 . Then ζ is a zero of the polynomial f (t) = t 10 + t 9 + t 8 + t 7 + t 6 + t 5 + t 4 + t 3 + t 2 + t + 1 This is irreducible by Lemma 3.22, so [Q(ζ ) : Q] = 10 6= 2k and the result follows. 7.17 Let ζ = e2πi/13 . Then ζ is a zero of the polynomial f (t) = t 12 + t 11 + t 10 + t 9 + t 8 + t 7 + t 6 + t 5 + t 4 + t 3 + t 2 + t + 1 This is irreducible by Lemma 3.22, so [Q(ζ ) : Q] = 12 6= 2k and the result follows. 7.18 For the regular 17-gon, the same method leads to an irreducible polynomial t 16 + t 15 + · · · + t + 1 with degree 16, which is a power of 2. 2π 7.19* The angle 2π 5 is constructible (regular pentagon). We claim that the angle 25 2π 2πi/25 is not. To prove this, let ζ = e . Constructing 25 is equivalent to constructing ζ . Now ζ is a zero of t 25 − 1. This is divisible by t 5 − 1 in Q[t], so ζ is a zero of the quotient t 25 − 1 g(t) = 5 = t 20 + t 15 + t 10 + t 5 + 1 t −1

32 Put t = u + 1: the resulting polynomial is of the form u20 + · · · where all coefficients except the first are divisible by 5 and the constant term is not divisible by 52 . By Eisenstein’s Criterion with p = 5, g(t) is irreducible. But the degree is 20, which is not a power of 2. 7.20 Let a/b ∈ Q, where a, b ∈ Z. Then a, b are constructible. To construct the angle θ with tan θ = a/b, draw the right-angled triangle with vertices at 0, a, a + ib. Then the angle between the real axis and the line from 0 to a + ib is θ . 7.21* Begin with the trigonometric identity tan 3φ =

3 tan φ − tan3 φ 1 − 3 tan2 φ

Let φ = θ /3, and x = tan φ . Clearly φ is constructible if and only if x is. Now x satisfies a 3x − x3 = 1 − 3x2 b or equivalently P(x) = bx3 − 3ax2 − 3bx + a = 0 If P is irreducible over Q it has degree 3, so the number x is constructible if and only p is reducible. Since it is a cubic, this is equivalent to P having a rational root, say m/n where m, n ∈ Z are coprime. Now a m3 − 3mn2 = b 3m2 n − n3 (a) We claim that when a + b is odd, m3 − 3mn2 and 3m2 n − n3 are coprime. Suppose not. Let the gcd of m3 − 3mn2 and 3m2 n − n3 be h. Then a=

m3 − 3mn2 h

b=

3m2 n − n3 h

so ah = m3 − 3mn2

bh = 3m2 n − n3

(14)

Suppose that a prime p divides h, hence divides both m3 − 3mn2 and 3m2 n − n3 . Then p divides at least one of m, m2 − 3n2 , and at least one of n, 3m2 − n2 . Since m, n are coprime, p does not divide both m and n. If p|m, 3m2 − n2 then also p|n, contradiction. Similarly if p|n, m2 − 3n2 we get a contradiction. Therefore p does not divide m or n, but it divides both m2 − 3n2 and 3m2 − n2 . Therefore it divides their difference 2m2 + 2n2 = 2(m2 + n2 ). One possibility is that p = 2. Otherwise p|m2 + n2 . But also p|3m2 − n2 , so it divides their sum 4m2 . Since p 6= 2, we must have p|m, but then also p|n, contradiction. Thus the only possibility is that p = 2, so h is even. If h > 2 there is a prime p that divides m2 + n2 (by the same argument), contradicting m, n coprime. So h = 2. By (14) (a + b)h = m3 − 3mn2 + 3m2 n − n3 = (m − n)3

7 Ruler-and-Compass Constructions

33

so m − n is even. Therefore (m − n)3 is divisible by 8, so a + b is even, contradiction. Now a, b > 0 so a = |m3 − 3mn2 | = |m(m2 − 3n2 )|

b = |3m2 n − n3 | = |n(3m2 − n2 )|

so a2 + b2 = (m6 − 6m4 n2 + 9m2 n4 ) + (9m4 n2 − 6m2 n4 + n6 ) = m6 + 3m4 n2 + 3m2 n4 + n6 = (m2 + n2 )3 Therefore a2 + b2 is a perfect cube. Conversely, suppose that a2 + b2 is a perfect cube, say c3 . Then c3 = a2 + b2 = (a + ib)(a − ib). Consider this for Gaussian integers Z[i] = {h + ik : h, k ∈ Z}. In this ring, prime factorisation is possible and unique. Moreover, a + ib and a − ib are coprime. (Any common factor p divides 2a and 2b, hence either divides 2 or a, b. But a + b is odd so p divides 2. But this makes c even, so a2 + b2 is even. Since a + b is odd, one of a, b is odd and the other even, so the same holds for their squares: contradiction.) Therefore each factor a ± ib is a perfect cube in Z[i]. Let a + ib = (m + in)3 , where m, n ∈ Z. Expanding: a = m3 − 3mn2

b = 3m2 n − n3

Therefore m/n is a rational root of the polynomial P above, so θ /3 is constructible. (b) Next, suppose a + b is even. Since a, b are coprime they are both odd. We can choose notation so that 0 < a < b. We know that an angle of π/4 can be trisected (start with an angle of π/3 and bisect twice) so θ can be trisected if and only if θ + π/4 can be trisected. Now tan θ + π/4 =

tan θ + 1 (b + a)/2 = 1 − tan θ (b − a)/2

Now (b + a)/2 and b − a)/2 are coprime positive integers with one odd and the other even. So we can apply part (a) to deduce that θ can be trisected if and only if 

b+a 2

2

 +

b−a 2

2

is a perfect cube. But this expression equals a2 + b2 2 (c) Observe that 22 + 112 = 125 = 53 and 2 + 11 is odd. Similarly 92 + 132 = 250 = 2 · 53 and 9 + 13 is even. 7.22 (a) T. (b) T. (c) T. (d) F. (e) T. (f) T. (g) T. (h) F. (i) F.

34

8 The Idea Behind Galois Theory

√ √ 8.1 Suppose p(i, −i, 5, − 5) = 0 where p ∈ Q[t1 ,t2 ,t3 ,t4 ] is a polynomial over Q in four indeterminates t j . √ As hinted, S acts as complex √ √conjugation. Since p has real coefficients, and ± 5 are real, we have p(−i, i, 5, − 5) = 0 and S preserves the relation p = 0. √ To √ deal with R, expand p in powers of t3 ,t4 . When we substitute t3 = 5, t4 = − 5, we can collect terms to get √ √ √ p(i, −i, 5, − 5) = a + b 5 √ where√a, b ∈ Q(i). So we may suppose that a + b 5 = 0. We have to prove that a − b 5 also equals 0. That is, a = b = 0. If b = 0 then also a = 0. If b 6= 0 we can solve to get √ 5 = −a/b ∈ Q(i) √ so 5 = c + di with c,√ d ∈ Q. Square to get 5 = c2 − d 2 + 2cdi. Then cd = 0, so either 2 2 5 = c or 5 = d . But 5 is irrational, contradiction. √ Therefore a = b = 0. But √now a − b 5 = 0 and R preserves the relation p = 0. 8.2 Suppose that K ⊆ Q(i, 5) is a subfield. We know ⊇ Q. Choose some √ that K √ element of K, which must be of the form k = p + q 5 + ri + si 5 for p, q, r, s ∈ Q. Since K ⊇ Q, we may assume p = 0. If q = r = s = 0 for all k then K = Q. Otherwise we can find k ∈ Q with q 6=√0 or r 6= 0 or s 6= 0. If r = √ s = 0 for all k ∈ K then K ⊆ Q( 5) and since the degree is 2 we must have K = Q( 5). Similarly √ if q = s = 0 for all k ∈ K then K = Q(i), and if q = r = 0 for all k ∈ K then K = Q(i 5). Therefore we may assume there exists k with at least two of q, r, s 6= 0. Suppose √ that s = 0. Then k = q 5 + ri with q, r 6= 0. Dividing by q we √ √ see that √ K contains 2 = 5 − a2 + 2ai 5, so i 5 ∈ K. Then 5 +√ ai for rational a = 6 0. Now K contains k √ √ √ √ also ( 5 + ai)i 5 = √ i − a 5 ∈ K. Now i −√a 5 + a( 5 + ai) = (1 + a2 )i ∈ K, so i ∈ K. Therefore also 5 ∈ K and K = Q(i, 5). The other two cases are similar and lead to the same result. 8.3 We use the notation sk for the sum of all products of k distinct α, β , γ. (a) α 2 + β 2 + γ 2 = (α + β + γ)2 − 2(αβ + αγ + β γ) = s21 − 2s2 (b) (α + β + γ)(αβ + αγ + β γ) = α 2 β + · · · + 3αβ γ Therefore (α 2 β + · · · ) = s1 s2 − 3s3

8 The Idea Behind Galois Theory

35

(c) (α + β + γ)3 = α 3 + β 2 + γ 3 + 3(α 2 β + · · · ) + 6αβ γ Therefore α 3 + β 3 + γ 3 = s31 − 3(α 2 β + · · · ) − 6s3 = s31 − 3(s1 s2 − 3s3 ) − 6s3 = s31 − 3s1 s2 + 3s3 (d) (α − β )2 + (α − γ)2 + (β − γ)2 = 2(α 2 + β 2 + γ 2 ) − 2s2 = 2(s21 − 2s2 ) − 2s2 = 2s21 − 6s2 8.4 If p contains a term axi y j with i 6= j ∈ N and a ∈ Q, and is symmetric in x, y, then it must also contain the term ax j yi . Therefore we can write p as a sum of terms of the form a(xi y j + x j yi ) or axi yi . Write Ui j = xi y j + x j yi and Vi = xi yi . We claim that all Ui j and Vi can be expressed in terms of x + y and xy. To prove this, let Wi = xi + yi . Observe that xi y j + x j yi = xi yi (x j−i + y j−i ) if i < j xi yi = (xy)i Thus Ui j = ViW j−i if i < j, and Ui j = ViWi− j otherwise. Therefore it suffices to show that the Vi ,Wi can be expressed in terms of x + y and xy. This we do by induction on i. To start the induction, note that V1 = xy and W1 = x + y. Now use the identity xi + yi = (x + y)(xi−1 + yi−1 ) − xy(xi−2 + yi−2 ) or equivalently Wi = (x + y)Wi−1 − xyWi−2 and induction to show that Wi can be expressed in terms of x + y and xy. 8.5* Every symmetric polynomial is a linear combination of ‘symmetrised monomials’, that is, sums of all distinct permutations of a given monomial. So it is enough to prove that every symmetrised monomial can be written as a polynomial in the s j . To build intuition, let n = 4, so that s1 = t1 + t2 + t3 + t4 s2 = t1t2 + t1t3 + t1t4 + t2t3 + t2t4 + t3t4 s3 = t1t2t3 + t1t2t4 + t1t3t4 + t2t3t4 s4 = t1t2t3t4 Consider the monomial t14t22t37t45 , which has the same symmetrisation as p0 = t12t24t35t47 . This has the highest rank among all permutations of p. We work on p0 , which (after symmetrisation) also deals with p.

36 Pull out the largest factor of p0 of the form s4j , which is s24 . So p0 = s24 (t22t33t45 ) Next, pull out the largest factor that is of the form (ti1 ti2 ti3 ) j with distinct im . This is (t2t3t4 )2 . Now p0 = s24 (t2t3t4 )2 (t3t43 ) Then pull out the largest factor that is of the form (ti1 ti2 ) j with distinct im . This is t3t4 . So p0 = s24 (t2t3t4 )2 (t3t4 )(t4 )2 Now build q in the same way, but using corresponding powers of s4 , s3 , s2 , s1 : q = (s24 )(s3 )2 (s2 )(s1 ) The rank of p0 − q is less than that of p0 . Starting with monomials of highest rank, we can successively eliminate their symmetrisations and replace them by polynomials in the s j . By induction on the rank, we can express any symmetrised monomial in terms of the s j , hence any symmetric polynomial in terms of the s j . For a very similar procedure in terms of a ‘lexicographic’ ordering of monomials, see Stewart and Tall (2002) Theorem 1.12. 8.6 In L[t] we can ’long divide’ f (t) by t − α j , obtaining f (t) = (t − α j )g j (t) where g j (t) = a1 + a2 (t + α j ) + · · · + an (t n−1 + α j t n−2 + · · · + α n−1 ) j To verify this claim, observe that (t − α j )g j (t) = a1 (t − α j ) + a2 (t 2 − α 2j ) + · · · + an (t n − α nj ) = f (t) − f (α j ) = f (t) as required. Writing f (t) = ∏ j (t − α j ) and differentiating with respect to t, we see that D f = g1 + · · · + gn (15) Define

λ j = α1j + · · · + αnj

By (15) n

a1 +2a2t +· · ·+nant n−1 =

)] ∑ [a1 +a2 (t +α j )+· · ·+an (t n−1 +α j t n−2 +· · ·+α n−1 j

j=1

Comparing the coefficients of t n−1 ,t n−2 , . . . in turn leads directly to the required

8 The Idea Behind Galois Theory

37

equations, except for the last. The last is a simple consequence of the equations f (α j ) = 0. The first system of equations shows that an−1 λ1 = − an and then

2an an−2 + a2n−1 2an−2 + an−1 λ1 =− an a2n and so on. Inductively this gets us as far as λn . Then the final formula allows us to continue up to any desired λn+k . 8.7 Every element γ ∈ An is the product of an even number of 2-cycles, and by splitting these products into pairs we can obtain γ as a product of elements of the form (a b)(c d), (a b)(a c), or (a b)(b a), where a, b, c, d are distinct. But (a b)(b a) = 1, (a b)(a c) = (a c b), and (a b)(c d) = (a b c)(b c d). So γ is a product of 3-cycles. 8.8 Every non-identity element of A5 is either a 5-cycle, a 3-cycle, or a product of two disjoint 2-cycles. But if a, b, c, d, e are distinct we have (a b)(c d) = (a e d c b)(a c b d e) and (a b c) = (a c e b d)(a e c b d). The identity is (1 2 3 4 5)(1 5 4 3 2), and any single 5-cycle (a b c d e) can be written as (a d b e c)(a d b e c). Therefore every element of A5 is a product of two 5-cycles. From this we prove that A5 is simple. Suppose that N / A5 . First we claim that N contains some 5-cycle. If not, every element is either a 3-cycle or a product of two disjoint 2-cycles. But any 3-cycle is conjugate in A5 to any other 3-cycle, so N contains all 3-cycles. But (1 2 3)(3 4 5) = (1 2 3 4 5) so A5 contains a 5-cycle. The 5-cycles split into two conjugacy classes in A5 . On contains (1 2 3 4 5), the other contains (2 1 3 4 5). But a similar argument to that for 3-cycles shows that N contains all products of two disjoint 2-cycles. In particular (2 1 3 4 5) = (2 3 5).(1 2)(3 5) ∈ N. Therefore N contains all 5-cycles. Since these generate A5 , we have N = A5 . 8.9 This is not a new calculation. It is ‘completing the square’ in disguise. Without loss of generality we divide by the leading coefficient and work with a monic polynomial. Suppose that f (t) = t 2 + at + b = 0 and that f (t) = (t − α1 )(t − α2 ). Then a = −(α1 + α2 ) and b = α1 α2 . We have (α1 − α2 )2 = α12 + α22 − 2α1 α2 = (α1 + α2 )2 √ − 4α1 α2 = a2 − 4b. Since this is a rational function of a, b, the expression α1 − α2 = a2 − 4b is a Ruffini radical. Therefore p α1 − α2 = a2 − 4b λ2 = −

α1 + α2 = −a so

√ √ −a − a2 − 4b −a + a2 − 4b α2 = 2 2 8.10 Let f (t) = t 3 + at 2 + bt + c = (t − α1 )(t − α2 )(t − α3 ). Consider the expressions α1 =

u1 = α1 + α2 + α3 u2 = α1 + ωα2 + ω 2 α3 u3 = α1 + ω 2 α2 + ωα3

38 We know that u1 = −a. By direct calculation u32 = α13 + α23 + α33 + 6α1 α2 α3 +3ω(α12 α2 + α22 α3 + α32 α1 ) + 3ω 2 (α22 α1 + α32 α2 + α12 α3 ) = S + T + ω 2 X + ωY say, where S, T are symmetric functions of the α j , hence rational functions of the coefficients of f . Similarly, u33 = α13 + α23 + α33 + 6α1 α2 α3 +3ω 2 (α12 α2 + α22 α3 + α32 α1 ) + 3ω(α22 α1 + α32 α2 + α12 α3 ) = S + T + ωX + ω 2Y Clearly X + Y is symmetric in the α j , and so is (X − Y )2 . Thus X,Y are Ruffini radicals and can be expressed in terms of a, b, c. We can then express u1 , u2 , u3 as Ruffini radicals. Finally, we solve the system of three linear equations to obtain the α j in terms of the uk . We omit the explicit formulas, which are equivalent to those in the text. 8.11 Suppose that α j,1 , . . . , α j,k( j) is a radical sequence for J j /I. Concatenate the first s of these sequences to get α1,1 , . . . , α1,k(1) , α2,1 , . . . , α2,k(2) , . . . , αs,1 , . . . , αs,k(s) Inductively, it is easy to check that this is a radical sequence for I(J1 , . . . , Js )/I. Setting s = r, and noting that I(J1 , . . . , Jr ) = J, the result follows. 8.12 The discriminant ∆ is 256a(a − 1)2 . One way to do this is to find the four zeros expliciltly and use the formula (8.7) for ∆. 8.13 (a) F. (b) T. (c) T. (d) F. (e) F. (f) F. (g) F. (h) F.

9 Normality and Separability 9.1 t 3 − 1 = (t − 1)(t 2 + t + 1) and the zeros of the quadratic are

√ −1±i 3 . 2

√ Therefore the splitting field is Q(i 3).

t 4 + 5t 2 + 6 = (t 2 + 2)(t 2 + 3) √ √ so the splitting field is Q(i 2, i 3). t 6 − 8 = (t 2 )3 − 23 √ so the splitting field is Q( √ 2, ζ ) where ζ is a primitive 6th root of unity. For a simpler formula, replace ζ by i 3.

9 Normality and Separability

39

9.2 Respectively, the degrees are 2, 4, 4. 9.3 Let f = fnt n + · · · + f0 and g = gnt n + · · · + g0 , adding leading terms of 0 if necessary. Then (a) D( f + g) = D(( fn + gn )t n + · · · + ( f0 + g0 )) = n( fn + gn )t n−1 + · · · + ( f1 + g1 ) = (n fnt n−1 + · · · + f1 ) + (ngnt n−1 + · · · + g1 ) = D f + Dg (c) It is convenient to take this next. It follows directly from the definition of D. (b) The crucial point is that if f = t m , g = t n , then f g = t m+n . Now (D f )g + f (Dg) = mtm−1t n + nt mt n−1 = (m + n)t m+n−1 = D( f g). Now use induction and property (a) to extend this relation to arbitrary f , g. 9.4 The stated formula comes from the identity gf g = f . Taking the formal derivative we require f f D( )g + Dg = D f g g Now solve for D( gf ). We verify that D( gf + hk ) = D( gf ) + D( hk ). The corresponding calculation for the product is similar. We have f h f k + gh + = g k hk so from the definition f k + gh f h ) D( + ) = D( g k hk D( f k + gh) · hk − ( f k + gh)D(hk) = h2 k 2 [D f · k + f Dk + Dg · h + gDh]hk − ( f k + gh)[Dh · k + hDk] = h2 k2 Similarly we calculate D( f k + gh) · hk − ( f k + gh) · D(hk) h2 k2 and routine manipulations show that the two expressions are equal. 9.5 (a) Normal. (b) Normal. (c) Not normal since α is real but other zeros are not. (d) Normal. (e) Same as C/R so normal. 9.6 Let K ⊆ C be a subfield and let L/K be an extension of degree 2. Then L = K(α) where the minimal polynomial of α over K has degree 2. Suppose that it is t 2 + at + b where a, b, ∈ K. Then α + β = −a, so β = a + α. Since a ∈ K, we have K(β ) = K(a + α) = K(α). A similar result does not hold for any other degree. Indeed, for d ≥ 3 let α be the

40 unique positive real dth root of 2. Then Q(α) ⊆ R. But the minimal polynomial of α is t d − 2, with roots ζ k α where ζ is a primitive dth root of unity. So the splitting field also contains ζ . But ζ 6∈ R when d ≥ 3. 9.7 The complex zeros of f are the same whether we consider the coefficients to lie in K or in L. Those zeros generate the splitting field. 9.8 Let ∂ f = n and let the zeros of f in C be α1 , . . . , αn . Argue by induction on n. The result is clearly true when n = 1. Observe that (a + b)! is divisible by a!b!. The reason is that the binomial coefficient   (a + b)! a+b = a a!b! is an integer. If f is reducible, say f = gh, then the zeros of f are those of g (say α1 , . . . , αa ) together with those of h (namely αa+1 , . . . , αn ). We have ∂ g = a, ∂ h = b = n − a. By induction, [K(α1 , . . . , αa ) : K] divides a! and [K(αa+1 , . . . , αn ) : K] divides b!. The Short Tower Law now implies that [Σ : K] divides a!b!, which divides (a + b)! = n!. Otherwise f is irreducible, so [K(αn ) : K] = n. Dividing f by (t − αn ) we get g, of degree n − 1. Inductively, [Σ : K(αn )] divides (n − 1)!. Now the Short Tower Law implies that [Σ : K] divides n(n − 1)! = n!. 9.9 (a) T. (b) T. (c) F. (d) T. (e) F. (f) T. (g) F.

10 Counting Principles 10.1 For motivation, consider the case n = 3. Let K = C(s1 , s2 , s3 )

L = C(t1 ,t2 ,t3 )

and let f (t) = (t − t1 )(t − t2 )(t − t3 ) ∈ K[t]. Consider the tower K(t1 ,t2 ,t3 )/K(t1 ,t2 )/K(t1 )/K The polynomial f is irreducible over K, so [K(t1 ) : K] = 3 Since t1 + t2 + t3 = s1 ∈ K, we also have [K(t1 ,t2 ,t3 ) : K(t1 ,t2 )] = 1 What about [K(t1 ,t2 ) : K(t1 )]? Certainly t2 is a zero of a quadratic polynomial g(t) = (t −t2 )(t −t3 ) over K(t1 ), so this degree is at most 2. It is equal to 2 if and only if g(t) is irreducible over K(t1 ). But if g(t) is reducible over K(t1 ) then either t2 ∈ K(t1 ) or t3 ∈ K(t1 ). But these statements contradict the t j being independent transcendentals, so g(t) is irreducible over K(t1 ).

