# Let $n>2$ prove that $[z+\frac{1}{z}:\mathbb{Q}]=\frac{1}{2}\varphi(n)$ for every primitive $n$-th root of unity $z$

Let $$n>2$$ prove that $$\text{deg}(m_{z+1/z}):=[z+\frac{1}{z}:\mathbb{Q}]=\frac{1}{2}\varphi(n)$$ for every primitive $$n$$-th root of unity $$z$$

So, since I know that with $$z$$ being a primitive $$n$$-th root of unity $$1/z$$ is one aswell. I suppose that for a correct proof I would have to look at the factors in

$$\varphi(n)=\prod\limits_{i=1}^{r}(p_{i}-1)p_{i}^{k_{i}-1}$$ with $$n=p_{1}^{k_{1}},...,p_{r}^{k_{r}}$$

However I'm really stuck with this problem.

• The notion $[z+1/z:\mathbb{Q}]$ has no meaning. Maybe you mean $\mathbb{Q}[z+z^{-1}]$? – ZFR Feb 13 at 17:37
• In my script it is defined as the degree of the minimal polynomial of $z+\frac{1}{z}$. I'll add that! – Christian Singer Feb 13 at 17:38
• Try finding a minimal polynomial of $z$ over $\mathbb{Q}[z+1/z]$. – i707107 Feb 13 at 17:54
• $z^n - 1 = (z-1)(1+z+z^2+...+z^{n-1})$ – Paul Sinclair Feb 14 at 2:20