M3P11/M4P11/M5P11. Galois Theory

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1 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

2 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 x+18 (iii) x 6 +x 5 +x 4 +x 3 +x 2 +x+1. (iv) x 4 +4 (v) x (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?

3 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

4 M3/4/5P 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 = (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

5 (v) Over the complexes, x has roots ζ, ζ 3, ζ 5 and ζ 7 with ζ = e 2πi/8. In particular x 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 = (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

6 (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 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

7 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

8 (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