BSc and MSci EXAMINATIONS (MATHEMATICS) May-June 2014 This paper is also taken for the relevant examination for the Associateship of the Royal College of Science. M3P11/M4P11/M5P11 Galois Theory Date: examdate Time: examtime Credit will be given for all questions attempted but extra credit will be given for complete or nearly complete answers. Calculators may not be used. c 2014 Imperial College London M3P11/M4P11/M5P11 Page 1 of 3
1. (a) Prove that if f Z[x] factors in Q[x] as f = gh, then f can be factored as f = g 1 h 1 with g 1, h 1 Z[x] and deg(g 1 ) = deg(g), deg(h 1 ) = deg(h). (b)whatdoesitmeanforanon-zeropolynomialf Q[x]tobeirreducible? StateEisenstein s criterion. (c) For each of the polynomials below in Q[x], establish (with proof) whether or not they are irreducible. (i) x 3 5x+6 (ii) x 6 +120x+18 (iii) x 6 +x 5 +x 4 +x 3 +x 2 +x+1. (iv) x 4 +4 (v) x 4 +1 2. (a) Let E F be fields. What is the degree [F : E] of F over E? State the tower law for extensions E F K of fields. (b) Let E F be fields. What does it mean for α F to be algebraic over E? Prove that if [F : E] is finite then every α F is algebraic over E. (c) Again say E F are fields, and furthermore say p(x) E[x] is a polynomial. What does it mean to say that F is a splitting field for p(x)? (d) Now say E = Q. Assuming any results you need from the course, compute (with proofs) the degrees (over Q) of the splitting fields of the following polynomials: (i) x 3 11 (ii) x 7 1 (iii) (x 2 8)(x 2 18). 3. (a) What does it mean for a finite extension F/E of fields to be normal? Prove that a finite normal extension is a splitting field. (b) Say E F K are fields, and [K : E] <. Give proofs or counterexamples to the following assertions (you may assume any standard results from the course unless you are explicitly asked to prove them): (i) If K/E is normal then K/F is normal. (ii) If K/E is normal and K/F is normal then F/E is normal. (iii) If [F : E] = 2 then F/E is normal. (c) Is Q((1+ 3) 1/3 ) a normal extension of Q?
4. In this question you may assume the Fundamental Theorem of Galois Theory and any other results you need from the course. SayE isafieldofcharacteristiczero, andf(x) E[x]isanirreduciblepolynomialofdegree3. Let F/E be a splitting field for f(x), and say α,β,γ are the three roots of f(x) in E[x]. (a) Briefly (one mark each!) explain why F/E is Galois and α,β,γ are distinct. (b) Set G = Gal(F/E). Prove that G must be isomorphic to either the group S 3 or the group A 3. (c) Prove that if G = S 3 then there is a unique subfield K of F with E K F and [K : E] = 2. (d) Set d = (α β)(β γ)(γ α) F. Prove that d 2 E, and that d E if and only if G = A 3. (e) Some painful algebraic manipulation, which you may assume, shows that if f(x) = x 3 + Ax + B then d 2 = 4A 3 27B 2. Given this fact, prove that the Galois group of the splitting field of x 3 + x + 1 over Q is isomorphic to S 3, and compute the quadratic subfield K of the splitting field as in (c). M3P11/M4P11/M5P11 Galois Theory (2014) Page 3 of 3
M3/4/5P11 2014 SOLUTIONS - MAIN EXAM (1) (a) Clear denominators in the equation f = gh to get Nf = g 0 h 0 with g 0,h 0, Z[x] and N Z 1. Clearly deg(g 0 ) = deg(g) anddeg(h 0 ) = deg(h). WenowdealwithN asfollows: ifn > 1 and p N is a prime divisor of N, then all coefficients of g 0 h 0 are multiples of p. If p divides all the coefficients of g 0 then we can divide both N and g 0 by p and we have made N smaller. Similarly if p divides all the coefficients of h 0. If however both g 0 and h 0 have coefficients prime to p then say a i x i is the term of g 0 with i as large as possible subject to p a i, and similarly let b j x j denote the term fo h of largest degree with p b j. An easy check now shows that p does not divide the coefficient of x i+j in Nf, a contradiction. Hence we can always divide out the factor of p from g 0 or h 0 and now induction on N shows that we can reduce the equation to f = g 1 h 1 as required. Seen,6 (b) f Q[x] is irreducible if it has positive degree and for any factorisation f = gh with g,h Q[x] either g or h is constant. Seen,1 Eisenstein s criterion: if f = n i=0 a ix i Z[x] and if p is a prime number with p a n, p a i for all i < n, and p 2 a 0, then f is irreducible in Q[x]. Seen,2 (c) (i) By (a) if x 3 5x + 6 were not irreducible it would be a Seen sim product of two factors in Z[x] of positive degree, and one of these factors would have to have degree 1. Hence there would be an integer n such that n 3 5n+6 = 0, that is, n(5 n 2 ) = 6. In particular n would have to be a divisor of 6 but one checks easily that no divisor of 6 works and hence the polynomial is irreducible. 