Assignment:
Q1. If [K:F] is finite and u is algebraic over K ,prove that [F(u):F] divides [K(u):F]
Hint:[F(u):F] and [K(u):F(u)] are finite by Theorems
Apply Theorem to F⊆F(u)⊆K(u)
Theorem 1 Let F,K and L be fields with F ⊆ K ⊆ L .If [K:F] and [L:K] are finite ,then L is a finite dimensional extension of F and [L:F] = [L:K] [K:F]
Theorem 2 Let K be an extension field of F and u ∈K an algebraic element over F with minimal polynomial p(x) of degree n.Then
F(u) ≅ F[x]/(p(x))
{1_F,u ,u^2 , ……………..,u^(n-1)} is a basis of the vector space F(u) over F
[F(u):F] = n
Theorem 3 Shows that when u is algebraic over F,then F(u) does not depend on K but is completely determined by F[x] and the minimal polynomial p(x).Consequently ,we sometimes say that F(u) is the field obtained by adjoining u to F
Theorem 4 If K is a finite-dimensional extension field of F,then K is an algebraic extension of F.
Q2. Assume that u,v ∈ K are algebraic over F,with minimal polynomial p(x) and q(x),respectively.
If deg p(x) = m and deg q(x) = n and (m,n) = 1, prove that [F(u,v):F] = mn.
Show by example that the conclusion of part (a) may be false if m and n are not relatively prime.
What is [Q(√2 ,?2):Q]?
Provide complete and step by step solution for the question and show calculations and use formulas.