By Seyfi Türkelli
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Sample text
Then, the elements ξ r α’s are the roots of the polynomial xn − a. This implies that αd ∈ k where d = [k(α) : k]. And, clearly, σ ∈ G defined by σ(α) = ξα generates G. Hence, all is proven. 3. (Artin-Schreier) Let k be a field of characteristic p. i. Let K/k be a cyclic Galois extension of degree p with the Galois group G. Then there exists α ∈ K such that K = k(α) and α satisfies the equation xp − x − a = 0 with some a ∈ k. ii. Conversely, given a ∈ k, the polynomial f (x) = xp − x − a either has one root in k, in which case all its roots are in k, or it is irreducible.
Let K/k be a cyclic extension of degree n. Then there exists α ∈ K such that K = k(α) and αn − a = 0 for some a ∈ k. ii. Conversely, if there exists α ∈ K with αn = a ∈ k then k(α) is cyclic over k of degree d and αd ∈ k. Proof. Let G = Gal(K/k), ξ ∈ k be the n-th root of unity and σ ∈ G be a generator. Let f : G → K ∗ be a continuous map with f (σ r ) = ξ r . One can see that f is a cocycle. , n, and so σ r (α) = ξ r α. Since ξ is a primitive n-th root of unity in k, the elements σ r (α) = ξ r α’s are distinct and [k(α) : k] ≥ n; Thus k(α) = K.
And, we have σ(αn ) = σ(α)n = (ξα)n = αn . As all the elements of G fixes αn , αn ∈ k and the first claim is proven. Let ξ ∈ k be again our primitive n-th root of unity. Then, the elements ξ r α’s are the roots of the polynomial xn − a. This implies that αd ∈ k where d = [k(α) : k]. And, clearly, σ ∈ G defined by σ(α) = ξα generates G. Hence, all is proven. 3. (Artin-Schreier) Let k be a field of characteristic p. i. Let K/k be a cyclic Galois extension of degree p with the Galois group G. Then there exists α ∈ K such that K = k(α) and α satisfies the equation xp − x − a = 0 with some a ∈ k.