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Math Help - 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..
    y^3 - y + 2 is irreducible over \mathbb{Q} and thus if \alpha is a root of the polynomial, then \alpha will not be constructible because [\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
    y^3 - y + 2 is irreducible over \mathbb{Q} and thus if \alpha is a root of the polynomial, then \alpha will not be constructible because [\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 r\in \mathbb{R} is constructible iff there exists a tower of fields F_0,F_1,...,F_n in \mathbb{R} where that F_j\subseteq F_{j+1} and [F_{j+1}:F_j] = 2 with F_0 = \mathbb{Q}, r\in F_n. Therefore, [F_n:\mathbb{Q}] = 2^m for some m (because [F_n:F_0] = \Pi_{j=0}^n [F_{j+1}:F_j]). Let r have degree d over \mathbb{Q} (meaning the minimal polynomial has degree d). We see that \mathbb{Q}\subseteq \mathbb{Q}(r) \subseteq F_n and so [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 f. Thus, minimal polynomials of constructible numbers must be power of twos.
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