1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

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

    How did you arrive at this conclusion?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook