Quadratic polynomial satisfied by $\zeta_5+\zeta_5^{-1}$

Question

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."?

Answer 1

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

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