10 Counting Principles

41

Applying the Tower Law, [K(t1 ,t2 ,t3 ) : K] = 1 · 2 · 3 = 3! as required. Now consider the general case, which we prove by induction. Let Lr = K(t1 , . . .tr ) so that L0 = K, Ln = L, and L0 ⊆ L1 ⊆ · · · ⊆ Ln . We claim that [Lr+1 : Lr ] = n − r for all r. We have Lr+1 = Lr (tr+1 ), and tr+1 is a zero over Lr of the polynomial g(t) =

f (t) (t − t1 ) · · · (t − tr )

which lies in Lr [t] and has degree n − r. So certainly [Lr+1 : Lr ] ≤ n − r. To prove equality, it suffices to prove g(t) irreducible over Lr . If not, some divisor lies in Lr [t], and this divisor must be of the form h(t) = (t − t j1 ) · · · (t − t jl ) for some subset { j1 , . . . , jl } ⊆ {r + 1 . . . , n}. Then the coefficient t j1 + · · · + t jl must lie in Lr , contrary to the t j being independent transcendentals. 10.2 Elements of the fixed field F of α1 , α4 satisfy p + qζ + rζ 2 + sζ 3 + tζ 4 = p + tζ + sζ 2 + rζ 3 + qζ 4 so we need t = q, r = s. (The relation 1 + ζ + ζ 2 + ζ 3 + ζ 4 = 0 does not change these conditions because they are automatically satisfied for that expression.) Therefore the fixed field is spanned over Q by ζ + ζ 4 and ζ 2 + ζ 3 . But (ζ + ζ 4 )2 = ζ 2 + ζ 3 + 2 so F = Q(ζ + ζ 4 ). Clearly ζ + ζ 4 6∈ Q since the only linear relation between powers of ζ is 1 + ζ + ζ 2 + ζ 3 + ζ 4 = 0, given by the minimum polynomial. It follows that [Q(ζ + ζ 4 ) : Q] = 2, with basis {1, ζ + ζ 4 }. This is the order of the group α1 , α4 , so Theorem 10.5 is verified for this example. 10.3 The number 4 changes to 6 and there are now six Q-automorphisms, which map p + qζ + rζ 2 + sζ 3 + tζ 4 + uζ 5 + vζ 6 to

p + qζ + rζ 2 + sζ 3 + tζ 4 + uζ 5 + vζ 6 p + tζ + qζ 2 + uζ 3 + rζ 4 + vζ 5 + sζ 6 p + uζ + sζ 2 + qζ 3 + vζ 4 + tζ 5 + rζ 6 p + rζ + tζ 2 + vζ 3 + qζ 4 + sζ 5 + uζ 6 p + sζ + vζ 2 + rζ 3 + uζ 4 + qζ 5 + tζ 6 p + vζ + uζ 2 + tζ 3 + sζ 4 + rζ 5 + qζ 6

42 bearing in mind the relation 1 + ζ + ζ 2 + ζ 3 + ζ 4 + ζ 5 + ζ 6 = 0. The calculations are similar to Exercise 10.2 but slightly more complicated. 10.4 Since 0, 1 must map to 0, 1 respectively, there is only one such monomorphism: the inclusion map. 10.5 Expanding, the determinant is 1.ω.ω + 1.ω 2 .1 + 1.1.ω 2 − 1.ω.1 − 1.1.ω − 1.ω 2 ω 2 = 3ω 2 − 3ω = 3(−1 − ω) − 3ω = −3 − 6ω 6= 0 10.6 The Q-automorphisms are: α1 α2 α3 α4

: p + qζ + rζ 2 + sζ 3 + tζ 4 : : :

7→ 7→ 7→ 7→

p + qζ + rζ 2 + sζ 3 + tζ 4 p + sζ + qζ 2 + tζ 3 + rζ 4 p + rζ + tζ 2 + qζ 3 + sζ 4 p + tζ + sζ 2 + rζ 3 + qζ 4

Suppose that β = aα1 + bα2 + cα3 + dα4 = 0. Apply β to the elements 1, ζ , ζ 2 , ζ 3 ∈ Q(ζ ). That is, successively set p, q, r, s = 1 and all other coefficients to 0. We get: 0 = a+b+c+d 0 = aζ + bζ 2 + cζ 3 + dζ 4 0 = aζ 2 + bζ 4 + cζ + dζ 3 0 = aζ 3 + bζ + cζ 4 + dζ 2 The determinant 1 ζ D = 2 ζ ζ3

1 ζ2 ζ4 ζ

1 ζ3 ζ ζ4

1 ζ 4 = −5ζ + 5ζ 2 + 5ζ 3 − 5ζ 4 ζ 3 ζ2

which is nonzero, so a = b = c = d = 0. 10.7 Let ζ = e2πi/6 be a primitive 6th root of unity. The image of ζ under a Qautomorphism is either ζ or ζ −1 = ζ 5 , so there are only two Q-automorphisms determined by: α : ζ 7→ ζ α : ζ 7→ ζ −1 Suppose that aα + bβ = 0. Apply this to 1 and ζ to get 0 = a+b 0 = aζ + bζ −1

11 Field Automorphisms

43

Then b = −a and b = −aζ 2 , so a = b = 0. √ Alternatively, observe that ζ = 12 + i 23 , so [Q(ζ ) : Q] = 2. Therefore Q(ζ ) = √ √ √ Q(i 3) and Q-automorphisms are determined by mapping i 3 to ±i 3. 10.8* Let ζ = e2πi/n be a primitive nth root of unity. The Q-automorphisms of Q(ζ ) are determined by ζ 7→ ζ m where 1 ≤ m < n and m is prime to n, because these are the primitive nth roots of unity. The Q-automorphisms are linearly independent over Q. 10.9 (a) T. (b) F. (c) T. (d) F. (e) T. (f) T.

11 Field Automorphisms 11.1 Let φ : L → L be a K-monomorphism. Then φ is a linear map from L to L as a vector space over K , and φ is injective. Therefore, if the image of φ is M, the dimension of M over K equals that of L over K. Since the dimension of L over K is finite, we must have M = L so φ is surjective. Therefore φ is a K-automorphism of L. This result fails when the degree [L : K] is infinite. For example, let L = C(t) and define φ : L → L by φ ( f (t) = f (t 2 ). Now φ is clearly a C-monomorphism L → L, but is not surjective, because the polynomial g(t) = t is not in the image of φ . So φ is not a K-automorphism of L. 11.2 (a) Q(α, ζ )/Q where ζ is a primitive fifth root of unity. (the zeros of the minimal polynomial of α are ζ r α, where 0 ≤ r ≤ 4.) (b) Q(β )/Q where ζ is a primitive seventh root of unity. √, ζ √ (c) Q( 2, √3)/Q since the extension is already normal. (d) Q(α, ζ , 2)/Q where ζ is a primitive cube root of unity. (e) Observe that if α is a zero of t 3 − 3t 2 + 3 then so is H(α) = 2α−3 α−1 . This is not obvious, but one way to guess is to solve the equation numerically and experiment. To prove it, observe that if y = 2x−3 x−1 then y3 − 3y2 + 3 = −

x3 − 3x2 + 3 x3 − 1

(expand and simplify). Repeating the transformation, we find that H(H(α)) = α−3 α−2 . In fact, if β is any zero then the others are H(β ) and H(H(β )). It is easy to show that these are distinct: any equality implies that α is a zero of a quadratic over Q, but t 3 − 3t 2 + 3 is irreducible by Eisenstein. It follows that for any zero β , the extension Q(β )/Q is normal. In particular, this is the required normal closure. 11.3 (a) The Galois group is trivial since α must map to α, because Q(α) ⊆ R and α is the only real fifth root of 3. (b) The Galois group is trivial since β must map to β , because Q(β ) ⊆ R and β is the only real seventh root of 2.

44 √ √ √ √ (c) The Q-automorphisms are generated by 2 7→ ± 2 and 3 7→ ± 3 with independent choices of sign. Therefore the Galois√group is√isomorphic to Z2 × Z2 . (d) The Q-automorphisms √ are generated by 2 7→ ± 2 and α 7→ α, again because α is real (and so is 2). Therefore the Galois group is isomorphic to Z2 . (e) The extension is normal, so the Galois group of t 3 − 3t 2 + 3 is cyclic of order 3, generated by H as in Exercise 1.2. 11.4 (a) The Q-automorphisms are generated by α 7→ ζ r α, where 0 ≤ r ≤ 4. Therefore the Galois group is isomorphic to Z5 . (a) The Q-automorphisms are generated by β 7→ ζ r β , where 0 ≤ r ≤ 6. Therefore the Galois group is isomorphic to Z7 . (c) The extension is normal, so again the Galois is isomorphic to Z2 × Z2 . √ group√ (d) The Q-automorphisms are generated by 2 7→ ± 2 and α 7→ ω r α, where ω is a primitive cube root of unity and 0 ≤ r ≤ 2. Therefore the Galois group is isomorphic to Z2 × Z3 , which is isomorphic to Z6 . (e) The extension is normal, so again the Galois group of t 3 − 3t 2 + 3 is cyclic of order 3, generated by H as in Exercise 1.2. √ 11.5 Consider the non-normal extension Q(α)/Q where α = 3 2 ∈ R. Then in particular Q(α) ⊆ R. The minimal polynomial of α over Q is t 3 − 2 with zeros α, ωα, ω 2 α. By Theorem 5.16 there is a Q-monomorphism τ : Q(α) → C such that τ(α) = ωα. Since τ(α) 6∈ R ⊇ Q(α) we must have τ(Q(α)) 6= Q(α). So Lemma 11.8 fails for non-normal extensions. The proof of Lemma 11.8 when we assume that N/K is normal. √ unchanged √ √works 11.6 We first prove that [Q( 3, 5, 7) : Q] = 8. We know by the tower law that √ √ √ [Q( 3, 5, 7) : Q] √ √ √ √ √ √ √ √ √ = [Q( 3, 5, 7) : Q( 3, 5)][Q( 3, 5) : Q( 3)][Q( 3) : Q] As in Example 6.8, we √ prove that each factor equals 2. (a) Certainly [Q( √ √ √ 3) : Q] = √ 2. √ √ √ (b)If 5 6∈ Q( 3 then [Q( 3, 5) : Q( 3)] = 2. So suppose that 5 ∈ Q( 3), implying that √ √ 5 = p + q 3 p, q ∈ Q Squaring,

√ 5 = (p2 + 3q2 ) + 2pq 3

so p2 + 3q2 = 5 3q2

pq = 0

2 If p = 0 then = 5, which is impossible by Exercise If q = 0 then √ 1.3. √ √ p √= 5, which √ is impossible for the same reason. Therefore 5 6∈ Q( 3), so [Q( 3, 5) : Q( 3)] = 2. √ √ √ (c) FInally we claim that 7 6∈ Q( 3, 5). Suppose to the contrary that √ √ √ √ 7 = p + q 3 + r 5 + s 15 p, q, r, s ∈ Q

45

12 The Galois Correspondence Squaring:

√ √ √ 7 = p2 + 3q2 + 5r2 + 15s2 + (2pq + 10rs) 3 + (2pr + 6qs) 5 + (2ps + 2qr) 15 whence

p2 + 3q2 + 5r2 + 15s2 pq + 5rs pr + 3qs ps + qr

= = = =

7 0 0 0

(16)

Observe that if (p, q, r, s) satisfies (16) then so do (p, q, −r, −s), (p, −q, r, −s), and (p, −q, −r, s). Therefore √ √ √ √ p + q 3 + r 5 + s 15 = 7 √ √ √ √ p + q 3 − r 5 − s 15 = ± 7 √ √ √ √ p − q 3 + r 5 − s 15 = ± 7 √ √ √ √ p − q 3 − r 5 + s 15 = ± 7 √ √ √ Adding the first two equations we get p + q √3 = 0 or p + = 7, each implying √q 3√ that p = q =√ 0. Adding √ the first and third, r 5 = 0 or r 5 = 7, so r = 0, Finally, s = 0 since s 15 = 7√ is impossible by Exercise 1.3. √ √ We deduce that [Q( 3, 5, 7) : Q] = 8. The extension is normal, so the order of the Galois group is also 8. (It is, of course, isomorphic to Z2 × Z2 × Z2 , generated by sign changes on each of the three square roots.) 11.7 (a) F. (b) T. (c) T. (d) T. (e) T. (f) F. (g) T. (h) F. (i) F.

12 The Galois Correspondence 12.1 Let

√ √ x = p + qi + r 5 + si 5

Then √ √ I(x) = p + qi + r 5 + si 5 √ √ R(x) = p − qi + r 5 − si 5 √ √ S(x) = p + qi − r 5 − si 5 √ √ T (x) = p − qi − r 5 + si 5 We know that R2 = S2 = T 2 = I, RS = T, RT = S, ST = R, and all of I, R, S, T commute. The subgroups of G = {I, R, S, T } are {I}

{I, R}

{I, S}

{I, T }

{I, R, S, T }

46 and the corresponding fixed fields are √ √ Q(i, 5) Q( 5)

Q(i)

√ Q(i 5)

Q

12.2 We have a sequence of fields K⊆N⊆M⊆L and the corresponding subgroups of the Galois group are 1 = L∗ ⊆ M ∗ ⊆ N ∗ ⊆ K ∗ where X ∗ = Γ[L : X]. Suppose that M/N is normal. Let x ∈ N, γ ∈ N ∗ , δ ∈ M ∗ . Then γ(x) = x. Since M/N is normal, Lemma 1.8 implies that γ(δ (x)) = δ (x). Now (δ −1 γδ )(x) = δ −1 (γ(δ (x))) = δ −1 (δ (x)) = x so δ −1 γδ ∈ N ∗ . Therefore N ∗ is a normal subgroup of M ∗ . The converse is essentially the same argument in reverse. Suppose that N ∗ / M ∗ . Again, let x ∈ N, γ ∈ N ∗ , δ ∈ M ∗ . We have (δ −1 γδ )(x) = x so γ(δ (x)) = δ (x) so δ (x) ∈ N ∗†

= N. By the converse of Lemma 11.8 (which follows because all zeros of the minimal polynomial of x are images of x under suitable monomorphisms) the extension M/N/ is normal. Finally, restriction to N determines a homomorphism M ∗ → Γ(N/M) with kernel ∗ N . p √ √ 12.3* Let γ = 2 + 2. Then γ 2 = 2 + 2, so (γ 2 − 2)2 = 2. Therefore γ is a zero of the polynomial p(t) = t 4 − 2t 2 + 2. This is irreducible over Q by Eisenstein’s criterion, so it is the minimal polynomial of γ. The zeros of p are q √ α = 2− 2 q √ β = − 2− 2 q √ γ = 2+ 2 q √ δ = − 2+ 2 Compute q αγ =

√ 2 √ 22 − 2 = 2 = 2 − α 2

so that γ=

2 −α α

12 The Galois Correspondence

47

Similarly 2 −γ γ 2 δ = −β β 2 α = −δ δ β =

Thus each zero is a rational function (over Q) of any other zero. Therefore the effect of any Q-automorphism φ on any element of the splitting field Σ of p is uniquely determined by its effect on α (or on β or γ or δ , for that matter). Define ψ to be the unique Q-automorphism of Σ such that ψ(α) = γ. Then ψ(γ) = ψ( α2 − α) = 2γ − γ = β . Similarly ψ(β ) = δ and ψ(δ ) = α. So ψ 4 = id and ψ generates a cyclic group Z4 . The powers ψ 0 , ψ 1 , ψ 2 , ψ 3 map α respectively to α, β , γ, δ . Therefore the powers of ψ √ exhaust the Galois group, which must be Z4 . Finally, observe that (one value of) 4 i is p r √ √ ! √ 2+ 2 1 2+ 2 +i (2 + 2) − 2 2 2 The rest is easy. √ 12.4* Let α be a positive real sixth root of 7, so α = 6 7. Let ζ = e2πi/6 be a primitive sixth root of unity. Then the polynomial f (t) = t 6 − 7 is irreducible over Q (Eisenstein) and its zeros are ζ k α for 0 ≤ k ≤ 5. The splitting field for f is Σ = Q(ζ , α). The minimal polynomial of ζ over Q is g(t) = t 2 − t + 1. We claim that f (t) is irreducible over Q(ζ ) (and not just over Q). This is a consequence of the factorisation f (t) = (t − α)(t − ζ α)(t − ζ 2 α)(t − ζ 3 α)(t − ζ 4 α)(t − ζ 5 α) over C. If f is reducible, there must be a nontrivial factor g of degree 1, 2, or 3. Consider the degree 3 case, and suppose that g(t) = (t − ζ p α)(t − ζ q α)(t − ζ r α) ∈ Q(ζ ) √ If so, the constant term ζ p+q+r α 3 ∈ Q(ζ ), so α 3 ∈ Q(ζ ). But α 3 = 7 ∈ R. Therefore √ 7 = a + bζ + cζ 2 + dζ 3 + eζ 4 + f ζ 5 a, b, c, d, e, f ∈ Q Take the complex conjugate, which maps ζ k to ζ −k : √ 7 = a + f ζ + eζ 2 + dζ 3 + cζ 4 + bζ 5 a, b, c, d, e, f ∈ Q so b = f , c = e.

48 However, ζ = ζ2 = ζ3 = ζ4 = ζ5 =

so

√ 1+i 3 2 √ −1 + i 3 2 −1 √ −1 − i 3 2√ 1−i 3 2

√ 7 = a−d +b−c ∈ Q

√ 7 is irrational. The other cases are similar, since all powers α k are real and irrational for 1 ≤ k ≤ 5. Therefore f (t) is irreducible over Q(ζ )as claimed. The Tower Law now implies that Σ has degree 2 · 6 = 12 over Q. Define Qautomorphisms φ jk by but

φ jk (ζ ) = ζ j

φ jk (α) = ζ k α

where j = ±1 and 0 ≤ k ≤ 5. These extend uniquely to Q-automorphisms of Σ by Theorem 11.3. The Galois group of f has order 12, and contains the 12 distinct φ jk , so it must equal the set of all φ jk . Let ψ = φ11 and δ = φ−1,0 . Easy computations show that ψ 6 = 1, δ 2 = 1, and δ ψδ = ψ −1 . Therefore the Galois group is generated by ψ, δ and is isomorphic to the dihedral group D6 . 12.5* The polynomial t 6 − 2t 3 − 1 has integer coefficients and is irreducible modulo 5 (prove this using a case by case analysis, showing that there are no linear, quadratic, or cubic factors). Therefore the Galois group Γ permutes the six zeros of f . To find the zeros define u = t 3 . Then f becomes u2 − 2u − 1, whose zeros are √ √ θ = 1+ 2 φ = 1− 2 Let α=

√ 3 θ

β=

p 3

φ

choosing the unique real cube roots. Let ω be a primitive cube root of unity. The six zeros of f are then α, ωα, ω 2 α,√ β , ωβ , ω 2 β . p 3 Observe that αβ = θ φ = 3 −1 = −1. Therefore β = −1/α. The splitting field Σ of f is therefore Q(α, ω) and the Q-automorphisms of Σ are defined by their effect on α, ω. Irreducibility of f shows that [Q(α) : Q] = 6. Since α ∈ R we deduce that

12 The Galois Correspondence

49

ω 6∈ Q(α). Therefore [Q(α, ω) : Q(α)] = 2. By the Tower Law, [Q(α, ω) : Q] = 12. In particular, |Γ| = 12. Under a Q-automorphism, there are 6 possible images for α and 2 for ω, so all 12 combinations must occur and they all define Q-automorphisms. Define three Q-automorphisms ρ, σ , τ by ρ(α) = ωα

ρ(ω) = ω

σ (α) = −1/α

σ (ω) = ω

τ(α) = α

τ(ω) = ω 2

Then all the elements of Γ can be written as ρ r σ s τ t where 0 ≤ r ≤ 2, 0 ≤ s ≤ 1, 0 ≤ t ≤ 1. Simple computations show that ρ 3 = σ 2 = τ 2 = 1, σ ρσ = ρ, τρτ = ρ 2 , σ τ = τσ . Therefore ρ, τ generate a group isomorphic to D3 . The element σ has order 2, does not lie in D3 , and commutes with both ρ and τ. Therefore Γ ∼ = D3 × Z2 . This is isomorphic to D6 . 12.6 Let ζ be a primitive 12th root of unity. (a) Factorise t 12 − 1: t 12 − 1 = (t − 1)(t + 1)(t 2 + 1)(t 2 − t + 1)(t 2 + t − 1)(t 4 − t 2 + 1) All factors except the last correspond to primitive dth roots of unity where d is a proper divisor of 12. So ζ is a zero of g(t) = t 4 − t 2 + 1 In particular, ζ 4 = ζ 2 − 1. The zeros of g(t) are the other primitive 12th roots of unity, namely the powers ζ j where j is prime to 12. These are therefore ζ , ζ 5 , ζ 7 , ζ 11 . (b) To prove that g(t) is irreducible over Q, observe that it has no linear factors over Q because ζ , ζ 5 , ζ 7 , ζ 11 6∈ R, hence 6∈ Q. So the only possibility is that it is a product of two quadratics over Q. Since the roots of a real quadratic are complex conjugates, these factors must be (t − ζ )(t − ζ 11 )

(t − ζ 5 )(t − ζ 7 )

We claim that h(t) = (t − ζ )(t − ζ 11 ) 6∈ Q(t). Expanding, h(t) = t 2 − 2 cos

π +1 6

√ But cos π6 = 3/2, which is irrational: contradiction. (c) Let Γ = Γ(Q(ζ )/Q). The elements of Γ maps zeros of g(t) to zeros of g(t), so they are of the form φ j : ζ 7→ ζ j j = 1, 5, 7, 11 Since g(t) is irreducible over Q, each choice determines a Q-automorphism by Proposition 11.4. Therefore Γ = {φ1 , φ5 , φ7 , φ11 }

50 (d) φ1 is the identity. (φ 5 )2 maps ζ to (ζ 5 )5 = ζ 25 = ζ , So φ5 has order 2. Similarly (φ 7 )2 maps ζ to (ζ 7 )7 = ζ 49 = ζ and (φ 11 )2 maps ζ to (ζ 11 )11 = ζ 121 = ζ , so these elements also have order 2. Therefore Γ ∼ = Z2 × Z2 . 12.7 The nontrivial proper subgroups of Z2 × Z2 are: G1 = {φ1 , φ5 }

G2 = {φ1 , φ7 }

G3 = {φ1 , φ11 }

and each is isomorphic to Z2 . We find the fixed fields G†j for j = 1, 2, 3. Because 12 is not prime, the powers 1, ζ , . . . , ζ 11 are not linearly independent over Q. For instance, ζ 6 = −1. We also have ζ 3 = i, ζ 9 = −i. Moreover, ζ 4 is a primitive cube root of unity and ζ 2 is a primitive sixth root of unity. A basis for Q(ζ ) over Q is {1, ζ , ζ 2 , ζ 3 }. We first compute G†1 , which is the fixed field of φ5 . The effect of φ5 on the basis is: 1 7→ 1 ζ 7→ ζ 5 = ζ ζ 4 = ζ (ζ 2 − 1) = ζ 3 − ζ ζ 2 7→ ζ 10 = −ζ 4 = 1 − ζ 2 ζ 3 7→ ζ 15 = ζ 3 We see that 1, ζ 3 are fixed. Since [Q(ζ 3 ) : Q] = 2, the Galois correspondence implies that G†1 = Q(ζ 3 ). We can check this directly. We have φ5 (a + bζ + cζ 2 + dζ 3 ) = a + b(ζ 3 − ζ ) + c(1 − ζ 2 ) + dζ 3 so a + bζ + cζ 2 + dζ 3 is fixed if and only if a = a+c

b = −b

c = −c

d = b+d

so b = c = 0 and the element is a + dζ 3 . Similarly, G†2 = Q(ζ 2 ). For G†3 , no power of ζ (except ζ 12k = 1 for any k) is fixed by φ11 . However, 11 ζ = ζ −1 = ζ . Therefore G†3 = Q(ζ ) ∩ R. A nonzero element is ζ + ζ −1 and this generates the fixed field, so G†3 = Q(ζ + ζ −1 ). 12.8 (a) T. (b) F. (c) T. (d) F. (e) F.

