Lemma Let f(x) K[x] be a separable polynomial of degree n. Then the Galois group is a subgroup of S n, the permutations of the roots.

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1 15 Cubics, Quartics ad Polygos It is iterestig to chase through the argumets of 14 ad see how this affects solvig polyomial equatios i specific examples We make a global assumptio that the characteristic is either 2 or 3 Lemma 151 Let f(x) K[x] be a separable polyomial of degree The the Galois group is a subgroup of S, the permutatios of the roots Proof Clear, sice ay automorphism of a splittig field is determied by its actio o the roots Now A S ad so H = G A G is either equal to G or of idex two If we have the latter, by the Fudametal Theorem, it follows that there is a quadratic extesio M/K Sice this is uiversally true, o matter which field we start with, we might well expect that there is some uiversal formula which determies M Defiitio 152 Let f(x) K[x] be a polyomial, i which f(x) splits as f(x) = λ(x α 1 )(x α 2 ) (x α) The discrimiat is the square of the product δ = i<j(α i α j ) Lemma 153 Let f(x) K[x] be a polyomial with splittig field L/K ad discrimiat L The K ad = 0 if ad oly if f(x) has a repeated root Moreover if 0 the x 2 splits i K[x] if ad oly if the Galois group is a subgroup of A Proof The secod statemet is immediate If 0, the f(x) is surely separable ad so L/K is Galois We already kow that δ is ivariat uder the actio of A ad that a arbitrary elemet of S fixes δ up to sig Thus = δ 2 lies i the fixed field of G, which by the Fudametal Theorem of Galois Theory is equal to K Fially x 2 splits i K if ad oly if δ K if ad oly if δ is ivariat uder G if ad oly if G A We tur to the calculatio of the discrimiat 1

2 Defiitio 154 Let K be a field ad let λ 1, λ 2,, λ be scalars The determiat λ 1 λ 2 λ 3 λ λ 2 1 λ 2 2 λ 2 3 λ 2 λ 1 1 λ2 1 λ 1 2 λ 1 is kow as the Vadermode determiat Lemma 155 The Vadermode determiat is equal to λ 1 λ 2 λ 3 λ λ 2 1 λ 2 2 λ 2 3 λ 2 λ 1 1 λ 1 2 λ 1 = i λ j ) i<j(λ 2 λ 1 Proof First ote that we may replace λ i by the variable x i I this case both sides are polyomials i x 1, x 2,, x ad so both sides are elemets of the rig R = K[x 1, x 2,, x ] By uique factorisatio ad cosideratios of degree it suffices to check that x i x j is a factor of the LHS ad that the costat coefficiets match up The latter is a easy check To check that x i x j divides the LHS, it suffices to check that the LHS vaishes whe λ i = λ j But this is clear, as the we are takig the determiat of a matrix with two equal colums Remark 156 The Vadermode determiat provides a slick way of checkig that A is a ormal subgroup The key poit to check is that a traspositio, actig o δ, switches the sig But this is clear, lookig at the LHS, sice the determiat chages sig, whe oe switches two colums 2

3 = δ 2 = δ δ α 1 α 2 α 3 α α = α1 2 α2 2 α3 2 α 2 1 α 2 α 3 α α1 2 α2 2 α3 2 α 2 α1 1 α2 1 α2 1 α 1 α1 1 α2 1 α2 1 α 1 1 α 1 α1 2 α 1 1 α 1 α 2 α 3 α 1 α = α1 2 α2 2 α3 2 α 2 2 α2 2 α α 3 α3 2 α 1 3, α1 1 α2 1 α2 1 α 1 1 α α 2 α 1 where we used the fact that takig trasposes does ot affect the determiat The last product ca be computed by first multiplyig the matrices together ad the computig the determiat, as det(ab) = det A det B Rather tha write dow the geeral formula, it is perhaps more iterestig to compute i some relatively simple cases If = 2 we get ( ) ( ) ( ) α 2 α + β = α β 1 β α + β α 2 + β 2 Suppose f(x) = x 2 + ax + b = (x α)(x β) Multiplyig out we get (x α)(x β) = x 2 (α + β)x + αβ = x 2 + ax + b, so that comparig coefficiets, we have α + β = a ad αβ = b 3

