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1 13. Radical and Cyclic Extensions The main purpose of this section is to look at the Galois groups of x n a. The first case to consider is a = 1. Definition Let K be a field. An element ω K is said to be a primitive nth root of unity if ω n = 1, but no smaller power is equal to one. Let L/K be a field in which x n 1 splits. polynomial is defined to be Φ n (x) = ω (x ω), The nth cyclotomic where the product runs over the primitive nth roots of unity. Definition Let L/K be a Galois extension with Galois group G. Let α be an element of L. We say that β is a conjugate of α if β lies in the same orbit as α under the natural action of G. Lemma Let L/K be a Galois extension with Galois group G. Suppose that f(x) L[x] splits in L. Then f(x) K[x] if and only if the set of roots of f(x) is a union of orbits of G. Furthermore f(x) is irreducible if and only if the set of roots is an orbit of G. In particular if α L, then the minimum polynomial of α is (x β), where the product runs over the conjugates of α. β Proof. It suffices to prove that f(x) K[x] is irreducible if and only if the set of roots of f(x) is an orbit of G, that is, it suffices to prove the last statement. Suppose that f(x) = β (x β), where the product runs over the conjugates of α L. Then f(x) is invariant under the action of the Galois group, since any element of the Galois group simply switches the factors and this won t change the product. But then each individual coefficient of f(x) is invariant under every element of G, that is, each coefficient lies in L G = K. Thus f(x) K[x]. 1

2 On the other hand G acts transitively on the roots of any irreducible polynomial. Lemma Φ n (x) lies in the groundfield K. Proof. x n 1 is certainly a polynomial with coefficients in K. Let L/K be a splitting field, with Galois group G. Now Φ n (x) certainly divides x n 1 in L. On the other hand, if ω is a primitive nth root of unity, then the other roots of x n 1 are simply given as the n powers of ω, ω, ω 2, ω 3,..., ω n. Thus L = K(ω). Thus the action of an element φ of G is determined by its effect on ω. As φ must send ω to another generator, the conjugates of ω are all primitive roots of unity. Now apply (13.3). Lemma x n 1 = d n Φ d (x). Proof. Easy, since any nth root of unity is a primitive root of unity for some unique d n. Example Φ 1 (x) = x 1. x 2 1 = Φ 1 (x)φ 2 (x). Thus Φ 2 (x) = x + 1 (also clear, since 1 is the unique primitive 2th root of unity). x 4 1 = (x 2 1)(x 2 + 1). Thus Φ 4 (x) = x Lemma In characteristic zero, Φ n (x) Z[x] and the content is one. Proof. We already know that Φ n (x) Q[x]. On the other hand, by induction x n 1 = Φ n (x)f(x) where, by induction and (13.5), f(x) is integral and the content is one. The result follows as in the proof of Gauss Lemma. Definition U n denotes the group of units in Z/nZ. Proposition Let K be a field whose characteristic is coprime to n and let L/K be a splitting field for x n 1. Then the Galois group G of L/K is naturally isomorphic to a subgroup of U n, with equality if and only if Φ n (x) is irreducible over K. 2

3 Proof. First note that x n 1 has no repeated roots, as its derivative is nx n 1 0. Thus L/K is Galois. Let ω be a primitive nth root of unity. Define a map f : G U n by sending σ to i, where σ(ω) = ω i. This makes sense, as σ must permute the roots of x n 1, and the roots are nothing more than the powers of ω. On the other hand, ω is a generator of L/K, so that ω i is also a generator of L/K. Thus ω i is also a primitive root of unity. Thus i is coprime to n, so that i is a unit modulo n. Further σ is determined by its action on ω, so that f is injective. Suppose that f(σ) = i and f(τ) = j. Then σ(ω) = ω i and τ(ω) = ω j. In this case (τ σ)(ω) = τ(σ(ω)) = τ(ω i ) = (τ(ω)) i = (ω j ) i = ω ij. Thus f(τσ) = ij and so f is a group homomorphism. The calculation above also shows that if we pick another primitive root of unity ω i, that τ also acts on ω i, by raising to the power j. Thus G is naturally isomorphic to a subset of U n. Now G = U n if and only if we can move ω to any other primitive root of unity. But, by (13.3), this is equivalent to saying that Φ n (x) is irreducible. Here is the main: Theorem Φ n (x) is irreducible over Q. Proof. Suppose not. Then by Gauss Lemma, we may write Φ n (x) = f(x)g(x), where f(x) and g(x) are monic integral polynomials of non-zero degree and f(x) is irreducible. Suppose that ω is a root of f(x). Let p be a prime that does not divide n. I claim that ω p is a root of f(x). Suppose not. Then ω p is a root of g(x). But then ω is a root of h(x) = g(x p ). Thus f(x) divides h(x), as f(x) is the minimum polynomial of ω. Thus we have h(x) = f(x)k(x), 3