13 A Worked Example 13.1 Denote the appropriate Galois group √ √by Γ. (a) By the usual methods [Q( 2, √5) : Q]√= 4. Any Q-automorphism of √ √ √ √ Q( 2, 5) must map 2 to ± 2 and 5 to ± 5. There are four combinations

13 A Worked Example

51

of signs, and since |Γ| = 4 all four combinations yield Q-automorphisms. The Galois group is therefore generated by σ , τ, where √ √ √ √ σ ( 2) = − 2 σ ( 5) = 5 √ √ √ √ τ( 2) = 2 τ( 5) = − 5 Since σ 2 = τ 2 = 1 and σ τ = τσ , we must have Γ ∼ = Z2 × Z2 . (b) The minimal polynomial of α is t 2 + t + 1 which is quadratic. The complex zeros are α, α 2 (with α 3 = 1). There are two elements in Γ: the identity and σ , where σ (α) = α 2 . Then σ 2 (α) = α 4 = α, so σ 2 = 1 and Γ ∼ = Z2 . (c) Let f (t) = t 4 − 3t 2 + 4. Trying factors modulo p, for various small p, fails to prove f irreducible. However, it is irreducible, for the following reason. If it is reducible over Q then it is reducible over Z by Gauss’s Lemma. Since it is monic, any factors over Z can be assumed monic. Any linear factor must be of the form t + a where a|4, and a direct check shows that none of these occurs. The only remaining possibility is that f (t) = (t 2 + at + b)(t 2 + ct + d) for integers a, b, c, d. Expanding: a+c = 0 ad + bc = 0 b + d + ac = −3 bd = 4 The last of these equations implies that b = ±1, ±2, ±3 with d = 4/b. We can now use the first two equations to solve for a, c and check that b + d + ac is never equal to −3. To find the zeros of f , let u = t 2 . Then u2 − 3u + 4 = 0, so √ 3±i 7 u= 2 and the four zeros of f (t) are s t =± with independent q ± signs. Let α =

√ 3+i 7 2

√ 3+i 7 2

(taking

√ 3±i 7 2

√ 7 > 0 and choosing one of the two possible square

roots of in C). Then the zeros of f are α, −α, α, −α, where the bar indicates the complex conjugate. q 2

The product αα is equal to some value of 3 4+7 = ±2. Since αα is real and positive, αα = 2. Therefore α = 2/α so the splitting field for f over Q is Q(α). Since f is irreducible, [Q(α) : Q] = 4 and Γ = 4.

52 There are Q-automorphisms mapping α to any of α, −α, α. Since these give four possibilities, they exhaust Γ. The Q-automorphism determine by α 7→ α is the identity. The other three Q-automorphisms all have order 2. Therefore Γ ∼ = Z2 × Z2 . 13.2 (a) The subgroups of Γ are 1, Γ, plus three cyclic subgroups of order 2 generated respectively by σ , τ, and σ τ, which we denote by hσ i, hτi, hσ τi. (b) The subgroups of Γ are 1, Γ. (c) If we define σ (α) = −α and τ(α) = α then the subgroups of Γ are 1, Γ, plus three cyclic subgroups of order 2, namely hσ i, hτi, hσ τi. 13.3 The fixed fields are: (a) √ √ 1 : Q( 2, 5) Γ :Q

√ hσ i : Q( 5) √ hτi : Q( 2) √ hσ τi : Q( 10) (b) 1 : Q(α) Γ: Q (c) 1 : Q(α) Γ :Q hσ i : Q(α 2 ) hτi : Q(α + α) hσ τi : Q(α − α) 13.4 In all three cases the group is abelian, so all subgroups are normal. 13.5 In all three cases the corresponding extensions are either Σ/Q where Σ is the splitting field, or they are of degree 2 and can be written as Q(β )/Q where β 2 ∈ Q. Any such extension is easily seen to be normal. 13.6 The only nontrivial cases are those of the form Q(β )/Q where β 2 ∈ Q, which arise only in cases (a, c). The Galois group of such an extension is isomorphic to Z2 . In both cases the corresponding quotient group is Z2 × Z2 /Z2 , which has order 2 and thus must be isomorphic to Z2 . 13.7*√It is easy to prove (see Lemma 15.6) that the splitting field Σ is generated by α = 6 7 ∈ R together with a primitive sixth root of unity ζ . That is, Σ = Q(α, ζ ) The degree is 12 and the Galois group is Γ ∼ = D6 , generated by ψ, δ where ψ 6 = 1 =

13 A Worked Example

53

δ 2 and δ ψδ = ψ −1 . Explicitly, ψ(α) = ζ α δ (α) = α

ψ(ζ ) = ζ δ (ζ ) = ζ −1

To find the intermediate fields, we classify all subgroups of D6 and apply the Galois correspondence. Let hαi = A, a cyclic subgroup of order 6 that is normal in D6 . There are three classes of subgroups of Γ: (a) Subgroups contained in A. (b) Subgroups that intersect A trivially. (c) Subgroups not of types (a, b). ∼ Z6 , so must be Zd where d|6, so The type (a) subgroups are subgroups of A = d = 1, 2, 3, 6. Every type (b) subgroup has order 2, so much be generated by an element that does not lie in A. These elements are precisely those of the form ψ j δ where 0 ≤ j ≤ 5. Each of these elements has order 2, so it does generate a type (b) subgroup. Type (c) subgroups most contain some nontrivial power ψ k of ψ, and some element of the form ψ j δ where 0 ≤ j ≤ 5. If k = ±1 we get the whole of Γ. If k = ±2 then without loss of generality k = 2. If j is odd we may take j = 1; if j is even we may take j = 0. If both parities occur we get the whole of Γ, so there are two cases: hψ 2 , δ i and hψ 2 , ψδ i. If k = 3 then j prime to 3 gives the whole of Γ. Therefore j = 0, 3 so without loss of generality j = 0 and we find one further subgroup hψ 3 , δ i. Summarising, we obtain precisely 14 subgroups, shown in Table (1). (The column listing the fixed fields will be computed below.) By the Galois correspondence, the intermediate fields are precisely the fixed fields of these 14 subgroups. It remains to identify these fixed fields. We first prove: Lemma S13.1 Let L/K be a field extension with Galois group Γ. If H is a subgroup of Γ and γ ∈ Γ then (a) γHγ −1 = γ(H † ) (b) If H † = K(α1 , α2 , . . .) then γ(H † ) = K(γ(α1 ), γ(α2 ), . . .) Proof. Suppose that x ∈ H † , so h(x) = x for all h ∈ H. For any h ∈ H, compute (γhγ −1 )(γ(x)) = γ(h(x)) = γ(x) That is, γ(x) ∈ (γHγ −1 )† . Therefore γ(H † ) ⊆ (γHγ −1 )† . A similar argument proves the reverse inclusion. (b) Follows easily since γ is a K-automorphism.

54 subgroup 1 2 3 4 5 6 7 8 9 10 11 12 13 14

generators hi hψ 3 i hψ 2 i hψi hδ i hψδ i hψ 2 δ i hψ 3 δ i hψ 4 δ i hψ 5 δ i hψ 3 , δ i hψ 2 , δ i hψ 2 , ψδ i hψ, δ i

isomorphism type 1 Z2 Z3 Z6 Z2 Z2 Z2 Z2 Z2 Z2 D2 D3 D3 D6

order 1 2 3 6 2 2 2 2 2 2 4 6 6 12

fixed field Q(α, ζ ) Q(α 2 , ζ ) Q(α 3 , ζ ) Q(ζ ) Q(α) Q(ζ α) Q(ζ 2 α) Q((1 + ζ )α) Q((ζ + ζ 2 )α) Q((ζ 2 − 1)α) Q(α 2 ) Q(α 3 ) Q((ζ + ζ 2 )α 3 ) Q

TABLE 1: Subgroups of the Galois group and fixed fields.

We use this lemma in a moment. There is a systematic ‘bare hands’ method for finding fixed fields, and we illustrate this approach first, before moving on to a more efficient method making use of Galois theory. A general element x ∈ L can be written as x = a0 + a1 α + a2 α 2 + a3 α 3 + a4 α 4 + a5 α 5 +b0 ζ + b1 αζ + b2 α 2 ζ + b3 α 3 ζ + b4 α 4 ζ + b5 α 5 ζ where a j , b j ∈ Q. Moreover, α 6 = 7 and ζ 2 − ζ + 1 = 0. Therefore ζ 2 = ζ − 1, ζ 3 = −1, ζ 4 = −ζ , ζ 5 = 1 − ζ . Consider a typical case, subgroup 3 generated by ψ 2 . The element x is fixed by this subgroup if and only if it is fixed by ψ 2 . So we compute ψ 2 (x) = a0 + a1 ζ 2 α + a2 ζ 4 α 2 + a3 α 3 + a4 ζ 2 α 4 + a5 ζ 4 α 5 +b0 ζ + b1 αζ 3 + b2 α 2 ζ 5 + b3 α 3 ζ + b4 α 4 ζ 3 + b5 α 5 ζ 5 = (a0 − a1 ) − b1 αζ + b2 α 2 ζ + a3 α 3 − a4 α 4 ζ + b5 α 5 ζ +b0 ζ + a1 αζ − (a2 + b2 )α 2 ζ + b3 α 3 ζ + (a4 − b4 )α 4 ζ − (a5 + b5 )α 5 ζ Comparing coefficients, we find that a0 , a3 , b0 , b3 can be arbitrary, while all the others must be zero. Therefore the fixed field is Q(α 3 , ζ ). A more efficient way to reach the same conclusion is to observe that both α 3 and ζ are fixed by ψ 2 , so certainly Q(α 3 , ζ ) is contained in the fixed field. Moreover, this field has degree 4 over Q. By the Galois correspondence the degree of the fixed field is 12/3 = 4. Therefore the fixed field must be Q(α 3 , ζ ). By similar methods the fixed fields of subgroups 1, 2, 4 in the table are Q(α, ζ ), Q(α 2 , ζ ), Q(ζ ) respectively.

13 A Worked Example

55

Let H be subgroup 5, generated by δ . We have δ 2 = 1, so the degree of the fixed field must be 12/2 = 6. Moreover, δ fixes α, so it fixes Q(α). Since this has degree 6, the fixed field of H is Q(α). By standard properties of dihedral groups (or simple computations) the subgroups numbered 5-10 form two conjugacy classes, depending on the parity of j. Specifically, subgroups 5, 7, and 9 ( j even) are conjugate, and subgroups 6, 8, and 10 ( j odd) are conjugate. We can use Lemma S13.1 to deduce the fixed fields of subgroups 7 and 9 from that for subgroup 5. In fact, ζ j α is fixed by ψ 2 j δ , so the respective fixed fields are Q(ζ α) and Q(ζ 2 α). (In each case the degree is 6 since all elements ζ j α have the same minimal polynomial t 6 − 7.) What about the other three zeros of f ? Do they give the fixed fields of subgroups 6, 8, and 10? No, because these subgroups are not conjugate to number 5. In fact, ζ 3 α = −α so ζ 3 α generates the same subfield as α, and similarly for ζ 4 α and ζ 5 α. We must therefore look elsewhere for the fixed field of subgroup 6, generated by ψδ . This element has order 2, and it maps α to ζ α. Therefore the sum α + ζ α = (1 + ζ )α is fixed by ψδ . The minimal polynomial of (1 + ζ )α has degree 6. Therefore the fixed field is Q(1 + ζ )α). Now we can find the fixed fields of subgroups 8 and 10, which are conjugate to subgroup 6. Now consider subgroup 11. This is generated by subgroup 2 together with δ , so its fixed field is the fixed field of δ acting on the fixed field of subgroup 2. This is easily seen to be Q(α 2 ) of degree 3 = 12/4 as expected. We can deal with subgroups 12, 13 in a similar way. Finally the fixed field of subgroup 14 is of course Q. To summarise, we list these fixed fields in the final column of the table. 13.8* Using Exercise 12.5 and methods like those employed in Exercise 13.7, √ we obtain the results summarised in Table (2). For subgroup 13√observe that −6 = (α 3 − 1)(2ω + 1) ∈ L, and for subgroups 14–16 observe that −3 = ω − ω 2 . √ 8 13.9 Let K = Q(i). Let α = i (any choice of 8th root) which is a primitive 32nd root of unity. This is not in K, so the splitting field Σ of f (t) = t 8 − i ∈ K[t] is generated over K by {α, ζ } where ζ is a primitive 8th root of unity. We can choose ζ = α 4 . The zeros of f are ζ k α for 0 ≤ k ≤ 7, and Σ = K(α, ζ ) = K(α) = Q(i, α) = Q(α) By Proposition 6.7 the degree [Σ : K] = 8. A K-basis for Σ is {1, α, α 2 , α 3 , α 4 , α 5 α 6 , α 7 } The minimal polynomial of α over K is f , so the Galois group Γ = Γ(Σ/K) is determined by K-automorphisms ψ j : α 7→ ζ j α = α 4 j+1 with 0 ≤ j ≤ 7. We have ψ j ψk (α) = α (4 j+1)(4k+1) = α 16 jk+4 j+4k+1 = α 4(4 jk+ j+k)+1 = ψ4 jk+4 j+4k A little experiment shows that the powers of ψ1 , in order from the 0th to the 7th, are ψ0 , ψ1 , ψ6 , ψ7 , ψ4 , ψ5 , ψ2 , ψ3

56 subgroup 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

generators hρ, τi hρi hτi hρτi hρ 2 τi hi hρ, τ, σ i hρ, σ i hτ, σ i hρτ, σ i hρ 2 τ, σ i hσ i hρ, τσ i hτσ i hρτσ i hρ 2 τσ i

isomorphism type D6 Z3 Z2 Z2 Z2 D6 D6 × Z2 Z3 × Z2 Z2 × Z2 Z2 × Z2 Z2 × Z2 Z2 D6 Z2 Z2 Z2

order 6 3 2 6 2 1 12 6 4 4 4 2 6 2 2 2

fixed field Q(α 3 ) Q(α 3 , ω) Q(α) Q(ω 2 α) Q(ωα) Q(α, ω) Q Q(ω) Q(α + β ) Q(ωα + ω 2 β ) Q(ω 2 α + ωβ ) Q(α √+ β , ω) Q(√−6) Q(√−3(α − β ) Q(√−3(ωα − ω 2 β )) Q( −3(ω 2 α − ωβ ))

TABLE 2: Subgroups of the Galois group and fixed fields.

which exhaust Γ. Therefore Γ ∼ = Z8 , generated by ψ1 . 13.10 We can factorise f (t) = t 8 + t 4 + 1 as f (t) = (t 2 + t + 1)(t 2 − t + 1)(t 4 − t 2 + 1) If ζ is a primitive 12th root of unity, then t 2 + t + 1 = (t − ζ 4 )(t − ζ 8 ) t 2 − t + 1 = (t − ζ 2 )(t − ζ 10 ) 4

t − t 2 + 1 = (t − ζ )(t − ζ 5 )(t − ζ 7 )(t − ζ 11 ) so the splitting field is Q(ζ ). This contains a subfield Q(i) = Q(ζ 3 ). So we are seeking automorphisms of Q that fix ζ 3 . The automorphisms are determined by their effect on ζ , and are given by 1 : ζ 7→ ζ

σ : ζ 7→ ζ 5

ρ : ζ 7→ ζ 7

Their effect on ζ 3 is: 1 : ζ 3 7→ ζ 3 ρ : ζ 3 7→ ζ 15 = ζ 3 σ : ζ 3 7→ ζ 21 = ζ 9 6= ζ 3 τ : ζ 3 7→ ζ 33 = ζ 9 6= ζ 3

τ : ζ 7→ ζ 11

13 A Worked Example

57

Therefore √ the Galois √ f over Q(i) is Z2 , generated by σ . √group of 13.11 Let 2 = α, 3 = β , 5 = γ, and let Σ = Q(α, β , γ). By Example 6.8, this is a normal (separable) extension of Q of degree 8. By Example 8.4 the Galois group is generated by Q-automorphisms ρ2 , ρ3 , ρ5 which we rename for convenience: ρ : α 7→ −α, β 7→ β , γ 7→ γ σ : α 7→ α, β 7→ −β , γ 7→ γ τ : α 7→ α, β 7→ β , γ 7→ −γ which are all of order 2 and commute. Therefore the Galois group is Γ ∼ = Z2 ×Z2 ×Z2 as stated. We can think of Γ as a 3-dimensional vector space over the field F2 ∼ = Z2 . The subgroups are precisely the vector subspaces. So we can use linear algebra to find the subgroups. It helps to introduce coordinates in Z2 = {0, 1} so that α is identified with (1, 0, 0), β is identified with (0, 1, 0), and γ is identified with (0, 0, 1). The vector subspaces are then: 0

with basis 0/

X1

with basis {(1, 0, 0)}

X2

with basis {(0, 1, 0)}

X3

with basis {(1, 1, 0)}

X4

with basis {(0, 0, 1)}

X5

with basis {(1, 0, 1)}

X6

with basis {(0, 1, 1)}

X7

with basis {(1, 1, 1)}

Y1

with basis {(1, 0, 0), (0, 1, 0)}

Y2

with basis {(1, 0, 0), (0, 0, 1)}

Y3

with basis {(1, 0, 0), (0, 1, 1)}

Y4

with basis {(0, 1, 0), (0, 0, 1)}

Y5

with basis {(0, 1, 0), (1, 0, 1)}

Y6

with basis {(1, 1, 0), (0, 0, 1)}

Y7

with basis {(1, 1, 0), (1, 0, 1)}

Z with basis {(1, 0, 0), (0, 1, 0), (0, 0, 1)} Case-by-case calculations show that the corresponding fixed fields are:

58

0 has fixed field Q(α, β , γ) X1

has fixed field Q(β , γ)

X2

has fixed field Q(α, γ)

X3 has fixed field Q(αβ , γ) X4

has fixed field Q(α, β )

X5 has fixed field Q(αγ, β ) X6 has fixed field Q(α, β γ) X7 has fixed field Q(αβ , β γ) Y1

has fixed field Q(γ)

Y2

has fixed field Q(β )

Y3

has fixed field Q(β γ)

Y4

has fixed field Q(α)

Y5

has fixed field Q(αγ)

Y6

has fixed field Q(αβ )

Y7 has fixed field Q(αβ γ) Z

has fixed field Q

As illustration, we do the case Y7 . The basis is {(1, 1, 0), (1, 0, 1)}, which grouptheoretically correspond to the Q-automorphisms ρσ , ρτ. The general element of the field is x = a0 + a1 α + a2 β + a3 γ + a4 αβ + a5 αγ + a6 β γ + a7 αβ γ where the a j ∈ Q. The Q-automorphism ρσ maps this to ρσ (x) = a0 − a1 α − a2 β + a3 γ + a4 αβ − a5 αγ − a6 β γ + a7 αβ γ so ρσ (x) = x if and only if a1 = a2 = a5 = a6 = 0. Therefore x = a0 + a3 γ + a4 αβ + a7 αβ γ Then ρτ(x) = a0 − a3 γ − a4 αβ + a7 αβ γ so ρτ(x) = x if and only if we also have a3 = a4 = 0. Therefore the fixed field of Y7 consists of the elements a0 + a7 αβ γ, and this is Q(αβ γ) as stated. 13.12 The Galois group G ∼ = Z6 , generated by σ = σ3 , which maps ζ to ζ 3 . The nontrivial proper subgroups are Z2 = hσ 3 i and Z3 = hσ 2 i. Also, σ 2 (ζ k ) = ζ 2k

σ 3 (ζ k ) = ζ −k

Consider a general element α = p0 + p1 ζ + p2 ζ 2 + p3 ζ 3 + p4 ζ 4 + p5 ζ 5 + p6 ζ 6 ∈ Q(ζ )

14 Solubility and Simplicity

59

Then σ 3 (α) = p0 + p1 ζ 6 + p2 ζ 5 + p3 ζ 4 + p4 ζ 3 + p5 ζ 2 + p6 ζ Therefore σ 3 (α) = α if and only if p1 = p6 , p2 = p5 , p3 = p4 and the fixed field of Z2 is spanned over Q by ζ + ζ 6 , ζ 2 + ζ 5 , ζ 3 + ζ 4 . But (ζ + ζ 6 )2 = ζ 2 + ζ 5 + 2 (ζ + ζ 6 )(ζ 2 + ζ 5 ) = ζ 3 + ζ 6 + ζ + ζ 4 so the fixed field is Q(ζ + ζ 6 ). For Z3 , we have σ 2 (α) = p0 + p1 ζ 5 + p2 ζ 3 + p3 ζ + p4 ζ 6 + p5 ζ 4 + p6 ζ 2 Therefore σ 2 (α) = α if and only if p1 = p2 = p4

p3 = p5 = p6

and the fixed field of Z2 is spanned over Q by ζ + ζ 2 + ζ 4 and ζ 3 + ζ 5 + ζ 6 . But 1 + (ζ + ζ 2 + ζ 4 ) + (ζ 3 + ζ 5 + ζ 6 ) = 0 so the fixed field is Q(ζ + ζ 2 + ζ 4 ). These two fields, together with Q and Q(ζ ), give all intermediate fields. 13.13* Using the hint, the zeros are √ √ √ √ √ √ 3 3 3 3 3 3 2 + i ω 2 + i ω2 2 + i 2 − i ω 2 − i ω2 2 − i where ω = e2πi/3 . The Galois group is generated by √ √ 3 3 α : 2 7→ ω 2 i 7→ i √ √ 3 3 β : 2 7→ 2 i 7→ −i Now α 3 = β 2 = id and β αβ = α 2 , so the group is isomorphic to D3 . 13.14 Since f (t) ∈ Q[t], complex conjugation α 7→ α¯ is a Q-automorphism of the splitting field of f . If any zero is non-real, α has order 2. Therefore all zeros are real. 13.15 (a) F. (b) F. (c) F. (d) T. (e) T. (f) F. (g) T.

