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Thread: prove a function has no constructible roots

  1. #1
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    prove a function has no constructible roots

    prove that x^6 - x^2 + 2 has no constructible roots

    First, i let y=x^2, so the function become y^3 - y + 2

    then use the theorem, if polynomial f(x) has integer coefficient and a ration root p/q. (p,q)=1, then p|a0 , q|an.

    I got p|2 and q|1

    g(2) does not equal to zero.

    so. the function does not have a rational root.

    How do I go on from here..
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  2. #2
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    Quote Originally Posted by joseph0177 View Post
    prove that x^6 - x^2 + 2 has no constructible roots

    First, i let y=x^2, so the function become y^3 - y + 2

    then use the theorem, if polynomial f(x) has integer coefficient and a ration root p/q. (p,q)=1, then p|a0 , q|an.

    I got p|2 and q|1

    g(2) does not equal to zero.

    so. the function does not have a rational root.

    How do I go on from here..
    $\displaystyle y^3 - y + 2$ is irreducible over $\displaystyle \mathbb{Q}$ and thus if $\displaystyle \alpha$ is a root of the polynomial, then $\displaystyle \alpha$ will not be constructible because $\displaystyle [\mathbb{Q}(\alpha):\mathbb{Q}]=3,$ which is not a power of 2.
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  3. #3
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    Quote Originally Posted by NonCommAlg View Post
    $\displaystyle y^3 - y + 2$ is irreducible over $\displaystyle \mathbb{Q}$ and thus if $\displaystyle \alpha$ is a root of the polynomial, then $\displaystyle \alpha$ will not be constructible because $\displaystyle [\mathbb{Q}(\alpha):\mathbb{Q}]=3,$ which is not a power of 2.
    can you please explain about last part, [Q (3) : Q] = 3 is not power of 2

    how do you know [Q (3) : Q] is 3 , and it has no constructible root becasue it is not power of 2
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  4. #4
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    Quote Originally Posted by joseph0177 View Post
    can you please explain about last part, [Q (3) : Q] = 3 is not power of 2

    how do you know [Q (3) : Q] is 3 , and it has no constructible root becasue it is not power of 2
    For $\displaystyle r\in \mathbb{R}$ is constructible iff there exists a tower of fields $\displaystyle F_0,F_1,...,F_n$ in $\displaystyle \mathbb{R}$ where that $\displaystyle F_j\subseteq F_{j+1}$ and $\displaystyle [F_{j+1}:F_j] = 2$ with $\displaystyle F_0 = \mathbb{Q}, r\in F_n$. Therefore, $\displaystyle [F_n:\mathbb{Q}] = 2^m$ for some $\displaystyle m$ (because $\displaystyle [F_n:F_0] = \Pi_{j=0}^n [F_{j+1}:F_j]$). Let $\displaystyle r$ have degree $\displaystyle d$ over $\displaystyle \mathbb{Q}$ (meaning the minimal polynomial has degree $\displaystyle d$). We see that $\displaystyle \mathbb{Q}\subseteq \mathbb{Q}(r) \subseteq F_n$ and so $\displaystyle [F_n:\mathbb{Q}] = [F_n:\mathbb{Q}(r)][\mathbb{Q}(r):\mathbb{Q}] \implies 2^m = [F_n :\mathbb{Q}(r)]d\implies d = 2^f$ for some $\displaystyle f$. Thus, minimal polynomials of constructible numbers must be power of twos.
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