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Minimal polynomial of N^(1/M)

  1. Apr 22, 2013 #1
    Assume that [itex]\sqrt[M]{N}[/itex] is irrational where [itex]N,M[/itex] are positive integers. I'm under belief that

    [tex]
    X^M-N
    [/tex]

    is the minimal polynomial of [itex]\sqrt[M]{N}[/itex] in [itex]\mathbb{Q}[X][/itex], but I cannot figure out the proof. Assume as an antithesis that [itex]p(X)\in\mathbb{Q}[X][/itex] is the minimal polynomial such that [itex]\partial p < M[/itex].

    We can divide [itex]X^M-N[/itex] by [itex]p(X)[/itex] and obtain

    [tex]
    X^M-N = q(X)p(X) + r(X)
    [/tex]

    where [itex]\partial r < \partial p[/itex]. According to the assumption, [itex]r=0[/itex] must hold.

    So I'm somehow supposed to use the information [itex]X^M-N=q(X)p(X)[/itex] and [itex]r=0[/itex] and find a contradiction.

    Doesn't look very obvious to me.
     
  2. jcsd
  3. Apr 22, 2013 #2

    mfb

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    [itex]\sqrt[4]{4}[/itex]?

    ##X^2-2=0##
     
  4. Apr 22, 2013 #3
    I had never really thought about this. All I know is that it will be true when M is a prime number, or if N is a number with a square-free prime factorization ( use eisenstein's ). I don't if there are easier conditions

    edit: also, I think it will be true if M is not a power of 2. Over Q, you can factorize your polynomial into irreducibles that are either linear factors of x^2 + n terms, where n is a positive integer. If M is not a power of 2, then you must have a linear factor in such a decomposition. Then, the polynomial must have a rational root. But it is easy to see that it cannot, since this is a monic polynomial with coefficients in Z. Any rational root of these polynomials must themselves be in Z
     
    Last edited: Apr 22, 2013
  5. Apr 23, 2013 #4

    HallsofIvy

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    No, being a power of 2 is not relevant. For example, the ninth root of 9, [tex]\sqrt[9]{9}[/tex] is a zero of [tex]x^3- 3[/tex]. This will be false whenever M is a composite number and N is an n power where n divides M.
     
  6. Apr 26, 2013 #5
    some linear combinations of irrationals

    Well at least I was careful enough to mention that I was "under belief"...

    I'll hijack my own thread to a new topic. I have so many questions with similar themes, so I guess I might as well concentrate them here...

    I want to prove that there does not exist coefficients [itex]a_1,a_2,a_3,a_4\in\mathbb{Q}[/itex] such that

    [tex]
    a_1 + a_2\sqrt{2} + a_3\sqrt{3} + a_4\sqrt{6} = 0
    [/tex]

    without all coefficients being zero. How does that happen?

    I have a feeling that the proof might have something to do with polynomials

    [tex]
    p(X_1,X_2) = (X_1^2-2)^2 + (X_2^2-3)^2
    [/tex]

    and

    [tex]
    a(X_1,X_2) = a_1 + a_2X_1 + a_3X_2 + a_4X_1X_2
    [/tex]

    which would both satisfy [itex]p(\sqrt{2},\sqrt{3})=0[/itex] and [itex]a(\sqrt{2},\sqrt{3})=0[/itex], but I'm not sure what's the final trick.
     
  7. Apr 26, 2013 #6

    mfb

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    I think I would consider ##(b_1+b_2\sqrt{2}+b_3\sqrt{3})^2##. It gives all 4 different terms, and if the bracket cannot be rational the square of it cannot be rational either.
     
  8. Apr 27, 2013 #7
    If I first fix arbitrary [itex]a_1,a_2,a_3,a_4\in\mathbb{Q}[/itex], you are not going to find [itex]b_1,b_2,b_3\in\mathbb{Q}[/itex] such that they would map to the [itex]a_1,a_2,a_3,a_4[/itex] through that square.
     