3 (ii) Irreducible by Eisenstein with p = 2. 1 (iii)irreduciblebyatrickinlectures: iff(x)isthepolynomial in the question then f(x) = (x 7 1)/(x 1) so if y = x 1 then f(x) = ((1+y) 7 1)/y which is an irreducible polynomial in y by Eisenstein with p = 7. 2 (iv) x 4 + 4 = (x 4 + 4x 2 + 4) 4x 2 is the difference of two squares and hence factors as (x 2 +2x+2)(x 2 2x+2), and in particular is reducible. 2 1
(v) Over the complexes, x 4 + 1 has roots ζ, ζ 3, ζ 5 and ζ 7 with ζ = e 2πi/8. In particular x 4 + 1 has no real roots and hence no rational roots, so if it were reducible over Q it would have to factor into two irreducible quadratic polynomials. The roots of these quadratics would be non-real and hence complex conjugates, so one of them would have to be (x ζ)(x ζ 7 ), but the coefficient of x in this polynomial is (ζ+ζ 7 ) which is easily checked to be 2 Q. Hence this polynomial is irreducible. 3 (2) (a) F is naturally a vector space over E and [F : E] is its dimension. The tower law says that [K : E] = [K : F][F : E]. Seen,1+1 (b)α F isalgebraicovere ifthereisanon-zeropolynomial o(x) E[x] such that p(α) = 0. Seen,1 Now say [F : E] = n < and α F. Consider the n + 1 elements 1,α,α 2,...,α n of F. This is n+1 elements in an n- dimensional vector space over E and hence there is an E-linear relation between them. If n i=0 λ iα i = 0 with λ i E, not all zero, then set p(x) = i λ ix i and we see p(x) 0 and p(α) = 0, as required. Seen,4 (c) F is a splitting field for p(x) if p(x) factors into linear factors c(x α 1 )(x α 2 ) (x α d ) in F[x] and furthermore that F = E(α 1,...,α d ), the field generated by the α i. Seen,1 (d) (i) First note x 6 22 is irreducible by Eisenstein with p = 2 Seen sim or p = 11. Next note that if α R is the positive real 6th root of 22, then the roots of x 6 22 are ζ i α with ζ = e 2πi/6 and 0 i 5. Hence the splitting field is F := Q(α,ζ). Now [Q(α) : Q] = 6 (as x 6 22 is irreducible), and ζ Q(α) (as it is not real). However ζ satisfies a degree two equation over Q(α) (namely x 2 = x 1), so [F : Q(α)] = 2 (as it is greater than 1 and at most 2). Hence [F : Q] = [F : Q(α)][Q(α) : Q] = 2 6 = 12. 3 (ii) The roots of x 7 1 are ζ i for ζ = e 2πi/7 and 0 i 6. In particular the splitting field is Q(ζ 7 ) and hence it has degree equal to the degree of the minimum polynomial of ζ 7 over Q. Theminimumpolynomialofζ 7 dividesx 7 1 = (x 1)(x 6 +x 5 + x 4 +x 3 +x 2 +x+1)andhenceitdividesx 6 +x 5 +x 4 +x 3 +x 2 +x+1. But this polynomial is irreducible (this is a result from the course, and in Q1) and hence must be the minimum polynomial of ζ. Hence the splitting field has degree 6 over Q. 4 2
(iii) The roots of x 2 8 are ±2 2 and the roots of x 2 18 are ±3 2, sothefieldgeneratedbytheserootsisq( 2), whichhas degree2overq(asx 2 2isirreducibleoverQbyEisenstein...). 2 (iv) x 3 + x + 1 is a cubic whose derivative is 3x 2 + 1 which is always positive, and hence the cubic has one real root. This real root, call it α, is not rational (as it would then have to be ±1 by Gauss lemma, and neither of ±1 are roots). So x 3 + x + 1 must be irreducible and [Q(α) : Q] = 3. If β and γ are the other two roots in C then β and γ are not real, and hence[q(β,α) : Q] 2. Butβ satisfiesadegree2equationover Q(α) (namely (x 3 +x+1)/(x α)) and hence [Q(β,α) : Q] = 2. Finally γ Q(β,α) because α+β +γ Q. By the tower law, the splitting field has degree 6 over Q. 3 (3) (a) A finite extension of fields F/E is normal if, for all irreducible p(x) E[x], if p(x) has a root in F then p(x) splits completely in F. Seen,1 Now say F/E is finite and normal. Let α 1,...,α n be an E- basis for F. For each i let