14 Solubility and Simplicity 14.1 Let G be the group concerned. The subgroup N generated by a is cyclic of order n, and is normal. The quotient G/N is generated by the coset Nb, and has order 2 since b2 = 1. Thus there is a series {1} / N / G

60 whose quotients are Zn , Z2 respectively. Both are abelian, so G is soluble. 14.2 If n ≥ 5 then Sn contains a subgroup isomorphic to S5 , namely all permutations that fix the elements 6, 7, . . . , n. This subgroup in turn contains a subgroup isomorphic to A5 . If Sn were soluble then the subgroup isomorphic to A5 would also be soluble — but we know this is simple and not cyclic. So Sn is not soluble for n ≥ 5. 14.3 Let N / G. If n ∈ N and g ∈ G then g−1 ng ∈ N because N is normal. Therefore the conjugacy class of n is contained in N. Since n can be any element of N, it follows that N is a union of conjugacy classes (of G). Conjugacy is an equivalence relation, so in fact N is a disjoint union of conjugacy classes. Two elements of A5 are conjugate in A5 if and only if (a) they have the same cycle type, and (b) they are conjugate via an even permutation. The cycle types in A5 are () (a b)(c d) (a b c) (a b c d e) where the first entry (empty cycle) is the cycle type of the identity, and all elements not in the stated cycles are fixed. Each type constitutes a single conjugacy class in A5 , with the exception of the 5-cycles (a b c d e) which split into two conjugacy classes, because there is no freedom to make the conjugating element an even permutation. For instance, consider (1 2 3 4 5) and (2 1 3 4 5), for which the only possible conjugating element is (1 2) which is odd. Thus there are five conjugacy classes, which respectively contain 1, 15, 20, 12, 12 elements. Any normal subgroup of A5 is a disjoint union of these classes. It must contain the identity, and its order must be a proper divisor of 60. However, no combination of the above numbers that contains 1 adds up to a divisor of 60, except 1 and 60. Therefore no union of conjugacy classes can be a nontrivial proper normal subgroup of A5 , so A5 is simple. 14.4 We know that Sn is generated by all 2-cycles (a b), so it is enough to write (a b) as a product of cycles of the form (1 k). But (1 a)(1 b)(1 a) = (a b), and we are done. 14.5 Each of the Galois groups concerned has order 2m for some m. We claim that any such group G, other than the trivial group, has a normal subgroup of order 2. To prove this, consider the class equation (14.3.2), which becomes 2m = 1 + |C2 | + · · · + |Cr | where all the |C j divide 2m so are powers of 2. When m > 0 some |C j | ( j ≥ 2) must equal 1, otherwise 1 is even, and the corresponding element z 6= 1 satisfies g−1 zg = z for all g ∈ G. That is, zg = gz. Therefore the centre of G (the set of all such z) is nontrivial. Being an abelian group of order 2l for l > 0, the group Z contains an element y of order 2. The group Y = hyi has order 2 and is normal in G because gy = yg for all g ∈ G. Now G/Y has order 2m−1 , so by induction on m we see that G is soluble. (A similar proof shows that any group of prime-power order is soluble. See Corollary 23.7.) 14.6 Let H be a subgroup of A5 of order 15. By Cauchy’s Theorem 14.15, H has an element a of order 3, which must be a 3-cycle, and an element b of order 5,

14 Solubility and Simplicity

61

which must be a 5-cycle. Without loss of generality b = (1 2 3 4 5). It is easy to see that no 3-cycle commutes with b. Therefore a−1 Ha 6= H, so it intersects H in the identity. Similarly a−2 Ha2 intersects both H and a−1 Ha in the identity. These there subgroups therefore contain a total of 4 + 4 + 4 = 12 elements of H of order 5. Let A be the cyclic subgroup of order 3 generated by a. The conjugate b−1 Ab is also cyclic of order 3, and is different from A by an easy computation. Between them, A and b−1 Ab contribute a further 4 elements of order 3. So there are at least 16 elements, a contradiction. 14.7 There are many solutions. A direct method is to show that the only elements that commute with the n-cycle (1 2 3 . . . n) are the powers of that cycle,and that none of those powers commutes with (1 2). To do this, let σ = (1 2 3 . . . n) and suppose that γσ = σ γ. Suppose that γ(1) = j. Then σ (γ(1)) = σ ( j) ≡ j + 1 (mod n). Moreover, γ(σ (1)) = γ(2). Inductively, σ (γ(k)) ≡ γ(k) + 1 (mod n) and γ(σ (k)) ≡ γ(k + 1) (mod n). Since these are equal, γ(k) ≡ k + a (mod n) for fixed a, so γ = σ a . Now (1 2)σ l (1) = (1 2)(l + 1) = l + 1 unless l = 1. On the other hand, l σ (1 2)(1) = σ l (2) = l + 2. So (1 2) and σ l do not commute except in trivial cases. Less directly, but more quickly: if the centre is nontrivial then it is a normal subgroup, so it must be An , but this is not abelian. 14.8 The elements of G = Dn are uniquely of the form ak or ak b for 0 ≤ k < n. To find the conjugates g−1 ak g of an element ak observe first that if g = al then we just get ak . So we can take g = al b. Then g−1 ak g = ba−l ak al b = a−k . When n is odd, ak and a−k are distinct except when k = 0; when n is even they are distinct except when k = 0, n/2. Next, we consider the conjugates g−1 ak bg of ak b. If g = al then g−1 ak bg = a−l ak bal = ak−l a−l b = ak−2l b. Thus when n is odd we obtain all possible elements of the form ak b as conjugates of b, but when n is even only those of the form a2k b are conjugates of b, whereas all a2k+1 b are conjugates of ab. To summarise: the conjugacy classes are: n odd : 1, {a±1 }, . . . , {a±(n−1)/2 }, {ak b : 0 ≤ k < n} n even : 1, {a±1 }, . . . , {a±(n−2)/2 }, {an/2 }, {ak b : k odd}, {ak b : k even} The cardinalities of these classes are respectively n odd : 1, 2, . . . , 2, n n even :

n n 1, 2, . . . , 2, 1, , 2 2 As an example of the centraliser calculations, which are all very similar, consider the case when n = 2m is even and the element is ab. The conjugacy class is {ak b : k odd}, with n/2 elements. So we predict that the centraliser CG (ab) has order 4. Now ak commutes with b if and only if a−k bak = b, so a−2k = 1, thus k = 0, m. Similarly ak b commutes with b if and only if k = 0, m. So the centraliser is {1, am , b, am b} and has order 4 as expected.

62 14.9 Let h ∈ CG (x), so hx = xh. Then (g−1 hg)(g−1 xg) = g−1 hxg = g−1 xhg = (g−1 xg)(g−1 hg) so (g−1 hg) ∈ CG ((g−1 xg)). That is, g−1CG (x)g ⊆ CG ((g−1 xg)) The reverse inclusion follows if we replace x by gxg−1 and then change g to g−1 . Alternatively, a similar calculation can be used. 14.10 Since V is abelian, G / V / S4 . But G is not normal in S4 , because (1 3)−1 .(1 2)(3 4).(1 3) = (3 2)(1 4) 6∈ G 14.11 (1) Consider the natural map φ : G → G/H, which is a homomorphism. Its restriction ψ : A → G/H is also a homomorphism. Its image is easily seen to be A ∼ HA AH/H and its kernel is H ∩ A. Therefore H∩A = A . (2) Let φ : G → G/H be the natural map, and let ψ : G/H → G/H A/H be the natural map. Each is a homorphism, so their composition is also a homomorphism. Its image G ∼ G/H is G/H A/H and its kernel is A. Therefore A = A/H . 14.12* The icosahedron has 12 vertices, and five faces meet at each vertex. Therefore its group of rotational symmetries has order 5 · 12 = 60. Its elements can be classified geometrically as 1 identity element 15 rotations of order 2 about an axis joining the midpoints of opposite edges 20 rotations of order 3 about an axis joining the midpoints of opposite faces 24 rotations of order 5 about an axis joining opposite vertices Rotations that take the same geometric form but are performed about different axes are conjugate (by a rotation sending one axis to the other). The first three types listed above form single conjugacy classes. The final type (order 5) consists of 12 4π rotations through ± 2π 5 and 12 rotations through ± 5 . With suitable orientations of the axes, the + signs apply. These therefore form two distinct conjugacy classes. Thus the sizes of the conjugacy classes are 1, 15, 20, 12, 12. The argument of Exercise 14.3 now shows that the rotation group G of the icosahedron is simple. It remains to show that G is isomorphic to A5 (which explains why the same proof works). To do this, we need to define five ‘configurations’ that are permuted under the action of G. There are 30 edges, and 30/5 = 6, suggesting that we should choose each configuration to be a suitable 6-element subset of the edges. To define one such configuration, as in Figure 13 (left), choose an edge A and consider the diametrically opposite edge B; join them to form a rectangle. There are precisely two further pairs of opposite edges such that the corresponding three rectangles all meet each other at right angles, as in Figure 13 (centre). There are 6 edges in each such configurations, and every edge lies in exactly one configuration. Therefore the set of edges partitions into 5 disjoint sets, each consisting of 6 edges of this kind. The action of G preserves angles, so it must permute

15 Solution by Radicals

63

FIGURE 13: Geometry of icosahedron. Left: A subset of six edges with no vertices in common. Centre: The three mutually orthogonal rectangles associated with such a subset. Right: Partition of the 30 edges into 5 subsets of this kind. theses 5 configurations. This yields a homomorphism from G to S5 . Since G is simple, this homomorphism either maps everything to the identity or is injective, and the first case clearly does not occur. Thus the image of G is a subgroup of S5 of order 60. But any subgroup of order 60 must be normal (its index is 2), and the only nontrivial normal subgroup of S5 is A5 . So G is isomorphic to A5 . 14.13 (a) T. (b) T. (c) F. (d) F. (e) F.

15 Solution by Radicals 15.1 Solutions to this question are not unique. Natural choices are: √ √ (a) Q ⊆ Q(α) ⊆ Q(α, β ) ⊆ Q(α, β , γ) where α 5 = 5, β 2 = 99, γ 5 = 11 − 7 23. (b) Q ⊆ Q(α) ⊆ Q(α, β ) where α 2 = 6, β 3 = 5. √ (c) Q ⊆ Q(α) ⊆ Q(α, β ) where α 2 = 11, β 7 = 23, γ 2 = 1 + 99. 15.2 Let ζ be a primitive pth root of unity. The minimal polynomial of ζ is f (t) = 1 + t + · · · + t p−1 , by Lemma 19.13, and the splitting field Σ of f over Q is Q(ζ ) since the zeros of f are ζ j for 1 ≤ j ≤ p − 1. The Q-automorphisms of Σ are determined by their effect on ζ , and must send ζ to a zero of f . Thus they are the Q-automorphisms ρ j sending ζ to ζ j for 1 ≤ j ≤ p − 1. Moreover, ρk ρl sends ζ to ζ k l, so ρk ρl = ρkl , and the Galois group is isomorphic to the multiplicative group Z∗p of nonzero elements of the field Z p . 15.3 (a) The polynomial f (t) = t 5 − 4t − 2 is irreducible over Q by Eisenstein’s Criterion, so there are no multiple zeros. A numerical drawing of its graph, Figure 14, suggests that it has precisely three real zeros. Therefore the argument used in Theorem 15.11 shows that f (t) = 0 is not soluble by radicals. It remains to prove rigorously that f has precisely three real zeros. The derivative

64

FIGURE 14: Graph of polynomial (a). q is f 0 (t) = 5t 4 − 4. The real zeros of f 0 lie at t = ± 4 45 . A quick computation shows q q that f (t) > 0 when t = − 4 45 , while f (t) < 0 when t = + 4 45 . As in the proof of q Theorem 15.11, there is exactly one real zero of f less than − 4 45 , one in the interval q q q between − 4 45 and + 4 45 , and one greater than + 4 45 . (b) (a) The polynomial f (t) = t 5 − 4t 2 + 2 is irreducible over Q by Eisenstein’s Criterion, so there are no multiple zeros. A numerical drawing of its graph, Figure 15, suggests that it has precisely three real zeros. Therefore the argument used in Theorem 15.11 shows that f (t) = 0 is not soluble by radicals.

FIGURE 15: Graph of polynomial (b). It remains to prove rigorously that f has precisely threeqreal zeros. The derivative

is f 0 (t) = 5t 4 − 4t. The real zeros of f 0 lie at t = 0 and t =

4 5 . A quick computation q shows that f (t) > 0 when t = 0, while f (t) < 0 when t = 3 45 . As in the proof of Theorem 15.11,qthere is exactly one realq zero of f less than 0, one in the interval 3

between 0 and 3 45 , and one greater than 3 45 . (c) The graph is shown in Figure 16, and the argument is very similar to case (b). (d) The graph of f (t) = t 7 − 10t 5 + 15t + 5 is shown in Figure 17 (left). An

15 Solution by Radicals

65

FIGURE 16: Graph of polynomial (c). enlargement of the region near the origin (Figure 17 (right)) suggests that there are 5 distinct real zeros.

FIGURE 17: Graph of polynomial (d). Left: large scale. Right: enlargement of region near the origin. The derivative is f 0 (t) = t 6 − 50t 4 + 15. We can write f 0 (t) as a cubic in u = t 2 , and this cubic can easily be shown to have two positive real zeros and one negative real zero. Thus f 0 (t) has precisely four real zeros. Therefore f can change sign at most four times, so it has at most five real zeros. Direct calculation shows that f changes sign at least once in each of the five intervals [−4, −2] [−2, 0 · 5] [0 · 5, 1] [1, 2] [2, 4] u3 − 50u + 15

Now argue as in Theorem 15.11 with p = 7. 15.4 The substitution u = t + 1/t converts the equation to g(u) = u3 − u2 − 2u + 1 = 0 which can be solved by radicals using Cardano’s formula. In detail: the Tschirnhaus 7 transformation u = v + 31 leads to v3 − 73 v + 27 = 0. So, in the standard notation of 7 7 Cardano’s formula, p = − 3 , q = 27 . Therefore the solutions for v are

66

s

s

r 7 −49 3 7 −49 v1 = + + − 54 108 54 108 s s r r −49 −49 3 7 2 3 7 v2 = ω + +ω − 54 108 54 108 s s r r −49 −49 3 7 3 7 v3 = ω 2 + +ω − 54 108 54 108 3

r

and the solutions for u are u j = v j + 13 . If a is any of these three zeros of g, the equation a = t + 1/t is equivalent to the quadratic t 2 − at + 1 = 0 which can be solved using the quadratic formula. 15.5 The substitution u = t + 1/t converts the equation to g(u) = u3 + 2u2 − 8u + 5 = √ −3± 29 0. The zeros of this are u = 1, 2 . If a is any of these three zeros of g, the equation a = t + 1/t is equivalent to the quadratic t 2 − at + 1 = 0 which can be solved using the quadratic formula. 15.6* Let K = Q and let L = K(ζ ) where ζ is a primitive seventh root of unity. Then L/K is radical and its degree is 6. Let α = ζ 2 + ζ 5 and consider the intermediate field M = K(α). The minimal polynomial of α is the polynomial whose zeros are the distinct images of α under the Galois group. We compute this polynomial as m(t) = (t − (ζ 2 + ζ 5 ))(t − (ζ 4 + ζ 3 ))(t − (ζ + ζ 6 )) = t 3 + t 2 − 2t + 1 Therefore [M : K] = 3. If M/K were radical, the only possible radical sequence would be to adjoin a cube root of some rational p that is not a perfect cube. That is, M must √ be of the form M = Q( 3 p) for some choice of the cube root (not necessarily the real one). We prove this is impossible. Since M/K is the splitting field of m(t), the extension √ √ M/K is normal. Therefore M contains not only 3 p but also ω 3 p) for a primitive cube root of unity ω. Hence ω ∈ M. We know that [K(ω) : K] = 2, but the Tower Law implies that [K(ω) : K] divides 3. This is a contradiction, so M/K is not radical. 15.7 Let β be a zero of p(t) ∈ K[t] and suppose that there is a radical sequence α1 , . . . , αm such that ∈ K(α1 , . . . , αm ). By adjoining a suitable root of unity ζ (such as an nth root where n = n1 . . . nm where n j is the radical degree of α j ) we can arrange for L/K(α1 , . . . , αm , ζ ) to be a normal extension of K. Moreover, L/K is clearly radical. Now the normal closure Σ of K(β )/K is contained in L, so every zero of p belongs to L. Since L/K is radical, every zero of p is expressible by radicals. 15.8* Consider the tower of extensions K ⊆ K(α) ⊆ K(α, β ). Clearly [K(α) : K] = 2. We claim that [K(α, β ) : K(α)] = 2. Equivalently, we claim that β 6∈ K(α). For a contradiction, suppose that β ∈ K(α). Then β = p + qα where p, q ∈ K. Squaring, b = β 2 = p2 + aq2 + 2pqα

15 Solution by Radicals

67

If pq 6= 0 then α=

b − p2 − aq2 ∈K 2pq

but this is false, so either p = 0 or q = 0. If p = 0 then b = aq2 , so ab = a2 q2 . This is impossible because ab is not a square in K. If q = 0 then b = p2 which is impossible because b is not a square in K. Therefore β 6∈ K(α) as claimed. By the Tower Law, [K(α, β ) : K] = 4. Moreover, the extension K(α, β )/K is normal (and separable) so the Galois correspondence is bijective, and the order of Γ = Γ(K(α, β )/K) is 4. A basis for K(α, β ) over K is {1, α, β , αβ }. The minimal polynomial of α is t 2 −a with zeros ±α, so any K-automorphism of K(α, β ) sends α to ±α. Similarly, any K-automorphism of K(α, β ) sends β to ±β . Bearing in mind that φ (αβ ) = φ (α)φ (β ) for any K-automorphism φ , we conclude that there are at most four K-automorphisms: 1 : p + qα + rβ + sαβ 7→ p + qα + rβ + sαβ ρ : p + qα + rβ + sαβ 7→ p − qα + rβ − sαβ σ : p + qα + rβ + sαβ 7→ p + qα − rβ − sαβ τ : p + qα + rβ + sαβ 7→ p − qα − rβ + sαβ for p, q, r, s ∈ K. Since |Γ| = 4 these must be precisely the K-automorphisms of K(α, β ) (which can also be proved directly without difficulty). Now ρ 2 = σ 2 = 1, ρσ = τ, so the Galois group is Z2 × Z2 . 15.9* Let f (t) = t 5 − N pt + p. By Eisenstein’s Criterion f is irreducible. We claim that f has precisely three real zeros: if so, we can argue as q in Theorem 15.11.

The derivative is f 0 (t) = 5t 4 −N p. It is zero when t = ± 4 N5p (remember, N > 0). Thus f has three real zeros. At t = −∞, 0, 1, ∞ the value of f (t) is respectively −∞, p, 1 + p − N p, ∞. Since N ≥ 2 we have 1 + p − N p < 0, so there are at least three (hence exactly three) sign changes. Now use the argument of Theorem 15.11. 15.10* A similar argument does not work. To see why: Since f is irreducible, there are no multiple zeros. Number the zeros so that the complex conjugate pairs are {1, 2} and {3, 4}, and the single real zero is 5. As in Lemma 15.10, the Galois group of f contains complex conjugation, which is now the permutation (1 2)(3 4), and some 5-cycle. The continuation now depends on which 5-cycle occurs. If, for example, it is (1 2 3 4 5) then we generate A5 and deduce that the equation is not soluble by radicals. However, if the 5-cycle is (13425) then the group H generated by {(1 2)(3 4), (1 3 4 2 5)} is isomorphic to D5 , because [(1 2)(3 4)](1 3 4 2 5)[(1 2)(3 4)] = (2 4 3 1 5) = (1 3 4 2 5)−1 and the cyclic subgroup generated by (1 3 4 2 5) is normal in H. Since D5 is soluble, we cannot conclude that the equation is not soluble by radicals in this case.