4 Now reduce modulo p, Z F p. We get h(x) = ḡ(x p ) = (ḡ(x)) p. Let q be an irreducible factor of f(x) F p [x]. Then q divides ḡ(x) and so q 2 divides fḡ = Φ n (x). But then Φ n (x) would have repeated roots, a contradiction as n is coprime to p. Thus ω p is also a root of f(x). But given any primitive nth root of unity, we may write it as ω m, for some m coprime to n. In this case m = p 1 p 2... p k, for primes p 1, p 2,..., p k coprime to n. Repeating the argument above k times, we get that ω m is a root of f(x). But then f(x) = Φ m (x), a contradiction. Example Let us calculate the Galois group of x 6 1 over Q. By (13.10) this is isomorphic to U 6. Of the numbers, 1, 2, 3, 4, 5, only 1 and 5 are coprime to 6. Thus the group has order two and is isomorphic to Z 2. Now we turn to the general problem. Lemma Let L/K be a splitting field for x n a K[x], a 0. Then there is an intermediary field M which is a splitting field for x n 1. Proof. If n = mp and a = b p, where p is the characteristic of K, then x n a = (x m b) p. Thus we may as well assume that n is coprime to the characteristic. In particular x n a has n distinct roots α 1, α 2,..., α n. Let α and β be two roots of x n a. Then α/β is a root of x n 1. Thus α i where 1 i n, α 1 are the n distinct roots of x n 1. Lemma Let K be a field in which x n 1 splits and let L/K be a Galois extension with Galois group G of degree coprime to the characteristic. Then L/K is the splitting field of an irreducible polynomial x n a if and only if the Galois G has order n. Proof. Suppose that L/K is the splitting field of x n a. Let α be a root of x n a and let ω K be a primitive nth root of unity. We have 4

5 already seen that if α is a root of x n a then the other roots are ζα, where ζ = ω i is an nth root of unity. Define a map f : G Z n, by sending σ to i, where f(α) = ω i α. As α generates L, the action of σ is determined by its action on α. Thus f is injective. Now the Galois group is transitive on the roots, as x n a is irreducible. Thus f is surjective. Suppose that σ and τ G and f(σ) = i and f(τ) = j. Thus σ(α) = ω i α and τ(α) = ω j α. In this case (τ σ)(α) = τ(σ(α)) = τ(ω i α) = ω i τ(α) = ω i ω j α = ω i+j α. Thus f is a group homomorphism and so f is an isomorphism. Now suppose that G is cyclic, with generator σ. Then the automorphisms 1, σ, σ 2,..., σ n 1, are distinct automorphisms and therefore independent over L. Thus 1 + ωσ + ω 2 σ ω n 1 σ n 1 0. Thus there we may find β L such that α = β + ωσ(β) + ω 2 σ 2 (β) + + ω n 1 σ n 1 (β) 0. Note that σ(α) = ω 1 α. Let a = α n. Then a is invariant under σ, whence G, so that a K = L G. Note that G acts transtively on the roots of x n a, which are ω i α, so that x n a is irreducible. Thus α has degree n over K, so that L = K(α). Corollary Let x p a K[x], where p is a prime number coprime to the characteristic and suppose that x p 1 splits in K. Then either x p a splits in K or x p a is irreducible. Proof. Let L/K be a splitting field for x p a. Then the Galois group G is a subgroup of Z p, since if α is a root of x p a, then the action of any element of G is to send α to ω i α, where i Z p as above. Thus either G is the trivial group, when x p a splits in K or G = Z p, when the Galois group acts transitively on the roots of x p a and x p a is irreducible. 5

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