# Determining the minimal polynomial of $\omega := e^{2πi/p}$ over $\mathbb{Q}[\omega + \omega^{-1}]$

Let $p ≠ 2$ be a prime number, and $\omega = e^{2πi/p}$. I now want to find the minimal polynomial of $\omega$ over the field $\mathbb{Q}[\omega + \omega^{-1}]$.

I must admit that I don't really know how to get started with this one. If this was over $\mathbb{Q}$, then I would probably choose a polynomial of which I know that it has $\omega$ as a root (I think $x^p - 1$ would be a fitting one to start with), and then try to split off factors, and see how far I can go with that.

But I don't really know how to approach this question when trying to find the minimal polynomial over a field like $\mathbb{Q}[\omega + \omega^{-1}]$. I was given the hint that I might be able to use the fact that $\omega + \omega^{-1} \in \mathbb{R}$ (of which I can easily see that it is true), but even that hint couldn't get me started so far.

• You can remove some factors of $\Phi_p(x) = \sum_{k=0}^{p-1} x^k$ with $(x-w)(x-\bar{w}) = x^2+|w|^2-x (w+\bar{w})$ and $2\cos^2(a) = 1+\cos(2a)$. But the conjugate approach might be better. – reuns Jul 8 '16 at 8:58

Hint: In fact, we have $[\mathbb{Q}(\omega):\mathbb{Q}(\omega + \omega^{-1})]=2$. Consider here the monic polynomial $$f(x)=x^2-(\omega+\omega^{-1})x+1=0.$$ What are the roots of this polynomial ?
• Exactly $\omega, \omega^{-1}$ are the roots if I'm not mistaken, hence it's the minimal polynomial. Thanks! However, I'm left wondering: is there any clever strategy on how to find these polynomials with fields like this, or is it simply a matter of trying out polynomials that look promising? – moran Jul 8 '16 at 11:40
• In general one can use some algebraic number theory, e.g., for cyclotomic fields. $\mathbb{Q}(\omega+\omega^{-1})$ is the maximal real subfield of the cyclotomic field $\mathbb{Q}(\omega)$. If the minimal degree has only degree $2$, it is no mistake to try by hand. – Dietrich Burde Jul 8 '16 at 11:48
• Given a Galois extension $L/K$, the minimal polynomial $m$ over $K$ of an element $\alpha \in L$ has the Galois conjugates of $\alpha$ as its roots, so $m(x) = \prod_{\sigma \in \text{Gal}(L/K)} (x - \sigma(\alpha))$. In this example, $\text{Gal}(L/K) = \{\text{id}, \sigma\}$ where $\sigma$ is complex conjugation, so $m(x) = \prod_{\sigma \in \text{Gal}(L/K)} (x - \sigma(\omega)) = (x - \omega)(x - \overline{\omega})$ and then $\overline{\omega} = \omega^{-1}$ because $\omega$ lies on the unit circle in $\mathbb{C}$. – André 3000 Jul 8 '16 at 15:47