volved in answering the questions. Instructors may need to ll in additional details

Introduction1 Introduction ThisSolutions Manualcontains solutions to all of the exercises in the Fifth Edi- tion ofGalois Theory.Many of the exercises have several different solutions, or can be solved using several different methods. If your solution is different from the one presented here, it may still be correct — unless it is the kind of question that has only one answer.The written style is informal, and the main aim is to illustrate the key ideas in- volved in answering the questions. Instructors may need to ll 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

• Classical Algebra

1.1Letu=x+iy(x;y);v=a+ib(a;b);w=p+iq(p;q). Then uv= (x;y)(a;b) = (xayb;xb+ya) = (axby;bx+ay) = (a;b)(x;y) =vu (uv)w= [(x;y)(a;b)](p;q) = (xayb;xb+ya)(p;q) = (xapybpxbqyaq;xaqybq+xbp+yap) = (x;y)(apbq;aq+bp) = (x;y)[(a;b)(p;q)] = (uv)w 1.2(1) Changing the signs ofa;bdoes not affect(a=b) 2 , so we may assumea;b>0.(2) Any non-empty set of positive integers has a minimal element. Sinceb>0 is an integer, the set of possible elementsbhas a minimal element.Galois Theory, 5e Ian Stewart Solution Manual all Chapters 1 / 4

2 (3) We know thata 2 =2b 2 . Then (2ba) 2 2(ab) 2 =4b 2 4ab+a 2 2(a 2 2ab+b 2 ) =2b 2 a 2 =0 (4) If 2bathen 4b 2 a 2 =2b 2 , a contradiction. Ifabthen 2a 2 2b 2 =a 2 , a contradiction.(5) Ifabbthena2bsoa 2 4b 2 =2a 2 , a contradiction. Now (3) contra- dicts the minimality ofb.Note on the Greek approach.The ancient Greeks did not use algebra. They expressed them same underlying idea in terms of a geometric gure, Figure 1.

FIGURE 1: Greek proof that

p

• is irrational.

Start with square ABCD and let CE = AB. Complete square AEFG. The rest of the gure leads to a point H on AF. Clearly AC=AB=AF=AE. In modern notation, let AB =b

, AC =a

. Since AB = HF = AB and BH = AC, we have AE =a

+b

=b, say, and AF =a

+2b

=a, say. Thereforea

+b

=b;b

=ab;and a b = a

b 0.If p

• is rational, we can makea;bintegers, in which casea

;b

are also integers, and the same process of constructing rationals equal to p

• 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 nitely many steps if and only if the initial ratio lies inQ. See Fowler (1987) pages 33–35.

1.3A nonzero rational can be writtenuniquely, up to order, as a produce of prime

powers (with a sign):

r=p m1 1 p m k k where themjare integers. So r 2 =p 2m1 1 p 2m k k 2 / 4

• Classical Algebra3

Now p q=rif and only ifq=r 2 , and all exponents 2mjare even.

1.4*Clearly 18 p

325=185

p

• A little experiment shows that

3 p 13 2 !3 =185 p 13 (The factor 1 2

is the only real surprise here: it occurs because 13 is of the form 4n+1,

but it would take us too far aeld to explain why.) At any rate, 3 q 185 p 13= 3 p 13 2 so that 3 q 18+5 p 13+ 3 q 185 p 13= 3+ p 13 2 + 3 p 13 2 = 3 2 + 3 2 =3 1.5LetKbe the set of allp+qa+ra 2 , wherep;q;r2Q. ClearlyKis closed under addition and subtraction. Sincea 3 =2 we also havea 4 =2a, and it follows easily thatKis closed under multiplication.Tedious but elementary calculations, or computer algebra, show that (p+qa+ra 2 )(p+qwa+rw 2 a 2 )(p+qw 2 a+rwa 2

• =p

3 +2(q 3 3pqr) +4r 3 (1) so that (p+qa+ra 2 ) 1 = (p+qwa+rw 2 a 2 )(p+qw 2 a+rwa 2 ) p 3 +2(q 3 3pqr) +4r 3 implying closure under inverses, hence division.However, it is necessary to check thatp 3 +2(q 3 3pqr) +4r 3 =0 in rational numbers impliesp=q=r=0. By (1)p 3 +2(q 3 3pqr) +4r 3 =0 implies that p+qa+ra 2 =0 orp+qwa+rw 2 a 2 =0 orp+qw 2 a+rwa 2 =0. The required result follows since 1;a;a 2 are linearly independent overQ.

1.6The map is one-to-one since it is linear in(p;q;r)andp+qw 2 a+rwa 2 =0 impliesp=q=r=0. Compute (p+qa+ra 2 )(a+ba+ca 2 ) = (pa+2qc+2rb) + (pb+qa+2rc)a+ (pc+qb+ra)a 2 and compare with (p+qwa+rw 2 a 2 )(a+bwa+cw 2 a 2 ) = (pa+2qc+2rb) + (pb+qa+2rc)wa+ (pc+qb+ra)w 2 a

2 3 / 4

4 The coefcients are there same in both formulas, so products are preserved as re- quired. Thus the map is a monomorphism.All maps are onto their image. But the image here is notQ(a)becauseQ(a)R, butw62R. So the map is not an automorphism.

1.7Observe that (2i) 3 =211i=2 p 121 and(2+i) + (2i) =4.

1.8The inequality 27pq 2 +4p 3 <0 implies thatp<0, so we can nda;bsuch that p=3a 2 ;q=a 2 b, and the cubic becomes t 3 3a 2 t=a 2 b The inequality becomesa>jbj=2. Substitutet=2acosq, and observe that t 3 3a 2 t=8a 3 cos 3 q6a 3 cosq=2a 3 cos3q The cubic thus reduces to cos3q= b 2a which we can solve using cos 1 becausej b 2a j 1, getting q= 1 3 cos 1 b 2a There are three possible values ofq, the other two being obtained by adding 2p 3 or 4p 3 . Finally, eliminateqto get t=2acos

1 3 cos 1 b 2a

wherea= q p 3 ;b= 3q p .

1.9By inspection one root ist=4. Factoring outt4 leads to a quadratic whose roots are2+ p

• and2

p 3.

1.10If you carry out the algebra, it turns out that trying to solve foraandbleads back to the original cubic equation. Unless the solutions are obvious (in which case the method is pointless) no progress is made.Specically, suppose we want to solve(u+ p v) 3 =a+ p bfor rationalu;vgiven rationala;b. Then assuming p b; p vare irrational, we are led to u 3 +3uv=a (3u 2 +v) p v= p b It follows easily that(u p v) 3 =a p b, whence u 2 v= 3 p a 2 b

• / 4