68 15.11* We work in the context of subfields of C. We prove something more general. namely the modern version of Theorem 8.14: Theorem S15.1 Let f ∈ K[t] and let L/K be a field extension. Let N/L be the splitting field of f over L, let α1 , . . . , αr be the zeros of f in N, and let M = K(α1 , . . . , αr ) be the splitting field of f over K. Let L0 be the fixed field of the Galois group Γ of f over L. If σ ∈ Γ then σ |M ∈ Γ(M/L0 ∩ M), and the map θ : σ 7→ σ |M is an isomorphism of Γ with the subgroup Γ(M/L0 ∩ M) of the Galois group of f over K. Proof. If σ ∈ Γ(N/L) then σ fixes L and permutes {α1 , . . . , αr }, so σ (M) ⊆ M. Thus σ |M is an automorphism of M, and it clearly fixes L0 ∩ M. It follows easily that θ is a group homomorphism Γ(N/L) → Γ(M/L0 ∩ M) ⊆ Γ(M/K) ⊆ ΓK ( f ). If θ (σ ) is the identity then σ fixes {α1 , . . . , αr }. Since σ is the identity on L, it follows that σ fixes N. Therefore θ is one-to-one. We claim that θ (Γ(N/L)) = Γ(M/L0 ∩ M). To prove this, let J be the fixed field of θ (Γ(N/L)). We know that L0 ∩ M ⊆ J. Suppose that x ∈ M and x 6∈ L0 ∩ M. Since L0 is the fixed field of Γ(N/L), there exists σ ∈ Γ(N/L) such that σ (x) 6= x. Therefore θ (σ (x)) 6= x, so x 6∈ J. Therefore J = L0 ∩ M, so θ (Γ(N/L)) = Γ(M/J) = Γ(M/L0 ∩ M)

We apply Theorem S15.1 to prove the classical Theorem on Natural Irrationalities, Theorem 8.14. Here L = C(t1 , . . . ,tn ) and K = C(s1 , . . . , sn ). We suppose that x ∈ L lies in some radical extension R of K. We must find a radical extension R0 of K with x ∈ R0 ⊆ L. Without loss of generality R is the splitting field over K of some polynomial f ∈ K[t]. (At this stage we are really working in the case of ‘abstract’ field extensions, which is treated in Chapter 17, but this exercise fits naturally into this chapter, so if this bothers you, you should read Chapter 17 first.) To see why, note that C contains all roots of unity, and apply Exercise 8.12. Let M = R in Theorem S15.1, and let N be the splitting field of f over L. The proof of the theorem shows that the Galois group Γ of f over L, which is equal to Γ(N/L), permutes the zeros α1 , . . . , αr of f . Without loss of generality x = α1 . There exist γ j ∈ Γ such that γ j (x) = α j . Theorem S15.1 and normality of L/K imply that the polynomial g(t) = (t − γ1 (x)) · · · (t − γr (x)) has coefficients in K and is the minimal polynomial of x over K. Now the splitting field R0 of g in L contains x and is a radical extension of K since it has soluble Galois group. 15.12 We have f (t) = t 5 + pt + q over C. To rewrite this as t 5 + t + a consider the scaling t = αu

69

16 Abstract Rings and Fields Now f (t) becomes α 5 u5 + α pu + q If we choose α = p1/4 so that α 4 = p this becomes α 5 (u5 + u + α −5 q)

The zeros of f are those of u5 + u + α −5 q, which is soluble in ultraradicals with a = α −5 q = p−5/4 q. Having solved for u we set t = αu = p1/4 u. 15.13 (a) T. (b) T. (c) F. (d) T. (e) T. (f) T. (g) T. (h) T. (i) F.

16 Abstract Rings and Fields 16.1 Suppose that 1 ∈ I. Let r ∈ R. Since I is an ideal, 1.r ∈ I. But this equals R, so R ⊆ I. Also I ⊆ R, so I = R. 16.2 Strangely, the answer is ‘yes’. Let R be the direct product R = Z × Z with operations (x, y) + (u, v) = (x + u, y + v) (x, y)(u, v) = (xu, yv) This is a ring with identity element (1, 1). The subset I = {0} × Z is easily seen to be an ideal. It is isomorphic to Z, which is a ring. This is possible despite Exercise 16.1, because the identity in Z × Z is (1, 1), which is not in I. The identity in I is (0, 1). 16.3 Z6 : +01 2 3 4 5 ×0 1 2 3 4 5 0 01 2 3 4 5 0 0 0 0 0 0 0 1 12 3 4 5 0 1 0 1 2 3 4 5 2 23 4 5 0 1 2 0 2 4 0 2 4 3 34 5 0 1 2 3 0 3 0 3 0 3 4 45 0 1 2 3 4 0 4 2 0 4 2 5 50 1 2 3 4 5 0 5 4 3 2 1 Z7 : + 0 1 2 3 4 5 6

0 0 1 2 3 4 5 6

1 1 2 3 4 5 6 0

2 2 3 4 5 6 0 1

3 3 4 5 6 0 1 2

4 4 5 6 0 1 2 3

5 5 6 0 1 2 3 4

6 6 0 1 2 3 4 5

× 0 1 2 3 4 5 6

0 0 0 0 0 0 0 0

1 0 1 2 3 4 5 6

2 0 2 4 6 1 3 5

3 0 3 6 2 5 1 4

4 0 4 1 5 2 6 3

5 0 5 3 1 6 4 2

6 0 6 5 4 3 2 1

70 Z8 : + 0 1 2 3 4 5 6 7

0 0 1 2 3 4 5 6 7

1 1 2 3 4 5 6 7 0

2 2 3 4 5 6 7 0 1

3 3 4 5 6 7 0 1 2

4 4 5 6 7 0 1 2 3

5 5 6 7 0 1 2 3 4

6 6 7 0 1 2 3 4 5

7 7 0 1 2 3 4 5 6

× 0 1 2 3 4 5 6 7

0 0 0 0 0 0 0 0 0

1 0 1 2 3 4 5 6 7

2 0 2 4 6 0 2 4 6

3 0 3 6 1 4 7 2 5

4 0 4 0 4 0 4 0 4

5 0 5 2 7 4 1 6 3

6 0 6 4 2 0 6 4 2

7 0 7 6 5 4 3 2 1

Only Z7 is an integral domain, and only Z7 is a field. In Z6 we have 2 · 3 = 0, and in Z8 we have 2 · 4 = 0. So these are not integral domains and not fields. In the multiplication table for Z7 there is a 1 in every row except row 0, so every nonzero element has a multiplicative inverse and Z7 is a field. Less strongly, it is also an integral domain. 16.4 A field is prime if and only if it is equal to its prime subfield. Since the prime subfields must be either Q or Z p for prime p, these are precisely the prime fields. 16.5 Respectively, the prime subfields are Q, Q, Q, Q, Q, Q, Z5 , Z17 . 16.6 The addition table is isomorphic to that of the group Z2 × Z2 , so the set {0, 1, α, β } is a group under addition. The multiplication table for the subset {1, α, β } of nonzero elements is isomorphic to that of the group Z3 , and 0a = 0 for all a. It remains to verify the distributive law a(b + c) = ab + ac. This is obvious if a = 0, 1. Case-by-case checks show that it is true whenever a = α, β . 16.7 Let R be an integral domain and F its field of fractions, which we construct as the set of all ordered pairs (r, s) with r, s ∈ R, s 6= 0 subject to the equivalence relation (r, s) ∼ (t, u) ⇐⇒ ru = st We write [r, s] for the equivalence class of (r, s) and define [r, s] + [t, u] = [ru + ts, su] and [r, s][t, u] = [rt, su]. (1) These operations are well-defined. Suppose (r, s) ∼ (r0 , s0 ) and (t, u) ∼ (t 0 , u0 ). Then rs0 = sr0 and tu0 = ut 0 . For the operation of addition, we must show that [r, s] + [t, u] = [r0 , s0 ] + [t 0 , u0 ]. That is, [ru + ts, su] = [r0 u0 + t 0 s0 , s0 u0 ], which is true provided (r, s) + (t, u) ∼ (r0 , s0 ) + (t 0 , u0 ), that is, (ru + ts)s0 u0 = (r0 u0 + t 0 s0 )su Expanding, we must show that rus0 u0 + tss0 u0 = r0 u0 su + t 0 s0 su Now rus0 u0 = rs0 uu0 = r0 su0 u = r0 u0 su. Similarly tss0 u0 = tu0 ss0 = ut 0 ss0 = t 0 s0 su. Adding, we obtain the required result. For the operation of multiplication, we must show that [r, s][t, u] = [r0 , s0 ][t 0 , u0 ]

16 Abstract Rings and Fields

71

That is, [rt, su] = [r0t 0 , s0 u0 ], which is true provided rts0 u0 = r0t 0 su. But rts0 u0 = rs0tu0 = r0 st 0 u = r0t 0 su, which is what we want. (2) They are operations of F. The point here is that we must check the condition ‘sˆ 6= 0’ for the second entry in the sum and product expressions [ˆr, s]. ˆ We know that s, u 6= 0. In both the sum and product the second entry is sˆ = su, which is nonzero since R is an integral domain. (3) To check that F is a field we must verify the appropriate axioms. All computations here follow the same general lines, and we illustrate them with the distributive law (because it involves both addition and multiplication). Suppose that [r, s], [t, u], [v, w] ∈ F so that in particular s, u, w 6= 0. We have [r, s]([t, u] + [v, w]) = [r, s][tw + uv, uw] = [rtw + ruv, suw] = [rtsw + rvsu, susw] = [rt, su] + [rv, sw] = [r, s][t, u] + [r, s][v, w] Note that [rtw + ruv, suw] = [rtsw + rvsu, susw] because (rtw + ruv, suw) ∼ (rtsw + rvsu, susw); that is, (rtw + ruv, suw)(susw) = (rtsw + rvsu)suw. (4) Let φ (r) = [r, 1]. Then φ (rs) = [rs, 1] = [r, 1][s, 1] = φ (r)φ (s) and φ (r ± s) = [r ± s, 1] = [r, 1] ± [s, 1] = φ (r) ± φ (s). It remains to prove that φ is one-to-one. Suppose that φ (r) = φ (s). Then [r, 1] = [s, 1], that is, r · 1 = s · 1. So r = s as required. (5) Compute: [r, s][s, 1] = [rs, s] = [r, 1] since (rs, s) ∼ (r, a) because rs · 1 = s · r. 16.8 Suppose we can extend φ to ψ where ψ is a monomorphism. By definition b(a/b) = a so we must have ψ(b)ψ(a/b) = ψ(a) which leads to ψ(a/b) = ψ(a)/ψ(b) Thus ψ is unique if it exists; that is, if the formula above defines a ring monomorphism. It is routine to verify that ψ is well-defined and preserves the ring operations. It remains to prove that ψ is one-to-one. Suppose that ψ(a/b) = ψ(c/d) Then φ (a)/φ (b) = φ (c)/φ (d) so φ (a)φ (d) = φ (b)φ (c)

72 Since φ is a monomorphism, we have φ (ad) = φ (bc) so ad = bc,whence a/b = c/d and we are done. Finally, we apply this result to the field of fractions. Suppose F 0 is also a field of fractions of D. The inclusion map D → F extends to a monomorphism ψ : F 0 → F. Therefore ψ(F 0 ) is isomorphic to a subring of F. But ψ(F 0 ) contains all fractions a/b for a, b ∈ D, so must equal F. Therefore ψ is an isomorphism and F is unique up to isomorphism. p(t) where p, q ∈ K[t], q 6= 0, and only terms of even 16.9 (a) The set of all fractions q(t) degree in t occur in p and q. (b) Equal to K(t). p(t) where p, q ∈ K[t], q 6= 0, and only terms of degree a multiple (c) The set of all q(t) of 5 occur in p and q. (d) The same as (a). 16.10 Let f (t) = ant n + · · · + a0 and g(t) = bmt m + · · · + b0 where an 6= 0 and bm 6= 0. Define a j = 0 for j > n and bk = 0 for k > m. Then f (t) + g(t) = ∑ c j t j where c j = a j + b j . If j > max(m, n) then c j = 0. Therefore ∂ ( f + g) ≤ max(∂ f , ∂ g). Over Z6 we have 3t · 2t = (3 · 2)t 2 = 0t 2 = 0. Now ∂ f = ∂ g = 1, but ∂ ( f g) = −∞ so the equality is false. If R is an integral domain then let f (t) = ant n +· · ·+a0 and g(t) = bmt m +· · ·+b0 where an 6= 0 and bm 6= 0. Then f g has leading term an bmt m+n and the coefficient an bm 6= 0. Therefore ∂ ( f g) = ∂ f + ∂ g. 16.11 (a) F. (b) T. (c) F. (d) T. (e) F.

17 Abstract Field Extensions 17.1 Let M be the set of all multiples of f . Let m, m2 ∈ M, so that m j = h j f ( j = 1, 2) for some h j ∈ K[t]. Then m1 ± m2 = (h1 ± h2 ) f , which lies in M. Now let g ∈ K[t]. Then gm1 = (gh1 f ) = (gh1 ) f ∈ M. So M is an ideal of K[t]. 17.2 The kernel I of φ is an ideal of K. Since K is a field, I is either {0} or K. If I = K then φ (1) = 0, but as remarked in the question, the conventions used in this book do not allow this. Therefore K = {0}, so φ is one-to-one. 17.3 Any finite-dimensional vector space V over a finite field K has finitely many elements v1 . . . , vm . Then V = R{v1 } ∪ · · · R{vm } where R{vi } is the subspace spanned by vi . In particular, if K = F2 = {0, 1} and V = K 2 then V = {(0, 0), (1, 0)} ∪ {(0, 0), (0, 1)} ∪ {(0, 0), (1, 1)} which is a union of three subspaces.

73

17 Abstract Field Extensions

17.4 Clearly β 2 = id . Direct calculation shows that γ 3 = id . It is easy to show that α = βγ

δ = βα

ε = βγ

so G is generated by β , γ. Moreover, direct calculations show that β γβ = δ = γ −1 . Therefore G ∼ = D3 . Routine calculations show that φ (t) is invariant under α and β . These generate G since γ = β α, so φ (t) is invariant under G. By the Galois correspondence, the field M = K(φ (t)) of all rational expressions in φ (t) lies in F. Thus [K(t) : M] ≥ 6. We claim that also [K(t) : M] ≤ 6. This follows if we exhibit a polynomial equation of degree 6 over K(φ (t) satisfied by t. This follows since (t 2 − t + 1)3 − φ (t).t 2 (t − 1)2 = 0 Therefore [K(t) : M] = 6. By the Galois correspondence M = F, the fixed field of G. Nontrivial proper intermediate fields are the fixed fields of nontrivial proper subgroups of G. There are four such groups: A = hαi

B = hβ i

C = hεi

D = hγi

The first three have order 2 and the fourth has order 3. The fixed field A† of A consists of all f (t) such that f (1 − t) = f (t). Let ψ(t) = t(1−t). Clearly ψ(t) ∈ A† . We now use a similar argument to that above, based on the Galois correspondence. Since ψ(t) satisfies the quadratic equation t(1−t)−ψ(t) = 0 and |A| = 2, the field A† consists of all g(ψ(t)) for g ∈ K(t). The fixed field B† of B consists of all f (t) such that f (1/t) = f (t). One such function is χ(t) = t + 1/t. Since this satisfies the quadratic t 2 − tχ(t) + 1 = 0 and |B| = 2, B† consists of all g(χ(t)) for g ∈ K(t). The fixed field C† of C consists of all f (t) such that f (t/(t − 1)) = f (t). One such function is ξ (t) = t + t/(t − 1). Since this satisfies the quadratic t 2 − (t − 1)ξ (t) = 0 and |C| = 2, B† consists of all g(ξ (t)) for g ∈ K(t). Finally, the fixed field D† of D consists of all f (t) such that f (1 − 1/t) = f (t). One such function is t + γ(t) + γ 2 (t), which is 1 t 3 − 3t + 1 1 )= η(t) = t + (1 − ) + (1 − 1 t t(t − 1) (1 − t ) Since this satisfies a cubic equation (t 3 − 3t + 1) − t(t − 1)η(t) = 0 and |D| = 3, D† consists of all g(η(t)) for g ∈ K(t). 17.5* Let L < M be algebraic closures of K. Then M, L are algebraically closed, and no proper subfield of L or M that contains K is algebraically closed. There exists an ordinal µ, a sequence of ordinals ρ < µ, and elements αρ of L such that if we define K0 = K Kρ+1 = Kρ (αρ ) Kλ =

[ ρ 0, and suppose that the Galois group Γ(L/K) is cyclic of order p. Then L is a splitting field over K of an irreducible polynomial of the form t p − t + α, where α ∈ K. Before proving the theorem, it is useful to isolate an intriguing polynomial identity that will be needed to complete the proof. Lemma S18.2 Over any field of prime characteristic p, t p − t = t(t − 1)(t − 2) · · · (t − (p − 1))

(19)

Proof. Both sides of (19) are monic polynomials of the same degree, so it suffices to prove that they have the same zeros (with the same multiplicities). The zeros of t(t − 1)(t − 2) · · · (t − (p − 1)) are 0, 1, 2, . . . , p − 1 ∈ Z p , each with multiplicity 1. So we have to show that each of 0, 1, 2, . . . , p − 1 is a zero of t p − t. We do this using induction and the Frobenius monomorphism (Lemma 17.14). Clearly 0, 1 are zeros of t p − t. Suppose inductively that n p − n = 0 for n ∈ Z p . Then (n + 1) p − (n + 1) = n p + 1 p − n − 1 using Frobenius, and by induction n p − 1 = 0, so (n + 1) p − (n + 1) = 0, completing the induction step. Proof of Theorem S18.1. Let σ : L → L be a K-automorphism of L that generates Γ = Γ(L/K), so that σ p = 1 where here 1 = id l . If θ ∈ L \ K then [K(θ ) : K] > 1 and divides p, so it must be p. Therefore K(θ ) = L. For any x ∈ L, define the trace T (x) = x + σ (x) + · · · + σ p−1 (x) (as in Exercise 18.6, but now for characteristic p and Γ ∼ = Z p .) Then σ (T (x)) = σ (x) + σ 2 (x) + · · · + σ p (x) = T (x)

(20)

since σ p = 1. Therefore T (x) lies in the fixed field of Γ, which is K since the extension is separable and normal. Since σ is a K-automorphism, we know that T is a K-linear map T : L → K as vector spaces over K. Define another K-linear map S : L → K by S(x) = σ (x) − x That is, S = σ − 1. Equation (20) implies im(S) ⊆ ker(T )

(21)

19 Finite Fields

83

where im, ker are the image and kernel. We claim that (21) is actually an equality. The key step is to show that T is onto. In characteristic zero this would be easy: T (1) = p 6= 0. But in characteristic p we have T (1) = p = 0 and the argument breaks down. However, if T is not onto we must have T (x) = 0 for all x ∈ L, so the distinct K-automorphisms 1, σ , σ 2 , . . . , σ p−1 satisfy 1 + σ + σ 2 + · · · + σ p−1 = 0 and are thus linearly dependent over K. But this contradicts Lemma 10.1 (generalised). The proof of this lemma given in Chapter 10 does not require any restrictions on the characteristic. Since T is onto, the kernel ker(T ) has dimension p − 1. Now ker(S) is the fixed field of Γ, since S(x) = 0 if and only if σ (x) = x, and σ generates Γ. Since the extension is separable and normal, this fixed field is K, so ker(S) = K. Therefore ker(S) has dimension 1 over K, implying that im(S) has dimension p − 1. But im(S) ⊆ ker(T ) by (21). Since both space have the same dimension, im(S) = ker(T ) as claimed. We have already noted that 1 ∈ ker(T ), indeed K ⊆ ker(T ). So there exists θ ∈ L such that S(θ ) = 1. That is, σ (θ ) − θ = 1 so σ (θ ) = θ + 1 Also θ 6∈ K since σ fixes K. Therefore L = K(θ ). Inductively, σ k (θ ) = θ + k (0 ≤ k ≤ p − 1) The minimal polynomial of θ is therefore m(t) = (t − θ )(t − σ (θ )) · · · (t − σ p−1 (θ )) = (t − θ )(t − θ − 1) · · · (t − θ − (p − 1)) By Lemma S18.2, m(t) = (t − θ ) p − (t − θ ) Using Frobenius, m(t) = t p − θ p − t + θ = t p − t + (θ − θ p ) Let α = θ − θ p , which is in K since by definition all coefficients of m(t) lie in K. Then m(t) = t p − t + α. Being a minimal polynomial, m(t) is irreducible over K. Its zeros generate L over K (because θ does), so L is a splitting field for m(t). 18.14 (a) T. (b) F. (c) T. (d) T. (e) F. (f) F. (g) F. (h) T. (i) F.

19 Finite Fields 19.1 We have to find the prime powers. They are: 2, 3, 4, 5, 17, 65536, 65537, 83521, 103823, 282589933 − 1

84 19.2 (a) One way to construct a field with 8 elements is to find an irreducible cubic f (t) over Z2 and form the quotient Z2 [t]/h f i. An example is f (t) = t 3 + t + 1. (If it were reducible, it would have a linear factor, so either 0 or 1 would be a zero. But f (0) = 1 = f (1).) Explicitly, this field has elements a + bθ + cθ 2 where a, b, c ∈ Z2 and we apply the relation θ 3 = 1 + θ when forming products (so also θ 4 = θ + θ 2 ). Explicitly: (a + bθ + cθ 2 ) + (p + qθ + rθ 2 ) = (a + p) + (b + q)θ + (c + r)θ 2 (a + bθ + cθ 2 )(p + qθ + rθ 2 ) = (ap + br + cq) + (aq + bp + br + cq + cr)θ +(ar + bq + cp + cr)θ 2 (b) To construct a field with 9 elements we look for an irreducible quadratic f (t) over Z3 , and form the quotient Z3 [t]/h f i. A convenient choice is f (t) = t 2 + 1. (If it were reducible, it would have a linear factor, so either 0, 1, or 2 would be a zero. But f (0) = 1, f (1) = f (2) = 2.) We can think of the resulting field as the set of ‘complex numbers’ a + bi where i2 = −1 and a, b ∈ Z3 . Another way to represent this construction is to form the Gaussian integers Z[i] = {a + bi : a, b ∈ Z} and quotient by the ideal h3i. (c) To constrict a field with 16 elements we look for an irreducible quartic f (t) over Z2 , and form the quotient Z2 [t]/h f i. A convenient choice is f (t) = t 4 +t + 1. To prove this irreducible, observe that it has no linear factors since f (0) = 1 = f (1). The only possible quadratic factors are t 2 + 1,t 2 + t + 1, and their products in pairs are (t 2 + 1)2 = t 4 + 1, (t 2 +t + 1)2 = t 4 +t 2 + 1, and (t 2 + 1)(t 2 +t + 1) = t 4 +t 3 +t 2 + 1. (This is a good way to find f : list the reducible quartics, and pick something else.) 19.3 Group-theoretically, we are being asked to find the order of the Frobenius map φ : K → K. Let K = Fq where q = pn , and p is prime. By Theorem 19.2, K is a splitting field for f (t) = t q −t over Z p . The Frobenius map satisfies φ (x) = x p for x ∈ K, and is the identity on Z p . A power φ k is the identity on K if and only if φ (θ ) = θ where θ is chosen so that K = Z p (θ ). k n Clearly φ k (θ ) = θ p . In particular φ n (θ ) = θ p = θ since f (θ ) = 0. Therefore m the order m of φ divides n. Suppose that m < n. Then θ p = θ so Theorem 19.1 implies that θ generates a field isomorphic to F pm . This cannot be the whole of K = F pn since m < n. Thus the order of φ is n. 19.4 Let M be a subfield of K = Fq where q = pn , and p is prime. Then K is a vector space over M of dimension d, say. Therefore |M|d = pn , so |M| = pn/d . Hence d|n so r = n/d divides n. Since |M| = pr we must have M ∼ = F pr . It remains to show that there is a unique subfield of this cardinality. We know that K = Z p (θ ) for some θ such that θ q = θ but θ p 6= θ if 1 ≤ r ≤ n − 1. That is, the multiplicative group generated by θ has order q − 1 and is cyclic. It follows that every non-zero element of K can be represented uniquely as θ s where 0 ≤ s ≤ q − 1. The subfield M must be equal to Z p (θ s ) for some such s. The theory of cyclic groups implies that gcd(s, q) = pn−r . But all such θ s generate the same subgroup, and M is the union of that subgroup and {0}. Therefore M is unique.