4 a 2 = (α + β) 2 = α 2 + β 2 + 2αβ = (α 2 + β 2 ) + 2b So α 2 + β 2 = a 2 2b Thus is equal to 2 a a a 2 2b = 2(a2 2b) a 2 = a 2 4b, which should look familiar Oe ca make a similar computatio for cubics I this case we have Propositio 157 Let f(x) K[x] be a irreducible cubic The the Galois group is isomorphic to A 3 if x 2 splits i K ad is equal to S 3 otherwise Proof The Galois group is a trasitive subgroup of S 3, of which there are oly two, A 3 ad S 3 But G A 3 if ad oly if x 2 splits i K This gives us a method to solve the cubic First compute the itermediate field M correspodig to A 3, that is, adjoi the square root of The resultig field extesio L/M has Galois group isomorphic to Z 3, thus there ought to be a expressio ivolvig δ ad the coefficiets of the cubic, for which we eed to take a cube root We ow apply a similar techique for quartics Propositio 158 Let f(x) K[x] be a irreducible quartic The the Galois group is isomorphic to oe of (1) S 4, (2) D 4, (3) Z 4 (4) A 4, or (5) V = {e, (1, 2)(3, 4), (1, 3)(2, 4), (1, 4)(2, 3)} The latter two occur if ad oly if x 2 splits i K Proof These are oly the oly trasitive subgroups of S 4 Oce agai, this ought to yield a method to solve the quartic First adjoi δ, the square root of, to reduce the Galois group to A 4 Now use the fact that V A 4 is a ormal subgroup, with quotiet Z 3, to fid a further field extesio, that is obtaied by adjoiig appropriate 4

5 cube roots This reduces the Galois group to Z 2 Z 2 The remaiig field extesio is obtaied by adjoiig successive square roots Thus the geeral form of a solutio to a quartic equatio, ivolves takig square roots ad cube roots oly I practice determiig these formulas is somewhat ivolved ad uispirig A much more iterestig questio is to determie those regular polygos which are costructible Lemma 159 The regular -go is costructible if ad oly if the agle 2π is costructible Proof Suppose the regular -go is costructible The the agle subteded at the cetre of the -go (which is surely costructible) by two adjacet vertices is 2π Coversely suppose we ca costruct the agle 2π The we ca costruct the agles a 2π Place poits o the uit circle with the above agles ad simply joi up the poits Lemma 1510 If the regular polygo with m sides is costructible the the regular polygos with sides ad m sides are costructible Further if m ad are coprime, the coverse holds Proof Oe directio is clear If you ca costruct the regular polygo with m sides, the you ca certaily costruct the regular polygo with ad m sides Now suppose that m ad are coprime The there are itegers a ad b such that 1 = am + b, so that 1 m = a + b m By assumptio we ca costruct the agles ad so we ca costruct 2π ad 2π m, 2π m = a2π b2π m But the the m-go is cotructible 5

6 Usig (1510), to aswer the questio of which -gos are costructible, we oly eed to cosider the case whe is a power of a prime Sice we ca bisect ay agle, we ca certaily costruct ay 2 k -go Now costructig the agle 2π/ is basically the same as showig that a primitive th root of uity has degree a power of two over Q Lemma 1511 The agle θ = 2π/ is cotructible if ad oly if the degree of the miium polyomial of ω = e 2πi/ is a power of two Proof The agle θ is costructible if ad oly if the legths α = cos θ ad si θ are costructible So if the agle θ is costructible, the the degree of the miimum polyomial of α is a power of two Now As ω = ω 1, we have So ω is a root of the polyomial ω + ω = 2α ω 2 2αω + 1 = 0 x 2 2αx + 1 Q(α)[x] Thus the degree of ω over Q(α) is either oe or two Now apply the Tower Law Note that ω is a primitive root of uity Lemma 1512 Let ω be a primitive p k -th root of uity, where p is a odd prime If ω has degree a power of two over Q the k = 1 ad p = 2 s + 1, for some s Proof Φ p (x) = x p 1 + x p So the degree of ω, i the case k = 1, is p 1 As this is a power of two, we have p = 2 s + 1 Now if k > 1, we may as well assume that k = 2 Now So x p2 1 = Φ 1 (x)φ p (x)φ p 2(x) p 2 = 1 + (p 1) + d, where d is the degree of ω Thus d = p 2 p = p(p 1), which is ever a power of two 6

7 Lemma 1513 Let p be a prime of the form The s is a power of two 2 s + 1 Proof Suppose ot The we could write s = ab, where a is odd Now But the would ot be prime x a + 1 = (x + 1)(x a 1 x a 2 + ) p = (2 b ) a + 1, Theorem 1514 The regular -go is costructible if ad oly if = 2 k p 1, p 2,, p m, where p 1, p 2,, p m are distict odd primes of the form 2 2k + 1 Proof By what we have already proved, it suffices to cosider the case is a odd prime, of the form 2 2k + 1, ad we oly eed to prove that the correspodig agle is costructible Cosider the Galois group G of x 1 This is abelia, isomorphic to U 2 2k The order of this group is 2 2k 2 2k 1 = 2 2k 1, a power of two Thus G is a 2-group ad we ca filter a splittig field Q(ω)/Q by itermediary fields, all of which are quadratic extesios of the previous field Thus we ca do the same for the subfields, Q(cos θ) ad Q(si θ) But the cos θ ad si θ are costructible, which is what we wat 7