Galois Theory, U Glasgow course by John B. Fraleigh

By John B. Fraleigh

Thought of a vintage by way of many, a primary path in summary Algebra is an in-depth, creation to summary algebra. serious about teams, earrings and fields, this article provides scholars a company beginning for extra really expert paintings by way of emphasizing an realizing of the character of algebraic buildings. The 6th version of this article maintains the culture of educating in a classical demeanour whereas integrating box idea and a revised bankruptcy 0. New routines have been written, and former workouts have been revised and converted.

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5 we know that for such a Galois extension E/K, [E : K] = (E : K) and also every monomorphism ϕ ∈ MonoK (E, K) maps E into itself, hence restricts to an automorphism of E which will be denoted ϕ|E . 49, every automorphism α ∈ AutK (E) extends to a monomorphism E −→ K fixing elements of K. 1) | AutK (E)| = (E : K) = [E : K]. 2. For a finite Galois extension E/K, the group Gal(E/K) = AutK (E) is called the Galois group of the extension or the Galois group of E over K. 2) | Gal(E/K)| = (E : K) = [E : K].

Iii) The extension K(t)/K is finite dimensional. 3. 2, ker εt = (minpolyK,t (X)) = (0), where minpolyK,t (X) ∈ K[X] is an irreducible monic polynomial called the minimal polynomial of t over K. The degree of minpolyK,t (X) is called the degree of t over K and denoted by degK t. 4. If t ∈ L is algebraic over K then [K(t) : K] = deg minpolyK,t (X) = degK t. Proof. 9(ii). 5. Suppose that t ∈ L is algebraic over K and p(X) ∈ ker εt with deg p(X) = deg minpolyK,t (X). Then minpolyK,t (X) | p(X) and so p(X) = u minpolyK,t (X) for some u ∈ K.

Uk ∈ K, then the multiplicities of the uj are equal. Hence in K[X], p(X) = c(X − u1 )m · · · (X − uk )m , where c ∈ K and m 1. Proof. Let u ∈ K be a root of p(X) and suppose that it has multiplicity m, so we can write p(X) = (X − u)m p1 (X) where p1 (X) ∈ K(u)[X] and p1 (u) = 0. Now let v ∈ K be any other root of p(X). 34, there is a monomorphism ϕv : K(u) −→ K for which ϕv (u) = v. When p(X) is viewed as an element of K(u)[X], the coefficients of p(X) are fixed by ϕv . Then ϕv ((X − u)m p1 (X)) = (X − u)m p1 (X), and so (X − v)m p1 (X) = (X − u)m p1 (X), where p1 (X) ∈ K[X] is obtained applying ϕv to the coefficients of p1 (X).

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