# primitive n-th root

• April 25th 2010, 07:23 PM
superman123
primitive n-th root
There was an answered question in the morning but it is erased. Could you help me on that question.

"Let $\zeta$ be a primitive n-th root of unity over $\mathbb{Q}$, where n>2 and let $\alpha=\zeta+{\zeta}^{-1}$. Prove that $\alpha$ is algebraic over $\mathbb{Q}$ of degree $\frac{\varphi(n)}{2}.$ "
• April 25th 2010, 08:48 PM
Bruno J.
Consider the polynomial $f(x)=(x-\zeta)(x-\zeta^{-1})=x-\alpha x + 1$. Its roots are $\zeta, \zeta^{-1}$ and therefore $\zeta$ is of degree $2$ over $K=\mathbb{Q}[\alpha]$. What do you conclude?
• April 25th 2010, 09:10 PM
Bruno J.
Here's another proof. Let $L=\mathbb{Q}(\zeta), K=\mathbb{Q}(\alpha)$. Let $G = \mbox{Gal }L/\mathbb{Q}$. It's easy to see that $K$ is the fixed field of the subgroup $\{1, \sigma\} \lhd G$ where $\sigma : \zeta \mapsto \zeta^{-1}$ (prove this!). Therefore $[K:\mathbb{Q}] = |G/\{1, \sigma\}| = \phi(n)/2$.
• April 25th 2010, 09:38 PM
superman123
Quote:

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

By $f(x)=x^2-\alpha x + 1$ do we have $Gal(\mathbb{Q}(\zeta)/K)$
• April 25th 2010, 09:54 PM
superman123
I am trying to understand splitting fields. Are these fields same?

$\mathbb{Q}(i,\sqrt{1+\sqrt{2}})$
$\mathbb{Q}(\sqrt{1+\sqrt{2}},\sqrt{1-\sqrt{2}})$
$\mathbb{Q}(i,\sqrt[4]{2})$
• April 25th 2010, 10:24 PM
Bruno J.
Quote:

Originally Posted by superman123
By $f(x)=x^2-\alpha x + 1$ do we have $Gal(\mathbb{Q}(\zeta)/K)$

That is not a question. " $Gal(\mathbb{Q}(\zeta)/K)$" is not a statement.
• April 25th 2010, 10:40 PM
superman123
Quote:

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

$[\mathbb{Q}(\zeta):\mathbb{Q}[\alpha]~]~[\mathbb{Q}[\alpha]:\mathbb{Q}]=[\mathbb{Q}(\zeta):\mathbb{Q}]$