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Math Help - Minimal Polynomial

  1. #1
    MHF Contributor chiph588@'s Avatar
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    Minimal Polynomial

    Without finding the minimal polynomial for  r=i\cdot 2^{\frac{1}{3}}+i , show the degree of said polynomial must be  6 . I know this is true because I was able to derive  r\text{'s} minimal polynomial (  f(x)=x^6+3x^4-9x^2+9), but in doing so I used the assumption that it was of degree  6 as opposed to  2 or  3 .
    Last edited by chiph588@; March 4th 2010 at 07:26 PM.
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    Quote Originally Posted by chiph588@ View Post
    Without finding the minimal polynomial for  r=i\cdot 2^{\frac{1}{3}}+i , show the degree of said polynomial must be  6 . I know this is true because I was able to derive  r\text{'s} minimal polynomial (  f(x)=x^6+3x^4-9x^2+9), but in doing so I used the assumption that it was of degree  6 as opposed to  2 or  3 .
    show that both i and \sqrt[3]{2} are in \mathbb{Q}(i\sqrt[3]{2} + i) and thus \mathbb{Q}(i\sqrt[3]{2} + i)=\mathbb{Q}(\sqrt[3]{2},i). we also have [\mathbb{Q}(\sqrt[3]{2},i):\mathbb{Q}]=[\mathbb{Q}(\sqrt[3]{2},i):\mathbb{Q}(\sqrt[3]{2})] \times [\mathbb{Q}(\sqrt[3]{2}): \mathbb{Q}]=2 \times 3 = 6.
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  3. #3
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by NonCommAlg View Post
    show that both i and \sqrt[3]{2} are in \mathbb{Q}(i\sqrt[3]{2} + i) and thus \mathbb{Q}(i\sqrt[3]{2} + i)=\mathbb{Q}(\sqrt[3]{2},i). we also have [\mathbb{Q}(\sqrt[3]{2},i):\mathbb{Q}]=[\mathbb{Q}(\sqrt[3]{2},i):\mathbb{Q}(\sqrt[3]{2})] \times [\mathbb{Q}(\sqrt[3]{2}): \mathbb{Q}]=2 \times 3 = 6.
    Let  \alpha=i\sqrt[3]{2}+i .

    It turns out  \sqrt[3]{2} = -\frac{1}{6}\alpha^4-\alpha^2+\frac{1}{2}
    and  i = \frac{1}{6}\alpha^5+\frac{2}{3}\alpha^3-\frac{1}{2}\alpha ,
    so indeed we know  \mathbb{Q}(\sqrt[3]{2},i) \subseteq \mathbb{Q}(\alpha) .

    But is there and easier way to show  \sqrt[3]{2},i\in \mathbb{Q}(\alpha) ?
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    Quote Originally Posted by chiph588@ View Post
    Let  \alpha=i\sqrt[3]{2}+i .

    It turns out  \sqrt[3]{2} = -\frac{1}{6}\alpha^4-\alpha^2+\frac{1}{2}
    and  i = \frac{1}{6}\alpha^5+\frac{2}{3}\alpha^3-\frac{1}{2}\alpha ,
    so indeed we know  \mathbb{Q}(\sqrt[3]{2},i) \subseteq \mathbb{Q}(\alpha) .

    But is there and easier way to show  \sqrt[3]{2},i\in \mathbb{Q}(\alpha) ?
    well, this is how i did it: \alpha^3=-3i(1+\sqrt[3]{2} + \sqrt[3]{4})=-3 \alpha - 3i \sqrt[3]{4} and so \beta=i \sqrt[3]{4} \in \mathbb{Q}(\alpha) and we're done because \sqrt[3]{2}=\frac{-\beta^2}{2} and i=\frac{-\beta^3}{4}.
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