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Thread: 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 $\displaystyle \sqrt[4]{2} \not\in \mathbb{Q}(\sqrt{2})$ where $\displaystyle \mathbb{Q}(\sqrt{2})$ = $\displaystyle \{a+\sqrt{2}b \colon a,b\in\mathbb{Q}\}$

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

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

    For $\displaystyle b)$, I can tell you that the minimal polynomial will be $\displaystyle x^2-\sqrt{2}=0$ 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 $\displaystyle \alpha$ is a primitive element, the $\displaystyle ord(\alpha) = p^n - 1$

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

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

    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 $\displaystyle \alpha$ is a primitive element, the $\displaystyle ord(\alpha) = p^n - 1$

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

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

    But this is of degree p*n and not n
    $\displaystyle \alpha$ is a root of the polynomial $\displaystyle f(x) = x^{p^n} - x$ 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 $\displaystyle GF(p^n)$ is the splitting field of f(x) over GF(p) and it is the simple extension over GF(p) such that $\displaystyle [GF(p^n):GF(p)]=n$, the degree of the minimal polynomial of $\displaystyle \alpha$ should be n.
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