19 Finite Fields

85

19.5 The degree [F pn : F p ] is equal to n. The extension is normal and separable. Therefore its Galois group has order n. Now φ 0 ∈ Γ and φ has order n by Exercise 19.3. Therefore φ generates Γ and Γ ∼ = Zn . By the theory of cyclic groups, Γ has a unique subgroup of each order r that divides n. By Exercise 20.4 there is a unique subfield of F pn having pn/r elements. So there is a one-to-one correspondence between subfields of F pn and subgroups Zr of Zn , which reverses inclusion. This clearly must be the Galois correspondence. By considering θ , defined as in Exercise 19.4, it is easy to check that the fixed field of Zr is F pn/r . 19.6 No. Write r = pk m where p is the smallest prime divisor of r with m prime to p. We have k ≥ 1. Since r is composite, either m > 1 or k > 1. The binomial coefficient   pk m(pk m − 1) · · · (pk m − p + 1) r = p 1·2··· p is divisible by pk−1 since the only factors divisible by p are pk m in the numerator and p in the denominator. But it is not divisible by pk since one factor p cancels. Therefore it is not divisible by r. Since m > 1 or k > 1, the number p lies in the permitted range 1 ≤ p ≤ r − 1. 19.7 The numbers n = 13, 17, 19, 23, 29, 31, 37, 41 are prime, so we are working in Zn . By trial and error, the smallest generator in these cases is respectively 2, 3, 2, 5, 2, 3, 2, 7. The numbers 9 and 49 are squares of primes, and 8 is the cube of a prime. For n = 8, 9 we can use Exercise 19.2. 8 elements: The multiplicative group has order 7, which is prime, so any element other than 1 generates the group. In particular, the element θ has this property. In fact, its powers from 0 to 6 are 1, θ , θ 2 , θ + 1, θ 2 + θ , θ 2 + θ + 1, θ 2 + 1 9 elements: The ‘complex number’ 1 + i over Z3 has order 8. 49 elements: We construct a field with 49 elements in a similar manner to that used in Exercise 19.2 for 9 elements. Namely, the polynomial f (t) = t 2 + 1 is irreducible over Z7 since −1 ≡ 6 is not the square of any element. So Z7 [t]/h f i is a field with 49 elements, and we can think of these as ‘complex numbers’ a + bi where i2 = −1 and a, b, ∈ Z7 . Now a bit of intelligent trial and error shows that 3 + i has order 48. 19.8 We know that F pn is a vector space over F p . The dimension is clearly n, so there is a basis with n elements. Each generates a group of order p under addition, and since they are linearly independent and span the space, the additive group of F pn is a direct sum of n groups isomorphic to Z p . 19.9 The image of Z2 (t) under φ is Z2 (t 2 ). So φ is not onto. 19.10* By definition e(Sn ) is the lcm of the orders of the elements of Sn . Every σ ∈ Sn is a product of disjoint cycles, and its order is the lcm of the cycle lengths. If p is prime and some power pk < n, then Sn contains a pk -cycle. It follows that e(Sn ) = ∏ pk(p) p

86 where p runs through the primes and k(p) is the largest integer such that pk(p) ≤ n. To gain intuition, compute e(Sn ) for low values of n: e(S1 ) = 1 e(S2 ) = 2 e(S3 ) = 2.3 = 6 e(S4 ) = 22 .3 = 12 e(S5 ) = 22 .3.5 = 60 e(S6 ) = 22 .3.5 = 60 e(S7 ) = 22 .3.5.7 = 420 e(S8 ) = 23 .3.5.7 = 840 In this list, the only n for which Sn has an element of order e(Sn ) are n = 1, 2. Thus S3 cannot have an element of order 6 since this requires a cycle decomposition into a 2-cycle and a 3-cycle, of total length 5 > 3. Similarly S4 cannot have an element of order 12 since 22 + 3 = 7 > 4, the group S5 cannot have an element of order 60 since 3 + 5 = 8 > 5, and so on. Finding inequalities that rule out n seems to become easier as n increases. So we conjecture that n = 1, 2 are the only cases for which Sn has an element of order e(Sn ). This is certainly true for n ≤ 8. Suppose that the interval [n/2, n] contains two distinct coprime numbers c, d. Then cd divides e(Sn ), but c + d > n so there do not exist disjoint cycles of lengths c, d. In these cases, Sn has no element of order e(Sn ). So we have to prove that such c, d always exist. Suppose first that n = 2m is even. By Bertrand’s postulate there exists a prime p ∈ [m, 2m] = [n/2, n]. Since n > 4 this prime is odd. Let c = p and let d be any even number in [m, 2m]; for example d = p + 1. Then c and d are coprime. When n = 2m+1 is odd, Bertrand’s postulate yields a prime p ∈ [m, 2m]. Possibly p = m, in which case there are two multiples of p in the range [m, 2m], namely m and 2m. Otherwise there is only one multiple of p in that range. Then we can take c = p and let d be any odd number in [m, 2m] other than p. 19.11* This has a star because it’s very difficult unless you find the right idea. If you do, it’s easy! Let K be a finite field, with k elements. We first ask: how many squares are there in K? As x runs through K, its square x2 runs through the squares. However, distinct values of x may lead to the same square. In fact, x2 = y2 if and only if (x + y)(x − y) = 0, so either y = x or y = −x. If the characteristic of K is 2, then y = x, so there are k distinct squares. Thus every element is a square, a = x2 . But this equals x2 + 02 . If the characteristic is not 2, then y = ±x and the two values are distinct unless x = 0. Therefore there are k+1 2 distinct squares. Let a ∈ K, and let S be the set of all squares in K. Let T be the set of all elements a − s where s ∈ S. Then both S and T have k+1 / since otherwise K would have at least k + 1 2 elements. Therefore S ∩ T 6= 0, elements. Hence there exist x, y ∈ K such that x2 = a − y2 , so a = x2 + y2 .

87

20 Regular Polygons 19.12 (a) F. (b) T. (c) F. (d) T. (e) T. (f) T. (g) T. (h) T. (i) F.

20 Regular Polygons

FIGURE 18: Construction of a parallel to AB through X. 20.1 See Figure 18. Given line AB and point X not on AB, take some point P on AB and draw PX. Draw a circle centre P cutting PX at F and AB at E. Draw a circle of the same radius centre X, cutting PX at G. Draw circle C with centre E, passing through F. Draw circle D, of same radius, with centre G. Let D cut the circle centre X at H. Then XH is parallel to AB and passes through X. 20.2 Numerical computations to 6 decimal places show that √ 4+ 5 cos √ 10 3−1 5 cos−1 10 −1 1 − cos−1 89 cos 2 √ −1 −1 4 + 5 tan 1 − tan 20 −1

≈ 0 · 897448 ≈ 0 · 698161 ≈ 0 · 571315 ≈ 0 · 483148

2π 7 2π 9 2π 11 2π 13

≈ 0 · 897598 ≈ 0 · 698132 ≈ 0 · 571199 ≈ 0 · 483322

Therefore the approximation is correct to 3 decimal places in each case, and error of at most 0 · 1%.

88 20.3 Factorise: 232 − 1 = (216 + 1)(216 − 1) = (216 + 1)(28 + 1)(28 − 1) = (216 + 1)(28 + 1)(24 + 1)(24 − 1) = (216 + 1)(28 + 1)(24 + 1)(22 + 1)(22 − 1) = (216 + 1)(28 + 1)(24 + 1)(22 + 1)(2 + 1)(2 − 1) = 65537 · 257 · 17 · 5 · 3 = 4294967295 The only known Fermat primes are 3, 5, 17, 257, 65537. The only known constructible regular n-gons, for n odd, are those for which n is a product of distinct Fermat primes. These are the divisors of 4294967295 = 232 − 1. 20.4 Let 18233954 F18233954 = 22 = 10a Taking logarithms, 218233954 log 2 = a log 10 so a = 218233954

log 2 ≈ 0 · 30103 × 218233954 log 10

Therefore log10 a ≈ 18233954 log10 2 + log10 (0 · 30103) ≈ 18233954 × 0 · 30103 − 0 · 52139 ≈ 5488970 and we obtain the estimate F18233954 ≈ 1010

5488970

This number has roughly 105488970 digits. If a sheet of paper can contain (say) 10,000 digits, we would need 105488970 sheets of paper. The known universe is roughly 1010 light years in radius, approximately 1026 metres. So the volume of the universe (assumed to be spherical) is roughly 4 × 1078 cubic metres. There are about 15 × 106 sheets of paper per cubic metre, so the maximum number of sheets that the universe can contain is 6 × 1085 . This is extremely small compared to 105488970 . So there isn’t room in the universe to write down all the decimal digits of F18233954 , let alone to calculate with them. 20.5 Working (mod 641) we have: 232 + 1 ≡ 24 228 + 1 ≡ (−54 )228 + 1 ≡ −(5 · 27 )4 + 1 ≡ −14 + 1 ≡ 0 so 641 divides 232 + 1.

89

20 Regular Polygons n+1

20.6 We have to show that 22 − 1 = Fn · · · F0 , and we argue by induction on n. When n = 0 this equation is 22 − 1 = F0 which is true since F0 = 3. Now n+1

22

n

n

− 1 = (22 + 1)(22 − 1) = Fn (Fn−1 · · · F0 )

by induction, and the result follows. Suppose that r divides both Fm , Fn where without loss of generality m < n. The formula implies that r divides 2. But r is odd because all Fermat numbers are odd, so r = 1 and we are done. Since distinct Fermat numbers are coprime, there are at least as many distinct primes as there are distinct Fermat numbers greater than 1. But this number is infinite. 20.7 The numbers we require are all multiples by powers of 2 of 1, 3, 5, 15, 17, 51, 85. These are 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 17, 20, 24, 30, 32, 34, 40, 48, 51, 60, 64, 68, 80, 85, 96 20.8 The construction is illustrated in Figure 19 (left). For comparison, Figure 19 (right) shows a primitive fifth root of unity ζ in the complex plane. In order for the construction to be correct, then when O is the origin and P0 = 1 we must have P1 = Re(ζ ).

FIGURE 19: Left: Construction of a regular pentagon. Right: Geometry of fifth root of unity ζ . It is well known that

√ 5−1 Re(ζ ) = 4 It remains to calculate the distance ON. Let ∠ODP0 = 2θ . Then tan 2θ = 2. If t = tan θ the trigonometry tells us that √ 2t t 2 + t − 1 = 0, so t = 5−1 . Therefore ON = tan 2θ = 1−t , so = 1. That is, t 2 2 1−t 2 1 2

tan θ =

√ 5−1 4 .

Since this equals Re(θ ), the proof is complete.

20.9* Let AB = 1 so AE = 21 . By Pythagoras, EB = Moreover, AH = AF.√ √ Now BH = 1 − 5−1 = 3−2 5 . 2

√ 5 2 .

Therefore AF =

√ 5−1 2 .

90 The area of the rectangle on AB and BH is

√ 2 ( 5−1 2 )

√ 3− 5 2 .

√ 3− 5 2 .

The area of the square on AH

is = Therefore H divides AB in extreme and mean ratio. 20.10 (a) T. (b) F. (c) T. (d) T. (e) T. (f) F. (g) T. (h) T. (i) T. (j) T. (k) F.

21 Circle Division 21.1 Replacing j by k in the definition of al we have p−2

αl =

k

∑ θ kl ζ a k=0

Therefore

p−2

θ − jl αl =

k

∑ θ (k− j)l ζ a k=0

Summing over l: p−2

p−2 p−2

∑ θ − jl αl =

∑ ∑ θ (k− j)l ζ a

l=0

l=0 k=0 p−2 p−2 ak

=

k

∑ ζ ∑ θ (k− j)l

k=0

l=0

We know that for 0 ≤ m ≤ p − 2 the sum 1 + θ m + · · · + θ (p−2)m is 0 unless m = 0, in which case its value is p − 1. Therefore the only terms that survive in the double k sum add to (p − 1)ζ a . Hence ! p−2 1 ak − jl ζ = ∑ θ αl p − 1 l=0 21.2 By (21.11) t 24 − 1 = Φ24 (t)Φ12 (t)Φ8 (t)Φ6 (t)Φ4 (t)Φ3 (t)Φ2 (t)Φ1 (t) The table of cyclotomic polynomials contains all of the factors on the right hand side except Φ24 (t). Factorise: t 24 − 1 = (t 12 + 1)(t 12 − 1) = (t 4 + 1)(t 8 − t 4 + 1)(t 6 + 1)(t 6 − 1) = (t 4 + 1)(t 8 − t 4 + 1)(t 2 + 1)(t 4 − t 2 + 1)(t 3 + 1)(t 3 − 1) = (t 4 + 1)(t 8 − t 4 + 1)(t 2 + 1)(t 4 − t 2 + 1)(t + 1)(t 2 − t + 1)(t − 1)(t 2 + t + 1)

21 Circle Division

91

Comparing and dividing common factors, we are left with Φ24 (t) = t 8 − t 4 + 1 21.3 The proof is that of the Vandermonde-Gauss Theorem, Theorem 21.2, except we must provide specific estimates for the radical degree. The case d = 2 requires no radicals. When d = 3 we need square roots, which is why we take the maximum of 2 and (d − 1)/2. Proceed by induction and observe that if m < (d − 2)/2 then certainly m < (d − 1)/2. We can therefore reduce the problem to the case when d = p is an odd prime > 3. Then (21.9) expresses ζ in terms of known quantities and (p − 1)th roots. But p − 1 is even, so we may use a square roots and a ((p − 1)/2)th root. This completes the proof. 21.4 Let θ = π6 , so that ζ = cos θ + i sin θ is a primitive 12th root of unity. By definition Φ12 (t) = (t − ζ )(t − ζ 5 )(t − ζ 7 )(t − ζ 11 ) = [(t − ζ )(t − ζ 11 )][(t − ζ 5 )(t − ζ 7 )] = (t 2 − 2 cos θ t + 1)(t 2 − 2 cos 5θ t + 1) = t 4 + (−2 cos θ − 2 cos 5θ )t 3 + (2 + 4 cos θ cos 5θ )t 2 +(−2 cos θ − 2 cos 5θ )t + 1 Now

√ 3 cos θ = 2

√ 3 cos 5θ = − 2

so cos θ + cos 5θ = 0 so the cubic and linear terms vanish. Finally, 3 2 + 4 cos θ cos 5θ = 2 − 4 = −1 2 and Φ12 (t) = t 4 − t 2 + t. 21.5 We know that Φ12 (t) = t 4 − t 2 + 1 = (t − ζ )(t − ζ 5 )(t − ζ 7 )(t − ζ 11 ) where ζ = e2πi/12 . Φ12 (t) has no linear factors over Q since ζ 6∈ Q. Therefore, if it is reducible, it must be the product of two quadratic factors over Q. Since Q ⊆ R the only possibility is that the factors are (t − ζ )(t − ζ 11 ) and (t − ζ 5 )(t − ζ 7 ). If one of these quadratics lies in Q(t) they both do, since their product does. It therefore suffices to consider √ (t − ζ )(t − ζ 11 ) = t 2 − 2Re(ζ ) + 1. However, Re(ζ ) = 23 6∈ Q. 21.6 For any primitive kth root of unity ζ we have the basic identity 1 + ζ + ζ 2 + · · · + ζ k−1 = 0

92 If l = 0 then the sum 1 + θ l + θ 2l + · · · + θ (p−2)l is p − 1. So we may assume 1 ≤ l ≤ p − 2. If θ is a primitive (p − 1)th root of unity, then ζ = θ l is a primitive ((p − 1)/d)th root of unity, where d = gcd (p − 1, l). Thus the sum 1 + θ l + θ 2l + · · · + θ (p−2)l is a rearrangement of d(1 + ζ + ζ 2 + · · · + ζ (p−1)/d−1 ), which is 0. 21.7 When p is a prime, Lemma 20.10 implies that Φ p (t) = t p−1 +t p−2 + · · · +t + 1. Here the coefficients are 0 or 1. 21.8 When pk is prime power, the argument of Lemma 20.11 implies that Φ pk (t) = k−1

k−1

k−1

t (p−1)p + t (p−2)p + · · · + t p + 1. Again the coefficients are 0 or 1. 21.9 The proof rests on a simple group-theoretic lemma: Lemma S21.1 Let G be a finite group and let N be a normal subgroup of G. Suppose that g ∈ G. Let the order of the coset Ng ∈ G/N be a, and let the order of ga in N be b. Then the order of g is ab. Proof. Let c be the order of g. We know that (ga )b = 1 so c|ab. We claim that in fact c = ab. The powers g, g2 , . . . , ga−1 do not lie in N by the definition of ‘order’ in G/N. However, ga ∈ N. Therefore gd ∈ N if and only if a|d. To see why, suppose that d = qa + r where 0 ≤ r ≤ a − 1. Then gd and gr lie in the same coset of N, because gd = gqa+r = (ga )q gr = 1gr = gr . Thus a|c. Let c = ma. Then gc = gma = (ga )m , which by the definition of ‘order’ is not 1 if m < b. So c ≥ ab, whence c = ab. Having proved the lemma, we return to the exercise. Suppose that m is odd. Then the Euler function φ (m) is even. We claim that ζ is a primitive mth root of unity if and only if −ζ is a primitive 2mth root of unity. Let G be the multiplicative group of 2mth roots of unity, and let N be the subgroup of mth roots of unity. Then N / G and |G/N| = 2. If ζ ∈ N has order m then (−ζ )m = −1 since m is odd, so the order of −ζ modulo N is 2. By Lemma S21.1 −ζ has order 2m, so is a primitive 2mth root of unity. For the converse, since m is odd, the only element of order 2 in G is −1. Therefore the order of ζ is half that of −ζ , namely m. Thus ζ is a primitive mth root of unity. This proves the claim. Now let ζ be some fixed primitive mth root of unity. We have m−1

Φm (t) =

∏ (t − ζ j ) j=0

and

m−1

Φ2m (t) =

∏ (t − (−ζ ) j ) j=0

But

m−1

Φm (−t) =

∏ (−t − ζ j ) j=0

21 Circle Division

93

Since m is odd, the result follows. 21.10 Let p > q be odd primes. Then (21.7.12) implies that t pq − 1 = Φ pq (t)Φ p (t)Φq (t)Φ1 (t) Therefore Φ pq (t) =

(t − 1)(t pq − 1) (t p − 1)(t q − 1)

Rewrite this in the form Φ pq (t) =

t p(q−1) + t p(q−2) + · · · + t p + 1 t q−1 + t q−2 + · · · + t + 1

It is not immediately obvious how to compute the quotient here, so we first try a special case — say p = 7, q = 5. By polynomial long division we find that t 28 + t 21 + t 14 + t 7 + 1 t4 + t3 + t2 + t + 1 24 = t − t 23 + t 19 − t 18 + t 17 − t 16 + t 14 − t 13 +t 12 − t 11 + t 10 − t 8 + t 7 − t 6 + t 5 − t + 1

Φ35 (t) =

The coefficients are indeed 0, ±1 in this case, but the pattern is not clear. However, there are some useful clues, which become clearer if you actually carry out the division by hand. Start with the pattern t 24 − t 23 + t 19 − t 18 which arises when we try to divide the leading term t 28 by t 4 + t 3 + t 2 + t + 1. Note that 19 = 24 − 5. The natural continuation of this pattern (the next few steps in the long division of t 28 ) would give a(t) = t 24 − t 23 + t 19 − t 18 + t 14 − t 13 + t 9 − t 8 + t 4 − t 3 The term t 14 − t 13 is encouraging, but the pattern does not continue. The reason is that other terms t 21 ,t 14 ,t 7 ‘interfere’ with the pattern. So we try the same pattern on those terms: b(t) = t 17 − t 16 + t 12 − t 11 + t 7 − t 6 + t 2 − t c(t) = t 10 − t 9 + t 5 − t 4 d(t) = t 3 − t 2 The sum a(t) + b(t) + c(t) + d(t) is very nearly equal to Φ35 (t). In fact, Φ35 (t) = a(t) + b(t) + c(t) + d(t) + 1 If we separate a(t), b(t), c(t), d(t) into positive and negative parts, we find that the positive parts have no common term. The negative parts are all degree 1 less, so the same is true for them. This implies that the coefficients are all ±1 or 0.

