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Qube root of 2, zero of second order polynomial

  1. Mar 4, 2012 #1
    How do you prove that there does not exist numbers [itex]a,b\in\mathbb{Q}[/itex] such that

    [tex]
    0 = a + b\sqrt[3]{2} + \sqrt[3]{2}^2
    [/tex]
     
  2. jcsd
  3. Mar 4, 2012 #2

    micromass

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    The polynomial [itex]X^2+aX+c[/itex] will have to divide [itex]X^3-2[/itex] in that case.
     
  4. Mar 4, 2012 #3
    I think we only know that [itex]X-\sqrt[3]{2}[/itex] must divide [itex]X^2+bX+a[/itex].

    [itex]X^2+bX+a[/itex] doesn't need to divide anything.
     
  5. Mar 4, 2012 #4

    micromass

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  6. Mar 4, 2012 #5
    I see.

    I have tried to read Galois theory earlier, and now I started remembering stuff. :cool: (Although that Wikipedia-page didn't help much...)

    The knowledge that [itex]X^2+bX+a[/itex] must divide [itex]X^3-2[/itex] is one possible way to the proof, but actually the idea of the minimal polynomial can be used in more primitive ways too. For example, simply write

    [tex]
    X^3 - 2 = (X^2 + bX + a)(X - b) + (b^2 - a)X + ab - 2
    [/tex]

    and proof starts to appear.
     
  7. Mar 4, 2012 #6
    What is the logic behind this statement? From a false premiss you can of course deduce anyhing you like, but I assume that is not your intended logic here. When it comes to minimal polynomials and such, the general result is usually the other way around: if g(x)[itex]\epsilon[/itex]Q[x] has a root [itex]\alpha[/itex], then the minimal polynomial of [itex]\alpha[/itex] divides g.
     
  8. Mar 4, 2012 #7

    Office_Shredder

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    If the cube root of 2 satisfies a quadratic polynomial, then x3-2 is NOT the minimal polynomial (since there's a lower degree polynomial that gives us zero). So there are two possibilities

    1) The minimal polynomial is degree 1 - obviously false
    2) The minimal polynomial is degree 2 - in this case, the quadratic polynomial we have must be the minimal polynomial
     
  9. Mar 4, 2012 #8
    With the anti-thesis assumption the [itex]X^2+bX+a[/itex] becomes the "new" minimal polynomial. Or at least that's one way to get to the proof. See my reponse #5 to see the essential. There are several ways to complete the proof then.
     
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