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Prove that an element is algebraic over a field extension

  1. Jan 29, 2012 #1
    1. The problem statement, all variables and given/known data
    Let F|K be a field extension. If v e F is algebraic over K(u) for some u e F and v is transcendental over K, then u is algebraic over K(v).


    2. Relevant equations
    v transcendental over K implies K(v) iso to K(x).
    Know also that there exists f e K(u)[x] with f(v) = 0.


    3. The attempt at a solution
    Want to show that there exists h e K(v)[x] with h(u) = 0.
    I'm trying to find this directly since I don't see a contrapositive proof working out. I feel like I should use v alg|K(u) to get that [itex]K(u)(v) \cong K(u)[x]/(f)[/itex] though I'm not sure how to get to that h in K(v)[x]. Somehow pass to that quotient field and show that u is a root of some remainder polynomial of f?
     
  2. jcsd
  3. Jan 29, 2012 #2

    morphism

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    Hint: K(u,v) is finite over K(u).
     
  4. Jan 29, 2012 #3
    Thanks for the help, morphism.

    So then I have [K(u,v):K] = [K(u,v):K(v)][K(v):K] = [K(u,v):K(u)][K(u):K] = n[K(u):K]. Since v trans|K, [K(v):K] is inf, and by equality [K(u):K] must be inf. But this means that u is trans|K, so there is an isomorphism from K(u) → K(v) mapping u → v. But this implies u is alg|K(v). Is that right, or did I get off track somewhere?
     
  5. Jan 29, 2012 #4

    morphism

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    My only objection is to the sentence "But this implies u is alg|K(v)."

    How did you arrive at this conclusion?
     
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