94 This example suggests the following general argument. (It can be made more elegant, but is probably clearer in its present form.) Define Ψ = t q−1 + t q−2 + · · · + t + 1, so that t p(q−1) + t p(q−2) + · · · + t p + 1 Φ pq (t) = Ψ We know that (t + 1)Ψ = t q − 1 so for any k ≥ 0 (t k+1 − t k )Ψ = t q+k − t k which we rewrite as t q+k = t k + (t k+1 − t k )Ψ Putting k = l − q we get t l = t l−q + [t l−q+1 − t l−q ]Ψ = t l−2q + [t l−q+1 − t l−q + t l−2q+1 − t l−2q ]Ψ ... = t l−sq + [t l−q+1 − t l−q + · · · + t l−sq+1 − t l−sq ]Ψ

provided sq < l. Set r(l) = l (mod q), interpreted as the remainder on dividing l by q, so that 0 ≤ r(l) ≤ q − 1. Then t l = t r(l) + [t l−q+1 − t l−q + · · · + t l−r(l)q+1 − t l−r(l)q ]Ψ Now specialise to the case l = j p (we will use j = q − 1, q − 2, . . . , 1). Then t j p = t r( j p) + [t j p−q+1 − t j p−q + · · · + t r( j p)+1 − t r( j p) ]Ψ so

t j p t r( j p) = + [t j p−q+1 − t j p−q + · · · + t r( j p)+1 − t r( j p) ] Ψ Ψ

That is, Φ pq (t) =

q−1 r( j p) t

∑

j=0

Ψ

q−1

+ ∑ [t j p−q+1 − t j p−q + · · · + t r( j p)+1 − t r( j p) ] j=0

Now, as j runs from 0 to q − 1 the remainders r( j p) run through the same numbers 0 to q − 1 exactly once, because p is prime to q. Therefore q−1

∑ t r( j p) = Ψ

j=0

95

21 Circle Division and the first sum is 1. Split the second sum into its positive and negative terms: q−1

q−1

Φ pq (t) = 1+ ∑ [t j p−q+1 +t j p−2q+1 +· · ·+t r( j p)+1 ]− ∑ [t j p−q +t j p−2q +· · ·+t r( j p) ] j=0

j=0

which we abbreviate to Φ pq (t) = 1 + Σ1 − Σ2 . We claim that all terms in Σ1 are distinct, and the same goes for Σ2 . (However, the same power of t may occur in both Σ j : this does not matter for the proof since they then cancel.) It is enough to prove the claim for Σ2 since Σ1 = tΣ2 . Suppose, therefore, that j1 p − aq = js p − bq. Then ( j1 − j2 )p = (a − b)q. Therefore j1 ≡ j2 (mod q). But 0 ≤ j1 , j2 ≤ q − 1, so j1 = j2 so a = b and the claim is proved. Finally, note that every term in Σ1 has degree ≥ 1, so the term 1 does not appear in Σ2 . But now Φ pq (t) is the difference between two polynomials, each of which has coefficients 0 or 1 only. The coefficients of the difference are therefore 1, 0, or −1. This completes the proof that the coefficients of Φ pq (t) are 0, ±1 when p, q are distinct odd primes. 21.11 The relation is k−1 Φ pk a (t) = Φ pa (t p ) k−1

To prove this, we claim that ζ is a primitive (pk a)th root of unity if and only if ζ p is a primitive (pa)th root of unity. This follows from Lemma S21.1 in Exercise 21.9, using a similar argument to the one in that exercise. The formula is then proved in exactly the same way as the formula in Exercise 21.9. 21.12 The cyclotomic polynomial Φ105 (t) has degree 48 and begins 1 + t + t 2 − t 5 − t 6 − 2t 7 · · · so has −2 as a coefficient. A. Migotti discovered this in 1883. Further investigation shows that Φ1155 (t) has 3 as a coefficient in several places, and Φ3.5.7.11.13 (t) has a term 23t 2294 . Around 1935, in a letter to Edmund Landau, Issai Schur proved that the coefficients of Φn (t) can be arbitrarily large for sufficiently large n. There is an accessible discussion in Wilson and Gray (2001) pages 96–99. 21.13 Let p be prime. A number m in the range 0 ≤ m < pk is not prime to p if and only if m is divisible by p. The number of such k is pk /p = pk−1 . Therefore φ (pk ) = pk − pk−1 = (p − 1)pk−1 . Now suppose that r, s are coprime. Suppose that k is prime to rs. We claim that there exist unique integers a, b with 0 ≤ a < r, 0 ≤ b < s, with a prime to r and b prime to s, such that k ≡ as + br (mod rs). This will prove the required result, since the number of such pairs (a, b) is φ (r)φ (s). First we prove the analogous result without the conditions that k is prime to rs, a is prime to r, and b prime to s. There are at most rs numbers of the form as + br. They are all distinct modulo rs, because if as + br = cs + dr (mod rs) then clearly a ≡ c (mod r) and ba ≡ d (mod s). Thus there are exactly rs numbers of the form as + br, all distinct modulo rs. But there are only rs possible values for k modulo rs, so every k occurs. It is also easy to see that as + br is prime to rs if and only if a is prime to r and b is prime to s. Indeed, if a, r have a common factor h then h|as + br and h|rs. A similar argument applies of b and s have a common factor. The result follows.

96 m

l 1 Finally, we put the two parts together. If n = pm 1 · · · pl then

l

m

∏ φ (p j j )

φ (n) =

j=1 l

m −1

∏ (p j − 1)p j j

=

j=1

21.14 Dividing the above equation by n we get φ (n) = n

l

l

m

m

∏ φ (p j j )/ ∏ p j j j=1

j=1

l

=

∏ (p j − 1)p−1 j j=1 l

  1 = ∏ 1− pj j=1 21.15 Z∗n is a group, of order φ (n). The order of any element of a finite group divides the order of the group. Therefore every a ∈ Z∗n satisfies the equation aφ (n) = 1. In ‘mod n’ language, this is what we have to prove. 21.16 Suppose that m is fixed and we want to solve φ (n) = m. Then l

m j −1

m = ∏ (p j − 1)p j j=1

For each j we have (p j − 1)|m, so p j ≤ m + 1 and only finitely many primes p j m −1

can occur. Then p j j |m so m j is bounded for each j. Therefore only finitely many solutions are possible. 21.17 Experiment: φ (1) = 1 φ (1) + φ (2) = 1 + 1 = 2 φ (1) + φ (2) + φ (3) = 1 + 1 + 1 = 3 φ (1) + φ (2) + φ (4) = 1 + 1 + 2 = 4 φ (1) + φ (5) = 1 + 4 = 5 φ (1) + φ (2) + φ (3) + φ (6) = 1 + 1 + 1 + 3 = 6 So it looks as though the theorem should be

∑ φ (d) = n d|n

Further experiment verifies this for larger n.

21 Circle Division

97

One way to prove the formula is to observe that it is multiplicative. That is, if we define ψ(n) = ∑ φ (d) d|n

then ψ(rs) = ψ(r)ψ(s) whenever r, s are coprime. This follows easily since d|rs if and only if d = kl where k|r and l|s (and k, l are coprime). Moreover, each divisor d of rs can be represented uniquely in this form. The conjectured theorem is true if n = p j is a prime power, because then ψ(n) = φ (1) + φ (p) + φ (p2 ) + · · · + φ (p j ) = 1 + (p − 1) + (p − 1)p + · · · + (p − 1)p j−1 = 1 + (p − 1)(1 + p + · · · + p j−1 ) pj −1 = 1 + (p − 1) p−1 = 1 + (p j − 1) = p j = n Now use induction on the number of distinct prime factors of n, using the multiplicativity of ψ. For another proof, see Graham, Knuth, and Patashnik (1994) pages 134–135. 21.18 Since n is odd, it is prime to 4. Therefore φ (4n) = φ (4)φ (n) = 2φ (n). 21.19 Checking is trivial. The theorem is presumably n

∑ d = 2 φ (n)

d⊥n

where d ⊥ n means ‘d is prime to n’. To prove it, we use a variant of the old trick for summing an arithmetic sequence. Note that d is prime to n if and only if n − d is prime to n. Therefore 2∑d = d⊥n

∑ d + ∑ (n − d) d⊥n

=

d⊥n

∑n d⊥n

= nφ (n) 21.20* The elements of Z∗24 are 1, 5, 7, 11, 13, 17, 19, 23. Their squares are 1, 25, 49, 121, 169, 289, 361, 529, all of which are congruent to 1 modulo 24. Suppose that n > 24 has this property. If any of 2, 3, 5 is prime to n then 4, 9, or 25 must be congruent to 1 modulo n, which is impossible. Therefore n is divisible by 2, 3, and 5. We claim inductively that n is divisible by the product of the first k primes p1 . . . pk . This is true for k = 3. Assume, therefore, that p1 . . . pk |n. By Bertrand’s postulate, pk+1 < 2pk . Either pk+1 |n and we are done, or p2k+1 < 4p2k but must be congruent to 1 modulo n, so

98 p2k+1 > n. We claim that n > 4p2k . This is equivalent to p2 . . . pk−1 > pk , and since p2 = 3 this follows by Bertrand’s postulate. Thus pk+1 |n as required. Therefore n is divisible by all primes, a contradiction since n is finite. Finally, by inspection of the numbers < 24, the only n with the same property as 24 are its divisors 2, 3, 4, 6, 8, 12. Clearly these have the required property. All odd numbers greater than 3 are ruled out since 22 = 4. The number 10 is ruled out since 32 = 9, and the same goes for all even numbers prime to 3. That leaves only 18, but 52 ≡ 7 (mod 18). 21.21 Let ζ be a primitive 19th root of unity. Define r1 = ζ + ζ 18 r4 = ζ 4 + ζ 15 r7 = ζ 7 + ζ 12

r2 = ζ 2 + ζ 17 r5 = ζ 5 + ζ 14 r8 = ζ 8 + ζ 11

r3 = ζ 3 + ζ 16 r6 = ζ 6 + ζ 13 r9 = ζ 9 + ζ 10

All r j are real. Experiment shows that 2 is a primitive root (mod 19). Its powers in order from 20 are: 1 2 4 8 16 13 7 14 9 18 17 15 11 3 6 12 5 10 Therefore, taking every third term, we define s1 = ζ + ζ 8 + ζ 7 + ζ 18 + ζ 11 + ζ 12 = r1 + r8 + r7 s2 = ζ 2 + ζ 16 + ζ 14 + ζ 17 + ζ 3 + ζ 5 = r2 + r3 + r5 s3 = ζ 4 + ζ 13 + ζ 9 + ζ 15 + ζ 6 + ζ 10 = r4 + r6 + r9 and again these are real since the r j are. We claim that the s j are the roots of a cubic equation over Z. This follows by computing the elementary symmetric polynomials in the s j as expressions in the r j , and then using the definition of the r j . We omit the details. Since this cubic has three real roots, it can be solved for the s j by angle-trisection. Computing the elementary symmetric polynomials in r1 , r8 , r7 we find that they are all expressible in terms of the s j . Again we omit the details. Since the r j are real, a further trisection solves for the r j . Finally, we obtain ζ by solving the quadratic ζ + ζ −1 = r1 . 21.22 The list from 100 to 1000 is: 109, 163, 193, 257, 433, 487, 577, 769 21.23 The list from 1000 to 1,000,000 is: 1153, 1297, 1459, 2593, 2917, 3457, 3889, 10369, 12289, 17497, 18433, 39367, 52489, 65537, 139969, 147457, 209953, 331777, 472393, 629857, 746497, 786433, 839809, 995329 21.24 (1) Suppose a = ck and b = dk where k > 1 is odd. Then 2a 3b + 1 = 2ck 3dk + 1 = (2c 3d )k + 1. But when k is odd, xk + 1 is divisible by x + 1.

22 Calculating Galois Groups

99

(2) Suppose that 2a 3b + 1 ≡ 0 (mod 5). Then 2a 3b ≡ −1

(mod 5)

−1 b

2 (2 ) ≡ −1

(mod 5)

a−b

(mod 5)

a

2

≡ −1

The powers of 2 modulo 5 are 1, 2, 4, 3 (repeat). So a − b ≡ 2 (mod 4). (3) Suppose that 2a 3b + 1 ≡ 0 (mod 7). The powers of 2 modulo 7 are 1, 2, 4 (repeat). Now 2a 3b ≡ −1 (mod 7) 2 (−22 )b ≡ −1 (mod 7) a

If b is even then 2a+2b ≡ −1 (mod 7) which is impossible. If b is odd then 2a+2b ≡ 1 (mod 7) which occurs if and only if a + 2b ≡ 0 (mod 3). (4) Similar calculations lead to: p = 11: the condition for divisibility by p is a + 8b ≡ 5 (mod 10). p = 13: the condition for divisibility by p is a + 4b ≡ 6 (mod 12). p = 17: the condition for divisibility by p is b − 2a ≡ 8 (mod 16). p = 19: the condition for divisibility by p is a + 13b ≡ 9 (mod 1). (5) Suppose that 2a 3b + 1 ≡ 0 (mod 23). Multiply by 3−b to get 2a + 3−b ≡ 0 (mod 23). However, the powers of 2 modulo 23 are: 1 2 4 8 16 9 18 13 3 6 12 and the powers of 3 modulo 23 are: 1 3 9 4 12 No pair of these numbers adds to a multiple of 23. 21.25 (a) T. (b) T. (c) F. (d) F. (e) T. (f) F. (g) T. (h) T. (i) T. For (g): 483729409 is prime. Subtract 1 to get 483729408, with prime factors 213 310 .

22 Calculating Galois Groups 22.1 In the notation of the chapter, ∆ = δ 2 . Every element of G permutes the zeros of f , so it must map δ to ±δ . If every γ ∈ G fixes δ then δ lies in the fixed field of G, which is K. Therefore some element of G maps δ to −δ . Identifying G with a group of permutations on the zeros of f , we know that An fixes δ , so it fixes K(δ ). Since some element of G maps δ to −δ , G 6⊆ An . Therefore G ∩ An is a normal subgroup of G of index 2.

100 If ∆ is not a square in K then δ 6∈ K. Since the characteristic is not 2, the polynomial t 2 − ∆ is separable. Since δ 2 = ∆ ∈ K we have [K(δ ) : K] = 2. Now G ∩ An fixes K(δ ), and by the Galois correspondence the fixed field cannot be larger than K(δ ). So A†n = K(δ ) as required. 22.2* Suppose that the quartic f (t) = t 4 − wt 3 + xt 2 − yt + z = 0 has roots a, b, c, d so that w = a+b+c+d x = ab + ac + ad + bc + bd + cd y = abc + abd + acd + bcd z = abcd Define P = ab + cd Q = ac + bd R = ad + bc and let g(t) = (t − P)(t − Q)(t − R) = t 3 − Xt 2 +Y t − Z where X = P+Q+R Y = PQ + PR + QR Z = PQR The discriminant of f is ∆ = [(a − b)(a − c)(a − d)(b − c)(b − d)(c − d)]2 and the discriminant of g is ∆0 = [(X −Y )(X − Z)(Y − Z)]2 Explicit calculation (easily done using computer algebra) shows that ∆ = ∆0 . This is not quite the miracle it seems, since f and g are very closely related. There is an explanation of this using Galois theory, but we shall not give it here. At any rate, we can find ∆ if we can express X,Y, Z in terms of w, x, y, z, by using the formula for the discriminant of a cubic. That formula is ∆0 = X 2Y 2 − 4X 3 Z + 18XY Z − 4Y 3 − 27Z 2

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22 Calculating Galois Groups

101

See Edwards (1984) or use a Tschirnhaus transformation to eliminate the coefficient of t 2 , compute the discriminant, and invert the Tschirnhaus transformation. Calculation of X: Clearly X = x. Calculation of Y : It is useful to write Σ(ai b j ck . . .) to denote the sum of all permutations of the monomial ai b j ck . . .. Then Y = Σ(a2 bc) It is easy to see that Σ(abc)Σ(a) = Σ(a2 bc) + 4Σ(abcd) so Y = Σ(a2 bc) = Σ(abc)Σ(a) − 4Σ(abcd) = yw − 4z Calculation of Z: A short calculation shows that Z = Σ(a2 b2 c2 ) + Σ(a3 bcd) Now Σ(a3 bcd) = Σ(abcd)Σ(a2 ) = (w2 − 2x)z because Σ(a2 ) = (Σ(a))2 − 2Σ(ab) = w2 − 2x Also Σ(a2 b2 c2 ) = (Σ(abc))2 − 2Σ(a2 b2 cd) = y2 − 2Σ(abcd)Σ(ab) = y2 − 2xz Therefore Z = y2 − 2xz + (w2 − 2x)z Substitute in (22) to get ∆ = w2 x2 y2 − 4x3 y2 − 4w3 y3 + 18wxy3 − 27y4 − 4w2 x3 z + 16x4 z +18w3 xyz − 80wx2 yz − 6w2 y2 z + 144xy2 z − 27w4 z2 + 144w2 xz2 −128x2 z2 − 192wyz2 + 256z3 If you wish, you may change to signs of w, y to convert this to a formula for the discriminant of t 4 + wt 3 + xt 2 + yt + z. 22.3 Let f (t) = (t − α1 )(t − α2 )(t − α3 ) ∈ Q[t]) be irreducible over Q, where the α j ∈ C. Let Σ = Q(α1 , α2 , α3 ) be the splitting field of f . Define φ = α12 α2 + α22 α3 + α32 α1 ψ = α1 α22 + α2 α32 + α3 α12

102 Since f is irreducible, the Galois group of f (considered as as permutation group on the α j and hence identified with a subgroup of S3 ) is either A3 or S3 . Thus in either case the fixed field notation A†3 makes sense. We have to prove that A†3 = Q(φ , ψ). Since A3 cycles the indices 1, 2, 3 it is clear that φ , ψ lie in A†3 . Even permutations of the indices fix φ and ψ, whereas odd permutations swap them. Thus (as discussed in the text) both φ + ψ and φ ψ lie in Q. Therefore [Q(φ , ψ) : Q] ≤ 2. The extension Q(φ , ψ)/Q is normal, so Q(φ , ψ)∗ / Γ. There are two cases: either the Galois group Γ of f over Q is A3 , or it is S3 . If Γ = A3 the only possibility is Q(φ , ψ)∗ = A3 . By the Galois correspondence, † A3 = Q(φ , ψ)∗† = Q(φ , ψ) as required. Alternatively, Γ = S3 . Now Q(φ , ψ)∗ = A3 or S3 , so either Q(φ , ψ) = A†3 or Q(φ , ψ) = S†3 . We claim that the second case cannot occur. Suppose, for a contradiction, that it does. The odd permutations map φ to ψ, but now they also fix φ and ψ, so φ = ψ. A quick calculation shows that δ = (α1 − α2 )(α2 − α3 )(α3 − α1 ) = ψ − φ so δ = 0. Therefore αi = α j for distinct i, j, and f has a multiple zero. This contradicts irreducibility of f . Therefore the second case does not occur, and again Q(φ , ψ)∗ = A3 . 22.4 Call the determinant V and think of it as a polynomial in the α j . For the purposes of the proof (we are verifying a general polynomial identity) we may consider the α j to be independent trascendentals (prove this case, then substitute). Clearly V has two equal columns if αi = α j for i 6= j, so it vanishes, which means that V is divisible by αi − α j . Therefore V is divisible by δ . But V and δ have the same total degree, so V = qδ for some q ∈ K. The main diagonal of the determinant contributes a term α2 α32 . . . αnn−1 , and we compare coefficients to deduce that q = ±1. The precise sign does not matter, but it can easily be deduced; in fact we did this in Exercise 2.5, and it is (−1)−n(n+1)/2 . If W is the determinant of the transposed matrix then W = ±δ , so VW = ∆ whatever the sign. The product of the two matrices is clearly n λ1 . . . λn−1 λ1 λ2 . . . λn .. .. . . .. . . . . λn−1 λn . . . λ2n−2 as required. Note that the n at top left is ‘really’ λ0 . When n = 2 we have 2 λ1 = 2λ2 − λ12 ∆ = λ1 λ2

22 Calculating Galois Groups

103

Exercise 8.5 implies that a1 + a2 λ1 = 0 2a0 + a1 λ1 + a2 λ2 = 0 and routine computations show that ∆=

a21 − 4a0 a2 a22

as we expect. The cases n = 3, 4 are similar, but the calculations are more extensive. We summarise the results. n=3 The equations that determine the λ j are: a2 + a3 λ1 = 0 2a1 + a2 λ1 + a3 λ2 = 0 3a0 + a1 λ1 + a2 λ2 + a3 λ3 = 0 a0 λ1 + a1 λ2 + a2 λ3 + a3 λ4 = 0 Solve the first for λ1 , then the second for λ2 , and so on to get λ1 = − λ2 =

a22 2a1 − a3 a23

λ3 = − λ4 =

a2 a3

a32 3a1 a2 3a0 + 2 − a3 a3 a33

a42 4a1 a22 2a21 4a0 a2 − + + 3 2 a3 a4 a3 a4 a3 a4 a3 a4

Finally, 3 λ1 λ2 ∆ = λ1 λ2 λ3 λ2 λ3 λ4 leading to ∆=

2a62 14a1 a42 27a21 a22 12a0 a32 8a31 + − − + 3 a43 a43 a3 a63 a53 +

2a6 14a1 a4 42a0 a1 a2 27a20 − 2 + 52 − 4 2 3 a3 a3 a4 a3 a3 a4

+

28a21 a22 8a0 a32 12a31 24a0 a1 a2 + 3 − 2 − a3 a4 a23 a4 a33 a4 a3 a4

104 n=4 The equations that determine the λ j are: a3 + a4 λ1 = 0 2a2 + a3 λ1 + a4 λ2 = 0 3a1 + a2 λ1 + a3 λ2 + a4 λ3 = 0 4a0 + a1 λ1 + a2 λ2 + a3 λ3 + a4 λ4 = 0 a0 λ1 + a1 λ2 + a2 λ3 + a3 λ4 + a4 λ5 = 0 a0 λ2 + a1 λ3 + a2 λ4 + a3 λ5 + a4 λ6 = 0 Solve the first for λ1 , then the second for λ2 , and so on to get λ1 = − λ2 =

a23 2a2 − a4 a24

λ3 = −

λ4 =

a33 3a2 a3 3a1 + 2 − a4 a4 a34

a43 4a2 a23 2a22 4a1 a3 4a0 − 3 + 2 + 2 − a4 a44 a4 a4 a4

λ5 = − λ6 =

a3 a4

a53 5a2 a33 5a22 a3 5a1 a23 5a1 a2 5a0 a3 + 4 − 3 − 3 + 2 + 2 a4 a4 a4 a4 a4 a54

a63 6a2 a43 9a2 a23 6a1 a33 2a32 12a1 a2 a3 − 5 + 4 + 4 − 3 − a4 a4 a4 a34 a64 a4 −

6a0 a23 3a21 6a0 a2 + 2 + 2 a4 a4 a34

Finally, 3 λ ∆ = 1 λ2 λ3

λ1 λ2 λ3 λ4

λ2 λ3 λ4 λ5

λ3 λ4 λ5 λ6

leading to ∆=

a21 a22 a23 4a0 a32 a23 4a31 a33 18a0 a1 a2 a33 27a20 a43 − − 6 + − a64 a64 a4 a64 a64 −

4a21 a32 16a0 a42 18a31 a2 a3 80a31 a2 a3 6a0 a21 a32 144a20 a2 a23 + + − − + a54 a54 a54 a54 a54 a54

−

27a41 144a0 a21 a2 128a20 a22 192a20 a1 a3 256a30 + − − + a44 a44 a44 a44 a34

22 Calculating Galois Groups

105

22.5* We start from Exercise 22.4. Using Exercise 8.6 and induction, we find that λ1 = λ2 = · · · = λn−2 = 0 λn−1 = (1 − n)a λn = −nb λn+1 = λn+2 = · · · = λ2n−3 = 0 Now all we have to do is substitute these values in the determinant VW and expand: the only nonzero terms lead to the stated result. In detail: let A = −(n − 1)p Then

B = −nq

n 0 ∆ = ... 0 A

C = (n − 1)p2

A B .. . A · · · 0 0 B ··· 0 C 0 0 .. .

