Download Algebra II - Noncommunicative Rings, Identities by A. I. Kostrikin, I. R. Shafarevich PDF

By A. I. Kostrikin, I. R. Shafarevich

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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.

Noting that the action of G on M{1} (k + ) is defined by (λf )(x) := f (xλ), it is trivial to verify that T is a G−module monomorphism, and it is, in fact, an isomorphism of G−modules since G is finite and so f (σ −1 ) can take any value in k for any σ ∈ G. Hence, G we have M{1} (k + ) ∼ = K + and, for r > 0 G H r (K/k; K + ) = H r (K/k; M{1} (k + )) ∼ = H r (1; k + ) = 0. 31 Now, in order to prove our second claim, let f : G → K ∗ be a continuous map representing the cohomology class u ∈ H 1 (K/k; K ∗ ).

Proof. Let σ be a generator of G. Clearly, the automatically-continuous map f : G → K + defined by f (σ r ) = r is a cocyle. , p − 1 or, in other words, σ r (α) = α + r. Therefore, α has p distinct conjugates and [k(α) : k] ≥ p; Thus K = k(α). Noting that σ(αp − α) = σ(α)p − σ(α) = (α + 1)p − (α + 1) = αp − α, we get that αp − α is fixed by G and lies in k. Hence, by taking a = αp − α, we have proved the first claim. If α is a root of f then (α + i) s are all the roots of f . Then, it follows from a simple analysis that f is irreducible when α (or equivalently f has no root) is not in k.

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