p i (x) be the minimum polynomial of α i over E; then p i (x) E[x] is irreducible over E and has a root in F, so splits completely in F. Hence p(x) = i p i(x) also splits completely in F, and F must hence be the splitting field of p(x) because the α i are amongst the roots of p(x) and the α i generate F as a vector space over E and hence as a field over E. Seen,3 (b) (i) This is true. For if p(x) F[x] is irreducible and has a root α K then let q(x) denote the minimum polynomial of α over E; then q(x) is irreducible over E, and K/E is normal so q(x) splits completely in K. However q(α) = 0 and q(x) F[x], so by a fundamental property of minimum polynomials we know p(x) q(x) and in particular every root of p(x) is a root of q(x) and hence in K. In particular p(x) splits completely in K, which is what we wanted. Seen,4 (ii) This is true. We know E F and F has dimension 2 over E, and it is now easy to check that F has an E-basis of the form {1,α} with α F and α E. We must have α 2 = a+bα with a,b E and so the polynomial p(x) = x 2 bx a has α as a root and hence the other root is b α. In particular p(x) has no roots in E so it is irreducible over E as it has degree 2, and F is a splitting field of p(x) over E and hence F/E is normal. Seen sim,5 (iii) This is not true. For example if E = Q, F = Q( 2) and K = Q(2 1/4 ) then [F : E] = 2 and [K : E] = 4 (because x 2 2 3
and x 4 2 are irreducible by Eisenstein) and hence [K : F] = 2 by the tower law. By part (ii) we see that F/E and K/F are normal. However K/E is not x 4 2 is irreducible over Q and has a root in K but does not have all roots in K, as K R and x 4 2 has two non-real roots. Seen,3 (c) It is not. If α = (1+ 3) 1/3 R then α 3 1 = 3 so α is a root of the polynomial (x 3 1) 2 3 = 0, and this polynomial is x 6 2x 3 2 which is irreducible by Eisenstein with p = 2. However this polynomial has a root in Q(α) R but does not have all its roots in Q(α) as ωα is easily checked to be another root, with ω = e 2πi/3, and ωα R so ωα Q(α). Unseen,4 (4) (a) F/E is finite and normal because it s the splitting field of a polynomial, and it s separable because the characteristic of E is zero. Seen,1 The roots of f are distinct because if two coincided then f would have a root in common with its derivative and hence would not be coprime to its derivative; however its derivative hasdegree2asweareincharacteristiczero, andf isirreducible; this is a contradiction. Seen,1 (b) G is naturally a subgroup of the permutation group on the set {α,β,γ}, because any field automorphism of F sends a root of f(x) to a root of f(x), and this gives a natural group homomorphism from G to the permutation group. Moreover any automorphism which fixes α, β and γ fixes the field they generate,whichisf bydefinition,sothemapisinjective,meaning that G is a subgroup of S 3. Finally, E E(α) F and [E(α) : E] = 3 as f is irreducible. So by the tower law [F : E] is a multiple of 3 and hence G has order a multiple of 3. It is now straightforward to check that the only subgroups of S 3 with order a multiple of 3 are S 3 and A 3. Seen sim, 5 (c) This follows immediately from the Fundamental Theorem of Galois Theory and the fact that S 3 has a unique subgroup of order 3, namely A 3. Seen sim,2 (d) d 2 = (α β)(β α)(β γ)(γ β)(α γ)(γ α) is left invariant under any permutation of α,β,γ and hence is fixed by every element of G. By the fundamental Theorem, we deduce d 2 E and hence d 2 E. Unseen,4 If σ S 3 is a transposition then σ(d) = d d, but one checks that all 3-cycles fix d. So if G = S 3 then d is not fixed by all of G and hence d E, whereas if G = A 3 then d is fixed by all of G and so d E. Unseen,3 4
(e) First we note that x 3 +x+1 is irreducible, as if it were reducible then by Gauss Lemma it would have an integer root, and this integer would have to divide 1 but neither ±1 is a root. Part (b) now shows that the Galois group of the splitting field is either S 3 or A 3. We know that for d as in part (d) we have d 2 = 4 27 = 31 and hence the splitting field contains 31; by (d) we have G = S3 and becasue 31 is in the splitting field, we must have K = Q( 31). Unseen,4 5