··· ··· .. .

0 A .. .

Expand ∆ along the top row to obtain 0 0 ··· A B . .. . . .. .. . . . . n+1 .. ∆= + (−1) A 0 A ··· 0 0 A B ··· 0 C

A .. . A · · · 0 B ··· 0 0 ··· .. . . . .

which we write as n∆1 +(−1)n+1 A∆2 , where the determinants ∆1 , ∆2 are respectively obtained from ∆ by deleting row 1 and column 1, and row 1 and column n. Recall that a general n × n determinant can be expressed as |ai j | =

∑

sgn(σ )a1,σ (1) a2,σ (2) · · · an,σ (n)

σ ∈Sn

Each term a1,σ (1) a2,σ (2) · · · an,σ (n) contains exactly one entry from each row and one from each column. We claim that the only such terms that do not vanish are An−2C and Bn−1 for ∆1 , and An−1 for ∆2 . To see this, look at column n − 1 in ∆1 . Either we choose C or B. If we choose C, then deleting row n − 1 and column n − 1 the only choice in column 2 is A, and so on. If we choose B, then deleting row 1 and column n − 1 the only choice in column n − 2 is B, and so on. For ∆2 start with column 1, where we must choose A, then delete row n − 1 and column 1 and repeat. In summary, ∆ = n[(±)1 An−2C(±)2 Bn−1 ] + (−1)n+1 (±)3 An where the signs (±)1 , (±)2 , (±)3 are yet to be worked out.

(23)

106 Next, we find these signs. The corresponding permutation σ ∈ Sn is:   1 2 3 ... n−2 n−1 (±)1 σ = n−2 n−3 n−4 ... 1 n−1   1 2 3 ... n−2 n−1 (±)1 σ = n−1 n−2 n−3 ... 2 1   1 2 3 ... n−2 n−1 (±)1 σ = n−1 n−2 n−3 ... 2 1 For general n let  σn =

1 2 3 ... n−2 n−1 n−1 n−2 n−3 ... 2 1



and let s(n) = sgn(σn ). Expressing σn as a product of transpositions and counting the parity, we find that  1 if n ≡ 0, 1 (mod 4) s(n) = −1 if n ≡ 2, 3 (mod 4) Now (±)1 = s(n − 2) and (±)2 = (±)3 = s(n − 1), so (23) becomes ∆ = n[s(n − 2)An−2C + s(n − 1)Bn−1 ] + (−1)n+1 + s(n − 1)An Now consider separately the four cases n ≡ 0, 1, 2, 3 (mod 4). For instance, if n ≡ 0 (mod 4) then (−1)n+1 = −1, s(n − 2) = −1, s(n − 1) = −1, so ∆ = n[−an−2C − Bn−1 ] + An = n[−(n − 1)n−2 pn−2 (n − 1)p2 + nn−1 qn−1 ] = nn qn−1 − n(n − 1)n−1 pn + (n − 1)n pn = nn qn−1 − (n − 1)n−1 pn The other three cases are similar. 22.6* Clearly all the subgroups listed are transitive. Any conjugate of a transitive subgroup is transitive, so we may consider subgroups up to conjugacy. The only conjugacy classes not included in the list (some extensive but routine work needed here) are those corresponding to 1 Z2 = h(12)i Z2 = h(12)(34)i Z3 = h(123)i D3 = h(123), (12)i None of these is transitive.

23 Algebraically Closed Fields

107

22.7* The transitive subgroups of S4 are S4 , A4 , D4 , V, Z4 . Theorem 22.7 implies that Γ ⊆ A4 if and only if ∆ is a square (of a rational number). Thus: If ∆ is not square then Γ = S4 , D4 , Z4 If ∆ is square then Γ ∼ = A4 , V Case 1: ∆ is a square. If g is irreducible, the presence of elements with cubic minimal polynomial implies that |Γ| is divisible by 3. Therefore Γ = A4 . On the other hand, if g is reducible, then |Γ| is not divisible by 3. Therefore Γ = V. This deals with (b, d). Case 1: ∆ is not a square. If g is irreducible, the presence of elements with cubic minimal polynomial implies that |Γ| is divisible by 3. Therefore Γ = S4 . This is case (a), provided we observe in addition that if g is reducible then |Γ| is not a multiple of 3. We are left with the groups Γ = D4 , Z4 , and must distinguish these cases. The order of Γ is 8, 4 respectively. Both groups contain a 4-cycle. If f is irreducible over M then of square roots leads to an extra Q-automorphism determined √ the presence √ by m 7→ − m where m ∈ M is not square. If f is reducible over M then no such Q-automorphism occurs. Therefore |Γ| is 8 in the first case and 4 in the second. This deals with cases (c, e) and completes the classification. 22.8 Let ρ = (1 2 3), σ = (4 5 6), τ = (1 4). It is enough to exhibit elements mapping 1 to 1, 2, 3, 4, 5, 6. Now ρ(1) = 2, ρ 2 (1) = 3, τ(1) = 4, σ τ(1) = 5, σ 2 τ(1) = 6. 22.9 (a) T. (b) F. (c) F. (d) T. (e) T. (f) T.

23 Algebraically Closed Fields 23.1 Suppose for a contradiction that A5 has a subgroup H of order 15. Then H contains elements of order 3 and 5, which respectively are a 3-cycle and a 5-cycle, which we can take to be (1 2 3 4 5). Without loss of generality the 3-cycle is (1 a b). Squaring if necessary we can assume that a < b, so the possibilities are (1 2 3), (1 2 4), (1 2 5), (1 3 4), (1 3 5), (1 4 5). Suppose for example that the 3-cycle is (1 2 4). Then H contains (1 2 3 4 5)−1 (1 2 4) = (1 5)(3 4) of order 2. This is a contradiction. Similar calculations deal with the other cases. A slicker alternative is to use Lemma 25.2 from Chapter 25, noting that the index of H is 4. 23.2 Let G be a p-group, so that |G| = pn . Let H be a subgroup. Then |H| divides |G|, so |H| = pm where m ≤ n. Therefore H is a p-group. Suppose that N / G, so |N| = pm where m ≤ n. Then |G/N| = pn−m and G/N is a p-group.

108 Suppose that G is a normal subgroup of J and J/G is a p-group. Then |J/G| = pr for some r. So |J| = pn+r and J is a p-group. 23.3 The centre is a proper normal subgroup, so must be An . If n ≥ 3 then An is not abelian. If n = 3 then A3 does not commute with the 2-cycle (1 2). 23.4 Let |G| = p2 . The centre Z of G is nontrivial. If Z = G then G is abelian. Otherwise |Z| = p so Z is cyclic. Since |G/Z| = p we can find g ∈ G \ Z such that g generates G modulo Z. Then every element of G is of the form ga z for z ∈ Z. But all such elements commute. There are two possibilities: G ∼ = Z p × Z p , Z p2 . 23.5 Suppose that K is algebraically closed and L/K is algebraic. Let α ∈ L, with minimal polynomial m over K. Since K is algebraically closed, ∂ m = 1, so α ∈ K. Conversely, suppose that L/K algebraic implies that L = K. Let f ∈ K[t] be irreducible over K. If we adjoin zero α of f to K then K(α)/K is algebraic of degree ∂ f . By assumption K(α) = K, so α ∈ K and ∂ f = 1. This every irreducible polynomial over K splits in K, so K is algebraically closed. 23.6 Let α be algebraic over R with minimal polynomial m. By the Fundamental Theorem of Algebra, f splits over C. Either ∂ m = 1 and its zero lies on R, or ∂ m = 2 and R(α)/R is isomorphic to L/R where L ⊆ C and L 6= R. Thus L = C. 23.7 Suppose that C has an ordered field structure. Then i 6= 0 so either i > 0 or i < 0. Either way, i2 > 0. That is, −1 > 0, so 0 > 1, a contradiction. We can change the field operations to give C and ordered field structure (though not a very interesting one). Since C has the same cardinality as R, there is a bijection ˙ on C, and an ordering n, K > 0, and suppose that | qp − x| < Kq−r . Write r = n + s where s ≥ 1. We have p M −1 qn ≤ − x < Kq−r = Kq−n q−s q so that M

−1

p ≤ − x qn < Kq−s ≤ K q

25 What Did Galois Do or Know?

111

Therefore both | qp − x| and qn are bounded, so only finitely many q, hence finitely many p, can occur. −n! . We claim α is transcendental. 24.7 Let α = ∑∞ n=1 10 If not, then α is algebraic, of degree n, say. Exercise 24.6* implies that for any r > n and any K > 0 the number of rationals qp such that p − α < Kq−r q is finite. We obtain a contradiction, as follows. Write α= Then ql = 10l! and

l

∞

n=1

n=l+1

∑ 10−n! + ∑

10−n! =

pl + εl ql

|εl | < 2.10−(l+1)! < 2q−(l+1)

Take K = 2 and let r = n > 1. Then for all (in particular, infinitely many) l ≥ r we have p − α = |εl | < Kq−(l+1) ≤ Kq−r q which is a contradiction. 24.8 Suppose that z = x + iy where both x, y are algebraic. We prove the logically equivalent statement that z is algebraic. First note that iy is algebraic, being the product of two algebraic numbers i and y. There is a double extension Q ⊆ Q(x) ⊆ Q(x, iy) and both [Q(x) : Q)] and [Q(x, iy) : Q(x)] are finite. By the Tower Law, [Q(x, iy) : Q] is finite. Therefore x + iy is algebraic. 24.9 (a) T. (b) F. (c) T. (d) T. (e) T. (f) F. (g) T. (h) F. (i) T. (j) F. (k) T.

25 What Did Galois Do or Know? 25.1 An element g ∈ ker(φ ) if and only if φ (g) is the identity on H ; that is, g−1 h−1 Hhg = h−1 Hh ∀h ∈ H For any given h this equation states that g normalises h−1 Hh; that is, g ∈ NG (h−1 Hh)

112 This holds for all h ∈ H if and only if g∈

\

NG (h−1 Hh)

h∈H

as required. Its is easy to see that NG (h−1 Hh) = h−1 NG (H)h, and the second statement is proved. 25.2 Suppose for a contradiction that G is simple of order 5pk with k ≥ 2 and p ≥ 5. Then |G| ≥ 125. Since 5! = 120 < 125, Lemma 25.2 implies that G has no proper subgroups of index ≤ 5. If p = 5 then G is a p-group, so cannot be simple. If p > 5 then 5, pk are coprime. So every proper subgroup has order pl or 5pl where l < k. Their indices are all divisible by p, so Lemma 25.1(3) implies that G is not simple. 25.3* Let G be simple of order 84 = 22 .3.7. By Cauchy’s Theorem there exist subgroups of orders 2, 3, 7 (we are not using Sylow’s Theorem). A proper subgroup of index ≤ 6 implies the order divides 6!, which is false, so every proper subgroup has index ≥ 7. Therefore the possible orders of nontrivial proper subgroups are 2, 3, 4, 6, 7, 12. Let H7 be a subgroup of order 7. Its normaliser N(H7 ) has order divisible by 7, so from the above list it must equal 7. The index of N(H7 ) is 12, so H7 has 12 distinct conjugates. They intersect each other in the identity only, since 7 is prime. This gives 12.6 = 72 elements of order 7. Let H3 be a subgroup of order 3. Its normaliser N(H3 ) has order divisible by 3, so from the above list it must equal 3, 6, or 12. The index of N(H3 ) is therefore 28, 14, or 7. The number of elements of order 3 is at least 2.7 = 14. Therefore G contains at least 72 + 14 = 86 elements. But |G| = 84, contradiction. 25.4* Here’s a start . . . We can immediately eliminate: Prime order: 61, 67, 71, 73, 79, 83, 89, 97. Theorem 25.3 eliminates: Order pk , k ≥ 2: 64, 81. Order pq: 62, 65, 69, 74, 77, 82, 85, 86, 87, 91, 93, 94, 95 Order 2pk , k ≥ 2: 98 Order 3pk , k ≥ 2: 75, 96 Order 4pk , k ≥ 2, p ≥ 7: 68, 76, 92. The only orders not accounted for are 63, 66, 70, 72, 78, 80, 84, 88, 90, 96, 99. The case 84 is done in Exercise 25.3. Many of the rest can be handled fairly easily by special methods: 63 = 32 .7. Proper subgroups can have orders 1, 3, 7, 9 only. In particular there is an element of order 7 generating a Z7 subgroup. Its normaliser has order 7, so there are 9 conjugates, any two intersecting only in the identity. Thus there are at least 9.6 = 54 elements of order 7. There is also an element of order 3. Its normaliser has order 3 or 9, so there are at least 7 conjugates, giving at least 2.7 = 14 elements of order 3. But 14+54 = 68 > 63, contradiction.

25 What Did Galois Do or Know?

113

66 = 2.3.11. Proper subgroups can have orders 1, 2, 3, 6, 11 only. In particular there is an element of order 11 generating a Z11 subgroup. Its normaliser has order 11, so there are 6 conjugates, any two intersecting only in the identity. Thus there are at least 10.6 = 60 elements of order 11. There is also an element of order 3. Its normaliser has order 3 or 6, so there are at least 11 conjugates, giving at least 2.11 = 22 elements of order 3. But 22 + 60 = 82 > 66, contradiction. 70 = 2.5.7. Nontrivial proper subgroups can have orders 2, 5, 7, 10, or 14. Indices (especially of centralisers) are then 35, 14, 10, 7, or 5. So the class equation takes the form 1 + 35a + 14b + 10c + 7d + 5e = 70 Modulo 7 this gives 1 + 5(2c + e) ≡ 0 (mod 7) so 2c + e ≡ 4 (mod 7). That is, 2c + e = 4, 11, . . .. Modulo 5 this gives 1 + 7(2b + d) ≡ 0 (mod 5) so 2b + d ≡ 2 (mod 7). That is, 2b + d = 2, 9, . . .. In fact we must have 2c + e = 4, since 2c + e = 11 contributes 7(2c + e) = 77 to the class equation, which is too large. Similarly 2c + e = 2. So the class equation becomes 1 + 35a + 28 + 10 = 70 This has no solutions for a ≥ 0. 78 = 2.3.13. Nontrivial proper subgroups can have orders 2, 3, 6, 13. There exist elements (hence cyclic subgroups) of orders 2, 3, and 13. Any Z13 subgroup has normaliser Z13 , of index 6, so conjugates give at least 6.12 = 72 elements of order 13. Any Z3 subgroup has normaliser or order 3 or 6, index at least 13, so conjugates give at least 6.2 = 12 elements of order 3. But 72 + 12 > 78, contradiction. 88 = 23 .11 Nontrivial proper subgroups can have orders 2, 4, 8, 11. Indices are 44, 22, 11, 8. If there is no order-8 subgroup then this contradicts Lemma 25.1(3) with p = 2. So there is an order-8 subgroup. By Cauchy’s Theorem there is an order11 element (hence subgroup) Z11 . The normaliser of Z11 must be Z11 . Therefore it has at least 8 distinct conjugates, which all intersect trivially since 11 is prime, giving at least 10.8 = 80 elements of order 11. Since only 8 elements remain, the order-8 subgroup is unique, contrary to Lemma 25.1(4). 99 = 32 .11. Nontrivial proper subgroups can have orders 3, 9, 11. There is a Z11 subgroup, which must equal its normaliser, so it has 9 conjugates. These all intersect in 1, so there are at least 9.10 = 90 order-11 elements. There is also a Z3 subgroup, whose normaliser has order 3 or 9, so it has at least 11 conjugates, giving at least 22 order-3 elements. But 90 + 22 > 99, contradiction. The remaining orders 72, 80, 90 seem to be harder, and I leave them as a challenge. 25.5 Count the possible columns in the matrix. There are 23 − 1 = 7 possibilities for

114 the first column. The second column must be linearly independent from this, giving 23 − 2 = 6 possibilities. Finally, the third column must be linearly independent from both of these. Since they span a space with four elements, this gives 23 − 4 = 4 possibilities. The total is therefore 7.6.4 = 168. 25.6 To compute the order, start with GL2 (F7 ). Count the possible columns in the matrix. There are 72 − 1 = 48 possibilities for the first column, then 72 − 7 = 42 possibilities for the second column. Restricting to SL2 (F7 ), matrices of determinant 1, we must divide by 7 − 1 = 6. Finally, to get PSL2 (F7 ), we identify I and −I, which divides the total by 2. The result is 48.42/(6.2) = 168. 25.7 Any normal subgroup is a union of conjugacy classes. It must contain the identity, and its order divides 168. We therefore seek sums of the numbers 1, 21, 24, 24, 42, 56 that include 1 and divide 168. Intelligent trial and error shows that the only such sums are 1 and 1+21+24+24+42+56 = 168. These correspond to the identity subgroup and the whole group, so it is simple. 25.8* No answer since this exercise is a project.

26 Further Directions 26.1 Let G be a finite group of order n. Let L/K be a finite normal separable extension with Galois group Sn . For example, use K(t1 , . . . ,tn )/K(s1 , . . . , sn ) as in Section ??. Embed G in Sn using Cayley’s result. Let M be the fixed field of G. Then the Galois group Γ(L/M) ∼ = G. 26.2 We have to prove that p(x, y, z, w) = xw − yz − 1 is irreducible over C. Assume for a contradiction that this is false. Then p is a product of two linear factors: xw − yz − 1 = (a0 + a1 x + a2 y + a3 z + a4 w)(b0 + b1 x + b2 y + b3 z + b4 w) and by scaling we can assume a0 = 1. Equating coefficients of 1, x, y, z, w we get a0 b0 = −1 a1 b0 + a0 b1 = 0 a2 b0 + a0 b2 = 0 a3 b0 + a0 b3 = 0 a4 b0 + a0 b4 = 0 which imply that b0 = −1

b1 = a1

b2 = a2

b3 = a3

The coefficients of x2 , y2 , z2 , w2 imply: a21 = 0

a22 = 0

a23 = 0

a24 = 0

b4 = a4

115

26 Further Directions

so the first linear factor reduces to 1. This is a contradiction. 26.3 This is a bit of a scramble. Let Γ = SL2 (C). We prove that the derived group Γ0 = Γ. Therefore Γ is not soluble, because any soluble group except 1 has a proper normal subgroup N such that Γ/N is abelian, which implies that Γ0 ⊆ N. The derived group is generated by all commutators [g, h] = ghg−1 h1 of elements g, h ∈ Γ. Moreover, any conjugate of a commutator is a commutator. We collect enough commutators and products of commutators to give the whole of Γ. Among the elements of Γ are:       a 0 1 0 1b Da = L = U = c b c 1 0 1 0 a−1 Routine calculations show that 

a 0 [Lc , Da ] = c − ca−2 1



and any element of C can be expressed as c − ca−2 , so all elements Lc lie in Γ0 . Similarly all elements Ub lie in Γ0 . Now   a + bc a + b + bc Ub LcU1 = c 1+c We claim that suitable choices of a, b, c can make this equal to any unimodular matrix. We must solve     a + bc a + b + bc p q = c 1+c r s Since both matrices are unimodular, it is enough to solve for the p, q, r entries; the entry s = 1 + c follows automatically. It is now easy to see that the solution is a = p − rp + rq

b = p−q

c=r

Therefore every matrix in Γ lies in Γ0 , so Γ0 = Γ. 26.4 The prime factorisation of r = 355/113 is r = 5.71.113−1 . Therefore: (1) p = 3 implies |r| p = 30 = 1. (2) p = 5 implies |r| p = 5−1 . (3) p = 7 implies |r| p = 70 = 1. (4) p = 71 implies |r| p = 71−1 . (5) p = 113 implies |r| p = 113. 26.5 Equivalently, we must find the primes that divide the stated numbers. Factorising: (1) r = 12345 = 3.5.823 so p = 3, 5, 823. (2) r = 123456 = 26 .3.643 so p = 2, 3, 643. (3) r = 1234567 = 127.9721 so p = 127, 9721. (4) r = 12345678 = 2.32 .47.14593 so p = 2, 3, 47, 14593. (5) r = 123456789 = 32 .3607.3803 so p = 3, 3607, 3803. 26.6 We have to prove |x + y| p ≤ max(|x| p , |y| p ) (24)

116 Let x = pk a/b where k, a, b ∈ Z and p does not divide a or b. Let y = pl c/d where l.c.d ∈ Z and p does not divide c or d. Then |x| p = p−k

|y| p = p−l

We may assume that k ≥ l. Then |x| p = p−k ≤ p−l = |y| p so max(|x| p , |y| p ) = |y| p = p−l . Now pk a pl c x+y = + = pl b d



pk−l a c + b d



so |x + y| p ≤ p−l = |y| p = max(|x| p , |y| p ). 26.7 Among others: Section 3.5, proof of Theorem 8.10, Theorem 16.9, Section 20.5, Section 21.2, Section 21.3, Section 21.4, Section 21.6, Section 21.7, Section 25.3, Section 26.3.

References

117

References Edwards, H.M. (1984) Galois Theory, Springer, New York. Fowler, D.H. (1987) The Mathematics of Plato’s Academy, Clarendon Press, Oxford. Graham, R.L., Knuth, D.E., and Patashnik, O. (1994) Concrete Mathematics, Addison-Wesley, Reading. Neumann, P.M., Stoy, G.A, and Thompson, E.C. (1994) Groups and Geometry, Oxford University Press, Oxford. Potter, M.D. (1990) Sets: an Introduction, Oxford University Press, Oxford. Slomson, A. (1991) An Introduction to Combinatorics, Chapman and Hall/CRC Press, London. Stewart, I. and Tall, D. (2015) The Foundations of Mathematics (2nd ed.), Oxford University Press, Oxford. Stewart, I. and Tall, D. (2002) Algebraic Number Theory and Fermat’s Last Theorem, A.K. Peters, Natick. Wilson, R. and Gray, J. (eds.) (2001) Mathematical Conversations, Springer, New York.