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Math Help - Splitting field for X^3-5

  1. #1
    Ant
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    Splitting field for X^3-5

    Hi,

    Let  f=X^3 -5 , the \Sigma_f = \mathbb{Q}[\alpha,\omega] with \alpha real cube root of 5 and \omega primitive root of 1.

    [\mathbb{Q}(\alpha):\mathbb{Q}]=3 because  X^3-5 is irreducible.


    But what is [\mathbb{Q}(\alpha,\omega):\mathbb{Q}(\alpha)]?


    [\mathbb{Q}(\alpha,\omega):\mathbb{Q}(\alpha)] >1 because  \omega \in \mathbb{C} \setminus \mathbb{R} and yet  \mathbb{Q}(\alpha) is a subfield of  \mathbb{R} . Also the degree is  \leq 3 because  X^3 -1 has \omega as a root. So the possibilities are 2 or 3.


    If it's 2. Then because the min poly has to be monic and \omega^2 + \omega = \sqrt{2}

    I think that we have to express \sqrt{2} as a linear combination (with rational coefficients) of \alpha and \alpha^2 .

    I can't see how do to this?

    Thanks for any help!
    Last edited by Ant; May 18th 2013 at 04:02 AM.
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Splitting field for X^3-5

    The roots of X^3-1 are X_1=1, X_2=e^{\frac{2 \pi i}{3}}, X_3 = e^{\frac{4 \pi i}{3}} whereby X_2 and X_3 have minimal polynomial T^2+T+1
    Thanks from Ant
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  3. #3
    Ant
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    Re: Splitting field for X^3-5

    Of course, I made a stupid mistake:

     \omega^2 + \omega = -1 !

    thanks
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