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Math Help - Minimal Polynomials

  1. #1
    Member Haven's Avatar
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    Minimal Polynomials

    I am having a little troubles with this problem, I think there is something subtle that I am missing

    a) Show that \sqrt[4]{2} \not\in \mathbb{Q}(\sqrt{2}) where \mathbb{Q}(\sqrt{2}) = \{a+\sqrt{2}b \colon a,b\in\mathbb{Q}\}

    b) Find a minimal polynomial m_{\sqrt[4]{2}}(x) in \mathbb{Q}(\sqrt{2})[x] where m_{\sqrt[4]{2}}(\sqrt[4]{2}) = 0

    c) Show if \alpha is a primitive element in GF(p^{n}), the degree of the minimal polynomial of \alpha is n
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    For a). Suppose 2^{1/4}=a+b\sqrt{2}. Then \sqrt{2}=(a+b\sqrt{2})^2=a^2-2b^2+2ab\sqrt{2}. Now we must have a^2-2b^2=0 and 2ab=1. But a^2-2b^2=0 is impossible (why?).

    For b), I can tell you that the minimal polynomial will be x^2-\sqrt{2}=0 but I'll let you prove it.
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  3. #3
    Member Haven's Avatar
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    Okay, for c) i have, since \alpha is a primitive element, the ord(\alpha) = p^n - 1

    So, \alpha^{p^n - 1} = 1 \rightarrow \alpha^{p^n} = \alpha

    So \alpha is a root of the polynomial f(x) = (x^{p})^{n} - x

    But this is of degree p*n and not n
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  4. #4
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    Quote Originally Posted by Haven View Post
    Okay, for c) i have, since \alpha is a primitive element, the ord(\alpha) = p^n - 1

    So, \alpha^{p^n - 1} = 1 \rightarrow \alpha^{p^n} = \alpha

    So \alpha is a root of the polynomial f(x) = (x^{p})^{n} - x

    But this is of degree p*n and not n
    \alpha is a root of the polynomial f(x) = x^{p^n} - x over GF(p), but f(x) is reducible over GF(p) (check the case with p=2, n=2 s.t p^2=4).

    Since GF(p^n) is the splitting field of f(x) over GF(p) and it is the simple extension over GF(p) such that [GF(p^n):GF(p)]=n, the degree of the minimal polynomial of \alpha should be n.
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