Solvability by radicals Rodney Coleman, Laurent Zwald To cite this version: Rodney Coleman, Laurent Zwald. Solvability by radicals. 2020. hal-03066602 HAL Id: hal-03066602 https://hal.archives-ouvertes.fr/hal-03066602 Preprint submitted on 15 Dec 2020 HAL is a multi-disciplinary open access L’archive ouverte pluridisciplinaire HAL, est archive for the deposit and dissemination of sci- destinée au dépôt et à la diffusion de documents entific research documents, whether they are pub- scientifiques de niveau recherche, publiés ou non, lished or not. The documents may come from émanant des établissements d’enseignement et de teaching and research institutions in France or recherche français ou étrangers, des laboratoires abroad, or from public or private research centers. publics ou privés. Solvability by radicals Rodney Coleman, Laurent Zwald December 10, 2020 Abstract In this note we present one of the fundamental theorems of algebra, namely Galois’s theorem concerning the solution of polynomial equations. We will begin with a study of cyclic extensions, before moving onto radical extensions. 1 Cyclic extensions We say that a finite extension E of a field F is cyclic if the Galois group Gal(E=F ) is cyclic. In this section we aim to present some elementary properties of such extensions. We begin with a preliminary result known as Hilbert’s theorem 90. EXTSOLVcycth1 Theorem 1 Let E=F be a finite cyclic Galois extension of degree n. We suppose that σ is a generator of the Galois group Gal(E=F ). If α 2 E, then NE=F (α) = 1 if and only if there exists ∗ β β 2 E such that α = σ(β) . β proof If α = σ(β) , then βσ(β)σ2(β) ··· σn−1(β) N (α) = ασ(α) ··· σn−1(α) = = 1; E=F σ(β)σ2(β) ··· σn−1(β)σn(β) because σn(β) = β. n−1 Now suppose that NE=F (α) = 1, i.e., ασ(α) ··· σ (α) = 1. We define a finite sequence n−1 (δi)i=0 as follows: 2 n−1 δ0 = α; δ1 = ασ(α); δ2 = ασ(α)σ (α); : : : ; δn−1 = ασ(α) ··· σ (α) = NE=F (α) = 1: REDUCdedcor1 From Result 1 the characters 1; σ; : : : ; σn−1 form an independant set and so there exists γ 2 E such that n−2 n−1 δ0γ + δ1σ(γ) + ··· + δn−2σ (γ) + σ (γ) 6= 0: We note this sum β. Then 2 n−1 n σ(β) = σ(δ0)σ(γ) + σ(δ1)σ (γ) + ··· + σ(δn−2)σ (γ) + σ (γ) −1 2 n−1 n = α δ1σ(γ) + δ2σ (γ) + ··· + δn−1σ (γ) + σ (γ); −1 n because α δi = σ(δi−1). As σ = idE, we have n −1 σ (γ) = γ = α δ0γ; hence σ(β) = α−1β. 2 The theorem which we have just proved is often refered to as the ’multiplicative’ version of Hilbert’s theorem 90 to distinguish from the ’additive’ version, which we will now present. 1 EXTSOLVcycth2 Theorem 2 If E=F is a finite cyclic Galois extension of degree n, then TE=F (α) = 0 if and only if there exists β 2 E such that α = β − σ(β). n proof Let σ be a generator of the Galois group G = Gal(E=F ), so that σ = idn. If α = β − σ(β), then n−1 n−1 n X k X k X k TE=F (α) = σ (α) = σ (β) − σ (β) = 0; k=0 k=0 k=1 because σn(β) = β. Conversely, suppose that TE=F (α) = 0. As TE=F 6= 0, we may find an element x such that TE=F (x) 6= 0. Let w = ασ(x) + (α + σ(α)) σ2(x) + ··· + α + σ(α) + ··· + σn−2(α) σn−1(x): Then σ(w) = σ(α)σ2(x) + σ(α) + σ2(α) σ3(x) + ··· + σ(α) + σ2(α) + ··· + σn−1(α) σn(x): Since TE=F (α) = 0, we have −α = σ(α) + ··· + σn−1(α); so the last summand in the expression for σ(w) is −αx, hence 2 n−1 w − σ(w) = α x + σ(x) + σ (x) + ··· + σ (x) = αTE=F (x): Setting β = w , we obtain TE=F (x) w σ(w) w − σ(w) β − σ(β) = − = ; TE=F (x) σ(TE=F (x)) TE=F (x) because TE=F (x) 2 F . Therefore αT (x) β − σ(β) = E=F = α; TE=F (x) as required. 