Extension Field : Kronecker's Theorem

Related BrainMass Solutions

• Finite Extension Field and Isomorphism

  Theorem 7.12 from Fraleigh: A finite extension field E of a field F is an algebraic extension of F Algebraic extension: every element of E is algebraic over F.

• Splitting Fields : Let F be an extension field of K of degree 2, then F is the splitting field over K for some polynomial.

  Proof: Theorem: If F is a finite dimensional extension field of K, then F is finitely generated and algebraic over K. According to this theorem, since [F:K]=2 is finite, then F is algebraic over K.

• Algebraic Extensions

  Theorem 10.4 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 10.7 Let K be an extension field of F and u an algebraic element

• Galois Theory : Use the method found in the proof below to find a primitive element for the extension Q(i, 5^(1/4)) over Q.

  *Recall that the extension field F of K is called a simple extension if there exists an element In this case, u is called a primitive element. **Theorem: Let F be a finite extension of the field K.

• multiplication of degrees theorem

  393070 Multiplication of Degrees Theorem Let u be an algebraic element of K (field extension of F) whose minimal polynomial in F[x] has prime degree.

• Show that a finite extension of fields K/k which have prime degree have only trivial subextensions.

  If F is some intermediate field between k and K, then [K:k] = [K:F][F:k] (where [] denotes the degree) by a well-known and basic theorem. Now, [K:k] is prime, so one of the degrees [K:F] or [F:k] must be 1. and in that case F=K or F=k.

• Proposition

  We now introduce the following theorem. Theorem 1. A nonzero polynomial f in F[x] is separable if and only if it's relatively prime to its derivative f^' in F[x].

• Solvable Roots and Solvable Group

  (i) The roots of f are solvable; (ii) The splitting field K of f is contained in some normal radical extension of F; (iii) The Galois group of f is solvable. (i)⇒(ii): The field K obviously is a normal extension of F.

• Transcendence basis

  Proof: From the condition, we know that is an extension field of that is an extension field of . In another word, . has transcendence degree over and is algebraic over . First, I claim that is algebraically independent over .