prove a function has no constructible roots

**joseph0177** · April 5th 2009, 08:54 PM

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..

**NonCommAlg** · April 6th 2009, 07:39 AM

Originally Posted by joseph0177

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.

**joseph0177** · April 7th 2009, 09:04 PM

Originally Posted by NonCommAlg

$y^3 - y + 2$ 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

**ThePerfectHacker** · April 7th 2009, 09:53 PM

Originally Posted by joseph0177

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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