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Field theory

  1. Feb 13, 2008 #1
    [SOLVED] field theory

    1. The problem statement, all variables and given/known data
    Assume pi is transcendental over Q. Find a subfield F of the reals such that pi is algebraic of degree 3 over F.


    2. Relevant equations



    3. The attempt at a solution
    Umm...the only subfield I know of the reals is the rationals. Is the answer Q(pi^(1/3))? Do people understand the that simple extension notation? How would you read Q(pi^(1/3)) in English?
     
    Last edited: Feb 13, 2008
  2. jcsd
  3. Feb 13, 2008 #2

    NateTG

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    I think a field containing [itex]\pi^{\frac{1}{3}}[/itex] would also contain [itex]\pi[/itex], so [itex]\pi[/itex] would end up being algebraic of degree zero.

    [itex]\mathbb{Q}(\sqrt[3]{\pi})[/itex] can be read as "the rational numbers adjoin the cube root of pi."
     
  4. Feb 13, 2008 #3
    Do you mean degree 1?

    What should I adjoin to Q then? pi^3?
     
  5. Feb 13, 2008 #4
    Yes, and you use the fact that pi is transcendental to show that pi is not contained in Q(pi^3).
     
  6. Feb 13, 2008 #5
    It just seems obvious that there is now way you get pi with a polynomial over the rationals evaluated at pi^3. Is there a better explanation?

    Also, I understand why adjoining pi^(1/3) makes no sense. But can you explain the thought-process that gave away why adjoining pi^3 does give you a third-degree polynomial that has pi as a zero and why there is no first or second degree polynomial in Q(pi^3) that will have pi as a zero?
     
  7. Feb 14, 2008 #6

    NateTG

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    The first part is easy - the polynomial with [itex]\pi[/itex] as a solution will be:
    [tex]x^3-\pi^3[/tex]

    For the second part, the existance of such polynomials would require the field to contain [itex]\pi[itex] or [itex]\pi^2[/itex].
     
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