Math 5111 (Algebra 1) Lecture #22 of 24 ∼ November 30th, 2020 Cyclotomic and Abelian Extensions, Galois Groups of Polynomials Cyclotomic and Abelian Extensions Constructible Polygons Galois Groups of Polynomials Symmetric Functions and Discriminants Cubic Polynomials This material represents x4.3.4-4.4.3 from the course notes. Cyclotomic and Abelian Extensions, 0 Last time, we defined the general cyclotomic polynomials and showed they were irreducible: Theorem (Irreducibility of Cyclotomic Polynomials) For any positive integer n, the cyclotomic polynomial Φn(x) is irreducible over Q, and therefore [Q(ζn): Q] = '(n). We also computed the Galois group: Theorem (Galois Group of Q(ζn)) The extension Q(ζn)=Q is Galois with Galois group isomorphic to × (Z=nZ) . Explicitly, the elements of the Galois group are the × a automorphisms σa for a 2 (Z=nZ) acting via σa(ζn) = ζn . Cyclotomic and Abelian Extensions, I By using the structure of the Galois group we can in principle compute all of the subfields of Q(ζn). In practice, however, this tends to be computationally difficult × when the subgroup structure of (Z=nZ) is complicated. The simplest case occurs when n = p is prime, in which case ∼ × (as we have shown already) the Galois group G = (Z=pZ) is cyclic of order p − 1. In this case, let σ be a generator of the Galois group, with a × σ(ζp) = ζp where a is a generator of (Z=pZ) . Then by the Galois correspondence, the subfields of Q(ζp) are the fixed fields of σd for the divisors d of p − 1. Cyclotomic and Abelian Extensions, II We may compute an explicit generator for each of these fixed fields by exploiting the action of the Galois group on the basis 2 p−1 fζp; ζp ; : : : ; ζp g for Q(ζp)=Q. p−2 This set is obtained from the standard basis f1; ζp; : : : ; ζp g p−1 p−2 using the relation ζp + ζp + ··· + ζp + 1 = 0 from the minimal polynomial of ζp. Since all of these basis elements are Galois conjugates, the action of any element of the Galois group permutes them. Cyclotomic and Abelian Extensions, III P For any subgroup H of G, define the element αH = σ2H σ(ζp). We claim that αH is a generator for the fixed field of H. To see this, observe first that if τ 2 H, then τ(αH ) = αH because τ merely permutes the elements σ(ζp) for σ 2 H. Conversely, because the elements σ(ζp) for σ 2 G form a basis, if τ 2 G has τ(αH ) = αH then τ(ζp) must equal σ(ζp) −1 for some σ 2 H. But then τσ acts as the identity on ζp and hence on all of Q(ζp), so it must be the identity element: thus, τ = σ 2 H. We conclude that the automorphisms fixing αH are precisely the elements of H, and so Q(αH ) is the fixed field of H. Cyclotomic and Abelian Extensions, IV Example: Find generators for each of the subfields of Q(ζ7). × We know that G = Gal(Q(ζ7)=Q) is isomorphic to (Z=7Z) . × By trial and error we can see that 3 has order 6 in (Z=7Z) , so it is a generator. The corresponding automorphism 3 generating G is the map σ with σ(ζ7) = ζ7 . The subgroups of G are then hσi = fe; σ; σ2; σ3; σ4; σ5g, σ2 = fe; σ2; σ4g, σ3 = fe; σ3g, and σ6 = feg. A generator of the fixed field of hσi is given by 2 3 4 5 ζ7 + σ(ζ7) + σ (ζ7) + σ (ζ7) + σ (ζ7) + σ (ζ7) = 3 2 6 4 5 ζ7 + ζ7 + ζ7 + ζ7 + ζ7 + ζ7 . Similarly, the fixed field of σ2 is generated by 2 4 2 4 ζ7 + σ (ζ7) + σ (ζ7) = ζ7 + ζ7 + ζ7 , while the fixed field of 3 3 6 σ is generated by ζ7 + σ (ζ7) = ζ7 + ζ7 . Cyclotomic and Abelian Extensions, V Example: Find generators for each of the subfields of Q(ζ7). We can also use the Galois action to compute the minimal polynomials of each of these elements, since we may compute all of these elements' Galois conjugates. 2 4 For example, the element ζ7 + ζ7 + ζ7 has one other Galois 3 5 6 conjugate inside Q(ζ7), namely ζ7 + ζ7 + ζ7 . Then their common minimal polynomial is 2 4 3 5 6 2 m(x) = [x − (ζ7 + ζ7 + ζ7 )] · [x − (ζ7 + ζ7 + ζ7 )] = x + x + 2, as follows from multiplying out and simplifying the coefficients. Solving the quadraticp yields an explicit formula ζ + ζ2 + ζ4 = −1− −7 , and thus the corresponding fixed 7 7 7 2 p 2 4 field Q(ζ7 + ζ7 + ζ7 ) = Q( −7). Cyclotomic and Abelian Extensions, VI Example: Find generators for each of the subfields of Q(ζ7). 6 Similarly, the element ζ7 + ζ7 = 2 cos(2π=7) has two other 2 5 Galois conjugates, namely ζ7 + ζ7 = 2 cos(4π=7) and 3 4 ζ7 + ζ7 = 2 cos(6π=7). Their common minimal polynomial is 6 2 5 3 4 m(x) = [x − (ζ7 + ζ7 )] · [x − (ζ7 + ζ7 )][x − (ζ7 + ζ7 )] = x3 + x2 − 2x − 1. Our analysis indicates that the splitting field of this 6 2 5 3 4 polynomial is Q(ζ7 + ζ7 ) = Q(ζ7 + ζ7 ) = Q(ζ7 + ζ7 ), that it has degree 3 over Q, and that its Galois group is cyclic of order 3. is cyclic of order 3. Cyclotomic and Abelian Extensions, VII For other n, we can perform similar computations, although there is not usually as convenient a basis available. In general, the primitive nth roots of unity form a basis for Q(ζn) precisely when n is squarefree. When n is prime, the main computational requirement is × finding a generator of (Z=pZ) . For other n, we can simplify some of these computations by writing Q(ζn) as a composite of smaller cyclotomic fields. Cyclotomic and Abelian Extensions, VIII We can essentially reduce most computations down to working in the individual prime-power cyclotomic fields: Proposition (Composites of Cyclotomic Extensions) If a and b are relatively prime integers, then the composite of Q(ζa) and Q(ζb) is Q(ζab), the intersection is Q, and ∼ Gal(Q(ζab)=Q) = Gal(Q(ζa)=Q) × Gal(Q(ζb)=Q). a1 a2 ak In particular, if the prime factorization of n is n = p1 p2 ··· pk , a then Q(ζn) is the composite of the fields Q(ζp i ) for 1 ≤ i ≤ k, ∼ i and Gal(Q(ζn)=Q) = Gal(Q(ζ a1 )=Q) × · · · × Gal(Q(ζ ak )=Q). p1 pk More generally, for any a and b, the composite of Q(ζa) and Q(ζb) is Q(ζlcm(a;b)) and the intersection is Q(ζgcd(a;b)). Cyclotomic and Abelian Extensions, IX Proof: b a Observe that ζab = ζa and ζab = ζb, so both ζa and ζb are in Q(ζab): thus, the composite field is contained in Q(ζab). Also, since a and b are relatively prime, there exist integers s s t as+bt and t with sa + tb = 1. Then ζb · ζa = ζab = ζab, and so ζab is contained in the composite field of Q(ζa) and Q(ζb). Hence the composite field is Q(ζab). Also since [Q(ζab): Q] = '(ab) = '(a)'(b) = [Q(ζa): Q] · [Q(ζb): Q], by the formula for the degree of a composite extension we must have [Q(ζa) \ Q(ζb): Q] = 1 so Q(ζa) \ Q(ζb) = Q. The statement about the Galois group of Q(ζab)=Q follows immediately from our result on the Galois group of a composite of Galois extensions. The second statement then follows by a trivial induction by breaking n into the individual prime power factors. Cyclotomic and Abelian Extensions, X By using this decomposition of Gal(Q(ζn)=Q), we can show that every abelian group appears as a Galois group over Q: Theorem (Abelian Galois Groups over Q) If G is an abelian group, then there exists an extension K=Q with Galois group isomorphic to G. For general finite groups G, it is still an open problem whether G is the Galois group of some extension K=Q. The problem of computing which groups occur as Galois groups over Q, or more generally over an arbitrary field F , is known as the inverse Galois problem. Cyclotomic and Abelian Extensions, XI Proof: By the classification of finite abelian groups, G is isomorphic to a direct product of cyclic groups, say as ∼ G = (Z=m1Z) × · · · × (Z=mk Z). By a theorem of Dirichlet, for any positive integer m there exist infinitely many primes congruent to 1 modulo m. In particular, we may choose distinct primes pi such that pi ≡ 1 mod mi for each i. Then since mi divides jGal(Q(ζpi )=Qj = pi − 1 and Gal(Q(ζpi )=Q is cyclic, there exists a subgroup of index mi . Cyclotomic and Abelian Extensions, XII Proof (continued): If Ki represents the corresponding fixed field, then Ki =Q is Galois (since Gal(Q(ζpi )=Q is abelian, so every subgroup is normal) and by the fundamental theorem of Galois theory we see that its Galois group is cyclic of order mi . By our results on cyclotomic fields, since the pi are distinct primes, the intersection of any two of the fields Q(ζpi ) is Q, so the same holds for the fields Ki . Hence by our results on Galois groups of composites, we see that the Galois group of K = K1K2 ··· Kk over Q is ∼ isomorphic to Gal(K1=Q) × Gal(K2=Q) × · · · × Gal(Kk =Q) = ∼ (Z=m1Z) × · · · × (Z=mk Z) = G, as desired.