1. ## primitive n-th root

There was an answered question in the morning but it is erased. Could you help me on that question.

"Let $\displaystyle \zeta$ be a primitive n-th root of unity over $\displaystyle \mathbb{Q}$, where n>2 and let $\displaystyle \alpha=\zeta+{\zeta}^{-1}$. Prove that $\displaystyle \alpha$ is algebraic over $\displaystyle \mathbb{Q}$ of degree $\displaystyle \frac{\varphi(n)}{2}.$ "

2. Consider the polynomial $\displaystyle f(x)=(x-\zeta)(x-\zeta^{-1})=x-\alpha x + 1$. Its roots are $\displaystyle \zeta, \zeta^{-1}$ and therefore $\displaystyle \zeta$ is of degree $\displaystyle 2$ over $\displaystyle K=\mathbb{Q}[\alpha]$. What do you conclude?

3. Here's another proof. Let $\displaystyle L=\mathbb{Q}(\zeta), K=\mathbb{Q}(\alpha)$. Let $\displaystyle G = \mbox{Gal }L/\mathbb{Q}$. It's easy to see that $\displaystyle K$ is the fixed field of the subgroup $\displaystyle \{1, \sigma\} \lhd G$ where $\displaystyle \sigma : \zeta \mapsto \zeta^{-1}$ (prove this!). Therefore $\displaystyle [K:\mathbb{Q}] = |G/\{1, \sigma\}| = \phi(n)/2$.

4. Originally Posted by Bruno J.
Consider the polynomial $\displaystyle f(x)=(x-\zeta)(x-\zeta^{-1})=x-\alpha x + 1$. Its roots are $\displaystyle \zeta, \zeta^{-1}$ and therefore $\displaystyle \zeta$ is of degree $\displaystyle 2$ over $\displaystyle K=\mathbb{Q}[\alpha]$. What do you conclude?
By $\displaystyle f(x)=x^2-\alpha x + 1$ do we have $\displaystyle Gal(\mathbb{Q}(\zeta)/K)$

5. I am trying to understand splitting fields. Are these fields same?

$\displaystyle \mathbb{Q}(i,\sqrt{1+\sqrt{2}})$
$\displaystyle \mathbb{Q}(\sqrt{1+\sqrt{2}},\sqrt{1-\sqrt{2}})$
$\displaystyle \mathbb{Q}(i,\sqrt[4]{2})$

6. Originally Posted by superman123
By $\displaystyle f(x)=x^2-\alpha x + 1$ do we have $\displaystyle Gal(\mathbb{Q}(\zeta)/K)$
That is not a question. "$\displaystyle Gal(\mathbb{Q}(\zeta)/K)$" is not a statement.

7. Originally Posted by Bruno J.
Consider the polynomial $\displaystyle f(x)=(x-\zeta)(x-\zeta^{-1})=x-\alpha x + 1$. Its roots are $\displaystyle \zeta, \zeta^{-1}$ and therefore $\displaystyle \zeta$ is of degree $\displaystyle 2$ over $\displaystyle K=\mathbb{Q}[\alpha]$. What do you conclude?
$\displaystyle [\mathbb{Q}(\zeta):\mathbb{Q}[\alpha]~]~[\mathbb{Q}[\alpha]:\mathbb{Q}]=[\mathbb{Q}(\zeta):\mathbb{Q}]$