1
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.