  9. Apr 27, 2013 #8
    My question is equivalent to asking that how do I prove that [itex]\mathbb{Q}(\sqrt{2})(\sqrt{3})[/itex] is two dimensional vector space with respect to the scalar field [itex]\mathbb{Q}(\sqrt{2})[/itex]. Does this remark help those who know the theory?

    [tex]
    \mathbb{Q}(\sqrt{2})(\sqrt{3}) = \big\{x_1+x_2\sqrt{3}\;\big|\;x_1,x_2\in\mathbb{Q}(\sqrt{2})\big\}
    [/tex]

    It looks like two dimensional, but it's not obvious that non-zero [itex]x_1,x_2\in\mathbb{Q}(\sqrt{2})[/itex] wouldn't exist so that [itex]x_1+x_2\sqrt{3}=0[/itex].
     
  10. Apr 27, 2013 #9

    mfb

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    a_1 does not have to match, as my expression is not supposed to be 0. You can write it on the RHS.
     
  11. Apr 27, 2013 #10
    I invented one proof on my own now. I can assume that [itex]a_1,a_2,a_3,a_4\in\mathbb{Z}[/itex]. Equations

    [tex]
    (a_1 + a_2\sqrt{2})^2 = (a_3\sqrt{3} + a_4\sqrt{6})^2
    [/tex]

    and

    [tex]
    (a_1 + a_3\sqrt{3})^2 = (a_2\sqrt{2} + a_4\sqrt{6})^2
    [/tex]

    eventually imply

    [tex]
    2a_2^2 = 3a_3^2
    [/tex]

    which then implies [itex]a_2=\pm\infty[/itex] and [itex]a_3=\pm\infty[/itex] when you work out how [itex]2^n[/itex] divides these coeffients with arbitrarily large [itex]n[/itex].
     
    Last edited: Apr 27, 2013
  12. Apr 28, 2013 #11
    No, no, mfb, you'r idea aint working.

    [tex]
    (b_1 + b_2\sqrt{2} + b_3\sqrt{3})^2 = b_1^2 + 2b_2^2 + 3b_3^2 + 2b_1b_2\sqrt{2} + 2b_1b_3\sqrt{3} + 2b_2b_3\sqrt{6}
    [/tex]

    So you were thinking, that once [itex]a_2,a_3,a_4\in\mathbb{Q}[/itex] were fixed, you could find [itex]b_1,b_2,b_3\in\mathbb{Q}[/itex] such that

    [tex]
    a_2 = 2b_1b_2
    [/tex]
    [tex]
    a_3 = 2b_1b_3
    [/tex]
    [tex]
    a_4 = 2b_2b_3
    [/tex]

    These equations actually fix [itex]b_1,b_2,b_3[/itex] uniquely, and they can be solved. The answer is

    [tex]
    b_1=\sqrt{\frac{a_2a_3}{2a_4}}
    [/tex]
    [tex]
    b_2=\sqrt{\frac{a_2a_4}{2a_3}}
    [/tex]
    [tex]
    b_3=\sqrt{\frac{a_3a_4}{2a_2}}
    [/tex]

    We see they can be irrational too.
     
  13. Apr 28, 2013 #12
    Actually

    [tex]
    (\sqrt[9]{9})^3 = 3^{\frac{2}{3}} \neq 3
    [/tex]

    :wink:

    [itex]\sqrt[9]{27}[/itex] would be a zero of [itex]X^3-3[/itex].

    Anyway, it's clear that [itex]\sqrt[M]{N}=\sqrt[M']{N'}[/itex] often holds while [itex]M\neq M'[/itex] and [itex]N\neq N'[/itex]. It still seems reasonable, that [itex]X^M-N[/itex] will turn out to be the minimal polynomial of [itex]\sqrt[M]{N}[/itex] if there does not exist [itex]M'<M[/itex] and [itex]N'<N[/itex] such that [itex]\sqrt[M]{N}=\sqrt[M']{N'}[/itex].
     
  14. Apr 28, 2013 #13

    mfb

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    Okay. I did not check it in detail, it was just an idea.

    Your proof is nice.
     
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