2 In the following we consider extensions E=F of a particular sort, namely where the field F contains a primitive nth root of unity, for some positive integer n. EXTSOLVcycth3 Theorem 3 Let E=F be a cyclic Galois extension of degree n, where F contains a primitive nth root of unity ζ. Then there exists an irreducible polynomial f(X) = −α + Xn 2 F [X] such that E is a splitting field of f. proof Let σ be a generator of the Galois group Gal(E=F ). Since ζ 2 F , we have NE=F (ζ) = ζn = 1 and so, by Hilbert’s theorem 90 (multiplicative version), there is an element β 2 E∗ such β that ζ = σ(β) . Then σ(β) = ζ−1β σ2(β) = σ(ζ)−1σ(β) = ζ−1(ζ−1β) = ζ−2β σ3(β) = σ(ζ)−2σ(β) = ζ−2(ζ−1β) = ζ−3β 2 and generally σm(β) = ζ−mβ, for all m 2 N∗. As the elements σi(β), for 0 ≤ m < n are distinct, n−1 the automomorphisms idjF (β); σjF (β); : : : ; σjF (β) form a set of n distinct elements of the Galois group Gal(F (β)=F ). Then [F (β): F ] ≥ n and n = jGal(E=F )j = [E : F ] = [E : F (β)][F (β): F ] =) [F (β): F ] ≤ n: Therefore [F (β): F ] = n, which implies that E = F (β). We claim that βn 2 F . Now σ(βn) = σ(β)n = ζ−nβn = βn; and it follows that βn belongs to the fixed field of Gal(E=F ), i.e., F . Thus there exists α 2 F such that βn = α. We have shown that E = F (β), where β is a root of the polynomial f(X) = −α + Xn. The roots of f have the form ζiβ, for 0 ≤ i < n and ζ 2 F by hypoth- esis, so E contains all the roots of f and hence is a splitting field of the polynomial f. Since [F (β): F ] = n, the degree of the minimal polynomial m(β; F ) is n, hence f = m(β; F ); it follows that f is irreducible. 2 The theorem which we have just proved has a converse. EXTSOLVcycth4 Theorem 4 Let F be a field containing an nth primitive root of unity ζ and E a splitting field of the polynomial f(X) = −α + Xn 2 F [X], with α 6= 0. Then E = F (β), where β is a root of f, and the Galois group G = Gal(E=F ) is cyclic of order dividing n. The order of G is equal to n if and only if f is irreducible. proof If β is root of f, then the roots of f have the form ζiβ, for 0 ≤ i < n. As ζ 2 F , all the roots of f lie in F (β), so F (β) is a splitting field of f and so E = f(β). Let σ 2 G = Gal(E=F ). As σ permutes the roots of f, we may write σ(β) = ζk(σ)β, where k(σ) is uniquely determined modulo n. We thus obtain a mapping φ : G −! (Zn; +); σ 7−! [k(σ)]; which is clearly a group homomorphism. If φ(σ) = [0], then σ(β) = β, which implies that σ is the identity on F (β) and so φ is a monomorphism. Since G is isomorphic to a subgroup of the cyclic group (Z ; +), G is cyclic of order dividing n. n thGALPOLYirred2 If f is irreducible, then G acts transitively on the roots of f (Result 4). Thus there exists σ 2 G such that σ(β) = ζiβ, for 0 ≤ i < n. It follows that φ is surjective, so G is isomorphic to (Zn; +). Now suppose that f is reducible. As f has no multiple roots, f has two distinct irreducible propGALPOLYirred1 factors, so G does not act transitively on the roots of f (Result 5). Thus there exist roots ζiβ and ζjβ, with j > i, for which there exists no element σ 2 G such that σ(ζiβ) = ζjβ, i.e., σ(β) = ζj−iβ. It follows that φ is not surjective and so jGj < n. 2 Corollary 1 Let p be prime, F a field containing a primitive pth root of unity and E a splitting field of the polynomial f(X) = −α + Xp 2 F [X], with α 6= 0. Then either f is irreducible and Gal(E=F ) ' Zp or f splits in F and GalE=F ) = fidF g. EXTSOLVcycth4 proof The order of Gal(E=F ) = p or 1. In the first case, from Theorem 4, f is irreducible and Gal(E=F ) ' Zp, because Gal(E=F ) is cyclic. Suppose now that the order of Gal(E=F ) is 1. Then [E : F ] = 1. There exists β 2 E such that E = F (β), so [F (β): F ] = 1, which implies that β 2 F . It now follows that f splits in F . 2 3 2 Radical extensions Suppose that E=F is a field extension such that E = F (α) for some α 2 E. If there exists n 2 N∗ such that αn 2 F , then we say that E is a pure extension of type n.