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I got one problem from Dummit Foote stating that determine the quadratic polynomial satisfied by the period $\alpha=\zeta_5+\zeta_5^{-1}$ of the the $5th$ root of unity $\zeta_5$. Determine the quadratic equation satisfied by $\zeta_5$ over $\Bbb Q(\alpha)$ and use this to explicitly solve for the $5th$ root of unity.

I am not getting even to find out the quadratic polynomial as $x=\zeta_5+\zeta_5^{-1}$ iff $x^2=\zeta_5^2+\zeta_5^{-2}+2$ not in $\Bbb Q$. Moreover, if the base field is $\Bbb R$ then $\zeta_5+\zeta_5^{-1}$ is in $\Bbb R$. So probably I am not understanding the question correctly. Please help or give me some hints steps..

Edit With the help of the comments

  1. "determine the quadratic polynomial satisfied by the period $\zeta_5+\zeta_5^{-1}$ of the the $5th$ root of unity $\zeta_5$". The answer is $x^2+x-1$.
  2. " Determine the quadratic equation satisfied by $\zeta_5$ over $\Bbb Q(\alpha)$". The answer is $x^2-\alpha x +1$

What about the last part? "use this to explicitly solve for the $5th$ root of unity."?

Ri-Li
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    $1+\zeta + \zeta^2 + \zeta^3 + \zeta^4 = 0.$ Divide by $\zeta^2, $ you get $\zeta^2 + \zeta + 1 + \frac{1}{\zeta} + \frac{1}{\zeta^2} = 0$ – Will Jagy Feb 04 '19 at 02:46
  • I have solved the first two parts thanks for the help. What is the meaning of the last one? – Ri-Li Feb 04 '19 at 03:11
  • what is $\alpha$? – David Holden Feb 04 '19 at 03:19
  • Sorry I have edited $\alpha=\zeta_5+\zeta_5^{-1}$ – Ri-Li Feb 04 '19 at 03:21
  • Well, now you have a quadratic polynomial of which $\alpha$ is a root, so solve that quadratic equation! – Gerry Myerson Feb 04 '19 at 03:26
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    Which one? the second one? Yes I got it. It should be the 2nd one thanks.. :) – Ri-Li Feb 04 '19 at 03:28
  • Use this to explicitly solve for the 5th root of unity. I guess that the idea is to express a primitive 5-th root $\zeta$ of 1 in terms of radicals. The quadratic equation in 2. gives $\zeta=(\alpha \pm \sqrt {\alpha ^2 -1})/2$. The one in 1. gives $\alpha=(-1 \pm \sqrt 5 ) /2$. – nguyen quang do Feb 04 '19 at 16:48
  • Take a look at this thread for a bit more general discussion of when and how rewriting everything in terms of the new variable $z=x+1/x$ helps. Summary: Even degree palindromic polynomials in $x$ can be rewritten as a lower degree polynomials in $z$ by utilizing the palindromic symmetry. – Jyrki Lahtonen Feb 06 '19 at 21:28

1 Answers1

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You have found that

  • $\alpha = \zeta_5+\zeta_5^{-1}$ satisfies $x^2+x-1$
  • $\zeta_5$ satisfies $x^2-\alpha x+1$

we can solve for $\alpha$ using the quadratic equation: $\alpha = \frac{-1 + \sqrt{5}}{2}$ (and note $\alpha^2 = \frac{6 - 2 \sqrt{5}}{4} = \frac{3 - \sqrt{5}}{2}$)

and similarly we can solve for $\zeta_5 = \frac{\alpha + \sqrt{\alpha^2 - 4}}{2}$

putting these together and finding the right choices of signs we have

$$\zeta_5 = \frac{\frac{-1 + \sqrt{5}}{2} + \sqrt{-\frac{5 + \sqrt{5}}{2}}}{2}$$